CHAPTER 1: INTRODUCTION
1.3-1Þ
Answers will vary.
1.3-2.
Answers will vary.
1.3-3.
By using operations research (OR), FedEx managed to survive crises that could drive it
out of business. The new planning system provided more flexibility in choosing the
destinations that it serves, the routes and the schedules. Improved schedules yielded into
faster and more reliable service. OR applied to this complex system with a lot of
interdependencies resulted in an efficient use of the assets. With the new system, FedEx
maintained a high load factor while being able to service in a reliable, flexible and
profitable manner. The model also enabled the company to foresee future risks and to
take measures against undesirable outcomes. The systematic approach has been effective
in convincing investors and employees about the benefits of the changes. Consequently,
"today FedEx is one of the nation's largest integrated, multi-conveyance freight carriers"
[p. 32].
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CHAPTER 2: OVERVIEW OF THE OPERATIONS RESEARCH
MODELING APPROACH
2.1-1.
(a) The rise of electronic brokerage firms in the late 90s was a threat against full-service
financial service firms like Merrill Lynch. Electronic trading offered very low costs,
which were hard to compete with for full-service firms. With banks, discount brokers and
electronic trading firms involved, the competition was fierce. Merrill Lynch needed an
urgent response to these changes in order to survive.
(b) "The group's mission is to aid strategic decision making in complex business
situations through quantitative modeling and analysis" [p.8].
(c) The data obtained for each client consisted of "data for six categories of revenue, four
categories of account type, nine asset allocation categories, along with data on number of
trades, mutual fund exchanges and redemptions, sales of zero coupon bonds, and
purchases of new issues" [p. 10].
(d) As a result of this study, two main pricing options, viz., an asset-based pricing option
and a direct online pricing option were offered to the clients. The first targeted the clients
who want advice from a financial advisor. The clients who would choose this option
would be charged at a fixed rate of the value of their assets and would not pay for each
trade. The latter pricing option was for the clients who want to invest online and who do
not want advice. These self-directed investors would be charged for every trade.
(e) "The benefits were significant and fell into four areas: seizing the marketplace
initiative, finding the pricing sweet spot, improving financial performance, and adopting
the approach in other strategic initiatives" [p.15].
2.1-2.
(a) This study arose from GM's efforts to survive the competition of the late 80s. Various
factors, including the rise of foreign imports, the increase in customer expectations and
the pricing constraints, forced GM to close plants and to incur large financial losses.
While trying to copy Japanese production methods directly, GM was suffering from
"missing production targets, working unscheduled overtime, experiencing high scrap
costs, and executing throughput-improvement initiatives with disappointing results" [p.
7]. The real problems were not understood and the company was continuously losing
money while the managers kept disagreeing about solutions.
(b) The goal of this study was "to improve the throughput performance of existing and
new manufacturing systems through coordinated efforts in three areas: modeling and
algorithms, data collection, and throughput-improvement processes" [p. 7].
(c) The data collection was automated by using programmable logic controllers (PLCs).
The software kept track of the production events including "machine faults and blocking
and starving events" [p. 13] and recorded their duration. The summary of this data was
then transferred to a centralized database, which converted this to workstationperformance characteristics and used in validating the models, determining the bottleneck
processes and enhancing throughput.
(d) The improved production throughput resulted in more than $2.1 billion in
documented savings and increased revenue.
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2.1-3.
(a) The San Francisco Poice Department has a total police force of 1900, with 850
officers on patrol. The total budget of SFPD in 1986 was $176 million with patrol
coverage cost of $79 million. This brings out the importance of the problem .
Like most police departments, SFPD was also operated with manually designed
schedules. It was impossible to know if the manual schedules were optimal in serving
residents' needs. It was difficult to evaluate alternative policies for scheduling and
deploying officers. There was also the problem of poor response time and low
productivity, pressure of increasing demands for service with decreasing budgets. The
scheduling system was facing the problem of providing the highest possible correlation
between the number of officers needed and the number actually on duty during each
hour. All these problems led the Task Force to search for a new system and thus
undertake this study.
(b) After reviewing the manual system, the Task Force decided to search for a new
system. The criteria it specified included the following six directives:
-- the system must use the CAD (computer aided dispatching ) system, which
provides a large and rich data base on resident calls for service. The CAD
system was used to dispatch patrol officers to call for service and to maintain
operating statistics such as call types, waiting times, travel time and total time
consumed in servicing calls. The directive was to use this data on calls for
service and consumed times to establish work load by day of week and hour
of day.
-- it must generate optimal and realistic integer schedules that meet management
policy guidelines using a computer.
-- it must allow easy adjustment of optimal schedules to accommodate human
considerations without sacrificing productivity.
-- it must create schedules in less than 30 minutes and make changes in less than
60 seconds.
-- it must be able to perfonn both tactical scheduling and strategic policy testing in
one integrated system
--the user interface must be flexible and easy, allowing the users (captains) to
decide the sequence of functions to be executed instead of forcing them to
follow a restrictive sequence.
2.1-4.
(a) Taking all the statistics of AIDS cases into account it was inferred that just one-third
of all cases nation-wide involved some aspect of Injection Drug Use(IDU). But in
contrast to this national picture, over 60% of 500 cases reported in New Haven ,
Connecticut was traced to drug use. Though it was realized previously, by 1987 it was
clear that the dominant mode of HIV transmission in New Haven was the practice of
needle sharing for drug injection.
This was the background of the study and in 1987 a street outreach program was
implemented which included a survey of drug addicts with partial intent to detennine
why lDUs continued to share needles given the threat of HIV infection and AIDS. h was
claimed by the survey respondents that IDUs shared needles since they were scared and
feared arrest for possessing a syringe without prescription which was forbidden by law in
Connecticut. Respondents also pointed out difficulties involved in entering drug
treatment program . The officials recognized that logical intervention was needle
exchange whereby IDUs exchanged their used needles for clean ones. This would remove
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infectious drug injection equipment from circulation and also ease access to clean
needles. Further, contacts made as a result of needle exchange m ight lead some active
lDUs to consider counseling or enter drug treatment. After a lot of lobbying finally the
bill for the first legal needle exchange program became effective on July 1, 1990.
(b) The design for the needle exchange program was achieved over the summer of 1980.
The relevant committee decided that IDUs would be treated with respect and so no
identification information was asked of program clients. The program began operating on
November 13, 1990.
The needle exchange operate on an outreach basis. A van donated by Yale
university visits neighborhoods with high concentration of IDUs. Outreach staff members
try to educate the clients over there by different means like distributing literature
documenting risks of HIV infection, dispensing condoms , clean packets, etc.
The primary goal of needle exchange is to reduce incidence of new HIV infection
among IDUs. While studies showed consistent self-reported reductions in risky behavior
among lDUs participating in needle exchange programs the studies were not convincing.
So the mechanics of needle exchange require that the behavior of needles must change.
What was required was to reduce the time needles spend circulating in the population. As
needles circulate for shorter period of time, needles share fewer people which lower the
number of infected needles in the pool of circulating needles which in effect lowers
chances of an IDU becoming infected being injected with a previously infected needle.
To use this theory required invention of new data collection system which is as follows.
A syringe tracking and testing is a system developed to interview the needles
returned to the program. All clients participating in the needle exchange are given unique
code names and every needle distributed receives a code. Every time a client exchanges
needles, an outreach worker records the date and location of exchange. He also records
the code name of the client receiving the needles alongside the codes of the needles.The
client then places the returned needles in a canister to which the worker puts a label with
the date and location of exchange and code name of client.
All returned needles are brought to a laboratory at Yale University where a
technician collates the information on the canister labels with the tracking numbers on
the returned needles . For non-program or street needles returned to needle exchange, the
location, date, and client code are recorded. A sample of the returned needles are tested
for HIV.
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2.2-1.
The financial benefits that resulted from this study include savings of $40 million in 2001
and of $5 million in 2002. The savings for any major disruption have been between $1
and $5 million. The new system enabled Continental Airlines to operate in an efficient
and cost-effective manner in case of disruptions. The time to recover and the costs
associated with disruptions are reduced. What-if analysis allowed the company to
evaluate various scenarios in short periods of time. Since the complete reliable data can
be generated quickly, the company reacts to facts rather than forecasts. These
improvements in handling irregularities resulted in better and more reliable service and
hence happier customers.
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2.2-2.
(a) Swift & Company operates in an industry that involves highly skilled labor, many
production pathways and perishable products. To generate profit, the company needs to
make an efficient use of every single animal procured. Before this study, Swift was not
able to meet the shipping deadlines and as a result of this, it was forced to offer
discounts. The consequences of this practice included highly reduced profits, inaccurate
forecasts and very low reliability. The company had to find a way to come up with the
best product mix and to survive in this business defined by volatility and velocity.
(b) The purpose of the scheduling models is "to fix the production schedule for the next
shift and to create a projection of short order" [p. 74]. They generate shift-level and daily
schedule for 28 days. The capable-to-promise (CTP) models "determine whether a plant
can ship a requested order-line-item quantity on the requested date and time given the
availability of cattle and constraints on the plants' capacity during the 90-day model
horizon" [p. 75]. The starting inventory, committed orders, and production schedule
generated by the CTP models are inputs to the available-to-promise (ATP) models. Every
15 minutes, the ATP models determine the unsold production of each shift and alert the
salespeople to undesirable inventory levels.
(c) The company now uses 45 optimization models.
(d) As a result of this study, the key performance measure, namely the weekly percentsold position has increased by 22%. The company can now allocate resources to the
production of required products rather than wasting them. The inventory resulting from
this approach is much lower than what it used to be before. Since the resources are used
effectively to satisfy the demand, the production is sold out. The company does not need
to offer discounts as often as before. The customers order earlier to make sure that they
can get what they want by the time they want. This in turn allows Swift to operate even
more efficiently. The temporary storage costs are reduced by 90%. The customers are
now more satisfied with Swift. With this study, Swift gained a considerable competitive
advantage. The monetary benefits in the first years was $12.74 million, including the
increase in the profit from optimizing the product mix, the decrease in the cost of lost
sales, in the frequency of discount offers and in the number of lost customers. The main
nonfinancial benefits are the increased reliability and a good reputation in the business.
2.2-3.
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2.2-4.
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2.2-5.
(a) The objective was to simultaneously maximize the company's use of its aircraft, crew,
and facilities.
(b) Network optimization, linear programming, integer programming, nonlinear
programming, and dynamic programming.
(c) Since the inception of the study, it had generated savings in excess of $54 million
with projected additional savings of $27 million annually.
2.3-1.
(a) Towards the end of 90s, Philips Electronics faced challenges in coordinating its
supply chains. Decentralized short-term planning was no longer very reliable. The spread
of the information to various branches of the global supply chains was taking a lot of
time and the information was distorted while it was being transferred. To deal with the
uncertainty, the companies had to keep high inventory levels.
(b) The ultimate purpose of this study was "to improve competitiveness by improving
customer service, increasing sales and margins, and reducing obsolescence and
inventories" [p. 38]. To achieve this, the project team aimed at designing a collaborativeplanning (CP) process that would improve trust and collaboration between partners and
accelerate decision making.
(c) "The algorithm can generate feasible plans within seconds. In fact, the calculation of
the plan is hardly noticeable to the people participating in the weekly CP meeting. The
speed of the algorithm also allows planners to compute multiple plans during the
meeting, creating an interactive planning environment. The software environment also
provides strong problem-solving support, used extensively during the CP meetings. One
such capability is called backward pegging. It exploits the one-to-one relationship
between the storage of an end item in some future period and a constraining stock on
hand or scheduled receipt of one or more upstream items. Thus, the backward-pegging
mechanism makes the actual material bottlenecks in the network visible" [p. 41-42].
(d) The four steps of the collaborative-planning process are gathering data, deciding,
escalating and deploying.
(e) This study allowed the companies to solve complex problems quickly, to exploit
profitable opportunities and to enhance trust within the supply chain. The information is
now conveyed to other parties in a shorter time and more accurately. As a result of this,
the companies can have accurate information about the availability of material at
different stages. This results in the reduction of inventory and obsolescence as well as the
ability to respond promptly to the changes in market conditions. The benefit from
decreasing inventory and obsolescence is around $5 million per year in total.
Nonfinancial benefits include enhanced flexibility and reliability throughout the chain.
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2.3-2.
(a)
The role of evaluating a model is to extract information from it. It entails two,
often simultaneous activities-- identifying alternatives and calculating objectives.
The most known technique for identifying alternatives is optimization. The
process yields a single solution which maximizes or minimizes a single objective
function. The most prevalent technique used for identifying multiple alternatives is
sensitivity analysis. The process can show how the optimum changes when model
parameters change or can provide near-optimal alternative solutions.
The author views that optimization should not be the sole goal, not just because
models are abstractions of real world but because they do not provide adequate
information for making decisions. Its objective is to find only one solution. But the
decision maker probably would prefer information on several alternatives. Though
sensitivity analysis increases effectiveness of optimization , it is deficient. It only yields
alternative solution near optimum. The decision maker rather needs unique solutions
which offer distinct alternatives.
So the author opines that research should be devoted to identify multiple
alternatives. One may begin in the solution process itself. Each solution is a feasible
alternative, which the decision maker may choose over the optimum. New algorithms
may be designed to identify distinct alternatives.
The second step of evaluation should involve calculating quantifiable objective
for each alternative.
Thus summarizing, the author views that although optimization has dominated
research in MS/OR it is but one technique for addressing one part of MS/OR process. It
is deficient since it does not provide adequate information for making important
decisions. Complex decisions rather require information on many alternatives and also an
understanding of basic trade-offs and principles. Optimization alone cannot provide this
information.
(b) The key to MS/OR is not only possessing knowledge. Though different practitioners
take different approaches -- three key steps being
--modeling
-- evaluating
--deciding, which are all complementary.
In MS/OR systematized knowledge is reflected in better decisions. The key to good
decisions is knowledge and judgment. Modeling and evaluation form a systematized way
for acquiring knowledge; judgment is acquired through experience.
The problems which do not require judgment are the ones which can be
formulated with well-defined objective functions and solved automatically with
algorithms which are pretty efficient an example being the shortest path algorithm. On
the other hand, there are problems which are easy to formulate but difficult to solve. For
example, a carpet store owner would not argue with the objective of the cutting stock
problem but may not be happy with solutions provided by available software. He would
benefit from models that offer help in cutting the carpet. Combining knowledge from
modeling with judgment of store owner would give best result.
Generally, important questions facing management are not well-defined as
shortest path or cutting stock problem. Neither there are related well-defined problems
which can be optimized, example the facilities layout problem.
Thus the roles are all complementary. Most depend on both judgment of decision
maker and knowledge gained from modeling and evaluating.
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2.3-3.
(a) The problem is to design and schedule the company's product line containing over
5000 products serving a wide variety of markets.
(b) The algorithm is a genetic algorithm (the subject of Sec. 14.4), which is a particular
kind of metaheuristic (the subject of Chap. 14).
2.4-1.
The credibility of analyses and therefore the probability that policies based upon them
will be implemented depends on the perceived validity of the models.
The process of model validation though is a burden which helps to learn lessons
which may not lead to just improvements in the model but also to changes in the
scientific theory and public policy. This happened in PAWN with the Nutrient model and
eutrophication. When PAWN was started, the Dutch eutrophication control strategy was
to decrease phosphate discharges into surface water from point sources mostly sewage
treatment plants.
To find out how effective this strategy is the Algae Bloom model was applied to
some major Dutch lakes. It was revealed that in most cases this required enormous
percentage decrease in phosphate concentrations.
Next question was what was to be done to achieve a particular percentage
decrease in phosphate concentration. The Dutch strategy was based on the fact that large
amount of phosphates and other nutrients accumulated in bottom of the lakes was bound
permanently to the bottom and hence unavailable to support algae blooms. This was
contradicted both in the Nutrient Model calibration process and validation process.
Studies taken convinced them that nutrients particularly phosphate can be
liberated from bottom sediments both in normal steady mode and explosive mode. This
was widely accepted in the scientific community.
But the conclusion implied that use of a phosphate reduction program as the only
way to limit algae bloom would have hardly any immediate success. But analysis with
the Algae Bloom model suggests other tactics which could be effective and combination
of tactics should be tailored to individual lakes.
2.4-2.
The author feels that observation and experimentation are not emphasized in the MS/OR
literature or in the training of its workers as much as experience would lead one to
believe. As examples he has given some experiences with the US Air Force in early '50s
which strengthens his belief.
He opines that observing actual operations as part of the analysis process provides
a required base for understanding what is going on in a problem situation. They can help
to point out difficulties being encountered, suggest hypothesis and theories that may
account for problems and offer evidence regarding the validity of the models built as part
of problem solving process.
lf a problem is in regard to a non-existing system or an operating system fulfills
an important function that must continue, so that controlled experiments with are not
possible-- one can build a theory about relevant phenomena and analyze the theory but
numerical results obtained in this way clearly can be viewed with suspicion.
Alternatively if a similar system exists, one can extrapolate from results with it to make
estimates about the prospective system. In fact, administrative emergencies or an
executive desire to try something new may cause the behavior of a system already in
existence to change. The analyst may then be able to collect data useful for analyzing
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how the system would operate under changed circumstances or for identifying problems
that might crop up under different operating regimes.
From his personal experiences he gives evidence to give substance to these
remarks of his.
If data was used from one system to predict performance of another he believes
that the parameter values form observing another similar system can be useful, and
incorporating such estimates in a crude study can be better than not doing a study at all.
Parameter values from one context to another cannot be expected to support detailed
findings, but even crude findings are enough to provide indispensable information on
which to base policy.
He has also analyzed the results of a continent wide Air Defense exercise. He
says here that analysis must be carefully planned, and planning must begin early. Early
work serves to put attention on the structure of the work and issues to be faced as well as
other responsibilities.
Thus, in nutshell, the author views that skills involved in observation and
experimentation are enumerous and should be part of the tool kit of many MS/OR
analysts. He views that discriminating observation and carefully planned experimentation
and analysis are central to MS/OR.
Observing actual operations and collection of data allow us to discern problems,
develop hypotheses and validate models needing skill.
Similarly, accurate and complete data are required to estimate validity. Program
evaluation brings together many of the issues of observation and experimentation.
Thus, issues of scientific and professional craft related to observation and
experimentation should occur as important pieces in experience, literature and training of
MS/OR workers.
2.4-3.
(a) The author views that analysts do not believe that a model can be completely
validated. He further opines that policy models can at best be invalidated. Thus, the
objective of validation or invalidation attempts is to increase the degree of confidence
that the events obtained from the model will take place under conditions assumed. After
trying all invalidation procedures, one will have a good understanding of strengths and
weaknesses of the model and will be able to meet criticisms of omissions. Knowing the
Iimitations of the model will enable one to express proper confidence on its results.
(b) Model Validity deals with correspondence of the model to the real world and related
to pointing out all stated and implied assumptions, identification and inclusion of all
decision variables and hypothesized relations among variables. Different assumptions are
made and the analyst compares each assumption and hypothesis to the internal and
external problem environments viewed by the decision maker and comments on the
extent of divergence.
Data validity deals with raw and structured data, where structured data is
manipulated raw data. Raw data validity is concerned with measurement problems and
determining if the data is accurate, impartial and representative. Structured data validity
needs review of each step of the manipulation and is a part of model verification.
Logical/mathematical validity deals with translating the model form into a
numerical, computer process that produces solutions. There is no standard method to
determine this. Approaches include comparing model outcomes with expected or
historical results and a close scrutiny of the model form and its numerical representation
on a flow chart.
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Predictive validity is analyzing errors between actual and predicted outcomes for
a model's components and relationships. Here one looks for errors and their magnitudes,
why they exist and if how they can be corrected.
Operational validity attempts to assess the importance of errors found under
technical validity. It must find out if the use of the model is appropriate for the observed
and expected errors. lt also deals with the fact whether the model can produce
unacceptable answers for proper ranges of parameter values.
Dynamic validity is concerned with determining how the model will be
maintained during its life cycle so it will continue to be an accepted representation of the
real system. The two areas of interest thus are update and review.
(c) Sensitivity analysis plays an important role in testing the operational validity of a
model. In this, values of model parameters are varied over some range of interest to
determine if and how the recommended solution changes. If the solution is sensitive to
certain parameter changes, the decision maker may want the model analysts to explore
further or justify in detail values of these parameters. Sensitivity analysis also involves
the relationship between small changes in parameter values and magnitude of related
changes in outputs.
(d) Validating a model tests the agreement between behavior of the model and the real
world system being modeled. Models of a non-existing system are the difficult to
validate. Three concepts apply here: face validity or expert opinion, variable -parameter
validity and sensitivity analysis and hypothesis validity. Though these concepts are
applicable to all models, models of real systems can be subjected to further tests. Validity
is measured by how well the real-system compares with model-generated data. The
model is replicatively valid if it matches data acquired from the real system. It is
predictively valid when it matches data before getting the data from the real system. A
model is structurally valid if it reproduces the observed real system behavior as well as
reflects the way in which real system works to produce this.
The author views that there is no validation methodology appropriate for all
models. He says that a decision-aiding model can never be completely validated as there
are never real data about the alternatives not implemented. Thus, analysts must be careful
in devising, implementing, interpreting and reporting validation tests for their models.
(e) Basic validation steps have been cited in page 616 of the article.
2.5-1.
(a) In the late 1970s, oil companies began to experience downward pressure on
profitability due to rapid and continuing changes in external environment. Partially in
response to these pressures Texaco's Computer Information Systems department
developed an improved on-line interactive gasoline blending system called OMEGA. It
was first installed in 1983 and is now used in all seven Texaco US refineries and in two
foreign plants.
(b) A simple interactive user interface makes OMEGA easy to use. All input data can be
entered by hand. OMEGA can interface with refinery data acquisition system. The user
can access stock qualities, stock availabilities, blend specification and requirements,
starting values and limits, optimization options, automatic stock selection, automatic
blend specification and several other options.
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Several features aid the user in performing planning functions. By choosing
appropriate options user can obtain optimization options. The user also has other options.
Each refinery uses different set of features depending on its availability of blending
stocks. These vary depending on the configuration of the refinery and particular crudes
being refined. Availability and easy use of OMEGA features has provided engineers and
blenders with powerful and easy tool.
(c) OMEGA is constantly being updated and extended. It had to be modified to take into
account EPA's regulation for a lead phase down for regular-leaded gasoline so that now
OMEGA could be more accurate for these lower lead levels.
OMEGA is continuously modified to reflect changes in refinery operations.
Differences in refineries required changes to the system.
When Texaco began installing OMEGA in their foreign refineries, additional
changes had to be made to handle different requirements of different countries.
Improvements to OMEGA are needed to enable it to answer the new and
unanticipated what-if questions often asked by refinery engineers.
(d) Each refinery uses OMEGA in varying degrees and for various purposes depending
on their needs, complexity and configuration. Below the typical usage of the system is
pointed out.
On a monthly basis, refineries use OMEGA to develop a gasoline blending plan
for the month. The refinery planning model's projected blending stock volumes are input
to OMEGA. The blending planner calculates 3 to 8 blends in a single OMEGA run. The
refinery planning model's blend compositions are input into OMEGA as initial values.
Once a reasonable blend is developed, the marketing department is contacted to discuss
resulting grade splits. After marketing department does their job a finalized blending plan
is developed for the month. The scheduler determines when each of the grades will be
blended. All these work are done by using OMEGA.
(e) OMEGA contributes to overall profitability. To measure actual benefit, a method tried
was comparing blend composition that blenders used with and without OMEGA. Here
OMEGA achieved as much as 30 percent increase in profit. Average increase in profit is
approximately 5 percent of gross gasoline revenue. If OMEGA is used to calculate
blending recipes fewer blends fail to meet their quality specification. OMEGA's more
reliable gasoline grade-split estimates provide significant aid to those developing
marketing strategies and refinery production targets. OMEGA is used for what-if case
studies performed for example for economic analysis of refinery improvement projects
and analysis of how proposed Government regulations would affect Texaco. OMEGA's
features have enabled Texaco with capacity to do things not possible with previous
blending system, for example, to deal with mix stocks, consider new grades of gasoline,
more control on inventory, etc. OMEGA's features make it easy and quick to explore new
avenues of profitability for a refinery.
2.5-2.
(a) Yellow Freight System, Inc. was founded in 1926 as a regional motor carrier serving
the Mid-West. Today it is one ofthe largest motor carriers in the country. From a mixed
operation in the 1970s, Yellow now predominantly serves the less-than-truckload (LTL)
portion of the freight market. The '80s were a difficult decade for the motor carrier
industry. Deregulation made the way for tremendous opportunity for growth but also
presented management with new and difficult challenges to manage these larger
operations more efficiently than before. After 1980, motor carriers were forced to
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compete on price, which led to a lot of pressure to cut costs. The result was decrease in
transportation rates. Between 1980 to 1990, transportation rates translated to a drop in
real terms of 29%. In addition to real rate decreases, the shipping community in response
to intense international competition, started to increase their expectation in service. For
many shippers, Yellow Freight is a full partner in their total quality management
programs. Another important component of the logistics system is timely delivery of
freight. Service reliability is also critical. This heightened emphasis on service was a
problem for some long-standing operating practices used by national LTL carriers. The
effect of these pressures can be seen in the tremendous attrition the industry suffered. Out
of top 20 revenue producing LTL carriers in 1979, only 6 are there today. In this period,
Yellow Freight grew from 248 to 630 terminals. This growth has had the effect of
creating an extremely large and complex operation. The large network also needs a
greater degree of coordination.
In 1986, Yellow initiated a project to improve its ability to manage a complex
system. Yellow was interested in using a modern network method to simulate and
optimize a large network. The project had a main goal -- improved service and servi.ce
reliability through better management control of the network. This goal was
supplemented by broader management objectives. There was also an expectation that
improved planning would lead to higher productivity level and lower costs.
Consequently, a project team was formed.
(b) The development effort at Yellow started with an existing model as a base and then
were modified. The result of this effort was SYSNET. SYSNET is more than 80,000
lines of FORTRAN code for performing sophisticated optimizations using modem
network tools. They developed an innovative, interactive optimization technology that
puts human beings in the loop, placing sophisticated, up-to-date optimization methods in
their hands. These methods were required in the development of a system that would
handle the entire network without resorting to heuristic methods to decrease the size of
the problem. As a result, user is able to analyze impacts of changes in the whole network
in a simple but interactive fashion. Projects can be completed earlier new with greater
precision. Decisions on shipment consolidations are now optimized taking into account
the system effect of each decision.
Yellow uses SYSNET for two sets of applications:
-- main use is tactical load planning, which involves monthly planning and
revision of set of instructions that govern handling and consolidation of
shipments through the network.
-- the second set of applications involve longer range planning of the network
itself. These problems cover the location and sizing of new facilities, and long
range decisions that govern the flow of freight between terminals.
At Yellow SYSNET is more than just a piece of code. It embodies an entire
planning methodology adopted by all levels of the company. From strategic planning
studies communicated to high-level management to network routing instructions sent
right to the field, SYSNET has become a comprehensive planning process that has
allowed management to maintain control of a large complex operation. In addition,
Yellow uses SYSNET as the central tool in the design and evaluation of projects of over
$10 million in annual savings.
(c) The interactive aspects of the code proved important in two respects:
-- the user was needed to guide the search for changes in the network. For
example, the user may know that freight levels are on the rise in the Midwest or a
2-15
particular breakbulk is facing problems with capacity. ln other cases, the user may know
that the current solution is a local minimum and a major change in the network is needed
to achieve an overall improvement. A human being can easily point out these spatial
patterns and test for promising configurations.
-- the second use proved critical to the adoption of the system and was the user's
capability to accept and reject suggestions made by the computer. SYSNET displays
suggested changes and allows the user to evaluate each one in terms of difficult to
quantify factors. Also local factors, such as work rules or special operating practices that
are not incorporated into the model can be accounted for by a knowledgeable user.
(d) For strategic planning, the outputs from SYSNET are a set of reports used to prepare
management summaries on different options. SYSNET is also used on a operational basis
to perform load planning. ln this role, SYSNET is used to maintain a file that determines
the actual routing of shipments through the service network. This file, which contains the
load planning, is accessed directly by systems that are used by every terminal manager in
the field. SYSNETS control of load planning and its capability to communicate these
instructions to the field is the most important accomplishment of the project.
(e) SYSNETs effect can be seen in four areas:
-- quality of planning practices and management culture
-- cost savings resulting directly from improvement in load planning
-- in analyzing projects
--improved service to customers from more reliable transportation
Qualitative changes includes the following:
--management had more control over network operations. SYSNET now allows
managers to have direct control. The new load pattern closely controls the
loading of directs and management can quickly change the load pattern in
response to changing needs.
-- it could set real istic performance standards. SYSNET allowed Yellow to set
direct loading standards based on anticipated freight levels, creating more
realistic performance expectations.
-- planners can better understand the total system now. Yel\ow can now evaluate
new projects and ideas based on their impact on the entire system
-- SYSNET allows managers to analyze projects formally before making
decisions
-- with SYSNET managers can analyze new options quickly in response to
changing situations
-- Analysts can now try new ideas on computers which ultimately leads to new
ideas in the field
-- because of SYSNET, Yellow is more open now to use of new information
technologies
-- the new system has reduced claims. SYSNET has had a substantial impact on
management culture at Yellow
Performance improvement due to better load planning include:
A study was undertaken to estimate savings that could be attributed to SYSNET.
Total cost savings for the system were estimated at over $7.3 million annually. Savings in
breakbulk handling costs also increased.
Besides this, reducing shipments handled in the long run may bring down
investments in fixed facilities. SYSNET brought down the cost of routing trailers in part
2-16
by identifying directs with lower transportation costs savings due to better routing of
trailers were estimated to be $1 million annually.
Ongoing projects include:
Operations planning uses SYSNET to scrutinize various projects with a wide
range from relocating breakbulks to realigning satellites with breakbulks.
Using SYSNET, operational planning now completes over 200 projects per
year, mostly on an informal, exploratory basis. SYSNETS speed in evaluating
different ideas is critical to this process.
In 1990, Yellow used SYSNET to identify over $10 million in annual savings
from different projects. SYSNET improved the speed with which such
analyses could be completed and expanded the scope of each project thus
allowing Yellow to study system impacts with more precision than before.
SYSNET thus has played a main role in identification, design and evaluation
of these projects.
Improved service includes :
Savings from SYSNET are substantial compared to the cost of its development
and implementation. Following the implementation of SYSNET management
can be better focused on improved service.
Yellow continues to use SYSNET for a number of planning projects and to
continuously monitor and improve the load planning system which is now
used directly within linehaul operations group responsible for day-to-day
management of tlows through the system. tn addition, Yellow is using
SYSNET as a foundation to expand the use of optimization methods for the
other aspects of its operations.
SYSNET is now very popular within the company for its capability to carry out
accurate, comprehensive network planning projects.
2.6-1.
(a) Implementing this major change in operations needed involvement and support of all
levels of the company. The process started with acceptance of system with operation
planning department. Operation planning was responsible for guiding the project and
managing with close cooperation from the information services department and all
aspects of the implementation. The systems acceptance was largely due to the use of
interactive optimization which gave users the support needed to optimize such a large
network while simultaneously keeping them in close control of the entire process. Users
could also analyze suggested changes to the network based on changes in flows and
costs, which could be compared against actual field totals.
The next step was to validate the cost model. They were able to compare both
total system costs and different subcategories against actual cost summaries for these
categories. The individual cost categories within SYSNET consistently match corporate
statistics within a few percent and total costs often match with l or 2 percent.
The validation of the cost model, both in totality and individual components,
played a vital role in gaining upper management's acceptance. The interactive reports and
features that convinced operations planning also played a strong role in winning support
of top management. They ran sessions for upper management to demonstrate how
SYSNET made suggestions and generated supporting reports to back-up the numbers.
They also demonstrated how standard operating practices could be detrimental and why
coordinating the entire network was important. By taking all these efforts, they gained
the needed confidence of upper management required to support a field implementation.
2-17
(b) With the support of upper management, they were able to develop an implementation
strategy. The controlled direct program changed operating philosophy so drastically that
a single corporate-wide transition was viewed as not safe. In implementing SYSNET,
Yellow made a systematic change in the way it loaded directs. SYSNET encourages a
greater pfoportion of directs to be loaded onto breakbulks. It was not possible to change
this operation methods so easily over the whole network. It was also difficult to do it in a
piecemeal fashion. To deal with this problem, they developed a phased implementation
strategy that started with smallest breakbulks in the system and went up to larger ones.
Careful planning made sure that no breakbulk would be over capacity during the
intermediate stages of the process. The entire implementation was so planned as to ensure
that no breakbulk would find itself over capacity during the transition period.
(c) To communicate the new concept to terminal managers in the field involved three
steps:
-- designing new support tools so that SYSNET routing instructions were easy to
follow
-- training terminal managers and dock personnel to use these new system and
most important
--convincing terminal managers that the new approach was a good idea. They
developed two new support tools to assist field operations :
-- first was a set of reports that managers or dock supervisors could access from
their local computer terminals which would give them immediate access to
SYSNET load pattern.
-- second, was a revised shipment movement bill. This provides a very high level
of control over the routing of individual shipments.·
The Operations Planning department handled training by organizing series of
visits to all 25 breakbulks. During each visit, the staff members explained the principles
behind the controlled direct program, new reports and use of new routing directions.
Follow-up was done by phone calls.
The most important task was to convince terminal managers of the logic behind
the new operations strategy. Terminal managers needed to understand that they had to
follow the load planning since it was designeed to coordinate different parts of the
system. They used examples to illustrate the effect their decisions could have on other
terminals. Generally, people in the field accepted the principle that their decisions should
be coordinated with those in the rest of the system
(d) Following the implementation of SYSNET, they developed a target that represented
the expected number of directs that they should be loading based on the SYSNET plan.
Yellow then measured terminal manager's performance based on how close they were to
this target. After some period, it deemed compliance with the plan so good that it now
measures terminal managers performance on other activities and Yellow continues to
monitor compliance with the load plan informally. It then contacts managers that appear
to be not in compliance to determine the reasons. In short, SYSNET has changed load
planning from a decentralized process that depended on local management incentives to a
centralized process that relies on monitoring and enforcement
2.6-2.
(a)
The information processing industry has experienced several decades of sustained
profitable growth. Recently, competition has intensified leading to quick advances in
computer technology. This in turn leads to proliferation of both-end products and
2-18
services. These trends are especially relevant for after-sales service. Maintaining a
service parts logistic system to support products installed in the field is essential to
competing in this industry.
Growth in both sales and scope of products offered has dramatically increased the
number of spare parts that must be maintained. For IBM, the number of installed
machines and the annual usage of spare parts have both increased. This growth has
increased the dollar value of service inventories, which are used to maintain the very high
levels of service expected by IBMs customers. IBM has developed an extensive multipleechelon logistic structure to provide ready service for the large population of installed
machines, which are distributed through the United States.
IBM developed a large and sophisticated inventory management system to
provide customers with prompt and reliable service. A fast changing business
environment and pressures to decrease investment in inventory led IBM to look for
improvements in its control system.
In response to these new needs, IBM initiated the development of a new planning
and control system for management of service parts. The result of this was the creation
and implementation of a system called Optimizer.
(b) The complicating factors faced by the OR team are as follows :
-- there are more than 15 million part-location combinations
-- there are more than 50000 product-location combinations
-- frequent updating (weekly) of system control parameters was a requirement in
response to changes in the service environment and installed base
-- success of the system is important to IBM's daily operations and so can have a
major impact on its future sales and revenues
-- employees could be expected to protest against any change since the existing
control system was working and sophisticated and overall parts logistics
problem was complex.
(c) The system developed in this phase had minimum interface to provide data inputs and
multi-echelon algorithm without any improvements. Most of the big changes from the
original design was in this phase.
They discovered that the echelon structure was in reality more complex than the
one used in the analytic model. Consequently, they had to develop extensions to the
demand pass-up methodology and incorporate them into the model.
The test was conducted in early 1986 and led to the finding that the value of the
total inventory generated by the new system was smaller than expected. lt was discovered
that the problem was due to differences in criticality of parts. The algorithm made
extensive use of inexpensive, non-functional parts to meet product-service objective.
Another problem found out at this stage was the churn (instability) in the recommended
stock levels every week. Although stock levels are expected to change periodically in
response to changing failure rates and to changes in the installed base, it is desirable to
keep the stock levels quasi-static in order to avoid logistic and supply problems. They
developed control procedures and changed the model to take care of this problem.
(d) In this phase, they completed all functions required for implementation and
developed a measurement system to monitor the field implementation test. After being
done with the system coding for this phase, they conducted an extensive user acceptance
test. Every program module was tested individually and jointly. Finally, a field
implementation test went live on 7 machine types in early 1987. The working of the
2-19
system fulfilled expectations. Scope of the field test was slowly expanded. Results were
monitored on a weekly and then monthly basis by the measurement system.
(e) In this phase, they completed the development and installation of all the functions
currently in place in Optimizer. The system was able to provide the specified service
performance for all parts and locations. Improvements were made. User acceptance
testing and integration of final system went smoothly. The project staging helped to
sustain support for the project by demonstrating concrete progress throughout the
implementation process. lt also helped to eradicate problems in formulation and
algorithm and programming bugs early. So very few problems occurred when the system
went live in a national basis. The final Optimizer system for national implementation
consisted of four major modules:
-- a forecasting system module
-- a data delivery system module
-- a decision system that solves multi-echelon stock control problem
-- the PIMS interface system
(f) The implementation of Optimizer yielded a variety of benefits:
-- a decrease in investment on inventory
-- improved services
-- enhanced flexibility in responding to changing service requirements
-- provision of a planning capability
-- improved understanding of the impact of parts operations
-- increased responsive of the control system
-- increased efficiency of NSD human resources
-- identifying the role of functional parts in providing product service is an
example of benefits derived from implementation of Optimizer
-- ability to run Optimizer on a weekly basis has increased responsiveness of
entire parts inventory system
-- for machines controlled by Optimizer, inventory analysts no longer have to
specify parts stocking lists for each echelon in order to make sure that service
objectives are attained. They can now focus on other critical management
issues.
Optimizer thus has proved to be an extremely valuable planning and operating
control tool.
2.6-3.
(a) The main objective is to teach optimization principles to key employees and to
acquaint them (at a high level) with the available optimization tools, without turning
them into mathematicians.
(b) Six three-day modules conducted over a period of two years, interspersed with small
group assignments, plus two days per week for six months to complete a master case
study.
(c) They are designated as supply chain masters.
2.7-1.
Answers will vary.
2-20
2.7-2.
Answers will vary.
2.7-3.
Answers will vary.
2-21
CHAPTER 3: INTRODUCTION TO LINEAR PROGRAMMING
3.1-1.
Swift & Company solved a series of LP problems to identify an optimal production
schedule. The first in this series is the scheduling model, which generates a shift-level
schedule for a 28-day horizon. The objective is to minimize the difference of the total
cost and the revenue. The total cost includes the operating costs and the penalties for
shortage and capacity violation. The constraints include carcass availability, production,
inventory and demand balance equations, and limits on the production and inventory. The
second LP problem solved is that of capable-to-promise models. This is basically the
same LP as the first one, but excludes coproduct and inventory. The third type of LP
problem arises from the available-to-promise models. The objective is to maximize the
total available production subject to production and inventory balance equations.
As a result of this study, the key performance measure, namely the weekly percent-sold
position has increased by 22%. The company can now allocate resources to the
production of required products rather than wasting them. The inventory resulting from
this approach is much lower than what it used to be before. Since the resources are used
effectively to satisfy the demand, the production is sold out. The company does not need
to offer discounts as often as before. The customers order earlier to make sure that they
can get what they want by the time they want. This in turn allows Swift to operate even
more efficiently. The temporary storage costs are reduced by 90%. The customers are
now more satisfied with Swift. With this study, Swift gained a considerable competitive
advantage. The monetary benefits in the first years was $12.74 million, including the
increase in the profit from optimizing the product mix, the decrease in the cost of lost
sales, in the frequency of discount offers and in the number of lost customers. The main
nonfinancial benefits are the increased reliability and a good reputation in the business.
3.1-2.
(a)
(b)
3-1
(c)
(d)
3.1-3.
(a)
(b)
Slope-Intercept Form
3-2
Slope
Intercept
3.1-4.
(a)
1
(b) The slope is
0
1
, the
intercept is 0.
(c)
3.1-5.
Optimal Solution:
and
3-3
3.1-6.
Optimal Solution:
and
3.1-7.
(a) As in the Wyndor Glass Co. problem, we want to find the optimal levels of two
activities that compete for limited resources. Let
be the number of wood-framed
windows to produce and be the number of aluminum-framed windows to produce. The
data of the problem is summarized in the table below.
Resource
Glass
Aluminum
Wood
Unit Profit
(b)
Resource Usage per Unit of Activity
Wood-framed Aluminum-framed
$
$
maximize
subject to
3-4
Available Amount
(c) Optimal Solution:
,
and
(d) From Sensitivity Analysis in IOR Tutorial, the allowable range for the profit per
wood-framed window is between
and infinity. As long as all the other parameters
are fixed and the profit per wood-framed window is larger than $
, the solution
found in (c) stays optimal. Hence, when it is $
instead of $
, it is still optimal to
produce wood-framed and
aluminum-framed windows and this results in a total
profit of $
. However, when it is decreased to $
, the optimal solution is to make
wood-framed and aluminum-framed windows. The total profit in this case is
$
.
(e)
maximize
subject to
The optimal production schedule consists of
windows, with a total profit of $
.
wood-framed and
aluminum-framed
3.1-8.
(a) Let
be the number of units of product to produce and
be the number of units
of product to produce. Then the problem can be formulated as follows:
maximize
subject to
3-5
(b) Optimal Solution:
,
and
3.1-9.
(a) Let
be the number of units on special risk insurance and
on mortgages.
be the number of units
maximize
subject to
,
(b) Optimal Solution:
,
and
(c) The relevant two equations are
,
and
.
3.1-10.
(a)
maximize
subject to
3-6
, so
and
(b) Optimal Solution:
,
and
3.1-11.
(a) Let
be the number of units of product produced for
maximize
0
0
subject to
,
,
(b)
3-7
.
3.1-12.
3-8
3.1-13.
First note that
satisfies the three constraints, i.e.,
is always feasible for any
value of . Moreover, the third constraint is always binding at
,
.
To check if
is optimal, observe that changing simply rotates the line that always
passes through
. Rewriting this equation as
, we see that the
slope of the line is
, and therefore, the slope ranges from to
.
As we can see,
is optimal as long as the slope of the third constraint is less than the
slope of the objective line, which is
. If
, then we can increase the objective by
3-9
traveling along the third constraint to the point
of
when
. For
,
is optimal.
, which has an objective value
3.1-14.
Case 1:
If
If
(vertical objective line)
, the objective value increases as
increases, so
, the opposite is true so that all the points on the line from
, are optimal.
If
, the objective function is
Case 2:
If
If
, point
,
If
If
the line
)
.
, point
, any point on the line
is optimal.
Case 3:
shifted down)
, line
.
, point
,
to
.
and every feasible point is optimal.
(objective line with slope
,
, point
.
is optimal. Similarly, if
(objective line with slope
, any point on
, objective value increases as the line is
If
, i.e.,
,
, point
.
If
, i.e.,
,
, point
If
, i.e.,
,
is any point on the line
.
3-10
.
3.2-1.
(a)
maximize
subject to
(b) Optimal Solution:
and
(c) We have to solve
and
from the first one, we obtain
we get
, hence
. By subtracting the second equation
. Plugging this in the first equation,
, so
.
3.2-2.
(a) TRUE (e.g., maximize
)
(b) TRUE (e.g., maximize
)
(c) FALSE (e.g., maximize
)
3.2-3.
(a) As in the Wyndor Glass Co. problem, we want to find the optimal levels of two
activities that compete for limited resources. Let
and
be the fraction purchased of
the partnership in the first and second friends venture respectively.
Resource
Fraction of partnership in 1st
Fraction of partnership in 2nd
Money
Summer work hours
Unit Profit
Resource Usage per Unit of Activity
1
2
$
$
$
$
3-11
Available Amount
$
(b)
maximize
0
0
subject to
,
(c) Optimal Solution: (
and
3.2-4.
Optimal Solutions: (
connecting these two points,
,
and all points lying on the line
3-12
3.2-5.
3.2-6.
(a)
3-13
(b) Yes. Optimal solution: (
and
(c) No. The objective function value rises as the objective line is slid to the right and
since this can be done forever, so there is no optimal solution.
(d) No, if there is no optimal solution even though there are feasible solutions, it means
that the objective value can be made arbitrarily large. Such a case may arise if the data of
the problem are not accurately determined. The objective coefficients may be chosen
incorrectly or one or more constraints might have been ignored.
3.3-1.
Proportionality: It is fair to assume that the amount of work and money spent and the
profit earned are directly proportional to the fraction of partnership purchased in either
venture.
3-14
Additivity: The profit as well as time and money requirements for one venture should not
affect neither the profit nor time and money requirements of the other venture. This
assumption is reasonably satisfied.
Divisibility: Because both friends will allow purchase of any fraction of a full
partnership, divisibility is a reasonable assumption.
Certainty: Because we do not know how accurate the profit estimates are, this is a more
doubtful assumption. Sensitivity analysis should be done to take this into account.
3.3-2.
Proportionality: If either variable is fixed, the objective value grows proportionally to the
increase in the other variable, so proportionality is reasonable.
Additivity: It is not a reasonable assumption, since the activities interact with each other.
For example, the objective value at
is not equal to the sum of the objective values
at
and
.
Divisibility: It is not justified, since activity levels are not allowed to be fractional.
Certainty: It is reasonable, since the data provided is accurate.
3.4-1.
In this study, linear programming is used to improve prostate cancer treatments. The
treatment planning problem is formulated as an MIP problem. The variables consist of
binary variables that represent whether seeds were placed in a location or not and the
continuous variables that denote the deviation of received dose from desired dose. The
constraints involve the bounds on the dose to each anatomical structure and various
physical constraints. Two models were studied. The first model aims at finding the
maximum feasible subsystem with the binary variables while the second one minimizes a
weighted sum of the dose deviations with the continuous variables.
With the new system, hundreds of millions of dollars are saved and treatment outcomes
have been more reliable. The side effects of the treatment are considerably reduced and
as a result of this, postoperation costs decreased. Since planning can now be done just
before the operation, pretreatment costs decreased as well. The number of seeds required
is reduced, so is the cost of procuring them. Both the quality of care and the quality of
life after the operation are improved. The automated computerized system significantly
eliminates the variability in quality. Moreover, the speed of the system allows the
clinicians to efficiently handle disruptions.
3.4-2.
(a) Proportionality: OK, since beam effects on tissue types are proportional to beam
strength.
Additivity: OK, since effects from multiple beams are additive.
Divisibility: OK, since beam strength can be fractional.
Certainty: Due to the complicated analysis required to estimate the data about radiation
absorption in different tissue types, sensitivity analysis should be employed.
(b) Proportionality: OK, provided there is no setup cost associated with planting a crop.
3-15
Additivity: OK, as long as crops do not interact.
Divisibility: OK, since acres are divisible.
Certainty: OK, since the data can be accurately obtained.
(c) Proportionality: OK, setup costs were considered.
Additivity: OK, since there is no interaction.
Divisibility: OK, since methods can be assigned fractional levels.
Certainty: Data is hard to estimate, it could easily be uncertain, so sensitivity analysis is
useful.
3.4-3.
(a) Reclaiming solid wastes
Proportionality: The amalgamation and treatment costs are unlikely to be proportional.
They are more likely to involve setup costs, e.g., treating 1,000 lbs. of material does not
cost the same as treating 10 lbs. of material 100 times.
Additivity: OK, although it is possible to have some interaction between treatments of
materials, e.g., if A is treated after B, the machines do not need to be cleaned out.
Divisibility: OK, unless materials can only be bought or sold in batches, say, of 100 lbs.
Certainty: The selling/buying prices may change. The treatment and amalgamation costs
are, most likely, crude estimates and may change.
(b) Personnel scheduling
Proportionality: OK, although some costs need not be proportional to the number of
agents hired, e.g., benefits and working space.
Additivity: OK, although some costs may not be additive.
Divisibility: One cannot hire a fraction of an agent.
Certainty: The minimum number of agents needed may be uncertain. For example, 45
agents may be sufficient rather than 48 for a nominal fee. Another uncertainty is whether
an agent does the same amount of work in every shift.
(c) Distributing goods through a distribution network
Proportionality: There is probably a setup cost for delivery, e.g., delivering 50 units one
by one does probably cost much more than delivering all together at once.
Additivity: OK, although it is possible to have two routes that can be combined to
provide lower costs, e.g., F2-DC
50, but the truck may be able to deliver 50
DC-W2
units directly from F2 to W2 without stopping at DC and hence saving some money.
Another question is whether F1 and F2 produce equivalent units.
Divisibility: One cannot deliver a fraction of a unit.
Certainty: The shipping costs are probably approximations and are subject to change. The
amounts produced may change as well.. Even the capacities may depend on available
3-16
daily trucking force, weather and various other factors. Sensitivity analysis should be
done to see the effects of uncertainty.
3.4-4.
Optimal Solution: (
and
3.4-5.
Optimal Solution: (
and
3-17
3.4-6.
The feasible region can be represented as follows:
Given
, various cases that may arise are summarized in the following table:
slope
optimal solution
,
8
8
8
and all points on the line connecting these two
4
4
,
and all points on the line connecting these two
4
3-18
3.4-7.
(a) Optimal Solution: (
and
(b) Optimal Solution: (
and
(c) Optimal Solution: (
and
3-19
3.4-8.
(a)
minimize
8
4
subject to
(b) Optimal Solution:
(
and
21.82
(c)
Cost per Serving
Carbohydrates
Protein
Fat
Solution
Steak
$8
Potatoes
$4
Grams of Ingredients per Serving
5
15
20
5
15
2
1.27
2.91
3-20
Totals
50
40
24.91
>=
>=
<=
Requirement (g)
50
40
60
Total Cost
$21.82
3.4-9.
(a) Let
be the amount of space leased for
.
months in month
minimize
subject to
,
and
(b)
11
Unit Cost $650
Month
1
2
3
4
5
Space Leased (sf)
12
13
14
15
$1,000 $1,350 $1,600 $1,900
21
$650
22
23
24
$1,000 $1,350 $1,600
31
$650
32
33
$1,000 $1,350
41
$650
42
$1,000
51
$650
Contribution Toward Required Amount
1
0
1
1
0
1
1
1
0
1
1
1
1
0
1
1
1
1
1
30000
1
0
1
1
0
1
1
1
0
1
1
1
1
0
1
10000
1
1
0
1
1
1
1
0
0
1
1
1
0
20000
Totals
$30,000
$30,000
$40,000
$30,000
$50,000
>=
>=
>=
>=
>=
Resource
Available
$30,000
$20,000
$40,000
$10,000
$50,000
Total Cost
$76,500,000
3.4-10.
(a) Let
number of full-time consultants working the morning shift (8 a.m.-4 p.m.),
number of full-time consultants working the afternoon shift (Noon-8 p.m.),
number of full-time consultants working the evening shift (4 p.m.-midnight),
number of part-time consultants working the first shift (8 a.m.-noon),
number of part-time consultants working the second shift (Noon-4 p.m.),
number of part-time consultants working the third shift (4 p.m.-8 p.m.),
number of part-time consultants working the fourth shift (8 p.m.-midnight).
minimize
subject to
3-21
(b)
FT1 FT2 FT3 PT1 PT2 PT3 PT5
Unit Cost $320 $320 $320 $120 $120 $120 $120
Time of Day
8am Noon
Noon 4pm
4pm 8pm
8pm Midnight
Contribution Toward Required Amount
1
1
1
1
1
1
1
1
1
1
Number Hired 2.667 2.667
4
1.333 2.667 3.333
Totals
4
8
10
6
>=
>=
>=
>=
Minimum
Required
4
8
10
6
FT
2.667
5.333
6.667
4
>=
>=
>=
>=
2
*PT
2.667
5.333
6.667
4
Total Cost
$4,107
2
Note that the optimal solution has fractional components. If the number of consultants
have to be integer, then the problem is an integer programming problem and the solution
is
with cost $
.
3.4-11.
(a) Let
be the number of units shipped from factory
to customer
minimize
subject to
and
,
and
(b)
Shipping
Customer 1
Cost
Factory 1
$600
Factory 2
$400
Customer 2
$800
$900
Customer 3
$700
$600
Units
Customer 1
Shipped
Factory 1
0
Factory 2
300
300
=
Order Size
300
Customer 2
200
0
200
=
200
Customer 3
200
200
400
=
400
3.4-12.
(a)
3-22
400
500
=
=
Output
400
500
Total Cost
$540,000
.
(b)
maximize
subject to
and
A
(c)
Unit Profit
A1
0
Year
1
2
3
4
5
A2
0
1
1
1.4
1.4
A3
0
A4 B1
1.4 0
B2 B3 C2
0 1.7 1.9
D5
1.3
R1 R2 R3 R4 R5
0 0 0 0 1
Contribution Toward Required Amount
1
1
1
1
1
1 1.7
1.4
1.7
1
Amount Invested $60,000 $0 $84,000 $0
$0
$0
1
1 1
1 1
1 1
1 1
Totals
$60,000
$0
$0
$0
$0
=
=
=
=
=
Required
Amount
$60,000
$0
$0
$0
$0
Total Profit
$152,880
$0 $0 $117,600 $0 $0 $0 $0 $0
3.4-13.
(a) Let
be the amount of Alloy used for
minimize
22
2
.
25
2
27
subject to
and
(b)
Cost per Pound
Requirement
% tin
% zinc
% lead
% total
Alloy 1
$22
Alloy 2
$20
Alloy 3
$25
Alloy 4
$24
Alloy 5
$27
Contribution Toward Required Amount
60
25
45
20
50
10
15
45
50
45
30
60
10
30
10
1
1
1
1
1
Proportion 0.0435
0.2826
0.6739
0
3-23
0
Totals
40
35
25
1
=
=
=
=
Required
Amount
40
35
25
1
Cost per Pound
$23.46
3.4-14.
(a) Let
be the number of tons of cargo type
F (front), C (center), B (back).
stowed in compartment
maximize
subject to
and
(b)
Volume (cf/ton)
Profit (per ton)
Cargo
Placement (tons)
Front
Center
Back
Total
Available (tons)
Percentage of Front Capacity
Percentage of Front Capacity
Cargo 1
500
$320
Cargo 2
700
$400
Cargo 3
600
$360
Cargo 4
400
$290
Cargo 1
0
0
10
10
<=
20
Cargo 2
0
6
0
6
<=
16
Cargo 3
11
0
0
11
<=
25
Cargo 4
1
12
0
13
<=
13
100%
100%
=
=
100%
100%
Total
Weight
12
18
10
<=
<=
<=
Weight
Capacity
12
18
10
Total Profit
$13,330
Percentage of Middle Capacity
Percentage of Back Capacity
3-24
Total
Volume
Volume
Capacity
7,000 <=
7,000
9,000 <=
9,000
5,000 <=
5,000
3.4-15.
(a) Let
,
,
be the number of hours operator is assigned to work on day
,
,
and
,
, ,
, .
minimize
subject to
,
,
,
,
,
,
,
,
,
,
,
,
for all , .
3-25
for
,
(b)
Wage Rate
$10.00
$10.10
$9.90
$9.80
$10.80
$11.30
Monday
6
0
4
5
3
0
Tuesday
0
6
8
5
0
0
Hours Worked
K.C.
D.H.
H.B.
S.C.
K.S.
N.K.
Hours Worked
Monday
2
0
4
5
3
0
14
=
14
Tuesday
0
2
7
5
0
0
14
=
14
K.C.
D.H.
H.B.
S.C.
K.S.
N.K.
Hours Needed
Hours Available
Wednesday Thursday
6
0
0
6
4
0
5
0
3
8
0
6
Hours Worked
Wednesday
4
0
4
5
1
0
14
=
14
<=
Thursday
0
6
0
0
3
5
14
=
14
Friday
6
0
4
5
0
2
Friday
3
0
4
5
0
2
14
=
14
Hours
Worked
9
8
19
20
7
7
>=
>=
>=
>=
>=
>=
Output
8
8
8
8
7
7
Total Cost
$710
Hours Available
3.4-16.
(a) Let
slices of bread,
tablespoons of peanut butter,
tablespoons of strawberry jelly,
graham crackers,
cups of milk, and
cups of juice.
minimize
subject to
and
(b)
Unit Cost (cents)
Total Calories
Vitamin C (mg)
Protein (g)
Calories from Fat
Contents (tbsp)
Bread
(slice)
5
70
0
3
10
Bread
(slice)
2
=
2
Peanut Butter
Total Liquid
Peanut
Butter
(tbsp.)
4
Strawberry Graham
Jelly
Cracker
(tbsp.)
(tbsp.)
7
8
Nutritional Contents
100
50
60
0
3
0
4
0
1
75
0
20
Peanut
Butter
(tbsp.)
0.575
0.575
1
Strawberry Graham
Jelly
Cracker
(tbsp.)
(tbsp.)
0.287
1.039
>=
>=
0.575
1
Milk
(cup)
15
Juice
(cup)
35
150
2
8
70
100
120
1
0
Milk
(cup)
0.516
Juice
(cup)
0.484
2
3-26
Level
Achieved
Minimum
400
>=
400
<=
60
>=
60
13.949 >=
12
120
<=
Maximum
600
120
30%
of Total Calories
Total Cost (cents/student)
47.31
Times Strawberry Jelly
3.5-1.
Upon facing problems about juice logistics, Welch's formulated the juice logistics model
(JLM), which is "an application of LP to a single-commodity network problem. The
decision variables deal with the cost of transfers between plants, the cost of recipes, and
carrying cost- all cost that are key to the common planning unit of tons" [p. 20]. The goal
is to find the optimal grape juice quantities shipped to customers and transferred between
plants over a 12-month horizon. The optimal quantities minimize the total cost, i.e., the
sum of transportation, recipe and storage costs. They satisfy balance equations, bounds
on the ratio of grape juice sold, and limits on total grape juice sold.
The JLM resulted in significant savings by preventing unprofitable decisions of the
management. The savings in the first year of its implementation were over $130,000.
Since the model can be run quickly, revising the decisions after observing the changes in
the conditions is made easier. Thus, the flexibility of the system is improved. Moreover,
the output helps the communication within the committee that is responsible for deciding
on crop usage.
3.5-2.
(a)
maximize
subject to
(b) Optimal Solution:
and
(c), (e), (f)
Contribution per unit
Resource 1
Resource 2
Resource 3
Activity 1
$20
Activity 2
$30
Resource Usage
per Unit of Activity
2
1
3
3
2
4
Activity 1
Level of Activity
3.333
Resource
Used
10
<=
20
<=
20
<=
Activity 2
3.333
Resource
Available
10
20
20
Total Contribution
$166.67
3-27
(d)
Feasible?
Yes
Yes
Yes
Yes
No
No
$
$
$
$
Best
3.5-3.
(a)
maximize
5
4
3
subject to
and
(b)
Unit Profit
Machine 1
Machine 2
Part A
$50
Part B
$40
Part C
$30
Processing Time (hours per unit)
0.02
0.03
0.05
0.05
0.02
0.04
Part A
Part B
Hours
Used
0 <=
0 <=
Part C
Hours
Available
40
40
Total Profit
$0.00
Production
(c) Many answers are possible.
Feasible?
No
Yes
Yes
$57 5
$6
Best
(d)
Unit Profit
Part A
$50
Part B
$40
Machine 1
Machine 2
Processing Time (hours per unit)
0.02
0.03
0.05
0.05
0.02
0.04
Production
Part A
363.636
Part B
1090.909
Part C
$30
Hours
Used
40 <=
40 <=
Part C
0
Hours
Available
40
40
Total Profit
$61,818.18
3-28
3.5-4.
(a)
minimize
subject to
and
(b) Optimal Solution:
and
(c), (e), (f)
Unit Cost
Activity 1
$60
Activity 2
$50
Benefit Contribution per
Unit of Each Activity
Benefit 1
5
3
Benefit 2
2
2
Benefit 3
7
9
Level of Activity
Activity 1
6.75
Minimum
Level
Acceptable
Achieved
Level
60
>=
60
31
>=
30
126
>=
126
Activity 2
8.75
Total Cost
$842.50
(d)
Feasible?
No
No
No
Yes
Yes
Yes
$
$
$
Best
3-29
3.5-5.
(a)
minimize
2.10
1.80
1.50
subject to
and
(b), (e), (f)
Unit Cost
(per kg)
Carbohydrates
Protein
Vitamins
Diet (kg)
Corn
$2.10
Tankage
$1.80
Alfalfa
$1.50
Nutritional Contents (per kg)
90
20
40
30
80
60
10
20
60
Corn
1.143
Tankage
0
Minimum
Level
Daily
Achieved
Requirement
200
>=
200
180
>=
180
157.1429 >=
150
Alfalfa
2.429
Total Cost
$6.04
(c)
is a feasible solution with a daily cost of $8.70. This diet will
provide 210 kg of carbohydrates, 310 kg of protein, and 170 kg of vitamins daily.
(d) Answers will vary.
3.5-6.
(a)
minimize
subject to
and
(b), (e), (f)
Year 5
Year 10
Year 20
Units Purchased
Income per Unit of Asset ($million)
Asset 1
Asset 2
Asset 3
2
1
0.5
0.5
0.5
1
0
1.5
2
Asset 1
100
Asset 2
200
Cash Flow
Minimum
Achieved
Required
400
>=
400
150
>=
100
300
>=
300
Asset 3
0
Total Cost
($million)
300
(c)
is a feasible solution. This would generate $400 million
in 5 years, $300 million in 10 years, and $550 million in 20 years. The total investment
will be $400 million.
(d) Answers will vary.
3-30
3.6-1.
(a) In the following, the indices
and
refer to products, months, plants,
processes and regions respectively. The decision variables are:
amount of product produced in month in plant
to be sold in region , and
using process and
amount of product stored to be sold in March in region
.
The parameters of the problem are:
demand for product
in month in region
,
unit production cost of product in plant
production rate of product in plant
using process ,
using process ,
selling price of product ,
transportation cost of product product in plant
to be sold in region
,
days available for production in month ,
storage limit,
storage cost per unit of product .
The objective is to maximize the total profit, which is the difference of the total revenue
and the total cost. The total cost is the sum of the costs of production, inventory and
transportation. Using the notation introduced, the objective is to maximize
subject to the constraints
for
February;
for
March;
;
;
for
for
February, March;
for
and
3-31
February, March
(b)
3-32
(c)
3-33
3-34
(d)
3-35
3-36
3.6-2.
(a)
(b)
3.6-3.
(a)
3-37
(b)
3.6-4.
(a)
3-38
3-39
(b)
3-40
3.6-5.
(a)
(b)
3-41
3.6-6.
(a)
(b)
3-42
3.6-7.
(a) The problem is to choose the amount of paper type to be produced on machine type
at paper mill and to be shipped to customer , which we can represent as
for
;
;
and
. The objective is to minimize
subject to
for
Note that
mill and
mill .
(b)
;
DEMAND
for
;
RAW MATERIAL
for
;
CAPACITY
for
;
is the total amount of paper type
is the total amount of paper type
;
shipped to customer
(c)
3-43
from paper
made on machine type at paper
functional constraints
decision variables
;
(d)
3.6-8
Answers will vary.
3.7-1.
Answers will vary.
3.7-2.
Answers will vary.
3-44
Case%3.1% %
!
a)! In!this!case,!we!have!two!decision!variables:!the!number!of!Family!Thrillseekers!
we!should!assemble!and!the!number!of!Classy!Cruisers!we!should!assemble.!!We!
also!have!the!following!three!constraints:!
!
1.!The!plant!has!a!maximum!of!48,000!labor!hours.!
2.!The!plant!has!a!maximum!of!20,000!doors!available.!
3.!The!number!of!Cruisers!we!should!assemble!must!be!less!than!or!equal!to!
3,500.!
!!
A
1
2
3
4
5
6
7
8
9
10
11
12
13
Unit Profit
Labor Hours
Doors
Production
B
C
Family
Thrillseeker
$3,600
Classy
Cruiser
$5,400
D
Resources
Used
48,000
20,000
Resource Requirements
6
10.5
4
2
Family
Thrillseeker
3,800
E
F
<=
<=
Resources
Available
48,000
20,000
Classy
Cruiser
2,400
<=
3,500
Demand
Total Profit
$26,640,000
!
!
4
5
6
7
!
!
!
!
D
Resources
Used
=SUMPRODUCT(B6:C6,Production)
=SUMPRODUCT(B7:C7,Production)
!
Range Name
Cells
ClassyCruisers
Demand
Production
ResourcesAvailable
ResourcesUsed
TotalProfit
UnitProfit
C11
C13
B11:C11
F6:F7
D6:D7
F11
B3:C3
F
10
Total Profit
11 =SUMPRODUCT(UnitProfit,Production)
Solver!Parameters!
Set%Objective%Cell:!TotalProfit!
To:!Max!
By%Changing%Variable%Cells:%
! Production!
Subject%to%the%Constraints:%
! ClassyCruisers!<=!Demand!
! ResourcesUsed!<=!Resources!
Available!
Solver%Options:%
% Make!Variables!Nonnegative!
3-45
!
!
!
!!!
Solving!Method:!Simplex!LP!
!
Rachel’s!plant!should!assemble!3,800!Thrillseekers!and!2,400!Cruisers!to!obtain!
a!maximum!profit!of!$26,640,000.!
!
b)! In!part!(a)!above,!we!observed!that!the!Cruiser!demand!constraint!was!not!
binding.!!Therefore,!raising!the!demand!for!the!Cruiser!will!not!change!the!
optimal!solution.!!The!marketing!campaign!should!not!be!undertaken.!
!
c)! The!new!value!of!the!rightZhand!side!of!the!labor!constraint!becomes!48,000!*!
1.25!=!60,000!labor!hours.!!All!formulas!and!Solver!settings!used!in!part!(a)!
remain!the!same.!
!
A
1
2
3
4
5
6
7
8
9
10
11
12
13
Unit Profit
Labor Hours
Doors
Production
B
C
Family
Thrillseeker
$3,600
Classy
Cruiser
$5,400
Resource Requirements
6
10.5
4
2
Family
Thrillseeker
3,250
Classy
Cruiser
3,500
<=
3,500
Demand
D
Resources
Used
56,250
20,000
E
F
<=
<=
Resources
Available
60,000
20,000
Total Profit
$30,600,000
!
Rachel’s!plant!should!now!assemble!3,250!Thrillseekers!and!3,500!Cruisers!to!
achieve!a!maximum!profit!of!$30,600,000.!
!
d)! Using!overtime!labor!increases!the!profit!by!$30,600,000!–!$26,640,000!=!
$3,960,000.!!Rachel!should!therefore!be!willing!to!pay!at!most!$3,960,000!extra!
for!overtime!labor!beyond!regular!time!rates.!
3-46
!
!
e)! The!value!of!the!rightZhand!side!of!the!Cruiser!demand!constraint!is!3,500!*!1.20!
=!4,200!cars.!!The!value!of!the!rightZhand!side!of!the!labor!hour!constraint!is!
48,000!*!1.25!=!60,000!hours.!!All!formulas!and!Solver!settings!used!in!part!(a)!
remain!the!same.!!Ignoring!the!costs!of!the!advertising!campaign!and!overtime!
labor,!!
!
A
1
2
3
4
5
6
7
8
9
10
11
12
13
Unit Profit
Labor Hours
Doors
Production
B
C
Family
Thrillseeker
$3,600
Classy
Cruiser
$5,400
Resource Requirements
6
10.5
4
2
Family
Thrillseeker
3,000
Classy
Cruiser
4,000
<=
4,200
Demand
D
Resources
Used
60,000
20,000
E
F
<=
<=
Resources
Available
60,000
20,000
Total Profit
$32,400,000
!
Rachel’s!plant!should!produce!3,000!Thrillseekers!and!4,000!Cruisers!for!a!
maximum!profit!of!$32,400,000.!!This!profit!excludes!the!costs!of!advertising!
and!using!overtime!labor.!
!
!
f)! The!advertising!campaign!costs!$500,000.!!In!the!solution!to!part!(e)!above,!we!
used!the!maximum!overtime!labor!available,!and!the!maximum!use!of!overtime!
labor!costs!$1,600,000.!!Thus,!our!solution!in!part!(e)!required!an!extra!
$500,000!+!$1,600,000!=!$2,100,000.!!We!perform!the!following!cost/benefit!
analysis:!
!
Profit!in!part!(e):!!
!
!
$32,400,000!
−!!Advertising!and!overtime!costs:!
$!!2,100,000!
!!
!
! !
!
!
$30,300,000!
!
We!compare!the!$30,300,000!profit!with!the!$26,640,000!profit!obtained!in!part!
(a)!and!conclude!that!the!decision!to!run!the!advertising!campaign!and!use!
overtime!labor!is!a!very!wise,!profitable!decision.!
3-47
!
g)! Because!we!consider!this!question!independently,!the!values!of!the!rightZhand!
sides!for!the!Cruiser!demand!constraint!and!the!labor!hour!constraint!are!the!
same!as!those!in!part!(a).!!We!now!change!the!profit!for!the!Thrillseeker!from!
$3,600!to!$2,800!in!the!problem!formulation.!!All!formulas!and!Solver!settings!
used!in!part!(a)!remain!the!same.!
A
1
2
3
4
5
6
7
8
9
10
11
12
13
Unit Profit
Labor Hours
Doors
Production
B
C
Family
Thrillseeker
$2,800
Classy
Cruiser
$5,400
Resource Requirements
6
10.5
4
2
Family
Thrillseeker
1,875
D
Resources
Used
48,000
14,500
E
F
<=
<=
Resources
Available
48,000
20,000
Classy
Cruiser
3,500
<=
3,500
Demand
Total Profit
$24,150,000
!
Rachel’s!plant!should!assemble!1,875!Thrillseekers!and!3,500!Cruisers!to!obtain!
a!maximum!profit!of!$24,150,000.!
!
!
h)! Because!we!consider!this!question!independently,!the!profit!for!the!Thrillseeker!
remains!the!same!as!the!profit!specified!in!part!(a).!!The!labor!hour!constraint!
changes.!!Each!Thrillseeker!now!requires!7.5!hours!for!assembly.!!All!formulas!
and!Solver!settings!used!in!part!(a)!remain!the!same.!!
!
A
1
2
3
4
5
6
7
8
9
10
11
12
13
Unit Profit
Labor Hours
Doors
Production
B
C
Family
Thrillseeker
$3,600
Classy
Cruiser
$5,400
Resource Requirements
7.5
10.5
4
2
Family
Thrillseeker
1,500
Classy
Cruiser
3,500
<=
3,500
Demand
D
Resources
Used
48,000
13,000
E
F
<=
<=
Resources
Available
48,000
20,000
Total Profit
$24,300,000
Rachel’s!plant!should!assemble!1,500!Thrillseekers!and!3,500!Cruisers!for!a!
maximum!profit!of!$24,300,000.!
3-48
!
!
i)! Because!we!consider!this!question!independently,!we!use!the!problem!
formulation!used!in!part!(a).!!In!this!problem,!however,!the!number!of!Cruisers!
assembled!has!to!be!strictly!equal!to!the!total!demand.!The!formulas!used!in!the!
problem!formulation!remain!the!same!as!those!used!in!part!(a).!!
!
A
1
2
3
4
5
6
7
8
9
10
11
12
13
Unit Profit
Labor Hours
Doors
Production
B
C
Family
Thrillseeker
$3,600
Classy
Cruiser
$5,400
Resource Requirements
6
10.5
4
2
Family
Thrillseeker
1,875
Classy
Cruiser
3,500
=
3,500
Demand
D
Resources
Used
48,000
14,500
E
F
<=
<=
Resources
Available
48,000
20,000
Total Profit
$25,650,000
!
The!new!profit!is!$25,650,000,!which!is!$26,640,000!–!$25,650,000!=!$990,000!
less!than!the!profit!obtained!in!part!(a).!!This!decrease!in!profit!is!less!than!
$2,000,000,!so!Rachel!should!meet!the!full!demand!for!the!Cruiser.!
3-49
!
!
j)! We!now!combine!the!new!considerations!described!in!parts!(f),!(g),!and!(h).!!In!
part!(f),!we!decided!to!use!both!the!advertising!campaign!and!the!overtime!
labor.!!The!advertising!campaign!raises!the!demand!for!the!Cruiser!to!4,200!
sedans,!and!the!overtime!labor!increases!the!labor!hour!capacity!of!the!plant!to!
60,000!labor!hours.!!In!part!(g),!we!decreased!the!profit!generated!by!a!
Thrillseeker!to!$2,800.!!In!part!(h),!we!increased!the!time!to!assemble!a!
Thrillseeker!to!7.5!hours.!The!formulas!and!Solver!settings!used!for!this!problem!
are!the!same!as!those!used!in!part!(a).!
!
A
1
2
3
4
5
6
7
8
9
10
11
12
13
Unit Profit
Labor Hours
Doors
Production
B
C
Family
Thrillseeker
$2,800
Classy
Cruiser
$5,400
Resource Requirements
7.5
10.5
4
2
Family
Thrillseeker
2,120
Classy
Cruiser
4,200
<=
4,200
Demand
D
Resources
Used
60,000
16,880
E
F
<=
<=
Resources
Available
60,000
20,000
Total Profit
$28,616,000
!
Rachel’s!plant!should!assemble!2,120!Thrillseekers!and!4,200!Cruisers!for!a!
maximum!profit!of!$28,616,000!–!$2,100,000!=!$26,516,000.!
3-50
!
Case%3.2%
!
a)! We!want!to!determine!the!amount!of!potatoes!and!green!beans!Maria!should!
purchase!to!minimize!ingredient!costs.!!We!have!two!decision!variables:!the!
amount!(in!pounds)!of!potatoes!Maria!should!purchase!and!the!amount!(in!
pounds)!of!green!beans!Maria!should!purchase.!!We!also!have!constraints!on!
nutrition,!taste,!and!weight.!
!
Nutrition!Constraints!
1.!!We!first!need!to!ensure!that!the!dish!has!180!grams!of!protein.!!We!are!told!
that!100!grams!of!potatoes!have!1.5!grams!of!protein!and!10!ounces!of!green!
beans!have!5.67!grams!of!protein.!!Since!we!have!decided!to!measure!our!
decision!variables!in!pounds,!however,!we!need!to!determine!the!grams!of!
protein!in!one!pound!of!each!ingredient.!
!
We!perform!the!following!conversion!for!potatoes:!
! 1.5 g protein $ ! 28.35 g $ ! 16 oz. $ 6.804 g protein
!!
!
=
#
&
" 100 g potatoes % " 1 oz. % " 1 lb. % 1 lb. of potatoes
!
We!perform!the!following!conversion!for!green!beans:!
! 5.67 g protein $ ! 16 oz.$
9.072 g protein
!!
!
=
#
&"
%
" 10 oz. green beans % 1 lb.
1 lb. of green beans
!
!
2.!We!next!need!to!ensure!that!the!dish!has!80!milligrams!of!iron.!!We!are!told!
that!100!grams!of!potatoes!have!0.3!milligrams!of!iron!and!10!ounces!of!green!
beans!have!3.402!milligrams!of!iron.!!Since!we!have!decided!to!measure!our!
decision!variables!in!pounds,!however,!we!need!to!determine!the!milligrams!of!
iron!in!one!pound!of!each!ingredient.!
!
We!perform!the!following!conversion!for!potatoes:!
! 0.3 mg iron $ ! 28.35 g $ ! 16 oz. $
1.361 mg iron
!!
!
=
#
&"
" 100g potatoes % 1 oz. % " 1 lb. % 1 lb. of potatoes
!
We!perform!the!following!conversion!for!green!beans:!
! 3.402 mg iron $ ! 16 oz.$
5.443 mg iron
!
!
=
#
&"
" 10 oz. green beans % 1 lb. % 1 lb. of green beans
3-51
!
!
3.!We!next!need!to!ensure!that!the!dish!has!1,050!milligrams!of!vitamin!C.!!We!
are!told!that!100!grams!of!potatoes!have!12!milligrams!of!vitamin!C!and!10!
ounces!of!green!beans!have!28.35!milligrams!of!vitamin!C.!!Since!we!have!
decided!to!measure!our!decision!variables!in!pounds,!however,!we!need!to!
determine!the!milligrams!of!vitamin!C!in!one!pound!of!each!ingredient.!
!
We!perform!the!following!conversion!for!potatoes:!
! 12 mg Vitamin C $ ! 28.35 g $ ! 16 oz.$ 54.432 mg Vitamin C
!!
!
=
#
&
" 100g potatoes % " 1 oz. % " 1 lb. %
1 lb. of potatoes
!
We!perform!the!following!conversion!for!green!beans:!
! 28.35 mg Vitamin C $ ! 16 oz. $ 45.36 mg Vitamin C
!!
!
=
#
&
" 10 oz. green beans % " 1 lb. % 1 lb. of green beans
!
!
Taste!Constraint!
Edson!requires!that!the!casserole!contain!at!least!a!six!to!five!ratio!in!the!weight!
of!potatoes!to!green!beans.!!We!have:!
!
pounds of potatoes
6
≥ !
!!
pounds of green beans 5
!
!!
5!(pounds!of!potatoes)!≥!6!(pounds!of!green!beans)!
!
Weight!Constraint!
Finally,!Maria!requires!a!minimum!of!10!kilograms!of!potatoes!and!green!beans!
together.!!Because!we!measure!potatoes!and!green!beans!in!pounds,!we!must!
perform!the!following!conversion:!
! 1000 g $ ! 1 lb $
10 kg of potatoes and green beans #
" 1 kg &% #" 453.6 g&% !
!!
= 22.046 lb of potatoes and green beans
3-52
!
!
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
B
Unit Cost (per lb.)
Protein (g)
Iron (mg)
Vitamin C (mg)
Quantity (lb.)
C
D
Potatoes
$0.40
Green Beans
$1.00
E
Total
Nutrition
194.87
80.00
1,251.27
Nutritional Data (per pound)
6.804
9.072
1.361
5.443
54.432
45.36
Potatoes
13.57
Green Beans
11.31
Minimum Weight (lb.)
5
Times Potatoes
Taste Constraint:
67.833
>=
F
G
>=
>=
>=
Nutritional
Requirement
180
80
1,050
Total Weight
25
>=
22.046
67.833
Total Cost
$16.73
6 Times Green Beans
!
!
3
4
5
6
7
8
9
Total Weight
!
10 =SUM(Quantity)
!
!
Range Name
BeanRatio
MinimumWeight
NutritionalRequirement
PotatoRatio
Quantity
TotalCost
TotalNutrition
TotalWeight
UnitCost
E
Total
Nutrition
=SUMPRODUCT(C5:D5,Quantity)
=SUMPRODUCT(C6:D6,Quantity)
=SUMPRODUCT(C7:D7,Quantity)
G
9
Total Cost
10 =SUMPRODUCT(UnitCost,Quantity)
!
A
14
15 5
B
Cells
E15
E12
G5:G7
C15
C10:D10
G10
E5:E7
E10
C2:D2
!
!
C
Taste Constraint:
Times Potatoes =A15*C10
D
E
>= =F15*D10
!
!
Solver!Parameters!
Set%Objective%Cell:!TotalCost!
To:!Min!
By%Changing%Variable%Cells:%
! Quantity!
Subject%to%the%Constraints:%
! PotatoRatio!>=!BeanRation!
! TotalNutrition!>=!
NutritionalRequirement!
! TotalWeight!<=!MinimumWeight!
Solver%Options:%
% Make!Variables!Nonnegative!
! Solving!Method:!Simplex!LP!
3-53
F
G
6 Times Green Beans
!
!!!!
!
Maria!should!purchase!13.57!lb.!of!potatoes!and!11.31!lb.!of!green!beans!to!
obtain!a!minimum!cost!of!$16.73.!
!
b)! The!taste!constraint!changes.!!The!new!constraint!is!now.!
pounds of potatoes
1
≥ !
!!
pounds of green beans 2
!
!!
2!(pounds!of!potatoes)!≥!1!(pounds!of!green!beans)!
!
The!formulas!and!Solver!settings!used!to!solve!the!problem!remain!the!same!as!
part!(a).!!
!
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
B
Unit Cost (per lb.)
Protein (g)
Iron (mg)
Vitamin C (mg)
Quantity (lb.)
C
D
Potatoes
$0.40
Green Beans
$1.00
Nutritional Data (per pound)
6.804
9.072
1.361
5.443
54.432
45.36
Potatoes
10.29
Green Beans
12.13
Minimum Weight (lb.)
2
Times Potatoes
Taste Constraint:
20.576
>=
E
Total
Nutrition
180.00
80.00
1,110.00
Total Weight
22
>=
22.046
12.125
F
G
>=
>=
>=
Nutritional
Requirement
180
80
1,050
Total Cost
$16.24
1 Times Green Beans
!
!
Maria!should!purchase!10.29!lb.!of!potatoes!and!12.13!lb.!of!green!beans!to!
obtain!a!minimum!cost!of!$16.24.!
3-54
!
c)! The!rightZhand!side!of!the!iron!constraint!changes!from!80!mg!to!65!mg.!!The!
formulas!and!Solver!settings!used!in!the!problem!remain!the!same!as!in!part!(a).!
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
B
Unit Cost (per lb.)
Protein (g)
Iron (mg)
Vitamin C (mg)
Quantity (lb.)
C
D
Potatoes
$0.40
Green Beans
$1.00
Nutritional Data (per pound)
6.804
9.072
1.361
5.443
54.432
45.36
Potatoes
15.80
Green Beans
7.99
Minimum Weight (lb.)
5
Times Potatoes
Taste Constraint:
79.001
>=
E
Total
Nutrition
180.00
65.00
1,222.51
F
G
>=
>=
>=
Nutritional
Requirement
180
65
1,050
Total Weight
24
>=
22.046
47.947
Total Cost
$14.31
6 Times Green Beans
!
!
Maria!should!purchase!15.80!lb.!of!potatoes!and!7.99!lb.!of!green!beans!to!obtain!
a!minimum!cost!of!$14.31.!
!
d)! The!iron!requirement!remains!65!mg.!!We!need!to!change!the!price!per!pound!of!
green!beans!from!$1.00!per!pound!to!$0.50!per!pound.!!The!formulas!and!Solver!
settings!used!in!the!problem!remain!the!same!as!in!part!(a).!
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
B
Unit Cost (per lb.)
Protein (g)
Iron (mg)
Vitamin C (mg)
Quantity (lb.)
C
D
Potatoes
$0.40
Green Beans
$0.50
Nutritional Data (per pound)
6.804
9.072
1.361
5.443
54.432
45.36
Potatoes
12.53
Green Beans
10.44
Minimum Weight (lb.)
5
Times Potatoes
Taste Constraint:
62.657
>=
E
Total
Nutrition
180.00
73.90
1,155.79
Total Weight
23
>=
22.046
62.657
F
G
>=
>=
>=
Nutritional
Requirement
180
65
1,050
Total Cost
$10.23
6 Times Green Beans
!
!
Maria!should!purchase!12.53!lb.!of!potatoes!and!10.44!lb.!of!green!beans!to!
obtain!a!minimum!cost!of!$10.23.!
3-55
!
e)! We!still!have!two!decision!variables:!!one!variable!to!represent!the!amount!(in!
pounds)!of!potatoes!Maria!should!purchase!and!one!variable!to!represent!the!
amount!(in!pounds)!of!lima!beans!Maria!should!purchase.!!To!determine!the!
grams!of!protein!in!one!pound!of!lima!beans,!we!perform!the!following!
conversion:!
! 22.68 g protein # ! 16 oz. # = 36.288 g protein !
!
" 10 oz. lima beens $ " 1 lb. $ 1 lb. of lima beans
!
To!determine!the!milligrams!of!iron!in!one!pound!of!lima!beans,!we!perform!the!
following!conversion:!
! 6.804 mg iron # ! 16 oz. # = 10.886 mg iron !
!
" 10 oz. lima beans $ " 1 lb. $ 1 lb. of lima beans
!
Lima!beans!contain!no!vitamin!C,!so!we!do!not!have!to!perform!a!measurement!
conversion!for!vitamin!C.!
!
We!change!the!decision!variable!from!green!beans!to!lima!beans!and!insert!the!
new!parameters!for!protein,!iron,!vitamin!C,!and!cost.!!The!formulas!and!Solver!
settings!used!in!the!problem!remain!the!same!as!in!part!(a).!
!
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
B
Unit Cost (per lb.)
Protein (g)
Iron (mg)
Vitamin C (mg)
Quantity (lb.)
C
D
Potatoes
$0.40
Lima Beans
$0.60
Nutritional Data (per pound)
6.804
36.288
1.361
10.886
54.432
0
Potatoes
19.29
Lima Beans
3.56
Minimum Weight (lb.)
5
Times Potatoes
Taste Constraint:
96.451
>=
E
Total
Nutrition
260.41
65.00
1,050.00
Total Weight
23
>=
22.046
21.356
F
G
>=
>=
>=
Nutritional
Requirement
180
65
1,050
Total Cost
$9.85
6 Times Lima Beans
!
!
Maria!should!purchase!19.29!lb.!of!potatoes!and!3.56!lb.!of!lima!beans!to!obtain!a!
minimum!cost!of!$9.85.!
!
f)! Edson!takes!pride!in!the!taste!of!his!casserole,!and!the!optimal!solution!from!
above!does!not!seem!to!preserve!the!taste!of!the!casserole.!!First,!Maria!forces!
Edson!to!use!lima!beans!instead!of!green!beans,!and!lima!beans!are!not!an!
ingredient!in!Edson’s!original!recipe.!!Second,!although!Edson!places!no!upper!
limit!on!the!ratio!of!potatoes!to!beans,!the!above!recipe!uses!an!over!five!to!one!
ratio!of!potatoes!to!beans.!!This!ratio!seems!unreasonable!since!such!a!large!
amount!of!potatoes!will!overpower!the!taste!of!beans!in!the!recipe.!
3-56
!
g)! We!only!need!to!change!the!values!on!the!rightZhand!side!of!the!iron!and!vitamin!
C!constraints.!!The!formulas!and!Solver!settings!used!in!the!problem!remain!the!
same!as!in!part!(a).!!The!values!used!in!the!new!problem!formulation!and!
solution!follow.!
!
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
B
Unit Cost (per lb.)
Protein (g)
Iron (mg)
Vitamin C (mg)
Quantity (lb.)
C
D
Potatoes
$0.40
Lima Beans
$0.60
Nutritional Data (per pound)
6.804
36.288
1.361
10.886
54.432
0
Potatoes
12.60
Lima Beans
9.45
Minimum Weight (lb.)
5
Times Potatoes
Taste Constraint:
62.988
>=
E
Total
Nutrition
428.58
120.00
685.72
Total Weight
22
>=
22.046
56.690
F
G
>=
>=
>=
Nutritional
Requirement
180
120
500
Total Cost
$10.71
6 Times Lima Beans
!
!
Maria!should!purchase!12.60!lb.!of!potatoes!and!9.45!lb.!of!lima!beans!to!obtain!a!
minimum!cost!of!$10.71.!
3-57
Case%3.3! !
!
a)! The!number!of!operators!that!the!hospital!needs!to!staff!the!call!center!during!
each!twoZhour!shift!can!be!found!in!the!following!table:!
!
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Work Shift
7am-9am
9am-11am
11am-1pm
1pm-3pm
3pm-5pm
5pm-7pm
7pm-9pm
B
C
Average
Number
of Calls
40
85
70
95
80
35
10
Average
Calls/hour
from English
Speakers
32
68
56
76
64
28
8
Percent English Speakers
D
E
F
Average
English Spanish
Calls/hour Speaking Speaking
from Spanish Agents Agents
Speakers
Needed Needed
8
6
2
17
12
3
14
10
3
19
13
4
16
11
3
7
5
2
2
2
1
80%
Calls Handled per hour
6
!
!
For!example,!the!average!number!of!phone!calls!per!hour!during!the!shift!from!
7am!to!9am!equals!40.!Since,!on!average,!80%!of!all!phone!calls!are!from!English!
speakers,!there!is!an!average!number!of!32!phone!calls!per!hour!from!English!
speakers!during!that!shift.!Since!one!operator!takes,!on!average,!6!phone!calls!
per!hour,!the!hospital!needs!32/6!=!5.333!EnglishZspeaking!operators!during!
that!shift.!The!hospital!cannot!employ!fractions!of!an!operator!and!so!needs!6!
EnglishZspeaking!operators!for!the!shift!from!7am!to!9am.!
3-58
!
b)! The!problems!of!determining!how!many!SpanishZspeaking!operators!and!
EnglishZspeaking!operators!Lenny!needs!to!hire!to!begin!each!shift!are!
independent.!Therefore!we!can!formulate!two!smaller!linear!programming!
models!instead!of!one!large!model.!We!are!going!to!have!one!model!for!the!
scheduling!of!the!SpanishZspeaking!operators!and!another!one!for!the!
scheduling!of!the!EnglishZspeaking!operators.!
!
Lenny!wants!to!minimize!the!operating!costs!while!answering!all!phone!calls.!
For!the!given!scheduling!problem!we!make!the!assumption!that!the!only!
operating!costs!are!the!wages!of!the!employees!for!the!hours!that!they!answer!
phone!calls.!The!wages!for!the!hours!during!which!they!perform!paperwork!are!
paid!by!other!cost!centers.!Moreover,!it!does!not!matter!for!the!callers!whether!
an!operator!starts!his!or!her!work!day!with!phone!calls!or!with!paperwork.!For!
example,!we!do!not!need!to!distinguish!between!operators!who!start!their!day!
answering!phone!calls!at!9am!and!operators!who!start!their!day!with!paperwork!
at!7am,!because!both!groups!of!operators!will!be!answering!phone!calls!at!the!
same!time.!And!only!this!time!matters!for!the!analysis!of!Lenny’s!problem.!
!
We!define!the!decision!variables!according!to!the!time!when!the!employees!have!
their!first!shift!of!answering!phone!calls.!For!the!scheduling!problem!of!the!
EnglishZspeaking!operators!we!have!7!decision!variables.!First,!we!have!5!
decision!variables!for!fullZtime!employees.!
!
The!number!of!operators!having!their!first!shift!on!the!phone!from!7am!to!9am.!
The!number!of!operators!having!their!first!shift!on!the!phone!from!9am!to!11am.!
The!number!of!operators!having!their!first!shift!on!the!phone!from!11am!to!1pm.!
The!number!of!operators!having!their!first!shift!on!the!phone!from!1pm!to!3pm.!
The!number!of!operators!having!their!first!shift!on!the!phone!from!3pm!to!5pm.!
!
In!addition,!we!define!2!decision!variables!for!partZtime!employees.!
!
The!number!of!partZtime!operators!having!their!first!shift!from!3pm!to!5pm.!
The!number!of!partZtime!operators!having!their!first!shift!from!5pm!to!7pm.!
!
The!unit!cost!coefficients!in!the!objective!function!are!the!wages!operators!earn!
while!they!answer!phone!calls.!!All!operators!who!have!their!first!shift!on!the!
phone!from!7am!to!9am,!9am!to!11am,!or!11am!to!1pm!finish!their!work!on!the!
phone!before!5pm.!They!earn!4*$10!=!$40!during!their!time!answering!phone!
calls.!All!operators!who!have!their!first!shift!on!the!phone!from!1pm!to!3pm!or!
3pm!to!5pm!have!one!shift!on!the!phone!before!5pm!and!another!one!after!5pm.!
They!earn!2*$10+2*$12!=!$44!during!their!time!answering!phone!calls.!The!
second!group!of!partZtime!operators,!those!having!their!first!shift!from!5pm!to!
7pm,!earn!4*$12!=!$48!during!their!time!answering!phone!calls.!
!
3-59
There!are!7!constraints,!one!for!each!twoZhour!shift!during!which!phone!calls!
need!to!be!answered.!The!rightZhand!sides!for!these!constraints!are!the!number!
of!operators!needed!to!ensure!that!all!phone!calls!get!answered!in!a!timely!
manner.!On!the!leftZhand!side!we!determine!the!number!of!operators!on!the!
phone!during!any!given!shift.!For!example,!during!the!11am!to!1pm!shift!the!
total!number!of!operators!answering!phone!calls!equals!the!sum!of!the!number!
of!operators!who!started!answering!calls!at!7am!and!are!currently!in!their!
second!shift!of!the!day!and!the!number!of!operators!who!started!answering!calls!
at!11am.!
!
The!following!spreadsheet!describes!the!entire!problem!formulation!for!the!
EnglishZspeaking!employees:!
!
A
1
English
2
Speaking
3
4
5
Unit Cost
6
7
Work Shift?
8
7am-9am
9
9am-11am
10
11am-1pm
11
1pm-3pm
12
3pm-5pm
13
5pm-7pm
14
7pm-9pm
15
16
17
18
19
20 Number Working
B
C
D
E
F
G
H
Full-Time
on Phone
7am-9am
11am-1pm
$40
Full-Time
on Phone
9am-11am
1pm-3pm
$40
Full-Time
on Phone
11am-1pm
3pm-5pm
$40
Full-Time
on Phone
1pm-3pm
5pm-7pm
$44
Full-Time
on Phone
3pm-5pm
7pm-9pm
$44
Part-Time
on Phone
3pm-7pm
$44
Part-Time
on Phone
5pm-9pm
$48
1
0
1
0
0
0
0
0
1
0
1
0
0
0
0
0
1
0
1
0
0
0
0
0
1
0
1
0
0
0
0
0
1
0
1
0
0
0
0
1
1
0
0
0
0
0
0
1
1
Full-Time
on Phone
7am-9am
11am-1pm
6
Full-Time
on Phone
9am-11am
1pm-3pm
13
Full-Time
on Phone
11am-1pm
3pm-5pm
4
Full-Time
on Phone
1pm-3pm
5pm-7pm
0
Full-Time
on Phone
3pm-5pm
7pm-9pm
2
Part-Time
on Phone
3pm-7pm
5
Part-Time
on Phone
5pm-9pm
0
I
Total
Working
6
13
10
13
11
5
2
!
!
I
Total
Working
8
9
10
11
12
13
14
=SUMPRODUCT(B8:H8,NumberWorking)
=SUMPRODUCT(B9:H9,NumberWorking)
=SUMPRODUCT(B10:H10,NumberWorking)
=SUMPRODUCT(B11:H11,NumberWorking)
=SUMPRODUCT(B12:H12,NumberWorking)
=SUMPRODUCT(B13:H13,NumberWorking)
=SUMPRODUCT(B14:H14,NumberWorking)
K
19
Total Cost
20 =SUMPRODUCT(UnitCost,NumberWorking)
!
!
Solver!Parameters!
Set%Objective%Cell:!TotalCost!
To:!Min!
By%Changing%Variable%Cells:%
! NumberWorking!
Subject%to%the%Constraints:%
! TotalWorking!>=!
3-60
Range Name
Cells
AgentsNeeded
NumberWorking
TotalCost
TotalWorking
UnitCost
K8:K14
B20:H20
K20
I8:I14
B5:H5
K
>=
>=
>=
>=
>=
>=
>=
Agents
Needed
6
12
10
13
11
5
2
Total Cost
$1,228
!
!
6
7
J
!
AgentsNeeded!
Solver%Options:%
% Make!Variables!Nonnegative!
! Solving!Method:!Simplex!LP!
!!!!!
!
!
The!linear!programming!model!for!the!SpanishZspeaking!employees!can!be!
developed!in!a!similar!fashion.!
!
A
1
Spanish
2
Speaking
3
4
5
Unit Cost
6
7
Work Shift?
8
7am-9am
9
9am-11am
10
11am-1pm
11
1pm-3pm
12
3pm-5pm
13
5pm-7pm
14
7pm-9pm
15
16
17
18
19
20 Number Working
!
B
C
D
E
F
Full-Time
on Phone
7am-9am
11am-1pm
$40
Full-Time
on Phone
9am-11am
1pm-3pm
$40
Full-Time
on Phone
11am-1pm
3pm-5pm
$40
Full-Time
on Phone
1pm-3pm
5pm-7pm
$44
Full-Time
on Phone
3pm-5pm
7pm-9pm
$48
1
0
1
0
0
0
0
0
1
0
1
0
0
0
0
0
1
0
1
0
0
0
0
0
1
0
1
0
0
0
0
0
1
0
1
Full-Time
on Phone
7am-9am
11am-1pm
2
Full-Time
on Phone
9am-11am
1pm-3pm
3
Full-Time
on Phone
11am-1pm
3pm-5pm
2
Full-Time
on Phone
1pm-3pm
5pm-7pm
2
Full-Time
on Phone
3pm-5pm
7pm-9pm
1
G
Total
Working
2
3
4
5
3
2
1
H
I
>=
>=
>=
>=
>=
>=
>=
Agents
Needed
2
3
3
4
3
2
1
Total Cost
$416
c)! Lenny!should!hire!25!fullZtime!EnglishZspeaking!operators.!Of!these!operators,!6!
have!their!first!phone!shift!from!7am!to!9am,!13!from!9am!to!11am,!4!from!11am!
to!1pm,!and!2!from!3pm!to!5pm.!Lenny!should!also!hire!5!partZtime!operators!
who!start!their!work!at!3pm.!In!addition,!Lenny!should!hire!10!SpanishZspeaking!
operators.!Of!these!operators,!2!have!their!first!shift!on!the!phone!from!7am!to!
9am,!3!from!9am!to!11am,!2!from!11am!to!1pm!and!1pm!to!3pm,!and!1!from!
3pm!to!5pm.!The!total!(wage)!cost!of!running!the!calling!center!equals!$1640!per!
day.!
3-61
!
!
d)! The!restriction!that!Lenny!can!find!only!one!EnglishZspeaking!operator!who!
wants!to!start!work!at!1pm!affects!only!the!linear!programming!model!for!
EnglishZspeaking!operators.!This!restriction!does!not!put!a!bound!on!the!number!
of!operators!who!start!their!first!phone!shift!at!1pm!because!those!operators!can!
start!work!at!11am!with!paperwork.!However,!this!restriction!does!put!an!upper!
bound!on!the!number!of!operators!having!their!first!phone!shift!from!3pm!to!
5pm.!The!new!worksheet!appears!as!follows.!
!
A
1
English
2
Speaking
3
4
5
Unit Cost
6
7
Work Shift?
8
7am-9am
9
9am-11am
10
11am-1pm
11
1pm-3pm
12
3pm-5pm
13
5pm-7pm
14
7pm-9pm
15
16
17
18
19
20 Number Working
21
22
B
C
D
E
F
G
H
Full-Time
on Phone
7am-9am
11am-1pm
$40
Full-Time
on Phone
9am-11am
1pm-3pm
$40
Full-Time
on Phone
11am-1pm
3pm-5pm
$40
Full-Time
on Phone
1pm-3pm
5pm-7pm
$44
Full-Time
on Phone
3pm-5pm
7pm-9pm
$44
Part-Time
on Phone
3pm-7pm
$44
Part-Time
on Phone
5pm-9pm
$48
1
0
1
0
0
0
0
0
1
0
1
0
0
0
0
0
1
0
1
0
0
0
0
0
1
0
1
0
0
0
0
0
1
0
1
Full-Time
on Phone
7am-9am
11am-1pm
6
Full-Time
on Phone
9am-11am
1pm-3pm
13
Full-Time
on Phone
11am-1pm
3pm-5pm
6
Full-Time
on Phone
1pm-3pm
5pm-7pm
0
Full-Time
on Phone
3pm-5pm
7pm-9pm
1
<=
1
0
0
0
0
1
1
0
0
0
0
0
0
1
1
Part-Time
on Phone
3pm-7pm
4
Part-Time
on Phone
5pm-9pm
1
I
Total
Working
6
13
12
13
11
5
2
J
K
>=
>=
>=
>=
>=
>=
>=
Agents
Needed
6
12
10
13
11
5
2
Total Cost
$1,268
!
!
Lenny!should!hire!26!fullZtime!EnglishZspeaking!operators.!Of!these!operators,!6!
have!their!first!phone!shift!from!7am!to!9am,!13!from!9am!to!11am,!6!from!11am!
to!1pm,!and!1!from!3pm!to!5pm.!Lenny!should!also!hire!4!partZtime!operators!
who!start!their!work!at!3pm!and!1!partZtime!operator!starting!work!at!5pm.!The!
hiring!of!SpanishZspeaking!operators!is!unaffected.!The!new!total!(wage)!costs!
equal!$1680!per!day.!
3-62
!
e)! For!each!hour,!we!need!to!divide!the!average!number!of!calls!per!hour!by!the!
average!processing!speed,!which!is!6!calls!per!hour.!The!number!of!bilingual!
operators!that!the!hospital!needs!to!staff!the!call!center!during!each!twoZhour!
shift!can!be!found!in!the!following!table:!
!
1
2
3
4
5
6
7
8
9
10
11
12
!
A
B
C
Work Shift
7am-9am
9am-11am
11am-1pm
1pm-3pm
3pm-5pm
5pm-7pm
7pm-9pm
Average
Number
of Calls
40
85
70
95
80
35
10
Agents
Needed
7
15
12
16
14
6
2
Calls Handled per hour
6
!
f)! The!linear!programming!model!for!Lenny’s!scheduling!problem!can!be!found!in!
the!same!way!as!before,!only!that!now!all!operators!are!bilingual.!(The!formulas!
and!the!solver!dialog!box!are!identical!to!those!in!part!(b).)!
!
A
1
Bilingual
2
3
4
5
Unit Cost
6
7
Work Shift?
8
7am-9am
9
9am-11am
10
11am-1pm
11
1pm-3pm
12
3pm-5pm
13
5pm-7pm
14
7pm-9pm
15
16
17
18
19
20 Number Working
B
C
D
E
F
G
H
Full-Time
on Phone
7am-9am
11am-1pm
$40
Full-Time
on Phone
9am-11am
1pm-3pm
$40
Full-Time
on Phone
11am-1pm
3pm-5pm
$40
Full-Time
on Phone
1pm-3pm
5pm-7pm
$44
Full-Time
on Phone
3pm-5pm
7pm-9pm
$44
Part-Time
on Phone
3pm-7pm
$44
Part-Time
on Phone
5pm-9pm
$48
1
0
1
0
0
0
0
0
1
0
1
0
0
0
0
0
1
0
1
0
0
0
0
0
1
0
1
0
0
0
0
0
1
0
1
0
0
0
0
1
1
0
0
0
0
0
0
1
1
Full-Time
on Phone
7am-9am
11am-1pm
7
Full-Time
on Phone
9am-11am
1pm-3pm
16
Full-Time
on Phone
11am-1pm
3pm-5pm
6
Full-Time
on Phone
1pm-3pm
5pm-7pm
0
Full-Time
on Phone
3pm-5pm
7pm-9pm
2
Part-Time
on Phone
3pm-7pm
6
Part-Time
on Phone
5pm-9pm
0
I
Total
Working
7
16
13
16
14
6
2
J
K
>=
>=
>=
>=
>=
>=
>=
Agents
Needed
7
15
12
16
14
6
2
Total Cost
$1,512
!
!
Lenny!should!hire!31!fullZtime!bilingual!operators.!Of!these!operators,!7!have!
their!first!phone!shift!from!7am!to!9am,!16!from!9am!to!11am,!6!from!11am!to!
1pm,!and!2!from!3pm!to!5pm.!Lenny!should!also!hire!6!partZtime!operators!who!
start!their!work!at!3pm.!The!total!(wage)!cost!of!running!the!calling!center!
equals!$1512!per!day.!
!
g)! The!total!cost!of!part!(f)!is!$1512!per!day;!the!total!cost!of!part!(b)!is!$1640.!
Lenny!could!pay!an!additional!$1640Z$1512!=!$128!in!total!wages!to!the!
bilingual!operators!without!increasing!the!total!operating!cost!beyond!those!for!
the!scenario!with!only!monolingual!operators.!The!increase!of!$128!represents!a!
percentage!increase!of!128/1512!=!8.47%.!
3-63
!
h)! Creative!Chaos!Consultants!has!made!the!assumption!that!the!number!of!phone!
calls!is!independent!of!the!day!of!the!week.!But!maybe!the!number!of!phone!calls!
is!very!different!on!a!Monday!than!it!is!on!a!Friday.!So!instead!of!using!the!same!
number!of!average!phone!calls!for!every!day!of!the!week,!it!might!be!more!
appropriate!to!determine!whether!the!day!of!the!week!affects!the!demand!for!
phone!operators.!As!a!result!Lenny!might!need!to!hire!more!partZtime!employees!
for!some!days!with!an!increased!calling!volume.!
!
Similarly,!Lenny!might!want!to!take!a!closer!look!at!the!length!of!the!shifts!he!has!
scheduled.!Using!shorter!shift!periods!would!allow!him!to!“fine!tune”!his!calling!
centers!and!make!it!more!responsive!to!demand!fluctuations.!!
!
Lenny!should!investigate!why!operators!are!able!to!answer!only!6!phone!calls!
per!hour.!Maybe!additional!training!of!the!operators!could!enable!them!to!
answer!phone!calls!quicker!and!so!increase!the!number!of!phone!calls!they!are!
able!to!answer!in!an!hour.!
!
Finally,!Lenny!should!investigate!whether!it!is!possible!to!have!employees!
switching!back!and!forth!between!paperwork!and!answering!phone!calls.!During!
slow!times!phone!operators!could!do!some!paperwork!while!they!are!sitting!
next!to!a!phone,!while!in!times!of!sudden!large!call!volumes!employees!who!are!
scheduled!to!do!paperwork!could!quickly!switch!to!answering!phone!calls.!
!
Lenny!might!also!want!to!think!about!the!installation!of!an!automated!answering!
system!that!gives!callers!a!menu!of!selections.!Depending!upon!the!caller’s!
selection,!the!call!is!routed!to!an!operator!who!specializes!in!answering!
questions!about!that!selection.!
3-64
%
Case%3.4! !
!
a)! In!this!case,!the!decisions!to!be!made!are!
!!
TV!=!number!of!commercials!on!television!
!!
M!=!number!of!advertisements!in!magazines!
!!
SS!=!number!of!advertisements!in!Sunday!supplements!
!
The!resulting!linear!programming!model!is!
Maximize!Exposures!=!1,300!TV!+!600!M!+!500!SS!
subject!to!
!!
Resource%Constraints%
%%
%
300!TV!+!150!M!+!100!SS!≤!4,000!(ad!budget!in!$1,000s)!
!!
!
90!TV!+!30!M!+!40!SS!≤!1,000!(planning!budget!in!$1,000s)!
!!
!
TV!≤!5!(television!spots!available)!
!!
Benefits%Constraints:%
%
%
1.2!TV!+!0.1!M!≥!5!(millions!of!young!children)!
!
!
0.5!TV!+!0.2!M!+!0.2!S!≥!5!(millions!of!parents)!
!
FixedERequirement%Constraints:%
%
%
40!TV!+!120!SS!=!5!(coupon!budget!in!$1,000s)!
!
Nonnegativity%Constraints:%
%
%
TV!≥!0,!M!≥!0,!S!≥!0.!
!
The!linear!programming!spreadsheet!solution!is!shown!below.!
3-65
%%
%
%
%
!
%
%
%
b)! The!violations!of!the!four!assumptions!of!LP:!
(1)! Proportionality%assumption:%the!advertisement! cost!may!not!be!
proportional! to!number! of!commercials! on!television! or!number! of!
advertisements! in!magzines.! The!marginal! cost!for!additional!
commercial! can!decrease.!
(2)! Additivity%assumption:%This!assumption! can!be!violated! for!benefit!
constraints!because! it!states!that!there! is!no!overlap!between!people!
who! see!the!commercial! on!television! or!see!the!advertisements! in!
magzine! or! Sunday!supplements.!
(3)! Divisibility%assumption:%The!decision!variables! in!this! case!are!
number! of!commercial! on!TV!or!advertisements! in!magzines! and!
Sunday! supplements!of!major!newspapers.! Naturally,! these! variables!
should!take!on!integer!values.!
(4)! Certainty%assumption:%Since!this!LP!model! is!formulated! to!select!
some!future!courses!of!actions,!the!parameters! used! in!this!case,!such!
as!Exposures!per!Ad!or!Number! Reached!per!Ad,!are!based! on!a!
prediction! of!future! situation,!which!inevitably! introduces! some!
degree!of!uncertainty.!
%
c)! Since!none!of!the!assumptions! appear!to!be!badly! violated,! LP!is!reasonable! at!
least!as!a!first!approximation.! Later!models,! such!as!IP!or!NLP!can!provide!
some!refinement.!
%
3-66
CHAPTER 4: SOLVING LINEAR PROGRAMMING PROBLEMS:
THE SIMPLEX METHOD
4.1-1.
(a) Label the corner points as A, B, C, D, and E in the clockwise direction starting from
2.
(b)
(c)
(d)
(e)
A:
B:
C:
D:
E:
and
2
2 and
3 and
2 and
and
A:
B:
C:
D:
E:
Corner Point
A
B
C
D
E
A and B:
B and C:
C and D:
D and E:
E and A:
3
2
2
1
2 1
2
Adjacent Points
E, B
A, C
B, D
C, E
D, A
2
3
2
4-1
4.1-2.
(a) Optimal solution:
with
Label the corner points as A, B, C, and D in the clockwise direction starting from
(b)
Corner Point
(c)
Corresponding Constraint Boundary Eq.s
and
and
and
and
Corner Point
Adjacent Corner Points
and
and
and
and
(d) Optimal Solution:
with
Corner Point
Profit
(e)
Corner Point
Profit
Next Step
Check and .
Move to .
Check .
Stop, is optimal.
The next corner point is A, which has already been checked.
4-2
and
and
and
and
.
4.1-3.
(a)
Corner Point
Optimal Solution:
Profit
with
$
(b) Initiated at the origin, the simplex method can follow one of the two paths:
or
.
Consider the first path. The origin
is not optimal, since
and
are
adjacent to
, both are feasible and they have better objective values.
is not
optimal because
, which is adjacent to it, is feasible and better.
is optimal
since both corner points that are adjacent to it are worse.
4.1-4.
(a)
4-3
(b)
CP Solution
A
B
C
D
E
F
G
H
I
J
K
L
M
Feasibility
Infeasible
Infeasible
Feasible
Feasible
Infeasible
Infeasible
Feasible
Infeasible
Feasible
Feasible
Infeasible
Infeasible
Feasible
Objective
6750
5400
4500
5625
6300
9000
6000
6300
5625
4500
5400
6750
0
The point G is optimal.
(c) Start at the origin M
. Both adjacent points C
and J
are
feasible and have better objective values,so one can choose to move to either one of
them. Suppose we choose C, which is not optimal since its adjacent CPF solution D is
better. The other corner point that is adjacent to C is B, but it is infeasible, so move to D.
Its adjacent G is feasible and better. The CPF solutions that are adjacent to G, namely D
and I both have lower objective values. Hence, G is optimal. If one chooses to proceed to
J instead of C after the starting point, then the simplex path follows the points M, J, I, G
and using similar arguments, one obtains the optimality of G.
4.1-5.
(a)
4-4
(b)
CP Solution
A
B
C
D
E
F
Feasibility
Infeasible
Feasible
Feasible
Feasible
Infeasible
Feasible
Objective
The point C is optimal.
(c) The starting point F is not optimal, since B and D have better objective values. The
objective value increases faster along the edge FB (
) than along the edge FD
(
), so we choose to move to point B. B is not optimal because the adjacent point
C does better. Note that A is adjacent to B as well, but it is infeasible. C is optimal since
the two CPF solutions adjacent to C, namely B and D have lower objective values.
4.1-6.
Corner Point
Profit
Next Step
Check
and
.
Move to
.
Check
.
Move to
. Check
Move to
. Check
Stop,
is optimal.
4-5
.
.
4.1-7.
Corner Point
1 6
21
18
Cost
102
105
126
Next Step
Check 21
and
18 .
Stop, 12 6 is optimal.
4.1-8.
(a) TRUE. Use optimality test. In minimization problems, "better" means smaller. To see
this, note that min
max
.
(b) FALSE. CPF solutions are not the only possible optimal solutions, there can be
infinitely many optimal solutions. This is indeed the case when there are more than one
optimal solution. For example, consider the problem
maximize
subject to
where
,
and
with
are all optimal solutions.
(c) TRUE. However, this is not always true. It is possible to have an unbounded feasible
region where an entire ray with only one CPF solution is optimal.
4.1-9.
(a) The problem may not have an optimal solution.
(b) The optimality test checks whether the current corner point is optimal. The iterative
step only moves to a new corner point.
4-6
(c) The simplex method can choose the origin as the initial corner point only when it is
feasible.
(d) One of the adjacent points is likely to be better, not necessarily optimal.
(e) The simplex method only identifies the rate of improvement, not all the adjacent
corner points.
4.2-1.
(a) Augmented form:
maximize
subject to
(b)
CPF Solution
BF Solution
Nonbasic Variables
Basic Variables
A
B
C
D
E
F
(c)
BF Solution A: Set
and solve
BF Solution B: Set
and solve
BF Solution C: Set
and solve
From the last two equations,
.
and from the first two,
4-7
BF Solution D: Set
and solve
BF Solution E: Set
and solve
BF Solution F: Set
and solve
4.2-2.
(a) Augmented form:
maximize
subject to
(b)
CPF Solution
BF Solution
Nonbasic Variables
A
B
C
D
(c)
BF Solution A: Set
and solve
BF Solution B: Set
and solve
BF Solution C: Set
and solve
From these two equations,
.
4-8
Basic Variables
BF Solution D: Set
and solve
(d)
CP Infeasible Sol.'n
Basic Infeasible Sol.'n
Nonbasic Var.'s
Basic Var.'s
E
F
(e)
Basic Infeasible Solution E: Set
and solve
Basic Infeasible Solution F: Set
and solve
4.3-1.
After the sudden decline of prices at the end of 1995, Samsung Electronics faced the
urgent need to improve its noncompetitive cycle times. The project called SLIM (short
cycle time and low inventory in manufacturing) was initiated to address this problem. As
part of this project, floor-scheduling problem is formulated as a linear programming
model. The goal is to identify the optimal values "for the release of new lots into the fab
and for the release of initial WIP from every major manufacturing step in discrete
periods, such as work days, out to a horizon defined by the user" [p. 71]. Additional
variables are included to determine the route of these through alternative machines. The
optimal values "minimize back-orders and finished-goods inventory" [p. 71] and satisfy
capacity constraints and material flow equations. CPLEX was used to solved the linear
programs.
With the implementation of SLIM, Samsung significantly reduced its cycle times and as
a result of this increased its revenue by $1 billion (in five years) despite the decrease in
selling prices. The market share increased from 18 to 22 percent. The utilization of
machines was improved. The reduction in lead times enabled Samsung to forecast sales
more accurately and so to carry less inventory. Shorter lead times also meant happier
customers and a more efficient feedback mechanism, which allowed Samsung to respond
to customer needs. Hence, SLIM did not only help Samsung to survive a crisis that drove
many out of the business, but it did also provide a competitive advantage in the business.
4-9
4.3-2.
Optimal Solution:
,
4-10
4.3-3.
(a)
maximize
subject to
Initialization:
,
,
, is not optimal since
the improvement rates are positive. Since it offers a rate of improvement of 2, choose to
increase , which becomes the entering basic variable for Iteration 1. Given
, the
highest possible increase in is found by looking at:
The minimum of these two bounds is
, so
can be raised to
the basis. Using Gaussian elimination, we obtain:
and
Again
is not optimal since the rate of improvement for is
be increased to . Consequently, becomes . By Gaussian elimination:
The current solution is optimal, since increasing
value. Hence
,
.
(b) Optimal Solution:
,
4-11
or
leaves
and
can
would decrease the objective
(c) The solution is the same.
4.3-4.
Optimal Solution:
,
4-12
4.3-5.
Optimal Solution:
1.58 1.68 ,
9.89
4.3-6.
(a) The simplest adaptation of the simplex method is to force
and
into the basis at
the earliest opportunity. One can also find the optimal solution directly by using Gaussian
elimination.
(b)
(i) Increase
Let
setting
and
.
minimum
.
4-13
(ii) Increase
setting
.
minimum
Let
and
Optimal Solution:
.
and
4.3-7.
(a) Because
in the optimal solution, the problem can be reduced to:
maximize
subject to
or equivalently
maximize
subject to
Since
and
in the optimal solution, they should be basic variables in the
optimal solution. Choosing these two as the first two entering basic variables will lead to
an optimal solution. The leaving basic variables will be determined by the minimum ratio
test.
(b) Optimal Solution:
and
4-14
4.3-8.
(a) FALSE. The simplex method's rule for choosing the entering basic variable is used
because it gives the best rate of improvement for the objective value at the given corner
point.
(b) TRUE. The simplex method's rule for choosing the leaving basic variable determines
which basic variable drops to zero first as the entering basic variable is increased.
Choosing any other one can cause this variable to become negative, so infeasible.
(c) FALSE. When the simplex method solves for the next BF solution, elementary
algebraic operations are used to eliminate each basic variable from all but one equation
(its equation) and to give it a coefficient of one in that equation.
4.4-1.
Optimal Solution:
and
4-15
4.4-2.
Optimal Solution:
and
4.4-3.
(a) Optimal Solution:
and
4-16
(b) Optimal Solution:
and
Corner Point
(c)
Iteration 1:
and
Increase
, set
(slack variables)
.
minimum
Let
and
.
4-17
Iteration 2:
is not optimal so increase
, set
.
minimum
Let
and
.
Optimal Solution:
and
(d) Optimal Solution:
and
(e) - (f)
The coefficients for
and
are negative so this solution is not optimal. Let
enter
the basis, since it offers largest improvement rate, so the column lying under
will be
the pivot column. To find out how much can be increased, use the ratio test:
:
:
so
minimum,
leaves the basis and its row is the pivot row.
4-18
The coefficient of
is still negative, so this solution is not optimal. Let
enter the
basis, its column is the pivot column. To find out how much
can be increased, use the
ratio test:
:
:
so
minimum
,
leaves the basis and its row is the pivot row.
All the coefficients in the objective row are nonnegative, so the solution
optimal with an objective value of .
(g)
4-19
is
4.4-4.
(a) Optimal Solution:
and
(b) Optimal Solution:
and
Corner Point
4-20
(c)
Iteration 1:
and
Increase
and set
(slack variables)
.
minimum
Let
Iteration 2:
and
.
is not optimal so increase
, set
minimum
Let
and
.
Optimal Solution:
and
(d) Optimal Solution:
and
4-21
.
(e) - (f)
The coefficients for
and
are negative so this solution is not optimal. Let
enter
the basis, since it offers largest improvement rate, so the column lying under
will be
the pivot column. To find out how much can be increased, use the ratio test:
:
:
so
minimum
,
leaves the basis and its row is the pivot row.
The coefficient of
is still negative, so this solution is not optimal. Let
enter the
basis, its column is the pivot column. To find out how much
can be increased, use the
ratio test:
:
:
so
minimum,
leaves the basis and its row is the pivot row.
All the coefficients in the objective row are nonnegative, so the solution
optimal with an objective value of .
(g)
4-22
is
Maximize
X1
2
X2
3
Constraint 1
Constraint 2
1
1
2
1
Solution
10
10
Totals
30
20
Limit
30
20
<=
<=
Objective
50
4.4-5.
(a) Set
.
(0)
2
(1)
4
3
80
(2)
8
60
(3)
60
40
40
Optimality Test: The coefficients of all nonbasic variables are negative, so the solution
8 60 40 is not optimal.
Choose
(1)
as the entering basic variable, since it has the largest coefficient.
8
(2)
8
60
(3)
60
40
We choose
26.67
15
40
minimum
40
as the leaving basic variable. Set
.
(0)
(1)
(2)
(3)
Optimality Test: The coefficient of
not optimal.
Let
is negative, so the solution
is
be the entering basic variable.
(1)
35
35
70
(2)
15
15
30
(3)
2
2
16.67
We choose
as the leaving basic variable. Set
(0)
(1)
4-23
.
min
(2)
(3)
Optimality Test: All of the coefficients are positive, so the solution
is optimal.
76.67.
(b) Optimal solution:
6.67 16.67 and
76.67
Bas|Eq|
Coefficient of
| Right
Var|No| Z|
X1
X2
X3
X4
X5
X6 | side
___|__|__|_____________________________________|______
| | |
|
Z | 0| 1|
-2
-4
-3
0
0
0 |
0
X4| 1| 0|
1
3
2
1
0
0 |
80
X5| 2| 0|
3
4*
2
0
1
0 |
60
X6| 3| 0|
2
1
2
0
0
1 |
40
Bas|Eq|
Coefficient of
| Right
Var|No| Z|
X1
X2
X3
X4
X5
X6 | side
___|__|__|_____________________________________|______
| | |
|
Z | 0| 1|
1
0
-1
0
1
0 |
60
X4| 1| 0|-1.25
0
0.5
1 -0.75
0 |
35
X2| 2| 0| 0.75
1
0.5
0 0.25
0 |
15
X6| 3| 0| 1.25
0
1.5*
0 -0.25
1 |
25
Bas|Eq|
Coefficient of
| Right
Var|No| Z|
X1
X2
X3
X4
X5
X6 | side
___|__|__|_____________________________________|______
| | |
|
Z | 0| 1|1.833
0
0
0 0.833 0.667 | 76.67
X4| 1| 0|-1.67
0
0
1 -0.67 -0.33 | 26.67
X2| 2| 0|0.333
1
0
0 0.333 -0.33 | 6.667
X3| 3| 0|0.833
0
1
0 -0.17 0.667 | 16.67
(c) Excel Solver
Maximize
X1
2
X2
4
X3
3
Constraint 1
Constraint 2
Constraint 3
1
3
2
3
4
1
2
2
2
Solution
0
6.67
16.67
4-24
Totals
53.33
60
40
<=
<=
<=
Limit
80
60
40
Objective
76.67
4.4-6.
(a) Optimal Solution:
and
(b) Optimal Solution:
and
4-25
(c)
Maximize
X1
3
X2
5
X3
6
Constraint 1
Constraint 2
Constraint 3
Constraint 4
2
1
1
1
1
2
1
1
1
1
2
1
Solution
0
1.33
1.33
Totals
2.67
4
4
2.67
Limit
4
4
4
3
Objective
14.67
4.4-7.
Optimal Solution:
<=
<=
<=
<=
and
4-26
4.4-8.
Optimal Solution:
and
4.5-1.
(a) TRUE. The ratio test tells how far the entering basic variable can be increased before
one of the current basic variables drops below zero. If there is a tie for which variable
should leave the basis, then both variables drop to zero at the same value of the entering
basic variable. Since only one variable can become nonbasic in any iteration, the other
will remain in the basis even though it will be zero.
(b) FALSE. If there is no leaving basic variable, then the solution is unbounded and the
entering basic variable can be increased indefinitely.
(c) FALSE. All basic variables always have a coefficient of zero in row 0 of the final
tableau.
(d) FALSE.
Example 1:
maximize
subject to
Clearly, any solution
for
with
is optimal. The
problem has infinitely many optimal solutions and the feasible region is not bounded.
Example 2: maximize
subject to
Any solution
with
is optimal.
4-27
4.5-2.
(a)
(b) Yes, the optimal solution is
10 with
10.
(c) No, the objective function value is maximized by sliding the objective function line to
the right. This can be done forever, so there is no optimal solution.
(d) No, there exist solutions that make the objective value arbitrarily large. This usually
occurs when a constraint is left out of the model.
4-28
(e) Let the objective function be
. Then, the initial tableau is:
Coefficient of
BV
Eq.
Right Side
1
30
30
1
The pivot column, the column of
, has all negative elements, so
is unbounded.
(f) The Solver tells that the Objective Cell values do not converge. There is no optimal
solution because a better solution can always be found.
Maximize
X1
1
X2
1
Constraint 1
Constraint 2
1
3
3
1
Solution
0
0
Totals
0
0
<=
<=
Limit
30
30
Objective
0
4.5-3.
(a)
4-29
(b) No. the objective function value is maximized by sliding the objective function line
upwards. This can be done forever, so there is no optimal solution.
(c) Yes, the optimal solution is
with
(d). No, there exist solutions that make
constraint is left out of the model.
.
arbitrarily large. This usually occurs when a
(e) Let the objective function be
. Then, the initial tableau is:
Coefficient of
BV
Eq.
The pivot column, the column of
Right Side
, has all elements negative, so
is unbounded.
(f) The Solver tells that the Objective Cell values do not converge. There is no optimal
solution because a better solution can always be found.
4-30
Maximize
X1
1
X2
1
Constraint 1
Constraint 2
2
1
1
2
Solution
0
Totals
0
0
<=
<=
Limit
20
20
Objective
0
0
4.5-4.
We can see from either the second or third iteration that because all of the constraint
coefficients of
are nonpositive, it can be increased without forcing any basic variable
to zero. From the third iteration,
is feasible for
any
and
is unbounded.
4-31
4.5-5.
(a) The constraints of any LP problem can be expressed in matrix notation as:
,
If
for
.
are feasible solutions and
, then
with
,
so
and
,
is also a feasible solution.
(b) This follows immediately from (a), since basic feasible solutions are feasible
solutions.
4.5-6.
(a) Suppose
is the value of the objective function for an optimal solution and
are optimal BF solutions. From Problem 4.5-5,
is feasible
for any choice of
satisfying
. The objective function
value at is:
,
so
is also an optimal solution.
(b) Consider any feasible solution that is not a weighted average of the optimal BF
solutions. Since
is feasible, it must be a weighted average of the basic feasible
solutions, which are not all optimal by assumption. Let
are the basic
feasible solutions that are not optimal. Then,
where
,
for some . The objective function value at
,
and
is:
.
Since
and
is not optimal,
,
for every . Because there is at least one positive
.
Hence,
cannot be optimal.
4-32
4.5-7.
(a)
(b)
Unit Profit (Prod.1)
Unit Profit (Prod.2)
Objective
Multiple Opt. Solutions
line segment between
&
line segment between
&
line segment between
&
line segment between
&
line segment between
&
(c)
Corner Point
Profit
Optimal Solution:
with
(d)
RS
RS
So the unique optimal solution is
with V
4-33
.
4.5-8.
Since the objective coefficients (row
optimal BF solutions.
Hence, the optimal BF solutions are
with objective function value .
) for
and
,
4-34
are zero, we can pivot to get other
,
, and
, all
4.6-1.
(a) Optimal Solution:
and
(b) Initial artificial BF solution:
(c) Optimal Solution:
and
4-35
4.6-2.
(a) - (b) Initial artificial BF solution:
Optimal Solution:
and
(c) - (d) - (e) - (f) Initial artificial BF solution:
4-36
Optimal Solution:
and
(g) The basic solutions of the two methods coincide. They are artificial BF solutions for
the revised problem until both artificial variables
and
are driven out of the basis,
which in the two-phase method is the end of Phase 1.
(h)
Maximize
X1
4
X2
2
X3
3
X4
5
Constraint 1
Constraint 2
2
8
3
1
4
1
2
5
Solution
0
0
50
50
Totals
300
300
<=
<=
Limit
300
300
Objective
400
4-37
4.6-3.
(a)
maximize
subject to
(b) Optimal Solution:
Pivoting
for
and
gives an alternate optimal BF solution,
(c) Optimal Solution:
and
4-38
.
Pivoting
for
gives an alternate optimal BF solution,
.
(d) The basic solutions of the two methods coincide. They are artificial BF solutions for
the revised problem until both artificial variables
and
are driven out of the basis,
which in the two-phase method is the end of Phase 1.
(e)
Minimize
X1
2
X2
3
X3
1
Constraint 1
Constraint 2
1
3
4
2
2
0
0.8
1.8
0
Solution
Totals
8
6
>=
>=
Limit
8
6
Objective
7
4.6-4.
Once all artificial variables are driven out of the basis in a maximization (minimization)
problem. Choosing an artificial variable to reenter the basis can only lower (raise) the
objective function value by an arbitrarily large amount depending on .
4-39
4.6-5.
(a)
(b) The Solver could not find a feasible solution.
Maximize
X1
90
X2
70
Constraint 1
Constraint 2
2
1
1
1
Solution
1
Totals
2
1
<=
>=
Limit
2
2
Objective
90
0
4-40
(c)
In the optimal solution, the artificial variable
the problem has no feasible solutions.
4-41
is basic and takes a positive value, so
(d)
Since the artificial variable
is not zero in the optimal solution of Phase I Problem, the
original model must have no feasible solutions.
4-42
4.6-6.
(a)
(b) The Solver could not find a feasible solution.
Unit Cost
Benefit 1
Benefit 2
Solution
X1
5000
2
1
0
X2
7000
1
2
0
Totals
0
0
Minimum
Level
>=
1
>=
1
Objective
0
(c)
4-43
(d)
4-44
4.6-7.
(a) Initial artificial BF solution:
(b) Optimal Solution:
and
4-45
(c) Initial artificial BF solution:
(d)
(e) - (f) Optimal Solution:
and
4-46
(g) The basic solutions of the two methods coincide. They are artificial basic feasible
solutions for the revised problem until both artificial variables and are driven out of
the basis, which in the two-phase method is the end of Phase 1.
(h)
Maximize
Constraint 1
Constraint 2
Solution
X1
2
1
2
0
X2
5
2
4
0
X3
3
1
1
Totals
50
50
Right Hand
Side
>=
20
=
50
Objective
150
50
4.6-8.
(a)
(b)
4-47
(c) Optimal Solution:
Pivoting
and
into the basis for
provides the alternative optimal BF solution
(d)
Minimize
X1
2
X2
1
X3
3
Constraint 1
Constraint 2
5
3
2
2
7
5
Solution
70
35
0
Totals
420
280
Right Hand
Side
=
420
>=
280
Objective
175
4.6-9.
(a) Optimal Solution:
and
4-48
.
(b) Optimal Solution:
and
(c) In both the Big-M method and the two-phase method, only the final tableau represents
a feasible solution for the original problem.
4-49
(d)
Minimize
Constraint 1
Constraint 2
Solution
X1
3
2
3
0
X2
2
1
3
15
X3
4
3
5
Totals
60
120
Right Hand
Side
=
60
>=
120
Objective
90
15
4.6-10.
(a) Optimal Solution:
and
(b) Optimal Solution:
and
4-50
(c) Only the final tableau for the Big-M method and the two-phase method represent
feasible solutions to the original problem.
(d)
Minimize
X1
3
X2
2
X3
7
Constraint 1
Constraint 2
1
2
1
1
0
1
Solution
20
30
0
Totals
10
10
Right Hand
Side
=
10
>=
10
Objective
120
4.6-11.
(a) FALSE. The initial basic solution for the artificial model is not feasible for the
original model.
(b) FALSE. If at least one of the artificial variables is not zero, then the real problem is
infeasible.
(c) FALSE. The two methods are basically equivalent, so they should take the same
number of iterations.
4.6-12.
(a) Substitute
, where both
aximize
subject to
4
4
and
are nonnegative.
2
4
5
2
4-51
0
(b) Optimal Solution:
Note that
,
,
and
, and
are renamed as
,X ,
and
(c)
Maximize
X1
1
X2
4
X3
2
Constraint 1
Constraint 2
4
1
1
1
1
2
Solution
1
9
>=
0
0
>=
0
Totals
5
10
Right Hand
Side
<=
5
<=
10
Objective
35
4-52
respectively.
4.6-13.
(a) Optimal Solution:
(b) Let
OLD
and
and
OLD
.
maximize
subject to
(c) Optimal Solution:
and
Optimal solution for the revised problem:
with
4-53
4.6-14.
(a) Let
OLD
,
OLD
, and
OLD
.
maximize
subject to
(b)
Optimal solution for the revised problem:
Optimal solution for the original problem:
and
4-54
(c)
Maximize
X1
1
X2
2
X3
1
Constraint 1
Constraint 2
Constraint 3
0
1
3
3
1
1
1
4
2
Solution
45
55
45
Totals
120
80
100
Right Hand
Side
<=
120
<=
80
<=
100
Objective
110
4.6-15.
(a) In order to decrease the objective function value in the simplex method, choose the
nonbasic variable that has the (largest) positive coefficient in the objective row, as the
entering basic variable. The ratio test is conducted the same way as in the maximization
problem to determine the leaving basic variable.
(b) Optimal Solution:
and
4-55
4.6-16.
(a)
maximize
subject to
(b)
(c)
4-56
(d)
X1
2
Maximize
X2
1
X3
4
X4
3
Constraint 1
Constraint 2
Constraint 3
Constraint 4
1
1
2
1
1
0
1
2
3
1
0
1
2
1
0
2
Solution
4
0
>=
0
0
>=
0
3
>=
0
Totals
2
1
8
2
<=
>=
<=
=
Right Hand
Side
4
1
2
2
Objective
17
4.6-17.
Reformulation:
maximize
subject to
Since this is the optimal tableau for Phase 1 and the artificial variable
problem is infeasible.
4-57
, the
4.7-1.
The CP solution
remains feasible and optimal if the constraint
is changed to
with
. However, if
, then this solution ceases to be feasible and
the optimal solution becomes
. This agrees with the allowable range (allowable
increase: E
, allowable decrease: ) for this constraint given in Figure 4.10.
4-58
Now, suppose instead that the constraint
is replaced by
. Then, the
intersection of the lines
and
can be expressed as
. This CP solution is feasible as long as
or equivalently
. In that case, provided that the objective function is the same, this solution is
optimal. Hence, the right-hand side of this constraint can be increased or decreased by .
If the third constraint is
becomes
or equivalently
also
, as given in Figure 4.10.
, then the CP solution determined by this and
. This point is feasible and optimal as long as
, so the allowable change for this constraint is
4.7-2.
(a)
Constraint (1):
:
and
,
Constraint (2):
:
and
and
,
and
(b) From (a), we see that the right-hand sides
and
are sensitive parameters.
The graph in part (a) shows that both constraints are active (binding) at the optimal
solution, so all the coefficients
,
,
, and
are sensitive
parameters, too. As will be seen in (c), the objective coefficients
and
are
not sensitive parameters.
4-59
(c) Observe that the optimal solution remains the same for
fixed) and
(with
fixed)
(with
(d) The dashed lines "- - -" in the graph below suggest that the CP solution ranges from
to
when
. Outside this range, the CP solution becomes
infeasible. The dashed lines "- -" represent the second constraint for different right-hand
side values. They suggest that the CP solution ranges from
to
when
. Hence, the allowable ranges are
and
.
4-60
(e)
4.7-3.
(a) Optimal Solution:
Corner Point
8 3
8 3 and
38
4-61
38
(b)
Increasing resource 1 to
units increases
to
.
Increasing resource 2 to
units increases
.
The third constraint is not binding, so
.
, so
to
(c) To increase by , resource 1 should be increased by
problem with resource 1 set to
returns the result
4.7-4.
(a) Optimal Solution:
and
4-62
, so
. Solving the LP
.
(b) The shadow prices for the three resources are given by the reduced costs (in the
objective function) for the corresponding slack variables. These values are circled in the
table above. The shadow prices for resources 1, 2 and 3 are ,
and respectively.
They represent the rate at which the objective function value
increases as the
corresponding resource is increased. For instance, increasing resource 3 by one unit
increases by , provided that no other constraints cause any trouble.
(c)
Maximize
X1
1
X2
7
X3
3
Totals
3.5
2
3
Constraint 1
Constraint 2
Constraint 3
2
4
3
1
3
2
1
0
1
Solution
0.5
0
4.5
Cell
Name
$B$10 Solution X1
$C$10 Solution X2
$D$10 Solution X3
Right Hand
Side
<=
4
<=
2
<=
3
Objective
14
Final Reduced Objective Allowable Allowable
Value
Cost
Coefficient Increase
Decrease
0.5
0
1
7.33333
10
0
-5.5
-7
5.5
1E+30
4.5
0
3
22
3
Constraints
Final Shadow Constraint Allowable Allowable
Cell
Name
Value
Price
R.H. Side
Increase
Decrease
$E$5 Constraint 1 Totals
-3.5
0
4
1E+30
7.5
$E$6 Constraint 2 Totals
2
2.5
2
1E+30
2
$E$7 Constraint 3 Totals
3
3
3
1E+30
4.5
4-63
4.7-5.
(a) Optimal Solution:
and
(b) The shadow prices are
,
of resources 1, 2 and 3 respectively.
and
. They are the marginal values
(c)
Maximize
Constraint 1
Constraint 2
Constraint 3
Solution
X1
2
1
2
1
0
X2
2
1
1
1
1
X3
3
Totals
4
2
10
1
1
3
Right Hand
Side
<=
4
<=
2
<=
12
Objective
7
3
4-64
Variable Cells
Cell
Name
$B$10 Solution X1
$C$10 Solution X2
$D$10 Solution X3
Final Reduced Objective Allowable Allowable
Value
Cost
Coefficient Increase
Decrease
0
-2.5
2
2.5
1E+30
1
0
-2
1.6667
1
3
0
3
1E+30
1
Constraints
Final Shadow Constraint Allowable Allowable
Cell
Name
Value
Price
R.H. Side
Increase
Decrease
$E$5 Constraint 1 Totals
4
0.5
4
1
2
$E$6 Constraint 2 Totals
2
2.5
2
2
6
$E$7 Constraint 3 Totals
10
0
12
1E+30
2
4.7-6.
(a) Optimal Solution:
(b) The shadow prices are
resources 1 and 2 respectively.
and
and
4-65
. They are the marginal values of
(c)
Maximize
X1
5
X2
4
X3
1
X4
3
Resource 1
Resource 2
3
3
2
3
3
1
1
3
Solution
11
0
3
0
Final
Value
11
0
3
0
Reduced
Cost
Totals
24
36
Right Hand
Side
<=
24
<=
36
Objective
52
Variable Cells
Cell
$B$9
$C$9
$D$9
$E$9
Name
Solution X1
Solution X2
Solution X3
Solution X4
0
-0.33333
0
-0.66667
Objective Allowable
Allowable
Coefficient Increase
Decrease
5
1E+30
0.3636
4
0.33333
1E+30
-1
2.66667
1.33333
3
0.66667
1E+30
Constraints
Final
Cell
Name
Value
$F$5 Resource 1 Totals
24
$F$6 Resource 2 Totals
36
Shadow
Constraint Allowable
Allowable
Price
R.H. Side
Increase
Decrease
0.66667
24
12
132
1
36
1E+30
12
4-66
4.9-1.
4-67
4.9-2.
4-68
Case%4.1%
!
a)! The!fixed!design!and!fashion!costs!are!sunk!costs!and!therefore!should!not!be!
considered!when!setting!the!production!now!in!July.!Since!the!velvet!shirts!have!
a!positive!contribution!to!covering!the!sunk!costs,!they!should!be!produced!or!at!
least!considered!for!production!according!to!the!linear!programming!model.!Had!
Ted!raised!these!concerns!before!any!fixed!costs!were!made,!then!he!would!have!
been!correct!to!advise!against!designing!and!producing!the!shirts.!With!a!
contribution!of!$22!and!a!demand!of!6000!units,!maximum!expected!profit!will!
be!only!$132,000.!This!amount!will!not!be!enough!to!cover!the!$500,000!in!fixed!
costs!directly!attributable!to!this!product.!
!
b)! The!linear!programming!spreadsheet!model!for!this!problem!is!shown!below.!
!
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
B
C
Price
L&M Cost
Material Cost
Net Contribution
Wool
Acetate
Cashmere
Silk
Rayon
Velvet
Cotton
Cost of
Material
$9.00
$1.50
$60.00
$13.00
$2.25
$12.00
$2.50
D
E
Wool
Cashmere
Slacks
Sweater
$300
$450
$160
$150
$30.00
$90.00
$110.00 $210.00
F
G
H
I
J
K
Silk
Silk
Tailored
Wool
Velvet
Cotton
Cotton
Blouse Camisole Skirt
Blazer
Pants Sweater Miniskirt
$180
$120
$270
$320
$350
$130
$75
$100
$60
$120
$140
$175
$60
$40
$19.50
$6.50
$6.75
$24.75
$39.00
$3.75
$1.25
$60.50 $53.50 $143.25 $155.25 $136.00 $66.25 $33.75
L
Velvet
Shirt
$200
$160
$18.00
$22.00
M
ButtonDown
Blouse
$120
$90
$3.38
$26.63
Material Requirements
2.5
1.5
1.5
2
3
2
1.5
1.5
0.5
2
1.5
3
1.5
1.5
Wool
Cashmere
Slacks
Sweater
Items Produced 4,200
4,000
<=
<=
Demand Forecast 7,000
4,000
>=
Minimum Production 4,200
60%
of demand
Silk
Silk
Tailored
Blouse Camisole Skirt
7,000
15,000
8,067
<=
<=
12,000 15,000
>=
2,800
Wool
Velvet
Blazer
Pants
5,000
0
<=
<=
5,000
5,500
>=
3,000
60%
of demand
0.5
Cotton
Cotton
Sweater Miniskirt
0
60,000
Velvet
Shirt
6,000
<=
6,000
ButtonDown
Blouse
9,244
N
O
P
Material
Used
25,100
28,000
6,000
18,000
30,000
9,000
30,000
<=
<=
<=
<=
<=
<=
<=
Material
Available
45,000
28,000
9,000
18,000
30,000
20,000
30,000
Total
Contribution
$6,862,933
Fixed Cost $8,960,000
Total Profit -$2,097,067
Also:
Silk Camisole >= Silk Blouse
Cotton Miniskirt >= Cotton Sweater
!
B
C
D
Material Cost =SUMPRODUCT(CostOfMaterial,C11:C17) =SUMPRODUCT(CostOfMaterial,D11:D17)
Net Contribution =Price-LMCost-MaterialCost
=Price-LMCost-MaterialCost
6
7
!
Range Name
CostOfMaterial
FixedCost
ItemsProduced
LMCost
MaterialAvailable
MaterialCost
MaterialRequirements
MaterialUsed
NetContribution
Price
TotalContribution
TotalProfit
Cells
B11:B17
P23
C22:M22
C5:M5
P11:P17
C6:M6
C11:M17
N11:N17
C7:M7
C4:M4
P22
P24
!
!
N
Material
Used
=SUMPRODUCT(C11:M11,ItemsProduced)
=SUMPRODUCT(C12:M12,ItemsProduced)
=SUMPRODUCT(C13:M13,ItemsProduced)
=SUMPRODUCT(C14:M14,ItemsProduced)
=SUMPRODUCT(C15:M15,ItemsProduced)
=SUMPRODUCT(C16:M16,ItemsProduced)
=SUMPRODUCT(C17:M17,ItemsProduced)
!
P
Total
Contribution
=SUMPRODUCT(NetContribution,ItemsProduced)
Fixed Cost 8960000
Total Profit =TotalContribution-FixedCost
!
9
10
11
12
13
14
15
16
17
!
O
!
!
!
!
20
21
22
23
24
4-69
!
!
!
TrendLine!should!produce!4,200!Wool!Slacks,!4,000!Cashmere!Sweaters,!7,000!
Silk!Blouses,!15,000!Silk!Camisoles,!8,067!Tailored!Skirts,!5,000!Wool!Blazers,!
40,000!Cotton!Minis,!6,000!Velvet!Shirts,!and!9,244!ButtonRDown!Blouses.!The!
total!net!contribution!of!all!clothing!items!is!$6,862,933.!However,!with!the!total!
fixed!cost!of!$860,000!+!3($2,700,000)!or!$8,960,000,!TrendLines!actually!loses!
$2,097,067.!
!
c)! If!velvet!cannot!be!sent!back!to!the!textile!wholesaler,!then!the!whole!quantity!
will!be!considered!as!a!sunk!cost!and!therefore!added!to!the!fixed!costs.!The!
objective!function!coefficients!of!items!using!velvet!will!no!longer!include!the!
material!cost.!The!net!contribution!of!the!velvet!pants!and!shirts!are!now!$175!
and!$40,!respectively.!The!revised!spreadsheet!model!is!as!follows.!
!
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
B
Price
L&M Cost
Material Cost
Net Contribution
Wool
Acetate
Cashmere
Silk
Rayon
Velvet
Cotton
Cost of
Material
$9.00
$1.50
$60.00
$13.00
$2.25
$12.00
$2.50
C
D
E
F
G
H
I
J
Wool
Cashmere
Silk
Silk
Tailored
Wool
Velvet
Cotton
Slacks
Sweater Blouse Camisole
Skirt
Blazer
Pants Sweater
$300
$450
$180
$120
$270
$320
$350
$130
$160
$150
$100
$60
$120
$140
$175
$60
$30.00
$90.00
$19.50
$6.50
$6.75
$24.75
$3.75
$110.00 $210.00 $60.50 $53.50 $143.25 $155.25 $175.00 $66.25
K
Cotton
Miniskirt
$75
$40
$1.25
$33.75
L
M
ButtonDown
Blouse
$120
$90
$3.38
$40.00 $26.63
Material Requirements
2.5
1.5
1.5
2
3
2
1.5
1.5
0.5
2
1.5
3
Wool
Cashmere
Silk
Silk
Tailored
Wool
Slacks
Sweater Blouse Camisole
Skirt
Blazer
Items Produced 4,200
4,000
7,000
15,000
3,178
5,000
<=
<=
<=
<=
<=
Demand Forecast 7,000
4,000
12,000 15,000
5,000
>=
>=
>=
Minimum Production 4,200
2,800
3,000
60%
60%
of demand
of demand
Velvet
Pants
3,667
<=
5,500
1.5
1.5
0.5
Cotton
Sweater
0
Cotton
Miniskirt
60,000
N
O
P
Material
Used
25,100
28,000
6,000
18,000
30,000
20,000
30,000
<=
<=
<=
<=
<=
<=
<=
Material
Available
45,000
28,000
9,000
18,000
30,000
20,000
30,000
Velvet
Shirt
$200
$160
Velvet
Shirt
6,000
<=
6,000
ButtonDown
Blouse
15,763
Total
Contribution
$7,085,822
Original Fixed Cost $8,960,000
Velvet Sunk Cost
$240,000
Total Profit -$2,114,178
Also:
Silk Camisole >= Silk Blouse
Cotton Miniskirt >= Cotton Sweater
!
!
O
20
21
22
23
24
25
P
Total
Contribution
=SUMPRODUCT(NetContribution,ItemsProduced)
Original Fixed Cost 8960000
Velvet Sunk Cost =B16*P16
Total Profit =TotalContribution-FixedCost-VelvetSunkCost
!
!
!
The!production!plan!changes!considerably.!TrendLines!should!produce!3,178!
tailored!skirts!(down!from!8,067),!3,667!velvet!pants!(up!from!0),!60,000!cotton!
minis!(up!from!40,000),!and!15,763!buttonRdown!blouses!(up!from!9,244).!The!
production!decisions!for!all!other!items!are!unaffected!by!the!change.!The!total!
net!contribution!of!all!clothing!items!equals!$840,000!+!$1,226,00!+!$!2,025,000!
+!$2,983,822.22!=!$7,085,822.!The!sunk!costs!now!include!the!material!cost!for!
velvet!and!totals!$9,200,000.!The!loss!now!equals!$2,114,178.!
4-70
!
d)! When!TrendLines!cannot!return!the!velvet!to!the!wholesaler,!the!costs!for!velvet!
cannot!be!recovered.!These!cost!are!no!longer!variable!cost!but!now!are!sunk!
cost.!As!a!consequence!the!increased!net!contribution!of!the!velvet!items!makes!
them!more!attractive!to!produce.!This!way!the!revenues!from!selling!these!items!
can!contribute!to!the!recovery!of!at!least!some!of!the!fixed!costs.!Instead!of!zero!
TrendLines!now!produces!3,667!velvet!pants.!These!pants!also!require!some!
acetate!and!thus!their!production!affects!the!production!plan!for!all!other!items.!
Since!it!is!not!optimal!to!make!full!use!of!the!ordered!velvet!in!part!(b)!it!comes!
as!no!surprise!that!the!loss!in!part!(c)!is!even!bigger!than!in!part!(b).!
!
e)! The!unit!contribution!of!a!wool!blazer!changes!to!$75.25.!
!
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
B
C
D
E
F
G
Wool
Cashmere
Silk
Silk
Tailored
Slacks
Sweater Blouse Camisole
Skirt
Price $300
$450
$180
$120
$270
L&M Cost $160
$150
$100
$60
$120
Material Cost $30.00
$90.00
$19.50
$6.50
$6.75
Net Contribution $110.00 $210.00 $60.50 $53.50 $143.25
Wool
Acetate
Cashmere
Silk
Rayon
Velvet
Cotton
Cost of
Material
$9.00
$1.50
$60.00
$13.00
$2.25
$12.00
$2.50
H
Wool
Blazer
$320
$220
$24.75
$75.25
I
J
Velvet
Cotton
Pants Sweater
$350
$130
$175
$60
$39.00
$3.75
$136.00 $66.25
K
Cotton
Miniskirt
$75
$40
$1.25
$33.75
L
M
ButtonVelvet Down
Shirt Blouse
$200
$120
$160
$90
$18.00 $3.38
$22.00 $26.63
Material Requirements
2.5
1.5
1.5
2
3
2
1.5
1.5
0.5
2
1.5
3
Wool
Cashmere
Silk
Silk
Tailored
Wool
Velvet
Slacks
Sweater Blouse Camisole
Skirt
Blazer
Pants
Items Produced 4,200
4,000
7,000
15,000
10,067
3,000
0
<=
<=
<=
<=
<=
<=
Demand Forecast 7,000
4,000
12,000 15,000
5,000
5,500
>=
>=
>=
Minimum Production 4,200
2,800
3,000
60%
60%
of demand
of demand
1.5
1.5
0.5
Cotton
Sweater
0
Cotton
Miniskirt
60,000
Velvet
Shirt
6,000
<=
6,000
N
O
P
Material
Used
20,100
28,000
6,000
18,000
30,000
9,000
30,000
<=
<=
<=
<=
<=
<=
<=
Material
Available
45,000
28,000
9,000
18,000
30,000
20,000
30,000
ButtonDown
Blouse
6,578
Fixed Cost
Total Profit
Total
Contribution
$6,527,933
$8,960,000
-$2,432,067
Also:
Silk Camisole >= Silk Blouse
Cotton Miniskirt >= Cotton Sweater
!
!
TrendLines!should!produce!10,067!skirts!(up!from!8,067),!the!minimum!of!
3,000!wool!blazers!(down!from!5,000),!and!6,578!buttonRdown!blouses!(down!
from!9,244).!The!production!decisions!for!all!other!items!are!unaffected!by!the!
change.!The!total!net!contribution!of!all!clothing!items!is!$6,527,933.33.!The!
total!loss!is!$2,432,067.!
4-71
!
f)! The!available!acetate!changes!from!28,000!to!38,000!square!yards.!The!resulting!
spreadsheet!solution!is!shown!below.!
!
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
B
Price
L&M Cost
Material Cost
Net Contribution
Wool
Acetate
Cashmere
Silk
Rayon
Velvet
Cotton
Cost of
Material
$9.00
$1.50
$60.00
$13.00
$2.25
$12.00
$2.50
C
D
E
F
G
H
I
J
Wool
Cashmere
Silk
Silk
Tailored
Wool
Velvet
Cotton
Slacks
Sweater Blouse Camisole
Skirt
Blazer
Pants Sweater
$300
$450
$180
$120
$270
$320
$350
$130
$160
$150
$100
$60
$120
$140
$175
$60
$30.00
$90.00
$19.50
$6.50
$6.75
$24.75 $39.00
$3.75
$110.00 $210.00 $60.50 $53.50 $143.25 $155.25 $136.00 $66.25
K
Cotton
Miniskirt
$75
$40
$1.25
$33.75
L
M
ButtonVelvet Down
Shirt Blouse
$200
$120
$160
$90
$18.00 $3.38
$22.00 $26.63
Material Requirements
2.5
1.5
1.5
2
3
2
1.5
1.5
0.5
2
1.5
3
Wool
Cashmere
Silk
Silk
Tailored
Wool
Velvet
Slacks
Sweater Blouse Camisole
Skirt
Blazer
Pants
Items Produced 4,200
4,000
7,000
15,000
14,733
5,000
0
<=
<=
<=
<=
<=
<=
Demand Forecast 7,000
4,000
12,000 15,000
5,000
5,500
>=
>=
>=
Minimum Production 4,200
2,800
3,000
60%
60%
of demand
of demand
1.5
1.5
0.5
Cotton
Sweater
0
Cotton
Miniskirt
60,000
Velvet
Shirt
6,000
<=
6,000
N
O
P
Material
Used
25,100
38,000
6,000
18,000
30,000
9,000
30,000
<=
<=
<=
<=
<=
<=
<=
Material
Available
45,000
38,000
9,000
18,000
30,000
20,000
30,000
ButtonDown
Blouse
356
Fixed Cost
Total Profit
Total
Contribution
$7,581,267
$8,960,000
-$1,378,733
Also:
Silk Camisole >= Silk Blouse
Cotton Miniskirt >= Cotton Sweater
!
!
TrendLines!should!produce!14,733!skirts!(up!from!8,067)!and!356!buttonRdown!
blouses!(down!from!9,244).!The!production!decisions!for!all!other!items!are!
unaffected!by!the!change.!The!total!net!contribution!of!all!clothing!items!is!
$7,581,267.!The!loss!is!$1,378,733.!
4-72
!
g)! We!need!to!include!new!decision!variables!representing!the!number!of!clothing!
items!that!are!sold!during!the!November!sale.!The!new!spreadsheet!model!is!
shown!below.!
!
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
B
C
D
E
Price
L&M Cost
Material Cost
Net Contribution (Sept-Oct)
Wool
Slacks
$300
$160
$30.00
$110.00
Cashmere
Sweater
$450
$150
$90.00
$210.00
Silk
Blouse
$180
$100
$19.50
$60.50
Nov Discount
Price (Nov)
Net Contribution (Nov)
40%
$180
-$10.00
$270
$30.00
$108
-$11.50
Cost of
Material
$9.00
$1.50
$60.00
$13.00
$2.25
$12.00
$2.50
Wool
Acetate
Cashmere
Silk
Rayon
Velvet
Cotton
F
G
H
Silk
Tailored
Wool
Camisole
Skirt
Blazer
$120
$270
$320
$60
$120
$140
$6.50
$6.75
$24.75
$53.50 $143.25 $155.25
$72
$5.50
$162
$35.25
$192
$27.25
I
J
K
Velvet
Cotton
Cotton
Pants
Sweater Miniskirt
$350
$130
$75
$175
$60
$40
$39.00
$3.75
$1.25
$136.00 $66.25 $33.75
$210
-$4.00
$78
$14.25
$45
$3.75
L
Velvet
Shirt
$200
$160
$18.00
$22.00
M
ButtonDown
Blouse
$120
$90
$3.38
$26.63
1.5
1.5
0.5
2
1.5
3
1.5
1.5
Sept-Oct Sales
Demand Forecast
Nov Sales
Total Sales
Minimum Production
Wool
Slacks
4,200
<=
7,000
Cashmere
Sweater
4,000
<=
4,000
Silk
Blouse
7,000
<=
12,000
0
4,200
>=
4,200
60%
of demand
2,000
6,000
0
7,000
Silk
Tailored
Camisole
Skirt
15,000
8,067
<=
15,000
0
15,000
0
8,067
>=
2,800
Wool
Blazer
5,000
<=
5,000
Velvet
Pants
0
<=
5,500
0
5,000
>=
3,000
60%
of demand
0
0
0.5
Cotton
Cotton
Sweater Miniskirt
0
60,000
0
0
0
60,000
O
P
Material
Used
25,100
28,000
9,000
18,000
30,000
9,000
30,000
<=
<=
<=
<=
<=
<=
<=
Material
Available
45,000
28,000
9,000
18,000
30,000
20,000
30,000
$120
$72
-$58.00 -$21.38
Material Requirements
2.5
1.5
1.5
2
3
2
N
Velvet
Shirt
6,000
<=
6,000
0
6,000
ButtonDown
Blouse
9,244
Fixed Cost
Total Profit
Total
Contribution
$6,922,933
$8,960,000
-$2,037,067
0
9,244
Also:
Silk Camisole >= Silk Blouse
Cotton Miniskirt >= Cotton Sweater
!
!
9
10
11
B
C
Nov Discount 0.4
Price (Nov) =(1-NovDiscount)*Price
Net Contribution (Nov) =PriceNov-LMCost-MaterialCost
D
=(1-NovDiscount)*Price
=PriceNov-LMCost-MaterialCost
!
Range Name
CostOfMaterial
FixedCost
LMCost
MaterialAvailable
MaterialCost
MaterialRequirements
MaterialUsed
NetContribution
NetContributionNov
NovDiscount
NovSales
Price
PriceNov
SeptOctSales
TotalContribution
TotalProfit
TotalSales
!
Cells
B15:B21
P27
C5:M5
P15:P21
C6:M6
C15:M21
N15:N21
C7:M7
C11:M11
C9
C30:M30
C4:M4
C10:M10
C26:M26
P26
P28
C31:M31
!
!
13
14
15
16
17
18
19
20
21
N
Material
Used
=SUMPRODUCT(C15:M15,TotalSales)
=SUMPRODUCT(C16:M16,TotalSales)
=SUMPRODUCT(C17:M17,TotalSales)
=SUMPRODUCT(C18:M18,TotalSales)
=SUMPRODUCT(C19:M19,TotalSales)
=SUMPRODUCT(C20:M20,TotalSales)
=SUMPRODUCT(C21:M21,TotalSales)
!
!
O
24
25
26
27
28
P
Total
Contribution
=SUMPRODUCT(NetContribution,SeptOctSales)+SUMPRODUCT(NetContributionNov,NovSales)
Fixed Cost 8960000
Total Profit =TotalContribution-FixedCost
!
!
It!only!pays!to!produce!2,000!more!Cashmere!sweaters.!The!production!plan!for!
all!other!items!is!the!same!as!in!part!(b).!The!sale!of!the!Cashmere!sweaters!
increases!the!total!net!contribution!by!$60,000!to!$6,922,933,!and!reduces!the!
loss!to!$2,037,066.67.!
4-73
Case%4.2%
!
a)! We!define!12!decision!variables,!one!for!each!age!group!surveyed!in!each!region.!
Rob's!restrictions!are!easily!modeled!as!constraints.!For!example,!his!condition!
that!at!least!20!percent!of!the!surveyed!customers!have!to!be!from!the!first!age!
group!requires!that!the!sum!of!the!variables!for!the!age!group!"18!to!25"!across!
all!three!regions!is!at!least!400.!All!his!other!requirements!are!modeled!similarly.!
Finally,!the!sum!of!all!variables!has!to!equal!2000,!because!that!is!the!number!of!
customers!Rob!wants!to!have!interviewed.!
!
A
B
1 Cost of Survey
2
3
Silicon Valley
4
Region
Big Cities
5
Small Towns
6
7
8 Number to Survey
9
Silicon Valley
10
Region
Big Cities
11
Small Towns
12
Total in A.G.
13
14
Required in A.G.
15
Percentage Required in A.G.
16
17
18
19
20
C
D
18 to 25
$4.75
$5.25
$6.50
E
Age Group
26 to 40
41 to 50
$6.50
$6.50
$5.75
$6.25
$7.50
$7.50
18 to 25
600
150
100
850
>=
400
20%
Age Group
26 to 40
41 to 50
0
0
550
0
0
300
550
300
>=
>=
550
300
27.5%
15%
F
G
H
I
J
Required
in Region
300
700
400
Percentage
Required
in Region
15%
35%
20%
51 and over
$5.00
$6.25
$7.25
51 and over
300
0
0
300
>=
300
15%
Total
in Region
900
>=
700
>=
400
>=
Total Surveys
Required Surveys
2000
=
2000
Total Cost
$11,200
Profit Margin
Bid
15%
$12,880
!
!
Range Name
CostOfSurvey
NumberToSurvey
PercentageRequiredInAG
PercentageRequiredInRegion
RequiredInAG
RequiredInRegion
RequiredSurveys
TotalCost
TotalInAG
TotalInRegion
TotalSurveys
!
Cells
C3:F5
C9:F11
C15:F15
J9:J11
C14:F14
I9:I11
G
H
I
J15
7
Total
Required
J17
8
in Region
in Region
C12:F12
9 =SUM(C9:F9)
>= =J9*RequiredSurveys
G9:G11
10 =SUM(C10:F10) >= =J10*RequiredSurveys
J13
! 11 =SUM(C11:F11) >= =J11*RequiredSurveys !
B
C
D
Total in A.G. =SUM(C9:C11)
=SUM(D9:D11)
>=
>=
Required in A.G. =C15*RequiredSurveys =D15*RequiredSurveys
12
13
14
!
!
!
!
!
13
14
15
16
17
!
I
J
Total Surveys =SUM(NumberToSurvey)
=
Required Surveys 2000
Total Cost =SUMPRODUCT(CostOfSurvey,NumberToSurvey)
4-74
!
!
The!cost!of!conducting!the!survey!meeting!all!constraints!imposed!by!AmeriBank!
incurs!cost!of!$11,200.!The!mix!of!customers!is!displayed!in!the!spreadsheet!
above.!Note!that!there!are!multiple!optimal!solutions!that!all!lead!to!a!total!cost!
of!$11,200.!
!
b)! Sophisticated!Surveys!will!submit!a!bid!of!(1.15)($11,200)!=!$12,880.!
!
c)! We!need!to!include!the!new!lowerRbound!constraint!(Minimum!to!Survey!in!
C19:F21)!on!all!variables:!NumberToSurvey!(C9:F11)!≥!MinimumToSurvey!
(C19:F21)!
!
A
B
C
D
E
F
1 Cost of Survey
Age Group
2
18 to 25
26 to 40
41 to 50
51 and over
3
Silicon Valley
$4.75
$6.50
$6.50
$5.00
4
Region
Big Cities
$5.25
$5.75
$6.25
$6.25
5
Small Towns
$6.50
$7.50
$7.50
$7.25
6
7
Age Group
8 Number to Survey
18 to 25
26 to 40
41 to 50
51 and over
9
Silicon Valley
600
50
50
200
10
Region
Big Cities
50
450
150
50
11
Small Towns
200
50
100
50
12
Total in A.G.
850
550
300
300
13
>=
>=
>=
>=
14
Required in A.G.
400
550
300
300
15
Percentage Required in A.G.
20%
27.5%
15%
15%
16
17
Age Group
18 Minimum to Survey
18 to 25
26 to 40
41 to 50
51 and over
19
Silicon Valley
50
50
50
50
20
Region
Big Cities
50
50
50
50
21
Small Towns
50
50
50
50
22
23
(Number to Survey >= Minimum to Survey)
!
!
The!new!requirement!increases!the!bid!to!$13,096.!
4-75
G
H
Total
in Region
900
>=
700
>=
400
>=
I
J
Required
in Region
300
700
400
Percentage
Required
in Region
15%
35%
20%
Total Surveys
Required Surveys
2000
=
2000
Total Cost
$11,388
Profit Margin
Bid
15%
$13,096
!
d)! We!include!upper!bounds!on!the!total!number!of!people!surveyed!in!Silicon!
Valley!and!from!the!age!group!of!18!to!25!yearRolds:!G9!≤!MaxInSiliconValley!
(L9)!and!C12!≤!MaxIn18to25!(C17).!
!
A
B
C
D
E
F
1 Cost of Survey
Age Group
2
18 to 25
26 to 40
41 to 50
51 and over
3
Silicon Valley
$4.75
$6.50
$6.50
$5.00
4
Region
Big Cities
$5.25
$5.75
$6.25
$6.25
5
Small Towns
$6.50
$7.50
$7.50
$7.25
6
7
Age Group
8 Number to Survey
18 to 25
26 to 40
41 to 50
51 and over
9
Silicon Valley
100
50
50
450
10
Region
Big Cities
400
450
50
50
11
Small Towns
100
50
200
50
12
Total in A.G.
600
550
300
550
13
>=
>=
>=
>=
14
Required in A.G.
400
550
300
300
15
Percentage Required in A.G.
20%
27.5%
15%
15%
16
<=
17
MaxIn18to25
600
18
19
Age Group
20 Minimum to Survey
18 to 25
26 to 40
41 to 50
51 and over
21
Silicon Valley
50
50
50
50
22
Region
Big Cities
50
50
50
50
23
Small Towns
50
50
50
50
24
25
(Number to Survey >= Minimum to Survey)
G
H
I
Total
in Region
650
>=
950
>=
400
>=
Required
in Region
300
700
400
Total Surveys
Required Surveys
J
K
Percentage
Required
in Region
15%
<=
35%
20%
L
Max in
Silicon
Valley
650
2000
=
2000
Total Cost
$11,575
Profit Margin
Bid
15%
$13,311
!
!
The!new!requirements!increase!the!bid!to!$13,311.!
!
e)! The!three!cost!factors!for!the!age!group!"18!to!25"!are!changed.!
!
A
B
C
D
E
F
1 Cost of Survey
Age Group
2
18 to 25
26 to 40
41 to 50
51 and over
3
Silicon Valley
$6.50
$6.50
$6.50
$5.00
4
Region
Big Cities
$6.75
$5.75
$6.25
$6.25
5
Small Towns
$7.00
$7.50
$7.50
$7.25
6
7
Age Group
8 Number to Survey
18 to 25
26 to 40
41 to 50
51 and over
9
Silicon Valley
50
50
50
500
10
Region
Big Cities
100
600
200
50
11
Small Towns
250
50
50
50
12
Total in A.G.
400
700
300
600
13
>=
>=
>=
>=
14
Required in A.G.
400
550
300
300
15
Percentage Required in A.G.
20%
27.5%
15%
15%
16
<=
17
MaxIn18to25
600
18
19
Age Group
20 Minimum to Survey
18 to 25
26 to 40
41 to 50
51 and over
21
Silicon Valley
50
50
50
50
22
Region
Big Cities
50
50
50
50
23
Small Towns
50
50
50
50
24
25
(Number to Survey >= Minimum to Survey)
G
H
Total
in Region
650
>=
950
>=
400
>=
Required
in Region
300
700
400
Total Surveys
Required Surveys
!
!
With!the!new!cost!factors!the!bid!increases!to!$13,829.!
4-76
I
J
K
Percentage
Required
in Region
15%
<=
35%
20%
2000
=
2000
Total Cost
$12,025
Profit Margin
Bid
15%
$13,829
L
Max in
Silicon
Valley
650
!
f)! We!eliminate!all!lower!and!upper!bounds!on!the!age!groups!and!regions!and!
replace!them!with!Rob's!strict!requirements.!These!requirements!also!ensure!
that!exactly!2000!people!are!surveyed!so!that!we!can!drop!that!constraint!too.!
!
A
B
C
D
E
F
1 Cost of Survey
Age Group
2
18 to 25
26 to 40
41 to 50
51 and over
3
Silicon Valley
$6.50
$6.50
$6.50
$5.00
4
Region
Big Cities
$6.75
$5.75
$6.25
$6.25
5
Small Towns
$7.00
$7.50
$7.50
$7.25
6
7
Age Group
8 Number to Survey
18 to 25
26 to 40
41 to 50
51 and over
9
Silicon Valley
50
50
50
250
10
Region
Big Cities
50
600
300
50
11
Small Towns
400
50
50
100
12
Total in A.G.
500
700
400
400
13
=
=
=
=
14
Required in A.G.
500
700
400
400
15
Percentage Required in A.G.
25%
35%
20%
20%
16
17
Age Group
18 Minimum to Survey
18 to 25
26 to 40
41 to 50
51 and over
19
Silicon Valley
50
50
50
50
20
Region
Big Cities
50
50
50
50
21
Small Towns
50
50
50
50
22
23
(Number to Survey ≥ Minimum to Survey)
G
H
I
Required Surveys
Total
in Region
400
1000
600
=
=
=
Required
in Region
400
1000
600
J
2,000
Percentage
Required
in Region
20%
50%
30%
Total Cost
$12,475
Profit Margin
Bid
15%
$14,346
!
!
Rob's!strict!requirements!increase!the!cost!of!the!survey!by!$450.!The!new!bid!of!
Sophisticated!Surveys!is!$14,346.25.!
4-77
Case%4.3%
!
a!&!b)!
A
1 Data:
2
3
Area
4
1
5
2
6
3
7
4
8
5
9
6
10
11
12 Solution:
13
14
Area 1
15
Area 2
16
Area 3
17
Area 4
18
Area 5
19
Area 6
20
Total In School
21
22
Capacity
23
24
25 Grade Constraints:
26
27
28
6th Graders
29
7th Graders
30
8th Graders
31
32
B
Percentage
in 6th
Grade
32%
37%
30%
28%
39%
34%
C
Percentage
in 7th
Grade
38%
28%
32%
40%
34%
28%
D
Percentage
in 8th
Grade
30%
35%
38%
32%
27%
38%
E
330
<=
368.56
362.11
369.33
<=
396
G
Bussing Cost ($/Student)
School 1
School 2
School 3
$300
$0
$700
$400
$500
$600
$300
$200
$200
$500
$0
$400
$500
$300
$0
Number of Students Assigned
School 1
School 2
School 3
0
450
0
0
422.22
177.78
0
227.78
322.22
350
0
0
366.67
0
133.33
83.33
0
366.67
800
1,100
1,000
<=
<=
<=
900
1,100
1,000
240
<=
269.33
288.00
242.67
<=
288
F
Total
From Area
450
600
550
350
500
450
=
=
=
=
=
=
Number of
Students
450
600
550
350
500
450
Total
Bussing
Cost
$555,556
300
<=
339.11
300.89
360.00
<=
360
30%
of total in school
36%
of total in school
!
!
Range Name
BussingCost
Capacity
NumberOfStudents
PercentageInGrade
Solution
TotalBussingCost
TotalFromArea
TotalInSchool
Cells
E4:G9
B22:D22
G14:G19
B4:D9
B14:D19
G24
E14:E19
B20:D20
!
20
!
12
13
14
15
16
17
18
19
E
Total
From Area
=SUM(B14:D14)
=SUM(B15:D15)
=SUM(B16:D16)
=SUM(B17:D17)
=SUM(B18:D18)
=SUM(B19:D19)
!!
G
21
Total
22
Bussing
23
Cost
24 =SUMPRODUCT(BussingCost,Solution)
A
B
C
D
Total In School =SUM(B14:B19) =SUM(C14:C19) =SUM(D14:D19)
!
A
25 Grade Constraints:
26
27
28
6th Graders
29
7th Graders
30
8th Graders
31
32
!
B
C
=$E$26*TotalInSchool
<=
=SUMPRODUCT(B14:B19,B4:B9)
=SUMPRODUCT(B14:B19,C4:C9)
=SUMPRODUCT(B14:B19,D4:D9)
<=
=$E$32*TotalInSchool
=$E$26*TotalInSchool
<=
=SUMPRODUCT(C14:C19,B4:B9)
=SUMPRODUCT(C14:C19,C4:C9)
=SUMPRODUCT(C14:C19,D4:D9)
<=
=$E$32*TotalInSchool
!
!
!
4-78
!
D
E
=$E$26*TotalInSchool
0.3
<=
=SUMPRODUCT(D14:D19,B4:B9)
=SUMPRODUCT(C4:C9,D14:D19)
=SUMPRODUCT(D14:D19,D4:D9)
<=
=$E$32*TotalInSchool
0.36
!
c)! The!recommendation!to!the!school!board!is!to!assign!students!to!schools!as!
shown!in!the!above!solution!section!of!the!spreadsheet.!!Quantities!that!are!not!
integers!must!be!rounded!since!partial!students!cannot!be!sent.!
!
d)! The!following!solution!decreases!total!bussing!costs!by!over!$135,000!but!
violates!the!grade!constraints!that!were!imposed.!!Solutions!will!vary!and!those!
than!satisfy!the!grade!constraints!will!increase!the!total!bussing!costs.!
!
A
1 Data:
2
3
Area
4
1
5
2
6
3
7
4
8
5
9
6
10
11
12 Solution:
13
14
Area 1
15
Area 2
16
Area 3
17
Area 4
18
Area 5
19
Area 6
20
Total In School
21
22
Capacity
23
24
25 Grade Constraints:
26
27
28
6th Graders
29
7th Graders
30
8th Graders
31
32
B
Percentage
in 6th
Grade
32%
37%
30%
28%
39%
34%
C
Percentage
in 7th
Grade
38%
28%
32%
40%
34%
28%
D
Percentage
in 8th
Grade
30%
35%
38%
32%
27%
38%
Number of Students Assigned
School 1
School 2
School 3
0
450
0
0
600
0
0
0
550
350
0
0
500
0
0
0
0
450
850
1,050
1,000
<=
<=
<=
900
1,100
1,000
255
<=
293.00
310.00
247.00
<=
306
315
<=
366.00
339.00
345.00
<=
378
4-79
300
<=
318.00
302.00
380.00
<=
360
E
F
G
Bussing Cost ($/Student)
School 1
School 2
School 3
$300
$0
$700
$400
$500
$600
$300
$200
$200
$500
$0
$400
$500
$300
$0
Total
From Area
450
600
550
350
500
450
=
=
=
=
=
=
Number of
Students
450
600
550
350
500
450
Total
Bussing
Cost
$420,000
30%
of total in school
36%
of total in school
!
!
e)! The!number!of!students!assigned!from!each!area!to!each!school!changes!to!the!
solution!shown!below!and!the!total!bussing!cost!is!reduced!by!almost!$162,000.!
!
A
1 Data:
2
3
Area
4
1
5
2
6
3
7
4
8
5
9
6
10
11
12 Solution:
13
14
Area 1
15
Area 2
16
Area 3
17
Area 4
18
Area 5
19
Area 6
20
Total In School
21
22
Capacity
23
24
25 Grade Constraints:
26
27
28
6th Graders
29
7th Graders
30
8th Graders
31
32
B
Percentage
in 6th
Grade
32%
37%
30%
28%
39%
34%
C
Percentage
in 7th
Grade
38%
28%
32%
40%
34%
28%
D
Percentage
in 8th
Grade
30%
35%
38%
32%
27%
38%
Number of Students Assigned
School 1
School 2
School 3
0
450
0
0
600
0
0
0
550
350
0
0
318.18
0
181.82
131.82
50
268.18
800
1,100
1,000
<=
<=
<=
900
1,100
1,000
240
<=
266.91
285.09
248.00
<=
288
330
<=
383.00
353.00
364.00
<=
396
4-80
300
<=
327.09
312.91
360.00
<=
360
E
F
G
Bussing Cost ($/Student)
School 1
School 2
School 3
$300
$0
$700
$400
$500
$600
$300
$0
$0
$500
$0
$400
$500
$300
$0
Total
From Area
450
600
550
350
500
450
=
=
=
=
=
=
Number of
Students
450
600
550
350
500
450
Total
Bussing
Cost
$393,636
30%
of total in school
36%
of total in school
!
!
f)! The!number!of!students!assigned!from!each!area!to!each!school!changes!to!the!
solution!shown!below!and!the!total!bussing!cost!is!reduced!by!over!$215,000.!
!
A
1 Data:
2
3
Area
4
1
5
2
6
3
7
4
8
5
9
6
10
11
12 Solution:
13
14
Area 1
15
Area 2
16
Area 3
17
Area 4
18
Area 5
19
Area 6
20
Total In School
21
22
Capacity
23
24
25 Grade Constraints:
26
27
28
6th Graders
29
7th Graders
30
8th Graders
31
32
!
B
Percentage
in 6th
Grade
32%
37%
30%
28%
39%
34%
C
Percentage
in 7th
Grade
38%
28%
32%
40%
34%
28%
Number of Students Assigned
School 1
School 2
School 3
38.71
411.29
0
0
236.56
363.44
0
77.96
472.04
350
0
0
435.48
0
64.52
75.81
374.19
0
900
1,100
900
<=
<=
<=
900
1,100
1,000
270
<=
306.00
324.00
270.00
<=
324
330
<=
369.75
352.25
378.00
<=
396
270
<=
301.25
274.75
324.00
<=
324
E
F
Option!
Cost!
current!
1!
2!
$555,556!
$393,636!
$340,054!
#!students!
walking!1!to!1.5!
miles!
0!
900!
900!
h)! Answers!will!vary.!
!
4-81
G
Bussing Cost ($/Student)
School 1
School 2
School 3
$0
$0
$700
$400
$500
$600
$0
$0
$0
$500
$0
$400
$500
$0
$0
Total
From Area
450
600
550
350
500
450
=
=
=
=
=
=
Number of
Students
450
600
550
350
500
450
Total
Bussing
Cost
$340,054
30%
of total in school
36%
of total in school
g)!
!
!
D
Percentage
in 8th
Grade
30%
35%
38%
32%
27%
38%
#!students!walking!
more!than!1.5!miles!
0!
0!
491!
!
CHAPTER 5: THE THEORY OF THE SIMPLEX METHOD
5.1-1.
(a) Optimal Solution: ÐB‡" ß B‡# Ñ œ Ð#ß #Ñ and ^ ‡ œ "!
(c)
maximize
subject to
^œ
$B"  #B#
#B"  B#  B$
œ'
B"  #B#
 B% œ '
B" ß B# ß B$ ß B%
!
(b) - (d)
(e)
Step
"
#
$
CPF Sol.'n Deleted Defining Eq.
Ð!ß !Ñ
B" œ !
Ð$ß !Ñ
B# œ !
Ð#ß #Ñ OPTIMAL
Added Defining Eq.
#B"  B# œ '
B"  #B# œ '
5-1
Deleted Ind.Var.
B"
B#
Added Ind.Var.
B$
B%
5.1-2.
(a) Optimal Solution: ÐB‡" ß B‡# Ñ œ Ð$ß *Ñ and ^ ‡ œ #"!
(c)
maximize
subject to
^œ
"!B"  #!B#
 B"  #B#  B$
B"  B#
 B%
&B"  $B#
 B&
B" ß B# ß B$ ß B% ß B&
5-2
œ "&
œ "#
œ %&
!
(b) - (d)
(e)
Step
"
#
$
CPF Sol.'n Deleted Defining Eq.
Ð!ß !Ñ
B# œ !
Ð!ß (Þ&Ñ
B" œ !
Ð$ß *Ñ OPTIMAL
Added Defining Eq.
B"  #B# œ "&
B"  B# œ "#
5-3
Deleted Ind.Var.
B#
B"
Added Ind.Var.
B$
B%
5.1-3.
(a) Optimal Solution: ÐB‡" ß B‡# Ñ œ Ð$ß %Ñ and ^ ‡ œ ")
(b) The corner point Ð$ß %Ñ has the best objective value "), so is optimal.
CPF Sol.'n
Ð!ß !Ñ
Ð!ß "Ñ
Ð!Þ'ß #Þ)Ñ
Ð$ß %Ñ
Ð$Þ$$ß $Þ$$Ñ
Ð#Þ&ß !Ñ
Defining Equations
B" œ !ß B# œ !
B" œ !ß $B"  B# œ "
$B"  B# œ "ß B"  #B# œ &
B"  #B# œ &ß %B"  #B# œ #!
%B"  #B# œ #!ß %B"  B# œ "!
%B"  B# œ "!ß B# œ !
BF Solution
Ð!ß !ß "ß #!ß "!ß &Ñ
Ð!ß "ß !ß ")ß ""ß $Ñ
Ð!Þ'ß #Þ)ß !ß "#ß "!Þ%ß !Ñ
Ð$ß %ß 'ß !ß #ß !Ñ
Ð$Þ$$ß $Þ$$ß (Þ'(ß !ß !ß "Þ'(Ñ
Ð#Þ&ß !ß )Þ&ß "!ß !ß (Þ&Ñ
NB Var.'s
B" ß B#
B" ß B$
B$ ß B'
B% ß B'
B% ß B&
B# ß B&
D
!
$
*Þ'
")
"'Þ'(
&
(c) All sets yield a solution.
CP Infeas. Sol.'n
Ð "$ ß !Ñ
Ð&ß !Ñ
Ð!ß "!Ñ
Ð!ß &# Ñ
Ð *# ß $#
& Ñ
Ð""ß $%Ñ
$!
Ð #&
( ß ( Ñ
Ð&ß !Ñ
Ð!ß "!Ñ
Defining Equations
$B"  B# œ "ß B# œ !
B"  #B# œ &ß B# œ !
%B"  #B# œ #!ß B" œ !
B"  #B# œ &ß B" œ !
%B"  #B# œ #!ß $B"  B# œ "
$B"  B# œ "ß %B"  B# œ "!
%B"  B# œ "!ß B"  #B# œ &
%B"  #B# œ #!ß B# œ !
%B"  B# œ "!ß B" œ !
5.1-4.
(a) ÐB" ß B# ß B$ Ñ œ Ð"!ß !ß !Ñ
(b) B# œ !ß B$ œ !ß B"  B#  #B$ œ "!
5-4
Basic Infeas. Solutions
Ð "$ ß !ß !ß #" "$ ß "" "$ ß % #$ Ñ
Ð&ß !ß "%ß %!ß $!ß !Ñ
Ð!ß "!ß *ß !ß #!ß "&Ñ
Ð!ß &# ß  $# ß "&ß "# "# ß !Ñ
%'
Ð *& ß $#
& ß !ß !ß & ß 'Ñ
Ð""ß $%ß !ß *#ß !ß &#Ñ
$! &#
#!
Ð #&
( ß ( ß ( ß  ( ß !ß !Ñ
Ð&ß !ß "'ß !ß "!ß "!Ñ
Ð!ß "!ß ""ß %!ß !ß #&Ñ
NB Var.'s
B# ß B $
B# ß B'
B" ß B%
B" ß B '
B$ ß B%
B$ ß B&
B& ß B'
B# ß B%
B" ß B&
5.1-5.
(a)
CPF Sol.'n
Ð!ß !ß !Ñ
Ð%ß !ß !Ñ
Ð%ß #ß !Ñ
Ð#ß %ß !Ñ
Ð!ß %ß !Ñ
Ð!ß %ß #Ñ
Ð#ß %ß $Ñ
Ð%ß #ß %Ñ
Ð%ß !ß %Ñ
Ð!ß !ß #Ñ
Defining Equations
B" œ !ß B# œ !ß B$ œ !
B" œ %ß B# œ !ß B$ œ !
B" œ %ß B"  B# œ 'ß B$ œ !
B# œ %ß B"  B# œ 'ß B$ œ !
B" œ !ß B# œ %ß B$ œ !
B" œ !ß B# œ %ß B"  #B$ œ %
B"  B# œ 'ß B# œ %ß B"  #B$ œ %
B"  B# œ 'ß B" œ %ß B"  #B$ œ %
B# œ !ß B" œ %ß B"  #B$ œ %
B# œ !ß B" œ !ß B"  #B$ œ %
(b) B"  B# œ 'ß B# œ %ß B"  #B$ œ %
(c) B" œ %ß B" œ !ß B# œ ! Ê inconsistent system
5.1-6.
(a) - (b)
Defining Equations
B" œ !ß B# œ !
B" œ !ß #B"  B# œ "!
B" œ !ß  $B"  #B# œ '
B" œ !ß B"  B# œ '
B# œ !ß #B"  B# œ "!
B# œ !ß  $B"  #B# œ '
B# œ !ß B"  B# œ '
#B"  B# œ "!ß $B"  #B# œ '
#B"  B# œ "!ß B"  B# œ '
$B"  #B# œ 'ß B"  B# œ '
CP
Ð!ß !Ñ
Ð!ß "!Ñ
Ð!ß $Ñ
Ð!ß 'Ñ
Ð&ß !Ñ
Ð#ß !Ñ
Ð'ß !Ñ
Ð#ß 'Ñ
Ð%ß #Ñ
Ð"Þ#ß %Þ)Ñ
Feas.?
No
No
No
No
No
No
Yes
Yes
Yes
Yes
Basic Solution
Ð"!ß !ß "!ß 'ß 'Ñ
Ð!ß "!ß !ß "%ß %Ñ
Ð!ß $ß (ß !ß $Ñ
Ð!ß 'ß %ß 'ß !Ñ
Ð&ß !ß !ß #"ß ""Ñ
Ð#ß !ß "%ß !ß )Ñ
Ð'ß !ß #ß #%ß !Ñ
Ð#ß 'ß !ß !ß )Ñ
Ð%ß #ß !ß "%ß !Ñ
Ð"Þ#ß %Þ)ß #Þ)ß !ß !Ñ
NB Var.'s
B" ß B#
B" ß B$
B" ß B%
B" ß B&
B# ß B$
B# ß B%
B# ß B&
B$ ß B%
B$ ß B&
B% ß B&
5.1-7.
(a) - (b)
Defining Equations
B" œ !ß B# œ !
B" œ !ß B# œ "!
B" œ !ß #B"  &B# œ '!
B" œ !ß B"  B# œ ")
B" œ !ß $B"  B# œ %%
B# œ !ß B# œ "!
B# œ !ß #B"  &B# œ '!
B# œ !ß B"  B# œ ")
B# œ !ß $B"  B# œ %%
B# œ "!ß #B"  &B# œ '!
B# œ "!ß B"  B# œ ")
B# œ "!ß $B"  B# œ %%
#B"  &B# œ '!ß B"  B# œ ")
#B"  &B# œ '!ß $B"  B# œ %%
B"  B# œ ")ß $B"  B# œ %%
CP
Ð!ß !Ñ
Ð!ß "!Ñ
Ð!ß "#Ñ
Ð!ß ")Ñ
Ð!ß %%Ñ
No Solution
Ð$!ß !Ñ
Ð")ß !Ñ
Ð"%Þ'(ß !Ñ
Ð&ß "!Ñ
Ð)ß "!Ñ
Ð""Þ$$ß "!Ñ
Ð"!ß )Ñ
Ð"#Þ$1ß (Þ!)Ñ
Ð"$ß &Ñ
Feas.?
Yes
Yes
No
No
No
Basic Solution
Ð!ß !ß "!ß '!ß ")ß %%Ñ
Ð!ß "!ß !ß "!ß )ß $%Ñ
Ð!ß "#ß #ß !ß 'ß $#Ñ
Ð!ß ")ß )ß $!ß !ß #'Ñ
Ð!ß %%ß $%ß "'!ß #'ß !Ñ
No
No
Yes
Yes
No
No
Yes
No
Yes
Ð$!ß !ß "!ß !ß "#ß %'Ñ
Ð")ß !ß "!ß #%ß !ß "!Ñ
Ð"%Þ'(ß !ß "!ß $!Þ'(ß $Þ$$ß !Ñ
Ð&ß "!ß !ß !ß $ß "*Ñ
Ð)ß "!ß !ß 'ß !ß "!Ñ
Ð""Þ$$ß "!ß !ß "#Þ'(ß $Þ$$ß !Ñ
Ð"!ß )ß #ß !ß !ß 'Ñ
Ð"#Þ$"ß (Þ!)ß #Þ*#ß !ß "Þ$)ß !Ñ
Ð"$ß &ß &ß *ß !ß !Ñ
5-5
NB Var.'s
B" ß B#
B" ß B$
B" ß B%
B" ß B&
B" ß B'
B# ß B$
B# ß B%
B# ß B&
B# ß B'
B$ ß B%
B$ ß B&
B$ ß B '
B% ß B&
B% ß B'
B& ß B '
5.1-8.
(a) If the feasible region is unbounded, then there may be no optimal solution.
(b) There may be multiple optimal solutions, in which case the weighted average of any
optimal CPF solutions is optimal, too.
(c) An adjacent CPF solution may have an equal objective function value, then all the
points that lie on the line segment between these two corner points are optimal.
5.1-9.
(a) FALSE. (p.5-10) Property 1: (a) If there is exactly one optimal solution, then it must
be a CPF solution. (b) If there are multiple optimal solutions, then at least two of them
must be adjacent CPF solutions. An optimal solution that is not a CPF solution can be
obtained by taking a convex combination of two optimal CPF solutions.
(b) FALSE. (p.5-12) The number of CPF solutions is at most 
78
œ
8 
Ð78Ñx
7x8x .
(c) FALSE. (p.5-13) The adjacent CPF solution that has a better objective function value
than the initial CPF solution may be adjacent to another CPF solution that has an even
better objective function value.
5.1-10.
(a) TRUE. By Property 1(a), there must be multiple solutions, since this optimal solution
is not a CPF solution. But then, there must be infinitely many optimal solutions, namely
any convex combination of optimal solutions.
(b) TRUE. Any point B on the line segment connecting B‡ and B‡‡ can be expressed as
B œ αB‡  Ð"  αÑB‡‡ with α − Ò!ß "Ó. Both B‡ and B‡‡ have the optimal objective value
^ ‡ . The objective function value at B is
^ œ - X ÐαB‡  Ð"  αÑB‡‡ Ñ œ α^ ‡  Ð"  αÑ^ ‡ œ ^ ‡ ,
so B is optimal. Since the feasible region is convex, any such point is feasible.
(c) FALSE. The simultaneous solution of any set of 8 constraint boundary equations may
be infeasible or may not even exist.
5.1-11.
(a) TRUE. If there are no optimal solutions, then either the problem is infeasible or the
objective value is unbounded (Chapter 3). The former is not the case by assumption of
the problem. Also by assumption again, the feasible region is bounded, so the objective
value is bounded, so the latter cannot be the case. Hence, there must be at least one
optimal solution.
(b) FALSE. If a solution is optimal, it need not be a BF solution. A convex combination
of two optimal BF solutions is optimal even though it is not a BF solution. This follows
from Property 1, since BF solutions are CPF solutions.
(c) TRUE. Since BF solutions correspond to CPF solutions, this follows directly from
Property 2.
5-6
5.1-12.
B" œ !ß #B"  B#  $B$ œ '!ß $B"  $B#  &B$ œ "#! Ê ÐB" ß B# ß B$ Ñ œ Ð!ß "&ß "&Ñ
5.1-13.
Since B#  ! and B$  !, B# œ ! and B$ œ ! cannot be part of the three boundary equations, so the boundary equations are B" œ !ß #B"  B#  B$ œ #!ß $B"  B#  #B$ œ $!.
Then, the optimal solutions is ÐB" ß B# ß B$ Ñ œ Ð!ß "!ß "!Ñ.
5.1-14.
(a)
(b) The simplex method follows this path because moving along the chosen edges
provides the greatest increase in the objective value for a unit move in the chosen
direction among all possible edges at each vertex/decision point.
(c)
Edge
"
#
Constraint Boundary Equations
B# œ !ß B" œ !
#B"  B#  #B$ œ %ß B" œ !
End Points
Ð!ß !ß !Ñß Ð!ß !ß #Ñ
Ð!ß !ß #Ñß Ð!ß #ß "Ñ
Additional Constraints
B$ œ !ß #B"  B#  #B$ œ %
B# œ !ß B"  B#  B$ œ $
(d) - (e)
CP
Ð!ß !ß !Ñ
Ð!ß !ß #Ñ
Ð!ß #ß "Ñ
Defining Equations
B" œ !ß B# œ !ß B$ œ !
B" œ !ß B# œ !ß #B"  B#  #B$ œ %
B" œ !ß #B"  B#  #B$ œ %ß B"  B#  B$ œ $
BF Solution
Ð!ß !ß !ß %ß $Ñ
Ð!ß !ß #ß !ß "Ñ
Ð!ß !ß #ß !ß "Ñ
NB Var.'s
B" ß B# ß B $
B" ß B# ß B%
B" ß B% ß B&
The nonbasic variables having value zero are equivalent to indicating variables. They
indicate that their associated inequality constraints are actually equalities. The associated
equalities are the defining equations.
5-7
5.1-15.
(a)
(b) The simplex method follows this path because moving along the chosen edges
provides the greatest increase in the objective value for a unit move in the chosen
direction among all possible edges at each vertex/decision point.
(c)
Edge
"
#
Constraint Boundary Equations
B" œ !ß B$ œ !
B$ œ !ß B"  #B#  B$ œ $!
End Points
Ð!ß !ß !Ñß Ð!ß "&ß !Ñ
Ð!ß "&ß !Ñß Ð"!ß "!ß !Ñ
Additional Constraints
B# œ !ß B"  #B#  B$ œ $!
B" œ !ß B"  B#  B$ œ #!
(d) - (e)
CP
Ð!ß !ß !Ñ
Ð!ß "&ß !Ñ
Ð"!ß "!ß !Ñ
Defining Equations
B" œ !ß B# œ !ß B$ œ !
B" œ !ß B$ œ !ß B"  #B#  B$ œ $!
B$ œ !ß B"  #B#  B$ œ $!ß B"  B#  B$ œ #!
BF Solution
Ð!ß !ß !ß #!ß $!Ñ
Ð!ß "&ß !ß &ß !Ñ
Ð"!ß "!ß !ß !ß !Ñ
NB Var.'s
B" ß B# ß B $
B" ß B$ ß B&
B$ ß B% ß B&
The nonbasic variables having value zero are equivalent to indicating variables. They
indicate that their associated inequality constraints are actually equalities. The associated
equalities are the defining equations.
5.1-16.
(a) When the objective is to maximize ^ œ B$ , both corner points Ð%ß #ß %Ñ and Ð%ß !ß %Ñ
are optimal, with ^ ‡ œ %.
(b) When the objective is to maximize ^ œ B"  #B$ , all the corner points Ð!ß !ß #Ñ,
Ð%ß !ß %Ñ, Ð%ß #ß %Ñ, Ð#ß %ß $Ñ and Ð!ß %ß #Ñ are optimal, with ^ ‡ œ %.
5-8
5.1-17.
(a) Geometrically, each constraint is a plane and the points that are feasible for a given
(inequality) constraint form a half-space. The line segment defined by any two feasible
points must lie entirely on the feasible side of the plane and therefore, all the points on
the line segment are feasible, implying that the set of solutions for any one constraint is a
convex set.
(b) Because the points in the feasible region of the LP problem satisfy all the constraints
simultaneously, it must be the case that for any two feasible points, the points on the line
segment joining them must also satisfy each constraint (from (a)). Hence, the set of
solutions that satisfy all the constraints simultaneously is a convex set.
5.1-18.
To maximize ^ œ $B"  %B#  $B$ , starting at the origin Ð!ß !ß !Ñ, one first chooses to
move to Ð!ß %ß !Ñ because this edge offers the best rate of improvement among all edges at
the origin. From Ð!ß %ß !Ñ, the edge that increases the objective function fastest is the one
that connects to either Ð!ß %ß #Ñ or Ð#ß %ß !Ñ. From either one these, the edge that gives the
best rate of increase connects to Ð#ß %ß $Ñ. Then, the only edge that provides an
improvement in ^ connects to the optimal solution Ð%ß #ß %Ñ.
5.1-19.
(a)
Original Constraint
B" !
B# !
B$ !
B"  B% œ %
B#  B& œ %
B"  B#  B' œ '
B"  #B$  B( œ %
Boundary Equation
B" œ !
B# œ !
B$ œ !
B" œ %
B# œ %
B"  B# œ '
B"  #B$ œ %
Indicating Variable
B"
B#
B$
B%
B&
B'
B(
(b)
CPF Sol.'n
Ð#ß %ß $Ñ
Ð%ß #ß %Ñ
Ð!ß %ß #Ñ
Ð#ß %ß !Ñ
Defining Equations
B"  B# œ 'ß B# œ %ß B"  #B$ œ %
B"  B# œ 'ß B"  #B$ œ %ß B" œ %
B" œ !ß B# œ %ß B"  #B$ œ %
B$ œ !ß B"  B# œ 'ß B# œ %
BF Solution
Ð#ß %ß $ß #ß !ß !ß !Ñ
Ð%ß #ß %ß !ß #ß !ß !Ñ
Ð!ß %ß #ß %ß !ß #ß !Ñ
Ð#ß %ß !ß #ß !ß !ß 'Ñ
NB Var.'s
B& ß B' ß B(
B% ß B' ß B(
B" ß B& ß B(
B$ ß B& ß B'
(c) Because the sets of defining equations of Ð%ß #ß %Ñ, Ð!ß %ß #Ñ and Ð#ß %ß !Ñ differ from
the set of defining equations of Ð#ß %ß $Ñ by only one equation, they are adjacent to
Ð#ß %ß $Ñ. On the other hand, the sets of defining equations of Ð%ß #ß %Ñ, Ð!ß %ß #Ñ and
Ð#ß %ß !Ñ differ by more than one equation, they are not adjacent to each other. The same
statement is true if we substitute "nonbasic variables" for "defining equations" and
"variable" for "equation."
5-9
5.1-20.
(a) B& enters.
(b) B% leaves.
(c) Ð%ß #ß %ß !ß #ß !ß !Ñ
5.1-21.
5.2-1.
 B$ 
(a) Optimal Solution: B" œ F " , œ
 B& 
^ œ -B œ  )
%
'
(b) Shadow prices: -F F
"
)
(
%
'
$
œ
"
#(  '
!
"
Iteration 0: F œ F " œ 
!
-F œ  !
 "" $
' *
 # $
 $! 
 ! 
 
*  &!  œ **!
 
!
 &! 
5.2-2.
- œ &
"
#(
 "" $
*  ' *
 # $
)
! , E œ 
#
$
$
&
$
%
#
#
"
Revised B# coefficients: 
!
)
(
%
"   "Þ$$ 
$ œ
"
"!   #Þ'( 
#
%
!
B'
"
,B œ
œ
"  F  B(   !
! , - œ  &
"  ")!   &! 
$
#(! œ $!

"!
")!   &! 
"
!
!
#!
,,œ 
"
$!
!
#!
#!
œ
"  $!   $! 
'
!
! , so B# enters.
!
$
$
œ
, so B( leaves.
"  &   & 
5-10
Iteration 1:
"
Fnew
"
œ
!
BF œ 
$
&
"
œ
$Î&
,
"Î& 
"
!
B'
" $Î&
#!
#
œ
œ  , -F œ  !



B#
!
"Î&
$!
'
)
Revised row 0:  ! )Î& 
- & ) ( % ' ! ! 
$ & % # % ! "
#
$
$
#
#
"
!
œ  "Î& ! $Î& %Î& #Î& ! )Î& , so B% enters.
"
Revised B% coefficients: 
!
Iteration 2:
"
Fnew
#
œ
#
BF œ 
$
&
"
œ
B%
&Î%
œ
B#   "Î#
$Î&
#
%Î&
œ
, so B' leaves.


"Î&
#
#Î& 
&Î%
"Î#
$Î%
"Î# 
$Î%
#!
&Î#
œ
, - œ %
"Î#  $!   &  F
)
- & ) ( % ' ! ! 
Revised row 0:  " " 
$ & % # % ! "
#
- œ $
#
!
#
"
"
CP Ð!ß !Ñ: F œ F " œ 
!
Row 0:  $
#
"
BF œ 
"
#
#
"
!
!
'
,,œ 
"
'
"
!
!
B$
"
,B œ
œ
"  F  B%   !
#
! !
!
"Î#
, F " œ 
"
"Î#
!
'
'
œ
"  '   ' 
!
"
B"
"Î# !
'
$
œ
œ
, - œ $
B%   "Î# "  '   $  F
Row 0:  $Î#
CP Ð#ß #Ñ: F œ 
#
œ  ! ! ! ! ! " " , so the current solution is optimal.
œ Ð!ß &ß !ß &Î#ß !Ñ and ^ ‡ œ &!
! , E œ 
CP Ð$ß !Ñ: F œ 
$
ÐB‡" ß B‡# ß B‡$ ß B‡% ß B‡& Ñ
Optimal Solution:
5.2-3.
$
#
"
! 
#
"
"
#
"
#Î$
, F " œ 
#
"Î$
# 
#
"
"
#
#
!
! œ !
"Î#
"Î$
#Î$ 
"Î$
'
#
œ
, - œ $
#Î$  '   #  F
B
#Î$
BF œ  "  œ 
B#
"Î$
Row 0:  $
!
 $
"
"
!
!
"
!
!
 $
"
Optimal Solution: ÐB‡" ß B‡# Ñ œ Ð#ß #Ñ and ^ ‡ œ "!
5-11
#
!
! œ !
!
#
"Î$
$Î#
"Î$ 
!
5.2-4.
- œ "
#
!
! , E œ 
"
"
"
Iteration 0: F œ F " œ 
!
-F œ  !
$
"
!
)
,,œ 

"
%
"
!
!
B
"
, BF œ  $  œ 

"
B%
!
! , Row 0:  "
#
"
Revised B# coefficients: 
!
Iteration 1:
"
Fnew
$
œ
"
BF œ 
!
"
"
œ
!
)
)
œ 


"
%
%
! , so B# enters the basis.
!
!
$
$
œ  , so B$ leaves the basis.


"
"
"
!
"
"Î$
"Î$
B#
"Î$ !
)
)Î$
œ
œ
, - œ #



B%
"Î$ "
%
%Î$  F
Revised row 0:  #Î$
! 
œ  "Î$
Revised B" coefficients: 
$
"
Iteration 2: Fnew
œ
"
"
"
"
œ
B
"Î#
BF œ  #  œ 
B"
"Î#
Revised row 0:  "Î#
œ !
"
"
$
"
!
#Î$
!
 "
"
#
!
!
! , so B" enters the basis.
"Î$
"Î$
!
"
"Î$
œ
, so B% leaves.


"
"
#Î$ 
"Î#
"Î#
"Î#
$Î# 
"Î#
)
#
œ
, - œ #
$Î#  %   #  F
"Î# 
!
"
!
!
"
"
$
"
"
!
!
 "
"
#
!
"
!
"Î# , so the current solution is optimal.
"Î#
Optimal Solution: ÐB‡" ß B‡# Ñ œ Ð#ß #Ñ and ^ ‡ œ '
5.2-5.
- œ &
%
"
$
!
! , E œ 
"
Iteration 0: F œ F " œ 
!
-F œ  !
$
$
#
$
$ " " !
#%
,,œ 
" $ ! "
$'
!
B&
"
,B œ
œ
"  F  B'   !
! , Row 0:  &
Revised B" coefficients: 
"
!
%
"
$
!
!
#%
#%
œ
"  $'   $' 
! , so B" enters the basis.
!
$
$
œ
, so B& leaves the basis.
"  $   $ 
5-12
Iteration 1:
$
œ
$
"
Fnew
BF œ 
!
"
"
œ
!
"
"Î$
"
!
#%
)
œ  , -F œ  &


"
$'
"#
B"
"Î$
œ

B'
"
Revised row 0:  &Î$ ! 
$ $
$
œ !
#
#Î$
%
"Î$
Revised B$ coefficients: 
"
Iteration 2:
$
œ
$
"
Fnew
BF œ 
$
"
"
œ
$
"
"
$
"
!
%Î$
!
 &
"
&Î$
!
%
"
Revised row 0:  #Î$ " 
$ $
"Î$
#
!
"Î%
"Î% 
$
"
#Î$
"
$
"
!
#Î$
!
 &
"
%
" 
"
(a) F
"
"
œ !
"
!
"
!
$
"
#
Final constraint columns for ÐB" ß B# ß B$ Ñ:
"
F Eœ !
"
"
-F œ  "
$
"
#
#
!
!  #
"
"
!  "
#
$ & " !
" " œ # ! !
"
" % ! "
Final objective coefficients for ÐB" ß B# ß B$ Ñ:
-F F " E  - œ  "
Right-hand side:
"
F ,œ !
"
"
$
"
#
!
#
& " !
# ! !  "
% ! "
"
!  &   "% 
"
$ œ
&
and D œ  "




!
#
""
5-13
$
!
!
" , so current solution is optimal.
Optimal Solution: ÐB‡" ß B‡# ß B‡$ ß B‡% Ñ œ Ð""ß !ß $ß !Ñ and ^ ‡ œ &#
5.3-1.
!
! , so B$ enters the basis.
B"
"Î"# "Î%
#%
""
œ
œ
, - œ &
B$   "Î% "Î%  $'   *  F
œ !
!
!
$
"
œ
, so B' leaves.


"
"
%
"Î"#
"Î%
$
$
# œ #
!
!
!
 "% 
# &
œ)
 "" 
Final tableau:
(b) Defining equations: #B"  #B#  $B$ œ &ß B"  B#  B$ œ $ß B" œ !
5.3-2.
" "
(a) F " œ 
"
#
Final constraint columns for ÐB" ß B# ß B$ ß B% Ñ:
F " E œ 
-F œ  $
" "
%
"
#  $
#
#
"
"
#
"
"
œ
" #
" !
$ "
"
!
Final objective coefficients for ÐB" ß B# ß B$ ß B% Ñ:
-F F " E  - œ  $
# 
"
#
"
!
" !
 %
$ "
Right-hand side:
F " , œ 
" "
&
"
and ^ œ  $
œ
"
#  %   $ 
$
"
# œ $
!
#
!
"
#   œ *
$
Final tableau:
(b) Defining equations: %B"  #B#  B$  B% œ &ß $B"  B#  #B$  B% œ %ß B" œ !ß
B$ œ !
5-14
5.3-3.
F
"
#
 " "
%
œ # !
 " ! " 
Final constraint columns for ÐB" ß B# ß B$ Ñ:
F " E œ
-F œ  !
 "
#
 "
#
"
!
!
'
#  #
%
%
"  "
"Î#   !
$Î# œ !
"Î#   "
#
#
#
%
%
!
!
"
!
Final objective coefficients for ÐB" ß B# ß B$ Ñ:
-F F " E  - œ  !
#
!
!
' ! % "
"
%
!
!
 '
"
# œ !
Right-hand side:
F " , œ
#  #   ( 
 " "
# !
%
$ œ ! and ^ œ  !
 " ! "  "   " 
#
(
!
(
' ! œ '
"
Final tableau:
5.3-4.
(a) F "
 $Î"'
 "Î%
œ
$Î)
 !
"Î) ! ! 
"Î# ! ! 

"Î% " !
!
! "
Current constraint columns for ÐB" ß B# ß B$ Ñ:
 $Î"'
 "Î%
F " E œ 
$Î)
 !
-F œ  #!
'
!
"Î) ! !  ) # $   " ! *Î"' 
"Î# ! !  % $ !   ! " $Î% 

œ

"Î% " !
# ! "
! ! "Î)
!
! "  ! ! "   ! !
" 
!
5-15
Current objective coefficients for ÐB" ß B# ß B$ Ñ:
-F F " E  - œ  #!
'
!
"
!
! 
!
!
!
"
!
!
*Î"' 
$Î% 
   #!
"Î)
" 
'
) œ !
!
&Î% 
!
 #& 
 ! 
!   œ &!!
!
 #! 
Right-hand side:
 $Î"' "Î) ! !  #!!   #& 
"Î# ! !  "!!   ! 
 "Î%
F " , œ 

 œ   and ^ œ  #!
$Î)
"Î% " !
&!
!
 !
!
! "  #!   #! 
'
Current tableau:
(b) The revised simplex method would generate the reduced costs for row 0 and then the
revised column for B$ .
(c) Defining equations: )B"  #B#  $B$ œ #!!ß %B"  $B# œ "!!, B$ œ !
Note that #B"  B$ œ &! is also binding at the current solution.
5.3-5.
(a)
 -"
-" 
-#
""
&
-$
œ
(
"'
ã !
!
ã !    $&
%
&
 "
#
#
"
"
$
" ! ã ,
(
œ  "'
! " ã #, 
!
Ê -" œ $# , -#  # œ ! Ê -# œ #, -$  $ œ ! Ê -$ œ $
$Î&
(b) F " œ 
"Î&
"Î&
$Î&
, F " , œ ,‡ Í 

#Î&
"Î&
(c) Using (a): ^ ‡ œ -F ,‡ œ  -#
Using (b): ^ ‡ œ - F , œ  $Î&
"
- $   œ  #
$
%Î& 
5-16
$
&
%
&
D‡ 
"Î&
,
"
œ Í,œ&


#Î&
#,
$
"
$   œ ""
$
,
œ  $Î&
#, 
!
%Î& 
&
œ ""
"! 
5.3-6.
Iteration 1: Multiply row 2 by &Î# and add to row 0, i.e., premultiply E! by
 ! &Î# !  and add to row 0, where
" ! ã " ! ! ã % 
E! œ ! # ã ! " ! ã "# .
 $ # ã ! ! " ã ") 
Iteration 2: Add row 3 to row 0, i.e., premultiply E" by  !
where
!
"  and add to row 0,
! ! ã % "
! !
" ! ã "
E" œ ! " ã ! "Î# ! ã ' œ ! "Î# ! E! .
 $ ! ã ! " " ã '   ! " " 
Therefore, the final row 0 is: initial row 0   !
œ  $
&
œ  $
&
ã ! ! ! ã !    !
ã ! ! ! ã !  !
&
#
$
#
&Î#
! E!   !
!  ! ! "
" E!
!
" E" ,
! !
"
! "Î# ! E!
 ! " " 
5.3-7.
(a) Use the columns corresponding to artificial variables in exactly the same way as a
slack variable would have been used. Note that the shadow price of this column may be
positive or negative.
(b) For the reversed inequalities, use the negative of the column corresponding to the
slack variable in exactly the same formulae. The artificial column may be discarded.
(c) Same as (b).
(d) No change, use slack and artificial variables as above.
5-17
5.3-8.
Q aximize
^œ
*!B"  (!B#
 Q B&
#B"  B#  B$
œ#
B"  B#
 B%  B& œ #
B" ß B# ß B$ ß B% ß B& !
subject to
Initial Tableau:
BV
^
B$
B&
Eq
!
"
#
^
"
!
!
B"
*!  Q
#
#
Coefficient of
B#
B$
(!  Q !
"
"
"
!
B%
Q
!
"
B&
!
!
"
RS
#Q
#
#
The columns that will contain W ‡ for applying the fundamental insight in the final tableau
are those associated with B$ and B& , since those columns form the # ‚ # identity matrix
in the initial tableau.
5.3-9.
$Î"!
(a) F " œ 
#Î"!
"Î"!
#Î& 
Final constraint columns for ÐB" ß B# ß B$ ß B% ß B' Ñ:
F " E œ 
$Î"!
#Î"!
-F œ  'Q  $
"Î"!
"
#Î&  $
%
#
%Q  # 
#
!
"
!
!
!
œ
"   "
"
!
$Î&
#Î&
"Î"!
#Î& 
$Î"!
#Î"!
Final objective coefficients for ÐB" ß B# ß B$ ß B% ß B' Ñ:
-F F " E  - œ   'Q  $
  %Q #
%Q  # 
'Q $
!
"
"
!
#Q #
$Î&
#Î&
Q
"Î"!
#Î& 
$Î"!
#Î"!
Q  œ !
!
Right-hand side:
F " , œ 
#Î"!
$Î"!
"Î"!
)
*Î&
œ



'
%Î& 
#Î&
D œ "%Q  -F BF œ "%Q   'Q  $
%Q  # 
5-18
*Î&
œ(
%Î& 
"
"Î#
"Î# 
Final tableau:
(b) The constraints in the original tableau can be expressed as  E ã M ã M ã , 
with the second identity matrix corresponding to the artificial variables. Premultiply this
matrix by M to get:
 E‡
ã W‡
ã , ‡  œ M E
ã P‡
where M œ W ‡ œ P‡ œ 
Final tableau: >‡ œ >  @X  E
œ  ^ ‡ -
ã C‡
œ  -/X EM
ã M
$
>‡ œ >  @X  E
œ  ^ ‡ œ -
ã !
œ  -@X E
#
ã M/X @
! M
ã M
! M
ã M
ã C‡
ã ^‡ 
(d) Defining equations:
 "# .
ã M ã M ã M, ,
ã ,
ã M
ã M/X ,  @X , 
 "# M .
! œ -
ã !
ã M/ X
ã M
ã M
ã ^‡ 
ã M/ X ,   @ X  E
ã M/X @
Hence, @ œ C‡ œ   "#
ã @
ã ,
ã M/X C‡
ã M/ X
ã @
ã M
ã ! ã  M/ X ,   @ X  E
ã M/ X
Hence, @ œ C‡  M/X œ   "# M
(c) > œ  #
ã ,  œ  ME
ã ! ã  M/ X , 
ã M/ X
ã M/X C‡
œ  -/X EM@X E
ã M
"Î"!
.
#Î& 
$Î"!
"Î&
Original row 0: > œ  -/X EM
ã M
ã M/ X ,  @ X , 
ã M
ã M/ X , 
ã ,
B œ M, Í M" B œ ,
B"  %B#  #B$ œ )ß $B"  #B# œ 'ß B$ œ !
5-19
ã ,
5.3-10.
(a)
#B"  #B#  B$  B%
$B"  B#  B$
œ "! (i)
 B& œ #! (ii)
Multiply (i) by "Þ& and add to (ii).
%B#  "# B$  $# B%  B& œ $& (iii)
Divide (*) by # and add to (iii).
B"  $B#  B%  B& œ $! (iv)
Multiply (iii) by #.
)B#  B$  $B%  #B& œ (! (v)
Optimal Solution: ÐB‡" ß B‡# ß B‡$ Ñ œ Ð$!ß !ß (!Ñ and ^ ‡ œ #$!
(b)
(original objective)$(iv)  #(v)
$B"  (B#  #B$
$B"  *B#  $B$
 $B&
"'B#  #B$  'B%  %B&
Ê
")B#  $B$  'B%  (B&
Hence, the shadow prices are * and (.
(c) Defining equations: #B"  #B#  B$ œ "!ß $B"  B#  B$ œ #!ß B# œ !
(d) F œ 
#
"
"
, F " œ 
$ " 
$
C‡ œ  $
# 
"
$
"
œ *
#
"
" "
"!
$!
,B œ
œ
#  F  $ #  #!   (! 
(
# #
" " !
Revised row 0:  * (  $ " " ! "    $ ( # ! !  œ  ! ") ! * ( ,
so the current solution is optimal.
(e) Final tableau:
5-20
5.4-1.
"
F œ F " œ 
!
Iteration 0:
!
"
"
Revised B# coefficients: 
!
!
$
$
œ 


"
&
&
B# enters and B( leaves.
(œ
Iteration 1:
"
Fnew
 ++"###
"
+##
œ

 $&
"
&
"
"  !
&
"
œ
!
"
!
œ

"
!
 $&
 $&
"
&

%
#
&
"  #  œ  # 
&
&
"
Revised B% coefficients: 
!
 $&
B% enters and B' leaves.
(œ
Iteration 2:

"
+w""
+w
  +#%w""


"
Fnew
œ  %"
#
&
œ
#
Iteration 0:
%
!
FœF
BF œ
! , E œ
!
"
"
œ !
!
$
"
#
!
"
!
 B%   "
B& œ !
 B'   !
-F œ  !
!

&
"  $&
!
%
œ
" 
 "
"  !
&
#
5.4-2.
- œ "
&
%
 "#
"
%
!
!
!
"
!
"
!
&
"
#
"
!
!
!
"
!
 $%
"
#

!
 "! 
! ,,œ
)


"
( 
!  "!   "! 
!
)
)
œ



"
(
( 
! , Row 0:  "
B$ enters the basis.
"
Revised B$ coefficients: !
!
!
"
!
B% leaves the basis.
5-21
#
%
!
!
!
!  &   & 
!
" œ "
"  #   # 
 &
( œ   "& 
 # 
"
Iteration 1:
&
 &
œ  "
! ! "

" ! !
! "  !
"
"
Fnew
 #
&
&
 B$   &"
BF œ B& œ   &
 B'    #
!
!
!
œ  (&
"
!
!
$

"
!
#
"
%
!
&
"
#
 '&
%
&
!
!
!
!
"
!
B# enters the basis.
 &
Revised B# coefficients:   "&
 #
&
B& leaves.
"
  "* 
& 
( œ  "*
 # 
! !
" !
! "
"
&
Revised row 0:
 %&
!  &
! œ   "&
"   #
&
! !  "!
  #

)
" !
œ '


$
(
! "
"
-F œ  %
!
"
!
!
! "
"
#
%
!
!
!
"
! ! "
& 
   "*



% œ
" !
&
! "  !    #& 
"
Iteration 2:
"*
"
Fnew
"
œ!
"
 "*
&
"*
#
"*
!
!  "&
!   "&
"   #&
 B$   "*
"
BF œ B# œ   "*
 B'    )
%
-F œ  %
#
!
"*
"
 "*
&
"*
#
"*
5-22
%
! !   "*
"
" !  œ   "*
)
! "    "*
"
 "*
$#
!  "!

   "*
$! 


)
œ "*
!


 '* 
" (
"*
&
"*
#
"*
!
!
"
Revised row 0:

"%
"*
$
! "
#
'
"*
œ  #*
"*
!
"
%
!
"%
"*
!
&
"
#
'
"*
"
!
!
!
! "
"
!
"
!
!
#
%
!
!
!
The current solution is optimal.
$#
‡
Optimal Solution: ÐB‡" ß B‡# ß B‡$ Ñ œ !ß $!
"* ß "*  and ^ œ
5.4-3.
- œ #
#
Iteration 0:
$
!
FœF
!
"
! , E œ
"
œ !
!
!
"
!
 "
#
 "
 B%   "
BF œ B& œ !
 B'   !
-F œ  !
!
!
!
"
!
"
!
" " " ! !
 % 
" " ! " ! , , œ
#


" $ ! ! "
"# 
!  %   % 
!
#
#
œ
"  "#   "# 
! , Row 0:  #
B$ enters the basis.
Revised B$ coefficients:
"))
"*
"
!
!
!
"
!
B& leaves the basis.
5-23
#
$
!
!
!
!  "   " 
!
" œ "

"
$ $
Iteration 1:
 " 
"
"
" , Fnew œ !
(œ
 $ 
!
 B%   "
BF œ B$ œ !
 B'   !
-F œ  !
" !  %   # 
" !
#
œ #
$ "  "#   ' 
!
$
" ! 
" !
$ " 
Revised row 0:
!
$
œ %
 "
!
#
 "
"
"
"
"
!
!
"
"
$
$
"
!
!
!
B# enters the basis.
"
Revised B# coefficients: !
!
!
!  #
"
!
"
!
 "Î# 
 "Î#
"
( œ "Î# , Fnew œ "Î#
 # 
 #
BF œ
 B#   "Î#
B$ œ "Î#
 B'   #
-F œ  #
!
$
Revised row 0:
 "Î#
&Î#
œ  &Î#
!
 "
!
#
 "
!
"Î#
$
"Î#
"Î#
"
"Î#
"Î#
"
!
!
!
!
"
!  %   " 
!
#
œ $


#
"
"#
" " " ! !
" " ! " !   #
" $ ! ! "
&Î#
!
" !  "   # 
" !
" œ "

$ "
"  %
B% leaves.
Iteration 2:
#
!
The current solution is optimal.
Optimal Solution: ÐB‡" ß B‡# ß B‡$ Ñ œ Ð!ß "ß $Ñ and ^ ‡ œ (
5-24
#
$
!
!
!
5.4-4.
- œ  "!
#!
Iteration 0:
!
 "
! , E œ
"
 &
!
FœF
BF œ
"
"
œ !
!
!
! ,
"
!
"
!
 B$   "
B% œ !
 B&   !
-F œ  !
# " ! !
 "& 
" ! " ! , , œ "#
 %& 
$ ! ! "
!
"
!
!  "&   "& 
!
"# œ "#

"
%&   %& 
! , Row 0:  "!
!
B# enters the basis.
"
Revised B# coefficients: !
!
!
"
!
B$ leaves the basis.
Iteration 1:
 "Î# 
 "Î#
"
( œ "Î# , Fnew œ "Î#
 $Î# 
 $Î#
BF œ
!
!
!
!
!  #   # 
!
" œ "

"
$ $
! !
" !
! "
 B#   "Î# ! !  "&   (Þ& 
B% œ "Î# " !
"# œ
%Þ&
 B&   $Î# ! "  %&   ##Þ& 
-F œ  #!
!
!
Revised row 0:
 "!
#!
!
œ  #!
 "
!
"
 &
!
"!
# " ! !
" ! " !   "!
$ ! ! "
!
B" enters the basis.
Revised B" coefficients:
!
 "Î#
"Î#
 $Î#
B% leaves.
5-25
#!
!
!
!
! !  "   "Î# 
" !
" œ
$Î#
! "  &   "$Î# 
Iteration 2:

"Î$ 
 "Î$
"
#Î$ , Fnew œ "Î$
(œ
 "$Î$ 
 #Î$
 B#   "Î$
BF œ B" œ "Î$
 B&   #Î$
-F œ  #!
"!
!
Revised row 0:
 "!Î$
œ !
%!Î$
!
!
"!Î$
 "
"
 &
%!Î$
"Î$
#Î$
"$Î$
"Î$
#Î$
"$Î$
!
!
"
!  "&   * 
!
"# œ $
"  %&   $ 
# " ! !
" ! " !   "!
$ ! ! "
!
The current solution is optimal.
Optimal Solution: ÐB‡" ß B‡# Ñ œ Ð$ß *Ñ and ^ ‡ œ #"!
5-26
#!
!
!
!
CHAPTER 6: DUALITY THEORY
6.1-1.
(a)
(b)
minimize
subject to
"&C"  "#C#  %&C$
C"  C#  &C$
#C"  C#  $C$
C" ß C# ß C$ !
minimize
subject to
%C"  #C#  "#C$
C"  #C#  C$
C"  C#  C$
C"  C#  $C$
C" ß C# ß C$ !
"!
#!
#
#
$
6.1-2.
(a)
C"
C#
C$
B"
"
%
#
&Ÿ
minimize
subject to
B#
#
'
$
"Ÿ
B$
%
&
$
$Ÿ
B%
$
%
)
%Ÿ
Ÿ #!
Ÿ %!
Ÿ &!
#!C"  %!C#  &!C$
C"  %C#  #C$
#C"  'C#  $C$
%C"  &C#  $C$
$C"  %C#  )C$
C" ß C# ß C$ !
&
"
$
%
(b) The dual problem has no feasible solution.
6.1-3.
(a) Apply the simplex method to the dual of the problem, since the dual has fewer
constraints (not including nonnegativity constraints). We expect that the simplex method
will go through fewer basic feasible solutions.
(b) Apply the simplex method to the primal problem, since it has fewer constraints (not
including nonnegativity constraints). We expect that the simplex method will go through
fewer basic feasible solutions.
6.1-4.
(a)
minimize
subject to
"#C" 
C" 
C" 
#C" 
C" ß C#
C#
C#
C#
C#
"
#
"
!
(b) It is clear from the dual problem that ÐC" ß C# Ñ œ Ð!ß !Ñ is the optimal dual solution. By
strong duality, ^ œ ! Ÿ !.
6-1
6.1-5.
(a)
minimize
subject to
$C"  &C#
C"
C#
C"  #C#
C" ß C#
#
'
*
!
(b) Optimal Solution: ÐC"‡ ß C#‡ Ñ œ Ð#ß 'Ñ, so shadow prices for resources 1 and 2 are # and
' respectively.
(c)
6-2
6.1-6.
(a)
minimize
subject to
'C"  %C#
#C"
"
#C"  C# $
#C"  #C#
#
C" ß C# !
(b) Optimal Solution: ÐC"‡ ß C#‡ Ñ œ Ð"Î#ß $Î#Ñ, so shadow prices for resources 1 and 2 are
"Î# and $Î# respectively.
(c)
6-3
6.1-7.
(a) The feasible region is empty.
(b)
minimize
subject to
2C"  %C#
C"  %C# "
C"  C# #
C" ß C# !
(c) As the objective line is dragged up and to the left the objective value decreases. This
can be done forever, so the objective function value is unbounded.
6-4
6.1-8.
Primal: maximize
subject to
B"  B#
B"  B# Ÿ "
B"  B# Ÿ !
B" ß B# !
Let B" œ B# œ - Ä ∞, ^ œ #- is unbounded.
Dual: minimize
subject to
C"
C"  C# "
C"  C# "
C" ß C# !
The dual problem is infeasible.
6-5
6.1-9.
Primal: maximize
subject to
B"
B"  B# Ÿ !
B"  B# Ÿ "
B" ß B# !
 C#
Dual: minimize
subject to
C"  C# "
C"  C# !
C" ß C# !
Neither the primal nor the dual is feasible. They have the same two constraints, which
contradict each other, so their feasible region is empty.
6.1-10.
Primal: maximize
subject to
B"  B#
B"
Ÿ "
B"  B# Ÿ !
B" ß B# !
The primal problem is clearly infeasible.
Dual: minimize
subject to
C"
C"  C# "
C# "
C" ß C# !
Let - Ä ∞ in the feasible solution Ð-ß "Ñ, so the objective function value is unbounded.
6-6
6.1-11.
Let B! and C! be a primal and a dual feasible point respectively. By weak duality,
∞  -B! Ÿ C! ,  ∞.
Furthermore, for any primal feasible point B and any dual feasible point C,
-B Ÿ C! , and -B! Ÿ C,.
This means that the primal problem cannot be unbounded, as it is bounded above by C! ,
and similarly, the dual problem cannot be bounded as it is bounded below by -B! .
Therefore, since the primal problem (and the dual problem) has a feasible solution and
the objective function value is bounded, it must have an optimal solution.
6.1-12.
(a) From the primal, EB Ÿ ,, B
CX E  - X
,  EB
!, B
!, C
In other words, CX EB
duality.
! and from the dual, CX E
! Ê ÐCX E  - X ÑB
! Ê CX Ð,  EBÑ
- X B and CX ,
-X , C
!, so
!
!.
CX EB, so CX ,
CX EB
- X B, which is weak
(b) There are many ways to prove this. The simplest is by contradiction. Assume the
primal objective ^ can be increased indefinitely and the dual does have a feasible
solution. By weak duality, - X B Ÿ CX , for all primal feasible B, given C is a dual feasible
solution. This means that ^ is bounded above, which contradicts the assumption. Hence,
if ^ is unbounded, then the dual must be infeasible.
6.1-13.
Primal: maximize
^ œ -B
Dual: minimize
subject to
EB Ÿ ,
B !
subject to
[ œ C,
CE C !
Since changing , to , keeps the dual feasible region unchanged, C‡ must be feasible for
the new problem. Let C be the optimal solution for the new dual, then clearly C, Ÿ C‡ ,,
since C is optimal. Furthermore, by strong duality, -B œ C, Ÿ C‡ ,.
6.1-14.
(a) TRUE. If E is an 8 ‚ 7 matrix, then in standard form, the number of functional
constraints is 8 for the primal and 7 for the dual. The number of variables is 7 in the
primal and 8 in the dual. Hence, for both, the sum of the number of constraints and
variables is 7  8.
(b) FALSE. This cannot be true since the weak and strong duality theorems imply that
the primal and the dual objective function values are the same only at optimality.
(c) FALSE. If the primal problem has an unbounded objective function value, the dual
problem must be infeasible, since by weak duality, if the dual has a feasible solution C,
the primal objective value is ^ œ -B Ÿ C, .
6-7
6.2-1.
(a)
Iteration 0: Since all coefficients are zero, at the current solution Ð!ß !Ñ, the three
resources (production time per week at plant 1, 2 and 3) are free goods. This means
increasing them does not improve the objective value.
Iteration 1: Ð!ß &Î#ß !Ñ. Now resource 2 has been entirely used up and contributes
&Î# to profit per unit of resource. Since this is positive, it is worthwhile to continue fully
using this resource.
Iteration 2: Ð!ß $Î#ß "Ñ. Resources 2 and 3 are used up and contribute a positive
amount to profit. Resource 1 is a free good while resources 2 and 3 contribute $Î# and "
per unit of resource respectively.
(b)
Iteration 0: Ð$ß &Ñ. Both activities 1 and 2 (number of batches of product 1 and
2 produced per week) can be initiated to give a more profitable allocation of the
resources. The current contribution of the resources required to produce one batch of
product 1 or 2 to the profit is smaller than the unit profit per batch of product 1 or 2
respectively.
Iteration 1: Ð$ß !Ñ. Again activity 1 can be initiated to give a more profitable use
of resources, but activity 2 is already being produced (or the resources are being used just
as well in other activities).
Iteration 2: Ð!ß !Ñ. Both activities are being produced (or the resources are being
used just as profitably elsewhere).
(c)
Iteration 1: Since activities 1 and 2 can be initiated to increase the profit (give the
same amount of resources), we choose to increase one of these. We choose activity 2 as
the entering activity (basic variable), since it increases the profit by & for every unit of
product 2 produced (as opposed to $ for product 1).
Iteration 2: Only activity 1 can be initiated for more profit, so we do so.
Iteration 3: Both activity 1 and 2 are being used. Furthermore, since the
coefficients for B$ , B% and B& are nonnegative, it is not worthwhile to cut back on the use
of any of the resources. Thus, we must be at the optimal solution.
6.3-1.
(a)
minimize
[ œ #!C"  "!C#
subject to
&C"  C# '
#C"  #C# )
C" ß C# !
6-8
(b)
Primal:
ÐB" ß B# Ñ œ Ð&Î#ß "&Î%Ñ is optimal with ^ œ %&. Infeasible corner point solutions are
Ð!ß "!Ñ and Ð"!ß !Ñ.
Dual:
ÐC" ß C# Ñ œ Ð"Î#ß (Î#Ñ is optimal with [ œ %&. Infeasible corner point solutions are
Ð!ß %Ñ, Ð!ß !Ñ and Ð'Î&ß !Ñ.
6-9
(c)
Primal BS
Ð!ß &ß "!ß !Ñ
Ð!ß !ß #!ß "!Ñ
Ð%ß !ß !ß 'Ñ
Ð&Î#ß "&Î%ß !ß !Ñ
Ð!ß "!ß !ß "!Ñ
Ð"!ß !ß $!ß !Ñ
Feasible?
Yes
Yes
Yes
Yes
No
No
^
%!
!
#%
%&
)!
'!
Dual BS
Ð!ß %ß #ß !Ñ
Ð!ß !ß 'ß )Ñ
Ð'Î&ß !ß !ß #)Î&Ñ
Ð"Î#ß (Î#ß !ß !Ñ
Ð%ß !ß "%ß !Ñ
Ð!ß 'ß !ß %Ñ
(d)
Primal: Ð!ß !ß #!ß "!Ñ
Dual: Ð!ß !ß 'ß )Ñ
Primal: Ð!ß &ß "!ß !Ñ
Dual: Ð!ß %ß #ß !Ñ
Primal: Ð&Î#ß "&Î%ß !ß !Ñ
Dual: Ð"Î#ß (Î#ß !ß !Ñ
6.3-2.
(a)
minimize
[ œ )C"  %C#
subject to
C"  C# "
$C"  C# #
C" ß C# !
6-10
Feasible?
No
No
No
Yes
Yes
Yes
(b)
Primal:
ÐB" ß B# Ñ œ Ð#ß #Ñ is optimal with ^ œ '. Infeasible corner point solutions are Ð)ß !Ñ and
Ð!ß %Ñ.
Dual:
ÐC" ß C# Ñ œ Ð"Î#ß "Î#Ñ is optimal with [ œ '.
6-11
(c)
Primal BS
Ð%ß !ß %ß !Ñ
Ð!ß !ß )ß %Ñ
Ð!ß )Î$ß !ß %Î$Ñ
Ð#ß #ß !ß !Ñ
Ð!ß %ß %ß !Ñ
Ð)ß !ß !ß %Ñ
Feasible?
Yes
Yes
Yes
Yes
No
No
^
%
!
"'Î$
'
)
)
(d)
Primal: Ð!ß !ß )ß %Ñ
Dual: Ð!ß !ß "ß #Ñ
Primal: Ð!ß )Î$ß !ß %Î$Ñ
Dual: Ð#Î$ß !ß "Î$ß !Ñ
Primal: Ð#ß #ß !ß !Ñ
Dual: Ð"Î#ß "Î#ß !ß !Ñ
6-12
Dual BS
Ð!ß "ß !ß "Ñ
Ð!ß !ß "ß #Ñ
Ð#Î$ß !ß "Î$ß !Ñ
Ð"Î#ß "Î#ß !ß !Ñ
Ð!ß #ß "ß !Ñ
Ð"ß !ß !ß "Ñ
Feasible?
No
No
No
Yes
Yes
Yes
6.3-3.
NB Primal Var.
B" ß B#
B" ß B%
B% ß B&
B$ ß B&
B# ß B$
B" ß B&
B$ ß B%
B# ß B&
Assoc. Dual Var.
C% ß C&
C% ß C#
C# ß C$
C" ß C$
C& ß C"
C% ß C$
C" ß C#
C& ß C$
NB Dual Var.
C" ß C# ß C$
C" ß C$ ß C&
C" ß C% ß C&
C# ß C% ß C&
C# ß C$ ß C%
C" ß C# ß C&
C$ ß C% ß C&
C" ß C# ß C%
In all cases, complementary slackness holds: B" C% œ B# C& œ B$ C" œ B% C# œ B& C$ œ !.
6.3-4.
If either the primal or the dual has a degenerate optimal basic feasible solution, then the
other may have multiple solutions. For example, consider the problem:
maximize
subject to
$B"
+"" B"  B# œ !
#B"  B$ œ "
B" ß B# ß B$ !
If +""  !, we can pivot and get an alternative optimal solution to the dual problem. If
+"" Ÿ !, we cannot.
The converse is true, however. If a problem has multiple optimal solutions, then two of
them must be adjacent corner points. To move from the tableau of one solution to that of
the other requires exactly one pivot. Suppose B4 enters and B5 leaves. A partial tableau is:
B5
B4
-4
RS
+54
,5
+54 must be positive and ,5 !. If ,5  !, then - 4 or ^ would change with the pivot. If
,5 œ !, then B4 pivots in at value zero and the resulting tableau represents the same
corner point, contradicting the assumption that the two optimal solutions are distinct.
6.3-5.
(a)
Minimize
[ œ C"
subject to
C" 2
C" 4
C" !
The optimal solution is C" œ 2 and [ œ 2.
6-13
(b) ÐC" ß C# ß C$ Ñ œ Ð2ß !ß #Ñ is the optimal basic feasible solution for the dual. By
complementary slackness, C" B$ œ C# B" œ C$ B# œ !, so B# œ B$ œ !. Since
B"  B#  B$ œ ", ÐB" ß B# ß B$ Ñ œ Ð"ß !ß !Ñ is optimal for the primal.
(c) For -"  %, the dual is infeasible and the primal objective function is unbounded.
6.3-6.
(a)
minimize
[ œ "!C"  "!C#
subject to
C"  $C# #
#C"  $C# (
C"  #C# %
C" ß C# !
(b) Ð!ß &Î#Ñ is feasible for the dual problem. By weak duality,
[ œ "! † !  "! † &Î# œ #&
D,
so the optimal primal objective function value is less than #&.
(c)
The primal basic solution is ÐB" ß B# ß B$ ß B% ß B& Ñ œ Ð!ß "!ß "!ß !ß !Ñ, which is not feasible.
The dual basic solution is ÐC" ß C# ß D" -" ß D# -# ß D$ -$ Ñ œ Ð#ß "ß $ß !ß !Ñ.
6-14
(d)
ÐC" ß C# Ñ œ Ð!ß (Î$Ñ is optimal with [ œ (!Î$. From the dual solution, C# , C$ and C& are
basic; therefore, B$ , B& and B" are nonbasic primal variables, B# and B% are basic.
ÐB" ß B# ß B$ ß B% ß B& Ñ œ Ð!ß "!Î$ß !ß "!Î$ß !Ñ is the primal optimal basic solution with
^ œ (!Î$.
6.3-7.
(a)
minimize
[ œ 'C"  "&C#
subject to
C"  %C# #
$C"  'C# &
#C"  &C# $
$C"  (C# %
C"  C# "
C" ß C# !
6-15
(b) ÐC" ß C# Ñ œ Ð%Î$ß "Î'Ñ is optimal with [ œ #"Î#.
(c) ÐD" -" Ñ and ÐD# -# Ñ are nonbasic in the dual, so B" and B# must be basic in the
optimal primal solution.
(d)
ÐB" ß B# Ñ œ Ð$Î#ß $Î#Ñ is optimal with ^ œ #"Î#.
6-16
(e) The defining equations are:
B"  $B#  #B$  $B%  B&
%B"  'B#  &B$  (B%  B&
B$
B%
B&
œ'
œ "&
œ!
œ!
œ !,
which have the solution ÐB" ß B# ß B$ ß B% ß B& Ñ œ Ð$Î#ß $Î#ß !ß !ß !Ñ.
6.3-8.
(a)
minimize
subject to
[ œ "!C"  #!C#
#C"  $C# $
#C"  C# (
C"  C# #
C" ß C# !
(b) Because B# , B% and B& are nonbasic in the optimal primal solution, C" , C# and C% will
be basic in the optimal dual solution.
(c) The defining equations are:
#C"  $C#  C$
#C"  C#
 C%
C"  C#
 C&
C$
C&
œ$
œ(
œ#
œ!
œ !,
which have the solution ÐC" ß C# ß C$ ß C% ß C& Ñ œ Ð*ß (ß !ß ")ß !Ñ.
(d) ÐC" ß C# Ñ œ Ð*ß (Ñ is optimal with [ œ #$!.
6.3-9.
(a)
minimize
subject to
[ œ "!C"  '!C#  ")C$  %%C%
#C# 
C"  &C# 
C" ß C# ß C$ ß C%
C$  $C%
C$  C%
!
6-17
#
"
(b) The defining equations for ÐB" ß B# Ñ œ Ð"$ß &Ñ are:
B"  B# œ ") and $B"  B# œ %%.
Then C$ and C% must be basic in the optimal dual solution whereas C" , C# and C$ are nonbasic.
(c) The basic variables in the primal optimal solution are B" , B# , B$ and B% . Introduce B"
and B# into the basis.
ÐB" ß B# ß B$ ß B% ß B& ß B' Ñ œ Ð"$ß &ß &ß *ß !ß !Ñ is optimal with ^ œ $". The dual solution is
ÐC" ß C# ß C$ ß C% ß C& ß C' Ñ œ Ð!ß !ß "Î#ß "Î#ß !ß !Ñ.
(d) The defining equations are:
#C# 
C"  &C# 
C"
C#
C$  $C%
C$  C%
œ#
œ"
œ!
œ !,
which are satisfied by Ð!ß !ß "Î#ß "Î#ß !ß !Ñ.
6-18
6.3-10.
(a) The optimal dual solution corresponds to row 0 computed by the simplex method to
determine optimality.
(b) The complementary basic solution corresponds to row 0 as well.
6.4-1.
(a)
(b)
minimize
[ œ 10C"  2C#
subject to
C"  2C# œ 1
2C"  C# 1
C2 Ÿ ! (C1 unconstrained in sign)
Standard form:
Dual: minimize
subject to
maximize

^ œ B
"  B"  B#
subject to

B
"  B"  #B# Ÿ "!

B
"  B"  #B# Ÿ 10

#B"  #B
"  B# Ÿ #


B" ß B" ß B# !
[ œ "!C"  1!C#  #C$
C"  C#  #C$ 1
C"  C#  #C$ 1
#C"  #C#  C$
1
C" ß C# ß C$ !
Let C"w œ C"  C# (so C"w is unrestricted in sign) and C#w œ C$ Ðso C#w Ÿ !Ñ.
Then the dual is:
minimize
[ w œ 10C"w  2C#w
subject to
C"w  2C#w œ 1
2C"w  C#w
1
C2w Ÿ ! (C1w unconstrained in sign)
as given in part (a).
6.4-2.
(a) Since ÖEB œ ,× is equivalent to
E
,
 E B Ÿ  , ,
changing the primal functional constraints from EB Ÿ , to EB œ , changes the dual to:
minimize
subject to
[ œ  CX
,
?X 
, 
 CX
E
?X 
E 
Cß ?
!.
-
6-19
Let C œ C  ?.
minimize
[ œ C,
subject to
CE
-
C unrestricted in sign
Hence, the only change is the deletion of the nonnegativity constraints.
(b) ÖEB
,× is equivalent to ÖEB Ÿ ,×, so the dual of
maximize
^ œ -B
subject to
EB
B
,
!
is
minimize
[ œ CÐ,Ñ
subject to
CÐEÑ
C
-
!.
Let C œ C.
minimize
[ œ C,
subject to
CE
-
CŸ!
Hence, C
! is replaced by C Ÿ ! in the dual.
(c)
Primal: maximize
subject to
Dual: minimize
maximize
^ œ -B  -B
EB Ÿ ,
subject to
B unrestricted in sign
EB  EB Ÿ ,
B ß B !
[ œ C,
[ œ C,
^ œ -B
Í
Í
minimize
CE CÐEÑ C !
- is replaced by CE œ - .
subject to
CE œ C !
subject to
Hence, CE
6.4-3.
maximize
subject to
[ œ )C"  'C#
C"  $C# Ÿ #
%C"  #C# Ÿ $
#C"
Ÿ"
C" ß C# !
6-20
6.4-4.
(a)
maximize
subject to
[ œ C"  C#
2C"  C# Ÿ 1
C"  2C# Ÿ 2
C" ß C# !
(b)
Since [ can be increased indefinitely, the primal problem is infeasible, by weak duality.
6.4-5.
minimize
[ œ #Þ(C"  'C#  'C$w
subject to
Í
maximize
subject to
Í
maximize
subject to
Í
maximize
subject to
!Þ$C"  !Þ&C#  !Þ'C$w
!Þ%
w
!Þ"C"  !Þ&C#  !Þ%C$ !Þ&
C" !ß C$w Ÿ !ß C# unrestricted in sign
[ œ #Þ(C"  'C#  'C$w
!Þ$C"  !Þ&C#  !Þ'C$w
!Þ%
!Þ"C"  !Þ&C#  !Þ%C$w
!Þ&
w
C" !ß C$ Ÿ !ß C# unrestricted in sign
[ w œ #Þ(C"w  'C#w  'C$
!Þ$C"w  !Þ&C#w  !Þ'C$ !Þ%
!Þ"C"w  !Þ&C#w  !Þ%C$ !Þ&
C"w Ÿ !ß C$ !ß C#w unrestricted in sign
[ w œ #Þ(C"w  'C#w  'C$
!Þ$C"w  !Þ&C#w  !Þ'C$ Ÿ !Þ%
!Þ"C"w  !Þ&C#w  !Þ%C$ Ÿ !Þ&
C"w Ÿ !ß C$ !ß C#w unrestricted in sign
6-21
6.4-6.
(a)
maximize
subject to
Dual: minimize
subject to
(b)
maximize
subject to
Dual: minimize
subject to
^ œ #B"  &B#  $B$
B"  #B#  B$ #!
#B"  %B#  B$ œ &!
B" ß B# ß B$ !
[ œ #!C"  &!C#
C"  #C# #
#C"  %C# &
C"  C# $
C" Ÿ !ß C# unconstrained in sign
^ œ #B"  B#  %B$  $B%
B"  B#  $B$  #B% Ÿ %
B"
 B$  B% "
#B"  B#
Ÿ #
B"  #B#  B$  #B% œ #
B" unconstrained in signß B# ß B$ ß B%
!
[ œ %C"  C#  #C$  #C%
C"  C#  #C$  C% œ #
C"
 C$  #C%
"
$C"  C#
 C% %
#C"  C#
 #C%
$
C" ß C$ !ß C# Ÿ !ß C% unconstrained in sign
6.4-7.
(a)
minimize
subject to
(b)
maximize
subject to
Standard form:
[ œ $!!C"  $!!C#
#C"  )C# %
$C" 
C# #
%C" 
C# $
#C"  &C# &
C" ß C# unconstrained in sign
^ œ %B"  #B#  $B$  &B%
#B"  $B#  %B$  #B% œ $!!
)B"  B#  B$  &B% œ $!!
B" ß B# ß B$ ß B% !
maximize
subject to
^œ
%B"  #B#  $B$  &B%
#B"  $B#  %B$  #B% Ÿ $!!
#B"  $B#  %B$  #B% Ÿ $!!
)B"  B#  B$  &B% Ÿ $!!
)B"  B#  B$  &B% Ÿ $!!
B" ß B# ß B$ ß B% !
6-22
Dual: minimize
subject to
[ œ $!!C"  $!!C#  $!!C$  $!!C%
#C"  #C#  )C$ 
$C"  $C# 
C$ 
%C"  %C# 
C$ 
#C"  #C#  &C$ 
C" ß C# ß C$ ß C% !
)C%
C%
C%
&C%
%
#
$
&
Let C"w œ C"  C# and C#w œ C$  C% .
minimize
subject to
[ œ $!!C"w  $!!C#w
#C"w  )C#w
%
$C"w 
C#w
#
w
w
%C" 
C# $
w
#C"  &C#w
&
C"w ß C#w unconstrained in sign
6.4-8.
(a)
minimize
subject to
[ œ "#!C"  )!C#  "!!C$
C#  $C$ œ "
$C"  C# 
C$ œ #
C"  %C#  #C$ œ "
C" ß C# ß C$ !
(b)Standard form:
maximize
subject to
Dual: minimize
^ œ Bw"  B"ww  #B#w  #B#ww  B$w  B$ww
$Bw#  $B#ww  B$w  B$ww Ÿ "#!
Bw"  B"ww  B#w  B#ww  %B$w  %B$ww Ÿ )!
$Bw"  $B"ww  B#w  B#ww  #B$w  #B$ww Ÿ "!!
Bw" ß B"ww ß B#w ß B#ww ß B$w ß B$ww !
[ œ "#!C"  )!C#  "!!C$
subject to
C# 
C# 
$C"  C# 
$C"  C# 
C"  %C# 
C"  %C# 
C" ß C# ß C$ !
minimize
[ œ "#!C"  )!C#  "!!C$
subject to
C# 
$C"  C# 
C"  %C# 
C" ß C# ß C$ !
6-23
$C$
$C$
C$
C$
#C$
#C$
"
"
#
#
"
"
$C$ œ "
C$ œ #
#C$ œ "
6.4-9.
The dual problem for the Wyndor Glass Co. example:
[ œ %C"  "#C#  ")C$
maximize
C"
 $C$ Ÿ $
 #C#  #C$ Ÿ &
C" ß C# ß C$ !
subject to
The dual of the dual:
Í
minimize
^ œ $B"  &B#
subject to
B"
%
#B# "#
$B"  #B# ")
B" ß B# !
maximize
^ œ $B"  &B#
subject to
B"
Ÿ%
B# Ÿ "#
$B"  #B# Ÿ ")
B" ß B# !
6.4-10.
(a) The objective is unbounded below.
(b)
maximize
subject to
#C"  %C#
C"  C# Ÿ "
#C"  C# Ÿ $
C" ß C# Ÿ !
6-24
(c) The dual has no feasible solution.
6-25
CHAPTER 7: LINEAR PROGRAMMING UNDER UNCERTAINTY
7.1-1.
(a)
(b)
,
minimize
subject to
(c) Optimal Solution:
,
(d) Since the new dual constraint
current solution is no longer optimal.
(e) New
is violated by
, the
column:
(f) The new primal variable adds a constraint to the dual,
, which is not
satisfied by
, so the current solution is no longer optimal.
(g)
new
, new column:
7-1
7.1-2.
(a)
,
New Tableau:
The current basic solution
(b)
is infeasible and superoptimal.
,
New Tableau:
The current basic solution
is infeasible and superoptimal.
7-2
(c)
New Tableau:
The current basic solution
stays optimal.
(d)
New Tableau:
Proper Form:
The current basic solution
(e)
stays optimal.
,
7-3
New Tableau:
The current basic solution
(f)
is feasible and optimal.
,
New Tableau:
Proper Form:
The current basic solution
is feasible and optimal.
7-4
7.1-3.
(a)
,
New Tableau:
From the tableau, we see that the primal basic solution is feasible, but not optimal.
From the graph, we can see the current basic solution is feasible, but not optimal.
7-5
(b)
New Tableau:
Proper Form:
The primal basic solution is both feasible and optimal.
From the graph, we see that the current basic solution is feasible and optimal.
7-6
(c)
,
New Tableau:
The primal basic solution is infeasible, but satisfies the optimality criterion.
From the graph, the current basic solution is infeasible and superoptimal.
7-7
(d)
,
New Tableau:
Proper Form:
The primal basic solution is feasible and optimal.
From the graph, the current basic solution is feasible and optimal.
7-8
7.2-1.
The model Ep(x) is developed to identify a long-term management plan that satisfies the
legal requirements and optimizes PALCO's operations and profitability. The model
consists of a linear program with the objective of maximizing present net worth subject to
harvest-flow constraints, political and environmental constraints. Detailed sensitivity
analysis is performed to "determine the optimal mix of habitat types within each of
individual watersheds" [p. 93]. Many instances of the LP problem are run with varying
parameters.
The financial benefits of this study include an increase of over $398 million in present
net worth and of over $29 million in average yearly net revenues. Sustained-yield annualharvest levels have increased. The habitat mix is improved in accordance with political
and environmental regulations. A more profitable long-term plan paved the way for
improved short- and mid-term plans. Sensitivity analysis enabled PALCO to improve its
knowledge base of the ecosystem and to adjust its plans quickly when a change in costs
or in regulations occurs. Since its decisions are now justified through a systematic
approach, PALCO is able to obtain better terms from banks. The study did not only affect
PALCO and the habitat controlled by PALCO. It has also "shown that the forest product
industries can coexist with wildlife and contribute to their habitats" [p. 104] and
"increased quality of life for future generations" [p. 105].
7.2-2.
(a)
,
New Tableau:
The current basic solution is infeasible and superoptimal.
7-9
(b)
,
New Tableau:
The current basic solution is infeasible and superoptimal.
(c)
,
New Tableau:
The current basic solution is feasible and optimal.
7-10
(d)
New Tableau:
The current basic solution is feasible and optimal.
(e)
,
New Tableau:
The current basic solution is feasible and optimal.
7-11
(f)
,
New Tableau:
Proper Form:
The current basic solution is feasible, but not optimal.
(g)
,
New Tableau:
The current basic solution is feasible and optimal.
7-12
(h) New Tableau and Proper Form:
The current basic solution is infeasible and superoptimal.
(i)
,
,
,
7-13
New Tableau:
Proper Form:
7.2-3.
,
7.2-4.
7-14
(a)
,
The current basic solution is superoptimal, but infeasible.
(b)
,
7-15
The current basic solution is feasible, but not optimal.
(c)
,
The current basic solution is feasible and optimal.
7-16
(d)
,
The current basic solution is feasible and optimal.
(e)
7-17
The current basic solution is feasible, but not optimal.
(f)
New Tableau:
Proper Form:
The current basic solution is infeasible and superoptimal.
Tableau After Reoptimization:
7-18
(g)
,
,
The current basic solution is neither feasible nor optimal.
7-19
7.2-5.
,
is feasible if
7.2-6.
(a)
,
,
7-20
.
The current basic solution is feasible and optimal.
(b)
The current basic solution remains feasible and optimal.
(c)
The current basic solution is feasible and optimal.
7-21
(d)
The current basic solution remains feasible and optimal.
(e)
7-22
The current basic solution is superoptimal, but infeasible.
7-23
(f)
The current basic solution is feasible and optimal.
(g)
7-24
,
,
7-25
The current basic solution is feasible and optimal.
(h)
7-26
The current basic solution is infeasible and superoptimal.
Tableau After Reoptimization:
7.2-7.
(a) F2-DC, F2-W1 and DC-W2 have the smallest margins for error (100). The greatest
effort in estimating the unit shipping costs should be placed on these lanes.
(b)
Cost
Allowable Range
F1-DC
F2-DC
F1-W1
F2-W1
DC-W1
DC-W2
(c) The range of optimality for each unit shipping cost indicates how much that shipping
cost can change before the optimal shipping quantities change.
(d) Use the 100% rule for simultaneous changes in the objective function coefficients. If
the sum of the percentage changes does not exceed 100%, the optimal solution will
remain optimal. If it exceeds 100%, then it may or may not be optimal for the new
problem.
7-27
7.2-8.
(a)
The allowable range for
(b) Increasing by
tableau to become
is
and the one for
is
.
(
)causes the coefficient of
in row 0 of the final
. To make it , add
times row 2 to row 0:
.
For optimality, we need
and
, so
. Hence, the
allowable range for
is
. Similarly, increasing
by
(
)causes the coefficient of
in row 0 of the final tableau to become
. To make it , add
times row 1 to row 0:
.
For optimality, we need
the allowable range for is
and
, so
. Hence,
.
7-28
(c)
The allowable range for
is
.
The allowable range for
is
.
(d) If we increase
by
, the final right-hand side becomes:
.
In order to preserve feasibility,
Similarly, if is increased by
, so the allowable range for
, the final right-hand side becomes:
is
.
.
In order to preserve feasibility,
, so the allowable range for
7-29
is
.
(e) (in MPL)
7.2-9.
If we increase
by
, the final right-hand side becomes:
.
Assuming
,
must satisfy:
.
Assuming
,
must satisfy:
.
Assuming
,
must satisfy:
.
7-30
The allowable range for
is
The allowable range for
is
.
.
7-31
The allowable range for
is
.
7.2-10.
If we increment
tableau becomes
by
(
. Add
), the coefficient of
times row 1 to row 0 to get:
in row 0 of the final
.
For optimality, we need
for is
.
The allowable range for
is
optimal as long as
.
, so
. Hence, the allowable range
. No matter how large
7-32
gets,
stays
7.2-11.
If we increment
tableau becomes
by
(
. Add
), the coefficient of
times row 2 to row 0 to get:
in row 0 of the final
.
For optimality, we need
and
, so
, so the allowable range for is
. Looking at Figure 6.3, we see that if
,
lies exactly on the constraint boundary. Thus, if
is
decreased any more,
does not remain optimal and the optimal solution becomes
. On the other hand, as
increases, the objective function gets closer to the
horizontal line
, so for any
,
stays optimal.
7.2-12.
(a)
(b)
Row
Row
Row
Row
and
(c) (in MPL)
7-33
7.2-13.
(a)
,
,
7-34
(b) Incrementing
by
, the coefficient of
in row 0 of the final tableau becomes
. In order for the solution to remain optimal,
, so
.
Incrementing
by
, the coefficient of
in row 0 of the final tableau becomes
. Using row 2 to eliminate this coefficient, we get:
.
To keep optimality, we need:
and
.
(c) (in MPL)
7.2-14.
7-35
New Tableau:
Proper Form:
The current basic solution is feasible and optimal.
7.3-1.
(a)
Activity 1
Unit Profit $2.00
Resource 1
Resource 2
Activity 2
$5.00
Resource Usage
1
2
1
3
Solution
6
Totals
10
<=
12
<=
Available
10
12
Total Profit
$22.00
2
Variable Cells
Cell
Name
$B$9 Solution X1
$C$9 Solution X2
Final Reduced Objective Allowable Allowable
Value
Cost
Coefficient Increase
Decrease
6
0
2
0.5 0.333333333
2
0
5
1
1
Constraints
Final Shadow Constraint Allowable Allowable
Cell
Name
Value
Price
R.H. Side Increase
Decrease
$D$5 Constraint 1 Totals
10
1
10
2
2
$D$6 Constraint 2 Totals
12
1
12
3
2
7-36
(b) The optimal solution is
Activity 1
Unit Profit $1.00
Resource 1
Resource 2
Solution
0
Activity 1
Unit Profit $3.00
Solution
Activity 2
$5.00
Resource Usage
1
2
1
3
The optimal solution is
Resource 1
Resource 2
if the unit profit for Activity 1 is $1.
if the unit profit for Activity 1 is $3.
Activity 2
$5.00
Resource Usage
1
2
1
3
10
Activity 1
Unit Profit $2.00
Solution
The optimal solution is
Activity 1
Unit Profit $2.00
Resource 1
Resource 2
Solution
Totals
10
<=
10
<=
0
if the unit profit for Activity 2 is $2.50.
Activity 2
$2.50
Totals
10
<=
10
<=
Available
10
12
Total Profit
$20.00
0
if the unit profit for Activity 2 is $7.50.
Activity 2
$7.50
Resource Usage
1
2
1
3
0
Available
10
12
Total Profit
$30.00
Resource Usage
1
2
1
3
10
Available
10
12
Total Profit
$20.00
4
(c) The optimal solution is
Resource 1
Resource 2
Totals
8
<=
12
<=
Totals
8
<=
12
<=
Available
10
12
Total Profit
$30.00
4
7-37
(d)
Unit Profit for
Activity 1
$1.00
$1.20
$1.40
$1.60
$1.80
$2.00
$2.20
$2.40
$2.60
$2.80
$3.00
Unit Profit for
Activity 2
$2.50
$3.00
$3.50
$4.00
$4.50
$5.00
$5.50
$6.00
$6.50
$7.00
$7.50
Solution
Activity 1
Activity 2
0
4
0
4
0
4
0
4
6
2
6
2
6
2
6
2
10
0
10
0
10
0
Solution
Activity 1
Activity 2
10
0
10
0
10
0
6
2
6
2
6
2
6
2
0
4
0
4
0
4
0
4
Total Profit
$20.00
$20.00
$20.00
$20.00
$20.80
$22.00
$23.20
$24.40
$26.00
$28.00
$30.00
Total Profit
$20.00
$20.00
$20.00
$20.00
$21.00
$22.00
$23.00
$24.00
$26.00
$28.00
$30.00
The allowable range for the unit profit of Activity 1 is approximately between $1.60 and
$1.80 up to between $2.40 and $2.60. The allowable range for the unit profit of Activity
2 is between $3.50 and $4 up to between $5.50 and $6.
(e) The allowable range for the unit profit of Activity 1 is approximately between $1.67
and $2.50. The allowable range for the unit profit of Activity 2 is between $4 and $6.
(f) The allowable range for the unit profit of Activity 1 is approximately between $1.67
and $2.50. The allowable range for the unit profit of Activity 2 is between $4 and $6.
7-38
(g)
Total Profit
Unit Profit for Activity 2
Unit Profit for Activity 1
$2.50
$1.00
$11.00
$1.20
$12.20
$1.40
$14.00
$1.60
$16.00
$1.80
$18.00
$2.00
$20.00
$2.20
$22.00
$2.40
$24.00
$2.60
$26.00
$2.80
$28.00
$3.00
$30.00
$3.00
$12.00
$13.20
$14.40
$16.00
$18.00
$20.00
$22.00
$24.00
$26.00
$28.00
$30.00
$3.50
$14.00
$14.20
$15.40
$16.60
$18.00
$20.00
$22.00
$24.00
$26.00
$28.00
$30.00
$4.00
$16.00
$16.00
$16.40
$17.60
$18.80
$20.00
$22.00
$24.00
$26.00
$28.00
$30.00
$4.50
$18.00
$18.00
$18.00
$18.60
$19.80
$21.00
$22.20
$24.00
$26.00
$28.00
$30.00
$5.00
$20.00
$20.00
$20.00
$20.00
$20.80
$22.00
$23.20
$24.40
$26.00
$28.00
$30.00
$5.50
$22.00
$22.00
$22.00
$22.00
$22.00
$23.00
$24.20
$25.40
$26.60
$28.00
$30.00
$6.00
$24.00
$24.00
$24.00
$24.00
$24.00
$24.00
$25.20
$26.40
$27.60
$28.80
$30.00
$6.50
$26.00
$26.00
$26.00
$26.00
$26.00
$26.00
$26.20
$27.40
$28.60
$29.80
$31.00
$7.00
$28.00
$28.00
$28.00
$28.00
$28.00
$28.00
$28.00
$28.40
$29.60
$30.80
$32.00
$7.50
$30.00
$30.00
$30.00
$30.00
$30.00
$30.00
$30.00
$30.00
$30.60
$31.80
$33.00
(h) Keeping the unit profit of Activity 2 fixed, the unit profit of Activity 1 cannot be
changed to less than 1.67 or more than 2.5 without changing the optimal solution.
Similarly if the unit profit of Activity 1 is fixed at 1, the unit profit of Activity 2 needs to
stay between 4 and 6 so that the optimal solution remains the same. Otherwise, the
objective function line becomes either too flat or too steep and the optimal solution
becomes
or
.
7-39
7.3-2.
(a) The original model:
Activity 1
Unit Profit $2.00
Resource 1
Resource 2
Solution
Activity 2
$5.00
Resource Usage
1
2
1
3
6
Totals
10
<=
12
<=
Available
10
12
Total Profit
$22.00
2
With one additional unit of resource 1:
Unit Profit
Resource 1
Resource 2
Solution
Activity 1
$2.00
Activity 2
$5.00
Resource Usage
1
2
1
3
9
Totals
11
<=
12
<=
Available
11
12
Total Profit
$23.00
1
The shadow price (the increase in total profit) is $1.
(b) The shadow price of $1 is valid in the range of
Available
Resource 1
5
6
7
8
9
10
11
12
13
14
15
to
Total
Profit
$12.50
$15.00
$17.50
$20.00
$21.00
$22.00
$23.00
$24.00
$24.00
$24.00
$24.00
Solution
Activity 1
Activity 2
0
2.5
0
3
0
3.5
0
4
3
3
6
2
9
1
12
0
12
0
12
0
12
0
7-40
.
Incremental
Profit
$2.50
$2.50
$2.50
$1.00
$1.00
$1.00
$1.00
$0.00
$0.00
$0.00
(c) With one additional unit of resource 2:
Activity 1
Unit Profit $2.00
Resource 1
Resource 2
Solution
Activity 2
$5.00
Resource Usage
1
2
1
3
4
Totals
10
<=
13
<=
Available
10
13
Total Profit
$23.00
3
The shadow price (the increase in total profit) is $1.
(d) The shadow price of $1 is valid in the range of
Available
Resource 2
6
7
8
9
10
11
12
13
14
15
16
17
18
Solution
Activity 1
Activity 2
6
0
7
0
8
0
9
0
10
0
8
1
6
2
4
3
2
4
0
5
0
5
0
5
0
5
Total
Profit
$12.00
$14.00
$16.00
$18.00
$20.00
$21.00
$22.00
$23.00
$24.00
$25.00
$25.00
$25.00
$25.00
to
.
Incremental
Profit
$2.00
$2.00
$2.00
$2.00
$1.00
$1.00
$1.00
$1.00
$1.00
$0.00
$0.00
$0.00
(e) From the sensitivity report, the shadow prices for both constraints are $1. According
to the allowable increase and decrease, the allowable range for the right-hand side of the
first constraint is to . Similarly, the allowable range for the right-hand side of the
second constraint is
to .
Variable Cells
Final Reduced Objective Allowable Allowable
Cell
Name
Value
Cost
Coefficient Increase
Decrease
$B$9 Solution Resource Usage
6
0
2
0.5
0.333333333
$C$9 Solution Activity 2
2
0
5
1
1
Constraints
Cell
Name
$D$5 Resource 1 Totals
$D$6 Resource 2 Totals
Final Shadow Constraint Allowable
Value
Price
R.H. Side Increase
10
1
10
2
12
1
12
3
7-41
Allowable
Decrease
2
2
7. -3.
(a) Optimal Solution:
, with profit 6.
(b) When one unit is added to resource 1, the profit increases to 6.5, so the shadow price
for resource 1 is 0.5. When one unit is added to resource 2, the profit increases to 6.5, so
the shadow price for resource 2 is 0.5.
(c) The original model:
Activity 1
Unit Profit
1
Resource 1
Resource 2
Solution
Activity 2
2
Resource Usage
1
3
1
1
2
Totals
8
<=
4
<=
Available
8
4
Total Profit
6
2
7-42
The shadow price for resource 1 is 0.5.
Activity 1
Unit Profit
1
Resource 1
Resource 2
Solution
Activity 2
2
Resource Usage
1
3
1
1
1.5
Totals
9
<=
4
<=
Available
9
4
Total Profit
6.5
2.5
The shadow price for resource 2 is 0.5.
Activity 1
Unit Profit
1
Resource 1
Resource 2
Solution
Activity 2
2
Resource Usage
1
3
1
1
3.5
Totals
8
<=
5
<=
Available
8
5
Total Profit
6.5
1.5
(d) The allowable range for the right-hand side of the resource 1 constraint is
approximately between (or less) and .
Available
Resource 1
4
5
6
7
8
9
10
11
12
13
14
Solution
Activity 1
Activity 2
4
0
3.5
0.5
3
1
2.5
1.5
2
2
1.5
2.5
1
3
0.5
3.5
0
4
0
4
0
4
Total
Profit
4
4.5
5
5.5
6
6.5
7
7.5
8
8
8
Incremental
Profit
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0
0
The allowable range for the right-hand side of the resource 2 constraint is approximately
between and .
7-43
Available
Resource 2
0
1
2
3
4
5
6
7
8
9
10
Solution
Activity 1
Activity 2
0
0
0
1
0
2
0.5
2.5
2
2
3.5
1.5
5
1
6.5
0.5
8
0
8
0
8
0
Total
Profit
0
2
4
5.5
6
6.5
7
7.5
8
8
8
Incremental
Profit
2
2
1.5
0.5
0.5
0.5
0.5
0.5
0
0
(e) The shadow price for both resources is 0.5. The allowable range for the right-hand
side of the first resource is between and
and that of the second resource is between
and .
Variable Cells
Final Reduced Objective Allowable
Cell
Name
Value
Cost
Coefficient Increase
$B$9 Solution Resource Usage
2
0
1
1
$C$9 Solution Activity 2
2
0
2
1
Allowable
Decrease
0.33333
1
Constraints
Cell
Name
$D$5 Resource 1 Totals
$D$6 Resource 2 Totals
Final Shadow Constraint Allowable
Value
Price
R.H. Side Increase
8
0.5
8
4
4
0.5
4
4
Allowable
Decrease
4
1.33333
(f) These shadow prices tell management that for each additional unit of resource, the
profit increases by 0.5 (for small changes). Management is then able to evaluate whether
or not to change the available amount of resources.
7.3-4.
(a)
Unit Profit
Subassembly A
Subassembly B
Production
Toys
$3.00
Subassemblies
-$2.50
Resource Usage
2
-1
1
-1
Toys
2,000
Used
3,000
1,000
Subassemblies
1,000
7-44
<=
<=
Available
3,000
1,000
Total Profit
$3,500
(b)
The estimate of the unit profit for toys can be off by something between and
before the optimal solution changes. There is no change in the solution for an increase in
the unit profit for toys, at least for an increase up to $1.
(c)
The estimate of the unit profit for subassemblies can be off by something between and
before the optimal solution changes. There is no change in the solution for an
increase in the unit profit for subassemblies, at least for an increase up to $1.
7-45
(d) A parameter analysis report for the change in unit profit for toys as in (b):
Unit Profit for Toys
$2.00
$2.25
$2.50
$2.75
$3.00
$3.25
$3.50
$3.75
$4.00
Toys
1,000
1,000
1,000
2,000
2,000
2,000
2,000
2,000
2,000
Subassemblies
0
0
0
1,000
1,000
1,000
1,000
1,000
1,000
Total Profit
$2,000
$2,250
$2,500
$3,000
$3,500
$4,000
$4,500
$5,000
$5,500
A parameter analysis report for the change in unit profit for subassemblies as in (c):
Unit Profit for Subassemblies
-$3.50
-$3.25
-$3.00
-$2.75
-$2.50
-$2.25
-$2.00
-$1.75
-$1.50
Toys
1,000
1,000
1,000
2,000
2,000
2,000
2,000
2,000
2,000
Subassemblies
0
0
0
1,000
1,000
1,000
1,000
1,000
1,000
Total Profit
$3,000
$3,000
$3,000
$3,250
$3,500
$3,750
$4,000
$4,250
$4,500
(e) The unit profit for toys can vary between $2.50 and $5 before the solution changes.
For subassemblies, the unit profit can change between -$3 and -1.50 before the solution
changes.
(f) The allowable range of the unit profit for toys is $2.50 to $5 whereas that for
subassemblies is -$3 to -$1.50.
Variable Cells
Cell
Name
$B$9 Production Toys
$C$9 Production Subassemblies
Final Reduced Objective Allowable Allowable
Value
Cost
Coefficient Increase Decrease
2,000
0
3
2
0.5
1,000
0
-2.5
1
0.5
(g)
Total Profit
Unit Profit Toys
$2.00
$2.25
$2.50
$2.75
$3.00
$3.25
$3.50
$3.75
$4.00
-$3.50
$2,000
$2,250
$2,500
$2,750
$3,000
$3,250
$3,500
$4,000
$4,500
-$3.25
$2,000
$2,250
$2,500
$2,750
$3,000
$3,250
$3,750
$4,250
$4,750
Unit
-$3.00
$2,000
$2,250
$2,500
$2,750
$3,000
$3,500
$4,000
$4,500
$5,000
Profit for Subassemblies
-$2.75 -$2.50 -$2.25 -$2.00
$2,000 $2,000 $2,000 $2,000
$2,250 $2,250 $2,250 $2,500
$2,500 $2,500 $2,750 $3,000
$2,750 $3,000 $3,250 $3,500
$3,250 $3,500 $3,750 $4,000
$3,750 $4,000 $4,250 $4,500
$4,250 $4,500 $4,750 $5,000
$4,750 $5,000 $5,250 $5,500
$5,250 $5,500 $5,750 $6,000
7-46
-$1.75
$2,250
$2,750
$3,250
$3,750
$4,250
$4,750
$5,250
$0
$0
-$1.50
$2,500
$3,000
$3,500
$4,000
$4,500
$0
$0
$0
$0
(h) As long as the sum of the percentage change of the unit profit for subassemblies does
not exceed 100% (where the allowable range is given in part (f)), the solution does not
change.
7.3-5.
(a)
Unit Profit
Subassembly A
Subassembly B
Production
Toys
$3.00
Subassemblies
-$2.50
Resource Usage
2
-1
1
-1
Used
3,000
1,000
Toys
2,000
<=
2500
Subassemblies
1,000
Toys
$3.00
Subassemblies
-$2.50
<=
<=
Available
3,000
1,000
Total Profit
$3,500
(b)
Unit Profit
Subassembly A
Subassembly B
Production
Resource Usage
2
-1
1
-1
Toys
2,001
<=
2500
Used
3,001
1,000
<=
<=
Subassemblies
1,001
Available
3,001
1,000
Total Profit
$3,500.50
The shadow price for subassembly A is $0.50, which is the maximum premium that the
company should be willing to pay.
(c)
Unit Profit
Subassembly A
Subassembly B
Production
Toys
$3.00
Subassemblies
-$2.50
Resource Usage
2
-1
1
-1
Toys
1,999
<=
2500
Used
3,000
1,001
Subassemblies
998
<=
<=
Available
3,000
1,001
Total Profit
$3,502.00
The shadow price for subassembly B is $2, which is the maximum premium that the
company should be willing to pay.
7-47
(d)
Availabile
Subassembly A
3,000
3,100
3,200
3,300
3,400
3,500
3,600
3,700
3,800
3,900
4,000
Toys
2,000
2,100
2,200
2,300
2,400
2,500
2,500
2,500
2,500
2,500
2,500
Solution
Subassemblies
1,000
1,100
1,200
1,300
1,400
1,500
1,500
1,500
1,500
1,500
1,500
Total
Profit
$3,500
$3,550
$3,600
$3,650
$3,700
$3,750
$3,750
$3,750
$3,750
$3,750
$3,750
Incremental
Profit
$50
$50
$50
$50
$50
$0
$0
$0
$0
$0
The shadow price is still valid until the maximum supply of subassembly A is at least
3,500.
(e)
Availabile
Subassembly B
1,000
1,100
1,200
1,300
1,400
1,500
1,600
1,700
1,800
1,900
2,000
Toys
2,000
1,900
1,800
1,700
1,600
1,500
1,500
1,500
1,500
1,500
1,500
Solution
Subassemblies
1,000
800
600
400
200
0
0
0
0
0
0
Total
Profit
$3,500
$3,700
$3,900
$4,100
$4,300
$4,500
$4,500
$4,500
$4,500
$4,500
$4,500
Incremental
Profit
$200
$200
$200
$200
$200
$0
$0
$0
$0
$0
The shadow price is still valid until the maximum supply of subassembly A is at least
1,500.
(f)
Variable Cells
Final Reduced Objective Allowable Allowable
Cell
Name
Value
Cost
Coefficient Increase Decrease
$B$9 Production Toys
2000
0
3
2
0.5
$C$9 Production Subassemblies 1000
0
-2.5
1
0.5
Constraints
Cell
Name
$D$5 Subassembly A Used
$D$6 Subassembly B Used
Final Shadow Constraint Allowable Allowable
Value
Price
R.H. Side Increase Decrease
3000
0.5
3000
500
1000
1000
2
1000
500
500
As shown in the sensitivity report, the shadow price is $0.50 for subassembly A and $2
for subassembly B. According to the allowable increase and decrease, the allowable
range for the right-hand side of the subassembly A constraint is 2,000 to 3,500. The
allowable range for the right-hand side of the subassembly B constraint is 500 to 1,500.
7-48
7.3-6.
Original Solution:
Cost per Shift
Time Period
6am-8am
8am-10am
10am- 12pm
12pm-2pm
2pm-4pm
4pm-6pm
6pm-8pm
8pm-10pm
10pm-12am
12am-6am
Number Working
6am-2pm
Shift
$170
1
1
1
1
0
0
0
0
0
0
6am-2pm
Shift
48
8am-4pm
Shift
$160
Noon-8pm 4pm-midnight 10pm-6am
Shift
Shift
Shift
$175
$180
$195
Shift Works Time Period? (1=yes, 0=no)
0
0
0
1
0
0
1
0
0
1
1
0
1
1
0
0
1
1
0
1
1
0
0
1
0
0
1
0
0
0
8am-4pm
Shift
31
0
0
0
0
0
0
0
0
1
1
Total
Working
48
79
79
118
70
82
82
43
58
15
>=
>=
>=
>=
>=
>=
>=
>=
>=
>=
Noon-8pm 4pm-midnight 10pm-6am
Shift
Shift
Shift
39
43
15
Minimum
Needed
48
79
65
87
64
73
82
43
52
15
Total Cost
$30,610
(a) The optimal solution does not change.
Cost per Shift
Time Period
6am-8am
8am-10am
10am- 12pm
12pm-2pm
2pm-4pm
4pm-6pm
6pm-8pm
8pm-10pm
10pm-12am
12am-6am
Number Working
6am-2pm
Shift
$170
1
1
1
1
0
0
0
0
0
0
6am-2pm
Shift
48
8am-4pm
Shift
$165
Noon-8pm 4pm-midnight 10pm-6am
Shift
Shift
Shift
$175
$180
$195
Shift Works Time Period? (1=yes, 0=no)
0
0
0
1
0
0
1
0
0
1
1
0
1
1
0
0
1
1
0
1
1
0
0
1
0
0
1
0
0
0
8am-4pm
Shift
31
0
0
0
0
0
0
0
0
1
1
Noon-8pm 4pm-midnight 10pm-6am
Shift
Shift
Shift
39
43
15
7-49
Total
Working
48
79
79
118
70
82
82
43
58
15
>=
>=
>=
>=
>=
>=
>=
>=
>=
>=
Minimum
Needed
48
79
65
87
64
73
82
43
52
15
Total Cost
$30,765
(b) The optimal solution changes.
Cost per Shift
Time Period
6am-8am
8am-10am
10am- 12pm
12pm-2pm
2pm-4pm
4pm-6pm
6pm-8pm
8pm-10pm
10pm-12am
12am-6am
Number Working
6am-2pm
Shift
$170
1
1
1
1
0
0
0
0
0
0
6am-2pm
Shift
48
8am-4pm
Shift
$160
Noon-8pm 4pm-midnight 10pm-6am
Shift
Shift
Shift
$175
$170
$195
Shift Works Time Period? (1=yes, 0=no)
0
0
0
1
0
0
1
0
0
1
1
0
1
1
0
0
1
1
0
1
1
0
0
1
0
0
1
0
0
0
8am-4pm
Shift
31
0
0
0
0
0
0
0
0
1
1
Total
Working
48
79
79
112
64
82
82
49
64
15
>=
>=
>=
>=
>=
>=
>=
>=
>=
>=
Noon-8pm 4pm-midnight 10pm-6am
Shift
Shift
Shift
33
49
15
Minimum
Needed
48
79
65
87
64
73
82
43
52
15
Total Cost
$30,150
(c) The optimal solution changes.
Cost per Shift
Time Period
6am-8am
8am-10am
10am- 12pm
12pm-2pm
2pm-4pm
4pm-6pm
6pm-8pm
8pm-10pm
10pm-12am
12am-6am
Number Working
6am-2pm
Shift
$170
1
1
1
1
0
0
0
0
0
0
6am-2pm
Shift
48
8am-4pm
Shift
$165
Noon-8pm 4pm-midnight 10pm-6am
Shift
Shift
Shift
$175
$170
$195
Shift Works Time Period? (1=yes, 0=no)
0
0
0
1
0
0
1
0
0
1
1
0
1
1
0
0
1
1
0
1
1
0
0
1
0
0
1
0
0
0
8am-4pm
Shift
31
0
0
0
0
0
0
0
0
1
1
Noon-8pm 4pm-midnight 10pm-6am
Shift
Shift
Shift
33
49
15
7-50
Total
Working
48
79
79
112
64
82
82
49
64
15
>=
>=
>=
>=
>=
>=
>=
>=
>=
>=
Minimum
Needed
48
79
65
87
64
73
82
43
52
15
Total Cost
$30,305
(d) The optimal solution does not change.
Cost per Shift
Time Period
6am-8am
8am-10am
10am- 12pm
12pm-2pm
2pm-4pm
4pm-6pm
6pm-8pm
8pm-10pm
10pm-12am
12am-6am
Number Working
Shift
$166
1
1
1
1
0
0
0
0
0
0
6am-2pm
Shift
48
Shift
$164
Shift
$171
Shift
$184
Shift Works Time Period? (1=yes, 0=no)
0
0
0
1
0
0
1
0
0
1
1
0
1
1
0
0
1
1
0
1
1
0
0
1
0
0
1
0
0
0
8am-4pm
Shift
31
Shift
$199
0
0
0
0
0
0
0
0
1
1
Total
Working
48
79
79
118
70
82
82
43
58
15
>=
>=
>=
>=
>=
>=
>=
>=
>=
>=
Noon-8pm 4pm-midnight 10pm-6am
Shift
Shift
Shift
39
43
15
Minimum
Needed
48
79
65
87
64
73
82
43
52
15
Total Cost
$30,618
(e) The optimal solution does not change.
Cost per Shift
Time Period
6am-8am
8am-10am
10am- 12pm
12pm-2pm
2pm-4pm
4pm-6pm
6pm-8pm
8pm-10pm
10pm-12am
12am-6am
Number Working
6am-2pm
Shift
$173.40
1
1
1
1
0
0
0
0
0
0
6am-2pm
Shift
48
8am-4pm
Shift
$163.20
Noon-8pm 4pm-midnight 10pm-6am
Shift
Shift
Shift
$178.50
$183.60
$198.90
Shift Works Time Period? (1=yes, 0=no)
0
0
0
1
0
0
1
0
0
1
1
0
1
1
0
0
1
1
0
1
1
0
0
1
0
0
1
0
0
0
8am-4pm
Shift
31
0
0
0
0
0
0
0
0
1
1
Noon-8pm 4pm-midnight 10pm-6am
Shift
Shift
Shift
39
43
15
7-51
Total
Working
48
79
79
118
70
82
82
43
58
15
>=
>=
>=
>=
>=
>=
>=
>=
>=
>=
Minimum
Needed
48
79
65
87
64
73
82
43
52
15
Total Cost
$31,222
(f)
Variable Cells
Cell
$C$21
$D$21
$E$21
$F$21
$G$21
Name
Number Working Shift
Number Working Shift
Number Working Shift
Number Working Shift
Number Working Shift
Final
Value
48
31
39
43
15
Reduced Objective Allowable
Cost
Coefficient Increase
0
170
1E+30
0
160
10
0
175
5
0
180
1E+30
0
195
1E+30
Allowable
Decrease
10
160
175
5
195
Part (a): The optimal solution does not change (within the allowable increase of $10).
Part (b): The optimal solution does change (outside the allowable decrease of $5).
Part (c):
Percentage of allowable increase for shift 2:
%
Percentage of allowable decrease for shift 4:
%
Sum:
%
The optimal solution may or may not change.
Part (d):
Percentage of allowable decrease for shift 1:
%
Percentage of allowable increase for shift 2:
%
Percentage of allowable decrease for shift 3:
%
Percentage of allowable increase for shift 4:
%
Percentage of allowable increase for shift 5:
%
Sum:
%
The optimal solution does not change.
Part (e):
Percentage of allowable increase for shift 1:
Percentage of allowable increase for shift 2:
%
%
Percentage of allowable increase for shift 3:
%
Percentage of allowable increase for shift 4:
%
Percentage of allowable increase for shift 5:
%
Sum:
%
The optimal solution may or may not change.
7-52
(g)
Cost per Shift (6a-2p)
$155
$158
$161
$164
$167
$170
$173
$176
$179
$182
$185
6a-2p
54
54
48
48
48
48
48
48
48
48
48
8a-4p
25
25
31
31
31
31
31
31
31
31
31
12p-10p 4p-12a
39
43
39
43
39
43
39
43
39
43
39
43
39
43
39
43
39
43
39
43
39
43
10p-6a Total Cost
15
$29,860
15
$30,022
15
$30,178
15
$30,322
15
$30,466
15
$30,610
15
$30,754
15
$30,898
15
$31,042
15
$31,186
15
$31,330
Cost per Shift (8a-4p) 6a-2p 8a-4p 12p-10p 4p-12a 10p-6a Total Cost
$145
48
31
39
43
15
$30,145
$148
48
31
39
43
15
$30,238
$151
48
31
39
43
15
$30,331
$154
48
31
39
43
15
$30,424
$157
48
31
39
43
15
$30,517
$160
48
31
39
43
15
$30,610
$163
48
31
39
43
15
$30,703
$166
48
31
39
43
15
$30,796
$169
48
31
39
43
15
$30,889
$172
54
25
39
43
15
$30,970
$175
54
25
39
43
15
$31,045
Cost per Shift (12p-10p) 6a-2p 8a-4p 12p-10p 4p-12a 10p-6a Total Cost
$160
48
31
39
43
15
$30,025
$163
48
31
39
43
15
$30,142
$166
48
31
39
43
15
$30,259
$169
48
31
39
43
15
$30,376
$172
48
31
39
43
15
$30,493
$175
48
31
39
43
15
$30,610
$178
48
31
39
43
15
$30,727
$181
48
31
33
49
15
$30,838
$184
48
31
33
49
15
$30,937
$187
48
31
33
49
15
$31,036
$190
48
31
33
49
15
$31,135
7-53
Cost per Shift (4p-12a)
$165
$168
$171
$174
$177
$180
$183
$186
$189
$192
$195
6a-2p 8a-4p 12p-10p 4p-12a 10p-6a Total Cost
48
31
33
49
15
$29,905
48
31
33
49
15
$30,052
48
31
33
49
15
$30,199
48
31
33
49
15
$30,346
48
31
39
43
15
$30,481
48
31
39
43
15
$30,610
48
31
39
43
15
$30,739
48
31
39
43
15
$30,868
48
31
39
43
15
$30,997
48
31
39
43
15
$31,126
48
31
39
43
15
$31,255
Cost per Shift (10p-6a)
$180
$183
$186
$189
$192
$195
$198
$201
$204
$207
$210
6a-2p 8a-4p 12p-10p 4p-12a 10p-6a Total Cost
48
31
39
43
15
$30,385
48
31
39
43
15
$30,430
48
31
39
43
15
$30,475
48
31
39
43
15
$30,520
48
31
39
43
15
$30,565
48
31
39
43
15
$30,610
48
31
39
43
15
$30,655
48
31
39
43
15
$30,700
48
31
39
43
15
$30,745
48
31
39
43
15
$30,790
48
31
39
43
15
$30,835
7-54
7.3-7.
6am-2pm 8am-4pm
Shift
Shift
Cost per Shift $170
$160
Time Period
6am-8am
8am-10am
10am- 12pm
12pm-2pm
2pm-4pm
4pm-6pm
6pm-8pm
8pm-10pm
10pm-12am
12am-6am
1
1
1
1
0
0
0
0
0
0
Noon-8pm
Shift
$175
4pm-midnight 10pm-6am
Shift
Shift
$180
$195
Shift Works Time Period? (1=yes, 0=no)
0
0
0
1
0
0
1
0
0
1
1
0
1
1
0
0
1
1
0
1
1
0
0
1
0
0
1
0
0
0
6am-2pm 8am-4pm
Shift
Shift
Number Working
48
31
Noon-8pm
Shift
39
0
0
0
0
0
0
0
0
1
1
Total
Working
48
79
79
118
70
82
82
43
58
15
>=
>=
>=
>=
>=
>=
>=
>=
>=
>=
4pm-midnight 10pm-6am
Shift
Shift
43
15
Minimum
Needed
48
79
65
87
64
73
82
43
52
15
Total Cost
$30,610
Variable Cells
Cell
$C$21
$D$21
$E$21
$F$21
$G$21
Final Reduced Objective AllowableAllowable
Name
Value Cost Coefficient Increase Decrease
Number Working Shift 48
0
170
1E+30
10
Number Working Shift 31
0
160
10
160
Number Working Shift 39
0
175
5
175
Number Working Shift 43
0
180
1E+30
5
Number Working Shift 15
0
195
1E+30
195
Constraints
Cell
$H$8
$H$9
$H$10
$H$11
$H$12
$H$13
$H$14
$H$15
$H$16
$H$17
Final Shadow ConstraintAllowableAllowable
Name
Value Price R.H. Side Increase Decrease
6am-8am Working
48
10
48
6
48
8am-10am Working
79
160
79
1E+30
6
10am- 12pm Working
79
0
65
14
1E+30
12pm-2pm Working
118
0
87
31
1E+30
2pm-4pm Working
70
0
64
6
1E+30
4pm-6pm Working
82
0
73
9
1E+30
6pm-8pm Working
82
175
82
1E+30
6
8pm-10pm Working
43
5
43
6
6
10pm-12am Working
58
0
52
6
1E+30
12am-6am Working
15
195
15
1E+30
6
(a) The following shifts can be increased by the indicated amounts without increasing the
total cost:
Serve 10-12 a.m.
Serve 12-2 p.m.
Serve 2-4 p.m.
Serve 4-6 p.m.
Serve 10-12 p.m.
.
7-55
(b) For each of the following shifts, the total cost increases by the amount indicated per
unit increase. These costs hold for the indicated increases.
Shift
Serve 6-8 a.m.
Serve 8-10 a.m.
Serve 6-8 p.m.
Serve 8-10 p.m.
Serve 12-6 a.m.
Increased Cost
$10
$160
$175
$5
$195
Valid for Increase
6
8
8
6
8
(c)
Percentage of allowable increase for 6-8 a.m.:
Percentage of allowable increase for 8-10 a.m.:
Percentage of allowable increase for 6-8 p.m.:
Percentage of allowable increase for 8-10 p.m.:
Percentage of allowable increase for 12-6 a.m.:
Sum:
%
The shadow prices are still valid.
%
%
%
%
%
(d)
%
%
Percentage of allowable increase for 6-8 a.m.:
Percentage of allowable increase for 8-10 a.m.:
Percentage of allowable increase for 10-12 a.m.:
Percentage of allowable increase for 12-2 p.m.:
Percentage of allowable increase for 2-4 p.m.:
Percentage of allowable increase for 4-6 p.m.:
Percentage of allowable increase for 6-8 p.m.:
Percentage of allowable increase for 8-10 p.m.:
Percentage of allowable increase for 10-12 p.m.:
Percentage of allowable increase for 12-6 a.m.:
Sum:
%
The shadow prices are still valid.
(e) All numbers can be increased by
that the shadow prices remain valid.
%
%
%
%
%
%
%
%
hours before it is no longer definite
7.3-8.
(a) Let
and
be the number of grandfather and wall clocks produced respectively.
maximize
subject to
and
7-56
(b) Optimal Solution:
,
The unit profit for grandfather clocks is allowed to vary between $200 and $400, so if it
changed from $300 to $375, the optimal solution would remain the same, provided that
there are no other changes in the model. However, if in addition to this, the unit profit for
wall clocks is changed to $175, the optimal solution becomes
.
(c) Using Excel Solver:
Unit Profit
Assembly (David)
Carving (LaDeana)
Shipping (Lydia)
Production
Grandfather
Clock
$300
Wall
Clock
$200
Hours
Used
33
40
20
Time Required
6
4
8
4
3
3
Grandfather
Clock
3.33
Wall
Clock
3.33
<=
<=
<=
Hours
Available
40
40
20
Total Profit
$1,667
7-57
(d) Changing the unit profit of grandfather clocks to $375 does not change the optimal
solution.
Unit Profit
Assembly (David)
Carving (LaDeana)
Shipping (Lydia)
Production
Grandfather
Clock
$375
Wall
Clock
$200
Hours
Used
33
40
20
Time Required
6
4
8
4
3
3
Grandfather
Clock
3.33
<=
<=
<=
Wall
Clock
3.33
Hours
Available
40
40
20
Total Profit
$1,917
If we also change the unit profit of wall clocks to $175, then the optimal solution changes
to reflect the fact that it is now more profitable to produce only grandfather clocks.
Unit Profit
Assembly (David)
Carving (LaDeana)
Shipping (Lydia)
Production
Grandfather
Clock
$375
Wall
Clock
$175
Hours
Used
30
40
15
Time Required
6
4
8
4
3
3
Grandfather
Clock
5
Wall
Clock
0
<=
<=
<=
Hours
Available
40
40
20
Total Profit
$1,875
7-58
(e) From the parameter analysis report, the allowable range to stay optimal for the unit
profit of grandfather clocks is roughly in the interval [$
$
.
Unit Profit
Production
Production
for
of
of
Grandfather Clocks Grandfather Clocks Wall Clocks Total Profit
$150
0.00
6.67
$1,333
$170
0.00
6.67
$1,333
$190
0.00
6.67
$1,333
$210
3.33
3.33
$1,367
$230
3.33
3.33
$1,433
$250
3.33
3.33
$1,500
$270
3.33
3.33
$1,567
$290
3.33
3.33
$1,633
$310
3.33
3.33
$1,700
$330
3.33
3.33
$1,767
$350
3.33
3.33
$1,833
$370
3.33
3.33
$1,900
$390
3.33
3.33
$1,967
$410
5.00
0.00
$2,050
$430
5.00
0.00
$2,150
$450
5.00
0.00
$2,250
From the parameter analysis report, the allowable range to stay optimal for the unit profit
of wall clocks is roughly in the interval [$170 $2 .
Unit Profit
for
Wall Clocks
$50
$70
$90
$110
$130
$150
$170
$190
$210
$230
$250
$270
$290
$310
$330
$350
Production
Production
of
of
Grandfather Clocks Wall Clocks Total Profit
5.00
0.00
$1,500
5.00
0.00
$1,500
5.00
0.00
$1,500
5.00
0.00
$1,500
5.00
0.00
$1,500
5.00
0.00
$1,500
3.33
3.33
$1,567
3.33
3.33
$1,633
3.33
3.33
$1,700
3.33
3.33
$1,767
3.33
3.33
$1,833
3.33
3.33
$1,900
3.33
3.33
$1,967
0.00
6.67
$2,067
0.00
6.67
$2,200
0.00
6.67
$2,333
7-59
(f)
Total Profit
Unit Profit (Wall Clock)
$50
$100
$150
$200
$250
$300
$350
$150
$750
$833
$1,000
$1,333
$1,667
$2,000
$2,333
Unit Profit (Grandfather Clock)
$200
$250
$300
$350
$400
$1,000 $1,250 $1,500 $1,750 $2,000
$1,000 $1,250 $1,500 $1,750 $2,000
$1,167 $1,333 $1,500 $1,750 $2,000
$1,333 $1,500 $1,667 $1,833 $2,000
$1,667 $1,667 $1,833 $2,000 $2,167
$2,000 $2,000 $2,000 $2,167 $2,333
$2,333 $2,333 $2,333 $2,333 $2,500
$450
$2,250
$2,250
$2,250
$2,250
$2,333
$2,500
$2,667
(g) If David is available to work a maximum of 45 hours, the optimal solution and the
total profit do not change. Even when he is available for 40 hours, he is required to use
less.
Unit Profit
Assembly (David)
Carving (LaDeana)
Shipping (Lydia)
Production
Grandfather
Clock
$300
Wall
Clock
$200
Time Required
6
4
8
4
3
3
Grandfather
Clock
3.33
Hours
Used
33
40
20
<=
<=
<=
Wall
Clock
3.33
Hours
Available
45
40
20
Total Profit
$1,667
If LaDeana is available for 5 more hours every week, the optimal number of grandfather
clocks to be produced increases whereas the optimal number of wall clocks to be
produced decreases. The total profit increases by $125.
Unit Profit
Assembly (David)
Carving (LaDeana)
Shipping (Lydia)
Production
Grandfather
Clock
$300
Wall
Clock
$200
Time Required
6
4
8
4
3
3
Grandfather
Clock
4.58
Hours
Used
36
45
20
Wall
Clock
2.08
<=
<=
<=
Hours
Available
40
45
20
Total Profit
$1,792
Finally, if Lydia increases her availability by 5 hours, the optimal number of grandfather
clocks to be produced decreases whereas the optimal number of wall clocks to be
produced increases. The optimal total profit increases by $166, which is more than the
increase caused by increasing LaDeana's working hours by the same amount.
7-60
Unit Profit
Assembly (David)
Carving (LaDeana)
Shipping (Lydia)
Production
Grandfather
Clock
$300
Wall
Clock
$200
Time Required
6
4
8
4
3
3
Grandfather
Clock
1.67
Hours
Used
37
40
25
Hours
Available
40
40
25
<=
<=
<=
Wall
Clock
6.67
Total Profit
$1,833
Note that in each case, the binding constraints remain the same.
(h)
Assembly
Time
Available
35
37
39
41
43
45
Grandfather
Clocks
3.33
3.33
3.33
3.33
3.33
3.33
Wall
Clocks
3.33
3.33
3.33
3.33
3.33
3.33
Total
Profit
$1,667
$1,667
$1,667
$1,667
$1,667
$1,667
Carving
Time
Available
35
37
39
41
43
45
Grandfather
Clocks
2.08
2.58
3.08
3.58
4.08
4.58
Wall
Clocks
4.58
4.08
3.58
3.08
2.58
2.08
Total
Profit
$1,542
$1,592
$1,642
$1,692
$1,742
$1,792
Shipping
Time
Available
15
17
19
21
23
25
Grandfather
Clocks
5.00
4.33
3.67
3.00
2.33
1.67
Wall
Clocks
0.00
1.33
2.67
4.00
5.33
6.67
Total
Profit
$1,500
$1,567
$1,633
$1,700
$1,767
$1,833
7-61
(i) The unit profit for grandfather clocks should stay in the interval
and that for
wall clocks should stay in
for the optimal solution to remain unchanged.
Variable Cells
Cell
Name
$B$12 Production Clock
$C$12 Production Clock
Final Reduced Objective
Value
Cost
Coefficient
3.33
0.00
300
3.33
0.00
200
Allowable
Increase
100
100
Allowable
Decrease
100
50
Final Shadow
Value
Price
33
0
40
25
20
33.33
Allowable
Increase
1E+30
13.333
10
Allowable
Decrease
6.667
13.333
5
Constraints
Cell
$D$6
$D$7
$D$8
Name
Assembly (David) Used
Carving (LaDeana) Used
Shipping (Lydia) Used
Constraint
R.H. Side
40
40
20
Provided that the maximum number of hours David is available is more than 33.334, the
binding constraints stay the same. LaDeana's number of available hours can differ from
40 only by 13.333. Lydia's maximum number of hours is allowed to vary between 15 and
30.
(j) The constraint associated with Lydia has the highest shadow price, so Lydia should be
the one to increase the maximum number of hours available to work per week.
(k) The constraint associated with David is not binding in the optimal solution. In other
words, David is required to work less than the maximum number of hours he is available.
Hence increasing his availability does not improve the profit unless the other partners
offer more time as well, so the shadow price of his constraint is equal to zero.
(l) The allowable increase for Lydia's hours is 10, so this shadow price can be used for an
increase of 5. If Lydia increases her available hours from 20 to 25, the total profit is
improved by approximately
$
, which is pretty close to what was
found in part (g). The difference is due to rounding.
(m) Percentage of Lydia's allowable increase used
%.
Percentage of David's allowable decrease used
%.
The sum is 125%, so by the 100% rule, the shadow prices may or may not be valid, and
hence should not be used to determine the effect on total profit.
7-62
(n)
7-63
7.4-1
(a) Applying the procedure for robust optimization with independent parameters, the
model is as follows:
Maximize
subject to
and
0,
The optimal solution is
per week).
and
(a total profit of $29,971
(b) The resulting solution provides $4971 more profit per week than the solution of
and
. Assuming there is only a small chance that the production rates
would fall below the guaranteed minimum amounts, this additional profit would be worth
the small chance of needing to pay a $5000 penalty.
7.4-2
(a) The model using the estimates of the parameters is:
Maximize
subject to
and
0,
The optimal solution is
and
7-64
(b) Using robust optimization to formulate a conservative version, the model is:
Maximize
subject to
and
0,
The optimal solution is
change in the value of from part a
and
or a (5.45/20.33) = 26.8%
7.4-3
(a) The model using the estimates of the parameters is:
Maximize
subject to
and
0,
The optimal solution is
and
7-65
.
(b) Using robust optimization to formulate a conservative version, the model is:
Maximize
subject to
and
0,
The optimal solution is
reduction in the value of
and
, or a (39.8/110) = 36.2%
from part a .
7.4-4
(a) The model using the estimates of the parameters is:
Maximize
subject to
and
0,
The optimal solution is
and
.
( ) The model using a conservative version of the model is:
Maximize
subject to
and
0,
The optimal solution is
.
and
7-66
.
7.5-1
(a) The three chance constraints are
0.99
0.99
0.99
The deterministic equivalents of these chance constraints are
(b) The optimal solution is
week).
and
($31,700 profit per
7.5-2
(a)
The chance constraint is
The deterministic equivalent is
(b)
Adjusted RHS
7.5(a)
Lower bound
Upper bound
(b)
Lower bound
Upper bound
(c)
When
, the lower bound is
thus guaranteeing a
probability of at least 0.95 that the optimal solution will be feasible.
7-67
7.5(a) When
and
the left-hand-sides of the three constraints are
84, 140, and 168, respectively, yielding z-scores of
and
. This leads to
probabilities of 0.977, 0.953, and 0.908, respectively, that the three constraints will be
satisfied.
(b) The three chance constraints are
The deterministic equivalents of the three chance constraints are
The optimal solution is
and
(c) All three constraints are satisfied with equality, so the probability the optimal solution
will turn out to be feasible is (0.975)(0.95)(0.90) = 0.834.
7.6-1
The revised model is
Maximize
subject to
and
The optimal solution is
and
In words, the optimal plan
is to produce 2 batches of product 1 per week; produce 6 batches of product 2 only if
scenario 1 occurs; produce 2 batches of the modified version of product 2 per week only
if scenario 2 occurs.
7.6When the probability is 0.65 or more, the optimal plan presented in Sec. 7.6 is still
optimal. When the probability is less than 0.65, the optimal solution becomes
and
7-68
7.6-3
Let
= test market advertising
= national campaign advertising if test market is very favorable
= national campaign advertising if test market is barely favorable
= national campaign advertising if test market is unfavorable
(all decision variables in units of $millions)
Then the revised model is
Maximize
subject to
and
and
The optimal solution is
and
. In words,
they should spend $5 million in advertising for the test market. If the test market is very
favorable or barely favorable, then they should spend $95 million advertising the drink
nationally. If the test market is unfavorable, they should drop the product. The expected
net profit is $14.75 million.
7.6The stochastic programming model is
Minimize
subject to
and
and
The optimal solution is
and
In words,
they should do activity 1 at level 12, and then activity 2 at level 12, 9, or 36 under
scenario 1, 2, or 3, respectively.
7-69
Case%7.1%
!
a)!
!
!
Range Name
Cost
FractionUsed
MinimumReduction
OneHundredPercent
ReductionInEmission
TotalCost
TotalReduction
!
!
!
Cells
B4:G4
B15:G15
J8:J10
B17:G17
B8:G10
J15
H8:H10
H
5
Total
6
Reduction
7
(millions of lbs.)
8 =SUMPRODUCT(B8:G8,FractionUsed)
9 =SUMPRODUCT(B9:G9,FractionUsed)
10 =SUMPRODUCT(B10:G10,FractionUsed)
!
!
!
J
13
Total Cost
14
($million)
15 =SUMPRODUCT(Cost,FractionUsed)
!
!
Variable Cells
Cell
$B$15
$C$15
$D$15
$E$15
$F$15
$G$15
Name
Fraction Taller Smokestack (Blast)
Fraction Taller Smokestack (Open Hearth)
Fraction Filter (Blast)
Fraction Filter (Open Hearth)
Fraction Better Fuel (Blast)
Fraction Better Fuel (Open Hearth)
Final
Value
100%
62.27%
34.35%
100%
4.76%
100%
Reduced Objective
Cost
Coefficient
-34%
8
0.00%
10
0.00%
7
-182%
6
0.00%
11
-4%
9
Allowable
Increase
0.336
0.429
0.382
1.816
2.975
0.044
Allowable
Decrease
1E+30
0.667
2.011
1E+30
0.045
1E+30
Shadow
Price
0.111
0.127
0.069
Allowable
Increase
14.297
20.453
2.042
Allowable
Decrease
7.480
1.690
21.692
Constraints
Cell
$H$8
$H$9
$H$10
!
Final
Value
60
150
125
Name
Particulates (millions of lbs.)
Sulfur oxides (millions of lbs.)
Hydrocarbons (millions of lbs.)
Constraint
R.H. Side
60
150
125
b)! The!right,hand,side!of!each!constraint!with!a!non,zero!shadow!price!is!sensitive,!
since!changing!its!value!will!impact!the!total!cost.!All!three!required!reductions!
in!emission!rates!are!sensitive!parameters.!All!of!the!objective!coefficients!have!
an!allowable!range!to!stay!optimal!around!them,!and!thus!are!not!as!sensitive.!
However,!for!some,!the!allowable!change!is!small—in!particular,!the!cost!of!the!
two!better!fuel!options!(with!an!allowable!increase!of!only!0.045!and!an!
allowable!decrease!of!0.044,!respectively)!are!fairly!sensitive.!Thus,!all!five!of!
these!parameters!should!be!estimated!more!closely,!if!possible.!
7-70
!
!
!
c)! The!sensitivity!report!and,!in!particular,!the!allowable!range!for!the!objective!
coefficients!can!be!used!to!determine!whether!the!solution!will!change.!The!
following!table!shows!in!which!cases!the!optimal!solution!will!change.!
!
Abatement!
Method!
!
Current!
Value!
10%!
Less!
Value!
Taller!Smoke!(Blast)!
Taller!Smoke!(Open!H)!
Filter!(Blast)!
Filter!(Open!H)!
Better!Fuel!(Blast)!
Better!Fuel!(Open!H)!
8!
10!
7!
6!
11!
9!
7.2!
9!
6.3!
5.4!
9.9!
8.1!
!
Solution!
Changes
?!
No!
Yes!
No!
No!
Yes!
No!
10%!
More!
Value!
8.8!
11!
7.7!
6.6!
12.1!
9.9!
!
Solution!
Changes
?!
Yes!
Yes!
Yes!
No!
No!
Yes!
This!suggests!that!focus!should!be!put!on!estimating!all!of!the!costs!except!the!
filter!for!the!open!hearth!furnaces,!since!it’s!optimal!solution!will!not!change!
with!a!10%!increase!or!decrease.!!Special!consideration!should!be!given!to!the!
estimate!of!the!cost!of!the!taller!smokestack!for!the!open!hearth!furnaces,!since!
it!affects!the!optimal!solution!for!both!an!increase!and!a!decrease.!Special!
consideration!should!also!be!given!to!the!estimate!of!the!cost!of!the!better!fuel!
options,!since!the!allowable!decrease!(for!the!blast!furnace)!or!allowable!
decrease!(for!the!open!hearth!furnace)!is!so!small.!
!
d)! Below!is!the!corresponding!dual!problem.!
!
7-71
!
!
This!is!the!sensitivity!report!for!the!primal!(maximization!problem)!
The!dual!variables!are!the!shadow!prices!of!the!constraints.!If!the!primal!had!
been!left!in!minimization!form,!the!sign!of!the!dual!variables!would!be!the!
opposite.!The!dual!would!be!the!same!except!that!the!“sign”!constraints!on!the!
dual!variables!changes!from!≤!to!≥!and!vice!versa,!and!the!dual!functional!
constraints!all!change!from!!≤!to!≥.!It!would!also!be!a!maximization!problem!
instead!of!minimization.!
!
e)! !!
!
!
Pollutant!
Rate!that!
cost!
changes!
($million)!
Particulates!
0.111!
Sulfur!oxides!
0.127!
Hydrocarbons!
0.069!
Maximum!increase!
before!rate!changes!
(million!lb.)!
Maximum!decrease!
before!rate!changes!
(million!lb.)!
14.297!
20.453!
2.042!
7.480!
1.690!
21.692!
!
7-72
!
!
f)! Particulates!and!sulfur!oxides:!
For!each!unit!increase!in!particulate!reduction,!cost!will!increase!by!$0.111!
million.!
For!each!unit!decrease!in!sulfur!oxide!reduction,!cost!will!decrease!by!$0.127!
million.!
Thus,!cost!will!remain!equal!if!for!each!unit!increase!in!particulate!reduction,!the!
sulfur!oxide!reduction!is!reduced!by!$0.111!/!$0.127!=!0.874!units.!
!
Particulates!and!hydrocarbons:!
For!each!unit!increase!in!particulate!reduction,!cost!will!increase!by!$0.111!
million.!
For!each!unit!decrease!in!hydrocarbon!reduction,!cost!will!decrease!by!$0.069!
million.!
Thus,!cost!will!remain!equal!if!for!each!unit!increase!in!particulate!reduction,!the!
hydrocarbon!reduction!is!reduced!by!$0.111!/!$0.069!=!1.609!units.!
!
Particulates!and!both!sulfur!oxides!and!hydrocarbons:!
For!each!unit!increase!in!particulate!reduction,!cost!will!increase!by!$0.111!
million.!
For!each!simultaneous!unit!decrease!in!sulfur!oxide!and!hydrocarbon!reduction,!
cost!will!decrease!by!$0.127!+!$0.069!=!$0.196.!
Thus,!cost!will!remain!equal!if!for!each!unit!increase!in!particulate!reduction,!the!
sulfur!oxide!and!hydrocarbon!reduction!are!each!reduced!by!$0.111!/!$0.196!=!
0.566!units.!
%
Case%7.2%
!
a)! The!decisions!to!be!made!are!how!much!acreage!should!be!planted!in!each!of!the!
crops!and!how!many!cows!and!hens!to!have!for!the!coming!year.!!The!
constraints!on!these!decisions!are!amount!of!labor!hours!available,!the!
investment!funds!available,!the!number!of!acres!available,!the!space!available!in!
the!barn!and!chicken!house,!the!minimum!requirements!for!feed!to!be!planted.!!
The!overall!measure!of!performance!is!monetary!worth,!which!is!to!be!
maximized.!
7-73
!
b!&!c)!
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
B
C
D
E
Planting
Totals
537
736
$37,300
Plantings
Soybeans
1
1.4
$70
Corn
0.9
1.2
$60
Wheat
0.6
0.7
$40
450
30
>=
30
1
acre/cow
100
>=
100
0.05
acre/hen
Cows
10
2
$850
Hens
0.05
0
$4
Livestock
Totals
400
60
$34,000
Beginning Value (Current Livestock)
Decrease in Value per Year
End Value (Current Livestock)
$35,000
10%
$31,500
$5,000
25%
$3,750
$35,250
Cost of New Livestock
End Value (New Livestock)
$1,500
$1,350
$3
$2
$0
$0
30
0
30
<=
42
2,000
0
2,000
<=
5,000
Wage
W&S
$5
S&F
$5.50
Neighbor
Totals
$12,817.00
Hours Worked
1063
1364
2427
W&S Hours
S&F Hours
Acreage
Plantings
537
736
580
Livestock
2,400
2,400
60
$34,000
$35,250
$20,000
W&S Hours Required
S&F Hours Required
Net Value
Acres Planted
Livestock
Hours Required per Month
Grazing Land Required
Net Annual Cash Income
Current Livestock
New Livestock
Total Livestock
Barn/House Limits
Neighboring Farm Work
Totals
Net Income
End of Year Value
Leftover Investment Fund
Living Expenses
Total Monetary Worth
$37,300
F
G
580
<=
Investment
Fund
$20,000
Neighbor
1,063
1,364
0
Total
4,000
4,500
640
<=
<=
<=
$12,817
$84,117
$35,250
$20,000
-$40,000
$99,367
Available
4,000
4,500
640
!
!
This!model!predicts!that!the!family’s!monetary!worth!at!the!end!of!the!coming!
year!will!be!$99,!367.!
7-74
!
!
A
B
1 Plantings
2
Soybeans
3
W&S Hours Required 1
4
S&F Hours Required 1.4
5
Net Value 70
6
7
Acres Planted 450
8
9
10
11
C
D
0.9
1.2
60
0.6
0.7
40
E
Planting
Totals
=SUMPRODUCT(B3:D3,AcresPlanted)
=SUMPRODUCT(B4:D4,AcresPlanted)
=SUMPRODUCT(B5:D5,AcresPlanted)
30
100
=SUM(AcresPlanted)
>=
=C10*B28
1
acre/cow
>=
=D10*C28
0.05
acre/hen
Corn
Wheat
!
!
A
13 Livestock
14
15
Hours Required per Month
16
Grazing Land Required
17
Net Annual Cash Income
18
19
Beginning Value (Current Livestock)
20
Decrease in Value per Year
21
End Value (Current Livestock)
22
23
Cost of New Livestock
24
End Value (New Livestock)
25
26
Current Livestock
27
New Livestock
28
Total Livestock
29
30
Barn/House Limits
B
C
10
2
850
0.05
0
4.25
D
Livestock
Totals
=SUMPRODUCT(B15:C15,TotalLivestock)
=SUMPRODUCT(B16:C16,TotalLivestock)
=SUMPRODUCT(B17:C17,TotalLivestock)
35000
0.1
=(1-B20)*B19
5000
0.25
=(1-C20)*C19
=SUM(B21:C21)
1500
=(1-B20)*B23
3
=(1-C20)*C23
=SUMPRODUCT(B23:C23,NewLivestock)
=SUMPRODUCT(B24:C24,NewLivestock)
30
0
=CurrentLivestock+NewLivestock
<=
42
2000
0
=CurrentLivestock+NewLivestock
<=
5000
Cows
Hens
E
F
Investment
Fund
<= 20000
!
!
A
B
32 Neighboring Farm Work
33
W&S
34
Wage 5
35
36
Hours Worked 1063
C
S&F
5.5
D
Neighbor
Totals
=SUMPRODUCT(Wage,HoursWorked)
1364
=SUM(HoursWorked)
!
!
A
38 Totals
39
W&S Hours
40
S&F Hours
41
Acreage
42
43
Net Income
44
End of Year Value
45
Leftover Investment Fund
46
Living Expenses
47
Total Monetary Worth
B
Plantings
=E3
=E4
=E7
=E5
C
Livestock
=6*D15
=6*D15
=D16
=D17
=D21+D24
=InvestmentFund-D23
!
!
Range Name
AcresPlanted
Available
BarnHouseLimits
CurrentLivestock
HoursWorked
InvestmentFund
MonetaryWorth
NewLivestock
Total
TotalLivestock
Wage
Cells
B7:D7
G39:G41
B30:C30
B26:C26
B36:C36
F23
E47
B27:C27
E39:E41
B28:C28
B34:C34
!
7-75
D
E
Neighbor
Total
=B36
=SUM(B39:D39)
=C36
=SUM(B40:D40)
0
=SUM(B41:D41)
=D34
=SUM(B43:D43)
=SUM(B44:D44)
=SUM(B45:D45)
-40000
=SUM(E43:E46)
F
G
Available
<= 4000
<= 4500
<= 640
!
!
Variable Cells
Cell
$B$7
$C$7
$D$7
$B$27
$C$27
$B$36
$C$36
Name
Acres Planted Soybeans
Acres Planted Corn
Acres Planted Wheat
New Livestock Cows
New Livestock Hens
Hours Worked W&S
Hours Worked S&F
Final Reduced Objective Allowable Allowable
Value
Cost
Coefficient Increase Decrease
450
0
70
1E+30
8.4
30
0
60
8.4
1E+30
100
0
40
17.15
1E+30
0
-53
700
53
1E+30
0
-0.857
3.5
0.857
1E+30
1063
0
5
57.3
0.915
1364
0
5.5
34.5
0.930
Name
Acres Planted Corn
Acres Planted Wheat
Cost of New Livestock Totals
Total Livestock Cows
Total Livestock Hens
W&S Hours Total
S&F Hours Total
Acreage Total
Final Shadow Constraint Allowable Allowable
Value
Price
R.H. Side Increase Decrease
30
-8.4
0
450
30
100
-24.15
0
450
100
$0
$0
20000
1E+30
20000
30
0
42
1E+30
12
2000
0
5000
1E+30
3000
4000
5
4000
1E+30
1063
4500
5.5
4500
1E+30
1364
640
57.3
640
974.29
450
Constraints
Cell
$C$7
$D$7
$D$23
$B$28
$C$28
$E$39
$E$40
$E$41
!
d)! The!allowable!range!for!the!value!per!acre!planted!of!soybeans!is!61.6!to!∞.!
The!allowable!range!for!the!value!per!acre!planted!of!corn!is!–∞!to!68.4.!
The!allowable!range!for!the!value!per!acre!planted!of!wheat!is!–∞!to!57.15.!
7-76
!
!
e)! Drought!
!
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
B
C
D
Soybeans
1
1.4
-$10
Corn
0.9
1.2
-$15
Wheat
0.6
0.7
$0
0
42
>=
42
1
acre/cow
133.33
>=
133.33
0.05
acre/hen
Cows
10
2
$850
Hens
0.05
0
$4
Livestock
Totals
553.333333
84
$47,033
Beginning Value (Current Livestock)
Decrease in Value per Year
End Value (Current Livestock)
$35,000
10%
$31,500
$5,000
25%
$3,750
$35,250
Cost of New Livestock
End Value (New Livestock)
$1,500
$1,350
$3
$2
$20,000
$17,700
30
12
42
<=
42
2,000
667
2,667
<=
5,000
Wage
W&S
$5
S&F
$5.50
Hours Worked
562.2
Plantings
W&S Hours Required
S&F Hours Required
Net Value
Acres Planted
Livestock
Hours Required per Month
Grazing Land Required
Net Annual Cash Income
Current Livestock
New Livestock
Total Livestock
Barn/House Limits
Neighboring Farm Work
Totals
-$630
F
G
175.333
<=
Investment
Fund
$20,000
Neighbor
562
1,036
0
Total
4,000
4,500
259.333
<=
<=
<=
$8,510
$54,914
$52,950
$0
-$40,000
$67,864
Neighbor
Totals
$8,510.47
1036.267 1598.46665
Plantings Livestock
W&S Hours
117.8
3,320
S&F Hours 143.7333
3,320
Acreage 175.3333
84
Net Income
End of Year Value
Leftover Investment Fund
Living Expenses
Total Monetary Worth
E
Planting
Totals
117.8
143.733
-$630
$47,033
$52,950
$0
Available
4,000
4,500
640
!
!
In!a!drought,!the!model!predicts!(under!the!optimal!solution)!that!the!family’s!
monetary!worth!at!the!end!of!the!year!will!be!$67,864.!
7-77
!
!
Flood!
!
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
B
C
D
Soybeans
1
1.4
$15
Corn
0.9
1.2
$20
Wheat
0.6
0.7
$10
0
422.6667
>=
42
1
acre/cow
133.33
>=
133.33
0.05
acre/hen
Cows
10
2
$850
Hens
0.05
0
$4
Livestock
Totals
553.333333
84
$47,033
Beginning Value (Current Livestock)
Decrease in Value per Year
End Value (Current Livestock)
$35,000
10%
$31,500
$5,000
25%
$3,750
$35,250
Cost of New Livestock
End Value (New Livestock)
$1,500
$1,350
$3
$2
$20,000
$17,700
30
12
42
<=
42
2,000
667
2,667
<=
5,000
Wage
W&S
$5
S&F
$5.50
Neighbor
Totals
$4,285.07
Hours Worked
219.6
579.4667
799.067
Plantings Livestock
W&S Hours
460.4
3,320
S&F Hours 600.5333
3,320
Acreage
556
84
Plantings
W&S Hours Required
S&F Hours Required
Net Value
Acres Planted
Livestock
Hours Required per Month
Grazing Land Required
Net Annual Cash Income
Current Livestock
New Livestock
Total Livestock
Barn/House Limits
Neighboring Farm Work
Totals
Net Income
End of Year Value
Leftover Investment Fund
Living Expenses
Total Monetary Worth
$9,787
$47,033
$52,950
$0
E
Planting
Totals
460.4
600.533
$9,787
F
G
556
<=
Investment
Fund
$20,000
Neighbor
220
579
0
Total
4,000
4,500
640
<=
<=
<=
$4,285
$61,105
$52,950
$0
-$40,000
$74,055
Available
4,000
4,500
640
!
!
In!a!flood,!the!model!predicts!(under!the!optimal!solution)!that!the!family’s!
monetary!worth!at!the!end!of!the!year!will!be!$74,055.!
7-78
!
!
Early!Frost!
!
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
B
C
D
Soybeans
1
1.4
$50
Corn
0.9
1.2
$40
Wheat
0.6
0.7
$30
450
30
>=
30
1
acre/cow
100.00
>=
100.00
0.05
acre/hen
Cows
10
2
$850
Hens
0.05
0
$4
Livestock
Totals
400
60
$34,000
Beginning Value (Current Livestock)
Decrease in Value per Year
End Value (Current Livestock)
$35,000
10%
$31,500
$5,000
25%
$3,750
$35,250
Cost of New Livestock
End Value (New Livestock)
$1,500
$1,350
$3
$2
$0
$0
30
0
30
<=
42
2,000
0
2,000
<=
5,000
Wage
W&S
$5
S&F
$5.50
Neighbor
Totals
$12,817.00
Hours Worked
1063
1364
2427.000
Plantings
537
736
580
Livestock
2,400
2,400
60
$26,700
$34,000
$35,250
$20,000
Plantings
W&S Hours Required
S&F Hours Required
Net Value
Acres Planted
Livestock
Hours Required per Month
Grazing Land Required
Net Annual Cash Income
Current Livestock
New Livestock
Total Livestock
Barn/House Limits
Neighboring Farm Work
Totals
W&S Hours
S&F Hours
Acreage
Net Income
End of Year Value
Leftover Investment Fund
Living Expenses
Total Monetary Worth
E
Planting
Totals
537
736
$26,700
F
580
<=
Investment
Fund
$20,000
Neighbor
1,063
1,364
0
Total
4,000
4,500
640
<=
<=
<=
$12,817
$73,517
$35,250
$20,000
-$40,000
$88,767
!
!
In!an!early!frost,!the!model!predicts!(under!the!optimal!solution)!that!the!
family’s!monetary!worth!at!the!end!of!the!year!will!be!$88,767.!
7-79
G
Available
4,000
4,500
640
!
!
Drought!and!Early!Frost!
!
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
B
C
D
Soybeans
1
1.4
-$15
Corn
0.9
1.2
-$20
Wheat
0.6
0.7
-$10
0
42
>=
42
1
acre/cow
100.00
>=
100.00
0.05
acre/hen
Cows
10
2
$850
Hens
0.05
0
$4
Livestock
Totals
520
84
$44,200
Beginning Value (Current Livestock)
Decrease in Value per Year
End Value (Current Livestock)
$35,000
10%
$31,500
$5,000
25%
$3,750
$35,250
Cost of New Livestock
End Value (New Livestock)
$1,500
$1,350
$3
$2
$18,000
$16,200
30
12
42
<=
42
2,000
0
2,000
<=
5,000
Wage
W&S
$5
S&F
$5.50
Neighbor
Totals
$10,838.80
Hours Worked
782.2
1259.6
2041.800
Plantings
97.8
120.4
142
Livestock
3,120
3,120
84
-$1,840
$44,200
$51,450
$2,000
Plantings
W&S Hours Required
S&F Hours Required
Net Value
Acres Planted
Livestock
Hours Required per Month
Grazing Land Required
Net Annual Cash Income
Current Livestock
New Livestock
Total Livestock
Barn/House Limits
Neighboring Farm Work
Totals
W&S Hours
S&F Hours
Acreage
Net Income
End of Year Value
Leftover Investment Fund
Living Expenses
Total Monetary Worth
E
Planting
Totals
97.8
120.4
-$1,840
F
G
142
<=
Investment
Fund
$20,000
Neighbor
782
1,260
0
Total
4,000
4,500
226
<=
<=
<=
$10,839
$53,199
$51,450
$2,000
-$40,000
$66,649
Available
4,000
4,500
640
!
!
In!a!drought!and!early!frost,!the!model!predicts!(under!the!optimal!solution)!that!
the!family’s!monetary!worth!at!the!end!of!the!year!will!be!$66,649.!
7-80
!
!
Flood!and!Early!Frost!
!
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
B
C
D
Soybeans
1
1.4
$10
Corn
0.9
1.2
$10
Wheat
0.6
0.7
$5
0
37.33333
>=
37.33333
1
acre/cow
250.00
>=
250.00
0.05
acre/hen
Cows
10
2
$850
Hens
0.05
0
$4
Livestock
Totals
623.333333
74.6666667
$52,983
Beginning Value (Current Livestock)
Decrease in Value per Year
End Value (Current Livestock)
$35,000
10%
$31,500
$5,000
25%
$3,750
$35,250
Cost of New Livestock
End Value (New Livestock)
$1,500
$1,350
$3
$2
$20,000
$16,650
Plantings
W&S Hours Required
S&F Hours Required
Net Value
Acres Planted
Livestock
Hours Required per Month
Grazing Land Required
Net Annual Cash Income
Current Livestock
30
New Livestock 7.333333
Total Livestock 37.33333
<=
Barn/House Limits
42
Wage
Hours Worked 76.39999
Totals
$1,623
G
287.333
<=
Investment
Fund
$20,000
Neighbor
76
540
0
Total
4,000
4,500
362
<=
<=
<=
$3,353
$57,960
$51,900
$0
-$40,000
$69,860
S&F
$5.50
Neighbor
Totals
$3,353.10
540.2
616.600
Plantings Livestock
W&S Hours
183.6
3,740
S&F Hours
219.8
3,740
Acreage 287.3333 74.66667
Net Income
End of Year Value
Leftover Investment Fund
Living Expenses
Total Monetary Worth
F
2,000
3,000
5,000
<=
5,000
Neighboring Farm Work
W&S
$5
E
Planting
Totals
183.6
219.8
$1,623
$52,983
$51,900
$0
Available
4,000
4,500
640
!
!
In!a!flood!and!early!frost,!the!model!predicts!(under!the!optimal!solution)!that!
the!family’s!monetary!worth!at!the!end!of!the!year!will!be!$69,860.!
7-81
!
!
f)!
!
Opt.!Sol.!Used!
Good!Weather!
Drought!
Flood!
Early!Frost!
Drought!&!E.F.!
Flood!&!E.F.!
Family’s!monetary!worth!at!year’s!end!if!the!scenario!is!actually:!
Good!
Drought!
Flood!
Early!Frost! Drought&EF!
Flood&EF!
$99,367!
$57,117!
$70,417!
$88,767!
$53,717!
$67,367!
$76,348!
$67,864!
$70,668!
$74,174!
$66,321!
$69,581!
$94,962!
$57,929!
$74,055!
$85,175!
$54,482!
$69,162!
$99,367!
$57,117!
$70,417!
$88,767!
$53,717!
$67,367!
$75,009!
$67,859!
$70,329!
$73,169!
$66,649!
$69,409!
$80,476!
$67,676!
$71,483!
$77,230!
$64,990!
$69,860!
Answers!will!vary.!No!solution!is!clearly!best.!The!Good!Weather!solution!is!the!
riskiest,!with!the!highest!upside!and!downside.!The!Flood!solution!appears!to!be!
a!good!middle!ground.!The!Drought,!Drought&EF,!and!Flood&EF!solutions!are!
the!most!conservative.!
!
g!and!h)!
The!expected!net!value!for!each!of!the!crops!is!calculated!as!follows:!
!
Soybeans:!($70)(0.4)!+!(–$10)(0.2)!+!($15)(0.1)!+!($50)(0.15)!+!(–
$15)(0.1)!+!
!!
!
! !
($10)(0.05)!=!$34,!
!
Corn:!($60)(0.4)!+!(–$15)(0.2)!+!($20)(0.1)!+!($40)(0.15)!+!(–$20)(0.1)!+!
!!
!
! !
($10)(0.05)!=!$27.5,!
!
Wheat:!($40)(0.4)!+!($0)(0.2)!+!($10)(0.1)!+!($30)(0.15)!+!(–$10)(0.1)!+!
!!
!
! !
($5)(0.05)!=!$20.75.!
The!resulting!spreadsheet!solution!is!shown!below:!
7-82
!
!
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
B
C
D
Soybeans
1
1.4
$34.00
Corn
0.9
1.2
$27.50
Wheat
0.6
0.7
$20.75
414
42
>=
42
1
acre/cow
100.00
>=
100.00
0.05
acre/hen
Cows
10
2
$850
Hens
0.05
0
$4
Livestock
Totals
520
84
$44,200
Beginning Value (Current Livestock)
Decrease in Value per Year
End Value (Current Livestock)
$35,000
10%
$31,500
$5,000
25%
$3,750
$35,250
Cost of New Livestock
End Value (New Livestock)
$1,500
$1,350
$3
$2
$18,000
$16,200
30
12
42
<=
42
2,000
0
2,000
<=
5,000
Wage
W&S
$5
S&F
$5.50
Neighbor
Totals
$5,581.00
Hours Worked
368.2
680
1048.200
Plantings
511.8
700
556
Livestock
3,120
3,120
84
$17,306
$44,200
$51,450
$2,000
Plantings
W&S Hours Required
S&F Hours Required
Net Value
Acres Planted
Livestock
Hours Required per Month
Grazing Land Required
Net Annual Cash Income
Current Livestock
New Livestock
Total Livestock
Barn/House Limits
Neighboring Farm Work
Totals
W&S Hours
S&F Hours
Acreage
Net Income
End of Year Value
Leftover Investment Fund
Living Expenses
Total Monetary Worth
E
Planting
Totals
511.8
700
$17,306
F
G
556
<=
Investment
Fund
$20,000
Neighbor
368
680
0
Total
4,000
4,500
640
<=
<=
<=
$5,581
$67,087
$51,450
$2,000
-$40,000
$80,537
Available
4,000
4,500
640
!
!
This!model!predicts!that!the!family’s!monetary!worth!at!the!end!of!the!coming!
year!will!be!(on!average)!$80,537.!
7-83
!
!
Variable Cells
Cell
$B$7
$C$7
$D$7
$B$27
$C$27
$B$36
$C$36
Name
Acres Planted Soybeans
Acres Planted Corn
Acres Planted Wheat
New Livestock Cows
New Livestock Hens
Hours Worked W&S
Hours Worked S&F
Final
Value
414
42
100
12
0
368.2
680
Reduced
Cost
0
0
0.00
0
0
0
0
Objective
Coefficient
34
27.5
20.75
700
3.5
5
5.5
Allowable
Increase
7.5
4.9
0.4
1E+30
0.02
0.389
0.395
Allowable
Decrease
0.4
22.5
1E+30
22.5
1E+30
0.071
0.075
Final
Value
42
100.00
$18,000
42
2,000
4,000
4,500
640
Shadow
Price
-4.9
-7.40
$0
22.5
0
5
5
21.3
Constraint
R.H. Side
0
0
20000
42
5000
4000
4500
640
Allowable
Increase
414
414
1E+30
1.333
1E+30
1E+30
1E+30
368.2
Allowable
Decrease
42
100
2000
12
3000
368.2
680
414
Constraints
Cell
$C$7
$D$7
$D$23
$B$28
$C$28
$E$39
$E$40
$E$41
Name
Acres Planted Corn
Acres Planted Wheat
Cost of New Livestock Totals
Total Livestock Cows
Total Livestock Hens
W&S Hours Total
S&F Hours Total
Acreage Total
!
!
!
i)! The!shadow!price!for!the!investment!constraint!is!zero,!indicating!that!additional!
investment!funds!will!not!increase!their!total!monetary!worth!at!all.!Thus,!it!is!
not!worthwhile!to!obtain!a!bank!loan.!The!shadow!price!would!need!to!be!at!
least!$1.10!before!a!loan!at!10%!interest!would!be!worthwhile.!
!
j)! The!expected!net!value!for!soybeans!can!increase!up!to!$7.50!or!decrease!up!to!
$0.40;!for!corn!can!increase!up!to!$4.90!or!decrease!up!to!$22.50;!for!wheat!can!
increase!up!to!$0.40!or!decrease!any!amount!without!changing!the!optimal!
solution.!The!expected!net!value!for!soybeans!and!wheat!should!be!estimated!
most!carefully.!
!
The!solution!is!sensitive!to!decreases!in!the!expected!value!of!soybeans!and!
increases!in!the!expected!value!of!wheat.!If!the!cumulative!decrease!in!the!
expected!value!of!soybeans!and!increase!in!the!expected!value!of!wheat!exceeds!
$0.40,!then!the!100%!rule!will!be!violated,!and!the!solution!might!change.!
!
!
!
k)! Answers!will!vary.!
7-84
Case%7.3%
!
a)! !
A
1 Data:
2
3
Area
4
1
5
2
6
3
7
4
8
5
9
6
10
11
12 Solution:
13
14
Area 1
15
Area 2
16
Area 3
17
Area 4
18
Area 5
19
Area 6
20
Total In School
21
22
Capacity
23
24
25 Grade Constraints:
26
27
28
6th Graders
29
7th Graders
30
8th Graders
31
32
B
Percentage
in 6th
Grade
32%
37%
30%
28%
39%
34%
C
Percentage
in 7th
Grade
38%
28%
32%
40%
34%
28%
D
Percentage
in 8th
Grade
30%
35%
38%
32%
27%
38%
E
330
<=
368.56
362.11
369.33
<=
396
G
Bussing Cost ($/Student)
School 1
School 2
School 3
$300
$0
$700
$400
$500
$600
$300
$200
$200
$500
$0
$400
$500
$300
$0
Number of Students Assigned
School 1
School 2
School 3
0
450
0
0
422.22
177.78
0
227.78
322.22
350
0
0
366.67
0
133.33
83.33
0
366.67
800
1,100
1,000
<=
<=
<=
900
1,100
1,000
240
<=
269.33
288.00
242.67
<=
288
F
Total
From Area
450
600
550
350
500
450
Number of
Students
450
600
550
350
500
450
=
=
=
=
=
=
Total
Bussing
Cost
$555,556
300
<=
339.11
300.89
360.00
<=
360
30%
of total in school
36%
of total in school
!
!
Range Name
BussingCost
Capacity
NumberOfStudents
PercentageInGrade
Solution
TotalBussingCost
TotalFromArea
TotalInSchool
!
20
Cells
E4:G9
B22:D22
G14:G19
B4:D9
B14:D19
G24
E14:E19
B20:D20
!
12
13
14
15
16
17
18
19
E
Total
From Area
=SUM(B14:D14)
=SUM(B15:D15)
=SUM(B16:D16)
=SUM(B17:D17)
=SUM(B18:D18)
=SUM(B19:D19)
!!
G
21
Total
22
Bussing
23
Cost
24 =SUMPRODUCT(BussingCost,Solution)
A
B
C
D
Total In School =SUM(B14:B19) =SUM(C14:C19) =SUM(D14:D19)
!
!
A
B
C
D
E
25 Grade Constraints:
26
=$E$26*TotalInSchool
=$E$26*TotalInSchool
=$E$26*TotalInSchool
0.3
27
<=
<=
<=
28
6th Graders =SUMPRODUCT(B14:B19,B4:B9) =SUMPRODUCT(C14:C19,B4:B9) =SUMPRODUCT(D14:D19,B4:B9)
29
7th Graders =SUMPRODUCT(B14:B19,C4:C9) =SUMPRODUCT(C14:C19,C4:C9) =SUMPRODUCT(C4:C9,D14:D19)
30
8th Graders =SUMPRODUCT(B14:B19,D4:D9) =SUMPRODUCT(C14:C19,D4:D9) =SUMPRODUCT(D14:D19,D4:D9)
31
<=
<=
<=
32
=$E$32*TotalInSchool
=$E$32*TotalInSchool
=$E$32*TotalInSchool
0.36
7-85
!
F
of total in school
of total in school
!
!
b)!
Variable Cells
Cell
$B$14
$C$14
$D$14
$B$15
$C$15
$D$15
$B$16
$C$16
$D$16
$B$17
$C$17
$D$17
$B$18
$C$18
$D$18
$B$19
$C$19
$D$19
Name
Area 1 School 1
Area 1 School 2
Area 1 School 3
Area 2 School 1
Area 2 School 2
Area 2 School 3
Area 3 School 1
Area 3 School 2
Area 3 School 3
Area 4 School 1
Area 4 School 2
Area 4 School 3
Area 5 School 1
Area 5 School 2
Area 5 School 3
Area 6 School 1
Area 6 School 2
Area 6 School 3
Final
Value
0
450
0
0
422.22
177.78
0
227.78
322.22
350
0
0
366.67
0
133.33
83.33
0
366.67
Reduced
Cost
177.778
0
266.667
-800.000
0
0
11.111
0
0
0
366.667
-433.333
0
233.333
0
0
200
0
Objective
Coefficient
300
0
700
0
400
500
600
300
200
200
500
0
0
0
400
500
300
0
Allowable
Increase
1E+30
177.778
1E+30
1E+30
34.211
4.545
1E+30
4.545
34.211
366.667
1E+30
1E+30
16.667
1E+30
108.333
33.333
1E+30
166.667
Allowable
Decrease
177.778
1E+30
266.667
800.000
4.545
34.211
11.111
34.211
7.692
2.08E+17
366.667
433.333
108.333
233.333
16.667
166.667
200
33.333
Name
8th Graders <=
8th Graders <=
8th Graders <=
Total In School School 1
Total In School School 2
Total In School School 3
6th Graders <=
6th Graders <=
6th Graders <=
6th Graders <=
6th Graders <=
6th Graders <=
7th Graders <=
7th Graders <=
7th Graders <=
7th Graders <=
7th Graders <=
7th Graders <=
8th Graders <=
8th Graders <=
8th Graders <=
Area 1 From Area
Area 2 From Area
Area 3 From Area
Area 4 From Area
Area 5 From Area
Area 6 From Area
Final
Value
242.67
369.33
360.00
800
1,100
1,000
269.33
368.56
339.11
269.33
368.56
339.11
288.00
362.11
300.89
288.00
362.11
300.89
242.67
369.33
360.00
450
600
550
350
500
450
Shadow
Price
0.00
0.00
-6666.67
0
-178
-144
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
-2777.78
0.00
0.00
0.00
0.00
0.00
177.778
577.778
477.778
311.111
-55.556
277.778
Constraint
R.H. Side
0
0
0
900
1100
1000
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
450
600
550
350
500
450
Allowable
Increase
1E+30
1E+30
5.333
1E+30
36.364
42.105
29.333
38.556
39.111
1E+30
1E+30
1E+30
48
32.111
0.889
0.258
1E+30
1E+30
2.667
39.333
60
3.774
3.774
3.774
72.727
12.903
3.226
Allowable
Decrease
45.333
26.667
0.667
100
3.774
3.883
1E+30
1E+30
1E+30
18.667
27.444
20.889
1E+30
1E+30
1E+30
2.909
33.889
59.111
1E+30
1E+30
1E+30
36.364
36.364
36.364
6.452
145.455
36.364
Constraints
Cell
$B$30
$C$30
$D$30
$B$20
$C$20
$D$20
$B$28
$C$28
$D$28
$B$28
$C$28
$D$28
$B$29
$C$29
$D$29
$B$29
$C$29
$D$29
$B$30
$C$30
$D$30
$E$14
$E$15
$E$16
$E$17
$E$18
$E$19
7-86
!
!
c)! The!bussing!cost!from!area!6!to!school!1!can!increase!$33.33!before!the!current!
optimal!solution!would!no!longer!be!optimal.!The!new!solution!with!a!10%!
increase!($50)!is!shown!below.!
!
A
1 Data:
2
3
Area
4
1
5
2
6
3
7
4
8
5
9
6
10
11
12 Solution:
13
14
Area 1
15
Area 2
16
Area 3
17
Area 4
18
Area 5
19
Area 6
20
Total In School
21
22
Capacity
23
24
25 Grade Constraints:
26
27
28
6th Graders
29
7th Graders
30
8th Graders
31
32
B
Percentage
in 6th
Grade
32%
37%
30%
28%
39%
34%
C
Percentage
in 7th
Grade
38%
28%
32%
40%
34%
28%
D
Percentage
in 8th
Grade
30%
35%
38%
32%
27%
38%
Number of Students Assigned
School 1
School 2
School 3
0
450
0
0
600
0
72.73
50
427.27
350
0
0
318.18
0
181.82
59.09
0
390.91
800
1,100
1,000
<=
<=
<=
900
1,100
1,000
240
<=
264.00
288.00
248.00
<=
288
330
<=
381.00
355.00
364.00
<=
396
300
<=
332.00
308.00
360.00
<=
360
E
F
G
Bussing Cost ($/Student)
School 1
School 2
School 3
$300
$0
$700
$400
$500
$600
$300
$200
$200
$500
$0
$400
$550
$300
$0
Total
From Area
450
600
550
350
500
450
=
=
=
=
=
=
Number of
Students
450
600
550
350
500
450
Total
Bussing
Cost
$559,318
30%
of total in school
36%
of total in school
!
d)! The!bussing!cost!from!area!6!to!school!2!can!increase!any!amount!and!the!
optimal!solution!from!part!(a)!will!still!be!optimal.!
!
e)! If!the!bussing!costs!increase!1%!from!area!6!to!all!the!schools,!then:!
!
Percentage!of!allowable!increase!for!school!1!used!=!($505!–!$500)!/!$33.33!=!
15%.!
Percentage!of!allowable!increase!for!school!2!used!=!($303!–!$300)!/!∞!=!0%.!
Percentage!of!allowable!increase!for!school!3!used!=!($0!–!$0)!/!$166.67!=!0%.!
Sum!=!15%.!Therefore,!the!bussing!costs!from!area!6!can!increase!uniformly!by!
(100%/15%)(1%)!=!6.67%!before!100%!will!be!reached.!Beyond!that,!the!
solution!might!change.!
7-87
!
!
!
If!the!bussing!costs!increase!10%!from!area!6!to!all!schools,!the!new!solution!is:!
!
A
1 Data:
2
3
Area
4
1
5
2
6
3
7
4
8
5
9
6
10
11
12 Solution:
13
14
Area 1
15
Area 2
16
Area 3
17
Area 4
18
Area 5
19
Area 6
20
Total In School
21
22
Capacity
23
24
25 Grade Constraints:
26
27
28
6th Graders
29
7th Graders
30
8th Graders
31
32
B
Percentage
in 6th
Grade
32%
37%
30%
28%
39%
34%
C
Percentage
in 7th
Grade
38%
28%
32%
40%
34%
28%
D
Percentage
in 8th
Grade
30%
35%
38%
32%
27%
38%
Number of Students Assigned
School 1
School 2
School 3
0
450
0
0
600
0
72.73
50
427.27
350
0
0
318.18
0
181.82
59.09
0
390.91
800
1,100
1,000
<=
<=
<=
900
1,100
1,000
240
<=
264.00
288.00
248.00
<=
288
330
<=
381.00
355.00
364.00
<=
396
300
<=
332.00
308.00
360.00
<=
360
E
F
G
Bussing Cost ($/Student)
School 1
School 2
School 3
$300
$0
$700
$400
$500
$600
$300
$200
$200
$500
$0
$400
$550
$330
$0
Total
From Area
450
600
550
350
500
450
=
=
=
=
=
=
Number of
Students
450
600
550
350
500
450
Total
Bussing
Cost
$559,318
30%
of total in school
36%
of total in school
!
f)! The!shadow!price!for!school!1!is!zero.!Thus,!adding!a!temporary!classroom!at!
school!1!would!not!save!any!money,!and!thus!would!not!be!worthwhile.!
!
The!shadow!price!for!school!2!is!–$177.78.!Thus,!adding!a!temporary!classroom!
at!school!2!would!save!($177.78)(20)!=!$3,555.60!in!bussing!cost.!This!is!
worthwhile,!since!it!exceeds!the!$2500!leasing!cost.!
!
The!shadow!price!for!school!3!is!–$144.44.!Thus,!adding!a!temporary!classroom!
at!school!3!would!save!($144.44)(20)!=!$2,888.80!in!bussing!cost.!This!is!also!
worthwhile,!since!it!exceeds!the!$2500!leasing!cost.!
!
g)! For!school!2,!the!allowable!increase!for!school!capacity!is!36.!This!means!the!
shadow!price!is!only!valid!for!a!single!additional!portable!classroom.!
!
For!school!3,!the!allowable!increase!for!school!capacity!is!42.!This!means!the!
shadow!price!is!valid!for!up!to!two!additional!portable!classrooms.!
7-88
!
!
h)! The!following!combinations!do!not!violate!the!100%!rule:!
Portables!to!
add!
to!school!2!
1!
Portables!to!
add!
to!school!3!
0!
0!
1!
0!
2!
!
!
!
100%,rule!calculation!
(20/36)!+!(0/42)!=!
55.6%!
(0/36)!+!(20/42)!=!
47.6%!
(0/36)!+!(40/42)!=!
95.23%!
Each!combination!yields!the!following!total!savings!
Portables! Portables!
to!add!
to!add!
to!school!2! to!school!3!
1!
0!
0!
1!
0!
2!
!
!
Bussing!Cost!Savings!
($177.78)(20)!=!
$3555.60!
($144.44)(20)!=!
$2888.80!
($144.44)(40)!=!
$5777.60!
7-89
!
Lease!
Cost!
$2500!
!
Total!
Savings!
$1055.60!
$2500!
$388.80!
$5000!
$777.60!
!
Of!these!combinations,!adding!one!portable!to!school!2!is!best!in!terms!of!
minimizing!total!cost.!The!spreadsheet!solution!is!shown!below.!
!
A
1 Data:
2
3
Area
4
1
5
2
6
3
7
4
8
5
9
6
10
11
12 Solution:
13
14
Area 1
15
Area 2
16
Area 3
17
Area 4
18
Area 5
19
Area 6
20
Total In School
21
22
Capacity
23
24
25 Grade Constraints:
26
27
28
6th Graders
29
7th Graders
30
8th Graders
31
32
B
Percentage
in 6th
Grade
32%
37%
30%
28%
39%
34%
C
Percentage
in 7th
Grade
38%
28%
32%
40%
34%
28%
D
Percentage
in 8th
Grade
30%
35%
38%
32%
27%
38%
Number of Students Assigned
School 1
School 2
School 3
0
450
0
0
520
80
0
150
400
350
0
0
340
0
160
90
0
360
780
1,120
1,000
<=
<=
<=
900
1,120
1,000
234
<=
261.20
280.80
238.00
<=
280.8
336
<=
381.40
364.60
374.00
<=
403.2
7-90
300
<=
334.40
305.60
360.00
<=
360
E
F
G
Bussing Cost ($/Student)
School 1
School 2
School 3
$300
$0
$700
$400
$500
$600
$300
$200
$200
$500
$0
$400
$500
$300
$0
Total
From Area
450
600
550
350
500
450
=
=
=
=
=
=
Number of
Students
450
600
550
350
500
450
Bussing Cost
Leasing Cost
Total Cost
$552,000
$2,500
$554,500
30%
of total in school
36%
of total in school
!
!
i)! Adding!two!portables!to!school!2!yields!the!following!solution.!This!is!the!best!
plan.!
!
A
1 Data:
2
3
Area
4
1
5
2
6
3
7
4
8
5
9
6
10
11
12 Solution:
13
14
Area 1
15
Area 2
16
Area 3
17
Area 4
18
Area 5
19
Area 6
20
Total In School
21
22
Capacity
23
24
25 Grade Constraints:
26
27
28
6th Graders
29
7th Graders
30
8th Graders
31
32
B
Percentage
in 6th
Grade
32%
37%
30%
28%
39%
34%
C
Percentage
in 7th
Grade
38%
28%
32%
40%
34%
28%
D
Percentage
in 8th
Grade
30%
35%
38%
32%
27%
38%
Number of Students Assigned
School 1
School 2
School 3
0
450
0
0
600
0
0
90
460
350
0
0
318.95
0
181.05
95.26
0
354.74
764
1,140
996
<=
<=
<=
900
1,140
1,000
229.2631579
<=
254.78
275.12
234.32
<=
275.1157895
342
<=
393.00
367.80
379.20
<=
410.4
298.7368421
<=
329.22
308.08
358.48
<=
358.4842105
E
Total
From Area
450
600
550
350
500
450
=
=
=
=
=
=
Number of
Students
450
600
550
350
500
450
Bussing Cost
Leasing Cost
Total Cost
$549,053
$5,000
$554,053
30%
of total in school
36%
of total in school
Case%7.4%
a)! In!this!case,!the!decisions!to!be!made!are!
!
TV!=!number!of!units!of!advertising!on!television!
!
PM!=!number!of!units!of!advertising!in!the!printed!media!
!
The!resulting!linear!programming!model!is!
Minimize!Cost!=!1!TV!+!2PM!(in!millions!of!dollars)!
subject!to!
!
Stain!Remover:!!!!
1!PM!!
!
≥!3!(in!%)!
!
Liquid!Detergent:!!! 3!TV!!!!+!2PM! ≥!18!(in!%)!
!
Powder!Detergent:!! –1TV!!+!4!PM!! ≥!4!!(in!%)!
and!TV!≥!0,!PM!≥!0.!
7-91
G
Bussing Cost ($/Student)
School 1
School 2
School 3
$300
$0
$700
$400
$500
$600
$300
$200
$200
$500
$0
$400
$500
$300
$0
%
!
F
!
!
a)! Optimal!Solution:!4!units!of!television!advertising!and!3!units!of!print!media!
advertising,!with!a!total!cost!of!$10!million.!
!
!
%
!
b)! The!spreadsheet!model!is!shown!below.!Solver!finds!to!take!4!units!of!television!
advertising!and!3!units!of!print!media!advertising,!with!a!total!cost!of!$10!
million.!
B
3
4
5
6
7
8
9
10
11
12
13
14
Unit Cost ($millions)
Stain Remover
Liquid Detergent
Powder Detergent
Advertising Units
C
Television
1
D
Print Media
2
Increase in Sales per Unit of Advertising
0%
1%
3%
2%
-1%
4%
Television
4
7-92
Print Media
3
E
Increased
Sales
3%
18%
8%
F
G
>=
>=
>=
Minimum
Increase
3%
18%
4%
Total Cost
($millions)
10
!
!
!
c)! Increasing!the!required!minimum!increase!in!sales!for!Stain!Remover!by!1%!
changes!the!solution!to!3.33!units!of!television!advertising!and!4!units!of!print!
media!advertising,!and!increases!the!total!cost!by!$1.33!million!to!$11.33!million.!!
!
!
!
Increasing!the!required!minimum!increase!in!sales!for!Liquid!Detergent!by!1%!
changes!the!solution!to!4.33!units!of!television!advertising!and!3!units!of!print!
media!advertising,,!and!increases!the!total!cost!by!$0.33!million!to!$10.33!
million.!
!
!
7-93
!
!
Increasing!the!required!minimmum!increase!in!sales!for!Powder!Detergent!by!
1%!has!no!impact!on!the!solution!nor!the!total!cost.!
!
!
!
!
d)! Original!Solution:!
B
3
4
5
6
7
8
9
10
11
12
13
14
!
!
Unit Cost ($millions)
Stain Remover
Liquid Detergent
Powder Detergent
Advertising Units
C
Television
1
D
Print Media
2
Increase in Sales per Unit of Advertising
0%
1%
3%
2%
-1%
4%
Television
4
E
Increased
Sales
3%
18%
8%
F
G
>=
>=
>=
Minimum
Increase
3%
18%
4%
Total Cost
($millions)
10
Print Media
3
!
Increasing!the!required!minimum!increase!in!sales!for!Stain!Remover!by!1%!
increases!the!total!cost!by!$1.333!million.!!
B
3
4
5
6
7
8
9
10
11
12
13
14
Unit Cost ($millions)
Stain Remover
Liquid Detergent
Powder Detergent
Advertising Units
C
Television
1
D
Print Media
2
Increase in Sales per Unit of Advertising
0%
1%
3%
2%
-1%
4%
Television
3.333
7-94
Print Media
4
E
Increased
Sales
4%
18%
13%
F
G
>=
>=
>=
Minimum
Increase
4%
18%
4%
Total Cost
($millions)
11.333
!
!
!
Increasing!the!required!minimum!increase!in!sales!for!Liquid!Detergent!by!1%!
increases!the!total!cost!by!$0.333!million.!
B
3
4
5
6
7
8
9
10
11
12
13
14
Unit Cost ($millions)
Stain Remover
Liquid Detergent
Powder Detergent
Advertising Units
C
Television
1
D
Print Media
2
Increase in Sales per Unit of Advertising
0%
1%
3%
2%
-1%
4%
Television
4.333
E
Increased
Sales
3%
19%
8%
F
G
>=
>=
>=
Minimum
Increase
3%
19%
4%
Total Cost
($millions)
10.333
Print Media
3
!!
!
!
Increasing!the!required!minimmum!increase!in!sales!for!Powder!Detergent!by!
1%!has!no!impact!on!the!total!cost.!
B
3
4
5
6
7
8
9
10
11
12
13
14
!
Unit Cost ($millions)
Stain Remover
Liquid Detergent
Powder Detergent
Advertising Units
C
Television
1
D
Print Media
2
Increase in Sales per Unit of Advertising
0%
1%
3%
2%
-1%
4%
Television
4
e)!
!
!
7-95
Print Media
3
E
Increased
Sales
3%
18%
8%
F
G
>=
>=
>=
Minimum
Increase
3%
18%
5%
Total Cost
($millions)
10
!
!
!
!
!
!
!
!
!
7-96
!
!
!
!
f)! Sensitivity!Report:!
Variable Cells
Cell
$C$14
$D$14
Name
Advertising Units Television
Advertising Units Print Media
Final
Value
4
3
Reduced
Cost
0
0
Objective
Coefficient
1
2
Allowable
Increase
2
1E+30
Allowable
Decrease
1
1.333
Name
Stain Remover Sales
Liquid Detergent Sales
Powder Detergent Sales
Final
Value
3%
18%
8%
Shadow
Price
133.33
33.33
0
Constraint
R.H. Side
0.03
0.18
0.04
Allowable
Increase
0.06
0.12
0.04
Allowable
Decrease
0.008571429
0.12
1E+30
Constraints
Cell
$E$8
$E$9
$E$10
!
The!shadow!price!indicates!the!increase!in!total!cost!(in!$millions)!per!unit!
increase!in!the!right!hand!side!(i.e.,!per!100%!increase).!Thus,!a!1%!increase!in!
the!minimum!required!increase!in!sales!will!only!increase!the!total!cost!by!one!
hundredth!of!the!shadow!price,!or!$1.33!million!for!the!Stain!Remover,!$0.33!
million!for!the!Liquid!Detergent,!and!$0!million!for!the!Powder!Detergent.!
!
The!allowable!range!for!the!required!minimum!increase!in!sales!constraint!
for!Stain!Remover!is!2.15%!to!9%.!
!
The!allowable!range!for!the!required!minimum!increase!in!sales!constraint!
for!Liquid!Detergent!is!6%!to!30%.!
!
The!allowable!range!for!the!required!minimum!increase!in!sales!constraint!
for!Powder!Detergent!is!,∞%!to!8%.!
!
!
These!allowable!ranges!can!also!be!seen!in!the!results!from!part!(c).!For!Stain!
Remover,!the!incremental!cost!remains!$1.33!million!for!each!1%!change!above!
3%.!Similarly,!for!Liquid!Detergent,!the!incremental!cost!remains!$0.33!million!
for!each!1%!change!above!between!6%!and!30%.!For!Powder!Detergent,!the!
incremental!cost!remains!$0!million!for!each!1%!change!throughout!the!
parameter!analysis!report.!
!
g)! Suppose!that!each!of!the!original!numbers!in!MinimumIncrease!(G8:G10)!is!
increased!by!1%.!
!
Percent!of!allowable!increase!for!Stain!Remover!used!=!(4%!–!3%)!/!6%!=!
16.7%.!
Percent!of!allowable!increaes!for!Liquid!Detergent!used!=!(19%!–!18%)!/!12%!=!
8.3%.!
Percent!of!allowable!increase!for!Powder!Detergent!used!=!(5%!–!4%)!/!4%!=!
25%.!
Sum!=!50%.!
!
Thus,!if!each!of!the!original!numbers!in!MinimumIncrease!(G8:G10)!is!increased!
by!2%,!the!sum!will!be!100%.!By!the!100%!rule,!this!is!the!most!they!can!be!
increased!before!the!shadow!prices!may!no!longer!be!valid.!
!
h)! Answers!will!vary.!
7-97
CHAPTER 8: OTHER ALGORITHMS FOR LINEAR PROGRAMMING
8.1-1.
(a), (c)
(b) Optimal Solution: ÐB" ß B# Ñ œ Ð"ß $Ñ, ^ œ %
Iteration
!
"
#
BV
^
B$
B%
B&
^
B$
B#
B&
^
B$
B#
B"
Eq. #
!
"
#
$
!
"
#
$
!
"
#
$
^
"
!
!
!
"
!
!
!
"
!
!
!
B"
"
1
!
1
!
1
!
1
!
!
!
"
)-1
B#
1
"
"
"
!
!
"
!
!
!
"
!
B$
!
"
!
!
!
"
!
!
!
"
!
!
B%
!
!
"
!
1
"
"
"
#
#
"
1
B&
!
!
!
"
!
!
!
"
1
"
!
1
RS
!
8
3
2
$
&
3
1
4
4
3
1
8.1-2.
Iteration
!
"
#
BV
^
B%
B&
^
B%
B#
^
B"
B#
Eq. #
!
"
#
!
"
#
!
"
#
^
"
!
!
"
!
!
"
!
!
B"
&
$
'
"
"
#
!
"
!
B#
#
"
$
!
!
"
!
!
"
B$
%
#
&
#
$
 "$
&
$
"
$
"
$
"
B%
!
"
!
!
"
!
"
"
#
B&
!
!
"
B&
!
"
!
!
!
"
!
!
RS
!
%
"!
 #!
$
 #$
#
$
 "$
 "$
"
$
"
$
"!
$
 ##
$
#
$
"
#
B'
!
!
"
!
B(
!
!
!
"
!
!
!
"
Optimal Solution: ÐB" ß B# ß B$ Ñ œ Ð#Î$ß #ß !Ñ, ^ œ ##Î$
8.1-3.
Iteration
!
"
BV
^
B&
B'
B(
^
B&
B#
B(
Eq. #
!
"
#
$
!
"
#
$
^
"
!
!
!
"
!
!
!
B"
(
#
)
$
$
'
#
"$
B#
#
%
%
)
!
!
"
!
B$
&
(
'
"
#
"
$
#
""
Optimal Solution: ÐB" ß B# ß B$ ß B% Ñ œ Ð!ß #ß !ß !Ñ, ^ œ %
)-2
B%
%
"
%
%
#
$
"
%
"
#
"
 "%
#
RS
!
&
)
%
%
$
#
"#
8.1-4.
(a) Optimal Solution: ÐB" ß B# Ñ œ Ð$ß $Ñ, ^ œ "&
Iter.
!
"
#
BV
^
B$
B%
B&
^
B"
Eq. #
!
"
#
$
!
"
^
"
!
!
!
"
!
B"
3
$*
"
&
!
"
B%
B&
^
B"
B#
B&
#
$
!
"
#
$
!
!
"
!
!
!
!
!
!
"
!
!
B#
2
"
"
$
1
"
$
#‡
$
%
$
!
!
"
!
B$
!
"
!
!
1
"
$
 "$
 &$
"
#
"
#
 "#
B%
!
!
"
!
!
!
B&
!
!
!
"
!
!
RS
!
12
6
27
12
%
"
!
!
"
!
!
!
"
#
(
"&
$
$
$
$
#
 "#
$
#
"
#
Primal Solution
Ð!ß !ß 12Þ6ß 27Ñ
Dual Solution
Ð!ß !ß !ß 3ß 2Ñ
Ð4ß !ß !ß #ß (Ñ
Ð"ß !ß !ß !ß "Ñ
Ð$ß $ß !ß !ß $Ñ
Ð "# ß $# ß !ß !ß !Ñ
Primal Solution
Ð!ß !ß "#ß 'ß #(Ñ
Dual Solution
Ð!ß !ß !ß $ß #Ñ
Ð%ß !ß !ß #ß (Ñ
Ð"ß !ß !ß !ß "Ñ
Ð$ß $ß !ß !ß $Ñ
Ð "# ß $# ß !ß !ß !Ñ
(b) The dual problem is:
minimize
"#C"  'C#  #(C$
$C"  C#  &C$
C"  C#  $C$
C" ß C# ß C$
subject to
Iter.
!
"
#
BV
^
C%
C&
^
C"
C&
^
C"
C#
Eq. #
!
"
#
!
"
#
!
"
#
^
"
!
!
"
!
!
"
!
!
C"
"#
$ ‡
"
!
"
!
!
"
!
C#
'
"
"
#
C$
#(
&
$
(
"
$
 #$
#
$
 %$
!
!
"
$
"
#
$
#
!.
C%
!
"
!
%
 "$
 "$
$
 "#
#
C&
!
!
"
!
!
"
$
RS
!
$
#
"#
"
"
"&
"
#
 $#
"
#
$
#
Optimal Solution: ÐC" ß C# ß C$ Ñ œ Ð "# ß $# ß !Ñ, ^ œ "&Þ
The sequence of basic and complementary basic solutions is identical to that in part (a).
)-3
8.1-5.
Iteration
!
"
BV
^
B$
B#
B"
^
B$
B#
B%
^
"
!
!
!
"
!
!
!
Eq. #
!
"
#
$
!
"
#
$
B"
!
!
!
"
$
#
"
$
#
$
B#
!
!
"
!
!
!
"
!
B$
!
"
!
!
!
"
!
!
B%
$
#
"
$
"
#
 "$
!
!
!
"
B&
"
 "$
!
"
$
&
#
!
"
#
"
RS
&%
'
"#
#
%&
%
*
'
Optimal Solution: ÐB" ß B# ß B$ ß B% ß B& Ñ œ Ð!ß *ß %ß 'ß !Ñ, ^ œ %&
8.1-6.
Iteration
!
"
#
BV
^
B#
B&
^
B#
B$
^
B%
B$
Eq. #
!
"
#
!
"
#
!
"
#
^
"
!
!
"
!
!
"
!
!
B"
!
"
"'
"'
#$
)
B#
!
"
!
!
"
!
"!$
&
#$
&
'
&
"
&
 "&
#
&
B$
#
$
#
!
!
"
!
!
"
B%
&
"
%
"
&
#
!
"
!
Optimal Solution: ÐB" ß B# ß B$ ß B% ß B& Ñ œ Ð!ß !ß *ß $ß !Ñ, ^ œ ""(
)-4
B&
!
!
"
"
$
#
 "#
"$
"!
$
 "!
"
"!
RS
"&!
$!
$!
"#!
"&
"&
""(
$
*
8.2-1.
(a)
The solution Ð!ß &Ñ is optimal with ^ œ "#!. It remains optimal as long as
))
"
 #%#
) Ÿ  # Í ) Ÿ #,
at which point Ð"!Î$ß "!Î$Ñ becomes optimal. In turn, this solution remains optimal until
))
 #%#
) Ÿ # Í ) Ÿ ),
at which point Ð&ß !Ñ becomes optimal.
)
!Ÿ)Ÿ#
#Ÿ)Ÿ)
) Ÿ ) Ÿ "!
ÐB‡" ß B‡# Ñ
Ð!ß &Ñ
Ð"!Î$ß "!Î$Ñ
Ð&ß !Ñ
)-5
^ ‡ Ð) Ñ
"#!  "!)
Ð$#!  "!)ÑÎ$
%!  &)
(b)
Iteration
!
"
#
$
BV
^
B$
B%
^
B#
B%
^
B#
B"
^
B$
B"
Eq. #
!
"
#
!
"
#
!
"
#
!
"
#
^
"
!
!
"
!
!
"
!
!
"
!
!
B"
)  )
"
#
%  #)
"
#
$
#
!
!
"
!
!
"
B#
#%  #)
#
"
!
"
!
!
"
!
%!&)
#
$
#
"
#
B$
!
"
!
"#  )
"
#
 "#
%!&)
$
#
$
 "$
!
"
!
B%
!
!
"
!
!
"
)%)
$
 "$
#
$
))
#
 "#
"
#
RS
!
"!
"!
"#!  "!)
&
&
$#!"!)
$
"!
$
"!
$
%!  &)
&
&
The solutions found in iterations ", # and $ are optimal for ! Ÿ ) Ÿ #, # Ÿ ) Ÿ ) and
) Ÿ ) Ÿ "! respectively.
(c) The graph in part (b) suggests that ) œ ! is optimal. Since ^Ð)Ñ is convex in ), the
maximum is attained at ) œ ! or ) œ "!. Thus, only the linear programming problems
corresponding to ) œ ! and ) œ "! need to be solved.
)-6
8.2-2.
Iteration
!
"
#
$
BV
^
B%
B&
B'
^
B#
B&
B'
^
Eq. #
!
"
#
$
!
"
#
$
!
^
"
!
!
!
"
!
!
!
"
B"
#!  %)
$
)
'
"!  ()
"
#
&‡
!
B#
$!  $)
$‡
'
"
!
"
!
!
!
B#
B&
B"
^
B%
B&
B"
"
#
$
!
"
#
$
!
!
!
"
!
!
!
!
!
"
!
!
!
"
"
!
!
)!"")
$
&
#
"%
$
"
'
B$
&
"
%
"
&)
B%
!
"
!
!
"!  )
"
$
"
$
#
#
 "$
#
$
"!)
"&
"
&
#'
"&
#
"&
"%#)
$
"
#
)
$
"
'
ÐB‡" ß B‡# ß B‡$ Ñ
Ð!ß "!ß !Ñ
Ð(ß $ß !Ñ
Ð "&# ß !ß !Ñ
)
! Ÿ ) Ÿ "!
(
"!
)!
( Ÿ ) Ÿ ""
)!
"" Ÿ )
"'!##)
"&
#‡
&
 #)
"&
"
 "&
!
"
!
!
B&
!
!
"
!
!
!
"
!
!
!
"
!
!
!
"
!
B'
!
!
!
"
!
!
!
"
"!()
&
 "&
 #&
"
&
"!#)
$
 "#
 %$
"
'
RS
!
$!
(&
%&
$!!  $!)
"!
"&
$&
#$!  "*)
$
"
(
"&!  $!)
"&
#
"&
"&
#
^ ‡ Ð) Ñ
$!!  $!)
#$!  "*)
"&!  $!)
8.2-3.
(a) Starting with the optimal tableau for ) œ !, after two iterations, we get:
Iter.
!
"
#
BV
^
B#
B"
^
B#
B&
^
B$
B&
Eq. #
!
"
#
!
"
#
!
"
#
^
"
!
!
"
!
!
"
!
!
B"
!
!
"
)$)
#
"
#
"
#
"$%)
#
"
#
"
#
B#
!
"
!
!
"
!
!
&  )
!
)
! Ÿ ) Ÿ )Î$
)Î$ Ÿ ) Ÿ &
& Ÿ)
B$
&)
"
!
&)
"
!
!
"
!
ÐB‡" ß B‡# ß B‡$ Ñ
Ð"!ß "!ß !Ñ
Ð!ß "&ß !Ñ
Ð!ß !ß "&Ñ
)-7
B%
#  #)
"
"
"#)
#
"
#
 "#
(#)
#
"
#
 "#
B&
)  $)
"
#
!
!
"
!
!
"
^ ‡ Ð) Ñ
##!
")!  "&)
"!&  $!)
RS
##!
"!
"!
")!  "&)
"&
&
"!&  $!)
"&
&
(b) The dual problem is:
minimize
$!C"  #!C#
subject to
C"  C#
#C"  C#
#C"  C#
C" ß C#
"!  )
"#  )
(  #)
!.
Starting with the optimal tableau for ) œ !, after two iterations, we get:
Iter.
!
"
#
BV
^
C#
C"
C&
^
C$
C"
C&
^
C$
C"
C%
Eq. #
!
"
#
$
!
"
#
$
!
"
#
$
^
"
!
!
!
"
!
!
!
"
!
!
!
)
! Ÿ ) Ÿ )Î$
)Î$ Ÿ ) Ÿ &
& Ÿ)
C"
!
!
"
!
!
!
"
!
!
!
"
!
C#
!
"
!
!
&
 "#
"
#
!
&
 "#
"
#
!
C$
"!
#
"
!
!
"
!
!
!
"
!
!
C%
"!
"
"
"
"&
 "#
 "#
"
!
!
!
"
ÐC"‡ ß C#‡ Ñ
Ð#  #)ß )  $)Ñ
Ð'  !Þ&)ß !Ñ
Ð$Þ&  )ß !Ñ
The basic solutions are the same as those in part (a).
)-8
C&
!
!
!
"
!
!
!
"
"&
 "#
 "#
"
RS
##!
)  $)
#  #)
&)
")!  "&)
%  "Þ&)
'  !Þ&)
&)
"!&  $!)
'Þ&  #)
$Þ&  )
&  )
^ ‡ Ð)Ñ
##!
")!  "&)
"!&  $!)
! Ÿ ) Ÿ )Î$ : C‡ from Ð#ß )Ñ to Ð##Î$ß !Ñ
)Î$ Ÿ ) Ÿ & : C‡ from Ð##Î$ß !Ñ to Ð"(Î#ß !Ñ
&Ÿ)
: C‡ œ Ð$Þ&  )ß !Ñ
)-9
8.2-4.
)
!Ÿ)Ÿ"
"Ÿ)Ÿ&
& Ÿ ) Ÿ #&
ÐB‡" ß B‡# Ñ
Ð"!  #)ß "!  #)Ñ
Ð"!  #)ß "&  $)Ñ
Ð#&  )ß !Ñ
^ ‡ Ð) Ñ
$!  ')
$&  )
&!  #)
)-10
8.2-5.
Starting with the optimal tableau for ) œ !, after two iterations, we get:
)
! Ÿ)Ÿ'
' Ÿ ) Ÿ ""
"" Ÿ ) Ÿ $&
ÐB‡" ß B‡# ß B‡$ ß B‡% Ñ
Ð$!  )ß !ß !ß !Ñ
Ð$'ß !ß '  )ß !Ñ
Ð&#Þ&  "Þ&)ß !ß ##Þ&  #Þ&)ß !Ñ
^ ‡ Ð) Ñ
"&!  &)
"%%  %)
"'!Þ&  #Þ&)
) œ $! provides the largest value of the objective function: B‡ Ð$!Ñ œ Ð(Þ&ß !ß &#Þ&ß !Ñ,
^ ‡ Ð$!Ñ œ #$&Þ&.
)-11
8.2-6.
)
!
Ÿ ) Ÿ "!Î*
"!Î* Ÿ ) Ÿ (!Î#$
(!Î#$ Ÿ ) Ÿ *!
ÐB‡" ß B‡# ß B‡$ Ñ
Ð!ß #!  #)ß !Ñ
Ð!ß $&  ""Þ&) ß &  %Þ&)Ñ
Ð!ß !ß *  !Þ")Ñ
^ ‡ Ð) Ñ
"!!  "!)
""!  )
""(  "Þ$)
8.2-7.
(a) Let BÐ5Ñ be the 5 th optimal solution obtained as ) is increased from !. Each BÐ5Ñ is
optimal for some )-interval, say )5 Ÿ ) Ÿ )5" , and the objective function value
^Ð)Ñ œ α5  "5 ) for some α5 and "5 , so ^Ð)Ñ is linear in this interval. As the interval
changes, α5 and "5 change so that a different linear function is obtained for each interval.
(b) The problem is:
maximize
^Ð)Ñ œ  Ð-4  α4 )ÑB4
8
 +34 B4 Ÿ ,3 , 3 œ "ß #ß á ß 7
4œ"
8
subject to
4œ"
B4
!, 4 œ "ß #ß á ß 8.
Note that the feasible region does not depend on ). Consider )"  )# and let )$ œ -)" 
Ð"Ñ
Ð#Ñ
Ð$Ñ
Ð"  -Ñ)# for some ! Ÿ - Ÿ ". Let B4 , B4 and B4 be the optimal values of B4
Ð4 œ "ß #ß á ß 8Ñ for )" , )# and )$ respectively. Let ^Ð)ß BÑ œ 84œ" Ð-4  α4 )ÑB4 .
^ ‡ Ð)" Ñ œ ^Ð)" ß BÐ"Ñ Ñ
^Ð)" ß BÐ$Ñ Ñ Ê -^ ‡ Ð)" Ñ
^ ‡ Ð)# Ñ œ ^Ð)# ß BÐ#Ñ Ñ
^Ð)# ß BÐ$Ñ Ñ Ê Ð"  -Ñ^ ‡ Ð)# Ñ
)-12
-^Ð)" ß BÐ$Ñ Ñ
Ð"  -Ñ^Ð)# ß BÐ$Ñ Ñ
Ê -^ ‡ Ð)" Ñ  Ð"  -Ñ^ ‡ Ð)# Ñ
-^Ð)" ß BÐ$Ñ Ñ  Ð"  -Ñ^Ð)# ß BÐ$Ñ Ñ
œ - Ð-4  α4 )" ÑB4  Ð"  -Ñ Ð-4  α4 )# ÑB4
8
Ð$Ñ
4œ"
8
Ð$Ñ
4œ"
œ  -4  α4 Ð-)"  Ð"  -Ñ)# ÑB4
8
Ð$Ñ
4œ"
œ  Ð-4  α4 )$ ÑB4 œ ^Ð)$ ß BÐ$Ñ Ñ œ ^ ‡ Ð)$ Ñ
8
Ð$Ñ
4œ"
Hence, ^ ‡ Ð)Ñ is convex in ).
8.2-8.
(a) The same argument as in part (a) of problem 8.2-7 holds.
(b) The problem is:
maximize
^Ð)Ñ œ  -4 B4
8
 +34 B4 Ÿ ,3  α3 ), 3 œ "ß #ß á ß 7
4œ"
8
subject to
4œ"
B4
!, 4 œ "ß #ß á ß 8.
Ð"Ñ
Ð#Ñ
Consider )"  )# and let )$ œ -)"  Ð"  -Ñ)# for some ! Ÿ - Ÿ ". Let B4 , B4
Ð$Ñ
B4
be the optimal values of B4 Ð4 œ "ß #ß á ß 8Ñ for )" , )# and )$ respectively.
Ð"Ñ
Ð#Ñ
-^ ‡ Ð)" Ñ  Ð"  -Ñ^ ‡ Ð)# Ñ œ - -4 B4  Ð"  -Ñ -4 B4
8
8
4œ"
4œ"
œ  -4 Ð-B4  Ð"  -ÑB4 Ñ
8
Ð"Ñ
Ð#Ñ
4œ"
Ð"Ñ
Ð#Ñ
If Bw4 œ -B4  Ð"  -ÑB4 Ð4 œ "ß #ß á ß 8Ñ, then Bw is feasible for ) œ )$ , since
 +34 Bw4 œ - +34 BÐ"Ñ
 +34 B4Ð#Ñ œ -Ð,3  α3 )Ñ  Ð"  -ÑÐ,3  α3 ) Ñ
4  Ð"  -Ñ
8
8
8
4œ"
4œ"
4œ"
œ ,3  α3 ), 3 œ "ß #ß á ß 7.
Since BÐ$Ñ is optimal for )$ ,
Ð#Ñ
 -4 Ð-BÐ"Ñ
 -4 B4Ð$Ñ œ ^ ‡ Ð)$ Ñ.
4  Ð"  -ÑB4 Ñ Ÿ
8
8
4œ"
4œ"
Hence, ^ ‡ Ð)Ñ is concave in ).
)-13
and
8.2-9.
From duality theory,
^ ‡‡ œ minimum
 Ð,3  53 ÑC3
7
3œ"
 +34 C3
7
subject to
-4 , 4 œ "ß #ß á ß 8
3œ"
C3
!, 3 œ "ß #ß á ß 7.
‡
ÐC"‡ ß C#‡ ß á ß C7
Ñ is feasible for this problem, so
^ ‡‡ Ÿ  Ð,3  53 ÑC3‡ œ ^ ‡   53 C3‡ .
7
7
3œ"
3œ"
8.3-1.
(a) Optimal Solution: ÐB*" ß B*# Ñ œ Ð"!ß "!Ñ and ^ * œ $!
)-14
(b) ÐB" ß B# Ñ œ Ð"!ß "!Ñ is optimal with ^ œ $!Þ
(c) The upper-bound technique goes from Ð!ß !Ñ to Ð&ß !Ñ to Ð"!ß &Ñ to Ð"!ß "!ÑÞ
)-15
8.3-2.
BV
^
B%
B&
Eq.
!
"
#
^
"
!
!
B"
"
!
#
B#
$
"
"
BV
^
B#
B&
Eq.
!
"
#
^
"
!
!
B"
"
!
#
B#
!
"
!
BV
^
C#
B&
Eq.
!
"
#
^
"
!
!
B"
"
!
#
C#
!
"
!
BV
^
B$
B&
Eq.
!
"
#
^
"
!
!
B"
"
!
#
C#
#
BV
^
B$
B&
Eq.
!
"
#
^
"
!
!
C"
"
!
#
"
#
#
C#
#
"
#
#
B$
#
#
#
B%
!
"
!
B&
!
!
"
RS
!
"
)
B# Ÿ $
B# Ÿ "
B# Ÿ )
B$
%
#
%
B%
$
"
"
B&
!
!
"
RS
$
"
(
B$ Ÿ #
B$ Ÿ "
B$ Ÿ " $%
B$
%
#
%
B%
$
"
"
B&
!
!
"
RS
$
#
(
B$ Ÿ #
B$ Ÿ "
B$ Ÿ " $%
B$
!
"
!
B%
"
 "#
"
B&
!
!
"
RS
(
"
$
B" Ÿ "
B" Ÿ " "#
B$
!
"
!
B%
"
 "#
"
B&
!
!
"
RS
)
"
"
B'
!
!
"
ÐB" ß B# ß B$ Ñ œ Ð"ß $ß "Ñ is optimal with ^ œ ).
8.3-3.
Initial Tableau
BV
^
B&
B'
Eq.
!
"
#
^
"
!
!
B"
#
#
"
B#
$
#
#
B$
#
"
$
B%
&
#
%
B&
!
"
!
B$
!
!
"
C%
B&
B'
$
(
 "!
(
'
(
%
(
$
(
"
(
'
(
"
(
 #(
RS
!
&
&
Final Tableau (after five iterations)
BV
^
B"
B$
Eq.
!
"
#
^
"
!
!
B"
!
"
!
C#
"
(
 )(
#
(
RS
&%
(
#
(
$
(
ÐB" ß B# ß B$ ß B% Ñ œ Ð#Î(ß "ß $Î(ß "Ñ is optimal with ^ œ &%Î(.
)-16
8.3-4.
Initial Tableau
BV
^
B'
B(
Eq.
!
"
#
^
"
!
!
B"
#
"
%
B#
&
$
'
B$
$
#
&
B%
%
$
(
C$
C%
!
"
!
B&
"
"
"
B'
!
"
!
B(
!
!
"
RS
!
'
"&
B(
!
!
"
RS
"!
"
!
Final Tableau (after seven iterations)
BV
^
C%
B(
Eq.
!
"
#
^
"
!
!
C"
#
$
"
$
 &$
C#
"
"
"
"
$
#
$
 "$
B&
B'
"
$
 "$
 %$
%
$
 "$
 ($
ÐB" ß B# ß B$ ß B% ß B& Ñ œ Ð"ß "ß "ß !ß !Ñ is optimal with ^ œ "!.
7.3-5.
ÐB" ß B# ß B$ Ñ œ Ð"!ß &ß &Ñ is optimal with ^ œ '!.
)-17
8.4-1.
It.
0
1
2
X1
1
1.04605
0.93406
X2
3
4.95395
6.06594
X3
7
10.9539
13.0659
8.4-2.
(a)
The feasible corner point solutions are Ð!ß !Ñ, Ð!ß %Ñ and Ð%ß !Ñ. The last one is optimal
with ^ œ "#.
(b)
Iter.
!
"
#
$
%
B"
"
"Þ)(&
#Þ'*)"
$Þ$%$*'
$Þ''("
B#
"
"Þ"#&
!Þ)!"*
!Þ%!!*&
!Þ#!!%(
^
%
'Þ(&
)Þ)*'#"
"!Þ%$#)
""Þ#!")
(c)
)-18
)-19
)-20
8.4-3.
(a)
Iter.
!
"
#
$
%
&
'
(
)
*
B"
%
#
"
!Þ&
!Þ#&
!Þ"#&
!Þ!'#&
!Þ!$"#&
!Þ!"&'#
!Þ!!()"
B#
%
'
(
(Þ&
(Þ(&
(Þ)(&
(Þ*$(&
(Þ*')(&
(Þ*)%$)
(Þ**#"*
^
"#
"%
"&
"&Þ&
"&Þ(&
"&Þ)(&
"&Þ*$(&
"&Þ*'))
"&Þ*)%%
"&Þ**##
(b) The value of B" is halved at each step so subsequent trial solutions should be of the
form ÐB" ß B# Ñ œ Ð#3 ß )  #3 Ñ for 3 œ "ß #ß á .
(c) The smallest integer 3 such that #3  #Ð3"Ñ œ #Ð3"Ñ Ÿ !Þ!" is ', so ÐB" ß B# Ñ œ
Ð#( ß )  #( Ñ œ Ð!Þ!!()ß (Þ**##Ñ in iteration *.
8.4-4.
(a) Optimal Solution: ÐB" ß B# Ñ œ Ð3ß 3Ñ, ^ œ 6
(b) The gradient is Ð1ß "Ñ. Moving from the origin in the direction Ð1ß "Ñ, the first
boundary point encountered is the optimal solution Ð3ß 3Ñ.
)-21
(c) α œ !Þ&
(d) α œ !Þ*
)-22
8.4-5.
(a)
(b)
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)-24
SUPPLEMENT TO CHAPTER 8
LINEAR GOAL PROGRAMMING AND ITS SOLUTION PROCEDURES
8S-1.
(a)
(b) Let
be the coefficient of
and
be the one for
, so
.
8S-2.
(a)
minimize
sum of amounts under market share for product 1 and 2
subject to
(b)
minimize
subject to
(c)
Market Share per $million
Ad
Ad
Ad
Camp. 1 Camp. 2 Camp. 3
Goal 1 (M. Share of Prod. 1)
0.5%
0.2%
Goal 2 (M. Share of Prod. 2)
0.3%
0.2%
Ad
Camp. 1
Millions of Dollars Spent 13.33
Ad
Camp. 2
0
Ad
Camp. 3
41.67
>=
10
Goals
Level
Achieved
Goal
15.0%
>= 15%
8.33%
>= 10%
Total
55
<=
55
8S-3.
(a)
(b)
(c)
maximize
subject to
8S-1
Penalty
Weights
Goal 1
Goal 2
Deviations
Amount Amount
Over
Under
0.0%
0.0%
0.0%
1.67%
Over
Goal
Under
Goal
1
1
Constraints
Balance
(Level-Over+Under)
Goal
15%
= 15%
10%
= 10%
Weighted Sum
of Deviations
1.67%
(d)
Goal 1 (Total Profit)
Goal 2 (Employment Level)
Goal 3 (Earnings Next Year)
Production Rate
Unit Contribution of Product
Product
Product
Product
1
2
3
20
15
25
6
4
5
8
7
5
Product
1
0
Product
2
0
Goals
Level
Achieved
375
75
=
75
>=
Product
3
15
Goal
Max
50
75
Benefit
Goal 1
Goal 2
Goal 3
Deviations
Amount Amount
Over
Under
25
0
0
0
Over
Under
Goal
Goal
Level Achieved
-6
-6
-3
Constraints
Balance
(Level-Over+Under)
50.000
75.000
Goal
=
=
50
75
Measure of
Performance
225
8S-4.
(a) No, we would not expect the optimal solution to change. Goal 1 is already met, so
increasing the weight on that goal would not change anything. Goal 2 is already
exceeded, so decreasing the penalty weight for this goal would only decrease our desire
to avoid exceeding this goal.
Goal 1 (Profit)
Goal 2 (Employment)
Goal 3 (Investment)
Contribution per Unit Produced
Product 1
Product 2
Product 3
12
9
15
5
3
4
5
7
8
Product 1
Units Produced 8.333333333
Product 2
0
Product 3
1.666666667
Goals
Level
Achieved
Goal
125
>= 125
48.333333 =
40
55
<=
55
Penalty
Weights
Profit
Employment
Investment
Deviations
Amount Amount
Over
Under
0
0
8.33333
0
0
0
Over
Goal
1
3
Under
Goal
7
4
Constraints
Balance
(Level - Over + Under)
Goal
125
= 125
40
= 40
55
= 55
Weighted Sum
of Deviations
8.333333333
(b)
Goal 1 (Profit)
Goal 2 (Employment)
Goal 3 (Investment)
Units Produced
Contribution per Unit Produced
Product 1
Product 2
Product 3
12
9
15
5
3
4
5
7
8
Product 1
11.667
Product 2
0
Product 3
0
Goals
Level
Achieved
Goal
140
>= 140
58.333
=
40
58.333
<=
55
Penalty
Weights
Profit
Employment
Investment
Deviations
Amount Amount
Over
Under
0
0
18.333
0
3.333
0
Over
Goal
2
3
Under
Goal
5
4
Constraints
Balance
(Level - Over + Under)
Goal
140
= 140
40
= 40
55
= 55
Weighted Sum
of Deviations
46.667
(c)
Goal 1 (Profit)
Goal 2 (Employment)
Goal 3 (Investment)
Units Produced
Contribution per Unit Produced
Product 1
Product 2
Product 3
12
9
15
5
3
4
5
7
8
Product 1
11.667
Product 2
0
Product 3
0
Goals
Level
Achieved
Goal
140
>= 140
58.333
=
40
58.333
<=
55
Penalty
Weights
Profit
Employment
Investment
8S-2
Deviations
Amount Amount
Over
Under
0
0
18.333
0
3.333
0
Over
Goal
1
3
Under
Goal
7
4
Constraints
Balance
(Level - Over + Under)
Goal
140
= 140
40
= 40
55
= 55
Weighted Sum
of Deviations
28.333
8S-5.
(a)
minimize
(amount under foreign capital goal)
(amount under citizens fed goal)
(amount under goal for citizens employed)
(amount over goal for citizens employed)
(b)
minimize
subject to
M
M
M
M
(c)
Contribution per 1000 Acres
Crop 1
Crop 2
Crop 3
Goal 1 (Foreign Capital) $3,000
$5,000
$4,000
Goal 2 (Citizens Fed)
150
75
100
Goal 3 (Citizens Employed)
10
15
12
Thousands of Acres Planted
(d)
Crop 1
8,333
Crop 2
6,667
Crop 3
0
Goals
Level
Achieved
Goal
$58,333,333 >= $70,000,000
1,750,000 >= 1,750,000
183,333
=
200,000
Total
15,000
<=
15,000
Penalty
Weights
Goal 1
Goal 2
Goal 3
Deviations
Amount
Amount
Over
Under
$0
$11,666,667
0
0
0
16,667
Over
Goal
1
Under
Goal
0.01
1
1
Constraints
Balance
(Level-Over+Under)
Goal
$70,000,000
= $70,000,000
1,750,000
= 1,750,000
200,000
=
200,000
Weighted Sum
of Deviations
133,333
minimize
subject to
M
M
M
M
8S-3
(e) Optimal Solution:
thousand acres
.
(f) With only
in the objective function, we get
, so fix
and
bring
into the objective function. Now
. Fix
at this value
(remembering subtract from RHS) and optimize for the third priority. Then the solution
in part (c) is obtained:
.
8S-6.
(a)
minimize
subject to
(b) - (c)
Optimal Solution:
BV
,
E
RHS
8S-4
(d)
(e)
minimize
subject to
The feasible region is a shown in figure (i) of part (d). Fix
.
minimize
subject to
The feasible region is a shown in figure (ii) of part (d). Fix
minimize
subject to
The solution is
with
.
8S-7.
(a)
minimize
subject to
8S-5
.
(b) - (c)
Optimal Solution:
BV
,
E
RHS
(d)
8S-8.
If
(a)
, where
, then
.
minimize
subject to
,
,
(b)
minimize
subject to
,
,
,
8S-6
Case%8S.1%A%Cure%for%Cuba%
!
a)! We!need!to!develop!a!goal!programming!problem!whose!solution!characterizes!
Mr.!Baker's!shipping!policy.!The!decision!variables!are!the!number!(in!1000’s)!of!
basic,!advanced,!and!supreme!packages!to!send,!and!the!number!of!doctors!to!
send.!Note:!measuring!most!variables!in!1000's!greatly!improves!the!reliability!
of!!Solver.!
!
Mr.!Baker!faces!three!hard!constraints.!Because!of!the!size!limitation,!the!total!
number!of!package!must!not!exceed!40,000.!Second,!the!total!weight!can!not!
exceed!6!million!pounds.!Finally,!the!total!number!of!Supreme!packages!cannot!
exceed!100!times!the!number!of!doctors.!!These!constraints!are!included!in!the!
spreadsheet!as!follows.!
!
TotalPackages!(E14)!≤!SizeLimit!(E16)!
!
TotalWeight!(E10)!≤!WeightRestriction!(G10)!
!
SupremePackages!(D14)!≤!SafetyRestriction!(D16)!
!
In!addition,!we!need!to!include!three!constraints!for!Mr.!Baker's!goals.!We!
measure!the!deviations!from!the!goals!using!changing!cells!(Deviations!in!I4:J6),!
and!enforce!the!correct!value!of!these!changing!cells!with!the!constraints!in!
columns!L!through!N.!
!
Finally,!the!penalty!weights!are!entered!in!I15:J17,!and!the!weight!sum!of!
deviations!calculated!in!L15.!
!
The!spreadsheet!follows.!
A
!
!
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
B
C
D
E
F
<=
>=
>=
<=
G
H
Goals
Goal 1 (Cost)
Goal 2 (Packages Sent)
Goal 3 (Population Reached)
Weight
Packages Sent (thousands)
Doctors
Basic
$300
1
30
Advanced
$350
1
35
Supreme
$720
1
54
Level
Achieved
21,000
40
1,488
120
180
220
Total
Weight
6,000
Basic
28
Advanced
0
120
Supreme
12
<=
Safety
12
Restriction
0.1
per Doctor
Total
Packages
40
<=
40
Size Limit
Goal
20,000
3
2,200
I
J
Deviations
Amount Amount
Over
Under
1,000
0
37
0
0
712
K
L
M
Constraints
Balance
(Level-Over+Under)
20,000
=
3
=
2,200
=
N
O
Goal
20,000 $thousand
3
thousand
2,200 thousand
Weight
Restriction
6,000
thousand pounds
Penalty
Weights
Goal 1
Goal 2
Goal 3
Over
Goal
0.001
Under
Goal
1
0.07
Weighted Sum
of Deviations
50.84
!
!
Mr.!Baker!should!send!28,000!basic!packages!and!12,000!supreme!packages!
along!with!120!doctors!to!Cuba.!
Cost per Doctor ($thousand)
33
8S-7
!
b)! The!penalty!weight!for!being!under!goal!3!changes.!One\half!percent!of!the!
population!is!55,000.!Therefore,!the!new!penalty!weight!is!10!points!/!55!
(thousand!people)!=!0.182.!The!new!solution!follows.!!
!
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
B
C
D
E
F
<=
>=
>=
<=
G
H
Goals
Goal 1 (Cost)
Goal 2 (Packages Sent)
Goal 3 (Population Reached)
Weight
Packages Sent (thousands)
Basic
$300
1
30
Advanced
$350
1
35
Supreme
$720
1
54
Level
Achieved
21,000
40
1,488
120
180
220
Total
Weight
6,000
Basic
28
Advanced
0
Doctors
120
Cost per Doctor ($thousand)
33
Supreme
12
<=
Safety
12
Restriction
0.1
per Doctor
I
J
K
Deviations
Amount Amount
Over
Under
1,000
0
37
0
0
712
Goal
20,000
3
2,200
L
M
Constraints
Balance
(Level-Over+Under)
20,000
=
3
=
2,200
=
N
O
Goal
20,000 $thousand
3
thousand
2,200 thousand
Weight
Restriction
6,000
thousand pounds
Total
Packages
40
<=
40
Size Limit
Penalty
Weights
Goal 1
Goal 2
Goal 3
Over
Goal
0.001
Under
Goal
Weighted Sum
of Deviations
130.45
1
0.182
!
!
The!optimal!shipping!policy!did!not!change.!The!plan!appears!to!be!insensitive!to!
increases!in!the!penalty!weight!for!violating!the!goal!to!reach!at!least!20%!of!the!
Cuban!population.!
!
c)! The!doctors!needed!per!thousand!supreme!packages!changes!from!0.1!to!0.075.!
The!new!solution!follows.!!
!
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
B
C
D
E
F
<=
>=
>=
<=
G
H
Goals
Goal 1 (Cost)
Goal 2 (Packages Sent)
Goal 3 (Population Reached)
Weight
Packages Sent (thousands)
Basic
$300
1
30
Advanced
$350
1
35
Supreme
$720
1
54
Level
Achieved
22,320
40
1,488
120
180
220
Total
Weight
6,000
Basic
28
Advanced
0
Doctors
160
Cost per Doctor ($thousand)
33
Supreme
12
<=
Safety
12
Restriction
0.075
per Doctor
Total
Packages
40
<=
40
Size Limit
Goal
20,000
3
2,200
I
J
Deviations
Amount Amount
Over
Under
2,320
0
37
0
0
712
K
L
M
Constraints
Balance
(Level-Over+Under)
20,000
=
3
=
2,200
=
N
Goal
20,000 $thousand
3
thousand
2,200 thousand
Weight
Restriction
6,000
thousand pounds
Penalty
Weights
Goal 1
Goal 2
Goal 3
Over
Goal
0.001
Under
Goal
Weighted Sum
of Deviations
131.77
1
0.182
!
!
While!the!number!of!packages!Mr.!Baker!should!ship!has!not!changed,!the!
number!of!doctors!is!now!160.!
8S-8
O
!
d)! The!budget!restriction!is!now!a!hard!constraint!and!the!penalty!variables!for!the!cost!
goal!can!be!eliminated.!
!
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
B
C
D
E
F
<=
>=
>=
<=
G
H
Goals
Cost (Hard Constraint)
Goal 2 (Packages Sent)
Goal 3 (Population Reached)
Weight
Packages Sent (thousands)
Basic
$300
1
30
Advanced
$350
1
35
Supreme
$720
1
54
Level
Achieved
20,000
40
1,465
120
180
220
Total
Weight
6,000
Basic
27
Advanced
2.5
Doctors
105
Cost per Doctor ($thousand)
33
Supreme
10.5
<=
Safety
10.5
Restriction
0.1
per Doctor
Total
Packages
40
<=
40
Size Limit
I
J
K
Deviations
Amount Amount
Over
Under
Goal
20,000
3
2,200
37
0
L
M
Constraints
Balance
(Level-Over+Under)
0
735.5
3
2,200
Under
Goal
1
0.07
Weighted Sum
of Deviations
51.49
N
O
Goal
=
=
$thousand
3
thousand
2,200 thousand
Weight
Restriction
6,000
thousand pounds
Penalty
Weights
Goal 2
Goal 3
Over
Goal
!
!
Mr.!Baker!should!send!27,000!basic!packages,!2,500!advanced!packages,!and!
10,500!supreme!packages!along!with!105!doctors!to!Cuba.!
!
e)! We!start!by!minimizing!the!amount!over!goal!1!(total!cost!≤!$20!million).!
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
!
!
Goal 1 (Cost)
Goal 2 (Packages Sent)
Goal 3 (Population Reached)
Weight
Packages Sent (thousands)
%
D
E
Basic
$300
1
30
Advanced
$350
1
35
Supreme
$720
1
54
Level
Achieved
20,000
19.024
1,027
<=
>=
>=
Goal
20,000
3
2,200
120
180
220
Total
Weight
4,185
<=
Weight
Restriction
6,000
Basic
0
Advanced
0
Doctors
191
Cost per Doctor ($thousand)
33
A
!
C
F
Goals
G
H
I
J
Deviations
Amount Amount
Over
Under
0
0
16.024
0
0
1,173
K
L
M
N
O
Constraints
Balance
(Level-Over+Under)
Goal
20,000
= 20,000 $thousand
3
=
3
thousand
2,200
= 2,200 thousand
Minimize Over Goal 1
Total
Supreme
Packages
19.024
19.02361111
<=
<=
Safety
19.1
40
Restriction
0.1
Size Limit
per Doctor
!
Then,!since!goal!2!is!already!met,!we!move!on!to!goal!3.!We!minimize!the!amount!
under!goal!3!(population!reached!≥!20%),!while!constraining!(amount!over!goal!
1!=!0)!and!(amount!under!goal!2!=!0).!
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
!
B
B
C
D
E
F
G
<=
>=
>=
Goal
20,000
3
2,200
<=
Weight
Restriction
6,000
Goals
Goal 1 (Cost)
Goal 2 (Packages Sent)
Goal 3 (Population Reached)
Weight
Packages Sent (thousands)
Basic
$300
1
30
Advanced
$350
1
35
Supreme
$720
1
54
Level
Achieved
20,000
40
1,464
120
180
220
Total
Weight
6,000
Basic
27
Advanced
2.5
Doctors
105
Cost per Doctor ($thousand)
33
Supreme
10.5
<=
Safety
10.5
Restriction
0.1
per Doctor
H
I
J
K
Deviations
Amount Amount
Over
Under
0
0
37
0
0
735
L
M
Constraints
Balance
(Level-Over+Under)
20,000
=
3
=
2,200
=
N
O
Goal
20,000 $thousand
3
thousand
2,200 thousand
Minimize Under Goal 3
(Over Goal 1 = 0)
(Under Goal 2 = 0)
Total
Packages
40
<=
40
Size Limit
Mr.!Baker!should!send!27!thousand!basic!packages,!2,500!advanced!packages,!
and!10,500!supreme!packages,!along!with!105!doctors.!
%
8S-9
!
Case%8S.2%Airport%Security%
!
a)! The!two!decisions!to!be!made!are!how!much!to!spend!on!the!two!security!
systems.!Hence,!we!define!the!following!two!variables.!
!
Let! PS!=!thousands!of!dollars!spent!per!portal!system!
!
SS!=!thousands!of!dollars!spent!per!screening!system.!
!
b)! Preemptive!goal!programming!is!appropriate!because!there!is!a!clear!order!of!
priorities.!
!
Priority!1!is!met!by!all!possible!systems.!
!
Priority!2!(hereafter!referred!to!as!goal!1)!is!that!the!false!alarm!rate!should!not!
exceed!10%.!The!false!alarm!rate!of!the!two!systems!is!as!follows:!
!
Portal!System:!10%!–!(1%)(PS!–!90)!/!15!
!
Screening!System:!6%!–!(1%)(SS!–!60)!/!30!
Goal!1!is!thus!
!
[10%!–!(1%)(PS!–!90)!/!15]!+![6%!–!(1%)(SS!–!60)!/!30]!≤!10%!
!
Priority!3!(hereafter!referred!to!as!goal!2)!is!that!the!first!budgetary!guideline!
should!be!met!(total!expenditures!≤!$250,000).!Goal!2!is!thus!
!
PS!+!SS!≤!250!
!
Priority!4!(hereafter!referred!to!as!goal!3)!is!that!the!second!budgetary!guideline!
should!be!met!(average!total!maintenance!cost!$30,000).!The!maintenance!cost!
of!the!two!systems!is!as!follows:!
!
Portal!System:!15!+!(PS!–!90)!/!10!
!
Screening!System:!9!+!(SS!–!60)!/!25!
Goal!3!is!thus!
!
[15!+!(PS!–!90)!/!10]!+![9!+!(SS!–!60)!/!25]!≤!30!
8S-10
!
c)! !
Screening
System
($thousands)
150
Goal 1
Satisfied
120
Goal 3
Satisfied
90
Goal 2
Satisfied
60
90
120
150
180
210 Portal
System
($thousands)
!
Goal!1!is!satisfied!inside!the!rightmost!polygon.!Goal!2!is!satisfied!in!the!polygon!
in!the!middle.!The!small!triangle!with!vertices!at!(180,!60),!(170,!80),!(190,!60)!is!
the!only!area!where!both!goal!1!and!goal!2!are!satisfied.!
!
Applying!preemptive!goal!programming,!the!first!solution!will!be!somewhere!
inside!the!region!where!goal!1!is!satisfied.!!
!
The!second!solution!(minimizing!the!amount!over!goal!2!while!constraining!goal!
1!to!be!met)!will!give!a!solution!inside!the!small!triangle!where!both!goal!1!and!
goal!2!are!met.!
!
The!third!solution!(minimizing!the!amount!over!goal!3!while!constraining!goal!1!
and!2!to!be!met)!will!pick!the!solution!inside!the!small!triangle!(since!goal!1!and!
2!must!remain!to!be!met)!that!is!closest!to!meeting!goal!3.!This!occurs!at!(170,!
80).!That!is,!they!should!spend!$170!thousand!on!the!portal!system!and!$80!
thousand!on!the!screening!system.!
8S-11
!
d)! We!start!by!minimizing!the!amount!over!goal!1!(false!alarm!rate!≤!10%).!
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
!
!
!
B
Goal 1 (False Alarm Rate)
Goal 2 (Total Expenditure)
Goal 3 (Maintenance Cost)
Minimum
Expenditure ($thousand/system)
Maximum
False Alarm Rate
Base Rate
Minus 1% per ($x thousand)
Maintenance Cost ($thousand)
Base Rate
Plus $1 per $x
C
Goals
Level
Achieved
10%
250
32.8
<=
<=
<=
Portal
System
90
<=
170
<=
210
Screening
System
60
<=
80
<=
150
5%
10%
15
5%
6%
30
23
15
10
9.8
9
25
D
E
Goal
10%
250
30
F
G
Deviations
Amount Amount
Over
Under
0
0
0
0
2.8
0
H
I
Constraints
Balance
(Level-Over+Under)
10%
250
30
J
=
=
=
K
L
Goal
10%
250 ($thousand)
30 ($thousand)
Minimize Over Goal 1
!
Since!goal!2!is!already!met,!we!move!on!to!minimizing!the!amount!over!goal!3!
(maintenance!cost!≤!$30,000),!while!constraining!(amount!over!goal!1!=!0)!and!
(amount!over!goal!2!=!0).!
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
B
Level
Achieved
Goal 1 (False Alarm Rate)
10%
Goal 2 (Total Expenditure)
250
Goal 3 (Maintenance Cost)
32.8
Minimum
Expenditure ($thousand/system)
Maximum
False Alarm Rate
Base Rate
Minus 1% per ($x thousand)
Maintenance Cost ($thousand)
Base Rate
Plus $1 per $x
C
Goals
<=
<=
<=
Portal
System
90
<=
170
<=
210
Screening
System
60
<=
80
<=
150
5%
10%
15
5%
6%
30
23
15
10
9.8
9
25
D
Goal
10%
250
30
8S-12
E
F
G
Deviations
Amount Amount
Over
Under
0
0
0
0
2.8
0
H
I
Constraints
Balance
(Level-Over+Under)
10%
250
30
J
=
=
=
K
L
Goal
10%
250 ($thousand)
30 ($thousand)
Minimize Over Goal 3
(Over Goal 1 = 0)
(Over Goal 2 = 0)
!
!
e)! The!first!two!goals!are!now!hard!constraints,!and!we!minimize!the!amount!over!
goal!3.!
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
B
Goal 1 (False Alarm Rate)
Goal 2 (Total Expenditure)
Goal 3 (Maintenance Cost)
Minimum
Expenditure ($thousand/system)
Maximum
False Alarm Rate
Base Rate
Minus 1% per ($x thousand)
Maintenance Cost ($thousand)
Base Rate
Plus $1 per $x
C
Goals
D
Level
Achieved
10%
250
32.8
<=
<=
<=
Portal
System
90
<=
170
<=
210
Screening
System
60
<=
80
<=
150
5%
10%
15
5%
6%
30
23
15
10
9.8
9
25
E
Goal
10% Hard Constraint
250 Hard Constraint
30
F
G
Deviations
Amount Amount
Over
Under
2.8
0
H
I
J
Constraints
Balance
(Level-Over+Under)
30
=
K
L
Goal
30
($thousand)
($thousand)
Minimize Over Goal 1
!
If!the!linear!program!had!no!feasible!solution,!this!would!imply!that!it!is!not!
possible!to!meet!all!of!the!higher!priority!goals!that!were!turned!into!hard!
constraints.!
!
f)! We!no!longer!use!goal!programming.!The!goal!is!to!minimize!the!total!false!alarm!
rate!subject!to!meeting!the!first!budgetary!guideline!(total!expenditure),!but!
ignoring!the!second!budgetary!guideline!(maintenance!cost).!The!spreadsheet!
model!follows.!
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
B
Level
Achieved
Total False Alarm Rate
9%
Total Expenditure
250
Maintenance Cost
34
Minimum
Expenditure ($thousand/system)
Maximum
False Alarm Rate
Base Rate
Minus 1% per ($x thousand)
Maintenance Cost ($thousand)
Base Rate
Plus $1 per $x
C
D
<=
Maximum
Expenditure
250
Portal
System
90
<=
190
<=
210
Screening
System
60
<=
60
<=
150
3%
10%
15
6%
6%
30
25
15
10
9
9
25
!
!
The!total!false!alarm!rate!can!be!lowered!to!9%!by!ignoring!the!second!
budgetary!guideline!(maintenance!cost).!
!
g)! Further!what\if!analysis!might!look!at!how!low!the!false\alarm!rate!can!be!
lowered!by!ignoring!the!first!budgetary!guideline,!but!meeting!the!second.!Also,!
it!would!be!interesting!to!look!at!how!the!minimum!false!alarm!rate!changes!as!
both!of!the!budgetary!guidelines!are!varied.!
8S-13
CHAPTER 9: THE TRANSPORTATION AND ASSIGNMENT PROBLEMS
9.1-1.
While growing continuously as a global company, Procter & Gamble faced the need to
restructure for enhanced effectiveness. The goal was to optimize work processes and to
minimize expenses while maintaining customer satisfaction. Lowered transportation costs
due to changes in the trucking industry and reduced product packages suggested that the
total transportation costs could be decreased. In the meantime, shorter product life cycles
justified smaller number of plants. Consequently, P&G had to decide on where to locate
the plants, what and how much to produce in each. This would be impossible without
reviewing the distribution system. Hence, two problems for each product category needed
to be solved: a distribution-location problem and a product-sourcing problem.
First, optimal distribution center (DC) locations and optimal customer assignments are
found by solving an uncapacitated facility-location model. The objective in this problem
is to minimize the total cost of transportation and supply while the primary restriction is
to satisfy customer demand. Fixed costs involved in locating DCs are ignored. The total
number of DCs is determined beforehand subjectively. The solution of this problem is an
input to the product sourcing problem.
With fixed DC locations and their capacities, product sourcing is modeled as a
transportation problem. Sources are plants, destinations are DCs and customers. The
location and capacity of the plants are specified by the product-strategy teams. Decision
variables are the amounts of demand at each destination to be met from each source. The
objective is to minimize the total cost while satisfying the demand at each destination
without exceeding the capacity of each source. The costs consist of manufacturing,
warehousing and transportation costs. An out-of-kilter algorithm is used to solve this
problem for each product category.
The benefits of this study included a reduction in the number of plants in North America
by 20% and savings of over $200 million per year. The reduction in manufacturing costs,
due to lowered number of plants and personnel coupled with improved efficiency of the
supply chain, outweighs the increase in delivery costs. The gains from this study led
P&G to making OR/MS a part of its decision-making process.
9.1-2.
(a)
9-1
(b)
(c)
Shipments
1
Plant 2
3
Total Received
Demand
1
0
0
10
10
=
10
Distribution Center
2
3
0
2
9
8
1
0
10
10
=
=
10
10
4
10
0
0
10
=
10
Total Shipped
12
=
17
=
11
=
Supply
12
17
11
Total Cost
$20,200
9.1-3.
(a) Let
and
be the number of pints purchased from Dick today and tomorrow
respectively, and be the number of pints purchased from Harry today and tomorrow
respectively.
9-2
(b)
(c)
9-3
9.1-4.
(a)
(b)
9-4
9.1-5.
Variable Cells
Cell
$D$12
$E$12
$F$12
$G$12
$D$13
$E$13
$F$13
$G$13
$D$14
$E$14
$F$14
$G$14
Name
Bellingham Sacramento
Bellingham Salt Lake City
Bellingham Rapid City
Bellingham Albuquerque
Eugene Sacramento
Eugene Salt Lake City
Eugene Rapid City
Eugene Albuquerque
Albert Lea Sacramento
Albert Lea Salt Lake City
Albert Lea Rapid City
Albert Lea Albuquerque
Final Reduced Objective Allowable Allowable
Value Cost Coefficient Increase Decrease
0
15
464
1E+30
15
20
0
513
15
21
0
84
654
1E+30
84
55
0
867
21
351
80
0
352
15
1E+30
45
0
416
21
15
0
217
690
1E+30
217
0
21
791
1E+30
21
0
728
995
1E+30
728
0
351
682
1E+30
351
70
0
388
84
1E+30
30
0
685
351
84
Constraints
Cell
$D$15
$E$15
$F$15
$G$15
$H$12
$H$13
$H$14
Final Shadow Constraint Allowable Allowable
Name
Value Price
R.H. Side Increase Decrease
Total Received Sacramento
80
-418
80
45
0
Total Received Salt Lake City
65
-354
65
55
0
Total Received Rapid City
70
-297
70
30
0
Total Received Albuquerque
85
0
85
0
1E+30
Bellingham Total Shipped
75
867
75
0
55
Eugene Total Shipped
125
770
125
0
45
Albert Lea Total Shipped
100
685
100
0
30
These ranges tell the management how much each individual cost can be changed
without changing the optimal solution.
9-5
9.1-6.
(a) Introduce a dummy customer 5 to represent the excess amount sent to customer 3 and
a dummy plant 4 to represent the units that are sold to, but not received by customers 4
and 5.
(a), (c), (d)
Unit Profit
1
2
3
Dummy
Plant
Customer
1
2
3
$800
$700
$500
$500
$200
$100
$600
$400
$300
($9,999) ($9,999) ($9,999)
1
0
40
0
0
40
=
40
Customer
2
3
60
0
0
0
0
20
0
0
60
20
=
=
60
20
4
0
40
20
0
60
=
60
1
($800)
($500)
($600)
$9,999
Customer
2
3
($700) ($500)
($200) ($100)
($400) ($300)
$9,999 $9,999
4
($200)
($300)
($500)
$0
1
0
40
0
0
40
=
40
Customer
2
3
60
0
0
0
0
20
0
0
60
20
=
=
60
20
4
0
40
20
0
60
=
60
Shipments
1
Plant 2
3
Dummy
Total Received
Commitment
4
$200
$300
$500
$0
Dummy
$500
$100
$300
$0
Dummy Total Shipped
0
60
0
80
0
40
60
60
60
=
60
=
=
=
=
Supply
60
80
40
60
Total Profit
$90,000
(b), (e)
Unit Cost
1
Plant 2
3
Dummy
Shipments
1
Plant 2
3
Dummy
Total Received
Commitment
The profit is $
.
9-6
Dummy
($500)
($100)
($300)
$0
Dummy Total Shipped
0
60
0
80
0
40
60
60
60
=
60
=
=
=
=
Supply
60
80
40
60
Total Cost
-$90,000
9.1-7.
(a)
(b), (c)
Unit Cost
Plant
A
B
Shipments
Plant
A
B
Total Received
Demand
Distribution Center
1
2
3
$800
$700
$400
$600
$800
$500
Distribution Center
1
2
3
0
20
20
20
0
0
20
20
20
=
=
=
20
20
20
Dummy
$0
$0
Dummy Total Shipped
Supply
10
50
=
50
30
50
=
50
40
=
Total Cost
40
$34,000
9.1-8.
(a) Let destination
represent the demand of
represent the extra demand up to
shipped to center
below.
at center and destination
in the parameter table
(b), (c)
Distribution Center
1
1 Extra
2
2 Extra
3
3 Extra
Plant A $800
$800
$700
$700
$400
$400
Plant B $600
$600
$800
$800
$500
$500
Dummy $99,999
$0
$99,999
$0
$99,999
$0
Dummy
$0
$0
$999,999
Distribution Center
2
0
10
0
10
0
0
10
0
20
20
10
20
=
=
=
20
10
20
Dummy
10
30
0
40
=
40
Unit Cost
Shipments
Plant A
Plant B
Dummy
Total Received
Demand
1
0
10
0
10
=
10
3
10
0
0
10
=
10
9-7
20
0
0
20
=
20
Total Shipped
50
=
50
=
30
=
Supply
50
50
30
Total Cost
$31,000
9.1-9.
(a) Let source
be regular time production and be overtime production in month
. Let destination
represent the contracted sales for product 1 and
represent the contracted sales for product 2 in month
. Destination 7 is dummy.
(b)
Hence, the total cost is $
and no overtime is necessary.
9.2-1.
(a) Vogel's approximation method would choose
9-8
as the first basic variable.
(b) Russell's approximation method would choose
(c) Initial BF solution using northwest corner rule:
9-9
as the first basic variable.
9.2-2.
(a) Northwest Corner Rule
Cost:
(b) Vogel's Approximation Method
Cost:
9-10
(c) Russell's Approximation Method
Cost:
Note that Vogel's and Russell's approximation methods return an optimal solution.
9.2-3.
(a) Northwest Corner Rule
Cost:
9-11
(b) Vogel's Approximation Method
Cost:
Arbitrarily breaking the tie differently returns the solution below with cost
(c) Russell's Approximation Method
Cost:
9-12
.
9.2-4.
(a) All the supply and demand values are integers. By the integer solutions property, the
resulting basic feasible solutions will be integral. All the supplies and demands are one,
so the only possible values of the variables in a basic feasible solution are and . The
's indicate the assignment of a source to a destination.
(b) There are
degenerate.
basic variables in every basic feasible solution and
(d) The variables are chosen in the order
of them are
.
9-13
(c) - (e)
Optimal assignment (source,destination):
, cost:
9-14
9.2-5.
Cost: $
,
for all and , so the solution is optimal.
9.2-6.
9-15
The current solution is optimal:
, with cost
. The optimality condition
9-16
and
for all and is met.
9.2-7.
(a) Northwest Corner Rule
3 iterations are required to reach optimality.
9-17
(b) Vogel's Approximation Method
The solution is optimal, no iteration of network simplex is needed.
(c) Russell's Approximation Method
The solution is optimal, no iteration of network simplex is needed.
9.2-8.
(a)
9-18
(b)
(c)
9-19
9-20
Optimal Solution:
, cost: $
.
9.2-9.
(a) Since there is no limit on the electricity and natural gas available, let the supply of
electricity be the sum of demands for electricity, water and space heating and the supply
of natural gas be the sum of demands for water and space heating.
9-21
(b), (c)
9-22
9-23
The optimal solution is to meet 2 units of electricity with electricity, 1 units of space
heating with natural gas, 10 units of water heating with solar heating, and 2 units of
space heating with solar heating. The cost is $2,6 .
9-24
(d), (e) The initial basic feasible solution provides the same optimal solution as in (c).
(f) The initial basic feasible solution provides the same optimal solution as in (c).
(g) The initial BF solution using Vogel's and Russell's methods provides the same
optimal solution as in (c). The optimal solution obtained starting from each of the three
rules is the same. (c) required four iterations of the transportation simplex while (d) and
(f) required none..
9-25
9.2-10.
Vogel's Approximation Method
Optimal Solution:
Quantity
This schedule incurs a cost of
Production Month
million dollars.
9-26
Installation Month
9.2-11.
(a)
(b)
9-27
9-28
The optimal solution is to send shipments from plant 1 to center 3,
to center 4,
from plant 2 to center 2, to center 3,
from plant 3 to center 1 and to center 2. This
has a total cost of $
.
9.2-12.
The optimal solution is to purchase
today, with a cost $
.
pints from Dick tomorrow and
9-29
pints from Harry
9.2-13.
9-30
9-31
9-32
Optimal Solution:
Cost: $
9-33
9.2-14.
Using Russell's approximation method:
9-34
The optimal solution is to ship
units from plant 1 to customer 2,
from plant 2 to
customer 1,
from plant 2 to customer 4,
from plant 3 to customer 3 and 4. This
offers a profit of $
.
9-35
9.2-15.
(a) - (b) - (c)
Using northwest corner rule:
With northwest corner rule, it took
seconds to find the initial BF solution and its
objective value is % above the optimal cost. The two iterations took
seconds.
9-36
Using Vogel's approximation method:
With Vogel's approximation method, it took
seconds to find the initial BF solution and
its objective value is % above the optimal cost. One iteration took
seconds.
Using Russell's approximation method:
9-37
With Russell's approximation method, it took
seconds to find the initial BF solution
and its objective value is % above the optimal cost. The two iterations took
seconds.
Let
denote the initial BF solution. The results are summarized in the following table.
Method
NW Corner
Vogel's
Russell's
Time to Get
seconds
seconds
seconds
Opt. Gap of
%
%
%
9.2-16.
(a) - (b) - (c)
Using northwest corner rule:
9-38
No. Iter.'ns
Time Iter.'ns
seconds
seconds
seconds
Total Time
seconds
seconds
seconds
With northwest corner rule, it took
seconds to find the initial BF solution and its
objective value is % above the optimal cost. The seven iterations took minutes.
Using Vogel's approximation method:
With Vogel's approximation method, it took
seconds to find the initial BF solution and
its objective value is % above the optimal cost. The two iterations took minute.
Using Russell's approximation method:
With Russell's approximation method, it took
seconds to find the initial BF solution
and its objective value is % above the optimal cost. The five iterations took minutes.
Optimal Solution: cost
Let
denote the initial BF solution. The results are summarized in the following table.
Method
NW Corner
Vogel's
Russell's
Time to Get
seconds
seconds
seconds
Opt. Gap of
%
%
%
9-39
No. Iter.'ns
Time Iter.'ns
minutes
minute
minutes
Total Time
seconds
seconds
seconds
9.2-17.
(a) Initial solution using northwest corner rule:
Final tableau: cost
(b)
minimize
subject to
Hence, the transportation simplex method takes one iteration while the general simplex
method takes four iterations. The computation times vary.
9-40
9.2-18.
Let
.
minimize
subject to
Initial simplex tableau:
Simplex tableau:
variables and
Transportation tableau:
constraints
variables and
constraints
Even though the transportation tableau is larger, it requires less work than the simplex
tableau.
9.2-19.
If we multiply the demand constraints by
two nonzero entries, one
and one
equality:
supplies
demands
, each constraint column will have exactly
. Summing all these constraints gives the
,
since the total supply equals the total demand. Hence, there is a redundant constraint.
9-41
9.2-20.
In the initialization step, after selecting the next basic variable, the allocation made is
equal to either the (remaining) supply or demand for that row or column. Since these
quantities are known to be integer, the allocation will be integer.
Given a current BF solution that is integer, step 3 of an iteration adds and subtracts,
around the chain-reaction cycle, the current value of the leaving basic variable. Since we
know this is an integer, and all the other basic variables on the cycle began with integer
values, the new BF solution must be all integer.
During the initialization step, we can select the next basic variable for allocation
arbitrarily from among the rows and columns not already eliminated. Thus, by altering
our selections, we can construct any BF solution as our initial one. Because we have
shown that the initialization step gives integer solutions, all BF solutions must be integer.
9.2-21.
(a) Let
be the number of tons hauled from pit
(North, South) to site
minimize
subject to
Initial tableau:
(b) This table is much smaller than the simplex tableau and it stores the same information.
9-42
.
(c) The solution is not optimal, since
.
(d)
9-43
The optimal solution is to haul
tons from the north pit to site 1 and tons to site 3,
tons from the south pit to site 2 and tons from the south pit to site 3. This incurs a cost
of $
.
(e) From the reduced costs
in the final tableau, we see that
.
If the contractor can negotiate a hauling cost per ton of
or less from the north pit to
site 2, or of
or less from the south pit to site 1, a new solution using these options
would give a cost at least as small as the current optimal cost $
.
9-44
9.2-22.
Variable Cells
Cell
$C$11
$D$11
$E$11
$F$11
$C$12
$D$12
$E$12
$F$12
$C$13
$D$13
$E$13
$F$13
Final Reduced Objective Allowable Allowable
Name
Value Cost
Coefficient Increase Decrease
Colombo River Berdoo
0
0
160
1E+30
0
Colombo River Los Devils
5
0
130
20
1E+30
Colombo River San Go
0
10
220
1E+30
10
Colombo River Hollyglass
0
0
170
0
20
Sacron River Berdoo
2
0
140
0
1E+30
Sacron River Los Devils
0
20
130
1E+30
20
Sacron River San Go
2.5
0
190
10
10
Sacron River Hollyglass
1.5
0
150
20
0
Calorie River Berdoo
0
10
190
1E+30
10
Calorie River Los Devils
0
50
200
1E+30
50
Calorie River San Go
1.5
0
230
10
20
Calorie River Hollyglass
0
-190
0
1E+30
190
Constraints
Cell
$C$14
$D$14
$E$14
$F$14
$G$11
$G$12
$G$13
Final Shadow Constraint Allowable Allowable
Name
Value Price
R.H. Side
Increase Decrease
Total To City Berdoo
2
180
2
2.5
1.5
Total To City Los Devils
5
150
5
0
1.5
Total To City San Go
4
230
4
3.5
1.5
Total To City Hollyglass
1.5
190
1.5
2.5
1.5
Colombo River From River 5
-20
5
1.5
0
Sacron River From River
6
-40
6
1.5
2.5
Calorie River From River
1.5
0
5
1E+30
3.5
(a) The optimal solution would change because the decrease of $
allowable decrease of $ million.
million is outside the
(b) The optimal solution would remain the same, since the allowable increase is
(c) By the
same.
.
% rule for simultaneous changes, the optimal solution must remain the
: $
$
% of allowable decrease
: $
$
% of allowable decrease
These sum up to
(d) By the
remain valid.
%
%
%.
% rule for simultaneous changes, the shadow prices may or may not
:
$
:
$
These sum up to
$
$
% of allowable decrease
%
% of allowable decrease
%
%.
9-45
9.2-23.
(a)
,
The current feasible solution is feasible, but not optimal.
(b)
We can revise the tableau by changing
change to
, to
(reduced cost
(reduced cost
(reduced cost
(reduced cost
(reduced cost
(reduced cost
from
, and
to
to
(reduced cost
9-46
. This causes
.
to
The basic solution remains feasible and optimal.
(c)
The basic solution remains feasible and optimal.
9-47
(d)
and
This solution satisfies the optimality criterion, but it is infeasible.
9.3-1.
(a)
9-48
(b)
(c), (d)
Task
Unit Cost
A
Assignee B
C
D
1
$8
$6
$7
$6
2
$6
$5
$8
$7
1
0
0
0
1
1
=
1
2
1
0
0
0
1
=
1
Demand
4
$7
$4
$6
$6
3
0
0
1
0
1
=
1
4
0
1
0
0
1
=
1
Task
Assignments
A
Assignee B
C
D
Total Assigned
3
$5
$3
$4
$5
Total
Assignments
1
1
1
1
Supply
1
1
1
1
Total Cost
$20
9.3-2.
(a) Ships are assignees and ports are assignments.
(b)
Optimal Solution:
This incurs a cost of $
=
=
=
=
.
9-49
(c)
(d) - (e)
9-50
9-51
One optimal assignment is:
the second port.
, where the first entry is ship and
9-52
(f) Continuing to pivot where reduced costs are zero:
Alternative optimal matching:
Alternative optimal matching:
9-53
Alternative optimal matching:
9.3-3.
(a)
(b) The optimal cost is $34,960.
9-54
(c)
(d)
The initial solution from Vogel's approximation method is optimal. Plant 2 produces
product 3, plant 4 produces product 2, plant 5 produces product 1. This incurs a cost of
$34,960.
9-55
9.3-4.
(a) After adding a dummy stroke, which everyone can swim in zero seconds, the problem
becomes that of assigning swimmers to strokes. The optimal solution turns out to be
the following: David swims the backstroke, Tony swims the breaststroke, Chris swims
the butterfly, and Carl swims the freestyle.
(b) Cost:
9.3-5.
(a)
(b) - (c)
Since all the reduced costs are nonnegative, this solution is optimal.
9-56
(d)
This is identical to the table in (a) except that plants 1 and 2 have been split into two
plants each.
(e)
The basic feasible solution for the transformed problem above corresponds to that given
in part (c).
9-57
9.3-6.
9-58
This solution corresponds to that given in Section 9.3; although the set of basic variables
is different, the values of the variables are the same.
9-59
9.3-7.
(a) Let assignees 1 and 2 represent plant A, assignees 3 and 4 represent plant B, and the
tasks be the distribution centers.
(b) Cost:
(c)
(d)
(e)
9-60
(f)
9.3-8.
(a)
(b)
(c)
(d) A transportation problem of size
has
for the assignment problem, there are
assignments. Thus,
basic variables are degenerate,
problems are always highly degenerate. This can be seen
the OR Courseware.
(e)
and one of
basic and equal zero.
are nonbasic, too.
9-61
basic variables. Since
basic variables, but only
they equal zero. Assignment
using the interactive routine in
and one of
are
Dual variables:
Looking at
are:
, we see that the allowable ranges for this solution to stay optimal
.
9.3-9.
minimize
subject to
for
for
for
The table of constraint coefficients is identical to that for the transportation problem
(Table 9.6). The assignment problem has a more special structure because
and
for every .
9.4-1.
Start with:
9-62
Subtract the minimum element from each element in the column and continue the
algorithm.
One optimal solution is to assign ships
to ports
, with cost
.
9.4-2.
Subtract the minimum element in each row from each element in the row and continue
the algorithm.
One optimal solution is that David swims the backstroke, Tony the breaststroke, Chris
the butterfly and Carl the freestyle. The total time is
.
9-63
9.4-3.
Cost:
9.4-4.
Subtract the minimum element in each row from every element in the row and continue
the algorithm. This gives an optimal solution with cost .
9.4-5.
Subtract the minimum element in each column from every element in the column and
continue the algorithm.
An optimal assignment is
, with cost .
9-64
9.4-6. Subtracting the minimum element in each row from each element in the row, and
then continuing the algorithm, we get:
An optimal assignment is (A, 1), (B, 3), (C, 4), (D, 2). Cost = 16.information is needed to
determine this.
9-65
Case%9.1%
!
Option!1!(Shipping!by!Rail):!
!
!
!
Option!2!(Shipping!by!Ship):!
!
!
9-66
!
Option!3!(Shipping!by!Best!Available!for!each!Route):!
!
!
!
When!comparing!the!three!options,!it!is!best!to!use!the!combination!plan,!while!
shipping!entirely!by!rail!leads!to!the!highest!costs.!
!
If!costs!of!shipping!by!water!are!expected!to!rise!considerably!more!than!for!
shipping!by!rail,!stay!with!rail!and!use!Option!1.!!If!the!reverse!is!true,!then!use!
Option!2.!!If!the!cost!comparisons!will!remain!roughly!the!same,!use!Option!3.!!
Option!3!is!clearly!the!most!feasible!but!may!not!be!chosen!if!it!is!too!logistically!
cumbersome.!!More!knowledge!of!the!situation!is!necessary!to!determine!this.!
%
9-67
Case%9.2%
!
a)! $20!million!is!saved!in!comparison!with!the!results!in!Figure!6.13!by!shipping!20!
million!fewer!barrels!to!Charleston!and!20!million!more!to!St.!Louis.!
B
C
3
4 Unit Cost ($millions)
5
Texas
6
Oil
California
7
Fields
Alaska
8
Middle East
9
10
11 Shipment Quantity
12 (millions of barrels)
13
Texas
14
Oil
California
15
Fields
Alaska
16
Middle East
17
Total Received
18
19
Capacity
20
!
D
E
New Orleans
2
5
5
2
F
Refineries
Charleston
Seattle
4
5
5
3
7
3
3
5
G
St. Louis
1
4
7
4
New Orleans
0
0
20
80
100
<=
100
Refineries
Charleston
Seattle
0
0
0
0
0
80
40
0
40
80
<=
<=
60
80
St. Louis
80
60
0
0
140
<=
150
H
I
J
Total Shipped
80
60
100
120
=
=
=
=
Supply
80
60
100
120
Total Cost
($millions)
940
b)! $40!million!is!saved!in!comparison!with!the!results!in!Figure!6.17.!
B
C
3
4 Unit Cost ($millions)
5
New Orleans
6
Refineries
Charleston
7
Seattle
8
St. Louis
9
10
11 Shipment Quantity
12 (millions of barrels)
13
New Orleans
14 Refineries
Charleston
15
Seattle
16
St. Louis
17
Total Received
18
19
Demand
20
D
Pittsburgh
6.5
7
7
4
E
F
Distribution Center
Atlanta
Kansas City
5.5
6
5
4
8
4
3
1
G
San Francisco
8
7
3
5
Pittsburgh
60
0
0
40
100
=
100
Distribution Center
Atlanta
Kansas City
40
0
40
0
0
0
0
80
80
80
=
=
80
80
San Francisco
0
0
80
20
100
=
100
H
I
J
Total Shipped
100
40
80
140
=
=
=
=
Supply
100
40
80
140
Total Cost
($millions)
1,390
!
!
The!cost!of!shipping!both!crude!oil!and!finished!product!under!this!plan!is!$940!
million!+!$1,390!million!=!$2,330!million!or!$2.33!billion!—!a!savings!of!$60!
million!compared!to!the!original!results!in!Table!6.20.!
9-68
!
!
c)! $35!million!is!saved!in!comparison!with!the!results!in!part!(b).!!
$75!million!is!saved!in!comparison!with!the!results!in!Figure!6.17.!
B
C
3
4 Unit Cost ($millions)
5
New Orleans
6
Refineries
Charleston
7
Seattle
8
St. Louis
9
10
11 Shipment Quantity
12 (millions of barrels)
13
New Orleans
14 Refineries
Charleston
15
Seattle
16
St. Louis
17
Total Received
18
19
Demand
20
!
D
Pittsburgh
6.5
7
7
4
E
F
Distribution Center
Atlanta
Kansas City
5.5
6
5
4
8
4
3
1
G
San Francisco
8
7
3
5
Pittsburgh
50
0
0
50
100
=
100
Distribution Center
Atlanta
Kansas City
20
0
60
0
0
0
0
80
80
80
=
=
80
80
San Francisco
0
0
80
20
100
=
100
H
I
J
Total Shipped
70
60
80
150
<=
<=
<=
<=
Capacity
100
60
80
150
Total Cost
($millions)
1,355
d)! This!solution!costs!$40!million!more!than!the!solution!in!part!(a).!
This!solution!costs!$20!million!more!than!the!solution!is!Figure!6.13.!
B
C
3
4 Unit Cost ($millions)
5
Texas
6
Oil
California
7
Fields
Alaska
8
Middle East
9
10
11 Shipment Quantity
12 (millions of barrels)
13
Texas
14
Oil
California
15
Fields
Alaska
16
Middle East
17
Total Received
18
19
Demand
20
D
E
New Orleans
2
5
5
2
F
Refineries
Charleston
Seattle
4
5
5
3
7
3
3
5
G
St. Louis
1
4
7
4
New Orleans
0
0
10
60
70
=
70
Refineries
Charleston
Seattle
0
0
0
0
0
80
60
0
60
80
=
=
60
80
St. Louis
80
60
10
0
150
=
150
H
I
J
Total Shipped
80
60
100
120
=
=
=
=
Supply
80
60
100
120
Total Cost
($millions)
980
!
!
The!total!cost!of!shipping!both!crude!oil!and!finished!product!under!this!plan!is!
$1,355!million!+!$980!million!=!$2,335!million!or!$2.335!billion.!This!is!$5!
million!more!than!the!cost!of!the!combined!total!obtained!in!part!(b),!but!$55!
million!less!than!the!total!in!Table!6.20.!
9-69
!
!
e)! The!two!transportation!problems!(shipping!to!refineries!and!shipping!to!
distributions!centers)!are!combined!into!a!single!model.!The!amount!shipped!to!
the!refineries!is!constrained!to!be!no!more!than!capacity:!
TotalReceived(D16:G16)!≤!Capacity(D18:G18).!The!total!shipped!out!of!the!
refineries!is!constrained!to!equal!the!total!amount!shipped!in:!
ShippedOut(H31:H34)!=!ShippedIn(J31:J34).!The!goal!is!to!minimize!the!total!
combined!cost!(in!J45)!which!is!the!sum!of!the!two!intermediate!costs!(in!J20!
and!J39).!
!
A
B
C
D
E
F
1 Shipping to Refineries
2
Refineries
3
Unit Cost ($millions)
New Orleans Charleston
Seattle
4
Texas
2
4
5
5
Oil
California
5
5
3
6
Fields
Alaska
5
7
3
7
Middle East
2
3
5
8
9
10
Shipment Quantity
Refineries
11
(millions of barrels)
New Orleans Charleston
Seattle
12
Texas
0
0
0
13
Oil
California
0
0
0
14
Fields
Alaska
20
0
80
15
Middle East
80
30
0
16
Total Received
100
30
80
17
<=
<=
<=
18
Capacity
100
60
80
19
20 Shipping to Distribution Centers
21
22
Unit Cost ($millions)
Pittsburgh
23
New Orleans
6.5
24
Refineries Charleston
7
25
Seattle
7
26
St. Louis
4
27
28
29
Shipment Quantity
30
(millions of barrels)
Pittsburgh
31
New Orleans
100
32
Refineries Charleston
0
33
Seattle
0
34
St. Louis
0
35
Total Received
100
36
=
37
Demand
100
38
39
40
41
42
43
44
45
G
H
I
J
Total Shipped
80
60
100
120
=
=
=
=
Supply
80
60
100
120
St. Louis
1
4
7
4
St. Louis
80
60
0
10
150
<=
150
Cost
(Oil Fields --> Refineries)
($millions)
950
Distribution Center
Atlanta
Kansas City San Francisco
5.5
6
8
5
4
7
8
4
3
3
1
5
Distribution Center
Atlanta
Kansas City San Francisco Shipped Out
0
0
0
100
10
0
20
30
0
0
80
80
70
80
0
150
80
80
100
=
=
=
80
80
100
=
=
=
=
Shipped In
100
30
80
150
Cost
(Refineries --> D.C.'s)
($millions)
1,370
Combined
Total
Cost
($millions)
2,320
!
!
The!total!combined!cost!is!$2,320!million!or!$2.32!billion,!which!is!$10!million!
less!than!in!part!(b),!$15!million!less!than!in!part!(d),!and!$70!million!less!than!in!
Table!6.20.!
9-70
!
f)! If!the!Los!Angeles!refinery!is!chosen!instead,!then!the!combined!shipping!cost!is!
$2,450!million.!
!
A
B
C
1 Shipping to Refineries
2
3
Unit Cost ($millions)
4
Texas
5
Oil
California
6
Fields
Alaska
7
Middle East
8
9
10
Shipment Quantity
11
(millions of barrels)
12
Texas
13
Oil
California
14
Fields
Alaska
15
Middle East
16
Total Received
17
18
Capacity
19
20 Shipping to Distribution Centers
21
22
Unit Cost ($millions)
23
New Orleans
24
Refineries
Charleston
25
Seattle
26
Los Angeles
27
28
29
Shipment Quantity
30
(millions of barrels)
31
New Orleans
32
Refineries
Charleston
33
Seattle
34
St. Louis
35
Total Received
36
37
Demand
38
39
40
41
42
43
44
45
D
E
F
G
New Orleans
2
5
5
2
Refineries
Charleston
Seattle
4
5
5
3
7
3
3
5
Los Angeles
3
1
4
4
New Orleans
40
0
0
60
100
<=
100
Refineries
Charleston
Seattle
0
0
0
0
0
80
60
0
60
80
<=
<=
60
80
St. Louis
40
60
20
0
120
<=
150
H
I
J
Total Shipped
80
60
100
120
=
=
=
=
Supply
80
60
100
120
Cost
(Oil Fields --> Refineries)
($millions)
880
Pittsburgh
6.5
7
7
8
Distribution Center
Atlanta
Kansas City
5.5
6
5
4
8
4
6
3
San Francisco
8
7
3
2
Pittsburgh
80
0
20
0
100
=
100
Distribution Center
Atlanta
Kansas City
20
0
60
0
0
60
0
20
80
80
=
=
80
80
San Francisco
0
0
0
100
100
=
100
Shipped Out
100
60
80
120
=
=
=
=
Shipped In
100
60
80
120
Cost
(Refineries --> D.C.'s)
($millions)
1,570
Combined
Total
Cost
($millions)
2,450
9-71
!
!
!
If!the!Galveston!refinery!is!chosen!instead,!then!the!combined!shipping!cost!is!
$2,470!million.!
!
A
B
C
D
1 Shipping to Refineries
2
3
Unit Cost ($millions)
4
Texas
5
Oil
California
6
Fields
Alaska
7
Middle East
8
9
10
Shipment Quantity
11
(millions of barrels)
12
Texas
13
Oil
California
14
Fields
Alaska
15
Middle East
16
Total Received
17
18
Capacity
19
20 Shipping to Distribution Centers
21
22
Unit Cost ($millions)
23
New Orleans
24
Refineries
Charleston
25
Seattle
26
Galveston
27
28
29
Shipment Quantity
30
(millions of barrels)
31
New Orleans
32
Refineries
Charleston
33
Seattle
34
Galveston
35
Total Received
36
37
Demand
38
39
40
41
42
43
44
45
!
!
E
F
New Orleans
2
5
5
2
Galveston
1
3
5
3
New Orleans
0
0
10
90
100
<=
100
Refineries
Charleston
Seattle
0
0
0
0
0
80
30
0
30
80
<=
<=
60
80
Galveston
80
60
10
0
150
<=
150
H
I
J
Total Shipped
80
60
100
120
=
=
=
=
Supply
80
60
100
120
Cost
(Oil Fields --> Refineries)
($millions)
870
Pittsburgh
6.5
7
7
5
Distribution Center
Atlanta
Kansas City
5.5
6
5
4
8
4
4
3
San Francisco
8
7
3
6
Pittsburgh
100
0
0
0
100
=
100
Distribution Center
Atlanta
Kansas City
0
0
0
30
0
0
80
50
80
80
=
=
80
80
San Francisco
0
0
80
20
100
=
100
Shipped Out
100
30
80
150
=
=
=
=
Shipped In
100
30
80
150
Cost
(Refineries --> D.C.'s)
($millions)
1,600
Combined
Total
Cost
($millions)
2,470
!
!
!
Site!
Total!Cost!
of!Shipping!
Crude!Oil!
Total!Cost!
of!Shipping!
Finished!Product!
Operating!Cost!
for!New!
Refinery!
Total!
Variable!
Cost!
Los!Angeles!
$880!million!
$1.57!billion!
$620!million!
$3.07!billion!
Galveston!
870!million!
1.60!billion!
570!million!
3.12!billion!
St.!Louis!
950!million!
1.37!billion!
530!million!
2.92!billion!
!
!
G
Refineries
Charleston
Seattle
4
5
5
3
7
3
3
5
g)! Answers!will!vary.!
!
9-72
!
Case%9.3%
!
a)! Assign!one!scientist!to!each!of!the!five!projects!to!maximize!the!total!number!of!
bid!points.!
!
A
B
C
D
E
F
Dr. Kvaal
Dr. Zuner
Dr. Tsai
Dr. Mickey
Dr. Rollins
Project Up
100
0
100
267
100
Project Stable
400
200
100
153
33
Project Choice
200
800
100
99
33
Project Hope
200
0
100
451
34
Project Release
100
0
600
30
800
8
9 Assignment
10
Dr. Kvaal
11
Dr. Zuner
12
Dr. Tsai
13
Dr. Mickey
14
Dr. Rollins
15
Total Assigned
16
17
Demand
Project Up
0
0
1
0
0
1
=
1
Project Stable
1
0
0
0
0
1
=
1
Project Choice
0
1
0
0
0
1
=
1
Project Hope
0
0
0
1
0
1
=
1
Project Release
0
0
0
0
1
1
=
1
1
2
3
4
5
6
7
Bid
G
H
I
Total
Assignments
1
1
1
1
1
=
=
=
=
=
Supply
1
1
1
1
1
Total Bid Points
2551
!
!
To!maximize!the!scientists!preferences!you!want!to!assign!Dr.!Tsai!to!lead!
project!Up,!Dr.!Kvaal!to!lead!project!Stable,!Dr.!Zuner!to!lead!project!Choice,!Dr.!
Mickey!to!lead!project!Hope,!and!Dr.!Rollins!to!lead!project!Release.!
!
!
b)! Since!there!are!only!four!assignees,!we!introduce!a!dummy!assignee!with!
preferences!of!–1.!The!task!that!gets!assigned!the!dummy!assignee!will!not!be!
done.!!
!
!
!
Project!Up!would!not!be!done.!
9-73
!
c)! Since!two!of!the!assignees!can!do!two!tasks,!we!need!to!double!them.!We!include!
assignees!Zunerg1,!Zunerg2,!Mickeyg1,!and!Mickeyg2!into!the!problem.!In!order!
to!have!an!equal!number!of!assignees!and!tasks!we!also!need!to!include!one!
dummy!task.!In!order!to!ensure!that!neither!Dr.!Kvall!nor!Dr.!Tsai!can!get!
assigned!the!dummy!task!and!thus!no!project,!we!insert!a!large!negative!number!
as!their!point!bid!for!the!dummy!project.!!
!
!
Dr.!Kvaal!leads!project!Stable,!Dr.!Zuner!leads!project!Choice,!Dr.!Tsai!leads!
project!Release,!and!Dr.!Mickey!leads!the!projects!Hope!and!Up.!
!
!
d)! Under!the!new!bids!of!Dr.!Zuner!the!assignment!does!not!change:!
!
!
!
e)! Certainly!Dr.!Zuner!could!be!disappointed!that!she!is!not!assigned!to!project!
Stable,!especially!when!she!expressed!a!higher!preference!for!that!project!than!
the!scientist!assigned.!The!optimal!solution!maximizes!the!preferences!overall,!
but!individual!scientists!may!be!disappointed.!We!should!therefore!make!sure!to!
communicate!the!reasoning!behind!the!assignments!to!the!scientists.!
9-74
!
f)! Whenever!a!scientist!cannot!lead!a!particular!project!we!use!a!large!negative!
number!as!the!point!bid.!
!
!
!
Dr.!Kvaal!leads!project!Stable,!Dr.!Zuner!leads!project!Choice,!Dr.!Tsai!leads!
project!Release,!Dr.!Mickey!leads!project!Up,!and!Dr.!Rollins!leads!project!Hope.!
!
g)! When!we!want!to!assign!two!assignees!to!the!same!task!we!need!to!duplicate!
that!task.!
!
!
!
Project!Up!is!led!by!Dr.!Mickey,!Stable!by!Dr.!Kvaal,!Choice!by!Dr.!Zuner,!Hope!by!
Dr.!Arriaga!and!Dr.!Santos,!and!Release!by!Dr.!Tsai!and!Dr.!Rollins.!
!
h)! No.!Maximizing!overall!preferences!does!not!maximize!individual!preferences.!
Scientists!who!do!not!get!their!first!choice!may!become!resentful!and!therefore!
lack!the!motivation!to!lead!their!assigned!project.!For!example,!in!the!optimal!
solution!of!part!(g),!Dr.!Santos!clearly!elected!project!Release!as!his!first!choice,!
but!he!was!assigned!to!lead!project!Hope.!
!
In!addition,!maximizing!preferences!ignores!other!considerations!that!should!be!
factored!into!the!assignment!decision.!For!example,!the!scientist!with!the!highest!
preference!for!a!project!may!not!be!the!scientist!most!qualified!to!lead!the!
project.!
9-75
CHAPTER
: NETWORK OPTIMIZATION MODELS
10.2-1.
(a)
(b)
Directed path:
Undirected paths:
AD-DC-CE-EF (A
D
AD-FD (A
D
F)
CA-CE-EF (A
C
E
AD-ED-EF (A
D
E
C
E
F)
F)
F)
Directed cycles:
AD-DC-CA
DC-CE-ED
DC-CE-EF-FD
Undirected cycle that includes every node: CA-CE-EF-FD-DB-AB
(c) {CA, CE, DC, FD, DB} is a spanning tree.
(d)
10.3-1.
Prior to this study, Canadian Pacific Railway (CPR) used to run trains only after a
sufficient level of freight was attained. This policy resulted in unreliable delivery times,
so poor customer service. In order to improve customer service and utilization of
available resources, CPR designed the railway operating plan called Integrated Operating
Plan (IOP). "The problem of designing a railway operating plan is to satisfy a set of
customer requirements expressed in terms of origin-destination traffic movements, using
a blocking plan and a train plan. Thus, the primary variables are the blocks and trains.
The constraints are the capacities of the lines and yards, the customer-service
requirements, and the availability of various assets, such as crews and locomotives. The
objective function in an abstract sense is to maximize profits" [p. 8].
Developing the blocking plan, i.e., determining the group of railcars to move together at
some point during their trips, involves solving a series of shortest-path problems over a
directed graph. The train plan is based on the blocking plan. It includes departure and
arrival times for the trains, blocks they pick up and crew schedules. This problem is
solved for each train using heuristics. Following this, simulation models and locomotive
cycle plans are developed.
This study enabled CPR to save $170 million in half a year. "Total documented cost
savings through the end of 2002 have exceeded half a billion dollars" [p. 12]. More
savings are expected in following years. The improvements in CPR's profitability and
operations can be attributed to the decrease in transit and dwelling times, lowered fuel
consumption, reduction of the workforce and of the number of railcars, and balanced
workloads. CPR can now schedule the trains and the crew more efficiently and provide a
more reliable customer service. By allowing variability in the parameters of its plans,
CPR gained flexibility and agility. It can now respond to disruptions more effectively by
shifting resources quickly. These improvements earned CPR many awards and more
importantly a significant competitive advantage.
-1
10.3-2.
(a)
(b)
The shortest path from the origin to the destination is O
a total distance of 160 miles.
-2
A
B
E
D
T, with
(c)
From
Origin
Origin
Origin
A
A
B
B
B
B
C
C
C
D
D
D
D
E
E
To
On Route
A
1
B
0
C
0
B
1
D
0
A
0
C
0
D
0
E
1
B
0
N
0
E
0
A
0
B
0
E
0
Destination
1
D
1
Destination
0
Total Distance (miles)
Distance (miles)
40
60
50
10
70
10
20
55
40
20
20
50
70
55
10
60
10
80
Nodes Net Flow
Origin
1
A
0
B
0
C
0
D
0
E
0
Destination
-1
=
=
=
=
=
=
=
Supply/Demand
1
0
0
0
0
0
-1
160
(d) Yes.
(e) Yes.
10.3-3.
(a) The nodes represent the years. Let
be the cost (in thousand dollars) of using the
same tractor from the end of year to the end of year .
(b)
The minimum-cost strategy is to replace the tractor at the end of the first year and keep
the new one until the end of the third year. This incurs a total cost of $29 thousand.
-3
(c)
From
Year 0
Year 0
Year 0
Year 1
Year 1
Year 2
To
Year 1
Year 2
Year 3
Year 2
Year 3
Year 3
On Route
1
0
0
0
1
0
Cost
$8,000
$18,000
$31,000
$10,000
$21,000
$12,000
Nodes
Year 0
Year 1
Year 2
Year 3
Total Cost $29,000
10.3-4.
(a) Length of the shortest path: 16
(b) Length of the shortest path: 17
-4
Net Flow
1
0
0
-1
=
=
=
=
Supply/Demand
1
0
0
-1
10.3-5.
The shortest-path problem is a minimum cost flow problem with a unit supply at the
origin and a unit demand at the destination. Label the origin as node and the destination
as node . Then, the LP formulation is as follows:
minimize
subject to
, for
, for
.
10.3-6.
(a) The flying times play the role of "distances."
(b) Shortest path: SE
C
E
LN, with total flight time 11.3
-5
(c)
From
Seattle
Seattle
Seattle
A
A
B
B
B
C
C
D
E
F
To
On Route
A
0
B
0
C
1
D
0
E
0
D
0
E
0
F
0
E
1
F
0
London
0
London
1
London
0
Total Time (hours)
Time (hours)
4.6
4.7
4.2
3.5
3.4
3.6
3.2
3.3
3.5
3.4
3.4
3.6
3.8
Nodes Net Flow
Seattle
1
A
0
B
0
C
0
D
0
E
0
F
0
London
-1
=
=
=
=
=
=
=
=
Supply/Demand
1
0
0
0
0
0
0
-1
11.3
10.3-7.
(a) Let node
denote phase being completed with million dollars left to spent and
be the time to complete phase
if a cost of
million dollars is spent.
-6
(b) Shortest path:
30
Phase
Research
Development
Design
Production
21
Level
Crash
Priority
Crash
Priority
3
Cost
9
6
12
3
15
Time
2
3
3
2
-7
3
3
,0
0
T. Time= 10 months.
10.4-1.
(a) Length: 18
(b) Length: 26
10.4-2.
(a) The nodes represent the groves and the branches represent the roads.
(b) Length: 5.2
10.4-3.
(a) The nodes are Main Office, Branch 1, Branch 2, Branch 3, Branch 4, and Branch 5.
The branches are the phones lines.
(b)
10.5-1.
Maximum flow: 9
-8
10.5-2.
Let node
be the source and node
be the sink.
maximize
subject to
, for
, where
if
10.5-3.
(a)
(b) Maximum flow = 395
-9
is not a branch.
(c) Maximum flow = 395
From
R1
R1
R2
R2
R2
R3
R3
A
A
B
B
B
C
C
D
E
F
To
A
B
A
B
C
B
C
D
E
D
E
F
E
F
T
T
T
Maximum Flow
Ship
65
30
40
50
60
80
70
60
45
60
55
45
45
85
120
145
130
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
Capacity
75
65
40
50
60
80
70
60
45
70
55
45
70
90
120
190
130
Nodes Net Flow
R1
95
R2
150
R3
150
A
0
B
0
C
0
D
0
E
0
F
0
T
-395
395
10.5-4.
(a)
(b)
-10
Supply/Demand
=
=
=
=
=
=
0
0
0
0
0
0
(c)
(d)
-11
(e)
From
Texas
Texas
Texas
Texas
California
California
California
California
Alaska
Alaska
Alaska
Alaska
Middle East
Middle East
Middle East
Middle East
New Orleans
New Orleans
New Orleans
New Orleans
Charleston
Charleston
Charleston
Charleston
Seattle
Seattle
Seattle
Seattle
St. Louis
St. Louis
St. Louis
St. Louis
To
New Orleans
Charleston
Seattle
St. Louis
New Orleans
Charleston
Seattle
St. Louis
New Orleans
Charleston
Seattle
St. Louis
New Orleans
Charleston
Seattle
St. Louis
Pittsburgh
Atlanta
Kansas City
San Francisco
Pittsburgh
Atlanta
Kansas City
San Francisco
Pittsburgh
Atlanta
Kansas City
San Francisco
Pittsburgh
Atlanta
Kansas City
San Francisco
Ship
11
7
2
8
5
4
8
7
7
3
12
6
1
9
3
15
5
9
6
4
8
7
3
5
4
6
7
8
9
11
9
7
Maximum Flow
108
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
Capacity
11
7
2
8
5
4
8
7
7
3
12
6
8
9
4
15
5
9
6
4
8
7
9
5
4
6
7
8
12
11
9
7
Nodes
Net Flow
Texas
28
California
24
Alaska
28
Middle East
28
New Orleans
0
Charleston
0
Seattle
0
St. Louis
0
Pittsburgh
-26
Atlanta
-33
Kansas City
-25
San Francisco
-24
Supply/Demand
=
=
=
=
0
0
0
0
10.5-5.
For convenience, call the Faireparc station siding and the Portstown station siding
.
Let node
represent siding
at time
for
and
. Node
is the source and node
is the sink.
Arcs with unit capacity exist between nodes
and
if and only if a freight
train leaving siding at time could not be overtaken by a scheduled passenger train
before it reached siding
. Arcs with capacity
exist between nodes
and
for
. There are no other arcs. For example, if
and a scheduled passenger train could overtake a freight train leaving siding at time
before it reached siding
, the following is part of the network:
The maximum flow problem in this case maximizes the number of sent freight trains.
-12
10.5-6.
(a)
(b)
From
A
A
B
B
C
C
D
D
E
To
B
C
D
E
D
E
E
F
F
Maximum Flow
Ship
8
7
7
1
1
6
2
6
9
<=
<=
<=
<=
<=
<=
<=
<=
<=
Capacity
9
7
7
2
4
6
3
6
9
Nodes Net Flow
A
15
B
0
C
0
D
0
E
0
F
-15
Supply/Demand
=
=
=
=
0
0
0
0
15
10.5.7
Answers will vary.
10.5.8
Answers will vary.
10.6-1.
In this study, flight delay and cancellation problems faced by United Airlines (UA) are
modeled as minimum-cost-flow network models. The overall objective is to minimize a
weighted sum of various measures related to delay. These include the total number of
delay minutes for every passenger, the number of passengers affected by delays and the
number of aircraft swaps. Nodes represent "arriving and departing aircraft, spare aircraft,
and recovered aircraft" on a two-dimensional network, with time and airport being the
two dimensions. Arcs represent "scheduled flights, connections, and aircraft
substitutions" [p. 56]. Costs include the revenue loss, the costs from swapping aircraft
and from delaying aircraft.
The delay problem is solved for each airport separately as a minimum-cost-flow network
problem. The flow on each arc can be at most one. The solution is a set of arcs starting at
a supply node and ending at a demand node, which determines flight delays due to
shortage in aircraft. The cancellation model is a minimum-cost-flow network problem on
the entire network. Again, the flow on each arc cannot exceed one. The solution
determines which flight is canceled and what flight its aircraft is assigned to.
This study has saved UA over half a billion dollars in delay costs alone in less than a
year. Many potential delays were prevented and hence the number of flight delays was
reduced by 50%. Customer inconveniences due to delays and cancellations were reduced.
Additionally, developing an efficient way of addressing these problems helped UA
respond to changes in the conditions quickly.
-13
10.6-2.
10.6-3.
(a)
(b)
minimize
subject to
10.6-4.
-14
10.6-5.
(a)
(b)
(c) Total cost: $488,125
From
P1
P1
P2
P2
WH1
WH1
WH1
WH2
WH2
WH2
To
WH1
WH2
WH1
WH2
RO1
RO2
RO3
RO1
RO2
RO3
Ship
125
75
125
175
100
50
100
50
150
50
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
Capacity
125
150
175
200
100
150
100
125
150
75
Unit Cost
$425
$560
$510
$600
$470
$505
$490
$390
$410
$440
Total Cost $488,125
10.6-6.
(a)
-15
Nodes
P1
P2
WH1
WH2
RO1
RO2
RO3
Net Flow
200
300
0
0
-150
-200
-150
=
=
=
=
=
=
=
Supply/Demand
200
300
0
0
-150
-200
-150
(b)
(c)
Vendor
1
2
3
Price
$22,500
$22,700
$22,300
From
V1
V1
V2
V2
V3
V3
WH1
WH1
WH2
WH2
V1
V2
V3
To
WH1
WH2
WH1
WH2
WH1
WH2
F1
F2
F1
F2
D
D
D
Fixed
Per
Shipping
Mile
Charge Charge
$300
$0.40
$200
$0.50
$500
$0.20
Ship
0
6
6
0
0
4
6
0
4
6
4
4
6
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
Miles to
WH1
1600
500
2000
Miles to
WH2
400
600
1000
Capacity
6
6
6
6
6
6
6
6
6
6
Unit Cost
$23,440
$22,960
$23,150
$23,200
$23,200
$23,000
$200
$700
$400
$500
$0
$0
$0
Total Cost $374,460
10.7-1.
(a)
-16
Total
Cost to
WH1
$23,440
$23,150
$23,200
Total
Cost to
WH2
$22,960
$23,200
$23,000
Nodes
V1
V2
V3
WH1
WH2
F1
F2
D
Net Flow
10
10
10
0
0
-10
-6
-14
=
=
=
=
=
=
=
=
Supply/Demand
10
10
10
0
0
-10
-6
-14
(b) Compute
for nonbasic arcs:
BD
AD
CB
All of them are nonnegative, so this solution is optimal. Since
exist. Network simplex:
Optimal nonbasic solutions have AB
, where
and C
DE
, AC
B and B
AD
, multiple optima
, AD
, CE
D are nonbasic arcs.
(c) Start with:
Network simplex:
AC
entering arc
AB
BD
and BC is leaving arc (reverses). The next BF solution is:
From (b), we recognize this solution as optimal.
-17
, and
10.7-2.
(a)
(b) The final feasible spanning tree is:
The flow to which it corresponds is the same as in Prob. 10.5-6.
10.7-3.
(a) There are no reverse arcs in this solution.
(b) The optimal BF spanning tree is:
which corresponds to a real flow of:
with cost
.
-18
10.7-4.
Initial BF spanning tree:
Optimal BF spanning tree:
which has a real flow of:
with cost
.
10.7-5.
Initial BF spanning tree:
-19
Optimal BF spanning tree:
which corresponds to the optimal solution given in Sec. 8.1.
10.7-6.
(a)
(b) Initial BF spanning tree:
(c) Optimal BF spanning tree:
The sequence of basic feasible solutions is identical with the transportation simplex
method.
-20
10.7-7.
Optimal BF spanning tree:
which correspond to the real flow of:
with a total cost of 750.
10.8-1.
Length of Path
Activity to Crash
Crash Cost
$
$
$
$
$
$
$
-21
10.8-2.
(a) Let
and
be the reduction in
and
respectively, due to crashing.
minimize
subject to
and
Optimal Solution:
(b) Let
and
and
be the reduction in
.
and
respectively, due to crashing.
minimize
subject to
and
Optimal Solution:
and
.
-22
(c) Let
,
,
, and
respectively, due to crashing.
be the reduction in the duration of
,
,
, and
minimize
subject to
and
Optimal Solution:
and
.
(d) Let
be the reduction in the duration of activity
due to crashing for
. Also let
denote the start time of activity for
and FINISH
the project duration.
minimize
subject to
FINISH
FINISH
and
FINISH
FINISH
(e)
Normal
Crash
Time
Time
Activity (months) (months)
A
8
5
B
9
7
C
6
4
D
7
4
Normal
Cost
$25,000
$20,000
$16,000
$27,000
Crash
Cost
$40,000
$30,000
$24,000
$45,000
Maximum Time
Reduction
(months)
3
2
2
3
Crash Cost
per Month
Saved
$5,000
$5,000
$4,000
$6,000
Project Completion Time (months)
Start
Time
0
0
8
7
Time
Reduction
0
2
2
2
Finish
Time
8
7
12
12
12
<=
Max Time
12
Total Cost $118,000
-23
(f) The solution found using LINGO agrees with the solution in (e), i.e., it is optimal to
reduce the duration of activities , , and
by two months. Then the entire project
takes 12 months and costs
thousand dollars.
(g)
Deadline of
Activity
A
B
C
D
Normal
Time
(months)
8
9
6
7
months
Crash
Time
(months)
5
7
4
4
Normal
Cost
$25,000
$20,000
$16,000
$27,000
Crash
Cost
$40,000
$30,000
$24,000
$45,000
Maximum Time Crash Cost
Reduction
per Month
(months)
Saved
3
$5,000
2
$5,000
2
$4,000
3
$6,000
Project Completion Time (months)
Start
Time
0
0
7
7
Time
Reduction
1
2
2
3
Finish
Time
7
7
11
11
11
<=
Max Time
11
Start
Time
0
0
8
7
Time
Reduction
0
2
1
1
Finish
Time
8
7
13
13
13
<=
Max Time
13
Total Cost $129,000
Deadline of
Activity
A
B
C
D
Normal
Time
(months)
8
9
6
7
months
Crash
Time
(months)
5
7
4
4
Normal
Cost
$25,000
$20,000
$16,000
$27,000
Crash
Cost
$40,000
$30,000
$24,000
$45,000
Maximum Time Crash Cost
Reduction
per Month
(months)
Saved
3
$5,000
2
$5,000
2
$4,000
3
$6,000
Project Completion Time (months)
Total Cost $108,000
-24
10.8-3.
(a) $7,834 is saved by the new plan given below.
Length of Path
Activity to Crash
$
$
$
$
&
&
Activity
Crash Cost
Duration
weeks
weeks
weeks
weeks
weeks
Cost
$
$
$
$
$
(b)
Activity
A
B
C
D
E
Normal
Time
(weeks)
3
4
5
3
4
Crash
Time
(weeks)
2
3
2
1
2
Normal
Cost
$54,000
$62,000
$66,000
$40,000
$75,000
Crash
Cost
$60,000
$65,000
$70,000
$43,000
$80,000
Maximum Time Crash Cost
Reduction
per Week
(weeks)
Saved
1
$6,000
1
$3,000
3
$1,333
2
$1,500
2
$2,500
Project Completion Time (weeks)
Start
Time
0
4
3
9
8
Time
Reduction
0
0
0
0
0
Finish
Time
3
8
8
12
12
12
<=
Max Time
12
Start
Time
0
3
3
8
7
Time
Reduction
0
0
1
0
0
Finish
Time
3
7
7
11
11
11
<=
Max Time
11
Total Cost $297,000
Activity
A
B
C
D
E
Normal
Time
(weeks)
3
4
5
3
4
Crash
Time
(weeks)
2
3
2
1
2
Normal
Cost
$54,000
$62,000
$66,000
$40,000
$75,000
Crash
Cost
$60,000
$65,000
$70,000
$43,000
$80,000
Maximum Time Crash Cost
Reduction
per Week
(weeks)
Saved
1
$6,000
1
$3,000
3
$1,333
2
$1,500
2
$2,500
Project Completion Time (weeks)
Total Cost $298,333
-25
Activity
A
B
C
D
E
Normal
Time
(weeks)
3
4
5
3
4
Crash
Time
(weeks)
2
3
2
1
2
Normal
Cost
$54,000
$62,000
$66,000
$40,000
$75,000
Crash
Cost
$60,000
$65,000
$70,000
$43,000
$80,000
Maximum Time Crash Cost
Reduction
per Week
(weeks)
Saved
1
$6,000
1
$3,000
3
$1,333
2
$1,500
2
$2,500
Project Completion Time (weeks)
Start
Time
0
3
3
7
7
Time
Reduction
0
0
1
0
1
Finish
Time
3
7
7
10
10
10
<=
Max Time
10
Start
Time
0
3
3
7
7
Time
Reduction
0
0
1
1
2
Finish
Time
3
7
7
9
9
9
<=
Max Time
9
Start
Time
0
3
3
6
6
Time
Reduction
0
1
2
1
2
Finish
Time
3
6
6
8
8
8
<=
Max Time
8
Start
Time
0
2
2
5
5
Time
Reduction
1
1
2
1
2
Finish
Time
2
5
5
7
7
7
<=
Max Time
7
Total Cost $300,833
Activity
A
B
C
D
E
Normal
Time
(weeks)
3
4
5
3
4
Crash
Time
(weeks)
2
3
2
1
2
Normal
Cost
$54,000
$62,000
$66,000
$40,000
$75,000
Crash
Cost
$60,000
$65,000
$70,000
$43,000
$80,000
Maximum Time Crash Cost
Reduction
per Week
(weeks)
Saved
1
$6,000
1
$3,000
3
$1,333
2
$1,500
2
$2,500
Project Completion Time (weeks)
Total Cost $304,833
Activity
A
B
C
D
E
Normal
Time
(weeks)
3
4
5
3
4
Crash
Time
(weeks)
2
3
2
1
2
Normal
Cost
$54,000
$62,000
$66,000
$40,000
$75,000
Crash
Cost
$60,000
$65,000
$70,000
$43,000
$80,000
Maximum Time Crash Cost
Reduction
per Week
(weeks)
Saved
1
$6,000
1
$3,000
3
$1,333
2
$1,500
2
$2,500
Project Completion Time (weeks)
Total Cost $309,167
Activity
A
B
C
D
E
Normal
Time
(weeks)
3
4
5
3
4
Crash
Time
(weeks)
2
3
2
1
2
Normal
Cost
$54,000
$62,000
$66,000
$40,000
$75,000
Crash
Cost
$60,000
$65,000
$70,000
$43,000
$80,000
Maximum Time Crash Cost
Reduction
per Week
(weeks)
Saved
1
$6,000
1
$3,000
3
$1,333
2
$1,500
2
$2,500
Project Completion Time (weeks)
Total Cost $315,167
Crash to
weeks.
-26
10.8-4.
(a) Let
.
be the reduction in the duration of activity and
minimize
5
0
4
8
be the start time of activity
6
7
subject to
FINISH
FINISH
FINISH
(b) Finish Time:
Activity
A
B
C
D
E
F
G
H
Normal
Time
(weeks)
5
3
4
6
5
7
9
8
weeks, total cost: $217 million.
Crash
Normal
Crash Maximum Time
Time
Cost
Cost
Reduction
(weeks) ($million) ($million)
(weeks)
3
20
30
2
2
10
20
1
2
16
24
2
3
25
43
3
4
22
30
1
4
30
48
3
5
25
45
4
6
30
44
2
Crash Cost
per Week
Saved
($million)
5
10
4
6
8
6
5
7
Project Completion Time (weeks)
Total Cost ($million)
Start
Time
0
0
3
3
2
2
7
9
Time
Reduction
2
1
0
0
0
0
1
2
Finish
Time
3
2
7
9
7
9
15
15
15
<=
Max Time
15
217
10.8-5.
(a) Let
.
be the reduction in the duration of activity and
be the start time of activity
minimize
subject to
FINISH
FINISH
FINISH
FINISH
-27
(b) Finish Time:
Activity
A
B
C
D
E
F
G
H
I
J
Normal
Time
(weeks)
32
28
36
16
32
54
17
20
34
18
weeks, total crashing cost: $
Crash
Time
(weeks)
28
25
31
13
27
47
15
17
30
16
Normal
Crash Maximum Time
Cost
Cost
Reduction
($million) ($million)
(weeks)
160
180
4
125
146
3
170
210
5
60
72
3
135
160
5
215
257
7
90
96
2
120
132
3
190
226
4
80
84
2
million, total cost: $
Crash Cost
per Week
Saved
($million)
5
7
8
4
5
6
3
4
9
2
Project Completion Time (weeks)
Total Cost ($million)
10.9-1.
Answers will vary.
10.9-2.
Answers will vary.
-28
billion.
Start
Time
0
0
32
25
26
25
41
58
58
76
Time
Reduction
0
3
0
0
0
3
0
0
0
2
Finish
Time
32
25
68
41
58
76
58
78
92
92
92
<=
Max Time
92
1,388
Case%10.1%
!
a)! There!are!three!supply!nodes!–!the!Yen!node,!the!Rupiah!node,!and!the!Ringgit!
node.!!There!is!one!demand!node!–!the!US$!node.!!Below,!we!draw!the!network!
originating!from!only!the!Yen!supply!node!to!illustrate!the!overall!design!of!the!
network.!!In!this!network,!we!exclude!both!the!Rupiah!and!Ringgit!nodes!for!
simplicity.!
-9.6 mil
!
!
b)! Since!all!transaction!limits!are!given!in!the!equivalent!of!$1000!we!define!the!
flow!variables!as!the!amount!in!thousands!of!dollars!that!Jake!converts!from!one!
currency!into!another!one.!His!total!holdings!in!Yen,!Rupiah,!and!Ringgit!are!
equivalent!to!$9.6!million,!$1.68!million,!and!$5.6!million,!respectively!(as!
calculated!in!cells!I16:K18!in!the!spreadsheet).!So,!the!supplies!at!the!supply!
nodes!Yen,!Rupiah,!and!Ringgit!are!Q$9.6!million,!Q$1.68!million,!and!Q$5.6!
million,!respectively.!The!demand!at!the!only!demand!node!US$!equals!$16.88!
million!(the!sum!of!the!outflows!from!the!source!nodes).!The!transaction!limits!
are!capacity!constraints!for!all!arcs!leaving!from!the!nodes!Yen,!Rupiah,!and!
Ringgit.!The!unit!cost!for!every!arc!is!given!by!the!transaction!cost!for!the!
currency!conversion.!
10-29
!
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
A
B
C
D
From
Yen
Yen
Yen
Yen
Yen
Yen
Yen
Rupiah
Rupiah
Rupiah
Rupiah
Rupiah
Rupiah
Rupiah
Ringit
Ringit
Ringit
Ringit
Ringit
Ringit
Ringit
Can$
Can$
Can$
Can$
Euro
Euro
Euro
Euro
Pound
Pound
Pound
Pound
Peso
Peso
Peso
Peso
To
Rupiah
Ringit
US$
Can$
Euro
Pound
Peso
Yen
Ringit
US$
Can$
Euro
Pound
Peso
Yen
Rupiah
US$
Can$
Euro
Pound
Peso
US$
Euro
Pound
Peso
US$
Can$
Pound
Peso
US$
Can$
Euro
Peso
US$
Can$
Euro
Pound
Convert
($thousands)
0
0
2,000
2,000
2,000
2,000
1,600
0
0
200
200
1,000
280
0
0
0
1,100
0
2,500
1,000
1,000
2,200
0
0
0
5,500
0
0
0
3,280
0
0
0
2,600
0
0
0
Total Cost
83.38
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
E
Transaction
Limit
($thousands)
5,000
5,000
2,000
2,000
2,000
2,000
4,000
5,000
2,000
200
200
1,000
500
200
3,000
4,500
1,500
1,500
2,500
1,000
1,000
F
G
Unit
Cost
0.50%
0.50%
0.40%
0.40%
0.40%
0.25%
0.50%
0.50%
0.70%
0.50%
0.30%
0.30%
0.75%
0.75%
0.50%
0.70%
0.70%
0.70%
0.40%
0.45%
0.50%
0.05%
0.20%
0.10%
0.10%
0.10%
0.20%
0.05%
0.50%
0.10%
0.10%
0.05%
0.50%
0.10%
0.10%
0.50%
0.50%
H
I
J
K
Nodes
Yen
Rupiah
Ringit
Can$
Euro
Pound
Peso
US$
Net Flow
($thousands)
9,600
1,680
5,600
0
0
0
0
-16,880
=
=
=
=
=
=
=
=
Supply/Demand
($thousands)
9,600
1,680
5,600
0
0
0
0
-16,880
Yen
Rupiah
Ringgit
Starting
Supply
(thousands)
1,200,000
10,500,000
28,000
Conversion
($ per)
0.008
0.00016
0.2
Starting
($thousands)
9,600
1,680
5,600
!
!
Range Name
Conversion
Convert
From
NetFlow
Nodes
StartingSupply
SupplyDemand
To
TotalCost
TransactionLimit
UnitCost
!
Cells
J16:J18
C4:C40
A4:A40
I4:I11
H4:H11
I16:I18
K4:K11
B4:B40
C42
E4:E24
F4:F40
!
2
3
4
5
6
7
8
9
10
11
I
Net Flow
($thousands)
=SUMIF(From,Nodes,Convert)-SUMIF(To,Nodes,Convert)
=SUMIF(From,Nodes,Convert)-SUMIF(To,Nodes,Convert)
=SUMIF(From,Nodes,Convert)-SUMIF(To,Nodes,Convert)
=SUMIF(From,Nodes,Convert)-SUMIF(To,Nodes,Convert)
=SUMIF(From,Nodes,Convert)-SUMIF(To,Nodes,Convert)
=SUMIF(From,Nodes,Convert)-SUMIF(To,Nodes,Convert)
=SUMIF(From,Nodes,Convert)-SUMIF(To,Nodes,Convert)
=SUMIF(From,Nodes,Convert)-SUMIF(To,Nodes,Convert)
H
13
14
15
16
17
18
!
!
42
I
Starting
Supply
(thousands)
Yen 1200000
Rupiah 10500000
Ringgit 28000
B
C
Total Cost =SUMPRODUCT(UnitCost,Convert)
!
10-30
!
J
J
=
=
=
=
=
=
=
=
K
Supply/Demand
($thousands)
=K16
=K17
=K18
0
0
0
0
=-SUM(K16:K18)
!
K
Conversion
Starting
($ per)
($thousands)
0.008
=StartingSupply*Conversion
0.00016
=StartingSupply*Conversion
0.2
=StartingSupply*Conversion
!
!
!
Jake!should!convert!the!equivalent!of!$2!million!from!Yen!to!each!US$,!Can$,!
Euro,!and!Pound.!He!should!convert!$1.6!million!from!Yen!to!Peso.!Moreover,!he!
should!convert!the!equivalent!of!$200,000!from!Rupiah!to!each!US$,!Can$,!and!
Peso,!$1!million!from!Rupiah!to!Euro,!and!$80,000!from!Rupiah!to!Pound.!
Furthermore,!Jake!should!convert!the!equivalent!of!$1.1!million!from!Ringgit!to!
US$,!$2.5!million!from!Ringgit!to!Euro,!and!$1!million!from!Ringgit!to!each!Pound!
and!Peso.!Finally,!he!should!convert!all!the!money!he!converted!into!Can$,!Euro,!
Pound,!and!Peso!directly!into!US$.!Specifically,!he!needs!to!convert!into!US$!the!
equivalent!of!$2.2!million,!$5.5!million,!$3.08!million,!and!$2.8!million!Can$,!
Euro,!Pound,!and!Peso,!respectively.!Assuming!Jake!pays!for!the!total!transaction!
costs!of!$83,380!directly!from!his!American!bank!accounts!he!will!have!
$16,880,000!dollars!to!invest!in!the!US.!
10-31
!
c)! We!eliminate!all!capacity!restrictions!on!the!arcs.!
!
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
A
B
C
From
Yen
Yen
Yen
Yen
Yen
Yen
Yen
Rupiah
Rupiah
Rupiah
Rupiah
Rupiah
Rupiah
Rupiah
Ringit
Ringit
Ringit
Ringit
Ringit
Ringit
Ringit
Can$
Can$
Can$
Can$
Euro
Euro
Euro
Euro
Pound
Pound
Pound
Pound
Peso
Peso
Peso
Peso
To
Rupiah
Ringit
US$
Can$
Euro
Pound
Peso
Yen
Ringit
US$
Can$
Euro
Pound
Peso
Yen
Rupiah
US$
Can$
Euro
Pound
Peso
US$
Euro
Pound
Peso
US$
Can$
Pound
Peso
US$
Can$
Euro
Peso
US$
Can$
Euro
Pound
Convert
($thousands)
0
0
0
0
0
9,600
0
0
0
0
1,680
0
0
0
0
0
0
0
5,600
0
0
1,680
0
0
0
5,600
0
0
0
9,600
0
0
0
0
0
0
0
Total Cost
67.48
D
E
Unit
Cost
0.50%
0.50%
0.40%
0.40%
0.40%
0.25%
0.50%
0.50%
0.70%
0.50%
0.30%
0.30%
0.75%
0.75%
0.50%
0.70%
0.70%
0.70%
0.40%
0.45%
0.50%
0.05%
0.20%
0.10%
0.10%
0.10%
0.20%
0.05%
0.50%
0.10%
0.10%
0.05%
0.50%
0.10%
0.10%
0.50%
0.50%
F
G
H
Net Flow
Nodes ($thousands)
Yen
9,600
Rupiah
1,680
Ringit
5,600
Can$
0
Euro
0
Pound
0
Peso
0
US$
-16,880
Yen
Rupiah
Ringgit
Starting
Supply
(thousands)
1,200,000
10,500,000
28,000
I
J
=
=
=
=
=
=
=
=
Supply/Demand
($thousands)
9,600
1,680
5,600
0
0
0
0
-16,880
Conversion
($ per)
0.008
0.00016
0.2
Starting
($thousands)
9,600
1,680
5,600
!
!
Jake!should!convert!the!entire!holdings!in!Japan!from!Yen!into!Pounds!and!then!
into!US$,!the!entire!holdings!in!Indonesia!from!Rupiah!into!Can$!and!then!into!
US$,!and!the!entire!holdings!in!Malaysia!from!Ringgit!into!Euro!and!then!into!
US$.!Without!the!capacity!limits!the!transaction!costs!are!reduced!to!$67,480.!
10-32
!
d)! We!multiply!all!unit!cost!for!Rupiah!by!6.!
!
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
A
B
C
From
Yen
Yen
Yen
Yen
Yen
Yen
Yen
Rupiah
Rupiah
Rupiah
Rupiah
Rupiah
Rupiah
Rupiah
Ringit
Ringit
Ringit
Ringit
Ringit
Ringit
Ringit
Can$
Can$
Can$
Can$
Euro
Euro
Euro
Euro
Pound
Pound
Pound
Pound
Peso
Peso
Peso
Peso
To
Rupiah
Ringit
US$
Can$
Euro
Pound
Peso
Yen
Ringit
US$
Can$
Euro
Pound
Peso
Yen
Rupiah
US$
Can$
Euro
Pound
Peso
US$
Euro
Pound
Peso
US$
Can$
Pound
Peso
US$
Can$
Euro
Peso
US$
Can$
Euro
Pound
Convert
($thousands)
0
0
0
0
0
9,600
0
0
0
0
1,680
0
0
0
0
0
0
0
5,600
0
0
1,680
0
0
0
5,600
0
0
0
9,600
0
0
0
0
0
0
0
Total Cost
92.68
D
E
Unit
Cost
0.50%
0.50%
0.40%
0.40%
0.40%
0.25%
0.50%
3.00%
4.20%
3.00%
1.80%
1.80%
4.50%
4.50%
0.50%
0.70%
0.70%
0.70%
0.40%
0.45%
0.50%
0.05%
0.20%
0.10%
0.10%
0.10%
0.20%
0.05%
0.50%
0.10%
0.10%
0.05%
0.50%
0.10%
0.10%
0.50%
0.50%
F
G
H
Net Flow
Nodes ($thousands)
Yen
9,600
Rupiah
1,680
Ringit
5,600
Can$
0
Euro
0
Pound
0
Peso
0
US$
-16,880
Yen
Rupiah
Ringgit
Starting
Supply
(thousands)
1,200,000
10,500,000
28,000
I
J
=
=
=
=
=
=
=
=
Supply/Demand
($thousands)
9,600
1,680
5,600
0
0
0
0
-16,880
Conversion
($ per)
0.008
0.00016
0.2
Starting
($thousands)
9,600
1,680
5,600
!
!
The!optimal!routing!for!the!money!doesn't!change,!but!the!total!transaction!costs!
are!now!increased!to!$92,680.!
!
e)! In!the!described!crisis!situation!the!currency!exchange!rates!might!change!every!
minute.!Jake!should!carefully!check!the!exchange!rates!again!when!he!performs!
the!transactions.!
!
The!European!economies!might!be!more!insulated!from!the!Asian!financial!
collapse!than!the!US!economy.!To!impress!his!boss!Jake!might!want!to!explore!
other!investment!opportunities!in!safer!European!economies!that!provide!higher!
rates!of!return!than!US!bonds.!
10-33
Case%10.2%
!
a)! The!network!showing!the!different!routes!troops!and!supplies!may!follow!to!
reach!the!Russian!Federation!appears!below.!
!
PORTS
Hamburg
Boston
Rotterdam
St.
Petersburg
Napoli
AIRFIELDS
Moscow
London
Jacksonville
Berlin
Rostov
Istanbul
!
10-34
!
b)! The!President!is!only!concerned!about!how!to!most!quickly!move!troops!and!
supplies!from!the!United!States!to!the!three!strategic!Russian!cities.!Obviously,!
the!best!way!to!achieve!this!goal!is!to!find!the!fastest!connection!between!the!US!
and!the!three!cities.!We!therefore!need!to!find!the!shortest!path!between!the!US!
cities!and!each!of!the!three!Russian!cities.!
!
The!President!only!cares!about!the!time!it!takes!to!get!the!troops!and!supplies!to!
Russia.!It!does!not!matter!how!great!a!distance!the!troops!and!supplies!cover.!
Therefore!we!define!the!arc!length!between!two!nodes!in!the!network!to!be!the!
time!it!takes!to!travel!between!the!respective!cities.!For!example,!the!distance!
between!Boston!and!London!equals!6,200!km.!The!mode!of!transportation!
between!the!cities!is!a!Starlifter!traveling!at!a!speed!of!400!miles!per!hour!*!
1.609!km!per!mile!=!643.6!km!per!hour.!The!time!is!takes!to!bring!troops!and!
supplies!from!Boston!to!London!equals!6,200!km!/!643.6!km!per!hour!=!9.6333!
hours.!Using!this!approach!we!can!compute!the!time!of!travel!along!all!arcs!in!
the!network.!
!
By!simple!inspection!and!common!sense!it!is!apparent!that!the!fastest!
transportation!involves!using!only!airplanes.!We!therefore!can!restrict!ourselves!
to!only!those!arcs!in!the!network!where!the!mode!of!transportation!is!air!travel.!
We!can!omit!the!three!port!cities!and!all!arcs!entering!and!leaving!these!nodes.!
!
The!following!six!spreadsheets!find!the!shortest!path!between!each!US!city!
(Boston!and!Jacksonville)!and!each!Russian!city!(St.!Petersburg,!Moscow,!and!
Rostov).!
!
!
Boston!to!St.!Petersburg:!
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
A
From
Boston
Boston
Boston
Jacksonville
Jacksonville
Jacksonville
London
London
London
Berlin
Berlin
Berlin
Istanbul
Istanbul
Istanbul
B
C
D
E
F
To
On Route
Distance (km) Time (hours)
London
1
6,200
9.63
Berlin
0
7,250
11.26
Istanbul
0
8,300
12.90
London
0
7,900
12.27
Berlin
0
9,200
14.29
Istanbul
0
10,100
15.69
St. Petersburg
1
1,980
3.08
Moscow
0
2,300
3.57
Rostov
0
2,860
4.44
St. Petersburg
0
1,280
1.99
Moscow
0
1,600
2.49
Rostov
0
1,730
2.69
St. Petersburg
0
2,040
3.17
Moscow
0
1,700
2.64
Rostov
0
990
1.54
Total Time
12.71
!!
!
10-35
G
H
I
Node
Net Flow
Boston
1
Jacksonville
0
London
0
Berlin
0
Istanbul
0
St. Petersburg
-1
Moscow
0
Rostov
0
Travel Speed (mph)
km/hr
Travel Speed (km/hr)
400
1.609
643.6
J
=
=
=
=
=
=
=
=
K
Supply/Demand
1
0
0
0
0
-1
0
0
!
!
Boston!to!Moscow:!
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
A
From
Boston
Boston
Boston
Jacksonville
Jacksonville
Jacksonville
London
London
London
Berlin
Berlin
Berlin
Istanbul
Istanbul
Istanbul
B
C
D
E
F
To
On Route
Distance (km) Time (hours)
London
1
6,200
9.63
Berlin
0
7,250
11.26
Istanbul
0
8,300
12.90
London
0
7,900
12.27
Berlin
0
9,200
14.29
Istanbul
0
10,100
15.69
St. Petersburg
0
1,980
3.08
Moscow
1
2,300
3.57
Rostov
0
2,860
4.44
St. Petersburg
0
1,280
1.99
Moscow
0
1,600
2.49
Rostov
0
1,730
2.69
St. Petersburg
0
2,040
3.17
Moscow
0
1,700
2.64
Rostov
0
990
1.54
Total Time
G
H
I
Node
Net Flow
Boston
1
Jacksonville
0
London
0
Berlin
0
Istanbul
0
St. Petersburg
0
Moscow
-1
Rostov
0
Travel Speed (mph)
km/hr
Travel Speed (km/hr)
J
=
=
=
=
=
=
=
=
K
Supply/Demand
1
0
0
0
0
0
-1
0
400
1.609
643.6
13.21
!
!
!
!
Boston!to!Rostov:!
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
A
From
Boston
Boston
Boston
Jacksonville
Jacksonville
Jacksonville
London
London
London
Berlin
Berlin
Berlin
Istanbul
Istanbul
Istanbul
B
C
D
E
F
To
On Route
Distance (km) Time (hours)
London
0
6,200
9.63
Berlin
1
7,250
11.26
Istanbul
0
8,300
12.90
London
0
7,900
12.27
Berlin
0
9,200
14.29
Istanbul
0
10,100
15.69
St. Petersburg
0
1,980
3.08
Moscow
0
2,300
3.57
Rostov
0
2,860
4.44
St. Petersburg
0
1,280
1.99
Moscow
0
1,600
2.49
Rostov
1
1,730
2.69
St. Petersburg
0
2,040
3.17
Moscow
0
1,700
2.64
Rostov
0
990
1.54
Total Time
G
H
I
Node
Net Flow
Boston
1
Jacksonville
0
London
0
Berlin
0
Istanbul
0
St. Petersburg
0
Moscow
0
Rostov
-1
Travel Speed (mph)
km/hr
Travel Speed (km/hr)
J
=
=
=
=
=
=
=
=
K
Supply/Demand
1
0
0
0
0
0
0
-1
400
1.609
643.6
13.95
!
!
!
!
Jacksonville!to!St.!Petersburg:!
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
A
From
Boston
Boston
Boston
Jacksonville
Jacksonville
Jacksonville
London
London
London
Berlin
Berlin
Berlin
Istanbul
Istanbul
Istanbul
B
C
D
E
F
To
On Route
Distance (km) Time (hours)
London
0
6,200
9.63
Berlin
0
7,250
11.26
Istanbul
0
8,300
12.90
London
1
7,900
12.27
Berlin
0
9,200
14.29
Istanbul
0
10,100
15.69
St. Petersburg
1
1,980
3.08
Moscow
0
2,300
3.57
Rostov
0
2,860
4.44
St. Petersburg
0
1,280
1.99
Moscow
0
1,600
2.49
Rostov
0
1,730
2.69
St. Petersburg
0
2,040
3.17
Moscow
0
1,700
2.64
Rostov
0
990
1.54
Total Time
15.35
!
!
10-36
G
H
I
Node
Net Flow
Boston
0
Jacksonville
1
London
0
Berlin
0
Istanbul
0
St. Petersburg
-1
Moscow
0
Rostov
0
Travel Speed (mph)
km/hr
Travel Speed (km/hr)
400
1.609
643.6
J
=
=
=
=
=
=
=
=
K
Supply/Demand
0
1
0
0
0
-1
0
0
!
!
Jacksonville!to!Moscow:!
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
A
From
Boston
Boston
Boston
Jacksonville
Jacksonville
Jacksonville
London
London
London
Berlin
Berlin
Berlin
Istanbul
Istanbul
Istanbul
B
C
D
E
F
To
On Route
Distance (km) Time (hours)
London
0
6,200
9.63
Berlin
0
7,250
11.26
Istanbul
0
8,300
12.90
London
1
7,900
12.27
Berlin
0
9,200
14.29
Istanbul
0
10,100
15.69
St. Petersburg
0
1,980
3.08
Moscow
1
2,300
3.57
Rostov
0
2,860
4.44
St. Petersburg
0
1,280
1.99
Moscow
0
1,600
2.49
Rostov
0
1,730
2.69
St. Petersburg
0
2,040
3.17
Moscow
0
1,700
2.64
Rostov
0
990
1.54
Total Time
G
H
I
Node
Net Flow
Boston
0
Jacksonville
1
London
0
Berlin
0
Istanbul
0
St. Petersburg
0
Moscow
-1
Rostov
0
Travel Speed (mph)
km/hr
Travel Speed (km/hr)
J
=
=
=
=
=
=
=
=
K
Supply/Demand
0
1
0
0
0
0
-1
0
400
1.609
643.6
15.85
!
!
!
!
Jacksonville!to!Rostov:!
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
A
From
Boston
Boston
Boston
Jacksonville
Jacksonville
Jacksonville
London
London
London
Berlin
Berlin
Berlin
Istanbul
Istanbul
Istanbul
B
C
D
E
F
To
On Route
Distance (km) Time (hours)
London
0
6,200
9.63
Berlin
0
7,250
11.26
Istanbul
0
8,300
12.90
London
1
7,900
12.27
Berlin
0
9,200
14.29
Istanbul
0
10,100
15.69
St. Petersburg
0
1,980
3.08
Moscow
0
2,300
3.57
Rostov
1
2,860
4.44
St. Petersburg
0
1,280
1.99
Moscow
0
1,600
2.49
Rostov
0
1,730
2.69
St. Petersburg
0
2,040
3.17
Moscow
0
1,700
2.64
Rostov
0
990
1.54
Total Time
16.72
!
!
10-37
G
H
I
Node
Net Flow
Boston
0
Jacksonville
1
London
0
Berlin
0
Istanbul
0
St. Petersburg
0
Moscow
0
Rostov
-1
Travel Speed (mph)
km/hr
Travel Speed (km/hr)
400
1.609
643.6
J
=
=
=
=
=
=
=
=
K
Supply/Demand
0
1
0
0
0
0
0
-1
!
!
The!spreadsheets!contain!the!following!formulas:!
!
Range Name
Distance
From
NetFlow
Node
OnRoute
SupplyDemand
Time
To
TotalTime
TravelSpeed
Cells
E2:E16
A2:A16
I2:I9
H2:H9
C2:C16
K2:K9
F2:F16
B2:B16
C18
I16
!
1
2
3
4
5
6
7
8
9
!
F
Time (hours)
=Distance/TravelSpeed
=Distance/TravelSpeed
=Distance/TravelSpeed
=Distance/TravelSpeed
=Distance/TravelSpeed
=Distance/TravelSpeed
=Distance/TravelSpeed
=Distance/TravelSpeed
!
B
Total Time
14
15
16
!
1
2
3
4
5
6
7
8
9
H
I
Travel Speed (mph) 400
km/hr 1.609
Travel Speed (km/hr) =I14*I15
!
!
I
Net Flow
=SUMIF(From,Node,OnRoute)-SUMIF(To,Node,OnRoute)
=SUMIF(From,Node,OnRoute)-SUMIF(To,Node,OnRoute)
=SUMIF(From,Node,OnRoute)-SUMIF(To,Node,OnRoute)
=SUMIF(From,Node,OnRoute)-SUMIF(To,Node,OnRoute)
=SUMIF(From,Node,OnRoute)-SUMIF(To,Node,OnRoute)
=SUMIF(From,Node,OnRoute)-SUMIF(To,Node,OnRoute)
=SUMIF(From,Node,OnRoute)-SUMIF(To,Node,OnRoute)
=SUMIF(From,Node,OnRoute)-SUMIF(To,Node,OnRoute)
! !
C
=SUMPRODUCT(OnRoute,Time)
!
!
Comparing!all!six!solutions!we!see!that!the!shortest!path!from!the!US!to!Saint!
Petersburg!is!Boston!→!London!→!Saint!Petersburg!with!a!total!travel!time!of!
12.71!hours.!The!shortest!path!from!the!US!to!Moscow!is!Boston!→!London!→!
Moscow!with!a!total!travel!time!of!13.21!hours.!The!shortest!path!from!the!US!to!
Rostov!is!Boston!→!Berlin!→!Rostov!with!a!total!travel!time!of!13.95!hours.!!The!
following!network!diagram!highlights!these!shortest!paths.!
18
!
!
!
10-38
!
c)! The!President!must!satisfy!each!Russian!city’s!military!requirements!at!
minimum!cost.!!Therefore,!this!problem!can!be!solved!as!a!minimumQcost!
network!flow!problem.!The!two!nodes!representing!US!cities!are!supply!nodes!
with!a!supply!of!500!each!(we!measure!all!weights!in!1000!tons).!The!three!
nodes!representing!Saint!Petersburg,!Moscow,!and!Rostov!are!demand!nodes!
with!demands!of!–320,!Q440,!and!–240,!respectively.!All!nodes!representing!
European!airfields!and!ports!are!transshipment!nodes.!We!measure!the!flow!
along!the!arcs!in!1000!tons.!For!some!arcs,!capacity!constraints!are!given.!All!
arcs!from!the!European!ports!into!Saint!Petersburg!have!zero!capacity.!All!truck!
routes!from!the!European!ports!into!Rostov!have!a!transportation!limit!of!
2,500*16!=!40,000!tons.!Since!we!measure!the!arc!flows!in!1000!tons,!the!
corresponding!arc!capacities!equal!40.!An!analogous!computation!yields!arc!
capacities!of!30!for!both!the!arcs!connecting!the!nodes!London!and!Berlin!to!
Rostov.!For!all!other!nodes!we!determine!natural!arc!capacities!based!on!the!
supplies!and!demands!at!the!nodes.!We!define!the!unit!costs!along!the!arcs!in!the!
network!in!$1000!per!1000!tons!(or,!equivalently,!$/ton).!For!example,!the!cost!
of!transporting!1!ton!of!material!from!Boston!to!Hamburg!equals!$30,000!/!240!
=!$125,!so!the!costs!of!transporting!1000!tons!from!Boston!to!Hamburg!equals!
$125,000.!
!
The!objective!is!to!satisfy!all!demands!in!the!network!at!minimum!cost.!The!
following!spreadsheet!shows!the!entire!linear!programming!model.!
10-39
!
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
From
Boston
Boston
Boston
Boston
Boston
Boston
Jacksonville
Jacksonville
Jacksonville
Jacksonville
Jacksonville
Jacksonville
Berlin
Hamburg
Istanbul
London
Rotterdam
Napoli
Berlin
Hamburg
Istanbul
London
Rotterdam
Napoli
Berlin
Hamburg
Istanbul
London
Rotterdam
Napoli
B
C
To
Berlin
Hamburg
Istanbul
London
Rotterdam
Napoli
Berlin
Hamburg
Istanbul
London
Rotterdam
Napoli
St. Petersburg
St. Petersburg
St. Petersburg
St. Petersburg
St. Petersburg
St. Petersburg
Moscow
Moscow
Moscow
Moscow
Moscow
Moscow
Rostov
Rostov
Rostov
Rostov
Rostov
Rostov
Ship
(thousand tons)
0
440
60
0
0
0
0
0
150
350
0
0
0
0
0
320
0
0
0
440
0
0
0
0
0
0
210
30
0
0
Total Cost ($thousand)
D
E
Capacity
(thousand tons)
<=
0
<=
<=
0
0
<=
<=
30
40
<=
<=
<=
30
40
40
F
Cost per
Vehicle
($thousand)
50
30
55
45
30
32
57
48
61
49
44
56
24
3
28
22
3
5
22
4
25
19
5
5
23
7
2
4
8
9
G
Vehicle
Starlifter
Transport
Starlifter
Starlifter
Transport
Transport
Starlifter
Transport
Starlifter
Starlifter
Transport
Transport
Starlifter
Truck
Starlifter
Starlifter
Truck
Truck
Starlifter
Truck
Starlifter
Starlifter
Truck
Truck
Starlifter
Truck
Starlifter
Starlifter
Truck
Truck
H
Maximum
Vehicles
0
0
0
200
2,500
200
2,500
2,500
I
Vehicle
Capacity
(tons)
150
240
150
150
240
240
150
240
150
150
240
240
150
16
150
150
16
16
150
16
150
150
16
16
150
16
150
150
16
16
J
Unit
Cost
($/ton)
$333.33
$125.00
$366.67
$300.00
$125.00
$133.33
$380.00
$200.00
$406.67
$326.67
$183.33
$233.33
$160.00
$187.50
$186.67
$146.67
$187.50
$312.50
$146.67
$250.00
$166.67
$126.67
$312.50
$312.50
$153.33
$437.50
$13.33
$26.67
$500.00
$562.50
412,867
!
!
L
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
M
Net Flow
Node
(thousand tons)
Boston
500
Jacksonville
500
Berlin
0
Hamburg
0
Istanbul
0
London
0
Rotterdam
0
Napoli
0
St. Petersburg
-320
Moscow
-440
Rostov
-240
Starlifter
Transport
Truck
N
=
=
=
=
=
=
=
=
=
=
=
O
Supply/Demand
(thousand tons)
500
500
0
0
0
0
0
0
-320
-440
-240
Capacity
(tons)
150
240
16
!
10-40
1
2
3
4
5
6
7
I
Vehicle
Capacity
(tons)
=VLOOKUP(G4,$L$24:$M$26,2)
=VLOOKUP(G5,$L$24:$M$26,2)
=VLOOKUP(G6,$L$24:$M$26,2)
=VLOOKUP(G7,$L$24:$M$26,2)
J
Unit
Cost
($/ton)
=1000*F4/I4
=1000*F5/I5
=1000*F6/I6
=1000*F7/I7
!
!
!
!
The!total!cost!of!the!operation!equals!$412.867!million.!The!entire!supply!for!
Saint!Petersburg!is!supplied!from!Jacksonville!via!London.!The!entire!supply!for!
Moscow!is!supplied!from!Boston!via!Hamburg.!Of!the!240!(=!240,000!tons)!
demanded!by!Rostov,!60!are!shipped!from!Boston!via!Istanbul,!150!are!shipped!
from!Jacksonville!via!Istanbul,!and!30!are!shipped!from!Jacksonville!via!London.!
The!paths!used!to!ship!supplies!to!Saint!Petersburg,!Moscow,!and!Rostov!are!
highlighted!on!the!following!network!diagram.!
!
!
d)! Now!the!President!wants!to!maximize!the!amount!of!cargo!transported!from!the!
US!to!the!Russian!cities.!In!other!words,!the!President!wants!to!maximize!the!
flow!from!the!two!US!cities!to!the!three!Russian!cities.!All!the!nodes!representing!
the!European!ports!and!airfields!are!once!again!transshipment!nodes.!The!flow!
along!an!arc!is!again!measured!in!thousands!of!tons.!The!new!restrictions!can!be!
transformed!into!arc!capacities!using!the!same!approach!that!was!used!in!part!
(c).!The!objective!is!now!to!maximize!the!combined!flow!into!the!three!Russian!
cities.!
!
10-41
!
!
The!linear!programming!spreadsheet!model!describing!the!maximum!flow!
problem!appears!as!follows.!
!
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
A
B
C
From
Boston
Boston
Boston
Boston
Boston
Boston
Jacksonville
Jacksonville
Jacksonville
Jacksonville
Jacksonville
Jacksonville
Berlin
Hamburg
Istanbul
London
Rotterdam
Napoli
Berlin
Hamburg
Istanbul
London
Rotterdam
Napoli
Berlin
Hamburg
Istanbul
London
Rotterdam
Napoli
To
Berlin
Hamburg
Istanbul
London
Rotterdam
Napoli
Berlin
Hamburg
Istanbul
London
Rotterdam
Napoli
St. Petersburg
St. Petersburg
St. Petersburg
St. Petersburg
St. Petersburg
St. Petersburg
Moscow
Moscow
Moscow
Moscow
Moscow
Moscow
Rostov
Rostov
Rostov
Rostov
Rostov
Rostov
Ship
(thousand tons)
45
19.2
45
75
21.6
46.4
75
0
105
90
0
0
75
0
0
150
0
0
45
11.2
15
0
9.6
24
0
8
135
15
12
22.4
Maximum Shipment
D
E
Capacity
(thousand tons)
<=
45
<=
<=
75
75
<=
75
<=
<=
105
90
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
75
0
0
150
0
0
45
11.2
15
30
9.6
24
0
8
135
15
12
22.4
J
K
Net Flow
Node
(thousand tons)
Boston
252.2
Jacksonville
270
Berlin
0
Hamburg
0
Istanbul
0
London
0
Rotterdam
0
Napoli
0
St. Petersburg
-225
Moscow
-104.8
Rostov
-192.4
Starlifter
Transport
Truck
L
M
Supply/Demand
(thousand tons)
=
=
=
=
=
=
Vehicle
Starlifter
Transport
Starlifter
Starlifter
Transport
Transport
Starlifter
Transport
Starlifter
Starlifter
Transport
Transport
Starlifter
Truck
Starlifter
Starlifter
Truck
Truck
Starlifter
Truck
Starlifter
Starlifter
Truck
Truck
Starlifter
Truck
Starlifter
Starlifter
Truck
Truck
G
Maximum
Vehicles
300
500
500
500
700
600
500
0
0
1,000
0
0
300
700
100
200
600
1,500
0
500
900
100
750
1,400
H
Vehicle
Capacity
(tons)
150
240
150
150
240
240
150
240
150
150
240
240
150
16
150
150
16
16
150
16
150
150
16
16
150
16
150
150
16
16
!
522.2
!
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
F
0
0
0
0
0
0
Capacity
(tons)
150
240
16
!
10-42
!
!
The!spreadsheet!shows!all!the!amounts!that!are!shipped!between!the!various!
cities.!The!total!supply!for!Saint!Petersburg,!Moscow,!and!Rostov!equals!225,000!
tons,!104,800!tons,!and!192,400!tons,!respectively.!The!following!network!
diagram!highlights!the!paths!used!to!ship!supplies!between!the!US!and!the!
Russian!Federation.!
!
!
!
!
!
!
10-43
!
e)! The!creation!of!the!new!communication!network!is!a!minimum!spanning!tree!
problem.!As!usual,!a!greedy!algorithm!solves!this!type!of!problem.!
!
!
!
!
Arcs!are!added!to!the!network!in!the!following!order!(one!of!several!optimal!
solutions):!
!
Rostov!–!Orenburg!
120!
Ufa!–!Orenburg!
75!
Saratov!–!Orenburg!
95!
Saratov!–!Samara!
100!
Samara!–!Kazan!
95!
Ufa!–!Yekaterinburg!
125!
Perm!–!Yekaterinburg!
85!
!
The!minimum!cost!of!reestablishing!the!communication!lines!is!$695!thousand.!
10-44
Case%10.3%
!
a)! A!diagram!of!the!project!network!appears!below.!
!
START
0
A
3
Evaluate the pr estige of eac h p otential un de rw riter
B
1.5
Select a syndicate of underw riters
C
D
N egotiate the commitmen t of each
memb er of the syndicate
N egotiate the spre ad for each
memb er of the syndicate
2
3
E
Pre par e the regi stration stateme nt
5
F
Submi t th e r egistration stateme nt to the SEC
G
H
Mak e prese ntations to
institutional investor s an d de velop
the inter est of pote ntial bu yers
6
1
J
I
D istribute the
re d herr ing
3
C alcul ate the
issue price
5
R eceive d eficienc y
me morandum from th e
SEC
3
K
A mend statement and
r esubmit to the S EC
1
L
R eceive r egistration
c onfir mation from the
SEC
M
N
C on firm that the
new issu e c omplies
w ith Òb lu e skyÓlaw s
1
O
A ppoint a
registrar
A ppoi nt a
tran sfer agent
3
P
Issue fin al p ros pectu s
!
!
!
!
10-45
3.5
Q
4.5
FINISH
!
2
Ph on e interested buyers
0
4
!
!
!
To!determine!the!project!schedule!and!which!activities!are!critical,!we!calculate!
the!early!start,!late!start,!early!finish,!late!finish,!and!slack!below.!
!
A
B
1
2 Activity Description
3
A
Evaluate prestige
4
B
Select syndicate
5
C
Negotiate commitment
6
D
Negotiate spread
7
E
Prepare registration
8
F
Submit registration
9
G
Present
10
H
Distribute red herring
11
I
Calculate price
12
J
Receive deficiency
13
K
Amend statement
14
L
Receive registration
15
M
Confirm blue sky
16
N
Appoint registrar
17
O
Appoint transfer
18
P
Issue prospectus
19
Q
Phone buyers
20
21
C
Time
(weeks)
3
1.5
2
3
5
1
6
3
5
3
1
2
1
3
3.5
4.5
4
D
E
F
G
LS
0
3
5.5
4.5
7.5
12.5
16
19
17
13.5
16.5
17.5
22
20
19.5
23
23.5
LF
3
4.5
7.5
7.5
12.5
13.5
22
22
22
16.5
17.5
19.5
23
23
23
27.5
27.5
Week
ES
0
3
4.5
4.5
7.5
12.5
13.5
13.5
13.5
13.5
16.5
17.5
19.5
19.5
19.5
23
23
EF
3
4.5
6.5
7.5
12.5
13.5
19.5
16.5
18.5
16.5
17.5
19.5
20.5
22.5
23
27.5
27
Project Duration
H
Slack
(weeks)
0
0
1
0
0
0
2.5
5.5
3.5
0
0
0
2.5
0.5
0
0
0.5
I
Critical?
Yes
Yes
No
Yes
Yes
Yes
No
No
No
Yes
Yes
Yes
No
No
Yes
Yes
No
!
27.5
!
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
Activity
A
B
C
D
E
F
G
H
I
J
K
L
M
N
O
P
Q
B
Description
Evaluate prestige
Select syndicate
Negotiate commitment
Negotiate spread
Prepare registration
Submit registration
Present
Distribute red herring
Calculate price
Receive deficiency
Amend statement
Receive registration
Confirm blue sky
Appoint registrar
Appoint transfer
Issue prospectus
Phone buyers
C
Time
(weeks)
3
1.5
2
3
5
1
6
3
5
3
1
2
1
3
3.5
4.5
4
D
E
F
G
H
Slack
EF
LS
LF
(weeks)
=ES+Time =LF-Time =MIN(F4)
=LF-EF
=ES+Time =LF-Time =MIN(F5,F6)
=LF-EF
=ES+Time =LF-Time =MIN(F7)
=LF-EF
=ES+Time =LF-Time =MIN(F7)
=LF-EF
=ES+Time =LF-Time =MIN(F8)
=LF-EF
=ES+Time =LF-Time =MIN(F9,F10,F11,F12) =LF-EF
=ES+Time =LF-Time =MIN(F15)
=LF-EF
=ES+Time =LF-Time =MIN(F15)
=LF-EF
=ES+Time =LF-Time =MIN(F15)
=LF-EF
=ES+Time =LF-Time =MIN(F13)
=LF-EF
=ES+Time =LF-Time =MIN(F14)
=LF-EF
=ES+Time =LF-Time =MIN(F15,F16,F17)
=LF-EF
=ES+Time =LF-Time =MIN(F18,F19)
=LF-EF
=ES+Time =LF-Time =MIN(F18,F19)
=LF-EF
=ES+Time =LF-Time =MIN(F18,F19)
=LF-EF
=ES+Time =LF-Time =ProjectDuration
=LF-EF
=ES+Time =LF-Time =ProjectDuration
=LF-EF
I
Week
ES
0
=MAX(E3)
=MAX(E4)
=MAX(E4)
=MAX(E5,E6)
=MAX(E7)
=MAX(E8)
=MAX(E8)
=MAX(E8)
=MAX(E8)
=MAX(E12)
=MAX(E13)
=MAX(E9,E10,E11,E14)
=MAX(E14)
=MAX(E14)
=MAX(E15,E16,E17)
=MAX(E15,E16,E17)
Critical?
=IF(Slack=0,"Yes","No")
=IF(Slack=0,"Yes","No")
=IF(Slack=0,"Yes","No")
=IF(Slack=0,"Yes","No")
=IF(Slack=0,"Yes","No")
=IF(Slack=0,"Yes","No")
=IF(Slack=0,"Yes","No")
=IF(Slack=0,"Yes","No")
=IF(Slack=0,"Yes","No")
=IF(Slack=0,"Yes","No")
=IF(Slack=0,"Yes","No")
=IF(Slack=0,"Yes","No")
=IF(Slack=0,"Yes","No")
=IF(Slack=0,"Yes","No")
=IF(Slack=0,"Yes","No")
=IF(Slack=0,"Yes","No")
=IF(Slack=0,"Yes","No")
Project Duration =MAX(EF)
!
Range Name
Activity
Critical?
Description
EF
ES
LF
LS
ProjectDuration
Slack
Time
!
Cells
A3:A19
I3:I19
B3:B19
E3:E19
D3:D19
G3:G19
F3:F19
E21
H3:H19
C3:C19
!
The!initial!public!offering!process!is!27.5!weeks!long.!!The!critical!path!is:!
START!→!A!→!B!→!D!→!E!→!F!→!J!→!K!→!L!→!O!→!P!→!FINISH!
10-46
!
b)! !
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
B
Activity
A
B
C
D
E
F
G
H
I
J
K
L
M
N
O
P
Q
C
D
Time
(weeks)
Normal
Crash
3
1.5
1.5
0.5
2
2
3
3
5
4
1
1
6
4
3
2
5
3.5
3
3
1
0.5
2
2
1
0.5
3
1.5
3.5
1.5
4.5
2
4
1.5
Description
Evaluate prestige
Select syndicate
Negotiate commitment
Negotiate spread
Prepare registration
Submit registration
Present
Distribute red herring
Calculate price
Receive deficiency
Amend statement
Receive registration
Confirm blue sky
Appoint registrar
Appoint transfer
Issue prospectus
Phone buyers
E
F
Cost
($thousand)
Normal
Crash
8
14
4.5
8
9
9
12
12
50
95
1
1
25
60
15
22
12
31
0
3
6
9
0
0
5
8.3
12
19
13
21
40
99
9
20
G
Maximum
Time Reduction
(weeks)
1.5
1
0
0
1
0
2
1
1.5
0
0.5
0
0.5
1.5
2
2.5
2.5
H
Crash Cost
per Week Saved
($thousand)
4
3.5
0
0
45
0
17.5
7
12.67
0
6
0
6.6
4.67
4
23.6
4.4
Project Finish Time (week)
Total Cost
I
Start
Time
(week)
0
1.5
3
2
5
10
11
14
12
11
14
14.5
17
16.5
16.5
18
18
J
Time
Reduction
(weeks)
1.5
1
0
0
0
0
0
0
0
0
0.5
0
0
1.5
2
0.5
0
K
Finish
Time
(week)
1.5
2
5
5
10
11
17
17
17
14
14.5
16.5
18
18
18
22
22
22
<=
Max Time
22
260.8
!!
!
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
G
Maximum
Time Reduction
(weeks)
=NormalTime-CrashTime
=NormalTime-CrashTime
=NormalTime-CrashTime
=NormalTime-CrashTime
=NormalTime-CrashTime
=NormalTime-CrashTime
=NormalTime-CrashTime
=NormalTime-CrashTime
=NormalTime-CrashTime
=NormalTime-CrashTime
=NormalTime-CrashTime
=NormalTime-CrashTime
=NormalTime-CrashTime
=NormalTime-CrashTime
=NormalTime-CrashTime
=NormalTime-CrashTime
=NormalTime-CrashTime
H
Crash Cost
per Week Saved
($thousand)
=(CrashCost-NormalCost)/MaxTimeReduction
=(CrashCost-NormalCost)/MaxTimeReduction
0
0
=(CrashCost-NormalCost)/MaxTimeReduction
0
=(CrashCost-NormalCost)/MaxTimeReduction
=(CrashCost-NormalCost)/MaxTimeReduction
=(CrashCost-NormalCost)/MaxTimeReduction
0
=(CrashCost-NormalCost)/MaxTimeReduction
0
=(CrashCost-NormalCost)/MaxTimeReduction
=(CrashCost-NormalCost)/MaxTimeReduction
=(CrashCost-NormalCost)/MaxTimeReduction
=(CrashCost-NormalCost)/MaxTimeReduction
=(CrashCost-NormalCost)/MaxTimeReduction
I
Start
Time
(week)
0
1.5
3
2
5
10
11
14
12
11
14
14.5
17
16.5
16.5
18
18
J
Time
Reduction
(weeks)
1.5
1
0
0
0
0
0
0
0
0
0.5
0
0
1.5
2
0.5
0
K
Finish
Time
(week)
=StartTime+NormalTime-TimeReduction
=StartTime+NormalTime-TimeReduction
=StartTime+NormalTime-TimeReduction
=StartTime+NormalTime-TimeReduction
=StartTime+NormalTime-TimeReduction
=StartTime+NormalTime-TimeReduction
=StartTime+NormalTime-TimeReduction
=StartTime+NormalTime-TimeReduction
=StartTime+NormalTime-TimeReduction
=StartTime+NormalTime-TimeReduction
=StartTime+NormalTime-TimeReduction
=StartTime+NormalTime-TimeReduction
=StartTime+NormalTime-TimeReduction
=StartTime+NormalTime-TimeReduction
=StartTime+NormalTime-TimeReduction
=StartTime+NormalTime-TimeReduction
=StartTime+NormalTime-TimeReduction
!
!
25
H
I
Total Cost =SUM(NormalCost)+SUMPRODUCT(CrashCostPerWeekSaved,TimeReduction)
!
Range Name
Activity
CrashCost
CrashCostPerWeekSaved
CrashTime
Description
FinishTime
MaxTime
MaxTimeReduction
NormalCost
NormalTime
ProjectFinishTime
StartTime
TimeReduction
TotalCost
Cells
A4:A20
F4:F20
H4:H20
D4:D20
B4:B20
K4:K20
K23
G4:G20
E4:E20
C4:C20
I23
I4:I20
J4:J20
I25
!
10-47
!
!
!
The!constraints!in!the!linear!programming!spreadsheet!model!were!as!follows:!
!
TimeReduction!≤!MaxTimeReduction!
!
ProjectFinishTime!≤!MaxTime!
!
BStart!≥!AFinish!
!
CStart!≥!BFinish!
!
DStart!≥!BFinish!
!
EStart!≥!CFinish!
!
EStart!≥!DFinish!
!
FStart!≥!EFinish!
!
GStart!≥!FFinish!
!
HStart!≥!FFinish!
!
IStart!≥!FFinish!
!
JStart!≥!FFinish!
!
KStart!≥!JFinish!
!
LStart!≥!KFinish!
!
MStart!≥!GFinish!
!
MStart!≥!HFinish!
!
MStart!≥!IFinish!
!
MStart!≥!LFinish!
!
NStart!≥!LFinish!
!
OStart!≥!LFinish!
!
PStart!≥!MFinish!
!
PStart!≥!NFinish!
!
PStart!≥!OFinish!
!
QStart!≥!MFinish!
!
QStart!≥!NFinish!
!
QStart!≥!OFinish!
!
ProjectFinishTime!≥!PFinish!
!
ProjectFinishTime!≥!QFinish!
!
Janet!and!Gilbert!should!reduce!the!time!for!step!A!(evaluating!the!prestige!of!
each!potential!underwriter)!by!1.5!weeks,!the!time!for!step!B!(selecting!a!
syndicate!of!underwriters)!by!1!week,!the!time!for!step!K!(amending!statement!
and!resubmitting!it!to!the!SEC)!by!0.5!weeks,!the!time!for!step!N!(appointing!a!
registrar)!by!1.5!weeks,!the!time!for!step!O!(appointing!a!transfer!agent)!by!two!
weeks,!and!the!time!for!step!P!(issuing!final!prospectus)!by!0.5!weeks.!!Janet!and!
Gilbert!can!now!meet!the!new!deadline!of!22!weeks!at!a!total!cost!of!$260,800.!
10-48
!
c)! We!use!the!same!model!formulation!that!was!used!in!part!(c).!!We!change!one!
constraint,!however.!!The!project!duration!now!has!to!be!lessQthanQorQequal!to!
24!weeks!instead!of!22!weeks.!!We!obtain!the!following!solution.!
!
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
Activity
A
B
C
D
E
F
G
H
I
J
K
L
M
N
O
P
Q
B
Description
Evaluate prestige
Select syndicate
Negotiate commitment
Negotiate spread
Prepare registration
Submit registration
Present
Distribute red herring
Calculate price
Receive deficiency
Amend statement
Receive registration
Confirm blue sky
Appoint registrar
Appoint transfer
Issue prospectus
Phone buyers
C
D
Time
(weeks)
Normal
Crash
3
1.5
1.5
0.5
2
2
3
3
5
4
1
1
6
4
3
2
5
3.5
3
3
1
0.5
2
2
1
0.5
3
1.5
3.5
1.5
4.5
2
4
1.5
E
F
Cost
($thousand)
Normal
Crash
8
14
4.5
8
9
9
12
12
50
95
1
1
25
60
15
22
12
31
0
3
6
9
0
0
5
8.3
12
19
13
21
40
99
9
20
G
Maximum
Time Reduction
(weeks)
1.5
1
0
0
1
0
2
1
1.5
0
0.5
0
0.5
1.5
2
2.5
2.5
H
Crash Cost
per Week Saved
($thousand)
4
3.5
0
0
45
0
17.5
7
12.67
0
6
0
6.6
4.67
4
23.6
4.4
I
Start
Time
(week)
0
1.5
3
2
5
10
12.5
11
13.5
11
14
14.5
18.5
16.5
16.5
19.5
20
J
Time
Reduction
(weeks)
1.5
1
0
0
0
0
0
0
0
0
0.5
0
0
0
0.5
0
0
K
Finish
Time
(week)
1.5
2
5
5
10
11
18.5
14
18.5
14
14.5
16.5
19.5
19.5
19.5
24
24
Project Finish Time (week)
24
<=
Max Time
24
Total Cost
236
!
!
Janet!and!Gilbert!should!reduce!the!time!for!step!A!(evaluating!the!prestige!of!
each!potential!underwriter)!by!1.5!weeks,!the!time!for!step!B!(selecting!a!
syndicate!of!underwriters)!by!1!week,!the!time!for!step!K!(amending!statement!
and!resubmitting!it!to!the!SEC)!by!0.5!weeks,!and!the!time!for!step!O!(appointing!
a!transfer!agent)!by!0.5!weeks.!!Janet!and!Gilbert!can!now!meet!the!new!deadline!
of!24!weeks!at!a!total!cost!of!$236,000.!
!
10-49
CHAPTER 11: DYNAMIC PROGRAMMING
11.2-1.
(a) The nodes of the network can be divided into "layers" such that the nodes in the 8th
layer are accessible from the origin only through the nodes in the Ð8  "Ñst layer. These
layers define the stages of the problem, which can be labeled as 8 œ "ß #ß $ß %. The nodes
constitute the states.
Let W8 denote the set of the nodes in the 8th layer of the network, i.e., W" œ ÖS×, W# œ
ÖEß Fß G×, W$ œ ÖHß I× and W% œ ÖX ×. The decision variable B8 is the immediate
destination at stage 8. Then the problem can be formulated as follows:
08‡ Ð=Ñ œ
‡
min Ò-=B8  08"
ÐB8 ÑÓ ´
B8 −W8"
min 08 Ð=ß B8 Ñ for = − W8 and 8 œ "ß #ß $
B8 −W8"
0%‡ ÐX Ñ œ !
(b) The shortest path is S  F  H  X .
(c) Number of stages: 3
=$
H
I
0$‡ Ð=Ñ
6
(
B‡$
X
X
=#
E
F
G
0# Ð=ß HÑ
""
"$

0# Ð=ß IÑ

"&
"$
0#‡ Ð=Ñ
""
"$
"$
="
S
0" Ð=ß EÑ
#!
0" Ð=ß FÑ
"*
0" Ð=ß GÑ
#!
B‡#
H
H
I
0"‡ Ð=Ñ
"*
Optimal Solution: B‡" œ F , B‡# œ H and B‡$ œ H.
11-1
B‡"
F
(d) Shortest-Path Algorithm:
8
1
#
$
%
&
Solved nodes
directly connected
to unsolved nodes
S
S
F
S
F
G
E
F
G
H
I
Closest
connected
unsolved node
F
G
H
E
H
I
H
H
I
X
X
total
distance
'
(
'  ( œ "$
*
'  ( œ "$
(  ' œ "$
*  & œ "%
'  ( œ "$
(  ' œ "$
"$  ' œ "*
"$  ( œ #!
8th
nearest
node
F
G
Distance to
8th nearest
node
'
(
E
*
SE
H
"$
FH
I
X
"*
GI
HX
Last
connection
SF
SG
The shortest-path algorithm required ) additions and ' comparisons whereas dynamic
programming required ( additions and $ comparisons. Hence, the latter seems to be more
efficient for shortest-path problems with "layered" network graphs.
11.2-2.
(a)
The optimal routes are S  E  J  X and S  G  L  X , the associated sales
income is "40. The route S  E  J  X corresponds to assigning ", #, and $
salespeople to regions ", #, and $ respectively. The route S  G  L  X corresponds
to assigning $, #, and " salespeople to regions ", #, and $ respectively.
11-2
(b) The regions are the stages and the number of salespeople remaining to be allocated at
stage 8 are possible states at stage 8, say =8 . Let B8 be the number of salespeople
assigned to region 8 and -8 ÐB8 Ñ be the increase in sales in region 8 if B8 salespeople are
assigned to it. Number of stages: 3.
=$
"
#
$
%
0$‡ Ð=$ Ñ
28
41
63
75
B‡$
"
#
$
%
="
'
0" Ð=" ß B" Ñ
"
#
$
140 132 140
=#
#
$
%
&
"
49
62
84
96
%
138
0# Ð=# ß B# Ñ
#
$



70
83
84
105 97
0"‡ Ð=" Ñ
140
%



98
0#‡ Ð=# Ñ
49
70
84
105
B‡#
1
#
1,$
#
B‡"
"ß $
The optimal solutions are ÐB‡" œ ", B‡# œ #, B‡$ œ $Ñ and ÐB‡" œ $, B‡# œ #, B‡$ œ "Ñ.
11.2-3.
(a) The five stages of the problem correspond to the five columns of the network graph.
The states are the nodes of the graph. Given the activity times >34 , the problem can be
formulated as follows:
‡
08‡ Ð=Ñ œ max Ò>=B8  08"
ÐB8 ÑÓ
B8
0'‡ Ð*Ñ œ !
(b) The critical paths are " Ä # Ä % Ä ( Ä * and " Ä # Ä & Ä ( Ä *.
11-3
(c) Interactive Deterministic Dynamic Programming Algorithm: Number of stages: 4
11.2-4.
(a) FALSE. It uses a recursive relationship that enables solving for the optimal policy for
stage 8 given the optimal policy for stage Ð8"Ñ [Feature 7, Section 11.2].
(b) FALSE. Given the current state, an optimal policy for remaining stages is
independent of the policy decisions adopted in previous stages. Therefore, the optimal
immediate decision depends on only the current state and not on how you got there. This
is the Principle of Optimality for dynamic programming [Feature 5, Section 11.2].
(c) FALSE. The optimal decision at any stage depends on only the state at that stage and
not on the past. This is again the Principle of Optimality [Feature 5, Section 11.2].
11.3-1.
The Military Airlift Command (MAC) employed dynamic programming in scheduling its
aircraft, crew and mission support resources during Operation Desert Storm. The primary
goal was to deliver cargo and passengers on time in an environment with time and space
constraints. The missions are scheduled sequentially. The schedule of a mission imposes
resource constraints on the schedules of following missions. A balance among various
objectives is sought. In addition to maximizing timely deliveries, MAC aimed at reducing
late deliveries, total flying time of each mission, ground time and frequency of crew
changes. Maximizing on-time deliveries is included in the model as a lower bound on the
load of the mission. The problem for any given mission is then to determine a feasible
schedule that minimizes a weighted sum of the remaining objectives. The constraints are
the lower bound constraints, crew and ground-support availability constraints. Stages are
the airfields in the network and states are defined as airfield, departure time, and
remaining duty day. The solution of the problem is made more efficient by exploiting the
special structure of the objective function.
The software developed to solve the problems cost around $2 million while the airlift
operation cost over $3 billion. Hence, even a small improvement in efficiency meant a
considerable return on investment. A systematic approach to scheduling yielded better
11-4
coordination, improved efficiency, and error-proof schedules. It enabled MAC not only
to respond quickly to changes in the conditions, but also to be proactive by evaluating
different scenarios in short periods of time.
11.3-2.
Let B8 be the number of crates allocated to store 8, :8 ÐB8 Ñ be the expected profit from
allocating B8 to store 8 and =8 be the number of crates remaining to be allocated to stores
‡
5 8. Then 08‡ Ð=8 Ñ œ max Ò:8 ÐB8 Ñ  08"
Ð=8 B8 ÑÓ. Number of stages: 3
!ŸB8 Ÿ=8
11-5
11.3-3.
Let B8 be the number of study days allocated to course 8, :8 ÐB8 Ñ be the number of grade
points expected when B8 days are allocated to course 8 and =8 be the number of study
days remaining to be allocated to courses 5 8. Then
08‡ Ð=8 Ñ œ
max
"ŸB8 ŸminÐ=8 ß%Ñ
‡
Ò:8 ÐB8 Ñ  08"
Ð=8 B8 ÑÓ.
Number of stages: 4
=%
1
2
3
4
0%‡ Ð=% Ñ
'
7
*
*
B‡%
1
2
3
4
=$
#
$
%
&
"
)
*
""
""
0$ Ð=$ ß B$ Ñ
#
$
%
  
"!  
"" "$ 
"$ "% "%
=#
$
%
&
'
"
"$
"&
")
"*
0# Ð=# ß B# Ñ
#
$
 
"$ 
"& "%
") "'
="
(
0" Ð=" ß B" Ñ
"
#
$
%
## #$ #" #!
Optimal Solution
"
%



"(
B‡"
#
0$‡ Ð=$ Ñ
)
"$
"$
"%
B‡$
"
#
$
$ß %
0#‡ Ð=# Ñ
"$
"&
")
"*
B‡#
"
"
1
"
0"‡ Ð=" Ñ
#$
B‡"
#
B‡#
"
B‡%
"
B‡$
$
11-6
11.3-4.
Let B8 be the number of commercials run in area 8, :8 ÐB8 Ñ be the number of votes won
when B8 commercials are run in area 8 and =8 be the number of commercials remaining
to be allocated to areas 5 8. Then
08‡ Ð=8 Ñ œ
‡
max Ò:8 ÐB8 Ñ  08"
Ð=8 B8 ÑÓ.
!ŸB8 Ÿ=8
Number of stages: 4
11-7
11.3-5.
Let B8 be the number of workers allocated to precinct 8, :8 ÐB8 Ñ be the increase in the
number of votes if B8 workers are assigned to precinct 8 and =8 be the number of
workers remaining at stage 8. Then
08‡ Ð=8 Ñ œ
‡
max Ò:8 ÐB8 Ñ  08"
Ð=8 B8 ÑÓ.
!ŸB8 Ÿ=8
Number of stages: 4
11-8
11.3-6.
Let &B8 be the number of jet engines produced in month 8 and =8 be the inventory at the
beginning of month 8. Then 08‡ Ð=8 Ñ is:
min
maxÐ<8 =8 ß!ÑŸB8 Ÿ78
‡
Ò-8 B8  .8 maxÐ=8  B8  <8 ß !Ñ  08"
ÐmaxÐ=8  B8  <8 ß !ÑÑÓ
and 0%‡ Ð=% Ñ œ -% maxÐ=%  <% ß !Ñ.
Using the following data adjusted to reflect that B8 is one fifth of the actual production,
Month
"
#
$
%
<8
#
$
&
%
78
&
(
'
#
-8
&Þ%!
&Þ&&
&Þ&!
&Þ'&
.8
!Þ!(&
!Þ!(&
!Þ!(&
!Þ!(&
the following tables are produced:
=%
#
$
%
=$
"
#
$
%
&
'
(
0%‡ Ð=% Ñ
""Þ$!
&Þ'&
!Þ!!
!






""Þ%&
=#
!
"
#
$
!




="
!
!

B‡%
#
"
!
"





"'Þ*&
""Þ$(&
0$ Ð=$ ß B$ Ñ
#
$
%








$$Þ%&

#(Þ*&
#(Þ)(&
##Þ%&
##Þ$(& ##Þ$!
"'Þ)(& "'Þ)!

""Þ$!


&

$)Þ*&
$$Þ$(&
#(Þ)!



'
%%Þ%&
$)Þ)(&
$$Þ$!




0# Ð=# ß B# Ñ
"
#
$
%
&
'



''Þ(#& ''Þ((& ''Þ)#&


'"Þ"(& '"Þ##& '"Þ#(& '"Þ%!

&&Þ'#& &&Þ'(& &&Þ(#& &&Þ)&
&&Þ*(&
&!Þ!(& &!Þ"#& &!Þ"(& &!Þ$!
&!Þ%#& &!Þ&&
0" Ð=" ß B" Ñ
"
#
$
%
&
0"‡ Ð=" Ñ B‡"
 ((Þ&#& ((Þ%& ((Þ$(& ((Þ$! ((Þ$! &
0$‡ Ð=$ Ñ
%%Þ%&
$)Þ)(&
$$Þ$!
#(Þ)!
##Þ$!
"'Þ)!
""Þ$!
(
''Þ*&
'"Þ&#&
&'Þ"!
&!Þ'(&
B‡$
'
'
'
&
%
$
#
0#‡ Ð=# Ñ
''Þ(#&
'"Þ"(&
&&Þ'#&
&!Þ!(&
B#‡
%
$
#
"
Hence, the optimal production schedule is to produce & † & œ #& units in the first month,
" † & œ & in the second, ' † & œ $! in the third and # † & œ "! in the last month.
11-9
11.3-7.
(a) Let B8 be the amount in million dollars spent in phase 8, =8 be the amount in million
dollars remaining, :" ÐB" Ñ be the initial share of the market attained in phase 1 when B" is
spent in phase 1, and :8 ÐB8 Ñ be the fraction of this market share retained in phase 8 if B8
is spent in phase 8, for 8 œ #ß $. Number of stages: 3
=$
!
"
#
$
0$‡ Ð=$ Ñ
!Þ$
!Þ&
!Þ'
!Þ(
B‡$
!
"
#
$
=#
!
"
#
$
0# Ð=# ß B# Ñ
!
"
#
!Þ!' 

!Þ"
!Þ"# 
!Þ"# !Þ#
!Þ"&
!Þ"% !Þ#% !Þ#&
="
%
0" Ð=" ß B" Ñ
" # $
%
& ' %Þ) $
0#‡ Ð=# Ñ
!Þ!'
!Þ"#
!Þ#
!Þ#&
$



!Þ")
0"‡ Ð=" Ñ
'
B‡#
!
"
"
#
B‡"
#
The optimal solution is B‡" œ #, B‡# œ ", and B‡$ œ ". Hence, it is optimal to spend two
million dollars in phase 1 and one million dollar in each one the phases 2 and 3. This will
result in a final market share of 6%.
(b) Phase 3:
=
!Ÿ=Ÿ%
0$‡ Ð=Ñ
!Þ'  !Þ!(=
B‡$
=
Phase 2: 0# Ð=ß B# Ñ œ Ð!Þ%  !Þ"B# ÑÒ!Þ'  !Þ!(Ð=  B# ÑÓ
œ !Þ!(B##  Ð!Þ!(=  !Þ!$#ÑB#  Ð!Þ#%  !Þ!#)=Ñ
`0# Ð=ßB# Ñ
`B#
œ !Þ!"%B#  !Þ!!(=  !Þ!$# œ ! Ê B‡# œ
=
#

"'
(
If = Ÿ #=  "'( : B‡# œ = because 0# Ð=ß B# Ñ is strictly increasing on the interval Ò!ß #= 
so on Ò!ß =Ó.
11-10
"'
( Ó,
If = 
=
#

"'
(:
B‡# œ
=
#

"'
(
because then the global maximizer is feasible.
We can summarize this result as:
B‡# Ð=Ñ œ min #= 
Now since ! Ÿ = Ÿ % Ÿ
$#
(,
=Ÿ
"'
( ß = .
=
#

"'
(,
so B‡# Ð=Ñ œ = and 0#‡ Ð=Ñ œ !Þ!'=  !Þ#%.
Phase 1: 0" Ð%ß B" Ñ œ Ð"!B"  B#" ÑÒ!Þ!'Ð%  B" Ñ  !Þ#%Ó
œ !Þ!'B$"  "Þ!)B#"  %Þ)B"
`0" Ð%ßB" Ñ
`B"
œ !Þ")B#" #Þ"'B"  %Þ) œ !
Ê B‡" œ
#Þ"'„#Þ"'# %Ð!Þ")ÑÐ%Þ)Ñ
#Ð!Þ")Ñ
œ #Þ*%& or *Þ!&&.
The derivative of 0" Ð%ß B" Ñ is nonnegative for B" Ÿ #Þ*%& and B" *Þ!&& and
nonpositive otherwise, so 0" Ð%ß B" Ñ is nonincreasing on the interval Ò#Þ*%&ß *Þ!&&Ó, and
nondecreasing else. Thus, 0" Ð%ß B" Ñ attains its maximum over the interval Ò!ß %Ó at
B‡" œ #Þ*%& with 0"‡ Ð%Ñ œ 'Þ$!#. Accordingly, it is optimal to spend #Þ*%& million dollars
in Phase 1, "Þ!&& in Phase 2 and Phase 3. This returns a market share of 6.302%.
11.3-8.
Let B8 be the number of parallel units of component 8 that are installed, :8 ÐB8 Ñ be the
probability that the component will function if it contains B8 parallel units, -8 ÐB8 Ñ be the
cost of installing B8 units of component 8, =8 be the amount of money remaining in
hundreds of dollars. Then
08‡ Ð=8 Ñ œ
max
B8 œ!ßáßminÐ$ßα=8 Ñ
‡
Ò:8 ÐB8 Ñ08"
Ð=8 -8 ÐB8 ÑÑÓ
where α=8 ´ maxÖα À -8 ÐαÑ Ÿ =8 ß α integer×.
=%
!ß "
#
$
% Ÿ =% Ÿ "!
0%‡ Ð=% Ñ
!
!Þ&
!Þ(
!Þ*
B‡%
!
"
#
$
11-11
0$ Ð=$ ß B$ Ñ œ P$ ÐB$ Ñ0%‡ Ð=$ -$ ÐB$ ÑÑ
=$
!
"ß #
$
%
&
'
(
) Ÿ =$ Ÿ "!
!
!
!
!
!
!
!
!
!
0$ Ð=$ ß B$ Ñ
"
#
$



!


!Þ$&
!

!Þ%*
!
!
!Þ'$ !Þ%!
!
!Þ'$ !Þ&' !Þ%&
!Þ'$ !Þ(# !Þ'$
!Þ'$ !Þ(# !Þ)"
0$‡ Ð=$ Ñ
!
!
!Þ$&
!Þ%*
!Þ'$
!Þ'$
!Þ(#
!Þ)"
B‡$
!
!ß "
"
"
"
"
#
$
0# Ð=# ß B# Ñ œ P# ÐB# Ñ0$‡ Ð=# -# ÐB# ÑÑ
=#
!ß "
#ß $
%
&
'
(
)
*
"!
!
!
!
!
!
!
!
!
!
!
0# Ð=# ß B# Ñ
"
#


!

!
!
!Þ#"!
!
!Þ#*%
!
!Þ$() !Þ#%&
!Þ$() !Þ$%$
!Þ%$# !Þ%%"
!Þ%)' !Þ%%"
0#‡ Ð=# Ñ
!
!
!
!Þ#"!
!Þ#*%
!Þ$()
!Þ$()
!Þ%%"
!Þ&!%
$



!
!
!
!Þ#)!
!Þ$*#
!Þ&!%
B‡#
!
!ß "
!ß "ß #
"
"
"
"
#
$
0" Ð=" ß B" Ñ œ P" ÐB" Ñ0#‡ Ð=" -" ÐB" ÑÑ
="
"!
!
!
0" Ð=" ß B" Ñ
"
#
!Þ## !Þ##(
$
!Þ$!#
0"‡ Ð=" Ñ
!Þ$!#
B‡"
$
The optimal solution is B‡" œ $, B‡# œ ", B‡$ œ " and B‡% œ $, yielding a system reliability
of 0.3024.
11.3-9.
The stages are 8 œ "ß # and the state is the amount of slack remaining in the constraint,
the goal is to find 0"‡ Ð%Ñ.
=#
!
"
#
$
%
0#‡ Ð=# Ñ
!
!
%
%
"#
B‡#
!
!
"
"
#
="
%
!
"#
0" Ð=" ß B" Ñ
" # $
' ) !
The optimal solution is B‡" œ ! and B‡# œ #.
11-12
%
"'
0"‡ Ð=" Ñ
"#
B‡"
!
11.3-10.
The stages are 8 œ "ß #ß $ and the state is the slack remaining in the constraint, the goal is
to find 0"‡ Ð""Ñ.
=$
!#
$&
')
*  ""
="
""
!
&!
0$‡ Ð=$ Ñ
!
"!
#!
$!
B‡$
!
"
#
$
0" Ð=" ß B" Ñ
"
#
$
%
&( '# '& ''
=#
!#
$
%&
'
(
)
*
"!
""
&
'&
0# Ð=# ß B# Ñ
!
"
#
!
 
"!  
"! #! 
#! #! 
#! $! 
#! $! %!
$! $! %!
$! %! %!
$! %! &!
0"‡ Ð=" Ñ
''
0#‡ Ð=# Ñ
!
"!
#!
#!
$!
%!
%!
%!
&!
B‡#
!
!
"
!ß "
"
#
#
"ß #
#
B‡"
%
The optimal solution is B‡" œ %, B‡# œ !, B‡$ œ " with an objective value D ‡ œ ''.
11.3-11.
Let =8 denote the slack remaining in the constraint.
0#‡ Ð=# Ñ œ max $'B#  $B#$ 
  ! for ! Ÿ B  #
#
œ $'  *B##  œ ! for B# œ #
  ! for B#  #
!ŸB# Ÿ=#
`0# Ð=ßB# Ñ
`B#
Ê B‡# œ 
=#
#
for ! Ÿ =#  #
for # Ÿ =# Ÿ $
0"‡ Ð$Ñ œ max Ò$'B"  *B#"  'B$"  0#‡ Ð$B" ÑÓ
 max Ò$'B  *B#  'B$  %)Ó
"
"
"
!ŸB" Ÿ"
œ max 
#
max Ò$'B"  *B"  'B"$  $'Ð$B" Ñ  $Ð$B" Ñ$ Ó
 "ŸB
" Ÿ$
!ŸB" Ÿ$

")ÐB#"  B"  #Ñ  !
for ! Ÿ B" Ÿ " Ê B"max œ "




  ! for " Ÿ B"  #  "$
`0" Ð$ßB" Ñ
œ

max
`B"

*ÐB#"  %B"  *Ñ œ ! for B" œ #  "$

 Ê B" œ #  "$




  ! for B"  #  "$
Since 0" Ð$ß "Ñ  0" Ð$ß #  "$Ñ, B‡" œ #  "$ ¶ "Þ'" and B‡# œ &  "$ ¶ "Þ$*
with the optimal objective value being 0"‡ Ð$Ñ ¶ *)Þ#$.
11-13
11.3-12.
08‡ Ð=8 Ñ œ
min
<8 ŸB8 Ÿ#&&
‡
Ò"!!ÐB8  =8 Ñ#  #!!!ÐB8  <8 Ñ  08"
ÐB8 ÑÓ
8 œ %:
=%
#!! Ÿ =% Ÿ #&&
0%‡ Ð=% Ñ
"!!Ð#&&  =% Ñ#
B‡%
#&&
8 œ $: 0$ Ð=$ ß B$ Ñ œ "!!ÐB$  =$ Ñ#  #!!!ÐB$  #!!Ñ  "!!Ð#&&  B$ Ñ#
` 0$ Ð=$ ßB$ Ñ
`B$
œ #!!ÐB$  =$ Ñ  #!!!  #!!Ð#&&  B$ Ñ
œ #!!Ò#B$  Ð=$  #%&ÑÓ œ ! Ê B$ œ
=$ #%&
#
If "&& Ÿ =$ Ÿ #'&, #!! Ÿ =$ #%&
Ÿ #&&, so B$ œ =$ #%&
is feasible for #%! Ÿ =$ Ÿ #&&
#
#
‡
#
#
and 0$ Ð=$ Ñ œ #&Ð#%&  =$ Ñ  #&Ð#'&  =$ Ñ  "!!!Ð=$  "&&Ñ.
0$‡ Ð=$ Ñ
#&Ð#%&  =$ Ñ#  #&Ð#'&  =$ Ñ#  "!!!Ð=$  "&&Ñ
=$
#%! Ÿ =$ Ÿ #&&
B‡$
=$ #%&
#
8 œ #: 0# Ð=# ß B# Ñ œ "!!ÐB#  =# Ñ#  #!!!ÐB#  #%!Ñ  0$‡ ÐB# Ñ
` 0# Ð=# ßB# Ñ
`B#
œ #!!ÐB#  =# Ñ  #!!!  &!Ð#%&  B# Ñ  &!Ð#'&  B# Ñ  "!!!
œ "!!Ò$B#  Ð#=#  ##&ÑÓ œ ! Ê B# œ
If #%(Þ& Ÿ =# Ÿ #&&, #%! Ÿ
#=# ##&
$
Ÿ #&&, so B‡# œ
#=# ##&
$
#=# ##&
$
and
0#‡ Ð=# Ñ œ "!! #=# ##&
 =#   #!!! #=# ##&
 #%!  0$‡  #=# ##&

$
$
$
#
œ
"!!
* ÒÐ##&
 =# Ñ#  Ð#&&  =# Ñ#  Ð#)&  =# Ñ#  '!Ð$=#  '"&ÑÓ.
If ##! Ÿ =# Ÿ #%(Þ&,
#=# ##&
$
Ÿ #%! Ÿ B# , so
` 0# Ð=# ßB# Ñ
`B#
! and hence B‡# œ #%! and
0#‡ Ð=# Ñ œ "!!Ð#%!  =# Ñ#  #!!!Ð#%!  #%!Ñ  0$‡ Ð#%!Ñ œ "!!Ð#%!  =# Ñ#  "!"ß #&!.
=#
##! Ÿ =# Ÿ #%(Þ&
#%(Þ& Ÿ =# Ÿ #&&
0#‡ Ð=# Ñ
"!!Ð#%!  =# Ñ#  "!"ß #&!
"!!
#
#
#
* ÒÐ##&  =# Ñ  Ð#&&  =# Ñ  Ð#)&  =# Ñ  '!Ð$=#  '"&ÑÓ
8 œ ": 0" Ð#&&ß B" Ñ œ "!!ÐB"  #&&Ñ#  #!!!ÐB"  ##!Ñ  0#‡ ÐB" Ñ
If ##! Ÿ B" Ÿ #%(Þ&:
` 0# Ð#&&ßB" Ñ
`B"
œ #!!ÐB"  %)&Ñ œ ! Ê B‡" œ #%#Þ&.
If #%(Þ& Ÿ B" Ÿ #&&:
` 0# Ð#&&ßB" Ñ
`B"
œ
)!!
$ ÐB"
 #%!Ñ  ! Ê B‡" œ #%(Þ&.
11-14
B‡#
#%!
#=# ##&
$
The optimal solution is B‡" œ #%#Þ& and
0"‡ Ð#&&Ñ œ "!!Ð#%#Þ&  #&&Ñ#  #!!!Ð#%#Þ&  ##!Ñ  0#‡ Ð#%#Þ&Ñ œ "'#ß &!!.
0"‡ Ð=" Ñ
"'#ß &!!
="
#&&
Summer
#%#Þ&
B‡"
#%#Þ&
Autumn
#%!
Winter
#%#Þ&
Spring
#&&
11.3-13.
Let =8 be the amount of the resource remaining at beginning of stage 8.
8 œ $:
max Ð%B$  B#$ Ñ
!ŸB$ Ÿ=$
`
`B$ Ð%B$
 B#$ Ñ œ %  #B$ œ ! Ê B‡$ œ #
`#
Ð%B$
`B#$
 B#$ Ñ œ #  ! Ê B‡$ œ # is a maximum.
=$
! Ÿ =$ Ÿ #
# Ÿ =$ Ÿ %
8 œ #:
0$* Ð=$ Ñ
%=$  =#$
%
B*$
=$
#
max Ò#B#  0$* Ð=#  B# ÑÓ
!ŸB# Ÿ=#
If ! Ÿ =#  B# Ÿ #:
max Ò#B#  %Ð=#  B# Ñ  Ð=#  B# Ñ# Ó
!ŸB# Ÿ=#
`
`B# Ò#B#
 %Ð=#  B# Ñ  Ð=#  B# Ñ# Ó œ #  #=#  #B# œ ! Ê B#‡ œ =#  "
`#
Ò#B#
`B##
 %Ð=#  B# Ñ  Ð=#  B# Ñ# Ó œ #  ! Ê B#‡ œ =#  " is a maximum.
0#* Ð=# Ñ œ #=#  ".
If # Ÿ =#  B# Ÿ %:
=#
! Ÿ =# Ÿ "
" Ÿ =# Ÿ %
8 œ ":
max Ð#B#  %Ñ, B‡# œ =#  # and 0#* Ð=# Ñ œ #=#  #=#  ".
!ŸB# Ÿ=#
0#* Ð=# Ñ
%=#  =##
#=#  "
B*#
!
=#  "
max Ò#B#"  0#* Ð%  #B" ÑÓ
!ŸB" Ÿ="
If ! Ÿ %  #B" Ÿ ": max Ò#B#"  %Ð%  #B" Ñ  Ð%  #B" Ñ# Ó œ Ð#B#"  )B" Ñ
!ŸB" Ÿ="
`
#
`B" Ð#B"
 )B" Ñ œ %B"  ) œ ! Ê B‡" œ #
`#
Ð#B#"
`B#"
 )B" Ñ œ %  ! Ê B‡" œ # is a maximum.
0" Ð%ß #Ñ œ ).
11-15
If " Ÿ %  #B" Ÿ %: max Ò#B#"  #Ð%  #B" Ñ  "Ó œ Ð#B#"  %B"  *Ñ
!ŸB" Ÿ="
`
#
`B" Ð#B"
 %B"  *Ñ œ %B"  % œ ! Ê B" œ "
`#
Ð#B#"
`B#"
 %B"  *Ñ œ %  ! Ê B" œ " is a minimum.
Corner points: " œ %  #B" Ê B" œ $Î#, 0" Ð%ß $Î#Ñ œ (Þ&
% œ %  #B" Ê B" œ !, 0" Ð%ß !Ñ œ * is maximum.
Hence, B‡" œ !, B‡# œ $, B‡$ œ " and 0"‡ Ð%Ñ œ *.
11.3-14.
min #B## Ê B‡# œ =# and 0#‡ Ð=# Ñ œ #=# ,
8 œ #:
B## =#
where =# represents the amount of # used by B## .
minÒB%"  0#‡ ÐÐ#  B#" Ñ ÑÓ œ ÒB"%  #Ð#  B"# Ñ Ó,
8 œ ":
B"
where Ð#  B#" Ñ œ maxÖ!ß #  B"# ×.
If B#" Ÿ #:
`
%
`B" ÐB"
 %  #B#" Ñ œ %B$"  %B" œ ! Ê B" œ !ß "ß ".
`#
ÐB%"
`B#"
 %  #B#" Ñ œ "#B#"  %
B" œ !,
`#
ÐB%"
`B#"
B" œ "ß ",
 %  #B#" Ñ œ %  !, so B" œ ! is a local maximum.
`#
ÐB%"
`B#"
 %  #B#" Ñ œ )  !, so B" œ "ß " are local minima
with D œ $.
If B#"
#:
B" œ ! and D œ %  $.
Hence, ÐB‡" ß B‡# Ñ − ÖÐ"ß "Ñß Ð"ß "Ñß Ð"ß "Ñß Ð"ß "Ñ×, all with D ‡ œ $.
11.3-15.
(a) Let =8 − Ö"ß #ß %× be the remaining factor % entering stage 8.
8 œ $:
8 œ #:
B‡$
"
#
$
=$
"
#
%
0$* Ð=$ Ñ
"'
$#
'%
="
%
0" Ð=" ß B" Ñ
"
#
%
)" %% )%
=#
"
#
%
0# Ð=# ß B# Ñ
"
#
%
#!  
$' $# 
') %) )!
0#‡ Ð=# Ñ
#!
$'
)!
8 œ ":
0"‡ Ð=" Ñ
)%
B‡"
%
The optimal solution is ÐB‡" ß B‡# ß B‡$ Ñ œ Ð%ß "ß "Ñ with D ‡ œ )%.
11-16
B‡#
"
"
%
(b) As in part (a), let =8 be the remaining factor (not necessarily integer) at stage 8.
0$‡ Ð=$ Ñ œ "'=$ and B‡$ œ =$
0#‡ Ð=# Ñ œ
max Ö%B##  0$‡ Ð=# ÎB# Ñ× œ
max Ö%B##  "'=# ÎB# ×
"ŸB# Ÿ=#
` 0# Ð=# ßB# Ñ
`B#
œ %B#  "'=# ÎB## and
` # 0# Ð=# ßB# Ñ
`B##
"ŸB# Ÿ=#
œ %  $#=# ÎB#$  !
when =# , B# !. Thus 0# Ð=# ß B# Ñ is convex in B# when =# , B#
occur at one of the endpoints.
!. The maximum should
B# œ ", 0# Ð=# ß "Ñ œ %  "'=#
B# œ =# , 0# Ð=# ß =# Ñ œ %=##  "'
%=##  "' Í Ð=#  $ÑÐ=#  "Ñ Ÿ ! Í " Ÿ =# Ÿ $
%  "'=#
B‡# œ 
"
=#
0"‡ Ð=" Ñ œ
%  "'=#
if " Ÿ =# Ÿ $
and 0#‡ Ð=# Ñ œ  #
if $ Ÿ =# Ÿ %
%=#  "'
if " Ÿ =# Ÿ $
if $ Ÿ =# Ÿ %
max ÖB"$  0#‡ Ð%ÎB" Ñ×
"ŸB" Ÿ%
œ
`#
B$"
`B#"
`#
B$"
`B#"
max max B$"  % "'
  "'ß
B#"
"ŸB" Ÿ%Î$
max B$"  %  "' B%" 
%Î$ŸB" Ÿ%
 % "'
  "' œ 'B"  #!%ÎB"%  ! when B"
B#
"
 %  "' B%"  œ 'B"  "#)ÎB"#  ! when B"
!
!
Hence, the maximum occurs at an endpoint.
B" œ ", 0" Ð=" ß "Ñ œ )"
B" œ %Î$, 0" Ð=" ß %Î$Ñ ¸ &%Þ$(
B" œ %, 0" Ð=" ß %Ñ œ )%
0"‡ Ð=" Ñ œ maxÖ)"ß &%Þ$(ß )%× œ )% and ÐB‡" ß B‡# ß B‡$ Ñ œ Ð%ß "ß "Ñ, just as when the
variables are restricted to be integers.
11.3-16.
Let =8 be the slack remaining in the constraint B"  B#  B$ Ÿ ", entering the 8th stage.
0$‡ Ð=$ Ñ œ
0#‡ Ð=# Ñ
œ
max B$ œ =$ and B‡$ œ =$
!ŸB$ Ÿ=$
max ÖÐ"  B# Ñ0$‡ Ð=#  B# Ñ× œ
=
# ŸB#
max ÖÐ"  B# ÑÐ=#  B# Ñ×
=# ŸB#
where =
# œ maxÖ=# ß !×.
` 0# Ð=# ßB# Ñ
`B#
` # 0# Ð=# ßB# Ñ
`B##
œ #B#  Ð=#  "Ñ œ ! Ê B# œ Ð"  =# ÑÎ#
œ #  !, so 0# Ð=# ß B# Ñ is concave in B# .
B# œ Ð=#  "ÑÎ#, 0# Ð=# ß Ð"  =# ÑÎ#Ñ œ Ð"  =# Ñ# Î%
11-17

B# œ =
# , 0# Ð=# ß =# Ñ œ 
Ð"  =# Ñ# Î#
!
=#
if =# Ÿ !
if =# !
maxÖ!ß =# ×
B# œ Ð"  =# ÑÎ# is feasible if and only if =
# Ÿ Ð"  =# ÑÎ#, equivalently when =#
0#‡ Ð=# Ñ œ 
0"‡ Ð=" Ñ œ
œ
B"
`
`B"  %
$
".
!
if =# Ÿ "
=
if =# Ÿ "
‡
# œ =#
B
œ
and

#
#
Ð"  =# ÑÎ# if =# "
Ð"  =# Ñ Î% if =# "
maxÖB" 0#‡ Ð"  B" Ñ× œ
B" !
max  %"  B#"  B" 
max max B"  %"  Ð"  B" Ñß !
!ŸB Ÿ#
B#
"
B$
!ŸB" Ÿ#
 B#"  B"  œ
$B#"
%
 #B"  " œ ! Ê B" œ
#„%$
$Î#
œ
%
$
„
#
$
Hence, ÐB‡" ß B‡# ß B‡$ Ñ œ Ð#Î$ß "Î$ß #Î$Ñ and D ‡ œ )Î#(.
11.4-1.
Let =8 be the current fortune of the player, E be the event to have $100 at the end and \8
be the amount bet at the 8th match.
0$‡ Ð=$ Ñ œ
max ÖPÖEl=$ ××
!ŸB$ Ÿ=$
! Ÿ =$  &!, 0$‡ Ð=$ Ñ œ !.
&! Ÿ =$  "!!, 0$‡ Ð=$ Ñ œ 
=$ œ "!!, 0$‡ Ð=$ Ñ œ 
=$  "!!, 0$‡ Ð=$ Ñ œ 
!
"
!
if B‡$ Á "!!  =$
"Î# if B‡$ œ "!!  =$
if B‡$  !
if B‡$ œ !
!
if B‡$ Á =$  "!!
"Î# if B‡$ œ =$  "!!
11-18
=$
! Ÿ =$  &!
&! Ÿ =$  "!!
=$ œ "!!
"!!  =$
0$‡ Ð=$ Ñ
!
"Î#
"
"Î#
B‡$
! Ÿ B‡$ Ÿ &!
"!!  =$
!
=$  "!!
0#‡ Ð=# Ñ
!
!
"Î%
"Î%
"Î#
"Î#
"Î%
"Î#
"Î%
"Î#
$Î%
"Î%
"Î#
$Î%
"Î#
"Î%
"
"Î#
"Î%
"Î#
$Î%
"Î#
"Î%
B‡#
! Ÿ B# Ÿ =#
! Ÿ B# Ÿ &!  =#
&!  =# Ÿ B# Ÿ =#
! Ÿ B#  &!
B# œ &!
! Ÿ B#  =#  &!
=#  &!  B#  "!!  =#
B# œ "!!  =#
"!!  =#  B# Ÿ =#
! Ÿ B#  #&
B# œ #&
#& Ÿ B# Ÿ (&
! Ÿ B#  "!!  =#
B# œ "!!  =#
"!!  =#  B# Ÿ =#  &!
=#  &!  B# Ÿ =#
B# œ !
!  B# Ÿ &!
&! Ÿ B# Ÿ "!!
! Ÿ B# Ÿ =#  "!!
B# œ =#  "!!
=#  "!!  B# Ÿ =#  &!
=#  &!  B# Ÿ =#
max  " 0$‡ Ð=#
!ŸB# Ÿ=# #
0#‡ Ð=# Ñ œ
=#
! Ÿ =#  #&
#& Ÿ =#  &!
=# œ &!
&!  =#  (&
=# œ (&
(&  =#  "!!
=# œ "!!
"!!  =#
 B# Ñ  "# 0$‡ Ð=#  B# Ñ
The entries in bold represent the maximum value in each case.
0"‡ Ð(&Ñ œ
max  " 0#‡ Ð(&
!ŸB" Ÿ(& #

$Î%



 &Î)
0" Ð(&ß B" Ñ œ  $Î%



 "Î#
 $Î)
="
(&
0"‡ Ð=" Ñ
$Î%
 B" Ñ  "# 0#‡ Ð(&  B" Ñ
if B" œ !
if !  B"  #&
if B" œ #&
if #&  B" Ÿ &!
if &!  B" Ÿ (&
B‡"
! or #&
11-19
Policy
"
#
B"
!
#&
won "st bet
#&
!
lost "st bet
#&
&!
won #nd bet
!
!
lost #nd bet
&!
!
11.4-2.
(a) Let B8 − Ö!ß Eß F× be the investment made in year 8, =8 be the amount of money on
hand at the beginning of year 8 and 08 Ð=8 ß B8 Ñ be the maximum expected amount of
money by the end of the third year given =8 and B8 .
‡
For ! Ÿ =8  &ß !!!, since one cannot invest less than $&!!!, 08 Ð=8 ß B8 Ñ œ 08"
Ð=8 Ñ and
B‡8 œ !.
For =8
&!!!,
 0 ‡ Ð=8 Ñ
if B8 œ !
8"
‡
‡
08 Ð=8 ß B8 Ñ œ  !Þ$08" Ð=8  "!ß !!!Ñ  !Þ(08" Ð=8  &!!!Ñ if B8 œ E
‡
‡
 !Þ*08"
Ð=8 Ñ  !Þ"08"
Ð=8  &!!!Ñ
if B8 œ F
=$
! Ÿ =$  &!!!
=$ &!!!
0$‡ Ð=$ Ñ
=$
=$  #!!!
=#
! Ÿ =#  &!!!
&!!! Ÿ =#  "!ß !!!
=# "!ß !!!
="
&!!!
B‡$
!
E
!
=#
=#  #!!!
=#  #!!!
0" Ð=" ß B" Ñ
!
E
F
)%!! *)!! )"&!
0# Ð=# ß B# Ñ
E
F


=#  $%!! =#  #&!!
=#  %!!! =#  #&!!
0"‡ Ð=" Ñ
*)!!
0#‡ Ð=# Ñ
=#
=#  $%!!
=#  %!!!
B‡#
!
E
E
B"‡
E
The optimal policy is to invest in E with an expected fortune after three years of $*)!!.
(b) Let B8 and =8 be defined as in (a). Let 08 Ð=8 ß B8 Ñ be the maximum probability of
having at least $#0,000 after 3 years given =8 and B8 .
=$
! Ÿ =$  5!!!
5!!! Ÿ =$  1!ß !!!
1!ß !!! Ÿ =$  15ß !!!
=$ 15ß !!!
=#
! Ÿ =#  5!!!
5!!! Ÿ =#  1!ß !!!
=# 1!ß !!!
0$ Ð=$ ß B$ Ñ
!
E F
! 

! !Þ( !Þ"
" !Þ( "
"
"
"
!
!
!Þ(
"
0$‡ Ð=$ Ñ
!
!Þ(
"
"
0# Ð=# ß B# Ñ
E
F


!Þ(
!Þ(3
!Þ(3
"
0#‡ Ð=# Ñ
!
!Þ(3
"
11-20
B$‡
!
E
!ß F
!ß Eß F
B‡#
!
F
!ß F
="
5!!!
0" Ð=" ß B" Ñ
!
E
F
!Þ73 !Þ( !Þ(57
0"‡ Ð=" Ñ
!Þ(57
B‡"
F
Hence, the optimal policies are (using the numbers on the arcs to represent the return on
investment indicated at the nodes):
and the maximum probability of having at least $10 thousand at the end of three years is
0.757.
11.4-3.
‡
‡
08 Ð"ß B8 Ñ œ OÐB8 Ñ  B8   "$  08"
Ð"Ñ  "   "$  08"
Ð!Ñ
B8
B8
‡
œ OÐB8 Ñ  B8   "$  08"
Ð"Ñ
B8
since 08‡ Ð!Ñ œ ! for every 8. 0$‡ Ð"Ñ œ "', 0$‡ Ð!Ñ œ ! and OÐB8 Ñ œ ! if B8 œ !,
OÐB8 Ñ œ $ if B8  !.
=#
!
"
="
"
!
!
"'
0# Ð=# ß B# Ñ
"
#
$



*Þ$$ 'Þ() 'Þ&*
%

(Þ#!
0#‡ Ð=# Ñ
!
'Þ&*
B‡#
!
$
!
'Þ&*
0" Ð=" ß B" Ñ
"
#
$
'Þ#! &Þ($ 'Þ#%
%
(Þ!)
0"‡ Ð=" Ñ
&Þ($
B"‡
#
The optimal policy is to produce two in the first run and to produce three in the second
run if none of the items produced in the first run is acceptable. The minimum expected
cost is $&(3.
11-21
11.4-4.
‡
‡
08‡ Ð=8 Ñ œ max "$ 08"
Ð=8  B8 Ñ  #$ 08"
Ð=8  B8 Ñ,
B8 !
with 0'‡ Ð=' Ñ œ ! for ='  & and 0'‡ Ð=' Ñ œ " for ='
=&
!
"
#
$
%
=&
=%
!
"
#
$
%
=%
=$
!
"
#
$
%
=$
=#
!
"
#
$
%
=#
="
#
&
&
&
&
0&‡ Ð=& Ñ
!
!
!
#Î$
#Î$
"
B&‡
!
!
!
B‡& #
B‡& "
B‡& Ÿ =&  &
!
!
!
!
#Î$
#Î$
"
0% Ð=% ß B% Ñ
"
#
$



!


%Î* %Î*

%Î* #Î$ #Î$
)Î* #Î$ #Î$



!
!
!
%Î*
#Î$
)Î*
"
0$ Ð=$ ß B$ Ñ
"
#


)Î#(

%Î*
"'Î#(
#!Î#( #Î$
)Î*
##Î#(


!
!
)Î#(
"'Î#(
#!Î#(
#%Î#(
"
!
%)Î)"
&.
%




#Î$

$



#Î$
#Î$

0# Ð=# ß B# Ñ
"
#


$#Î)"

%)Î)" %)Î)"
'%Î)" '#Î)"
(%Î)" (!Î)"


0" Ð=" ß B" Ñ
"
#
"'!Î#%$ "#%Î#%$
0%‡ Ð=% Ñ
!
!
%Î*
#Î$
)Î*
"
0$‡ Ð=$ Ñ
!
)Î#(
"'Î#(
#!Î#(
##Î#(
"
%




#Î$

$



#Î$
'#Î)"

B‡%
!
!
"ß #
!Þ#ß $
"
B‡% Ÿ =%  &
%




#Î$

0"‡ Ð=" Ñ
"'!Î#%$
B‡$
!
"
#
"
!ß "
B‡$ Ÿ =$  &
0#‡ Ð=# Ñ
!
$#Î)"
%)Î)"
'%Î)"
(%Î)"
"
B#‡
!
"
!ß "ß #
"
"
B‡# Ÿ =#  &
B‡"
"
The probability of winning the bet using the policy given above is "'!Î#%$ œ !Þ'&).
11-22
11.4-5.
Let B8 − ÖEß H× denote the decision variable of quarter 8 œ "ß #ß $, where E and H
represent advertising or discontinuing the product respectively. Let =8 be the level of
sales (in millions) above (=8 !) or below (=8 Ÿ !) the break-even point for quarter
Ð8  "Ñ. Let 08 Ð=8 ß B8 Ñ represent the maximum expected discounted profit (in millions)
from the beginning of quarter 8 onwards given the state =8 and decision B8 .
08 Ð=8 ß B8 Ñ œ $!  &=8 
"  ,8
,8 +8 +8 >.>

"  ,8 ‡
,8 +8 +8 08" Ð=8
 >Ñ>.>,
where +8 and ,8 are given in the table that follows.
8
"
#
$
+8
"
!
"
,8
&
%
$
For " Ÿ 8 Ÿ $,
08 Ð=8 ß EÑ œ $!  &=8 
+8 ,8
# 

"  ,8 ‡
,8 +8 +8 08" Ð=8
 >Ñ.>,
08 Ð=8 ß HÑ œ #!.
Note that once discontinuing is chosen the process stops.
08‡ Ð=8 Ñ œ maxÖ08 Ð=8 ß EÑß 08 Ð=8 ß HÑ×
8 œ %:
0%‡ Ð=% Ñ œ 
#!
%!=%
if =%  !
if =% !
8 œ $:
0$ Ð=$ ß HÑ œ #!
$
0$ Ð=$ ß EÑ œ $!  &Ð=$  "Ñ  "% " 0%‡ Ð=$  >Ñ.>,
For $ Ÿ =$ Ÿ ",
=
$
0$ Ð=$ ß EÑ œ $!  &Ð=$  "Ñ  "% " $ #!.>  =$ %!Ð=$  >Ñ.> œ &Ð=$  %Ñ#  '&
0$‡ Ð=$ Ñ œ maxÖ&Ð=$  %Ñ#  '&ß #!× œ 
#!
if $ Ÿ =$ Ÿ ", and B‡$ œ H,
&Ð=$  %Ñ#  '& if " Ÿ =$ Ÿ ", and B‡$ œ E.
For " Ÿ =$ Ÿ &,
$
0$ Ð=$ ß EÑ œ $!  &Ð=$  "Ñ  "% " %!Ð=$  >Ñ.> œ "&  %&=$
0$‡ Ð=$ Ñ œ maxÖ"&  %&=$ ß #!× œ "&  %&=$ and B‡$ œ E.
11-23
0$‡ Ð=$ Ñ
#!
&Ð=$  %Ñ#  '&
"&  %&=$
=$
$ Ÿ =$ Ÿ "
" Ÿ =$ Ÿ "
" Ÿ =$ Ÿ &
B‡$
H
E
E
8 œ #:
0# Ð=# ß HÑ œ #!
$
0# Ð=# ß EÑ œ $!  &Ð=#  "Ñ  "% " 0$‡ Ð=#  >Ñ.>,
For $ Ÿ =# Ÿ ",
$ ‡
=# "
"=#
%
"
0$ Ð=#  >Ñ.> œ "
#!.>  =# "
Ò&Ð=#  >  %Ñ#  '&Ó.>  "=# Ò"&  %&Ð=#  >ÑÓ.>
0# Ð=# ß EÑ œ &% Ð *# =##  %(=# 
%#(
' Ñ
Observe that 0# Ð$ß EÑ œ ""!Î$  0# Ð=# ß HÑ œ #!  0# Ð"ß EÑ œ #"&Î', so we
need to find $ Ÿ =# Ÿ " such that 0# Ð=# ß EÑ œ 0# Ð=# ß HÑ.
& * #
% Ð # =#
 %(=# 
%#(
' Ñ
œ #! & $ Ÿ =# Ÿ " Ê =‡# œ
%()"!
*
œ #Þ%""
For " Ÿ =# Ÿ ",
$ ‡
"=
%
"
0$ Ð=#  >Ñ.> œ ! # Ò&Ð=#  >  %Ñ#  '&Ó.>  "=# Ò"&  %&Ð=#  >ÑÓ.>
0# Ð=# ß EÑ œ &%  "$ Ð=#  %Ñ$  *# Ð=#  %Ñ#  #!=# 
"!$
' 
Since 0# Ð"ß EÑ œ #"&Î' and 0# Ð=# ß EÑ is increasing in " Ÿ =# Ÿ ", B‡# œ E is the
optimal decision in this interval.
=#
$ Ÿ =# Ÿ =‡#
=‡#  =# Ÿ "
" Ÿ =# Ÿ "
0#‡ Ð=# Ñ
#!
& * #
% Ð # =#  %(=# 
&
"
%  $ Ð=#

%#(
' Ñ
%Ñ$  *# Ð=#
 %Ñ#  #!=# 
"!$
' 
8 œ ":
0" Ð%ß HÑ œ #!
&
0" Ð%ß EÑ œ $!  &Ð%  $Ñ  "% " 0#‡ Ð%  >Ñ.>
œ $&  "% "=# % #!.>  &% =$‡# % Ð *# Ð%  >Ñ#  %(Ð%  >Ñ 
‡
&
 &% $  $" >$  #* >#  #!Ð%  >Ñ 
="
%
1st Quarter
Advertise.
0"‡ Ð=" Ñ
%Þ((
"!$
' .>
%#(
' Ñ.>
œ %Þ((
B‡"
E
2nd Quarter
If =# Ÿ #Þ%"", discontinue.
If =#  #Þ%"", advertise.
3rd Quarter
If =$ Ÿ ", discontinue.
If =$  ", advertise.
11-24
B‡#
H
E
E
CHAPTER 12: INTEGER PROGRAMMING
12.1-1.
if the decision is to build a factory in city ,
otherwise
(a)
if the decision is to build a factory in city ,
otherwise
for
LA, SF, SD.
maximize
NPV
subject to
LA
LA
LA
SF
SF
LA
SF
SF
SD
SD
SF
SD
LA
SD
LA
SF
SF
SD
SD
SD
LA
LA
SF
SD
LA
SF
SD
binary
(b) - (c)
California Manufacturing Co. Facility Location Problem
NPV ($millions)
Warehouse
Factory
Capital Required
($millions)
Warehouse
Factory
Build?
Warehouse
Factory
Los Angeles
6
San Francisco
4
San Diego
5
8
5
7
Los Angeles
5
San Francisco
2
San Diego
3
6
3
4
Los Angeles
0
<=
0
San Francisco
0
<=
1
San Diego
1
<=
1
Total NPV ($millions)
12-1
17
Capital
Spent
10
<=
Capital
Available
10
Total
Maximum
Warehouses
Warehouses
1
<=
1
12.1-2.
(a)
if does marketing,
otherwise
if does cooking,
otherwise
if does dishwashing,
otherwise
if does laundry,
otherwise
for
E Eve S Steven .
min
T
st
E
E
E
E
E
E
E
S
S
S
S
E
E
E
S
S
S
S
S
S
E
E
E
S
S
S
E
S
E
S
E
S
binary
(b) - (c)
Time Needed (hours)
Marketing
Cooking
Eve
4.5
7.8
Steven
4.9
7.2
Dishwashing
3.6
4.3
Laundry
2.9
3.1
Does Task?
Marketing
Eve
1
Steven
0
Total
1
=
1
Dishwashing
1
0
1
=
1
Laundry
0
1
1
=
1
Cooking
0
1
1
=
1
12.1-3.
(a)
if the decision is to invest in project ,
otherwise
for
.
maximize
NPV
subject to
binary
12-2
Tasks
Performed
2
=
2
=
2
2
Total Time (hours)
18.4
(b) - (c)
Estimated Profit
($million)
Capital
Project 1 Project 2 Project 3 Project 4 Project 5
1
1.8
1.6
0.8
1.4
6
Capital Required for Project ($million)
12
10
4
8
Project 1 Project 2 Project 3 Project 4 Project 5
Undertake?
1
0
1
1
0
Capital
Spent
($million)
20
<=
Capital
Available
($million)
20
Total Profit
($million)
3.4
12.1-4.
if the decision is to invest in opportunity ,
otherwise
(a)
for
.
Let
denote the estimated profit of opportunity
opportunity in millions of dollars.
maximize
subject to
binary, for
(b) Solution: Invest in opportunities 1, 3 and 5.
12-3
and
the capital required for
12.1-5.
Best Time
Carl
Chris
David
Tony
Ken
Assignments
Carl
Chris
David
Tony
Ken
Total Assigned
Demand
Stroke
Backstroke Breaststroke Butterfly
37.7
43.4
33.3
32.9
33.1
28.5
33.8
42.2
38.9
37.0
34.7
30.4
35.4
41.8
33.6
Freestyle
29.2
26.4
29.6
28.5
31.1
Stroke
Backstroke Breaststroke Butterfly
0
0
0
0
0
1
1
0
0
0
1
0
0
0
0
1
1
1
=
=
=
1
1
1
Freestyle
1
0
0
0
0
1
=
1
Total
Assignments
1
1
1
1
0
<=
<=
<=
<=
<=
Supply
1
1
1
1
1
Total Time
126.2
Each swimmer can swim only one stroke and each stroke can be assigned to only one
swimmer.
12.1-6.
(a) Let be the number of tow bars produced and
produced.
maximize
subject to
integers
(b) Optimal Solution:
$
12-4
be the number of stabilizer bars
(c)
Unit Profit
Machine 1
Machine 2
Level of Activity
Tow Bars
$130
Stabilizer Bars
$150
Resource Usage
per Unit of Activity
3.2
2.4
2
3
Activity 1
0
Resource
Used
12
<=
15
<=
Activity 2
5
Resource
Available
16
15
Total Profit
$750.00
12.1-7.
(a) Let
be the number of trucks hauling from pit to site
tons of gravel hauled from pit to site , for
and
and
be the number of
.
minimize
subject to
, for
and
integers, for
and
(b)
Hauling Cost
per Ton
North
South
Tons of Gravel
North
South
Total
Trucks
North
South
Site 1
$100
$180
Site 2
$190
$110
Site 3
$160
$140
Site 1
10
0
10
>=
10
Site 2
0
5
5
>=
5
Site 3
1
9
10
>=
10
Site 1
2
0
Site 2
0
1
Site 3
1
2
Cost per Truck
Capacity per Truck (tons)
Total
11
14
$50
5
<=
<=
18
14
Tons of Gravel
<=
Total Cost
$3,270
12.2-1.
Answers will vary.
12.2-2.
Answers will vary.
12.2-3.
Answers will vary.
12.2-4.
Answers will vary.
12-5
Truck Capacity
North
South
Site 1
10
0
Site 2
0
5
Site 3
5
10
12.3-1.
(a) Let
be a very large number, say
million.
max
st
, for
binary, for
(b)
Start-up Cost
Marginal Revenue
Product 1
$50,000
$70
Constraint 1
Constraint 2
Product 2
$40,000
$60
Product 3 Product 4
$70,000
$60,000
$90
$80
Resource
Used
6,000
<=
12,000 <=
Resource Used per Unit Produced
5
3
6
4
4
6
3
5
Product 1
0
<=
Only if Setup
0
Setup?
0
Units Produced
Product 2
2,000
<=
9,999
1
Product 3 Product 4
0
0
<=
<=
0
0
0
0
<=
<=
only if (1 or 2)
1
1
Which Constraint (0 = Constraint 1, 1 = Constraint 2):
0
Total
1
<=
Modified
Resource
Available
6,000
15,999
Max Products
2
Revenue $120,000
Setup Cost $40,000
Total Profit $80,000
12.3-2.
,
, for
12.3-3.
1.
binary
2.
binary, for
12-6
Resource
Available
6,000
6,000
.
12.3-4.
(a) Let
and
be binary variables that indicate whether or not toys 1 and 2 are
produced. Let
and
be the number of toys 1 and 2 that are produced. Also, let be
if factory 1 is used and if factory 2 is used.
maximize
subject to
integers
binary
(b)
Start-up Cost
Unit Profit
Factory 1
Factory 2
Units Produced
Only if Setup
Setup?
Toy 1
$50,000
$10
Toy 2
$80,000
$15
Hours Used per
Unit Produced
0.02
0.025
0.025
0.04
Toy 1
28,000
<=
99,999
1
Resource
Used
560
<=
700
<=
Toy 2
0
<=
0
0
Modified
Hours
Available
10,499
700
Hours
Available
500
700
Gross Profit $280,000
Setup Cost $50,000
Net Profit $230,000
Which Factory (0 = Factory 1, 1 = Factory 2)?
1
12.3-5.
(a) Let ,
, and
respectively.
be the number of long-, medium-, and short-range jets to buy
maximize
subject to
integers
12-7
(b)
Unit Profit ($million)
Money
Pilots
Maintenance
Level of Activity
(c)
Long-Range
4.2
Medium-Range
3
67
1
1.667
Resource Usage
per Unit of Activity
50
1
1.333
35
1
1
Long-Range
14
Medium-Range
0
Short-Range
16
min
min
min
maximize
subject to
binary, for
(d) Solution:
$
(same as in (b))
12-8
Short-Range
2.3
Resource
Used
1498 <=
30
<=
39.33333 <=
Resource
Available
1500
30
40
Total Profit
95.6
12.3-6.
(a)
maximize
subject to
binary, for
(b) Solution:
,
,
,
12.3-7.
(a) Let
be the number of units to produce of product
.
if product is produced,
otherwise
maximize
subject to
integer
integer
integer
binary
(b)
Customer 1
3
2
20%
3
Customer 2
2
3
40%
2
Customer 3
0
0.8
20%
5
Start Up?
0
1
1
Planes to Produce
0
<=
0
2
<=
2
1
<=
5
Startup Cost ($million)
Marginal net Revenue ($million)
Capacity Used per Plane
Maximum Order
Maximum Order (if start up)
12.4-1.
if
(i.e., produce units of ),
otherwise
(a)
for
and
.
max
st
binary
12-9
Capacity
Used
100%
<=
Total Startup Cost
Total Revenue
Total Profit
Capacity
Available
100%
2
6.8
4.8
($million)
(b) Solution:
except for
,
,
if
,
otherwise
(c)
for
and
.
max
st
binary
(d) Solution:
for
,
for
,
12.4-2.
Introduce the binary variables
.
and
and add constraints
,
,
12.4-3.
(a) Introduce the binary variables
production levels.
,
, and
to represent positive (nonzero)
maximize
subject to
,
,
binary
(b)
Unit Profit
Product 1
$50
Product 2
$20
Product 3
$25
Hours
Machine Hours Used per Unit Produced
Used
Milling machine
9
3
5
500
<=
Lathe
5
4
0
350
<=
Grinder
3
0
2
135.714 <=
Production Rate
Only if Produce
Produce?
Product 1
45.238
<=
999
1
Produce 3 Production Rate
Product 2
30.952
<=
999
1
Product 3
0
<=
0
0
Total
2
0
<=
20
12-10
Hours
Available
500
350
150
Total Profit
$2,881
<=
2
Sales Potential
12.4-4.
if
,
otherwise
(a)
for
and
.
Work out by hand the objective function contribution for
.
maximize
subject to
binary
(b) Solution:
,
,
if
,
otherwise
(c)
for
except
and
.
Work out by hand the objective function contribution for
.
maximize
subject to
binary
(d) Solution:
except
,
12-11
,
12.4-5.
if arc from node to node is in the shortest path,
otherwise
(a)
min
st
(1)
(2)
(3)
(4)
(5)
(6)
(7)
binary
(1), (2), (3) ensure that exactly one arc is used at each stage and they represent mutually
exclusive alternatives. (4), (5), (6) ensure that node is left only if it is entered and they
represent contingent decisions.
(b) Solution:
Shortest path: 1
except
2
5
,
6
12.4-6.
if route is chosen,
otherwise
(a)
Let
be the
. Let
th element of the location/route matrix, for
denote the cost of route , for
.
minimize
subject to
, for
binary, for
12-12
and
(b)
1
6
Time (hours)
Location
A
B
C
D
E
F
G
H
I
2
4
Route
5
6
4
6
7
5
8
3
1
2
0
3
0
1
1
1
1
Do Route?
4
5
Delivery Location on Route?
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
0
3
7
4
1
Route
5
6
1
0
7
0
8
1
9
7
1
1
1
10
6
1
1
1
9
0
10
0
Total on
Route
1
1
1
1
1
1
1
1
1
>=
>=
>=
>=
>=
>=
>=
>=
>=
1
1
1
1
1
1
1
1
1
Total
3
<=
3
Total Time (hours) 12
12.4-7.
if tract is assigned to station located in tract ,
otherwise
Let
be the response time to a fire in tract if that tract is served by a station located in
tract .
min
st
(1) Two fire stations have to be located.
, for
(2) Each tract needs to be assigned to a
station.
, for
and
(3) Tract can be assigned to the station
tract only if there is a station located in tract .
binary
(1) and (2) correspond to mutually exclusive alternatives and (3) represent contingent
decisions.
12-13
12.4-8.
if a station is located in tract ,
otherwise
(a)
minimize
2
2
4
30
5
subject to
binary
(b) Yes, this is a set covering problem. The activities are locating stations and the
characteristics are the fires.
is the set of all locations that could cover a fire in tract ,
e.g.,
. There has to be at least one station, so
for all .
(c) Solution:
,
$
thousand
12.4-9.
if district is chosen,
otherwise
Let be auxiliary variables that are zero for all , except for the index of the district with
largest that is chosen, is .
minimize
subject to
, for
, for
binary
This is a set partitioning problem with additional constraints.
12.5-1.
This study uses integer programming to model employee scheduling problem of Taco
Bell restaurants. In this integer program, the decision variables correspond to the number
of employees scheduled to start working at time and to work for time units. The
objective is to minimize the total payroll for the scheduling horizon. At any point in time,
12-14
the labor requirements in each store have to be met. The total number of employees is
bounded above. Without the upper bound, the problem could be solved efficiently as a
network flow problem using out-of-kilter algorithms, so the upper bound is eliminated
from the constraint set by using generalized Lagrange multipliers.
The new scheduling approach increased labor cost savings significantly. Additional
benefits include enhanced flexibility, elimination of variability among stores, improved
customer service and quality. Mathematical modeling served as a rational basis for the
evaluation of new ideas, buildings, equipment and menu items. It also allowed Taco Bell
to eliminate redundant tasks and to schedule balanced workloads. Consequently,
productivity is improved and Taco Bell saved $13 million each year in labor costs.
12.5-2.
(a) The dots represent the feasible solutions in the graph below.
Optimal Solution:
(b) The optimal solution of the LP relaxation is
nearest integer point is
, which is not feasible, since
Rounded Solutions
Violated Constraints
3rd
2nd and 3rd
none
none
Hence, none of the feasible rounded solutions is optimal for the IP problem.
12.5-3.
(a) The dots represent the feasible solutions in the graph below.
12-15
. The
.
Optimal Solution:
(b) The optimal solution of the LP relaxation is
nearest integer point is
, which is not feasible, since
Rounded Solutions
Violated Constraints
2nd
2nd and 3rd
none
none
Hence, none of the feasible rounded solutions is optimal for the IP problem.
12.5-4.
(a)
Solution
Feasible?
Yes
No
Yes
No
Optimal?
No
Yes
12-16
. The
.
Optimal Solution:
(b) The optimal solution of the LP relaxation is
. The nearest integer
point is
, which is not feasible. The other rounded solution is
,
which is not feasible either.
12-17
12.5-5.
(a)
Solution
Feasible?
Yes
Yes
No
Yes
Optimal?
No
No
Yes
Optimal Solution:
(b) The optimal solution of the LP relaxation is
. The nearest integer
point is
, which is not feasible. The other rounded solution is
,
which is feasible, but not optimal.
12-18
12.5-6.
(a) TRUE, Sec. 12.5.
(b) TRUE, Sec. 12.5.
(c) FALSE, the result need not be feasible, see Fig. 11.2 for a counterexample. Sec. 12.5,
11th paragraph explains this pitfall.
12.6-1.
Optimal Solution:
,
12-19
12.6-2.
Optimal Solution:
,
12.6-3.
Optimal Solution:
,
12-20
12.6-4.
Optimal Solution:
,
12.6-5.
12.6-6.
(a) FALSE. The feasible region for the IP problem is a subset of the feasible region for
the LP relaxation. It is called a relaxation because it relaxes the feasible region.
(b) TRUE. If the optimal solution for the LP relaxation is integer, then it is feasible for
the IP problem and since the solution for the latter cannot be better than the solution for
the former, it has to be optimal.
(c) FALSE. Figure 12.2 is a counterexample for this statement.
12-21
12.6-7.
(a) Initialization: Set
. Apply the bounding and fathoming steps and the optimality test as described below for the whole problem. If the whole problem is not
fathomed, then it becomes the initial subproblem for the first iteration below.
Iteration:
1. Branching: Choose the most recently created unfathomed subproblem (in case of a tie,
select the one with the smallest bound). Among the assignees not yet assigned for the
current subproblem, choose the first one in the natural ordering to be the branching
variable. Subproblems correspond to each of the possible remaining assignments for the
branching assignee. Form a subproblem for each remaining assignment by deleting the
constraint that each of the unassigned assignees must perform exactly one assignment.
2. Bounding: For each new subproblem, obtain its bound by choosing the cheapest
assignee for each remaining assignment and totaling the costs.
3. Fathoming: For each new subproblem, apply the two fathoming tests:
Test 1. bound
Test 2. The optimal solution for its relaxation is a feasible assignment (If this
solution is better than the incumbent, it becomes the new incumbent and Test 1 is
reapplied to all unfathomed subproblems with the new smaller ).
12-22
Optimality Test: Identical to the one given in the text.
(b) Matchings are indicated with the notation (assignee, assignment).
Optimal matching:
, with total cost
.
12.6-8.
(a)
Branch Step: Use the best bound rule.
Bound Step: Given a partial sequencing
of the first jobs, a lower
bound on the time for the setup of the remaining
jobs is found by adding the
minimum elements of the columns corresponding to the remaining jobs, excluding those
elements in rows "None",
.
Fathoming Step: see the summary of the Branch-and-Bound technique in Sec. 12.6.
12-23
(b) The optimal sequence is
, with a total setup time of
.
12.6-9.
Optimal Solution:
,
12.6-10.
(a) The only constraints of the Lagrangian relaxation are nonnegativity and integrality.
Since x is feasible for an MIP problem, it already satisfies these constraints, so it is
feasible for the corresponding Lagrangian relaxation.
(b) x is feasible for an MIP problem, so from (a), it has to be feasible for its Lagrangian
relaxation. Also, x
and
, so
x
x
x
.
12.7-1.
Prior to this study, Waste Management, Inc. (WM) encountered several operational
inefficiencies concerning the routing of its trucks. The routes served by different trucks
had overlaps and route planners or drivers determined in what order they were going to
visit the stops. The result was inefficient sequences and communication gaps between
customers and customer-service personnel. The problem is formulated as a mixed integer
program, or more specifically as a vehicle routing problem with time windows. The goal
is to obtain routes with minimum number of vehicles and travel time, maximum visual
attractiveness and a balanced workload. First, a network with nodes that represent actual
stops, landfills, lunch break and the depot is constructed. The binary variables
refer
12-24
to whether arc
is included in the route of vehicle or not. The integer variables
denote the number of disposal trips and the continuous variables
correspond to the
beginning time of service for node by vehicle . The objective function to be minimized
is the total travel time. The constraints make sure that each stop is served by exactly one
truck, each truck starts at the depot, the amount of garbage at the stops does not exceed
the vehicle capacity and each route includes a lunch break. An iterative two-phase
algorithm enhanced with metaheuristics is employed to solve the problem.
Financial benefits of this study include savings of approximately $18 million in 2003 and
estimated savings of $44 million in 2004. WM expects to save more and to increase its
cash flow by $648 million over a five-year interval. The savings in operational costs over
five years is expected to be $498 million. By using mathematical modeling, WM now
generates more efficient routes with minimal overlaps, a reduced number of vehicles and
cost-effective sequences. All these contribute to the decrease in operational costs. At the
same time, centralized routing made communication in the organization and with the
customers easier. Customer-service personnel can now address customer problems more
quickly, since they know the routes of the vehicles. As a result, WM provides a more
reliable customer service. Operational efficiency also affected the environment and the
employees positively. Emissions and noise are reduced. Finally, the benefits from this
study led WM to exploit operations research techniques in other operational areas, too.
12.7-2.
(a)
Corner Points
Optimal solution for the LP relaxation:
Optimal integer solution:
with
with
12-25
(b) LP relaxation of the entire problem:
Optimal Solution:
Branch
,
:
This subproblem is infeasible, so the branch is fathomed.
Branch
:
Optimal Solution:
,
, feasible for the original problem
Hence, the optimal solution for the original problem is
12-26
with
.
(c) Let
and
.
maximize
subject to
binary
(d) Optimal Solution:
in (a).
,
12-27
, so
and
as
12.7-3.
(a)
Corner Points
Optimal solution for the LP relaxation:
Optimal integer solution:
with
with
(b) LP relaxation of the entire problem:
Optimal Solution:
,
12-28
Branch
:
Optimal Solution:
Branch
,
, feasible for the original problem
:
Optimal Solution:
,
12-29
Branch
:
Optimal Solution:
Branch
,
:
Optimal Solution:
,
Hence, the optimal solution for the original problem is
(c) Let
and
with
.
.
minimize
subject to
binary
(d) Optimal Solution:
part (a).
,
12-30
, so
and
as in
12.7-4.
(a)
Optimal Solution:
Branch
: Infeasible
Branch
:
Optimal Solution:
,
,
, feasible for the original problem
Hence, the optimal solution for the original problem is
12-31
with
.
(b) Optimal Solution:
(c) Solution:
,
,
12.7-5.
12-32
12.7-6.
Optimal Solution: x
,
.
12.7-7.
(a) Let
be the number of ¼ units of product to be produced, for
maximize
subject to
integers
(b)
Optimal Solution:
,
12-33
.
(c)
Branch
: Infeasible
Branch
:
Optimal Solution:
,
, feasible for the original problem
Hence, the optimal solution for the original problem is
(d) Optimal Solution:
(e) Solution:
,
,
12-34
with
.
12.7-8.
Optimal Solution: x
,
12.7-9.
Optimal Solution: x
,
12-35
12.7-10.
Optimal Solution: x
and x
,
12.8-1.
(a)
(b)
(c)
12.8-2.
(a)
(b)
(c)
12.8-3.
From the first equation,
. Then, this equation becomes redundant. From the third
equation,
and
. Now, this equation is redundant, too. Since
, from
the second equation,
and this equation becomes redundant. Finally, the
fourth equation reduces to
. Consequently, all equations become redundant. The
solution is then fixed to
.
12-36
12.8-4.
(a) Redundant. Even if all the variables are set to their upper bounds,
(b) Not redundant. For example,
(c) Not redundant. For example
,
.
violates this constraint.
violates this constraint.
(d) Redundant. The least value of
the constraint is still satisfied.
is attained by
12.8-5.
for
12.8-6.
12.8-7.
for
12.8-8.
(a)
for
12-37
and it is
, so
(b)
for
12.8-9.
The minimum cover for the constraint
is
, so the resulting cutting
plane is
, which is the same constraint obtained using the tightening
procedure.
12.8-10.
12.8-11.
12.8-12.
12.8-13.
12-38
12.8-14.
(1)
(2)
and
(3)
and
(4)
and
(5)
and
Hence, the problem is reduced to finding binary variables
that
maximize
subject to
.
The objective is maximized when all variables with positive coefficients are set to their
upper bounds, so when
. This solution also satisfies the
constraints, so it is optimal.
Optimal Solution: x
,
12.9-1.
Since the variables
have different values,
take values from the set
and all the variables must
{4}. There are two feasible solutions
with
3 1 2 4 with
.
Either feasible solution is optimal.
12.9-2.
:
:
,
,
:
,
feasible solutions,
Hence,
, and
, but
, and
,
, and
,
with
, so this is not feasible.
, so this is not feasible.
, so this is feasible. There are two
and
with
.
is optimal.
12.9-3.
:
and
:
, but
feasible solutions,
so
is optimal.
, but
, so this is not feasible.
, so
with
and
and
12-39
. There are two
with
,
12.9-4.
Let
denote the task to which the assignee is assigned.
minimize
subject to
element
element
element
element
all-different
M
, for
12.9-5.
Relabel Carl, Chris, David, Tony and Ken as assignee
respectively. Relabel
Backstroke, Breaststroke, Butterfly, Freestyle and Dummy as tasks
respectively. Let be the task to which assignee is assigned.
minimize
subject to
element
element
element
element
element
all-different
, for
12.9-6.
Let
be the number of study days allocated to course for
minimize
subject to
.
element
element
element
element
, for
12.9-7.
Let
be the number of crates allocated to store for
minimize
subject to
element
element
element
, for
12-40
.
12.9-8.
minimize
subject to
, for
all-different
12.10-1.
Answers will vary.
12.10-2.
Answers will vary.
12-41
Case%12.1%
!
a)! With!this!approach,!we!need!to!formulate!an!integer!program!for!each!month!
and!optimize!each!month!individually.!!
!
!
In!the!first!month,!Emily!does!not!buy!any!servers!since!none!of!the!departments!
implement!the!intranet!in!the!first!month.!
!
!
In!the!second!month!she!must!buy!computers!to!ensure!that!the!Sales!
Department!can!start!the!intranet.!Emily!can!formulate!her!decision!problem!as!
an!integer!problem!(the!servers!purchased!must!be!integer.!Her!objective!is!to!
minimize!the!purchase!cost.!She!has!to!satisfy!to!constraints.!She!cannot!spend!
more!than!$9500!(she!still!has!her!entire!budget!for!the!first!two!months!since!
she!didn't!buy!any!computers!in!the!first!month)!and!the!computer(s)!must!
support!at!least!60!employees.!She!solves!her!integer!programming!problem!
using!the!Excel!solver.!
!
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
Original Cost
Discount
Unit Cost
B
Standard
Intel
$2,500
0%
$2,500
C
Enhanced
Intel
$5,000
0%
$5,000
D
SGI
Workstation
$10,000
10%
$9,000
E
Sun
Workstation
$25,000
25%
$18,750
F
Support
Number of Employees Server Supports
30
80
200
2,000
Total
Support
80
Budget
Budget Spent per Server Purchased
$2,500
$5,000
$9,000
$18,750
Budget
Spent
$5,000
Servers
Purchased
Standard
Intel
0
Enhanced
Intel
1
SGI
Workstation
0
Sun
Workstation
0
G
H
>=
Support
Needed
60
<=
Budget
Available
$9,500
Total
Cost
$5,000
!
!
Note,!that!there!is!a!second!optimal!solution!to!this!integer!programming!
problem.!For!the!same!amount!of!money!Emily!could!buy!two!standard!PC's!that!
would!also!support!60!employees.!However,!since!Emily!knows!that!she!needs!to!
support!more!employees!in!the!near!future,!she!decides!to!buy!the!enhanced!PC!
since!it!supports!more!users.!
12-42
!
!
For!the!third!month!Emily!needs!to!support!260!users.!Since!she!has!already!
computing!power!to!support!80!users,!she!now!needs!to!figure!out!how!to!
support!additional!180!users!at!minimum!cost.!She!can!disregard!the!constraint!
that!the!Manufacturing!Department!needs!one!of!the!three!larger!servers,!since!
she!already!bought!such!a!server!in!the!previous!month.!Her!task!leads!her!to!the!
following!integer!programming!problem!and!solution.!
!
A
1
2
3
4
5
6
7
8
9
10
11
12
Original Cost
Discount
Unit Cost
Support
Servers
Purchased
B
Standard
Intel
$2,500
0%
$2,500
C
Enhanced
Intel
$5,000
0%
$5,000
D
SGI
Workstation
$10,000
0%
$10,000
E
Sun
Workstation
$25,000
0%
$25,000
Number of Employees Server Supports
30
80
200
2,000
Standard
Intel
0
Enhanced
Intel
0
SGI
Workstation
1
F
G
Total
Support
200
>=
Sun
Workstation
0
H
Support
Needed
180
Total
Cost
$10,000
!
!
Emily!decides!to!buy!one!SGI!Workstation!in!month!3.!The!network!is!now!able!
to!support!280!users.!
!
!
In!the!fourth!month!Emily!needs!to!support!a!total!of!290!users.!Since!she!has!
already!computing!power!to!support!280!users,!she!now!needs!to!figure!out!how!
to!support!additional!10!users!at!minimum!cost.!This!task!leads!her!to!the!
following!integer!programming!problem:!
!
A
1
2
3
4
5
6
7
8
9
10
11
12
Original Cost
Discount
Unit Cost
Support
Servers
Purchased
B
Standard
Intel
$2,500
0%
$2,500
C
Enhanced
Intel
$5,000
0%
$5,000
D
SGI
Workstation
$10,000
0%
$10,000
E
Sun
Workstation
$25,000
0%
$25,000
Number of Employees Server Supports
30
80
200
2,000
Standard
Intel
1
Enhanced
Intel
0
!
!
SGI
Workstation
0
Sun
Workstation
0
F
G
Total
Support
30
>=
H
Support
Needed
10
Total
Cost
$2,500
Emily!decides!to!buy!a!standard!PC!in!the!fourth!month.!The!network!is!now!able!
to!support!310!users.!
12-43
!
!
Finally,!in!the!fifth!and!last!month!Emily!needs!to!support!the!entire!company!
with!a!total!of!365!users.!Since!she!has!already!computing!power!to!support!310!
users,!she!now!needs!to!figure!out!how!to!support!additional!55!users!at!
minimum!cost.!This!task!leads!her!to!the!following!integer!programming!
problem!and!solution.!
!
A
1
2
3
4
5
6
7
8
9
10
11
12
Original Cost
Discount
Unit Cost
Support
Servers
Purchased
B
Standard
Intel
$2,500
0%
$2,500
C
Enhanced
Intel
$5,000
0%
$5,000
D
SGI
Workstation
$10,000
0%
$10,000
E
Sun
Workstation
$25,000
0%
$25,000
Number of Employees Server Supports
30
80
200
2,000
Standard
Intel
0
Enhanced
Intel
1
SGI
Workstation
0
Sun
Workstation
0
F
G
Total
Support
80
>=
H
Support
Needed
55
Total
Cost
$5,000
!
!
Emily!decides!to!buy!another!enhanced!PC!in!the!fifth!month.!(Note!that!again!
she!could!have!also!bought!two!standard!PC's,!but!clearly!the!enhanced!PC!
provides!more!room!for!the!workload!of!the!system!to!grow.)!The!entire!
network!of!CommuniCorp!consists!now!of!1!standard!PC,!2!enhanced!PC's!and!1!
SGI!workstation!and!it!is!able!to!support!390!users.!The!total!purchase!cost!for!
this!network!is!$22,500.!
!
b)! Due!to!the!budget!restriction!and!discount!in!the!first!two!months!Emily!needs!
to!distinguish!between!the!computers!she!buys!in!those!early!months!and!in!the!
later!months.!Therefore,!Emily!uses!two!variables!for!each!server!type.!
!!
Emily!essentially!faces!four!constraints.!First,!she!must!support!the!60!users!in!
the!sales!department!in!the!second!month.!She!realizes!that,!since!she!no!longer!
buys!the!computers!sequentially!after!the!second!month,!that!it!suffices!to!
include!only!the!constraint!on!the!network!to!support!the!all!users!in!the!entire!
company.!This!second!constraint!requires!her!to!support!a!total!of!365!users.!
The!third!constraint!requires!her!to!buy!at!least!one!of!the!three!large!servers.!
Finally,!Emily!has!to!make!sure!that!she!stays!within!her!budget!during!the!
second!month.!
12-44
A
B
C
D
E
F
G
H
1
Standard
Enhanced
SGI
Sun
2
Intel
Intel
Workstation Workstation
3
Month 3-5 Cost
$2,500
$5,000
$10,000
$25,000
4
Month 2 Discount
0%
0%
10%
25%
5
Month 2 Cost
$2,500
$5,000
$9,000
$18,750
6
Total
Support
7 Support
Number of Employees Server Supports
Support
Needed
8
Month 2
30
80
200
2,000
200
>=
60
9
Month 3-5
30
80
200
2,000
400
>=
365
10
11
Budget
Budget
12 Budget
Budget Spent per Server Purchased
Spent
Available
13
Month 2
$2,500
$5,000
$9,000
$18,750
$9,000
<=
$9,500
14
15 Server Purchases
Standard
Enhanced
SGI
Sun
16
Intel
Intel
Workstation Workstation
17
Month 2
0
0
1
0
Month 2 Cost
$9,000
18
Month 3-5
0
0
1
0
Month 3-5 Cost $10,000
19
Total Purchases
0
0
2
0
Total Cost $19,000
20
21
22
Total Advanced Servers
2
>=
1
Advanced Servers Needed
!
!
!
!
Emily!should!purchase!a!discounted!SGI!workstation!in!the!second!month,!and!
another!regular!priced!one!in!the!third!month.!The!total!purchase!cost!is!
$19,000.!
!
c)! Emily's!second!method!in!part!(b)!finds!the!cost!for!the!best!overall!purchase!
policy.!The!method!in!part!(a)!only!finds!the!best!purchase!policy!for!the!given!
month,!ignoring!the!fact!that!the!decision!in!a!particular!month!has!an!impact!on!
later!decisions.!The!method!in!(a)!is!very!shortXsighted!and!thus!yields!a!worse!
result!that!the!method!in!part!(b).!
!
d)! Installing!the!intranet!will!incur!a!number!of!other!costs.!These!costs!include:!
!
Training!cost,!
Labor!cost!for!network!installation,!
Additional!hardware!cost!for!cabling,!network!interface!cards,!necessary!hubs,!
etc.,!
Salary!and!benefits!for!a!network!administrator!and!web!master,!
Cost!for!establishing!or!outsourcing!help!desk!support.!
!
e)! The!intranet!and!the!local!area!network!are!complete!departures!from!the!way!
business!has!been!done!in!the!past.!The!departments!may!therefore!be!
concerned!that!the!new!technology!will!eliminate!jobs.!For!example,!in!the!past!
the!manufacturing!department!has!produced!a!greater!number!of!pagers!than!
customers!have!ordered.!Fewer!employees!may!be!needed!when!the!
manufacturing!department!begins!producing!only!enough!pagers!to!meet!orders.!
The!departments!may!also!become!territorial!about!data!and!procedures,!fearing!
that!another!department!will!encroach!on!their!business.!Finally,!the!
departments!may!be!concerned!about!the!security!of!their!data!when!sending!it!
over!the!network.!
12-45
Case%12.2%
!
a)! We!want!to!maximize!the!number!of!pieces!displayed!in!the!exhibit.!!For!each!
piece,!we!therefore!need!to!decide!whether!or!not!we!should!display!the!piece.!!
Each!piece!becomes!a!binary!decision!variable.!!The!decision!variable!is!assigned!
1!if!we!want!to!display!the!piece!and!assigned!0!if!we!do!not!want!to!display!the!
piece.!
!
We!group!our!constraints!into!four!categories!–!the!artistic!constraints!imposed!
by!Ash,!the!personal!constraints!imposed!by!Ash,!the!constraints!imposed!by!
Celeste,!and!the!cost!constraint.!!We!now!step!through!each!of!these!constraint!
categories.!
!
!
Artistic!Constraints!Imposed!by!Ash!
!
!
Ash!imposes!the!following!constraints!that!depend!upon!the!type!of!art!that!is!
displayed.!!The!constraints!are!as!follows:!
!
!
1.!!Ash!wants!to!include!only!one!collage.!!We!have!four!collages!available:!!
“Wasted!Resources”!by!Norm!Marson,!“Consumerism”!by!Angie!Oldman,!“My!
Namesake”!by!Ziggy!Lite,!and!“Narcissism”!by!Ziggy!Lite.!!A!constraint!forces!us!
to!include!exactly!one!of!these!four!pieces!(D36=D38!in!the!spreadsheet!model!
that!follows).!
!
!
2.!!Ash!wants!at!least!one!wireXmesh!sculpture!displayed!if!a!computerX
generated!drawing!is!displayed.!!We!have!three!wireXmesh!sculptures!available!
and!two!computerXgenerated!drawings!available.!!Thus,!if!we!include!either!one!
or!two!computerXgenerated!drawings,!we!have!to!include!at!least!one!wireXmesh!
sculpture.!Therefore,!we!constrain!the!total!number!of!wireXmesh!sculptures!
(total)!to!be!at!least!(1/2)!time!the!total!number!of!computerXgenerated!
drawings!(L40!≥!N40).!
!
!
3.!!Ash!wants!at!least!one!computerXgenerated!drawing!displayed!if!a!wireXmesh!
sculpture!is!displayed.!!We!have!two!computerXgenerated!drawings!available!
and!three!wireXmesh!sculptures!available.!!Thus,!if!we!include!one,!two,!or!three!
wireXmesh!sculptures,!we!have!to!include!either!one!or!two!computerXgenerated!
drawings.!Therefore,!we!constraint!the!total!number!of!wireXmesh!sculptures!
(total)!to!be!at!least!(1/3)!times!the!total!number!of!computerXgenerated!
drawings!(L41!≥!N41).!
!
!
4.!!Ash!wants!at!least!one!photoXrealistic!painting!displayed.!!We!have!three!
photoXrealistic!paintings!available:!!“Storefront!Window”!by!David!Lyman,!
“Harley”!by!David!Lyman,!and!“Rick”!by!Rick!Rawls.!!At!least!one!of!these!three!
paintings!has!to!be!displayed!(G36!≥!G38).!
!
!
5.!!Ash!wants!at!least!one!cubist!painting!displayed.!!We!have!three!cubist!
paintings!available:!!“Rick!II”!by!Rick!Rawls,!“Study!of!a!Violin”!by!Helen!Row,!
and!“Study!of!a!Fruit!Bowl”!by!Helen!Row.!!At!least!one!of!these!three!paintings!
has!to!be!displayed!(H36!≥!H38).!
12-46
!
!
6.!!Ash!wants!at!least!one!expressionist!painting!displayed.!!We!have!only!one!
expressionist!painting!available:!!“Rick!III”!by!Rick!Rawls.!!This!painting!has!to!be!
displayed!(I36!≥!I38).!
!
!
7.!!Ash!wants!at!least!one!watercolor!painting!displayed.!!We!have!six!watercolor!
paintings!available:!!“Serenity”!by!Candy!Tate,!“Calm!Before!the!Storm”!by!Candy!
Tate,!“All!That!Glitters”!by!Ash!Briggs,!“The!Rock”!by!Ash!Briggs,!“Winding!Road”!
by!Ash!Briggs,!and!“Dreams!Come!True”!by!Ash!Briggs.!!At!least!one!of!these!six!
paintings!has!to!be!displayed!(J36!≥!J38).!
!
!
8.!Ash!wants!at!least!one!oil!painting!displayed.!!We!have!five!oil!paintings!
available:!!“Void”!by!Robert!Bayer,!“Sun”!by!Robert!Bayer,!“Beyond”!by!Bill!
Reynolds,!“Pioneers”!by!Bill!Reynolds,!and!“Living!Land”!by!Bear!Canton.!!At!
least!one!of!these!five!paintings!has!to!be!displayed!(K36!≥!K38).!
!
!
9.!!Finally,!Ash!wants!the!number!of!paintings!to!be!no!greater!than!twice!the!
number!of!other!art!forms.!!We!have!18!paintings!available!and!16!other!art!
forms!available.!!We!classify!the!following!pieces!as!paintings:!!“Serenity,”!“Calm!
Before!the!Storm,”!“Void,”!“Sun,”!“Storefront!Window,”!“Harley,”!“Rick,”!“Rick!II,”!
“Rick!III,”!“Beyond,”!“Pioneers,”!“Living!Land,”!“Study!of!a!Violin,”!“Study!of!a!
Fruit!Bowl,”!“All!That!Glitters,”!“The!Rock,”!“Winding!Road,”!and!“Dreams!Come!
True.”!!The!total!number!of!these!paintings!that!we!display!has!to!be!less!than!or!
equal!to!twice!the!total!number!of!other!art!forms!we!display!(L42!≤!N42).!
!
!
Personal!Constraints!Imposed!by!Ash!
1.!!Ash!wants!all!of!his!own!paintings!included!in!the!exhibit,!so!we!must!include!
“All!That!Glitters,”!“The!Rock,”!“Winding!Road,”!and!“Dreams!Come!True.”!(In!the!
spreadsheet!model,!we!constraint!the!total!number!of!Ash!paintings!to!equal!4:!
N36=N38.)!
!
!
2.!!Ash!wants!all!of!Candy!Tate’s!work!included!in!the!exhibit,!so!we!must!include!
“Serenity”!and!“Calm!Before!the!Storm.”!(In!the!spreadsheet!model,!we!constrain!
the!total!number!of!Candy!Tate!works!to!equal!2:!O36=O38.)!
!
!
3.!!Ash!wants!to!include!at!least!one!piece!from!David!Lyman,!so!we!have!to!
include!one!or!more!of!the!pieces!!“Storefront!Window”!and!“Harley”!(P36!≥!
P38).!
!
!
4.!!Ash!wants!to!include!at!least!one!piece!from!Rick!Rawls,!so!we!have!to!include!
one!or!more!of!the!pieces!!“Rick,”!“Rick!II,”!and!“Rick!III”!(Q36!≥!Q38)!
!
!
5.!!Ash!wants!to!display!as!many!pieces!from!David!Lyman!as!from!Rick!Rawls.!!
Therefore!we!constrain!the!total!number!of!David!Lyman!works!to!equal!the!
total!number!of!Rick!Rawls!works!!(L43!=!N43).!
12-47
!
!
6.!!Finally,!Ash!wants!at!most!one!piece!from!Ziggy!Lite!displayed.!!We!can!
therefore!include!no!more!than!one!of!!“My!Namesake”!and!“Narcissism”!(R36!
≤!R38).!
!
Constraints!Imposed!by!Celeste!
1.!!Celeste!wants!to!include!at!least!one!piece!from!a!female!artist!for!every!two!
pieces!included!from!a!male!artist.!!We!have!11!pieces!by!female!artists!
available:!!“Chaos!Reigns”!by!Rita!Losky,!“Who!Has!Control?”!by!Rita!Losky,!
“Domestication”!by!Rita!Losky,!“Innocence”!by!Rita!Losky,!“Serenity”!by!Candy!
Tate,!“Calm!Before!the!Storm”!by!Candy!Tate,!“Consumerism”!by!Angie!Oldman,!
“Reflection”!by!Angie!Oldman,!“Trojan!Victory”!by!Angie!Oldman,!“Study!of!a!
Violin”!by!Helen!Row,!and!“Study!of!a!Fruit!Bowl”!by!Helen!Row.!!The!total!
number!of!these!pieces!has!to!be!greaterXthanXorXequalXto!(1/2)!times!the!total!
number!of!pieces!by!male!artists!(L44!≥!N44).!
!
!
2.!!Celeste!wants!at!least!one!of!the!pieces!“Aging!Earth”!and!“Wasted!Resources”!
displayed!in!order!to!advance!environmentalism!(V36!≥!V38).!
!
!
3.!!Celeste!wants!to!include!at!least!one!piece!by!Bear!Canton,!so!we!must!include!
one!or!more!of!the!pieces!!“Wisdom,”!“Superior!Powers,”!and!“Living!Land”!to!
advance!Native!American!rights!(W36!≥!W38).!
!
!
4.!!Celeste!wants!to!include!one!or!more!of!the!pieces!!“Chaos!Reigns,”!“Who!Has!
Control,”!“Beyond,”!and!“Pioneers”!to!advance!science!(X36!≥!X38).!
!
!
5.!!Celeste!knows!that!the!museum!only!has!enough!floor!space!for!four!
sculptures.!!We!have!six!sculptures!available:!!“Perfection”!by!Colin!Zweibell,!
“Burden”!by!Colin!Zweibell,!“The!Great!Equalizer”!by!Colin!Zweibell,!“Aging!
Earth”!by!Norm!Marson,!“Reflection”!by!Angie!Oldman,!and!“Trojan!Victory”!by!
Angie!Oldman.!!We!can!only!include!a!maximum!of!four!of!these!six!sculptures!
(Y36!≤!Y38).!
!
!
6.!!Celeste!also!knows!that!the!museum!only!has!enough!wall!space!for!20!
paintings,!collages,!and!drawings.!!We!have!28!paintings,!collages,!and!drawings!
available:!!“Chaos!Reigns,”!“Who!Has!Control,”!“Domestication,”!“Innocence,”!
“Wasted!Resources,”!“Serenity,”!“Calm!Before!the!Storm,”!“Void,”!“Sun,”!
“Storefront!Window,”!“Harley,”!“Consumerism,”!“Rick,”!“Rick!II,”!“Rick!III,”!
“Beyond,”!“Pioneers,”!“Wisdom,”!“Superior!Powers,”!“Living!Land,”!“Study!of!a!
Violin,”!“Study!of!a!Fruit!Bowl,”!“My!Namesake,”!“Narcissism,”!“All!That!Glitters,”!
“The!Rock,”!“Winding!Road,”!and!“Dreams!Come!True.”!!We!can!only!include!a!
maximum!of!20!of!these!28!wall!pieces!(Z36!≤!Z38).!
!
!
7.!!Finally,!Celeste!wants!“Narcissism”!displayed!if!“Reflection”!is!displayed.!!So!if!
the!decision!variable!for!“Reflection”!is!1,!the!decision!variable!for!“Narcissism”!
must!also!be!1.!!However,!the!decision!variable!for!“Narcissism”!can!still!be!1!
even!if!the!decision!variable!for!“Reflection”!is!0!(L45!≥!N45).!
12-48
N
O
P
Painting?
Other Art Form?
Ash Briggs?
Candy Tate?
David Lyman?
Piece
Perfection
300
Burden
250
The Great Equalizer
125
Chaos Reigns
400
Who Has Control?
500
Domestication
400
Innocence
550
Aging Earth
700
Wasted Resources
575
Serenity
200
Calm before the Storm 225
Void
150
Sun
150
Storefront Window
850
Harley
750
Consumerism
400
Reflection
175
Trojan Victory
450
Rick
500
Rick II
500
Rick III
500
Beyond
650
Pioneers
650
Wisdom
250
Superior Powers
350
Living Land
450
Study of a Violin
400
Study of a Fruit Bowl
400
My Namesake
300
Narcissism
300
All That Glitters
50
The Rock
50
Winding Road
50
Dreams Come True
50
Total 3,950
<=
Budget 4,000
1
1
1
Q R
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
U
1
1
1
1
1
1
1
=
1
1
Wire Mesh Sculpture
Computer-Generated Drawing
Paintings
David Lyman Pieces
Female Artist Pieces
"Reflection"
1
1
1
1
10 5
1
1
1
1
1
1
1
1
1
1
10
1
5
1
1
1
1
1
1
1
4
=
4
2
=
2
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1 1 1 1 6 1
>= >= >= >= >=
1 1 1 1 1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1 1 0
>= >= <=
1 1 1
Z AA
1
1
1
1
1
1
1
1
Y
1
1
1
1
1
1
1
1
1
1
1
V W X
1
1
1
1
1
1
1
1
T
1
1
1
1
1
1
1
1
1
1
S
Include?
M
Hangs on Wall?
L
Sits on Floor?
K
Advances Science?
J
Advances Native American Rights?
I
Advances Environmentalism?
H
Male Artist?
G
Female Artist?
F
Ziggy Lite?
E
Rick Rawls?
D
Oil Painting?
Artist
Colin Zweibell
Colin Zweibell
Colin Zweibell
Rita Losky
Rita Losky
Rita Losky
Rita Losky
Norm Marson
Norm Marson
Candy Tate
Candy Tate
Robert Bayer
Robert Bayer
David Lyman
David Lyman
Angie Oldman
Angie Oldman
Angie Oldman
Rick Rawls
Rick Rawls
Rick Rawls
Bill Reynolds
Bill Reynolds
Bear Canton
Bear Canton
Bear Canton
Helen Row
Helen Row
Ziggy Lite
Ziggy Lite
Ash Briggs
Ash Briggs
Ash Briggs
Ash Briggs
C
Water-Color Painting?
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
B
Expressional Painting?
A
Cubist Painting?
The!problem!formulation!in!an!Excel!spreadsheet!follows.!
!
Photo-Realistic Painting?
!
Computer-Generated Drawing?
!
Wire-Mesh Sculpture?
Cost!Constraint!
The!cost!of!all!of!the!pieces!displayed!has!to!be!less!than!or!equal!to!$4!million!
(C36!≤!C38).!
Collage?
!
Price ($thousand)
!
1
1
1
1
1
1
5 10
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1 1 1 2 13
>= >= >= <= <=
1 1 1 4 20
0
0
1
1
0
0
0
0
1
1
1
1
0
0
1
0
1
0
0
0
1
0
0
1
0
0
0
1
0
0
1
1
1
1
15
>= 0.5 0.5 times Computer-Generated Drawing
>= 0.33 0.33 times Wire Mesh Sculpture
<= 10
2 times Other Art Forms
=
1 Rick Rawls Pieces
>= 5
0.5 times Male Artist Pieces
>= 0 "Narcissism"
!
40
41
42
43
44
45
K
Wire Mesh Sculpture
Computer-Generated Drawing
Paintings
David Lyman Pieces
Female Artist Pieces
"Reflection"
L
=E36
=F36
=L36
=P36
=S36
=AA18
M
>=
>=
<=
=
>=
>=
N
=O40*F36
=O41*E36
=O42*M36
=Q36
=O44*T36
=AA31
12-49
O
=1/SUM(F2:F35)
=1/SUM(E2:E35)
2
Rick Rawls Pieces
0.5
"Narcissism"
P Q
R
times
Computer-Generated Drawing
times
Wire Mesh Sculpture
times
Other Art Forms
times
Male Artist Pieces
!
!
!
In!the!optimal!solution,!15!pieces!are!displayed!at!a!cost!of!$3.95!million.!!The!
following!pieces!are!displayed:!
!
1.!“The!Great!Equalizer”!by!Colin!Zweibell!
2.!“Chaos!Reigns”!by!Rita!Losky!
3.!“Wasted!Resources”!by!Norm!Marson!
4.!“Serenity”!by!Candy!Tate!
5.!“Calm!Before!the!Storm”!by!Candy!Tate!
6.!“Void”!by!Robert!Bayer!(or!“Sun”!by!Robert!Bayer)!
7.!“Harley”!by!David!Lyman!
8.!“Reflection”!by!Angie!Oldman!
9.!“Rick!III”!by!Rick!Rawls!
10.!“Wisdom”!by!Bear!Canton!
11.!“Study!of!a!Fruit!Bowl”!by!Helen!Row!(or!“Study!of!a!Violin”)!
12.!“All!That!Glitters”!by!Ash!Briggs!
13.!“The!Rock”!by!Ash!Briggs!
14.!“Winding!Road”!by!Ash!Briggs!
15.!“Dreams!Come!True”!by!Ash!Briggs!
!
b)! The!formulation!of!this!problem!is!the!same!as!the!formulation!in!part!(a)!except!
that!the!objective!function!from!part!(a)!now!becomes!a!constraint!and!the!cost!
constraint!from!part!(a)!now!becomes!the!objective!function.!!Thus,!we!have!the!
new!constraint!that!we!need!to!select!20!or!more!pieces!to!display!in!the!exhibit.!!
We!also!have!the!new!objective!to!minimize!the!cost!of!the!exhibit.!
!
!
The!new!formulation!of!the!problem!in!an!Excel!follows.!
12-50
!
N
O
P
Photo-Realistic Painting?
Cubist Painting?
Expressional Painting?
Water-Color Painting?
Oil Painting?
Painting?
Other Art Form?
Ash Briggs?
Candy Tate?
David Lyman?
Piece
Perfection
300
Burden
250
The Great Equalizer
125
Chaos Reigns
400
Who Has Control?
500
Domestication
400
Innocence
550
Aging Earth
700
Wasted Resources
575
Serenity
200
Calm before the Storm 225
Void
150
Sun
150
Storefront Window
850
Harley
750
Consumerism
400
Reflection
175
Trojan Victory
450
Rick
500
Rick II
500
Rick III
500
Beyond
650
Pioneers
650
Wisdom
250
Superior Powers
350
Living Land
450
Study of a Violin
400
Study of a Fruit Bowl
400
My Namesake
300
Narcissism
300
All That Glitters
50
The Rock
50
Winding Road
50
Dreams Come True
50
Total 5,450
1
1
1
Q R
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
U
1
1
1
1
1
1
1
=
1
1
1
3
1
1
1
1
1 1 2 1 6 2
>= >= >= >= >=
1 1 1 1 1
Wire Mesh Sculpture
Computer-Generated Drawing
Paintings
David Lyman Pieces
Female Artist Pieces
"Reflection"
12-51
1
1
1
1
12 8
3
1
12
1
7
1
1
1
1
1
1
1
4
=
4
2
=
2
1 1 0
>= >= <=
1 1 1
Z AA
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
Y
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
V W X
1
1
1
1
1
1
1
1
1
1
1
T
1
1
1
1
1
1
1
1
1
1
S
Include?
M
Hangs on Wall?
L
Sits on Floor?
K
Advances Science?
J
Advances Native American Rights?
I
Advances Environmentalism?
H
Male Artist?
G
Female Artist?
F
Ziggy Lite?
E
Rick Rawls?
D
Computer-Generated Drawing?
Artist
Colin Zweibell
Colin Zweibell
Colin Zweibell
Rita Losky
Rita Losky
Rita Losky
Rita Losky
Norm Marson
Norm Marson
Candy Tate
Candy Tate
Robert Bayer
Robert Bayer
David Lyman
David Lyman
Angie Oldman
Angie Oldman
Angie Oldman
Rick Rawls
Rick Rawls
Rick Rawls
Bill Reynolds
Bill Reynolds
Bear Canton
Bear Canton
Bear Canton
Helen Row
Helen Row
Ziggy Lite
Ziggy Lite
Ash Briggs
Ash Briggs
Ash Briggs
Ash Briggs
C
Wire-Mesh Sculpture?
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
B
Collage?
A
Price ($thousand)
!
1
1
1
1
1
1
7 13
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1 1 1 4 16
>= >= >= <= <=
1 1 1 4 20
>= 0.5 0.5 times Computer-Generated Drawing
>= 1.00 0.33 times Wire Mesh Sculpture
<= 16
2 times Other Art Forms
=
1 Rick Rawls Pieces
>= 6.5 0.5 times Male Artist Pieces
>= 0 "Narcissism"
1
1
1
1
0
1
0
0
1
1
1
1
1
0
1
0
1
0
0
0
1
0
0
1
0
0
1
1
0
0
1
1
1
1
20
>=
20
!
!
!
In!the!optimal!solution,!exactly!20!pieces!are!displayed!at!a!cost!of!$5.45!million!
–!$1.45!million!more!than!Ash!decided!to!allocate!in!part!(a).!!All!pieces!from!
part!(a)!are!displayed!in!addition!to!the!following!five!new!pieces:!!
!
1.!“Perfection”!by!Colin!Zweibell!
2.!“Burden”!by!Colin!Zweibell!
3.!“Domestication”!by!Rita!Losky!
4.!“Sun”!(or!“Void”)!by!Robert!Bayer!
5.!“Study!of!a!Violin”!(or!“Study!of!a!Fruit!Bowl”)!by!Helen!Row!
!
c)! This!problem!is!also!a!cost!minimization!problem.!!The!problem!formulation!is!
the!same!as!that!used!in!part!(b).!!A!new!constraint!is!added,!however.!!The!
patron!wants!all!of!Rita’s!pieces!displayed.!!Rita!has!four!pieces:!!“Chaos!Reigns,”!
“Who!Has!Control?,”!“Domestication,”!and!“Innocence.”!!All!of!these!four!pieces!
must!be!displayed.!
!
The!problem!formulation!in!Excel!follows.!
12-52
!
O
P
Photo-Realistic Painting?
Cubist Painting?
Expressional Painting?
Water-Color Painting?
Oil Painting?
Painting?
Other Art Form?
Ash Briggs?
Candy Tate?
David Lyman?
Piece
Perfection
300
Burden
250
The Great Equalizer
125
Chaos Reigns
400
Who Has Control?
500
Domestication
400
Innocence
550
Aging Earth
700
Wasted Resources
575
Serenity
200
Calm before the Storm 225
Void
150
Sun
150
Storefront Window
850
Harley
750
Consumerism
400
Reflection
175
Trojan Victory
450
Rick
500
Rick II
500
Rick III
500
Beyond
650
Pioneers
650
Wisdom
250
Superior Powers
350
Living Land
450
Study of a Violin
400
Study of a Fruit Bowl
400
My Namesake
300
Narcissism
300
All That Glitters
50
The Rock
50
Winding Road
50
Dreams Come True
50
Total 5,800
1
1
1
Q R
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
T
U
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
=
1
1
1
1
1
2
1
1
1
1
2 1 1 1 6 2
>= >= >= >= >=
1 1 1 1 1
Wire Mesh Sculpture
Computer-Generated Drawing
Paintings
David Lyman Pieces
Female Artist Pieces
"Reflection"
12-53
1
1
1
1
11 9
2
2
11
1
8
1
1
1
1
1
1
1
4
=
4
2
=
2
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1 1 0 4
>= >= <= >=
1 1 1 4
Z AA AB
1
1
1
1
1
1
1
1
1
Y
1
1
1
1
1
1
1
1
1
1
V W X
1
1
1
1
1
1
1
1
1
1
S
1
1
1
1
1
1
8 12
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1 1 2 3 17
>= >= >= <= <=
1 1 1 4 20
>= 1
0.5 times Computer-Generated Drawing
>= 0.67 0.33 times Wire Mesh Sculpture
<= 18
2 times Other Art Forms
=
1 Rick Rawls Pieces
>= 6
0.5 times Male Artist Pieces
>= 0 "Narcissism"
Include?
N
Hangs on Wall?
M
Sits on Floor?
L
Advances Science?
K
Advances Native American Rights?
J
Advances Environmentalism?
I
Male Artist?
H
Female Artist?
G
Rita Losky?
F
Ziggy Lite?
E
Rick Rawls?
D
Computer-Generated Drawing?
Artist
Colin Zweibell
Colin Zweibell
Colin Zweibell
Rita Losky
Rita Losky
Rita Losky
Rita Losky
Norm Marson
Norm Marson
Candy Tate
Candy Tate
Robert Bayer
Robert Bayer
David Lyman
David Lyman
Angie Oldman
Angie Oldman
Angie Oldman
Rick Rawls
Rick Rawls
Rick Rawls
Bill Reynolds
Bill Reynolds
Bear Canton
Bear Canton
Bear Canton
Helen Row
Helen Row
Ziggy Lite
Ziggy Lite
Ash Briggs
Ash Briggs
Ash Briggs
Ash Briggs
C
Wire-Mesh Sculpture?
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
B
Collage?
A
Price ($thousand)
!
0
1
1
1
1
1
1
0
1
1
1
1
1
0
1
0
1
0
0
0
1
0
0
1
0
0
0
1
0
0
1
1
1
1
20
>=
20
!
!
%
!
In!the!optimal!solution,!exactly!20!pieces!are!displayed!at!a!total!cost!of!$5.8!
million.!!The!patron!has!to!pay!$1.8!million.!!The!following!pieces!are!displayed:!
!
1.!“Burden”!by!Colin!Zweibell!
2.!“The!Great!Equalizer”!by!Colin!Zweibell!
3.!“Chaos!Reigns”!by!Rita!Losky!
4.!“Who!Has!Control?”!by!Rita!Losky!
5.!“Domestication”!by!Rita!Losky!
6.!“Innocence”!by!Rita!Losky!
7.!“Wasted!Resources”!by!Norm!Marson!
8.!“Serenity”!by!Candy!Tate!
9.!“Calm!Before!the!Storm”!by!Candy!Tate!
10.!“Void”!by!Robert!Bayer!
11.!“Sun”!by!Robert!Bayer!
12.!“Harley”!by!David!Lyman!
13.!“Reflection”!by!Angie!Oldman!
14.!“Rick!III”!by!Rick!Rawls!
15.!“Wisdom”!by!Bear!Canton!
16.!“Study!of!a!Fruit!Bowl”!(or!“Study!of!a!Violin”)!by!Helen!Row!
17.!“All!That!Glitters”!by!Ash!Briggs!
18.!“The!Rock”!by!Ash!Briggs!
19.!“Winding!Road”!by!Ash!Briggs!
20.!“Dreams!Come!True”!by!Ash!Briggs!
%
12-54
Case%12.3%
a)! We!want!to!maximize!the!total!number!of!kitchen!sets,!so!each!of!the!20!
kitchen!sets!becomes!a!decision!variable.!!But!the!kitchen!sets!are!not!our!only!
decision!variables.!!Because!we!assume!that!any!particular!item!composing!a!
kitchen!set!is!replenished!immediately,!we!only!need!to!stock!one!of!each!item.!!
A!particular!item!may!compose!multiple!kitchen!sets.!!For!example,!tile!T1!is!part!
of!kitchen!sets!3,!7,!10,!and!17.!!So!a!kitchen!set!exists!when!all!of!the!items!
composing!that!kitchen!set!are!in!stock.!!Therefore,!each!of!30!items!also!
becomes!a!decision!variable.!!These!decision!variables!are!binary!decision!
variables.!!If!a!kitchen!set!or!item!is!in!stock,!the!decision!variable!is!1.!!If!a!
kitchen!set!or!item!is!not!in!stock,!the!decision!variable!is!0.!
!
A!handful!of!constraints!exist!in!this!problem.!
1.!We!cannot!indicate!that!a!kitchen!set!is!in!stock!unless!all!the!items!composing!
that!kitchen!set!are!also!in!stock.!!Thus,!a!kitchen!set!decision!variable!is!1!only!if!
all!the!decision!variables!for!the!items!composing!that!kitchen!set!are!also!1.!For!
example,!for!set!1!this!constraint!equals!(Set!1)!<=!
(T2+W2+L4+C2+O4+S2+D2+R2)!/!8.!
2.!Each!kitchen!set!requires!20!square!feet!of!tile.!!Thus,!if!a!particular!tile!is!in!
stock,!20!square!feet!of!that!tile!are!in!stock.!!The!warehouse!can!only!hold!50!
square!feet!of!tile,!so!only!a!maximum!of!two!different!styles!of!tile!can!be!in!
stock.!
3.!Each!kitchen!set!requires!five!rolls!of!wallpaper.!!Thus,!if!a!particular!style!of!
wallpaper!is!in!stock,!five!rolls!of!that!wallpaper!are!in!stock.!!The!warehouse!
can!only!hold!12!rolls!of!wallpaper,!so!only!a!maximum!of!two!different!styles!of!
wallpaper!can!be!in!stock.!
4.!A!maximum!of!two!different!styles!of!light!fixtures!can!be!in!stock.!
5.!A!maximum!of!two!different!styles!of!cabinets!can!be!in!stock.!
6.!A!maximum!of!three!different!styles!of!countertops!can!be!in!stock.!
7.!A!maximum!of!two!different!sinks!can!be!in!stock.!
8.!A!combination!of!four!different!styles!of!dishwashers!and!ranges!can!be!held!
in!stock.!
!
The!problem!formulated!in!an!Excel!spreadsheet!follows.!
12-55
!
O P Q
Cabinets
C1 C2 C3 C4
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
0 1 0
(>= 5 rolls)
1
0
1
0
1
1
Dwashers & Ranges
N
Sinks
1
J K L M
Light Fixtures
L1 L2 L3 L4
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
Countertops
G H I
Wallpaper
W1 W2 W3 W4
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
Cabinets
Item In Stock? 0 1 1 0
(>= 20 sq. ft.)
F
Light Fixtures
Set 1
Set 2
Set 3
Set 4
Set 5
Set 6
Set 7
Set 8
Set 9
Set 10
Set 11
Set 12
Set 13
Set 14
Set 15
Set 16
Set 17
Set 18
Set 19
Set 20
C D E
Floor Tile
T1 T2 T3 T4
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
Wallpaper
26
27
28
29
B
Floor Tile
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
0
Total Items 2 2 2 2 3 2 4
<= <= <= <= <= <= <=
Capacity 2 2 2 2 3 2 4
0
R S T U
Countertops
O1 O2 O3 O4
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
0
1
V W
Dwash
D1 D2
1
1
1
1
1
1
1
1
1
1
1
1
1
0
1
X
Y Z AA AB AC AD
Sinks
Ranges
S1 S2 S3 S4 R1 R2 R3
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
0
1
0
0
1
1
AE AF
R4
1
1
1
1
AG
Set In
Stock?
0
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
0
1
0
1
AH
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
AI
Fraction of
Items in Stock
0.625
0.75
0.625
0.5
0.375
0.625
0.375
1
0.625
0.5
0.75
0.75
0.375
0.286
1
0.429
0.143
1
0.429
1
1
Total Sets in Stock
4
!
!
b)! We!should!stock!the!following!items:!T2,!T3;!W1,!W3;!L1,!L3;!C1,!C2;!O1,!O2,!O4;!
D2;!S1,!S3;!and!R2,!R3,!and!R4.!This!combination!allows!four!different!kitchen!
sets!to!be!in!stock—set!8,!15,!18,!and!20.!
!!
Note!that!this!is!not!a!unique!solution.!!The!value!of!the!objective!function!is!
always!four!complete!kitchen!sets,!but!the!specific!items!and!kitchen!sets!stocked!
may!be!different.!!Throughout!this!solution,!we!will!refer!to!the!optimal!solution!
shown!above,!but!because!other!optimal!solutions!exist,!student!answers!may!
differ!from!the!solution!somewhat.!
12-56
!
c)! We!model!this!new!problem!by!changing!the!capacity!constraint!for!the!
dishwashers!and!ranges.!!Now,!instead!of!being!able!to!stock!a!combination!of!
only!four!different!styles!of!dishwashers!and!ranges,!we!can!stock!a!maximum!of!
two!different!styles!of!dishwashers!and!a!maximum!of!three!different!styles!of!
ranges.!!Because!we!only!have!two!different!styles!of!dishwashers!available,!we!
now!effectively!do!not!have!a!constraint!on!the!number!of!dishwashers!we!can!
carry.!
!
The!formulation!of!the!problem!in!Excel!follows:!
!
N
O P Q
Cabinets
C1 C2 C3 C4
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
0 1 0
(>= 5 rolls)
1
0
1
0
1
0
Ranges
1
J K L M
Light Fixtures
L1 L2 L3 L4
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
Sinks
G H I
Wallpaper
W1 W2 W3 W4
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
Countertops
Item In Stock? 0 1 1 0
(>= 20 sq. ft.)
F
Cabinets
Set 1
Set 2
Set 3
Set 4
Set 5
Set 6
Set 7
Set 8
Set 9
Set 10
Set 11
Set 12
Set 13
Set 14
Set 15
Set 16
Set 17
Set 18
Set 19
Set 20
C D E
Floor Tile
T1 T2 T3 T4
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
Light Fixtures
26
27
28
29
B
Wallpaper
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
Floor Tile
!
1
Total Items 2 2 2 2 3 2 3
<= <= <= <= <= <= <=
Capacity 2 2 2 2 3 2 3
0
R S T U
Countertops
O1 O2 O3 O4
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
0
V W
Dwash
D1 D2
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
X
Y Z AA AB AC AD
Sinks
Ranges
S1 S2 S3 S4 R1 R2 R3
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
0
1
0
1
0
1
AE AF
R4
1
1
1
1
AG
Set In
Stock?
0
0
0
1
0
0
0
1
0
0
1
0
0
0
1
0
0
0
0
1
AH
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
AI
Fraction of
Items in Stock
0.25
0.5
0.75
1
0.625
0.5
0.75
1
0.625
0.75
1
0.5
0.625
0.571
1
0.714
0.429
0.571
0.286
1
1
Total Sets in Stock
5
!
!
!
!
With!the!extra!space,!the!number!of!kitchen!sets!we!can!stock!increases!from!
four!to!five—now!sets!4,!8,!11,!15,!and!20.!To!keep!these!five!sets!in!stock,!we!
stock!the!following!items:!T2,!T3;!W1,!W3;!L1,!L3;!C1,!C3;!O1,!O2,!O3;!D1,!D2;!S1,!
S3;!R1,!R3,!and!R4.!(This!optimal!solution!is!not!unique.)!The!new!space!vacated!
by!the!nursery!department!provides!us!with!the!space!to!stock!the!new!range.!
12-57
d)! With!the!additional!space,!our!constraints!change.!!We!eliminate!the!constraints!
limiting!the!maximum!number!of!different!styles!of!sinks!and!countertops!we!
can!stock.!!Instead!of!stocking!two!of!the!four!styles!of!light!fixtures,!we!can!now!
stock!three!of!the!four!styles!of!light!fixtures.!!Finally,!instead!of!stocking!only!
two!of!the!four!cabinet!styles,!we!can!now!stock!three!of!the!four!cabinet!styles.!
!
The!problem!formulated!in!Excel!follows.!
!
1
26
27
28
29
N
O P Q
Cabinets
C1 C2 C3 C4
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
0 1 0
(>= 5 rolls)
1
0
1
1
1
0
Ranges
Item In Stock? 0 1 1 0
(>= 20 sq. ft.)
J K L M
Light Fixtures
L1 L2 L3 L4
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
Sinks
G H I
Wallpaper
W1 W2 W3 W4
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
Countertops
F
Cabinets
Set 1
Set 2
Set 3
Set 4
Set 5
Set 6
Set 7
Set 8
Set 9
Set 10
Set 11
Set 12
Set 13
Set 14
Set 15
Set 16
Set 17
Set 18
Set 19
Set 20
C D E
Floor Tile
T1 T2 T3 T4
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
Light Fixtures
B
Wallpaper
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
Floor Tile
!
1
Total Items 2 2 3 2 3 3 3
<= <= <= <= <= <= <=
Capacity 2 2 3 3 4 4 3
0
R S T U
Countertops
O1 O2 O3 O4
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
0
V W
Dwash
D1 D2
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
X
Y Z AA AB AC AD
Sinks
Ranges
S1 S2 S3 S4 R1 R2 R3
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
0
1
0
1
AE AF
R4
1
1
1
1
AG
Set In
Stock?
0
0
0
1
0
0
0
1
0
0
1
0
0
0
1
1
0
0
0
1
AH
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
AI
Fraction of
Items in Stock
0.5
0.5
0.75
1
0.75
0.5
0.875
1
0.75
0.75
1
0.5
0.75
0.714
1
1.000
0.429
0.571
0.571
1
1
Total Sets in Stock
6
!
!
With!the!extra!space,!we!are!now!able!to!stock!six!complete!kitchen!sets—set!4,!
8,!11,!15,!16,!and!20.!The!following!items!are!stocked:!T2,!T3;!W1,!W3,!L1,!L3,!L4;!
C1,!C3,!O1,!O2,!O3;!D1,!D2;!S1,!S2,!S3;!R1,!R3,!R4.!(Again,!this!optimal!solution!is!
not!unique.)!
!
%
e)! If!the!items!composing!a!kitchen!set!could!not!be!replenished!immediately,!we!
could!not!formulate!this!problem!as!a!binary!integer!program.!!We!would!have!to!
formulate!the!problem!as!an!integer!program!since!we!may!have!to!store!more!
than!one!kitchen!component!or!kitchen!set!to!ensure!that!we!meet!demand.!
!
The!assumption!of!immediate!replenishment!is!justified!if!the!average!time!to!
replenish!the!component!is!significantly!less!than!the!average!time!between!
demands!for!that!component.!
%
12-58
Case%12.4%
!
a)! Let!! xij!=!1!if!students!from!area!i!are!assigned!to!school!j;!0!if!not.!
!
Cij!=!bussing!cost!
!
Si!=!student!population!of!area!i!
!
Kj!=!capacity!of!school!j!
!
Pik!=!%!of!students!in!area!i!in!grade!k!
!
(for!i!=!1,!2,!3,!4,!5,!6!!!!j!=!1,!2,!3!!!!and!!!!k!=!6,!7,!8)!
!
and!xij.=!are!binary!variables!(for!i!=!1,!2,!3,!4,!5,!6!and!j!=!1,!2,!3).!
!
Note!x21!=!x43!=x52!=!0!due!to!infeasibility.!
!
b)! The!models!really!aren’t!too!different.!xij!are!binary!here,!which!amounts!to!
forcing!their!value!in!the!LP!of!Case!4.3!to!be!either!0!or!Si.!We!can!leave!out!the!
three!variables!known!to!be!0,!and!also!9!redundant!constraints.!The!LPX
relaxation!of!this!model,!with!0!≤!xij!≤!1!would!allow!us!to!interpret!xij!as!the!
fraction!of!students!from!area!i!to!be!assigned!to!school!j.!This!obviously!would!
be!a!more!general!model,!equivalent!to!that!in!Case!4.3.!
!
c)! Since!a!residential!area!cannot!be!split!across!multiple!schools,!the!decision!
becomes!which!school!to!send!each!area’s!students!to.!This!is!formulated!with!a!
binary!variable!for!each!area/school!combination,!representing!the!yesXorXno!
decision!of!whether!that!area!should!be!assigned!to!that!school.!Each!area!can!
only!be!sent!to!a!single!school,!represented!by!the!constraints!TotalAssignments!
(E14:E19)!=!Supply!(G14:G19).!
!
The!number!of!students!in!each!school!is!then!calculated!in!StudentAssignments!
(B24:D29)!based!upon!the!results!in!AreaAssignments!(B14:D19).!
!
12-59
!
!
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
B
Number
of Students
450
600
550
350
500
450
C
Percentage
in 6th
Grade
32%
37%
30%
28%
39%
34%
D
Percentage
in 7th
Grade
38%
28%
32%
40%
34%
28%
E
Percentage
in 8th
Grade
30%
35%
38%
32%
27%
38%
School 1
$300
–
$600
$200
$0
$500
School 1
1
0
0
0
0
1
School 2
0
1
0
1
0
0
School 3
0
0
1
0
1
0
Total
Assignments
1
1
1
1
1
1
=
=
=
=
=
=
School 1
450
0
0
0
0
450
900
<=
900
School 2
0
600
0
350
0
0
950
<=
1,100
School 3
0
0
550
0
500
0
1,050
<=
1,000
270
<=
297
297
306
<=
324
285
<=
320
308
322
<=
342
315
<=
360
346
344
<=
378
Data:
Area
1
2
3
4
5
6
Area Assignments
Area 1
Area 2
Area 3
Area 4
Area 5
Area 6
F
G
H
Bussing Cost ($/Student)
School 2
School 3
$0
$700
$400
$500
$300
$200
$500
–
–
$400
$300
$0
Supply
1
1
1
1
1
1
Student Assignments
Area 1
Area 2
Area 3
Area 4
Area 5
Area 6
Total In School
Capacity
Total
Bussing
Cost
$1,085,000
Grade Constraints:
6th Graders
7th Graders
8th Graders
30%
of total in school
36%
of total in school
!
!
!
d)! Without!prohibiting!the!splitting!of!residential!areas,!the!total!cost!was!
$555,556.!Thus,!adding!this!restriction!increases!the!cost!by!$1,085,000!X!
$555,556!=!$529,443.!
12-60
!
!
e)! As!shown!in!the!spreadsheet,!the!solution!remains!the!same,!but!the!bussing!
costs!are!reduced!to!$975,000.!
!
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
B
Number
of Students
450
600
550
350
500
450
C
Percentage
in 6th
Grade
32%
37%
30%
28%
39%
34%
D
Percentage
in 7th
Grade
38%
28%
32%
40%
34%
28%
E
Percentage
in 8th
Grade
30%
35%
38%
32%
27%
38%
School 1
$300
–
$600
$0
$0
$500
School 1
1
0
0
0
0
1
School 2
0
1
0
1
0
0
School 3
0
0
1
0
1
0
Total
Assignments
1
1
1
1
1
1
=
=
=
=
=
=
School 1
450
0
0
0
0
450
900
<=
900
School 2
0
600
0
350
0
0
950
<=
1,100
School 3
0
0
550
0
500
0
1,050
<=
1,000
270
<=
297
297
306
<=
324
285
<=
320
308
322
<=
342
315
<=
360
346
344
<=
378
Data:
Area
1
2
3
4
5
6
Area Assignments
Area 1
Area 2
Area 3
Area 4
Area 5
Area 6
F
G
H
Bussing Cost ($/Student)
School 2
School 3
$0
$700
$400
$500
$300
$0
$500
–
–
$400
$300
$0
Supply
1
1
1
1
1
1
Student Assignments
Area 1
Area 2
Area 3
Area 4
Area 5
Area 6
Total In School
Capacity
Total
Bussing
Cost
$975,000
Grade Constraints:
6th Graders
7th Graders
8th Graders
12-61
30%
of total in school
36%
of total in school
!
!
f)!! Again,!the!solution!remains!the!same,!but!the!bussing!costs!are!reduced!to!
$840,000.!
!
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
B
Number
of Students
450
600
550
350
500
450
C
Percentage
in 6th
Grade
32%
37%
30%
28%
39%
34%
D
Percentage
in 7th
Grade
38%
28%
32%
40%
34%
28%
E
Percentage
in 8th
Grade
30%
35%
38%
32%
27%
38%
School 1
$0
–
$600
$0
$0
$500
School 1
1
0
0
0
0
1
School 2
0
1
0
1
0
0
School 3
0
0
1
0
1
0
Total
Assignments
1
1
1
1
1
1
=
=
=
=
=
=
School 1
450
0
0
0
0
450
900
<=
900
School 2
0
600
0
350
0
0
950
<=
1,100
School 3
0
0
550
0
500
0
1,050
<=
1,000
270
<=
297
297
306
<=
324
285
<=
320
308
322
<=
342
315
<=
360
346
344
<=
378
Data:
Area
1
2
3
4
5
6
Area Assignments
Area 1
Area 2
Area 3
Area 4
Area 5
Area 6
F
G
H
Bussing Cost ($/Student)
School 2
School 3
$0
$700
$400
$500
$0
$0
$500
–
–
$400
$0
$0
Supply
1
1
1
1
1
1
Student Assignments
Area 1
Area 2
Area 3
Area 4
Area 5
Area 6
Total In School
Capacity
Total
Bussing
Cost
$840,000
Grade Constraints:
6th Graders
7th Graders
8th Graders
30%
of total in school
36%
of total in school
!
!
!
g)! For!all!three!options,!the!assignments!of!areas!to!schools!are!identical.!For!the!
current!alternative,!the!bussing!costs!are!$1,085,000.!For!option!1,!the!bussing!
costs!are!$975,000!(a!reduction!of!$110,000).!This!savings!results!from!the!fact!
that!students!from!area!3!would!no!longer!be!bussed!to!school!3.!For!option!2,!
the!bussing!costs!are!$840,000!(a!reduction!of!$135,000!over!option!1,!and!
$245,000!over!the!current!alternative).!This!additional!savings!results!from!the!
fact!that!students!would!no!longer!be!bussed!from!area!1!to!school!1.!
!
h)! Arguments!can!be!made!for!all!three!alternatives.!Answers!will!vary.!
!
12-62
CHAPTER 13: NONLINEAR PROGRAMMING
13.1-1.
In 1995, a number of factors including increased competition, the lack of quantitative
tools to support financial advices and the introduction of new regulations compelled
Bank Hapoalim to review its investment advisory process. Consequently, the OptiMoney system was developed as a tool to offer systematic financial advice. The
underlying mathematical model is a constrained nonlinear program with continuous or
discontinuous derivatives depending on the selected risk measure. The variables B3
denote the fraction of asset 3. The goal is to choose a portfolio that minimizes "risk"
among all portfolios with a fixed expected return. Opti-Money allows the investor to
choose among four risk measures, viz., symmetric return variability, asymmetric
downside risk, asymmetric return variability around more than one benchmark, and
classical Markowitz risk of a portfolio. Once the risk measure and the benchmark(s) are
specified, the objective function is formulated as a weighted sum of this risk measure and
a market-portfolio tracking term. Then the efficient frontier is constructed.
The Opti-Money system increased average monthly profit of Bank Hapoalim
significantly. The average annual return for customers has also increased. The excess
earnings using Opti-Money exceeds $200 million per year. The subsidiaries of the bank
like Continental Mutual Fund benefit from Opti-Money, too. The new system resulted in
"an organizational revolution in the investment advisory process at Bank Hapoalim" [p.
46]. As a result of this study, additional consultation-support systems are developed to
help the customer relations managers.
13.1-2.
#Î$
$Î%
"Î#
maximize
0 ÐBÑ œ "!!B"  "!B"  %!B#  &B#  &!B$  &B$
subject to
*B"  $B#  &B$ Ÿ &!!
&B"  %B#
Ÿ $&!
$B"
 #B$ Ÿ "&!
B$ Ÿ #!
B" ß B# ß B$
!
13-1
13.1-3.
Each term in the objective function changes (as above) from +34 B34 to
+34 B34  !Þ"+34 ÐB34  %!ÑWÐB34  %!Ñ
where +34 is the shipping cost from cannery 3 to warehouse 4 and
WÐBÑ œ 
!
"
if B  !
if B !.
The rest of the formulation is the same.
13.1-4.
Let W" and W# be the number of blocks of stock " and # to purchase respectively.
minimize
0 ÐW" ß W# Ñ œ %W"#  "!!W##  &W" W#
subject to
#!W"  $!W# Ÿ &!
&W"  "!W# minimum acceptable expected return
W " ß W#
!
13.2-1.
0 ÐBÑ œ 0" ÐB" Ñ  0# ÐB# Ñ  0$ ÐB$ Ñ
#Î$
$Î%
"Î#
with 0" ÐB" Ñ œ "!!B"  "!B" ß 0# ÐB# Ñ œ %!B#  &B# ß 0$ ÐB$ Ñ œ &!B$  &B$ .
.# 0" ÐB" Ñ
.B#"
œ  #!!
* B"
.# 0# ÐB# Ñ
.B##
œ  "#!
"' B#
.# 0$ ÐB$ Ñ
.B#$
œ  &!
% B$
%Î$
Ÿ ! for B"
!
&Î%
Ÿ ! for B#
!
$Î#
Ÿ ! for B$
!
0" , 0# and 0$ are concave on the nonnegative orthant so 0 is concave in the same region.
The constraints are linear. Hence, the problem is a convex programming problem.
13-2
13.2-2.
.# 0 ÐW" ßW# Ñ
.W"#
!ß .
œ)
.# 0 ÐW" ßW# Ñ .# 0 ÐW" ßW# Ñ
.W"#
.W##
#
0 ÐW" ßW# Ñ
.W##
#
0 ÐW" ßW# Ñ
!ß . .W
œ&
" .W#
œ #!!
0 ÐW" ßW# Ñ
  . .W
 œ "&(&
" .W#
#
#
!
!
Hence, 0 is convex everywhere.
13.2-3.
Objective function: ^ œ $B"  &B# Ê B# œ Ð$Î&ÑB"  Ð"Î&Ñ^ Ê slope: Ð$Î&Ñ
Constraint boundary: *B#"  &B## œ #"' Ê B# œ Ð"Î&ÑÐ#"'  *B#" Ñ
Ê
`B#
`B"
*B"
$
œ  &" Ð"Î&ÑÐ#"'*B
# Ñ œ  & for B" œ #
"
Hence, the objective function is tangent to this constraint at ÐB" ß B# Ñ œ Ð#ß 'Ñ.
13.2-4.
Constraint boundary: $B"  #B# œ ") Ê 1ÐB" Ñ œ B# œ  $# B"  * Ê
.1ÐB" Ñ
.B"
œ  $#
Objective function at Ð)Î$ß &Ñ: Ð*B#"  "#'B"  )&(Ñ  ")#B#  "$B## œ !
Ê 0 ÐB" Ñ œ B# œ
")###)'!"'$)B" ""(B#"
#'
Ê
.0 ÐB" Ñ
.B"
œ  $#
0 Ð)Î$Ñ œ 1Ð)Î$Ñ œ &
Hence, the objective function is tangent to this constraint at ÐB" ß B# Ñ œ Ð)Î$ß &Ñ.
13.2-5.
(a)
.0 ÐBÑ
.B
œ 48  12!B  $B# œ !
Ê B‡ œ
.# 0 ÐBÑ
.B#
œ !Þ%!41 or 39.596
œ 120  'B
.# 0 Ð!Þ%!41Ñ
.B#
#
120„120# %†$†48
6
. 0 Ð3*Þ&*6Ñ
.B#
œ 117.6 Ê 0 Ð!Þ%!41Ñ œ 2.475 is a local maximum.
œ 117.6 Ê 0 Ð3*Þ&*6Ñ œ 0.0253 is a local minimum.
(b) For B  3*Þ&*6,
.0 ÐBÑ
.B
 ! and
.# 0 ÐBÑ
.B#
œ 'B  12!  ! Ê 0 is unbounded above.
For B  !Þ%!41,
.0 ÐBÑ
.B
 ! and
.# 0 ÐBÑ
.B#
œ 'B  120  ! Ê 0 is unbounded below.
13-3
13.2-6.
(a)
. # 0 ÐBÑ
.B#
œ #  ! for all B Ê 0 is concave.
(b)
.# 0 ÐBÑ
.B#
œ "#B#  "#  ! for all B Ê 0 is convex.
(c)
. # 0 ÐBÑ
.B#
œ "#B'
(d)
.# 0 ÐBÑ
.B#
œ "#B#  #  ! for all B Ê 0 is convex.
(e)
. # 0 ÐBÑ
.B#
 ! for B  "Î#
Ê 0 is neither convex nor concave.
 ! for B  "Î#
œ 'B  "#B# 
!
!
for B  "Î# or B  !
Ê 0 is neither convex nor
for "Î#  B  !
concave.
13.2-7.
(a)
` # 0 ÐB" ßB# Ñ
`B#"
œ
` # 0 ÐB" ßB# Ñ
`B##
` # 0 ÐB" ßB# Ñ ` # 0 ÐB" ßB# Ñ
`B#"
`B##
œ #  ! for all ÐB" ß B# Ñ
0 ÐB" ßB# Ñ
  ` `B
 œ %  "# œ $  ! for all ÐB" ß B# Ñ
" `B#
#
#
Ê 0 is concave.
(b)
` # 0 ÐB" ßB# Ñ
`B#"
œ %  !ß `
` # 0 ÐB" ßB# Ñ ` # 0 ÐB" ßB# Ñ
`B#"
`B##
#
0 ÐB" ßB# Ñ
`B##
œ #  ! for all ÐB" ß B# Ñ
0 ÐB" ßB# Ñ
  ` `B
 œ )  ## œ %  ! for all ÐB" ß B# Ñ
" `B#
#
#
Ê 0 is convex.
(c)
` # 0 ÐB" ßB# Ñ
`B#"
œ #  !ß `
` # 0 ÐB" ßB# Ñ ` # 0 ÐB" ßB# Ñ
`B#"
`B##
#
0 ÐB" ßB# Ñ
`B##
œ %  ! for all ÐB" ß B# Ñ
0 ÐB" ßB# Ñ
  ` `B
 œ )  $# œ "  ! for all ÐB" ß B# Ñ
" `B#
#
#
Ê 0 is neither convex nor concave.
(d)
` # 0 ÐB" ßB# Ñ
`B#"
œ
` # 0 ÐB" ßB# Ñ
`B##
` # 0 ÐB" ßB# Ñ ` # 0 ÐB" ßB# Ñ
`B#"
`B##
œ
` # 0 ÐB" ßB# Ñ
`B" `B#
œ!
0 ÐB" ßB# Ñ
  ` `B
 œ!
" `B#
#
#
Ê 0 is both convex and concave.
(e)
` # 0 ÐB" ßB# Ñ
`B#"
œ
` # 0 ÐB" ßB# Ñ
`B##
` # 0 ÐB" ßB# Ñ ` # 0 ÐB" ßB# Ñ
`B#"
`B##
œ ! for all ÐB" ß B# Ñ
0 ÐB" ßB# Ñ
  ` `B
 œ !  "# œ "  ! for all ÐB" ß B# Ñ
" `B#
#
#
Ê 0 is neither convex nor concave.
13-4
13.2-8.
0 ÐBÑ œ 0" ÐB" Ñ  0# ÐB# Ñ  0$% ÐB$ ß B% Ñ  0&' ÐB& ß B' Ñ  0'( ÐB' ß B( Ñ
with 0" ÐB" Ñ œ &B" ß 0# ÐB# Ñ œ #B## ß 0$% ÐB$ ß B% Ñ œ B#$  $B$ B%  %B#% ß
0&' ÐB& ß B' Ñ œ B#&  $B& B'  $B#' ß 0'( ÐB' ß B( Ñ œ $B#'  $B' B(  B(# .
.# 0" ÐB" Ñ
.B#"
œ ! for all B" Ê 0" is convex (and concave).
.# 0# ÐB# Ñ
.B##
œ %  ! for all B# Ê 0# is convex.
.# 0$% ÐB$ ßB% Ñ
.B#$
œ #  !ß .
.# 0$% ÐB$ ßB% Ñ .# 0$% ÐB$ ßB% Ñ
.B#$
.B#%
#
0$% ÐB$ ßB% Ñ
.B#%
 .
#
œ )  ! for all ÐB$ ß B% Ñ
#
0$% ÐB$ ßB% Ñ

.B$ .B%
œ "'  $# œ (  ! for all ÐB$ ß B% Ñ
Ê 0$% is convex.
.# 0&' ÐB& ßB' Ñ
.B#&
œ #  !ß .
.# 0&' ÐB& ßB' Ñ .# 0&' ÐB& ßB' Ñ
.B#&
.B#'
#
0&' ÐB& ßB' Ñ
.B#'
 .
#
œ '  ! for all ÐB& ß B' Ñ
#
0&' ÐB& ßB' Ñ

.B& .B'
œ "#  $# œ $  ! for all ÐB& ß B' Ñ
Ê 0&' is convex.
0'( ÐB' ß B( Ñ œ 0&' ÐB( ß B' Ñ Ê 0'( is convex.
Hence, 0 is convex.
13.2-9.
(a)
maximize
0 ÐBÑ œ B"  B#
subject to
1ÐBÑ œ B#"  B## Ÿ "ß B
` # 0 ÐBÑ
`B#"
œ
` # 0 ÐBÑ
`B##
` # 1ÐBÑ
`B#"
œ
` # 1ÐBÑ
`B##
œ
` # 0 ÐBÑ
`B" `B#
!
0 ÐBÑ ` 0 ÐBÑ
` 0 ÐBÑ
œ !ß ` `B
  `B
 œ ! Ê 0 is concave (convex).
#
`B#
" `B#
#
#
"
#
#
#
1ÐBÑ ` 1ÐBÑ
` 1ÐBÑ
œ #  !ß ` `B
  `B
 œ %  !# œ %  ! Ê 1 is convex.
#
`B#
" `B#
#
#
"
#
#
#
The problem is a convex programming problem.
(b)
13-5
13.2-10.
(a)
Clearly, this is not a convex feasible region. For example, take the points Ð!ß #Ñ and
Ð!ß #Ñ, Ð!ß !Ñ œ "# Ð!ß #Ñ  "# Ð!ß #Ñ is not feasible.
(b) Feasible region: B#"  B## Ÿ #
Both 1" ÐB" Ñ œ B#" and 1# ÐB# Ñ œ B## are concave functions, so the feasible region need
not be convex.
.# 1" ÐB" Ñ
.B#"
œ
.# 1# ÐB# Ñ
.B##
œ "  !
To prove that the feasible region is not convex, one needs to find two feasible points C
and D , a scalar α − Ò!ß "Ó such that αC  Ð"  αÑD is not feasible. Such points are given in
part (a).
13.3-1.
Since the objective is to minimize a concave function, as shown in Problem 13.1-3, this is
a nonconvex programming problem.
13.3-2.
.0 ÐBÑ
.B
.# 0 ÐBÑ
.B#
œ 6  6B  6B# œ ! Ê B œ
œ '"#
!
!
for B 
for B 
6„6# %†$6
12
has no real solution
"
#
"
#
The slope of 0 increases from ' at B œ ! to  *# at B œ
thereafter. It is always negative, so B‡ œ ! is optimal.
13-6
"
#
and decreases for all B
13.3-3.
(a)
Linearly Constrained Convex Programming:
1" ÐB" ß B# Ñ œ #B"  B# and 1# ÐB" ß B# Ñ œ B"  #B# are linear.
` # 0 ÐB" ßB# Ñ
`B#"
œ "#B#"  %  !ß `
` # 0 ÐB" ßB# Ñ ` # 0 ÐB" ßB# Ñ
`B#"
`B##
#
0 ÐB" ßB# Ñ
`B##
œ )  ! for all ÐB" ß B# Ñ
0 ÐB" ßB# Ñ
  ` `B
 œ *'B#"  $#  ##  ! for all ÐB" ß B# Ñ
" `B#
#
#
Ê 0 is concave.
Geometric Programming:
0 ÐBÑ œ -" B+""" B#+"#  -# B"+#" B#+##  -$ B+"$" B+#$#  -% B"+%" B#+%#
where -" œ "ß +"" œ %ß +"# œ !
-# œ #ß +#" œ #ß +## œ !
-$ œ #ß +$" œ "ß +$# œ "
-% œ %ß +%" œ !ß +%# œ #
1" ÐBÑ œ -" B+""" B#+"#  -# B"+#" B#+##
where -" œ #ß +"" œ "ß +"# œ !
-# œ "ß +#" œ !ß +## œ "
1# ÐBÑ œ -" B+""" B#+"#  -# B"+#" B#+##
where -" œ "ß +"" œ "ß +"# œ !
-# œ #ß +#" œ !ß +## œ "
Fractional Programming:
0 w œ 0" Î0# where 0" œ 0 and 0# œ "
(b) Let C" œ B"  " and C# œ B#  ".
minimize
C"%  %C"$  )C"#  "!C"  #C" C#  %C##  "!C#
subject to
#C"  C# (
C"  #C# (
C" ß C# !
13.3-4.
(a) Let B" œ /C" and B# œ /C# .
minimize
0 ÐCÑ œ #/#C" C#  /C" #C#
subject to
1ÐCÑ œ %/C" C#  /#C" #C#  "# Ÿ !
/ C " ß / C#
! (true for any ÐC" ß C# Ñ)
13-7
(b)
` # 0 ÐC Ñ
`C"#
œ )/#C" C#  /C" #C#
` # 0 ÐC Ñ
`C##
œ #/#C" C#  %/C" #C#
` # 0 ÐC Ñ ` # 0 ÐC Ñ
`C"#
`C##
! for all ÐC" ß C# Ñ
! for all ÐC" ß C# Ñ
` 0 ÐC Ñ
  `C
 œ ")/$C" $C#
" `C#
#
#
! for all ÐC" ß C# Ñ
Ê 0 is convex.
` # 1ÐC Ñ
`C"#
œ
` # 1ÐC Ñ
`C##
` # 1ÐC Ñ ` # 1ÐC Ñ
`C"#
`C##
œ
` # 1ÐC Ñ
`C" `C#
œ %/C" C#  %/#C" #C#
! for all ÐC" ß C# Ñ
` 1ÐC Ñ
  `C
 œ ! for all ÐC" ß C# Ñ
" `C#
#
#
Ê 1 is convex.
Hence, this is a convex programming problem.
13.3-5.
(a)
maximize
subject to
"!C"  #!C#  "!>
C"  $C#  &!> Ÿ !
$C"  %C#  )!> Ÿ !
$C"  %C#  #!> œ "
C" ß C# ß > !
(b)
The variables Ð\"ß \#ß \$Ñ in this courseware solution correspond to the variables
ÐC" ß C# ß >Ñ in (a), so the optimal solution is ÐC" ß C# ß >Ñ œ Ð!ß !Þ"*#ß !Þ!"#Ñ with the
objective function value ^ œ $Þ*'#. Then, the optimal solution of the original problem is
ÐB" ß B# Ñ œ Ð!ß "'Þ'(Ñ with the optimal objective function value 0 ÐBÑ œ $Þ*'#.
13.3-6.
KKT conditions:
UB  EX ?  - œ C
EB  , œ @
Bß ?ß Cß @ !
BX ÐUB  EX ?  -Ñ  ?X ÐEB  ,Ñ œ !
This is the linear complementarity problem with:
B
U
^ œ  ß Q œ 
?
E
EX
UB  EX ?  ß; œ 
ßA œ 
.


,
!
EB  , 
13-8
13.4-1.
(a)
(b)
13.4-2.
(a)
13-9
(b)
13.4-3.
(a)
(b)
13-10
13.4-4.
(a)
0 ÐBÑ
œ B$  $!B  B'  #B%  $B#
(b)
0 ÐBÑ
œ B$  $!B  B'  #B%  $B#
0 w ÐBÑ
œ $B#  $!  'B&  )B$  'B
0 ww ÐBÑ œ 'B  $!B%  #%B#  '
Iteration 3
"
#
$
%
B3
"
"Þ#%!(
"Þ")')
"Þ"*#&
0 ÐB3 Ñ
#&
#'Þ"#'
#'Þ#))
#'Þ#)*
0 w ÐB3 Ñ
"$
&Þ(%))
!Þ$*&(
!Þ!!#$
0 ww ÐB3 Ñ
&%
"!'Þ'!
*#Þ#!"
*"Þ"#(
B3"
"Þ#%!(
"Þ")')
"Þ")#&
"Þ")#&
lB3  B3" l
!Þ#%!(
!Þ!&$*
!Þ!!%$
#I  !&
13.4-5.
(a) 0 w ÐBÑ œ %B$  #B  % Ê 0 w Ð!Ñ œ %ß 0 w Ð"Ñ œ #ß 0 w Ð#Ñ œ $#
Since 0 w ÐBÑ is continuous, there must be a point ! Ÿ B‡ Ÿ " such that 0 w ÐB‡ Ñ œ ! and
since 0 is a convex function (given that this is a convex programming problem), B‡ must
be the optimal solution. Hence, the optimal solution lies in the interval ! Ÿ B Ÿ ".
(b)
13-11
(c)
13.4-6.
(a) Consider the two cases:
Case 1:B8" œ B8 and B8" œ Bw8
Ê B8"  B8" œ B8  Bw8 œ B8  "# ÐB8  B8 Ñ œ "# ÐB8  B8 Ñ
Case 2:B8" œ Bw8 and B8" œ B8
Ê B8"  B8" œ B8w  B8 œ "# ÐB8  B8 Ñ  B8 œ "# ÐB8  B8 Ñ
In both cases: B8"  B8" œ "# ÐB8  B8 Ñ œ â œ
Ê lim ÐB8"  B8" Ñ œ lim
8Ä∞
8Ä∞ #
"
8"
"
#8" ÐB!
 B! Ñ
ÐB!  B! Ñ œ !
If the sequence of trial solutions selected by the midpoint rule did not converge to a
limiting solution, then there must be an %  ! such that regardless of what R is, there are
w
8 R and 7 R with lBw8  B7
l  %. In that case, choose R that satisfies
R
lBR  BR l œ # ÐB!  B! Ñ  %. Then for every 8 R , since B8w − ÒBR ß BR Ó:
w
lBw8  B7
l Ÿ lBR  BR l œ #R ÐB!  B! Ñ  %,
w
which contradicts that lBw8  B7
l  %. Hence, the sequence must converge.
13-12
(b) Let B be the limiting solution. Then, 0 w ÐBÑ ! for B  B and 0 w ÐBÑ Ÿ ! for B  B.
Suppose now that there exists an s
B with 0 ÐBÑ
s  0 ÐBÑ so that B is not a global maximum.
Case 1: s
B  B. By the Mean Value Theorem, there exists a D such that B
s  D  B and
w
0 ÐBÑ
s  0 ÐBÑ œ ÐB
s  BÑ0 ÐDÑ Ÿ !, so 0 ÐBÑ
s Ÿ 0 ÐBÑ.
Case #: s
B  B. By the Mean Value Theorem, there exists a D such that B
s  D  B and
w
0 ÐBÑ  0 ÐBÑ
œ
ÐB

BÑ0
ÐDÑ
!
,
so
0
ÐBÑ
Ÿ
0
ÐBÑ
.
s
s
s
Both cases give rise to a contradiction, so B must be a global maximum.
(c) The argument is the same as the one in part (b). Observe that D that is chosen between
B and B remains in the region where 0 is concave and the values B! and B! are given as
s
lower and upper bounds on the same global maximum.
(d) In the example illustrated in the graph below, the bisection method converges to B
rather than to B‡ , which is the global maximum.
(e) Suppose 0 w ÐBÑ  ! for all B and s
B is a global maximum. Then, by the Mean Value
Theorem, there exists a D such that s
B  D  B and 0 ÐBÑ
s  0 ÐBÑ œ ÐB
s  BÑ0 w ÐDÑ  !, so
w
0 ÐBÑ œ 0 ÐBÑ
s  ÐB
s  BÑ0 ÐDÑ  0 ÐBÑ
s . The objective function value can be strictly
increased by choosing smaller B values at any given point, so there exists no lower bound
B! on the global maximum, there is no global maximum indeed.
Suppose 0 w ÐBÑ  ! for all B and s
B is a global maximum. Then, by the Mean Value
Theorem, there exists a D such that B  D  s
B and 0 ÐBÑ  0 ÐBÑ
BÑ0 w ÐDÑ  !, so
s œ ÐB  s
0 ÐBÑ œ 0 ÐBÑ
BÑ0 w ÐDÑ  0 ÐBÑ
s  ÐB  s
s . The objective function value can be strictly
increased by choosing larger B values at any given point, so there exists no upper bound
B! on the global maximum, there is no global maximum indeed.
(f) Suppose 0 ÐBÑ is concave and there exists a lower bound B! on the global maximum.
In this case, 0 w ÐB! Ñ !, but 0 w ÐBÑ is monotone decreasing, so for B  B! , 0 w ÐBÑ !.
Hence, limBÄ∞ 0 w ÐBÑ !, so if limBÄ∞ 0 w ÐBÑ  !, there cannot be an B! .
Suppose 0 ÐBÑ is concave and there exists an upper bound B! on the global maximum.
In this case, 0 w ÐB! Ñ Ÿ !, but 0 w ÐBÑ is monotone decreasing, so for B  B! , 0 w ÐBÑ Ÿ !.
Hence, limBÄ∞ 0 w ÐBÑ Ÿ !, so if limBÄ∞ 0 w ÐBÑ  !, there cannot be an B! .
In either case, there is no global maximum, since one of the bounds does not exist.
13-13
13.4-7.
0 ÐBÑ œ 0" ÐB" Ñ  0# ÐB# Ñ
where 0" ÐB" Ñ œ $#B"  B%" and 0# ÐB# Ñ œ &!B#  "!B##  B$#  B#% .
.0" ÐB" Ñ
.B"
œ $#  %B$" œ ! Í B" œ #ß 0" Ð#Ñ œ %)
Bisection method with % œ !Þ!!" and initial bounds ! and % applied to 0# ÐB# Ñ gives
B# œ "Þ)!(' and 0# Ð"Þ)!('Ñ œ &#Þ*$', so 0 Ð#ß "Þ)!('Ñ œ "!!Þ*$'.
$B"  B# œ (Þ)!('  "" and #B"  &B# œ "$Þ!$)  "'
Since the optimal solution for the unconstrained problem is in the interior of the feasible
region for the constrained problem, it is also optimal for the constrained problem.
13.5-1.
(a)
(b) #B"  #B# œ ! and #B"  %B# œ " Ê B" œ B# œ !Þ& is optimal.
(c)
(d) Solution: ÐB" ß B# Ñ œ Ð!Þ&!)ß !Þ&!%Ñ, grad 0 ÐB" ß B# Ñ œ Ð)/$ß '/)Ñ
13-14
13.5-2.
The automatic routine (% œ !Þ!") gives
ÐB" ß B# Ñ œ Ð!Þ!!&ß !Þ!!$Ñ
f0 ÐB" ß B# Ñ œ Ð(/$ß $/)Ñ
f0 œ Ð%B#  %B" ß %B"  'B# Ñ œ ! Í ÐB" ß B# Ñ œ Ð!ß !Ñ is optimal.
13.5-3.
Solution: ÐB" ß B# Ñ œ Ð"Þ**(ß #Ñ, grad 0 ÐB" ß B# Ñ œ Ð!Þ!!#ß !Þ!!"Ñ
f0 œ Ð#B"  #B#  )ß #B"  %B#  "#Ñ œ ! Í ÐB" ß B# Ñ œ Ð#ß #Ñ is optimal.
13.5-4.
Solution: ÐB" ß B# Ñ œ Ð"Þ**%ß !Þ*)*Ñ, grad 0 ÐB" ß B# Ñ œ Ð!Þ!!$ß !Þ!"Ñ
f0 œ Ð%B"  #B#  'ß #B"  #B#  #Ñ œ ! Í ÐB" ß B# Ñ œ Ð#ß "Ñ is optimal.
13.5-5.
Iter. >w
0 Ð>Ñ
"
!Þ&
"%%
#
!Þ#& "%
‡
> œ !Þ"#&
‡
Ê B  > f0 ÐBÑ œ Ð!Þ&ß !Þ#&Ñ is the approximate solution.
Iter.
"
B8
Ð!ß !Ñ
f0 ÐB8 Ñ
Ð%ß #Ñ
0 ÐB8  f0 ÐB8 ÑÑ
#!>  #'>#  #&'>%
13.5-6.
(a) 0 ÐBÑ œ 0" ÐB" ß B# Ñ  0# ÐB# ß B$ Ñ
where 0" ÐB" ß B# Ñ œ $B" B#  B#"  $B## and 0# ÐB# ß B$ Ñ œ $B# B$  B#$  $B## .
Note that 0" ÐB$ ß B# Ñ œ 0# ÐB# ß B$ Ñ, so for any given B# , the maximizers of 0" and 0# are
the same, i.e., B" œ B$ . Hence, first maximize 0" (or 0# ) and obtain ÐB" ß B# Ñ. Then, set
B$ œ B" and 0 ÐBÑ œ #0" ÐB" ß B# Ñ.
13-15
(b)
Final Solution: ÐB" ß B# Ñ œ Ð!Þ!'*ß !Þ!$'Ñ Ê ÐB" ß B# ß B$ Ñ œ Ð!Þ!'*ß !Þ!$'ß !Þ!'*Ñ is an
approximate solution.
(c) Solution: ÐB" ß B# Ñ œ Ð!Þ!!%ß !Þ!!#Ñ, grad 0 ÐB" ß B# Ñ œ Ð#/$ß '/%Ñ
13.5-7.
Solution: ÐB" ß B# Ñ œ Ð!Þ**'ß "Þ**)Ñ, grad 0 ÐB" ß B# Ñ œ œ Ð!Þ!!'ß #/)Ñ
13.6-1.
KKT conditions:
(1) %B$  #B  %  ? Ÿ !
(2) BÐ%B$  #B  %  ?Ñ œ !
(3) B  # Ÿ !
(4) ?ÐB  #Ñ œ !
(5) B !
(6) ? !
If B œ #, from (2), %B$  #B  %  ? œ !, so ? œ $#, which violates (6). Hence,
B Á #, then from (4), ? œ !. From (2), either B œ ! or %B$  #B  % œ !. In the former
case, (1) is violated, so the latter equality must hold. This gives
$ "
$ "
&&
&&
 #"'
 #"'
Bœ

œ !Þ)$&"#.
# 
# 
13.6-2.
KKT conditions:
(1a) "  #?B" Ÿ !
(2a) B" Ð"  #?B" Ñ œ !
(3) B#"  B##  " Ÿ !
(4) ?ÐB#"  B##  "Ñ œ !
(5) B" !ß B# !
(6) ? !
("b) "  #?B# Ÿ !
(#b) B# Ð"  #?B# Ñ œ !
If B œ Ð"Î#ß "Î#Ñ, from (2a), ? œ "Î#. This solution satisfies all KKT conditions,
so it is optimal.
13-16
13.6-3.
KKT conditions:
(1a) %B$"  %B"  #B#  #?"  ?# Ÿ !
(2a) B" Ð%B$"  %B"  #B#  #?"  ?# Ñ œ !
(1b)  #B"  )B#  ?"  #?# Ÿ !
(2b) B# Ð  #B"  )B#  ?"  #?# Ñ œ !
(3a) #B"  B# "!
(4a) ?" Ð#B"  B#  "!Ñ œ !
(3b) B"  #B# "!
(4b) ?# ÐB"  #B#  "!Ñ œ !
(5) B" !ß B# !
(6) ?" !ß ?# !
If B œ Ð!ß "!Ñ, from (2b), ?"  #?# œ )! and from (4b), ?# œ !, so ?" œ )!. This
solution violates (1a), so it is not optimal.
13.6-4.
(a) KKT conditions: (1a) 24  #B"  ?" Ÿ !
(2a) B" Ð24  #B"  ?" Ñ œ !
(3a) B" Ÿ 8, B# Ÿ 7
(4a) ?" ÐB"  8Ñ œ !
(5) B" !ß B# !
(6) ?" !ß ?# !
(1b) 1!  #B#  ?# Ÿ !
(2b) B# Ð1!  #B#  ?# Ñ œ !
(4b) ?# ÐB#  7Ñ œ !
Consider B" œ 8. From (2a), ?" œ )Þ Then (1a), (3), (4a), (5) and (6) are satisfied.
Consider ?2 œ !. From (2b), B# œ &Þ Then (1b), (3), (4b), (5) and (6) are satisfied.
Thus, ÐB" ß B# Ñ œ Ð)ß &Ñ is optimal since this is a convex program.
(b)
Subproblem 1: maximize 0" ÐB" Ñ œ "#B"  B#" subject to ! Ÿ B" Ÿ "!
Subproblem 2: maximize 0# ÐB# Ñ œ &!B#  B## subject to ! Ÿ B" Ÿ "&
`0" ÐB" Ñ
`B"
œ 24  #B"  ! a ! Ÿ B" Ÿ )ß so B" œ ) is the maximum value over the feasible
region.
`0# ÐB# Ñ
`B#
œ 1!  #B# œ ! at B# œ & and
` # 0 ÐBÑ
`B##
œ # Ÿ ! so B# œ & is a global maximum.
13-17
13.6-5.
(a)
` # 0 ÐBÑ
`B#"
` # 0 ÐBÑ
`B##
"
œ  ÐB" "Ñ
# Ÿ ! for all ÐB" ß B# Ñ such that B" Á "
œ # Ÿ ! for all ÐB" ß B# Ñ
` # 0 ÐBÑ ` # 0 ÐBÑ
`B#"
`B##
` 0 ÐBÑ
  `B
 œ
" `B#
#
#
#
ÐB" "Ñ#
! for all ÐB" ß B# Ñ such that B" Á "
Ê 0 is concave.
Since also 1ÐBÑ œ B"  #B#  $ is linear, this is a convex programming problem.
(b) KKT conditions: (1a)
"
ÐB" "Ñ
?Ÿ!
(2a) B"  ÐB"""Ñ  ? œ !
(1b) #B#  #? Ÿ !
(2b) B# Ð#B#  #?Ñ œ !
(3) B"  #B# Ÿ $
(4) ?ÐB"  #B#  $Ñ œ !
(5) B" !ß B# !
(6) ? !
Consider ? Á !. From (4), B"  #B# œ $. Let B# œ !. Then, B" œ $ and from (2a),
? œ !Þ#&. This satisfies all the conditions, so ÐB" ß B# Ñ œ Ð$ß !Ñ is optimal.
(c) Since B## is monotonically strictly decreasing in B# ! and lnÐB"  "Ñ is monotonically strictly increasing in B" !, it is intuitively clear that one would like to increase B"
and decrease B# towards ! as much as possible, in order to maximize the objective
function. Let Y denote the set of feasible points. Then,
maxmin Y  œ minmax Y  œ ÖÐ$ß !Ñ×.
B"
B#
B#
B"
Hence, the solution Ð$ß !Ñ makes intuitive sense.
13.6-6.
KKT conditions:
(1a) $'  ")B"  ")B#"  ? Ÿ !
(1b) $'  *B##  ? Ÿ !
(2a) B" Ð$'  ")B"  ")B#"  ?Ñ œ ! (2b) B# Ð$'  *B##  ?Ñ œ !
(3) B"  B# Ÿ $
(4) ?ÐB"  B#  $Ñ œ !
(5) B" !ß B# !
(6) ? !
For ÐB" ß B# Ñ œ Ð"ß #Ñ, from (2b), ? œ ! and this violates (2a), so Ð"ß #Ñ is not optimal.
13-18
13.6-7.
(a) KKT conditions: (1a)
"
ÐB# "Ñ
?Ÿ!
(2a) B"  ÐB#""Ñ  ? œ !
"
(1b)  ÐB#B"Ñ
#  ? Ÿ !
"
(2b) B#  ÐB#B"Ñ
#  ? œ !
(3) B"  B# Ÿ #
(4) ?ÐB"  B#  #Ñ œ !
(5) B" !ß B# !
(6) ? !
For ÐB" ß B# Ñ œ Ð%ß #Ñ, from (2a), ? œ "Î$ and this violates (2b), so Ð%ß #Ñ is not optimal.
(b) Try B# œ ! and ? Á !. From (4), B" œ # and from (2a), ? œ ". This solution satisfies
all the conditions, so ÐB" ß B# Ñ œ Ð#ß !Ñ is optimal.
(c)
` # 0 ÐBÑ
`B#"
#
0 ÐBÑ
œ !ß ` `B
œ
#
#
#B"
ÐB# "Ñ#
#
` 0 ÐBÑ
"
!ß `B
œ  ÐB# "Ñ
# Ÿ ! for all B"
" `B#
!ß B#
!
Thus, 0 is not concave and this is not a convex programming problem.
(d) The function 0 ÐBÑ is monotonically strictly increasing in B" and monotonically
strictly decreasing in B# if B#  ". Any optimal solution in a bounded feasible region
with B#  " will have B" increased as much as possible and B# decreased toward " as
much as possible. The feasible region of the problem allows B" to be increased without
bound. However, then B# can only be decreased to the line B"  B# œ #.
0 ÐB#  #ß B# Ñ œ
B# #
B# "
Ä " as B# Ä ∞ and 0 ÐB#  #ß B# Ñ œ # at B# œ !
Conversely, if B# is decreased to !, B" can be increased to B" œ #. Hence, the optimal
solution is ÐB" ß B# Ñ œ Ð#ß !Ñ.
(e) maximize B"
subject to B"  B#  #> Ÿ !
B#  > œ "
B" ß B# ß > !
Í
maximize
subject to
ÐB" ß B# Ñ œ Ð#ß !Ñ is optimal.
13-19
B"
B"  B# Ÿ #
B# Ÿ "
B" ß B# !
13.6-8.
(a) KKT conditions: (1a) "  ? Ÿ !
(1b) #  $B##  ? Ÿ !
(2a) B" Ð"  ?Ñ œ ! (2b) B# Ð#  $B##  ?Ñ œ !
(3) B"  B# Ÿ "
(4) ?ÐB"  B#  "Ñ œ !
(5) B" !ß B# !
(6) ? !
The solution ÐB" ß B# ß ?Ñ œ Ð"  "Î$ß "Î$ß "Ñ satisfies all the conditions. Since this is
a convex programming problem, Ð"  "Î$ß "Î$Ñ is optimal.
(a) KKT conditions: (1a) #!  #?" B"  ?# Ÿ !
(2a) B" Ð#!  #?" B"  ?# Ñ œ !
(3a) B#"  B## Ÿ "
(4a) ?" ÐB#"  B##  "Ñ œ !
(5) B" !ß B# !
(6) ?" !ß ?# !
(1b) "!  #?" B"  #?# Ÿ !
(2b) B# Ð"!  #?" B"  #?# Ñ œ !
(3b) B"  #B# Ÿ #
(4b) ?# ÐB"  #B#  #Ñ œ !
The solution ÐB" ß B# ß ?Ñ œ Ð#Î&ß "Î&ß &&ß !Ñ satisfies all the conditions. Since this
is a convex programming problem, Ð#Î&ß "Î&Ñ is optimal.
13.6-9.
Í
minimize
subject to
0 ÐB Ñ
13 ÐBÑ ,3 for 3 œ "ß #ß á ß 7
B !
maximize
subject to
0 ÐBÑ
13 ÐBÑ Ÿ ,3 for 3 œ "ß #ß á ß 7
B !
3 ÐBÑ
(1)  ?3 `1`B

4
7
KKT conditions:
3œ"
`0 ÐBÑ
`B4
3 ÐBÑ
(2) B4   ?3 `1`B

4
7
3œ"
Ÿ ! for 4 œ "ß #ß á ß 8
`0 ÐBÑ
`B4 
œ ! for 4 œ "ß #ß á ß 8
(3) 13 ÐBÑ ,3 for 3 œ "ß #ß á ß 7
(4) ?3 Ð,3  13 ÐBÑÑ œ ! for 3 œ "ß #ß á ß 7
(5) B4 ! for 4 œ "ß #ß á ß 8
(6) ?3 ! for 3 œ "ß #ß á ß 7
13-20
13.6-10.
(a) An equivalent nonlinear programming problem is:
maximize
^ œ #B#"  B##
subject to
B"  B# Ÿ "!
B"  B# Ÿ "!
B" ß B#
!.
This problem can be fitted to the following problems.
- Linearly Constrained Optimization Problem: All constraints are linear.
- Quadratic Programming Problem: All constraints are linear and the objective function
involves only the squares of the variables.
- Convex Programming Problem: The objective function is concave and all constraints
are linear.
` # 0 ÐBÑ ` # 0 ÐBÑ
`B#"
`B##
` 0 ÐBÑ
  `B
 œ Ð%ÑÐ#Ñ  ! œ )
" `B#
#
#
! Ê 0 is concave.
- Geometric Programming Problem:
0 ÐB" ß B# Ñ œ -" T" ÐB" ß B# Ñ  -# T# ÐB" ß B# Ñ
with -" œ #, -# œ ", T" ÐB" ß B# Ñ œ B#" and T# ÐB" ß B# Ñ œ B##
1" ÐB" ß B# Ñ œ -" T" ÐB" ß B# Ñ  -# T# ÐB" ß B# Ñ
with -" œ -# œ ", T" ÐB" ß B# Ñ œ B" and T# ÐB" ß B# Ñ œ B#
1# ÐB" ß B# Ñ œ -" T" ÐB" ß B# Ñ  -# T# ÐB" ß B# Ñ
with -" œ -# œ ", T" ÐB" ß B# Ñ œ B" and T# ÐB" ß B# Ñ œ B#
- Fractional Programming Problem:
0 ÐB" ß B# Ñ œ
0" ÐB" ßB# Ñ
0# ÐB" ßB# Ñ
with 0" ÐB" ß B# Ñ œ #B#"  B## and 0# ÐB" ß B# Ñ œ "
(b) KKT conditions: (1a) %B"  ?"  ?# Ÿ !
(2a) B" Ð%B"  ?"  ?# Ñ œ !
(1b) #B#  ?"  ?# Ÿ !
(2b) B# Ð#B#  ?"  ?# Ñ œ !
(3a) B"  B#  "! Ÿ !
(4a) ?" ÐB"  B#  "!Ñ œ !
(3b) B"  B#  "! Ÿ !
(4b) ?# ÐB"  B#  "!Ñ œ !
(5) B" !ß B# !
(6) ?" !ß ?# !
(c) From (3a) and (3b), B"  B# œ "!, so (4a) and (4b) are automatically satisfied. Try
B" ß B# Á !. Then, (2a) and (2b) give %B"  ?"  ?# œ #B#  ?"  ?# œ !, so
B# œ #B" . Since B"  B# œ "!, B" œ "!Î$ and B# œ #!Î$. From (2a),
 ?"  ?# œ %!Î$. Let ?" œ ! and ?# œ %!Î$. Indeed, any Ð?" ß ?# Ñ œ Ð-ß -  %!Î$Ñ
with - ! works. This solution satisfies all the conditions, so ÐB" ß B# Ñ œ Ð"!Î$ß #!Î$Ñ is
optimal.
13-21
13.6-11.
(a) An equivalent nonlinear programming problem is:
maximize
subject to
0 ÐCÑ œ ÐC"  "Ñ$  %ÐC#  "Ñ#  "'ÐC$  "Ñ
C"  C#  C$ Ÿ #
C"  C#  C$ Ÿ #
C" ß C# ß C$
!.
(b) KKT conditions: (1a) $ÐC"  "Ñ#  ?"  ?# Ÿ !
(2a) C" Ð$ÐC"  "Ñ#  ?"  ?# Ñ œ !
(1b) )ÐC#  "Ñ  ?"  ?# Ÿ !
(2b) C# Ð)ÐC#  "Ñ  ?"  ?# Ñ œ !
(1c) "'  ?"  ?# Ÿ !
(2c) C$ Ð"'  ?"  ?# Ñ œ !
(3a) C"  C#  C$ Ÿ #
(4a) ?" ÐC"  C#  C$  #Ñ œ !
(3b) C"  C#  C$ Ÿ #
(4b) ?# ÐC"  C#  C$  #Ñ œ !
(5) C" !ß C# !ß C$ !
(6) ?" !ß ?# !
(c) If B œ Ð#ß "ß #Ñ, C œ Ð"ß !ß "Ñ. From (2a), ?"  ?# œ "#, which contradicts (2c), so
B œ Ð#ß "ß #Ñ is not optimal.
13.6-12.
(a) KKT conditions: (1a) '  #B"  ? Ÿ !
(2a) B" Ð'  #B"  ?Ñ œ !
(3) B"  B# Ÿ "
(4) ?ÐB"  B#  "Ñ œ !
(5) B" !ß B# !
(6) ? !
(1b) $  $B##  ? Ÿ !
(2b) B# Ð$  $B##  ?Ñ œ !
(b) For B œ Ð"Î#ß "Î#Ñ, (2a) gives ? œ &, which violates (2b), so this point is not
optimal.
(c) ÐB" ß B# ß ?Ñ œ Ð"ß !ß %Ñ satisfies all the conditions and since this is a convex
programming problem, Ð"ß !Ñ is optimal.
13.6-13.
(a) KKT conditions:
(1a) )  #B"  ? Ÿ !
(1b) #  $? Ÿ !
(1c) "  #? Ÿ !
(2a) B" Ð)  #B"  ?Ñ œ !
(2b) B# Ð#  $?Ñ œ ! (2c) B$ Ð"  #?Ñ œ !
(3) B"  $B#  #B$ Ÿ "#
(4) ?ÐB"  $B#  #B$  "#Ñ œ !
(5) B" !ß B# !ß B$ !
(6) ? !
For B œ Ð#ß #ß #Ñ, (2a) gives ? œ %, which violates (2b) and (2c), so it is not optimal.
(b) ÐB" ß B# ß B$ ß ?Ñ œ Ð""Î$ß #&Î*ß !ß #Î$Ñ satisfies all the conditions and since this is a
convex programming problem, Ð""Î$ß #&Î*ß !Ñ is optimal.
13-22
13.6-14.
KKT conditions:
(1a) #  #B" ? Ÿ !
(1b) $B##  %B# ? Ÿ !
(1c)  #B$  #B$ ? Ÿ !
(2a) B" Ð#  #B" ?Ñ œ !
(2b) B# Ð$B##  %B# ?Ñ œ ! (2c) B$ Ð  #B$  #B$ ?Ñ œ !
(3) B#"  #B##  B#$ %
(4) ?Ð%  B#"  #B##  B#$ Ñ œ !
(5) B" !ß B# !ß B$ !
(6) ? !
For B œ Ð"ß "ß "Ñ, (2a) gives ? œ ", which violates (2b), so it is not optimal.
13.6-15.
KKT conditions:
(1a) %B$"  #B" ? Ÿ !
(2a) B" Ð%B$"  #B" ?Ñ œ !
(3) B#"  B##  # Ÿ !
(4) ?ÐB#"  B##  #Ñ œ !
(5) B" !ß B# !
(6) ? !
(1b) %B#  #B# ? Ÿ !
(2b) B# Ð%B#  #B# ?Ñ œ !
For B œ Ð"ß "Ñ, (2a) gives ? œ #, and this satisfies all the conditions, so Ð"ß "Ñ is optimal.
13.6-16.
KKT conditions:
(1a) $#  %B$"  $?"  #?# Ÿ !
(2a) B" Ð$#  %B$"  $?"  #?# Ñ œ !
(1b) &!  #!B#  $B##  %B$#  ?"  &?# Ÿ !
(2b) B# Ð&!  #!B#  $B##  %B$#  ?"  &?# Ñ œ !
(3a) $B"  B# Ÿ ""
(4a) ?" Ð$B"  B#  ""Ñ œ !
(3b) #B"  &B# Ÿ "'
(4b) ?# Ð#B"  &B#  "'Ñ œ !
(5) B" !ß B# !
(6) ?" !ß ?# !
For B œ Ð#ß #Ñ, (4a) and (4b) give ?" œ ?# œ !, and this violates (2b), so Ð#ß #Ñ is not
optimal.
13.7-1.
(a)
` # 0 ÐBÑ
`B#"
0 ÐBÑ
0 ÐBÑ ` 0 ÐBÑ
` 0 ÐBÑ
œ %  !ß ` `B
œ )  !ß ` `B
  `B
 œ "'  !
#
#
`B#
" `B#
#
#
#
#
"
#
#
#
Ê 0 is strictly concave.
(b) BX UB œ %B#"  )B" B#  )B## œ %ÐB"  B# Ñ#  %B##  ! for all ÐB" ß B# Ñ Á Ð!ß !Ñ
Ê U is positive definite.
13-23
(c) KKT conditions:
(1a) "&  %B#  %B"  ? Ÿ !
(2a) B" Ð"&  %B#  %B"  ?Ñ œ !
(3) B"  #B# Ÿ $!
(4) ?ÐB"  #B#  $!Ñ œ !
(5) B" !ß B# !
(6) ? !
(1b) $!  %B"  )B#  #? Ÿ !
(2b) B# Ð$!  %B"  )B#  #?Ñ œ !
B œ Ð"#ß *Ñ with ? œ $ satisfies all these conditions.
13.7-2.
(a) KKT conditions: (1a) )  #B"  ? Ÿ !
(2a) B" Ð)  #B"  ?Ñ œ !
(3) B"  B# Ÿ #
(4) ?ÐB"  B#  #Ñ œ !
(5) B" !ß B# !
(6) ? !
(1b) %  #B#  ? Ÿ !
(2b) B# Ð%  #B#  ?Ñ œ !
B œ Ð#ß !Ñ with ? œ % satisfies all these conditions. Since this is a convex programming
problem, Ð#ß !Ñ is optimal.
(b) Objective function in vector notation:
maximize  )
Equivalent problem:
% 
B"
 " B
B#  # "
minimize
subject to
B# 
# !
B"

! #
B# 
D "  D#
#B"  ?  C"  D" œ )
#B#  ?  C#  D# œ %
B"  B#  @ œ #
B" !ß B# !
C" !ß C# !
? !ß @ !
D" !ß D# !
Complementarity constraint: B" C"  B# C#  ?@ œ !
13-24
(c)
Optimal Solution: ÐB" ß B# Ñ œ Ð#ß !Ñ with ? œ %
13-25
(d) Excel Solver Solution: ÐB" ß B# Ñ œ Ð#ß !Ñ
Solution
X1
2
X2
0
Sum
2 <=
Objective
12
2
13.7-3.
(a) Objective function in vector notation:
maximize  #!
5! 
B"
 " B
B#  # "
B# 
4! 2!
B"

2!
1!
B# 
Equivalent problem:
minimize
subject to
D "  D#
4!B"  2!B#  C"  C$  C%  D" œ #!
!B"  "!B#  C#  C$  %C%  D# œ &!
B"  B#  B$ œ '
B"  %B#  B% œ ")
B" !ß B# !ß B$ !ß B% !
C" !ß C# !ß C$ !ß C% !
D" !ß D# !
Enforced complementarity constraint: B" C"  B# C#  B$ C$  B% C% œ !
(b)
13-26
Optimal Solution: ÐB" ß B# Ñ œ Ð"Þ**$ß %Þ!$$Ñ with Ð?" ß ?# Ñ œ Ð"!ß "!Ñ
13.7-4.
(a) KKT conditions: (1a) #  #B"  ? Ÿ !
(2a) B" Ð#  #B"  ?Ñ œ !
(3) B"  B# Ÿ #
(4) ?ÐB"  B#  #Ñ œ !
(5) B" !ß B# !
(6) ? !
(1b) $  #B#  ? Ÿ !
(2b) B# Ð$  #B#  ?Ñ œ !
By plotting the points obtained, one observes that one optimal solution is on the
boundary , so B" Á !, B# Á ! and ? Á !. The point ÐB" ß B# Ñ œ Ð!Þ(&ß "Þ#&Ñ with ? œ !Þ&
satisfies all the conditions, so it is optimal.
13-27
(b)
minimize
D "  D#
subject to
#B"  ?  C"  D" œ #
#B#  ?  C#  D# œ $
B"  B#  @ œ #
B" !ß B# !
? !ß @ !
C" !ß C# !
D" !ß D# !
Enforced complementarity constraint: B" C"  B# C#  ?@ œ !
(c) Substitute ÐB" ß B# Ñ œ Ð!Þ(&ß "Þ#&Ñ and ? œ !Þ& in the constraints.
 C"  D" œ !
 C#  D# œ !
@ œ!
Enforced complementarity constraint: !Þ(&C"  "Þ#&C# œ !
Since C" ! and C# !, the unique solution of the complementarity constraint is C" œ
C# œ !, so D" œ D# œ !. Hence, ÐB" ß B# Ñ œ Ð!Þ(&ß "Þ#&Ñ is optimal.
(d)
13-28
Optimal Solution: ÐB" ß B# Ñ œ Ð!Þ(&ß "Þ#&Ñ with ? œ !Þ&
(e) Excel Solver Solution: ÐB" ß B# Ñ œ Ð!Þ(&ß "Þ#&Ñ
Solution
X1
0.75
X2
1.25
Sum
2 <=
Objective
3.125
2
13.7-5.
(a) KKT conditions: (1a) "#'  ")B"  ?"  $?$ Ÿ !
(2a) B" Ð"#'  ")B"  ?"  $?$ Ñ œ !
(1b) ")#  #'B#  #?#  #?$ Ÿ !
(2b) B# Ð")#  #'B#  #?#  #?$ Ñ œ !
(3a) B" Ÿ %
(4a) ?" ÐB"  %Ñ œ !
(3b) #B# Ÿ "#
(4b) ?# Ð#B#  "#Ñ œ !
(3c) $B"  #B# Ÿ ")
(4c) ?$ Ð$B"  #B#  ")Ñ œ !
(5) B"
!ß B#
!
(6) ?"
!ß ?#
!ß ?$
!
ÐB" ß B# Ñ œ Ð)Î$ß &Ñ with ? œ Ð!ß !ß #'Ñ satisfies these conditions, so it is optimal.
13-29
(b)
minimize
D "  D#
subject to
")B"  C"  C$  $C&  D" œ "#'
#'B#  C#  #C%  #C&  D# œ ")#
B"  B$ œ %
#B#  B% œ "#
$B"  #B#  B& œ ")
B" ß B# ß B$ ß B% ß B& !
C" ß C# ß C$ ß C% ß C& !
D" !ß D# !
Enforced complementarity constraint: B" C"  B# C#  B$ C$  B% C%  B& C& œ !
(c) Substitute ÐB" ß B# Ñ œ Ð)Î$ß &Ñ and ?$ œ C& œ #' in the constraints.
 C"  C$  D" œ !
 C#  #C%  D# œ !
B$ œ %Î$
B% œ #
B& œ !
Enforced complementarity constraint: Ð)Î$ÑC"  &C#  Ð%Î$ÑC$  #C% œ !
Since C3 ! for 3 œ "ß #ß á ß &, the complementarity constraint has the unique solutions
C" œ C# œ C$ œ C% œ !, so D" œ D# œ !. Hence, ÐB" ß B# Ñ œ Ð)Î$ß &Ñ is optimal.
13.7-6.
(a), (b), (c)
Price
Expected Return
Risk
Joint Risk
Stock 1
Stock 2
Stock 1
20
5
4
Stock 2
30
10
100
Stock 1
Stock 2
5
Stock 1
2.20
Stock 2
0.20
Price
Expected Return
Risk
Stock 1
20
5
4
Stock 2
30
10
100
Stock 1
Stock 2
5
Number of Blocks
<=
>=
50
13
Total
50
14
51.04
<=
>=
50
14
5
Number of Blocks
Joint Risk
Stock 1
Stock 2
Total
50
13
25.56
5
Stock 1
1.60
Stock 2
0.60
13-30
Price
Expected Return
Risk
Joint Risk
Stock 1
Stock 2
Stock 1
Stock 2
5
Stock 2
1.00
Price
Expected Return
Risk
Stock 1
20
5
4
Stock 2
30
10
100
Stock 1
Stock 2
5
5
&Þ!'
(Þ"%
"!Þ%%
"%Þ"#
.5
(Þ*%
'Þ)'
%Þ&'
"Þ))
Total
50
15
109.00
<=
>=
50
15
Total
50
16
199.44
<=
>=
50
16
5
Stock 1
1.00
Number of Blocks
.
"$
"%
"&
"'
Stock 2
30
10
100
Number of Blocks
Joint Risk
Stock 1
Stock 2
(d)
Stock 1
20
5
4
5
Stock 1
0.40
Stock 2
1.40
.  $5
 #Þ")
 (Þ%#
"'Þ$#
#'Þ$'
13.7-7.
(a) Let V" œ the production rate of product 1 per hour
V# œ the production rate of product 2 per hour
Maximize Profit = $200V"  $100V"#  $300V#  $100V##
subject to
V"  V# Ÿ # (maximum total production rate)
and V" !ß V# !Þ
(b), (c), (d)
Unit Profit = a(Rate) + b(Rate)2, where
Product 1
Product 2
a
$200
$300
b
-$100
-$100
Production Rate
Product 1
0.75
Product 2
1.25
13-31
Total
2
<=
2
Total Profit
$313
13.8-1.
(a)
(b) maximize $'!B""  $!B"#  #%!B#"  "#!B##  *!B#$  %&!B$"  $!!B$#  ")!B$$
subject to
B""  B"#  B#"  B##  B#$  B$"  B$#  B$$ Ÿ '!
$B""  $B"#  #B#"  #B##  #B#$ Ÿ #!!
B""  B"#  B$"  B$#  B$$ Ÿ (!
! Ÿ B"" Ÿ "&ß ! Ÿ B"#
! Ÿ B#" Ÿ #!ß ! Ÿ B## Ÿ #!ß ! Ÿ B#$
! Ÿ B$" Ÿ "!ß ! Ÿ B$# Ÿ &ß ! Ÿ B$$
where B" œ B""  B"# ß B# œ B#"  B##  B#$ ß B$ œ B$"  B$#  B$$ .
13-32
(c) Optimal solution using Solver:
Unit Profit
First Group
Second Group
Third Group
Resource 1
Resource 2
Resource 3
First Group
Second Group
Third Group
Total
Product 1
$360
$30
–
Product 2
$240
$120
$90
Product 3
$450
$300
$180
Resource Used per Unit Produced
1
1
1
3
2
0
1
0
2
Product 1
15
0
15
Units Produced
Product 2
Product 3
20
10
0
5
0
10
20
25
Total
Used
60
85
65
<=
<=
<=
Resource
Available
60
200
70
<=
<=
Product 1
15
30
Maximum
Product 2
20
20
Product 3
10
5
Total Profit
$18,000
Original variables: B" œ "&ß B# œ #!ß B$ œ #&
(d) The restriction on profit from products 1 and 2 can be modeled by introducing the
constraint: $'!B""  $!B"#  #%!B#"  "#!B##  *!B#$ 12,!!!.
13-33
(e) Optimal solution using Solver:
Unit Profit
First Group
Second Group
Third Group
Resource 1
Resource 2
Resource 3
First Group
Second Group
Third Group
Total
Product 1
$360
$30
–
Product 2
$240
$120
$90
Product 3
$450
$300
$180
Resource Used per Unit Produced
1
1
1
3
2
0
1
0
2
Product 1
15
0
15
Units Produced
Product 2
Product 3
20
10
15
0
0
0
35
10
Profit from Products 1&2
$12,000
>=
Total
Used
60
115
35
<=
<=
<=
Resource
Available
60
200
70
<=
<=
Product 1
15
30
Maximum
Product 2
20
20
Product 3
10
5
Total Profit
$16,500
$12,000
Original variables: B" œ "&ß B# œ $&ß B$ œ "!
13.8-2.
(a) KKT conditions: (1a) %  $B#"  ?"  &?# Ÿ !
(2a) B" Ð%  $B#"  ?"  &?# Ñ œ !
(1b) '  %B#  $?"  #?# Ÿ !
(2b) B# Ð'  %B#  $?"  #?# Ñ œ !
(3a) B"  $B# Ÿ )
(4a) ?" ÐB"  $B#  )Ñ œ !
(3b) &B"  #B# Ÿ "%
(4b) ?# Ð&B"  #B#  "%Ñ œ !
(5) B" !ß B# !
(6) ?" !ß ?# !
ÐB" ß B# Ñ œ Ð#Î&ß $Î#Ñ with ? œ Ð!ß !Ñ satisfies these conditions, so it is optimal with
^ œ (Þ&).
(b)
Profit data for doors when marketing costs are considered:
Production Rate Gross Profit Marketing Cost Net Profit
!
$!
$!
$!
"
$%!!
$"!!
$$!!
#
$)!!
$)!!
$!
$
$"#!!
$#(!!
$"*!!
$
H
$%H
$H
$%H  H$
Profit data for windows when marketing costs are considered:
Production Rate Gross Profit Marketing Cost Net Profit
!
$!
$!
$!
"
$'!!
$#!!
$%!!
#
$"#!!
$)!!
$%!!
$
$")!!
$")!!
$!
#
[
$'[
$#[
$'[  #[ #
13-34
Incremental Net Profit

$$!!
$$!!
$"*!!
Incremental Net Profit

$%!!
$!
$%!!
(c)
(d) Let B" œ B""  B"#  B"$ ß B# œ B#"  B##  B#$ ß 0" ÐB" Ñ œ %B"  B$" and 0# ÐB# Ñ œ
'B#  #B## .
0" Ð!Ñ œ !ß 0" Ð"Ñ œ $ß 0" Ð#Ñ œ !ß 0" Ð$Ñ œ "&
0# Ð!Ñ œ !ß 0# Ð"Ñ œ %ß 0# Ð#Ñ œ %ß 0# Ð$Ñ œ !
="" œ
$!
"!
œ $ß ="# œ
!$
#"
œ $ß ="$ œ
=#" œ
%!
"!
œ %ß ="# œ
%%
#"
œ !ß ="$ œ
"&!
$#
!%
$#
œ "&
œ %
Approximate linear programming model:
maximize
$B""  $B"#  "&B"$  %B#"  %B#$
subject to
B""  B"#  B"$  $B#"  $B##  $B#$ Ÿ )
&B""  &B"#  &B"$  #B#"  #B##  #B#$ Ÿ "%
! Ÿ B34 Ÿ " for 3 œ "ß # and 4 œ "ß #ß $
13-35
(e) Optimal solution using Solver:
Unit Profit ($hundred)
First
Second
Third
Doors
3
-3
-19
Windows
4
0
-4
Used per Unit Produced
Resource 1
1
3
Resource 2
5
2
First
Second
Third
Total
Units Produced
Power Saws Power Drills
1
1
0
0
0
0
1
1
Total
Used
4
7
<=
<=
<=
<=
<=
Resource
Available
8
14
Maximum
Doors
Windows
1
1
1
1
1
1
Total Profit
($hundred)
7
Original variables: B" œ "ß B# œ " (or B# œ #)
B"" œ ! Ê B"# œ ! Ê B"$ œ ! and B#" œ ! Ê B## œ ! Ê B#$ œ !
Hence, the special restriction for the model is satisfied. The approximate solutions Ð"ß "Ñ
and Ð"ß #Ñ are pretty close to the optimal solution Ð"Þ"&&ß "Þ&Ñ.
13.8-3.
(a)
(b) maximize 15!B""  5!B"#  10!B#"  75B##
subject to
B""  B"#  B#"  B## Ÿ 1!ß !!!
#B""  #B"#  B#"  B## Ÿ 15ß !!!
! Ÿ B"" Ÿ 3!!!ß ! Ÿ B"# Ÿ 2!!!
! Ÿ B#" Ÿ 5!!!ß ! Ÿ B## Ÿ 3!!!
13-36
(c) 3000 power saws and 7000 power drills should be produced in November.
Unit Profit
Power Saws Power Drills
Regular Time
$150
$100
Overtime
$50
$75
Used per Unit Produced
Power Supplies
1
1
Gear Assemblies
2
1
Units Produced
Power Saws Power Drills
Regular Time
3,000
5,000
Overtime
0
2,000
Total
3,000
7,000
Total
Used
10,000
13,000
<=
<=
<=
<=
Available
10,000
15,000
Maximum
Power Saws Power Drills
3,000
5,000
2,000
3,000
Total Profit
$1,100,000
13.8-4.
(a) Let B" œ B""  B"#  B"$ ß B# œ B#"  B##  B#$ ß 0" ÐB" Ñ œ $#B"  B%" and 0# ÐB# Ñ œ
&!B#  "!B##  B$#  B#% .
0" Ð!Ñ œ !ß 0" Ð"Ñ œ $"ß 0" Ð#Ñ œ %)ß 0" Ð$Ñ œ "&
0# Ð!Ñ œ !ß 0# Ð"Ñ œ %!ß 0# Ð#Ñ œ &#ß 0# Ð$Ñ œ '
="" œ $"ß ="# œ "(ß ="$ œ $$
=#" œ %!ß ="# œ "#ß ="$ œ %'
Approximate linear programming model:
maximize
$"B""  "(B"#  $$B"$  %!B#"  "#B##  %'B#$
subject to
$B""  $B"#  $B"$  B#"  B##  B#$ Ÿ ""
#B""  #B"#  #B"$  &B#"  &B##  &B#$ Ÿ "'
! Ÿ B34 Ÿ " for 3 œ "ß # and 4 œ "ß #ß $
(b) Optimal solution with the simplex method:
Original variables: B" œ #ß B# œ #
13-37
13.8-5.
 &B
Let 0" ÐB" Ñ œ  #  %B"
 "#  #B"
"
if ! Ÿ B" Ÿ #
%B
if # Ÿ B" Ÿ & and 0# ÐB# Ñ œ  #
*  B#
if & Ÿ B"
maximize
0" ÐB" Ñ  0# ÐB# Ñ
subject to
$B"  #B# Ÿ #&
#B"  B# Ÿ "!
B# Ÿ %
B" ß B# !
if ! Ÿ B# Ÿ $
.
if $ Ÿ B# Ÿ %
Possibly, the 03 ÐB3 Ñ's are piecewise-linear approximations of the original objective
function.
13.8-6.
(a) Assume that in the optimal solution of the linear program, there exists and B34 such
that B34  ?34 and B3Ð4"Ñ  !. Create a new solution with Bw34 œ minÖ?34 ß B34  B3Ð4"Ñ ×
and Bw3Ð4"Ñ œ maxÖ!ß B34  B3Ð4"Ñ  ?34 ×. This solution is feasible, since all the 13 's are
w
linear and B34  B3Ð4"Ñ œ Bw34  B3Ð4"Ñ
, but
w
=34 Bw34  =3Ð4"Ñ B3Ð4"Ñ
œ
=34 ÐB34  B3Ð4"Ñ Ñ
if B34  B3Ð4"Ñ Ÿ ?34
=34 ?34  =3Ð4"Ñ ÐB34  B3Ð4"Ñ  ?34 Ñ else.
Clearly, =34 ÐB34  B3Ð4"Ñ Ñ  =34 B34  =3Ð4"Ñ B3Ð4"Ñ , since =34  =3Ð4"Ñ .
Furthermore, Ð=34  =3Ð4"Ñ Ñ?34  Ð=34  =3Ð4"Ñ ÑB34 , since B34  ?34 .
Ê =34 ?34  =3Ð4"Ñ ÐB34  ?34 Ñ  =34 B34
Ê =34 ?34  =3Ð4"Ñ ÐB34  B3Ð4"Ñ  ?34 Ñ  =34 B34  =3Ð4"Ñ B3Ð4"Ñ
w
Ê =34 Bw34  =3Ð4"Ñ B3Ð4"Ñ
 =34 B34  =3Ð4"Ñ B3Ð4"Ñ
Thus, the original solution was not optimal.
(b) Make the same assumptions as in (a) and construct Bw from B in the same way. The
linear approximation of 13 is of the form â  +34 B34  +3Ð4"Ñ B3Ð4"Ñ  â Ÿ ,3 with
+34 Ÿ +3Ð4"Ñ , since 13 is convex. By the same analysis as the one in (a), it can be shown
that if the inequalities are reversed at appropriate places:
w
+34 Bw34  +3Ð4"Ñ B3Ð4"Ñ
 +34 B34  +3Ð4"Ñ B3Ð4"Ñ ,
w
so Bw is feasible. Furthermore, =34 Bw34  =3Ð4"Ñ B3Ð4"Ñ
 =34 B34  =3Ð4"Ñ B3Ð4"Ñ , so B was
not optimal.
13-38
13.8-7.
0" ÐB" Ñ œ 
0# ÐB# Ñ œ 
15B"
if ! Ÿ B" Ÿ 2!!!
25B"  2!ß !!! if 2!!! Ÿ B"
16B#
if ! Ÿ B# Ÿ 1!!!
24B#  8!!! if 1!!! Ÿ B#
D œ B"  B#
0" ÐB" Ñ  0# ÐB# Ñ Ÿ 6!ß !!!
! Ÿ B" Ÿ 3!!!
! Ÿ B# Ÿ 1&!!
maximize
subject to
S
(a) Let BV
3 and B3 denote the regular and overtime production at plant 3.
S
V
S
D œ BV
"  B"  B#  B#
V
S
V
15B"  25B"  16B#  24BS
# Ÿ 6!ß !!!
V
S
! Ÿ B" Ÿ 2!!!ß ! Ÿ B" Ÿ 1!!!
S
! Ÿ BV
# Ÿ 1!!!ß ! Ÿ B# Ÿ &!!
maximize
subject to
(b) Since overtime production is more expensive than regular time production, the
objective of maximizing the total production time will force the regular time to be used
first.
13.8-8.
(a) The objective function is linear, so concave.
` # 1" ÐBÑ ` # 1" ÐBÑ
`B#"
`B##
` # 1# ÐBÑ ` # 1# ÐBÑ
`B#"
`B##
ÐBÑ
  ``B1""`B
 œ % † !  !# œ !
#
#
#
ÐBÑ
  ``B1"#`B
 œ # † !  !# œ !
#
#
#
Ê 1" and 1# are convex.
(b) Let B" œ B""  B"#  B"$ . From the first constraint and B#
B" Ÿ "$Î# ¸ #Þ&&,
!,
so using an integer breakpoint requires $ linear pieces.
1"" ÐB" Ñ œ #B#" ß 1"# ÐB# Ñ œ B# ß 1#" ÐB" Ñ œ B#" ß 1## ÐB# Ñ œ B#
1"" Ð!Ñ œ !ß 1"" Ð"Ñ œ #ß 1"" Ð#Ñ œ )ß 1"" Ð$Ñ œ ")
1#" Ð!Ñ œ !ß 1#" Ð"Ñ œ "ß 1#" Ð#Ñ œ %ß 1#" Ð$Ñ œ *
=""ß" œ #ß =""ß# œ 'ß =""ß$ œ "!
=#"ß" œ "ß =#"ß# œ $ß =#"ß$ œ &
Approximate linear programming model:
maximize
&B""  &B"#  &B"$  B#
subject to
#B""  'B"#  "!B"$  B# Ÿ "$
B""  $B"#  &B"$  B# Ÿ *
! Ÿ B"" Ÿ "ß ! Ÿ B"# Ÿ "ß ! Ÿ B"$ ß ! Ÿ B#
We could have ! Ÿ B"$ Ÿ ", but the constraints will enforce the upper bound.
13-39
(c)
Original variables: B" œ "  "  ! œ #ß B# œ &
13.8-9.
(a) Let B" œ B""  B"#  B"$ and B# œ B#"  B##  B#$ .
0" ÐB" Ñ œ $#B"  B%" ß .
0# ÐB# Ñ œ %B#  B## ß .
#
#
0" ÐB" Ñ
.B#"
0# ÐB# Ñ
.B##
œ "#B#" Ÿ ! Ê 0" concave
œ #  ! Ê 0# concave
0" Ð!Ñ œ !ß 0" Ð"Ñ œ $"ß 0" Ð#Ñ œ %)ß 0" Ð$Ñ œ "&
0# Ð!Ñ œ !ß 0# Ð"Ñ œ $ß 0# Ð#Ñ œ %ß 0# Ð$Ñ œ $
="" œ $"ß ="# œ "&ß ="$ œ $$
=#" œ $ß =## œ "ß =#$ œ "
1"" ÐB" Ñ œ B#" ß .
#
1"" ÐB" Ñ
.B#"
œ #  ! Ê 1"" convex
1"# ÐB# Ñ œ B## ß .
#
1"# ÐB# Ñ
.B##
œ #  ! Ê 1"# convex
1"" Ð!Ñ œ !ß 1"" Ð"Ñ œ "ß 1"" Ð#Ñ œ %ß 1"" Ð$Ñ œ *
1#" Ð!Ñ œ !ß 1#" Ð"Ñ œ "ß 1#" Ð#Ñ œ %ß 1#" Ð$Ñ œ *
>""ß" œ "ß >""ß# œ $ß >""ß$ œ &
>#"ß" œ "ß >#"ß# œ $ß >#"ß$ œ &
Approximate linear programming model:
maximize
subject to
$"B""  "(B"#  $$B"$  $B#"  B##  B#$
B""  $B"#  &B"$  B#"  $B##  &B#$ Ÿ *
! Ÿ B"" Ÿ "ß ! Ÿ B"# Ÿ "ß ! Ÿ B"$ Ð Ÿ "Ñ
! Ÿ B#" Ÿ "ß ! Ÿ B## Ÿ "ß ! Ÿ B#$ Ð Ÿ "Ñ
13-40
(b) Solution with the simplex method:
Original variables: B" œ B# œ #
(c) KKT conditions: (1a) $#  %B$"  #B" ? Ÿ !
(2a) B" Ð$#  %B$"  #B" ?Ñ œ !
(3) B#"  B##  * Ÿ !
(4) ?ÐB#"  B##  *Ñ œ !
(5) B" !ß B# !
(6) ? !
(1b) %  #B#  #B# ? Ÿ !
(2b) B# Ð%  #B#  #B# ?Ñ œ !
For ÐB" ß B# Ñ œ Ð#ß #Ñ, from (4), ? œ !. This satisfies all the conditions, so is optimal to
the original problem.
13.8-10.
(a) 0 ÐBÑ œ 0" ÐB" Ñ  0# ÐB# Ñß 0" ÐB" Ñ œ $B#"  B$" ß 0# ÐB# Ñ œ &B##  B$#
.# 0" ÐB" Ñ
.B#"
œ '  'B"  ! if ! Ÿ B"  "
.# 0# ÐB# Ñ
.B##
œ "!  'B#  ! if ! Ÿ B#  &Î$
Neither 0" nor 0# is concave, so 0 is not concave. It is indeed enough to show one is not
concave.
(b) Let B" œ B""  B"#  B"$  B"% ß B# œ B#"  B## .
0" Ð!Ñ œ !ß 0" Ð"Ñ œ #ß 0" Ð#Ñ œ %ß 0" Ð$Ñ œ !ß 0" Ð%Ñ œ "'
0# Ð!Ñ œ !ß 0# Ð"Ñ œ %ß 0# Ð#Ñ œ "#
="" œ #ß ="# œ #ß ="$ œ %ß ="% œ "'
=#" œ %ß =## œ )
Special restrictions are needed:
(i)
B"# œ ! if B""  "
(ii)
B"$ œ ! if B"#  "
(iii)
B"% œ ! if B"$  "
(iv)
B## œ ! if B#"  ".
Since ="#  ="$  ="% , (ii) and (iii) are automatically satisfied upon optimization.
13-41
Approximate binary integer programming model:
maximize
#B""  #B"#  %B"$  "'B"%  %B#"  )B##
subject to
B""  B"#  B"$  B"%  #B#"  #B## Ÿ %
B""  B"# Ÿ !
 B#"  B## Ÿ !
B34 − Ö!ß "× for all 3ß 4
(c) Solution with BIP automatic routine:
B"" œ B"# œ B"$ œ B"% œ !ß B#" œ B## œ "ß D œ "#
Original variables: B" œ !ß B# œ #ß D œ "#
Alternate solution: B" œ #ß B# œ "ß D œ "#
13.9-1.
f0 ÐB" ß B# Ñ œ  B""" ß #B# 
Iteration 1: f0 Ð!ß !Ñ œ Ð"ß !Ñ
maximize
B"
subject to
B"  #B# Ÿ $
B" ß B# !
Ê B" œ $ß B# œ ! Ê BÐ"Ñ œ Ð!ß !Ñ  >Ð$ß !Ñ
>‡ œ " (0 ÐBÑ increases with >) Ê BÐ"Ñ œ Ð$ß !Ñ [solution found in Problem 13.6-5]
Iteration 2: f0 Ð$ß !Ñ œ Ð"Î%ß !Ñ
maximize
subject to
!Þ#&B"
B"  #B# Ÿ $
B" ß B# !
Ê B" œ $ß B# œ ! Ê BÐ"Ñ œ Ð$ß !Ñ  >Ð!ß !Ñ
Hence B œ Ð$ß !Ñ is optimal.
13.9-2.
f0 ÐB" ß B# Ñ œ Ð#B"  'ß $B##  $Ñ
B"  B# Ÿ "ß B" ß B# ! Ê B" ß B# Ÿ " Ê #B"  ' Ÿ %  $ Ÿ $B##  $
Resulting LP: maximize
subject to
-" B"  -# B#
B"  B# Ÿ "
B" ß B# !
where -"  -# , so Ð"ß !Ñ is always optimal.
Ð!Ñ
Ð!Ñ
Ð!Ñ
Ð!Ñ
Ê BÐ"Ñ œ ÐB" ß B# Ñ  >Ð"  B" ß B# Ñ
At >‡ œ ", BÐ"Ñ œ Ð"ß !Ñ is optimal.
13-42
13.9-3.
13.9-4.
maximize
"&B"  $!B#  %B" B#  #B#"  %B##
subject to
B"  #B# Ÿ $!
B" ß B# !
13.9-5.
(a)
13-43
(b)
13.9-6.
13.9-7.
13.9-8.
(a)
13-44
(b) KKT conditions: (1a) $  $B#"  ? Ÿ !
(2a) B" Ð$  $B#"  ?Ñ œ !
(3) B"  B# Ÿ "
(4) ?ÐB"  B#  "Ñ œ !
(5) B" !ß B# !
(6) ? !
(1b) %  #B#  ? Ÿ !
(2b) B# Ð%  #B#  ?Ñ œ !
ÐB" ß B# Ñ œ Ð"Î$ß #Î$Ñ with ? œ )Î$ satisfies these conditions, so the estimated solution
in part (a) is optimal.
13.9-9.
(a)
(b)
(c) KKT conditions: (1a) %  %B$"  %? Ÿ !
(2a) B" Ð%  %B$"  %?Ñ œ !
(3) %B"  #B# Ÿ &
(4) ?Ð%B"  #B#  &Ñ œ !
(5) B" !ß B# !
(6) ? !
(1b) #  #B#  #? Ÿ !
(2b) B# Ð#  #B#  #?Ñ œ !
ÐB" ß B# Ñ œ Ð!Þ)*$%ß !Þ("$"Ñ with ? œ !Þ&($( satisfies these conditions, so is optimal.
13-45
13.9-10.
(a) T ÐBà <Ñ œ $B"  %B#  B$"  B##  < "B"" B# 
(b)
fT ÐBà <Ñ œ
Ê fT Ð "%
 "%
"
%
"
 $  $B#"  < Ð"B" B# Ñ# 
 %  #B#  < Ð"B"B Ñ# 
"
#
"
%
à "Ñ œ 
  >fT Ð "%
"
%
"
B"

"
B# 
"

B#" 
" 

B##
"% "$
"'
"& "# 
à "Ñ œ  "%  "% "$
"' >
"
%
>‡ œ !Þ!!''!' Ê Bw œ  !Þ$%(* !Þ$&#% 
 "& "# > 
(c)
(d) True Solution: Ð"Î$ß #Î$Ñ
Percentage error in B" :
l"Î$!Þ$$"l
"Î$
œ !Þ(!%
Percentage error in B# :
l#Î$!Þ''$l
#Î$
œ !Þ&&%
Percentage error in 0 ÐBÑ:
l)'Î#($Þ"'*l
)'Î#(
œ !Þ&"%
13.9-11.
(a) T ÐBà <Ñ œ %B"  B%"  #B#  B##  < &%B"" #B# 
(b) fT ÐBà <Ñ œ
Ê fT Ð "#
 "#
"
#
"
#
%
 %  %B$"  < Ð&%B" #B# Ñ# 
 #  #B#  < Ð&%B##B Ñ# 
"
#
à "Ñ œ 
  >fT Ð "#
"
#
' "#
% "# 
à "Ñ œ  "#  ' "# >
>‡ œ !Þ!$"'( Ê Bw œ  !Þ(!&)
!Þ'%#& 
"
%
"

B#" 
"
B"
" 

B##
 % "# > 
13-46

"
B# 
(c)
13.9-12.
(a) T ÐBà <Ñ œ  B%"  #B#"  #B" B#  %B##  < #B" B" # "! 
(b) fT ÐBà <Ñ œ
Ê fT Ð &
&
#
  %B$"  %B"  #B#  < Ð#B" B# "!Ñ# 

 #B"  )B#  < Ð#B" B"# "!Ñ# 
& à "!!Ñ œ 
&   >fT Ð &
&"%
$% 
"
B" #B# "!

"
ÐB" #B# "!Ñ#
#
ÐB" #B# "!Ñ#


"

B##
& à "Ñ œ  &  &"%> &  $%> 
>‡ œ !Þ!!$&#* Ê Bw œ  $Þ")'# %Þ))!# 
(c)
minimize 0 ÐBÑ Ä maximize 0 ÐBÑ
1ÐBÑ
, Ä 1ÐBÑ Ÿ ,
13.9-13.
(a) KKT conditions: (1a) B#  %?B" Ÿ !
(2a) B" ÐB#  %?B" Ñ œ !
(3) B#"  B# Ÿ $
(4) ?ÐB#"  B#  $Ñ œ !
(5) B" !ß B# !
(6) ? !
(1b) B"  ? Ÿ !
(2b) B# ÐB"  ?Ñ œ !
ÐB" ß B# Ñ œ Ð"ß #Ñ with ? œ " satisfies these conditions.
13-47
"
B"

"
B# 
"

B#" 

(b)
13.9-14.
(a) T ÐBà <Ñ œ  #B"  ÐB#  $Ñ#  < B""$ 
(b) fT ÐBà <Ñ œ

"
 #  < ÐB" $Ñ
#
  #B#  ' 
"
< ÐB# $Ñ
#
$
Ê B" œ <Î#  $ß B# œ 
<Î#  $
<
"
"!#
"!%
"!'
B"
$Þ(!("
$Þ!(!(
$Þ!!("
$Þ!!!(

"
B# $ 
!
œ 
!
B#
$Þ(*$(
$Þ"("!
$Þ!$')
$Þ!!(*
Note that ÐB" ß B# Ñ Ä Ð$ß $Ñ as < Ä !, so Ð$ß $Ñ is optimal.
(c)
13.9-15.
T ÐBà <Ñ œ B#"  B##  B"  B#  B" B#  <ÎB#
13-48
13.9-16.
T ÐBà <Ñ œ #B"  $B#  B#"  B##  < #B"" B# 
"
B"

"
B# 
13.9-17.
"
T ÐBà <Ñ œ "#'B"  *B#"  ")#B#  "$B##  < %B

"
13.9-18.
(a) T ÐBà <Ñ œ B$"  %B##  "'B$  < B""" 
"
B# "

"
"##B#
"
B$ " 


"
")$B" #B#

Ð&B" B# B$ Ñ#
<
(b)
(c) Standard Excel Solver
Solution
X1
2.1943
>=
1
X2
1.8057
>=
1
X3
1
>=
1
Sum
5
13-49
=
Constraint
5
Minimize
39.6077
"
B"

"
B# 
(d) Evolutionary Solver
Solution
X1
2.1931
>=
1
X2
1.7964
>=
1
X3
1.0005
>=
1
Sum
4.990039
(e) LINGO
13-50
=
Constraint
5
Minimize
39.4648
13.10-1.
(a) Solving for the roots of B#  B  &!! œ !, one observes that B is feasible in the range

!ß " # #!!"  œ Ò!ß #"Þ)''Ó.
0 w ÐBÑ œ "!!!  )!!B  "#!B#  %B$
0 ww ÐBÑ œ  )!!  #%!B  "#B#
0 www ÐBÑ œ #%!  #%B
A rough sketch of 0 ÐBÑ:
The points that are marked as X correspond to a local minimum or maximum.
(b)
There is a local maximum near "Þ'!"' and a global maximum near #"Þ!&".
13-51
(c)
Local maximum: B œ "Þ'#%$
Local maximum: B œ #"Þ!("'
13-52
(d)
The first four iterations with initial trial solution B œ $, return B œ "Þ'#& with 0 ÐBÑ œ
($$Þ% as maximum. The next four iterations with initial trial solution B œ "&, return
B œ #"Þ!( with 0 ÐBÑ œ #!&'# as maximum. The global maximum is B œ #"Þ!(.
(e) B œ #"Þ!("'
X
X^2+X
Solution 21.0716 465.0834
<=
25
<=
Constraint
500
1000X ‐ 400X^2 + 40X^3 ‐X^4
Maximize 20561.7289
(f) B œ #"Þ!("'
X
X^2+X
Solution 21.0716 465.0838
<=
25
<=
Constraint
500
1000X ‐ 400X^2 + 40X^3 ‐X^4
Maximize 20561.7289
13-53
(g)
13.10-2.
(a) T ÐBà <Ñ œ $B" B#  #B#"  B##  < %B#"#B# 
"
#
"
B# #B"

"
B"

"
B# 

Ð#B" B## B#" B# Ñ#
<
(b)
(c) Evolutionary Solver
Solution
X1
0.8385
<=
2
X2
1.1825
<=
2
X1^2 + 2X^2 =
2X1 ‐ X2 =
X1*X2^2 + (X1^2)*X2 =
3.4998
0.4945
2.0039
<=
<=
=
Constraint
4
3
2
Maximize 3X1*X2 ‐ 2*X1^2 ‐ X2^2 0.1701
13-54
(d) Use global optimizer feature of LINGO.
13.10-3.
"
(a) T ÐBà <Ñ œ sin $B"  cos $B#  sin ÐB"  B# Ñ  < "B#"!B

#
"
"
"!!"!B" B##

"
B"

"
B# 
(b) SUMT can be used to obtain the global minimum if it is run with "enough" different
starting points. If a lattice of points over the feasible region is chosen so that the adjacent
points do not differ by more than #1Î$, then this set of points works for 0 ÐBÑ. Since sin
and cos have period #1, choosing lattice points with grid size not exceeding #1Î$ ensures
that the arguments of the sin and cos terms in 0 do not differ by more than #1 between
adjacent lattice points. Since the second constraint ensures B" Ÿ "! and B# Ÿ "!, at most
Ò"!ÎÐ#1Î$ÑÓ# ¸ #$ starting points are required if chosen correctly.
13-55
(c)
13.10-4.
(a)
x=
Profit =
=
0
<=
0.405
<=
5
x5 - 13x4 + 59x3 - 107x2 + 61x
10.735
(b)
x=
Profit =
=
0
<=
0.405
<=
5
x5 - 13x4 + 59x3 - 107x2 + 61x
10.735
13.10-5.
(a)
x=
Profit =
=
0
<=
3.184
<=
5
100x6 - 1,359x5 + 6,836x4 - 15,670x3 + 15,870x2 - 5,095x
906.902
13-56
(b)
0
<=
3.184
<=
5
x=
Profit =
=
100x6 - 1,359x5 + 6,836x4 - 15,670x3 + 15,870x2 - 5,095x
906.902
13.10-6.
City
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
Total
Democrat
152
81
75
34
62
38
48
74
98
66
83
86
72
28
112
45
93
72
1,319
Republican Total
62
214
59
140
83
158
52
86
87
149
87
125
69
117
49
123
62
160
72
138
75
158
82
168
83
155
53
81
98
210
82
127
68
161
98
170
1,321
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
District
3
4
8
6
5
5
7
1
7
9
6
9
10
2
2
1
4
10
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
<=
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
Min District Population
Max District Population
Number of Districts
District
1
2
3
4
5
6
7
8
9
10
150
350
10
Democrat Republican
Total
119
131
250
140
151
291
152
62
214
174
127
301
100
174
274
117
127
244
146
131
277
75
83
158
152
154
306
144
181
325
Total Republican Districts
Winner
Republican
Republican
Democrat
Democrat
Republican
Republican
Democrat
Republican
Republican
Republican
7
13.10-7.
(a)
Profit Per Batch ($000)
Doors
3
Windows
5
Hours Used Per Batch Produced
Plant 1
1
0
Plant 2
0
2
Plant 3
3
2
Batches Produced
Doors
2
Windows
6
Profit Per Batch ($000)
Doors
3
Windows
5
Hours
Used
2
12
18
<=
<=
<=
Hours
Available
4
12
18
Total Profit ($000)
36
(b)
Hours Used Per Batch Produced
Plant 1
1
0
Plant 2
0
2
Plant 3
3
2
Batches Produced
Doors
1.999987849
<=
4
Windows
6
,=
6
Hours
Used
1.99999 <=
12
<=
18
<=
Hours
Available
4
12
18
Total Profit ($000)
35.99996355
(c) The Standard Solver gives a better solution and finds it much more quickly. It is much
better suited to linear programs than the Evolutionary Solver.
13-57
13.10-8
Answers will vary.
13.11-1.
(a) Yes, this is a convex programming problem.
0 ÐBÑ œ 0" ÐB" Ñ  0# ÐB# Ñß 0" ÐB" Ñ œ %B"  B#" ß 0# ÐB# Ñ œ "!B#  B##
.# 0" ÐB" Ñ
.B#"
œ
.# 0# ÐB# Ñ
.B##
œ #  ! Ê 0 is concave.
1ÐBÑ œ 1" ÐB" Ñ  1# ÐB# Ñß 1" ÐB" Ñ œ B#" ß 1# ÐB# Ñ œ %B##
.# 1" ÐB" Ñ
.B#"
œ #  !ß .
#
1# ÐB# Ñ
.B##
œ )  ! Ê 1 is convex.
(b) No, this is not a quadratic programming problem because the constraints are
nonlinear.
(c) No, the Frank-Wolfe algorithm in Section 13.9 requires linear constraints, so it cannot
be applied to this problem.
(d) KKT conditions:
(1a) %  #B"  #B" ? Ÿ !
(2a) B" Ð%  #B"  #B" ?Ñ œ !
(3) B#"  %B##  "' Ÿ !
(4) ?ÐB#"  %B##  "'Ñ œ !
(5) B" !ß B# !
(6) ? !
(1b) "!  #B#  )B# ? Ÿ !
(2b) B# Ð"!  #B#  )B# ?Ñ œ !
Let B" œ B# œ ". Then from (2a), ? œ " and this violates (4), so it cannot be optimal.
(e) Let B" œ B""  B"#  B"$  B"% and B# œ B#"  B## .
0" ÐB" Ñ œ %B"  B#" ß 0# ÐB# Ñ œ "!B#  B##
0" Ð!Ñ œ !ß 0" Ð"Ñ œ $ß 0" Ð#Ñ œ %ß 0" Ð$Ñ œ $ß 0" Ð%Ñ œ !
0# Ð!Ñ œ !ß 0# Ð"Ñ œ *ß 0# Ð#Ñ œ "'
="" œ $ß ="# œ "ß ="$ œ "ß ="% œ $
=#" œ *ß =## œ (
1" ÐB" Ñ œ B#" ß 1# ÐB# Ñ œ %B##
1" Ð!Ñ œ !ß 1" Ð"Ñ œ "ß 1" Ð#Ñ œ %ß 1" Ð$Ñ œ *ß 1" Ð%Ñ œ "'
1# Ð!Ñ œ !ß 1# Ð"Ñ œ %ß 1# Ð#Ñ œ "'
>"" œ "ß >"# œ $ß >"$ œ &ß >"% œ (
>#" œ %ß >## œ "#
13-58
Approximate linear programming model:
maximize
$B""  B"#  B"$  $B"%  *B#"  (B##
subject to
B""  $B"#  &B"$  (B"%  %B#"  "#B## Ÿ "'
&B""  &B"#  &B"$  #B#"  #B##  #B#$ Ÿ "%
! Ÿ B34 Ÿ " for all 3ß 4
(f) Solution with the simplex method:
Original variables: B" œ "ß B# œ "Þ*"''(
(g) T ÐBà <Ñ œ %B"  B#"  "!B#  B##  < "'B"# %B# 
"
#
"
B"

"
B# 
(h)
(i) Standard Solver
Solution
X1
1.4104
<=
2
X2
1.8715
<=
2
X1^2 + 4X^2 =
16.0000
<=
Constraint
16
Maximize 4X1 ‐ X1^2 +10X2 ‐ X2^2 = 18.8652
(j) Evolutionary Solver
Solution
X1
1.4143
<=
2
X2
1.8708
<=
2
X1^2 + 4X^2 =
15.9999
<=
Constraint
16
Maximize 4X1 ‐ X1^2 +10X2 ‐ X2^2 = 18.8651
13-59
(k) LINGO Solver
13-60
CASES
Case 13.1 Savvy Stock Selection
(a) If Lydia wants to ignore the risk of her investment she should invest all her money
into the stock that promises the highest expected return. According to the predictions of
the investment advisors, the expected returns equal 20% for BB, 42% for LOP, 100% for
ILI, 50% for HEAL, 46% for QUI, and 30% for AUA. Therefore, she should invest 100%
of her money into ILI. The risk (variance) of this portfolio equals 0.333.
(b) Lydia should invest 40% of her money into the stock with the highest expected return,
40% into the stock with the second highest expected return, and 20% into the stock with
the third highest expected return. This intuitive solution can be found also by solving the
linear programming problem to
maximize
subject to
Expected Return
MaxExpectedReturn œ SUMPRODUCT(Portfolio, StockExpectedReturn)
Total œ OneHundredPercent
Portfolio Ÿ MaxInSingleStock.
BB
20%
Covariance Matrix
BB
(Variance on Diagonal)
BB 0.032
LOP 0.005
ILI 0.030
HEAL -0.031
QUI -0.027
AUA 0.010
Portfolio
Max in Single Stock
BB
0%
<=
40%
LOP
42%
ILI
100%
HEAL
50%
QUI
46%
AUA
30%
LOP
0.005
0.1
0.085
-0.07
-0.05
0.020
ILI
0.030
0.085
0.333
-0.11
-0.02
0.042
HEAL
-0.031
-0.07
-0.11
0.125
0.05
-0.060
QUI
-0.027
-0.05
-0.02
0.05
0.065
-0.020
AUA
0.010
0.020
0.042
-0.060
-0.020
0.08
LOP
0%
<=
40%
ILI
40%
<=
40%
HEAL
40%
<=
40%
QUI
20%
<=
40%
AUA
0%
<=
40%
Total
100% =
100%
Portfolio
Expected Return = 69.2%
Risk (Variance) = 0.04548
The total expected return of her new portfolio is 69.2% with a total variance of 0.04548.
(c) The risk of Lydia's portfolio is a quadratic function of her decision variables. We
apply quadratic programming to her decision problem.
(d) The expected return of Lydia's portfolio is no longer the objective function. It now
becomes part of a constraint:
PortfolioExpectedReturn(C21)
35%(MinimumExpectedReturn).
The objective is now to minimize the risk.
13-61
LOP
42%
ILI
100%
HEAL
50%
QUI
46%
AUA
30%
Covariance Matrix
BB
(Variance on Diagonal)
BB 0.032
LOP 0.005
ILI 0.030
HEAL -0.031
QUI -0.027
AUA 0.010
LOP
0.005
0.1
0.085
-0.07
-0.05
0.020
ILI
0.030
0.085
0.333
-0.11
-0.02
0.042
HEAL
-0.031
-0.07
-0.11
0.125
0.05
-0.060
QUI
-0.027
-0.05
-0.02
0.05
0.065
-0.020
AUA
0.010
0.020
0.042
-0.060
-0.020
0.08
BB
31.8%
Portfolio
<=
Max in Single Stock 40%
LOP
19.9%
<=
40%
ILI
0.0%
<=
40%
HEAL
16.8%
<=
40%
QUI
20.9%
<=
40%
AUA Total
10.6% 100% =
<=
40%
>=
Minimum
Expected
Return
35%
Expected Return
BB
20%
Portfolio
Expected Return = 35.9%
100%
Risk (Variance) = 0.00136
Lydia's optimal portfolio consists of 31.8% BB, 19.9% LOP, 16.8% HEAL, 20.9% QUI,
and 10.6% AUA. Her expected return equals 35.9% with a risk of 0.00136.
(e) Since the return constraint is not binding in the solution of part (d), decreasing the
right-hand-side will not affect the optimal solution. The minimum risk for a minimum
expected return of 25% is the same as the minimum risk for a minimum expected return
of 35%, which is 0.00136. However, for a minimum expected return of 40%, a new
portfolio is obtained.
LOP
42%
ILI
100%
HEAL
50%
QUI
46%
AUA
30%
Covariance Matrix
BB
(Variance on Diagonal)
BB 0.032
LOP 0.005
ILI 0.030
HEAL -0.031
QUI -0.027
AUA 0.010
LOP
0.005
0.1
0.085
-0.07
-0.05
0.020
ILI
0.030
0.085
0.333
-0.11
-0.02
0.042
HEAL
-0.031
-0.07
-0.11
0.125
0.05
-0.060
QUI
-0.027
-0.05
-0.02
0.05
0.065
-0.020
AUA
0.010
0.020
0.042
-0.060
-0.020
0.08
BB
Portfolio 22.9%
<=
Max in Single Stock 40%
LOP
21.0%
<=
40%
ILI
3.4%
<=
40%
HEAL
22.0%
<=
40%
QUI
18.8%
<=
40%
AUA Total
11.9% 100% =
<=
40%
>=
Minimum
Expected
Return
40%
Expected Return
BB
20%
Portfolio
Expected Return = 40.0%
Risk (Variance) = 0.00233
13-62
100%
Lydia's new optimal portfolio consists of 22.9% BB, 21% LOP, 3.4% ILI, 22% HEAL,
18.8% QUI, and 11.9% AUA. Her expected return equals 40% with a risk of 0.00233.
(f) Lydia's approach is very risky. She puts a lot of confidence in the advice of the two
investment experts. She cannot expect to find an optimal investment strategy with her
model if the estimates she uses for the input parameters are not accurate.
Case 13.2 International Investments
(a) When Charles sells a portion of his B-Bonds in a given year, the first DM 6100 of
interest are tax-free, but the interest earnings exceeding DM 6100 are levied a 30% tax.
Therefore, Charles encounters decreasing marginal returns and we can use separable
programming to solve this problem. Let NoTax5 and Tax5 be the base amount of BBonds Charles sells in the fifth year that yield untaxed interest and taxed interest
respectively. The variables NoTax6, Tax6, NoTax7, and Tax7 are defined in the same
way. The sum of the six variables must equal the total of DM 30,000 that Charles
invested at the beginning of the first year. When Charles sells B-Bonds with the base
amount NoTax5, he earns 50.01% of this amount as interest. In order for him not to pay
any taxes on this amount, the interest must not exceed DM 6100. This is included in the
model as a constraint. Any additional base amount of B-Bonds sold in year 5 yields
Charles only 0.7 ‚ 0.5001 œ 0.35007. A similar reasoning applies to other years. The
objective is to maximize Charles' interest income.
Net Interest
Tax Free
Taxed
Year 5
0.5001
0.3501
Year 6
0.6351
0.4446
Year 7
0.7823
0.5476
Bonds Sold (DM at Base Value)
Year 5
Year 6
Tax Free
0
9,605
Taxed
0
0
Year 7
7,798
12,598
Total Sold (DM)
Interest Earned
Tax Free
Taxed
Year 5
0
0
Year 6
6,100
0
30,000
Year 7
6,100
6,899
Total Interest (DM)
Tax Rate
30%
Total
Investment
=
30,000
<=
Maximum
6,100
19,099
(b) The optimal investment strategy for Charles is to sell a base amount of DM 9604.79
at the end of year 6 and the remaining DM 20395.21 at the end of year 7. His total aftertax interest income equals DM 19098.62.
(c) When Charles sells all B-Bonds in year 7, he must pay 30% of tax on the amount of
interest income exceeding DM 6100. This amount is earned interest not only from the last
year, but it also includes interest from all the previous years. Hence, Charles does not pay
30% tax on the 9% interest he earned last year, but he effectively pays tax on the total
interest of all the years. This tax payment decreases his after-tax interest so much that it
pays for him to sell some of his bonds in year 6 in order to take advantage of the yearly
tax-free income of DM 6100. Comparing the total amount of interest Charles earns if he
13-63
sells tax-free after year 6 and taxed after year 7, we see that in the former case his total
interest equals 63.51% while in the latter case it is only 54.761%. Therefore, it is better to
sell some bonds at the end of year 6 rather than to keep them until the end of the last
year.
(d) The following observation greatly simplifies the analysis of this problem: The interest
rate on the CD is much lower than the yearly interest rates on the B-Bonds. Therefore, it
can never be optimal for Charles to sell B-Bonds in year 5 in order to buy a CD for year 6
if he does not take advantage of the maximal tax-free amount of selling B-Bonds in year
6. In other words, Charles will only buy a CD for year 6 if he already plans to sell BBonds in year 6 to obtain at least the maximal tax-free amount of interest. The same
argument applies to year 7. Consequently, Charles will never earn untaxed interest on a
CD. Therefore, his yearly interest on the CD will always be 0.7 ‚ 0.04 œ 0.028 œ 2.8%.
To formulate the problem in Excel, let CD6 and CD7 be the amount invested in a CD in
year 6 and 7 respectively. The amount of money Charles can invest in a CD in year 6
equals the base amount of B-Bonds sold in year 5 plus the total after-tax interest earned
on the base amount. This gives the constraint CD6 œ 1.5001*NoTax5  1.35007*Tax5.
Similarly, for year 7, CD7 œ 1.6351*NoTax6  1.44457*Tax6  1.028*CD6.
Net Interest
Tax Free
Taxed
CD Tax Free
CD Taxed
Year 5
0.5001
0.3501
0.0400
0.0280
Bonds Sold (DM at Base Value)
Year 5
Tax Free
12,198
Taxed
0
Year 6
0.6351
0.4446
0.0400
0.0280
Year 7
0.7823
0.5476
Year 6
8,452
0
Year 7
6,118
3,232
Total Sold (DM)
30,000
CD's Purchased (DM)
Year 5
Tax Free
18,298
Taxed
0
Total
18,298
<=
Available Cash
18,298
Year 6
32,850
0
32,850
<=
32,850
Interest Earned
Tax Free
Taxed
Year 6
6,100
0
Year 5
6,100
0
Total Interest (DM)
Year 7
6,100
1,770
Tax Rate
30%
Total
Investment
=
30,000
<=
Maximum
6,100
20,070
Charles should sell the maximal base amount of B-Bonds in year 5 that yields tax-free
interest and then invest this money (base amount & interest) into a one-year CD for year
6. In year 6, he should sell again the maximal base amount of B-Bonds that yields taxfree interest and then invest this money (base amount & interest) and the money from his
CD into a one-year CD for year 7. In year 7, he should sell the remainder of the base
amount of B-Bonds. He again takes advantage of the maximum tax-free amount, but he
also sells a base amount of DM 400.13 for which he must pay taxes on the interest
earnings.
13-64
(e) The right-hand-side of the selling constraint should be changed.
Net Interest
Tax Free
Taxed
CD Tax Free
CD Taxed
Year 5
0.5001
0.3501
0.0400
0.0280
Year 6
0.6351
0.4446
0.0400
0.0280
Bonds Sold (DM at Base Value)
Year 5
Year 6
Tax Free
12,198
8,452
Taxed
0
0
Total Sold (DM)
CD's Purchased (DM)
Year 5
Tax Free
18,298
Taxed
0
Total
18,298
<=
Available Cash
18,298
Year 6
0
32,850
32,850
<=
32,850
Interest Earned
Tax Free
Taxed
Year 6
6,100
0
Year 5
6,100
0
Total Interest (DM)
Year 7
0.7823
0.5476
Year 7
7,798
21,553
50,000
Year 7
6,100
12,722
Tax Rate
30%
Total
Investment
=
50,000
<=
Maximum
6,100
31,022
The optimal investment strategy is similar to the previous one except that Charles must
now pay taxes on the interest earned from selling a base amount of DM 20400.13 in year
7.
(f) The maximum tax-free investment is changed to 12,200.
Net Interest
Tax Free
Taxed
CD Tax Free
CD Taxed
Year 5
0.5001
0.3501
0.0400
0.0280
Year 6
0.6351
0.4446
0.0400
0.0280
Bonds Sold (DM at Base Value)
Year 5
Year 6
Tax Free
0
15,719
Taxed
0
0
Total Sold (DM)
CD's Purchased (DM)
Year 5
Tax Free
0
Taxed
0
Total
0
<=
Available Cash
0
Year 6
25,702
0
25,702
<=
25,702
Interest Earned
Tax Free
Taxed
Year 6
9,983
0
Year 5
0
0
Total Interest (DM)
Year 7
0.7823
0.5476
Year 7
14,281
0
30,000
Year 7
12,200
0
Tax Rate
30%
Total
Investment
=
30,000
<=
Maximum
12,200
22,183
By getting married in year 5, Charles can increase his interest income by
22183  19998 œ DM 2185.
13-65
(g) Instead of maximizing his interest income, Charles now wants to maximize the
expected dollar amount he will have at the end of year 7. He considers exchanging marks
for dollars either at the end of year 5 or 7. Let CD-US be the amount of money in dollars
that Charles invests in a two-year CD at the end of year 5 and US be the amount of
money in dollars that Charles converts at the end of year 7. The total amount of money in
dollars Charles has at the end of year 7 equals (1.036)2 *CD-US  US; this is the new
objective function. At the end of year 5, $1 is assumed to be equal to DM 1.50, so
Charles can exchange marks for dollars at this rate in year 5. This is included as a
constraint. Similarly, we include a constraint for the currency conversion at the end of the
last year.
Net Interest
Tax Free
Taxed
CD Tax Free
CD Taxed
Municipal Bond
Conversion (DM per $)
Year 5
0.5001
0.3501
Year 6
0.6351
0.4446
0.0400
0.0280
0.0360
1.5
Year 7
0.7823
0.5476
0.0400
0.0280
0.0360
Tax Rate
30%
1.8
Bonds Sold (DM at Base Value)
Tax Free
Taxed
Year 5
0
0
Year 6
15,719
0
Total Sold (DM)
Year 7
14,281
0
30,000
Total
Investment
=
30,000
CD's Purchased (DM)
Tax Free
Taxed
Year 5
0
0
Year 6
25,702
0
American Municipal Bonds Purchased ($)
Year 5
Tax Free
0
Cost of Municipal Bond (DM)
0
Total Purchases (DM)
0
<=
0
25,702
<=
25,702
Year 5
0
0
Year 6
9,983
0
Available Cash (DM)
Interest Earned (DM)
Tax Free
Taxed
Total Interest (DM)
Year 7
12,200
0
<=
22,183
Ending Cash ($)
Year 5 Municipal Bond
$0
Year 5 Municipal Bond Interest
$0
German Bonds (Year 6 and 7) $16,667
Interest on German Bonds $12,324
Total $28,991
13-66
Maximum
12,200
Case 13.3 Promoting a Breakfast Cereal, Revisited
(a)
TV Spots
1
2
3
4
5
Sales
(millions)
1
1.75
2.45
2.8
3
Magazine Ads
5
10
15
20
25
Sales
(millions)
0.7
1.2
1.55
1.8
2
13-67
Ads in Sunday
Supplements
2
4
6
8
10
Sales
(millions)
1.2
2.2
3
3.5
3.75
(b)
TV Spots (polynomial of order 2)
3
2.5
2
Sales
(millions)
TV Spots
1
2
3
4
5
1.5
y = -0.1036x2 + 1.1264x - 0.04
1
Sales
(millions)
1
1.75
2.45
2.8
3
0.5
0
-1
1
3
5
TV Spots
TV Spots (polynomial of order 3)
3
2.5
2
Sales
(millions)
TV Spots
1
2
3
4
5
1.5
1
Sales
(millions)
1
1.75
2.45
2.8
3
y = -0.0083x3 - 0.0286x2 + 0.9298x + 0.1
0.5
0
-1
1
3
5
TV Spots
TV Spots (logarithmic form)
3
2.5
2
Sales
(millions)
TV Spots
1
2
3
4
5
Sales
(millions)
1
1.75
2.45
2.8
3
1.5
y = 1.2888ln(x) + 0.966
1
0.5
0
-1
1
3
TV Spots
13-68
5
Magazine Ads (polynomial of order 2)
2
1.5
Sales
(millions) 1
Magazine Ads
5
10
15
20
25
Sales
(millions)
0.7
1.2
1.55
1.8
2
y = -0.002x2 + 0.124x + 0.140
0.5
0
0
5
10
15
20
25
Magazine Ads
Magazine Ads (polynomial of order 3)
2
1.5
Sales
(millions)
Magazine Ads
5
10
15
20
25
Sales
(millions)
0.7
1.2
1.55
1.8
2
1
y = 0.000067x3 - 0.0050x2 + 0.1633x
0.5
0
0
5
10
15
20
25
Magazine Ads
Magazine Ads (logarithmic form)
2
1.5
Magazine Ads
5
10
15
20
25
Sales
(millions)
0.7
1.2
1.55
1.8
2
Sales
(millions) 1
y = 0.809ln(x) - 0.6267
0.5
0
0
5
10
15
Magazine Ads
13-69
20
25
Ads in Sunday Supplements (polynomial of order 2)
4
3.5
3
Sales
(millions)
Ads in Sunday
Supplements
2
4
6
8
10
2.5
2
y = -0.0321x2 + 0.706x - 0.09
1.5
Sales
(millions)
1.2
2.2
3
3.5
3.75
1
0.5
0
0
2
4
6
8
10
Ads in Sunday Supplements
Ads in Sunday Supplements (polynomial of order 3)
4
3.5
3
Sales
(millions)
Ads in Sunday
Supplements
2
4
6
8
10
Sales
(millions)
1.2
2.2
3
3.5
3.75
2.5
2
1.5
1
y = -0.000521x3 - 0.0228x2 + 0.657x - 0.02
0.5
0
0
2
4
6
8
10
Ads in Sunday Supplements
Ads in Sunday Supplements (logarithmic form)
4
3.5
3
Sales
(millions)
Ads in Sunday
Supplements
2
4
6
8
10
Sales
(millions)
1.2
2.2
3
3.5
3.75
2.5
2
y = 1.6331ln(x) + 0.0343
1.5
1
0.5
0
0
2
4
6
8
10
Ads in Sunday Supplements
In all three cases, the quadratic form is a close fit. The polynomial of order 3 is also a
good fit. The logarithmic form is not a bad fit, but not as closes as the polynomial forms.
We will use the quadratic form in the sequel.
13-70
(c) Let TV, M, and SS be the number of TV spots, magazine ads, and ads in Sunday
supplements respectively. Based on the results of part (b), using the quadratic form gives:
Sales œ 0.1036TV2  1.1264TV  0.04  0.002M2  0.124M  0.14  0.0321SS2  0.706SS  0.09
Cost of Ads œ 0.3TV  0.15M  0.1SS
Planning Cost œ 0.09TV  0.03M  0.04SS
Ê Profit œ $0.75 ‚ (Sales)  Cost of Ads  Planning Cost.
(d) The total sales generated are calculated in row 7 using the nonlinear equations from
part (b). Then, the gross profit from sales are calculated in H20. The TotalProfit (H23) is
the gross profit minus the cost of ads and of planning. The objective is to maximize this.
Sales per Ad = ax^2 + bx + k, where
a=
b=
k=
Sales Generated (millions)
Ad Budget
Planning Budget
Young Children
Parents of Young Children
Coupon Redemption per Ad
($thousands)
Number of Ads
Maximum TV Spots
TV Spots
-0.1036
1.1264
-0.0400
2.8296
Magazine Ads
-0.002
0.124
0.14
0.5600
SS Ads
-0.0321
0.706
-0.09
3.7903
Total
7.1799
Gross Profit per Sale
$0.75
Cost per Ad ($thousands)
150
100
30
40
Budget Spent
2,884
923
<=
<=
Budget Available
4,000
1,000
Number Reached per Ad (millions)
1.2
0.1
0
0.5
0.2
0.2
Total Reached
5.25
5.00
>=
>=
Minimum Acceptable
5
5
Total Redeemed
1,490
=
Required Amount
1,490
300
90
TV Spots
0
Magazine Ads
40
SS Ads
120
TV Spots
4.075
<=
5
Magazine Ads
3.596
SS Ads
11.218
13-71
Gross Profit
Cost of Ads
Planning Cost
Total Profit
5.385
2.884
0.923
1.578
($million)
(e) Separable programming formulation
Sales per Ad
Group 1
Group 2
Group 3
Group 4
Group 5
TV Spots
1
0.75
0.7
0.35
0.2
Magazine Ads
0.14
0.1
0.07
0.05
0.04
SS Ads
0.6
0.5
0.4
0.25
0.125
Ad Budget
Planning Budget
Cost per Ad ($thousands)
300
150
100
90
30
40
Budget Spent
3,156
938
<=
<=
Budget
Available
4,000
1,000
Young Children
Parents of Young Children
Number Reached per Ad (millions)
1.2
0.1
0
0.5
0.2
0.2
Total Reached
5.00
5.23
>=
>=
Min. Acceptable
5
5
Total Redeemed
1,490
=
Req. Amount
1,490
Coupon Redemption per Ad
($thousands)
Number of Ads
Group 1
Group 2
Group 3
Group 4
Group 5
Total
Maximum TV Spots
TV Spots
0
Magazine Ads
40
SS Ads
120
TV Spots
1.000
1.000
1.000
0.563
0.000
3.563
<=
5
Magazine Ads
5.000
2.250
0.000
0.000
0.000
7.250
SS Ads
2.000
2.000
2.000
2.000
2.000
10.000
<=
<=
<=
<=
<=
Maximum
TV Spots Magazine Ads
1
5
1
5
1
5
1
5
1
5
Total Sales
Gross Profit per Sale
Gross Profit
Cost of Ads
Planning Cost
Total Profit
SS Ads
2
2
2
2
2
7.3219
$0.75
5.491
3.156
0.938
1.397
($million)
(f) In part (d), 4.075 TV ads, 3.596 magazine ads, and 11.218 ads in Sunday supplements
are placed. In part (e), 3.563 TV ads, 7.25 magazine ads, and 10 ads in Sunday
supplements are placed. In Case 3.4, 3 TV ads, 14 magazine ads, and 7.75 ads in Sunday
supplements are placed. Unlike linear programming, nonlinear and separable
programming take into account the diminishing returns from repeated advertisements.
Since the solution is fairly different, it certainly appears that it was worthwhile to refine
the linear programming model used in Case 3.4.
13-72
CHAPTER 14: METAHEURISTICS
14.1-1.
(a)
Tours
1-2-3-4-5-1
1-2-3-5-4-1
1-2-4-3-5-1
1-2-4-5-3-1
1-2-5-3-4-1
1-2-5-4-3-1
Distance
34
34
36
31
30
25
Tours
1-3-2-4-5-1
1-3-2-5-4-1
1-3-4-2-5-1
1-3-5-2-4-1
1-4-2-3-5-1
1-4-3-2-5-1
Distance
32
26
28
28
37
31
Optimal Solution: 1-2-5-4-3-1 (or the reverse 1-3-4-5-2-1)
(b) Start with the initial trial solution 1-2-3-4-5-1. There are three possible sub-tour
reversals that improve upon this solution.
Reverse 2-3
Reverse 2-3-4
Reverse 3-4-5
1-2-3-4-5-1
1-3-2-4-5-1
1-4-3-2-5-1
1-2-5-4-3-1
Distance = 34
Distance = 32
Distance = 31
Distance = 25
Choose 1-2-5-4-3-1 as the next trial solution. There is no sub-tour reversal that improves
upon this solution. The tour 1-2-5-4-3-1 is optimal.
(c) Start with the initial trial solution 1-2-4-3-5-1. There are four possible sub-tour
reversals that improve upon this solution.
Reverse 4-3
Reverse 3-5
Reverse 2-4-3
Reverse 4-3-5
1-2-4-3-5-1
1-2-3-4-5-1
1-2-4-5-3-1
1-3-4-2-5-1
1-2-5-3-4-1
Distance = 36
Distance = 34
Distance = 31
Distance = 28
Distance = 30
Choose 1-3-4-2-5-1 as the next trial solution. There is only one possible sub-tour reversal
that improves upon this solution.
Reverse 2-5
1-3-4-2-5-1
1-3-4-5-2-1
Distance = 28
Distance = 25
Choose 1-3-4-5-2-1 as the next trial solution. There is no sub-tour reversal that improves
upon this. The solution 1-3-4-5-2-1 is optimal.
14-1
(d) Start with the initial trial solution 1-4-2-3-5-1. There are five possible sub-tour
reversals that improve upon this solution.
Reverse 2-4
Reverse 2-3
Reverse 3-5
Reverse 4-2-3
reverse 2-3-5
1-4-2-3-5-1
1-2-4-3-5-1
1-4-3-2-5-1
1-4-2-5-3-1
1-3-2-4-5-1
1-4-5-3-2-1
Distance = 37
Distance = 36
Distance = 31
Distance = 28
Distance = 32
Distance = 34
Choose 1-4-2-5-3-1 as the next trial solution. There is only one possible sub-tour reversal
that improves upon this solution.
Reverse 2-5
1-4-2-5-3-1
1-4-5-2-3-1
Distance = 28
Distance = 26
Choose 1-4-5-2-3-1 as the next trial solution. There is one possible sub-tour reversal that
improves upon this.
Reverse 4-5-2
1-4-5-2-3-1
1-2-5-4-3-1
Distance = 26
Distance = 25
Choose 1-2-5-4-3-1 as the next trial solution. There is no sub-tour reversal that improves
upon this. The solution 1-2-5-4-3-1 is optimal.
14.1-2.
(a) If the second reversal were chosen, the next trial solution would be 1-2-3-5-4-6-7-1
and there is no sub-tour reversal that gives an improvement.
(b) Start with the initial trial solution 1-2-4-5-6-7-3-1. There are two possible sub-tour
reversals that improve upon this solution.
Reverse 5-6
Reverse 2-4-5-6-7
1-2-4-5-6-7-3-1
1-2-4-6-5-7-3-1
1-7-6-5-4-2-3-1
Distance = 69
Distance = 66
Distance = 68
Choose 1-2-4-6-5-7-3-1 as the next trial solution. There is only one possible sub-tour
reversal that improves upon this.
Reverse 5-7
1-2-4-6-5-7-3-1
1-2-4-6-7-5-3-1
Distance = 66
Distance = 63
Choose 1-2-4-6-7-5-3-1 as the next trial solution. This is an optimal solution.
14-2
14.1-3.
(a)
Tours
1-2-3-4-5-6-1
1-2-3-4-6-5-1
1-2-3-6-4-5-1
1-2-4-4-6-5-1
1-2-5-4-3-6-1
Distance
64
59
67
64
67
Tours
1-2-6-3-4-5-1
1-5-2-3-4-6-1
1-5-2-4-3-6-1
1-6-2-3-4-5-1
1-6-3-2-4-5-1
Distance
69
56
61
63
66
Optimal Solution: 1-5-2-3-4-6-1 (or the reverse 1-6-4-3-5-2-1)
(b) Start with the initial trial solution 1-2-3-4-5-6-1. There are two possible sub-tour
reversals that improve upon this solution.
Reverse 5-6
Reverse 2-3-4-5
1-2-3-4-5-6-1
1-2-3-4-6-5-1
1-5-4-3-2-6-1
Distance = 64
Distance = 59
Distance = 63
Choose 1-2-3-4-6-5-1 as the next trial solution. There is no sub-tour reversal that
improves upon this solution.
(c) Start with the initial trial solution 1-2-5-4-3-6-1. There are two possible sub-tour
reversals that improve upon this solution.
Reverse 2-5
Reverse 5-4-3
1-2-5-4-3-6-1
1-5-2-4-3-6-1
1-2-3-4-5-6-1
Distance = 67
Distance = 61
Distance = 64
Choose 1-5-2-4-3-6-1 as the next trial solution. There is no sub-tour reversal that
improves upon this solution.
14.2-1.
Sears logistics services (SLS) provides delivery with its fleet of over 1,000 vehicles.
Sears product services (SPS) offers home service with its fleet of 12,500 vehicles and
technicians. A customer who asks for delivery or home service is given a day and a time
window based on customer preferences and working schedule in the region where the
customer is located. In either case, the goal is to generate efficient routes for the vehicles
and to provide customers with accurate and convenient time windows while minimizing
the operational costs. Both problems are instances of vehicle routing problem with time
windows (VRPTW). A basic VRPTW determines routes for Q vehicles, each starting at
the depot and returning to the depot after visiting a subset of customers in some order.
Every customer is visited by exactly one vehicle. The capacity constraints of the vehicles
and the time windows imposed by customers should be met. The objective is to minimize
the total cost. The problems faced by SLS and SPS differ from the basic VRPTW in that
they include additional constraints. For instance, in the case of SPS, technicians' skills
need to be considered in assigning service orders to them. In both cases, there may be
restrictions on total route times and travel times between any two locations. Hence, the
14-3
problem is a complex one and necessitates the use of a solution procedure that can
provide good solutions in acceptable time.
To solve the problem, first an initial route is found for each vehicle, then unassigned
stops are inserted into a route. This solution is improved using various local heuristic
techniques. In order not to be stuck at local optima, the procedure is enhanced with tabu
search technique. Once a stop in a route is relocated, the move is included in a tabu list
and remains prohibited for a number of iterations unless the objective function value it
offers exceeds the best value obtained up to that iteration.
Financial benefits of this study include $9 million in one-time savings and over $42
million in annual savings. The savings result from the reduction in travel times, mileage
and routing times. Sears now offers more timely delivery of merchandise and home
service, so more reliable customer service. The utilization of the fleets is improved. The
routing process became faster and the facility, equipment and personnel costs related to
routing decreased. Since the problem can be solved quickly, Sears can respond to
disruptions and adjust its schedules more efficiently.
14.2-2.
Start with the initial trial solution with links AB, AC, AE, CD, which costs 232.
Iteration 1:
Add
BC
BD
DE
Delete
AB
AC
AB
AC
CD
AC
AE
CD
Cost
138
246
56
164
268
152
240
256
Adding BD and deleting AB results in the lowest cost, so choose inserting links AC, AE,
BD. CD. In fact, this is the optimal solution.
14-4
14.2-3.
Start with the initial trial solution with links AB, AD, BE, CD, which costs 390.
Iteration 1: Minimum local search
Add
AC
CE
Delete
AD
CD
AB
AD
CD
BE
Cost
185
275
275
180
270
365
Current solution: AB, BE, CD, CE.
Tabu list: CE
Iteration 2: Minimum local search
Add
DE
Delete
CD
Cost
95
Current solution: AB, BE, CE, DE
Tabu list: CE, DE
Iteration 3: Minimum local search
Add
AC
Delete
BE
Cost
75
The solution AB, AC, CE, DE is optimal.
14.2-4.
Start with the initial trial solution with links OA, AB, BC, BE, ED, DT, which costs 314.
Iteration 1: Minimum local search
Add
ET
Delete
DE
Cost
122
Current solution: OA, AB, BC, BE, ET, DT
Tabu list: ET
Iteration 2: Minimum local search
Add
CE
Delete
BC
Cost
23
The solution OA, AB, CE, BE, ET, DT is optimal.
14-5
14.2-5.
14.2-6.
14-6
14.2-7.
(a)
14-7
(b)
(c)
14.3-1.
^- œ $!, X œ #
(a)
Maximization problem:
^8 œ #*, B œ Ð^8  ^- ÑÎX œ !Þ&, PÖacceptance× œ /B œ !Þ'!(
^8 œ $%, ^8  ^- ,
PÖacceptance× œ "
^8 œ $", ^8  ^- ,
PÖacceptance× œ "
^8 œ #%, B œ Ð^8  ^- ÑÎX œ $, PÖacceptance× œ /B œ !Þ!&
(b)
Minimization problem:
^8 œ #*, ^8  ^- ,
PÖacceptance× œ "
^8 œ $%, B œ Ð^-  ^8 ÑÎX œ #, PÖacceptance× œ /B œ !Þ"$&
^8 œ $", B œ Ð^-  ^8 ÑÎX œ !Þ&,
PÖacceptance× œ /B œ !Þ'!(
^8 œ #%, ^8  ^- ,
PÖacceptance× œ "
14.3-2.
Because of the randomness in the algorithm, the output will vary.
14-8
14.3-3.
(a) Initial trial solution: 1-2-3-4-5-1, ^- œ $4, X" œ !Þ2*^- œ 6.8
0.0000 - 0.3332
0.3333 - 0.6666
0.6667 - 0.9999
Sub-tour begins in slot 2.
Sub-tour begins in slot 3.
Sub-tour begins in slot 4.
The random number is 0.09656: choose a sub-tour that begins in slot 2. The sub-tour
needs to end either in slot 3 or slot 4.
0.0000 - 0.4999
0.5000 - 0.9999
Sub-tour ends in slot 3.
Sub-tour ends in slot 4.
The random number is 0.96657: choose a sub-tour that ends in slot 4.
Reverse 2-3-4 to obtain the new solution 1-4-3-2-5-1, ^8 œ $1. Since ^8  ^- , accept
this solution as the next trial solution.
(b) Because of the randomness in the algorithm, the output will vary.
14.3-4.
Because of the randomness in the algorithm, the output will vary.
14.3-5.
Because of the randomness in the algorithm, the output will vary.
14.3-6.
Because of the randomness in the algorithm, the output will vary.
14.3-7.
(a)
0 ÐBÑ œ B$  '!B#  *!!B  "!!
0 w ÐBÑ œ $B#  "#!B  *!! and 0 ww ÐBÑ œ 'B  "#!
Stationary Points:
End Points:
0 w ÐB‡ Ñ œ ! Ê B‡ is either "! or $! (stationary points of 0 ).
0 ww Ð"!Ñ œ '!  ! Ê B‡ œ "! is a local maximum.
0 ww Ð$!Ñ œ '!  ! Ê B‡ œ $! is a local minimum.
0 w Ð!Ñ œ *!!  ! Ê B œ ! is a local minimum.
0 w Ð$"Ñ œ '$  ! Ê B œ $" is a local minimum.
14-9
(b)
(c) B œ "&Þ&, 0 ÐBÑ œ ^- œ $&&)Þ*, X œ !Þ#^- œ '("Þ((&
P œ !, Y œ $", 5 œ ÐY  PÑÎ' œ &Þ"'(
The random number obtained from Table 20.3 is !Þ!*'&'. From Appendix 5,
PÖ] Ÿ "Þ$"&× ¶ !Þ!*'&',
with ] a standard Normal random variable, R Ð!ß &Þ"'(Ñ œ "Þ$"& † &Þ"'( œ 'Þ(*.
B œ "&Þ&  R Ð!ß &Þ"'(Ñ œ )Þ(", ^8 œ 0 ÐBÑ œ %!%(Þ'
Since ^8  ^- , accept B œ )Þ(" as the next trial solution.
(d) Because of the randomness in the algorithm, the output will vary.
14.3-8.
The nonconvex problem is to:
maximize
subject to
!Þ&B&  'B%  #%Þ&B$  $*B#  #!B
! Ÿ B Ÿ &.
(a) B œ #Þ&, 0 ÐBÑ œ ^- œ $Þ&"&', X œ !Þ#^- œ !Þ(!$"
P œ !, Y œ &, 5 œ ÐY  PÑÎ' œ !Þ)$$$
The random number obtained from Table 20.3 is !Þ!*'&'. From Appendix 5,
PÖ] Ÿ "Þ$"&× ¶ !Þ!*'&',
with
]
a
standard
R Ð!ß !Þ)$$$Ñ œ "Þ$"& † !Þ)$$$ œ "Þ!*&).
Normal
random
variable,
B œ #Þ&  R Ð!ß !Þ)$$$Ñ œ "Þ%!%#, ^8 œ 0 ÐBÑ œ "Þ&()#
Since Ð^8  ^- ÑÎX œ (Þ#%)), the probability of accepting B œ "Þ%!%# as the next trial
solution is PÖacceptance× œ /(Þ#%)) œ !Þ!!!(". From Table 20.3, the next random
number is !Þ*''&(  !Þ!!!(", so we reject B œ "Þ%!%# as the next trial solution.
(b) Because of the randomness in the algorithm, the output will vary.
14-10
14.3-9.
(a) B œ #&, 0 ÐBÑ œ ^- œ 6ß 640ß 625, X œ !Þ#^- œ 1,328,125
P œ !, Y œ &!, 5 œ ÐY  PÑÎ' œ )Þ$$$
The random number obtained from Table 20.3 is !Þ!*'&'. From Appendix 5,
PÖ] Ÿ "Þ$"&× ¶ !Þ!*'&',
with ] a standard Normal random variable, R Ð!ß )Þ$$$Ñ œ "Þ$"& † )Þ$$$ œ "!Þ*&).
B œ #&  R Ð!ß )Þ$$$Ñ œ "%Þ!%#, ^8 œ 0 ÐBÑ œ 7ß 995ß 655
Since ^8  ^- , accept the new solution.
(b) Because of the randomness in the algorithm, the output will vary.
14.3-10.
(a) ÐB" ß B# Ñ œ Ð")ß #&Ñ, 0 ÐB" ß B# Ñ œ ^- œ "$$ß &!*Þ&, X œ !Þ#^- œ #'ß (!"Þ*
P œ Ð!ß !Ñ, Y œ Ð$'ß &!Ñ
5" œ Ð$'  !ÑÎ' œ '
5# œ Ð&!  !ÑÎ' œ )Þ$$$
The random number obtained from Table 20.3 is !Þ!*'&'. From Appendix 5,
PÖ] Ÿ "Þ$"&× ¶ !Þ!*'&',
with ] a standard Normal random variable,
R Ð!ß 'Ñ œ "Þ$"& † ' œ (Þ)*
B" œ ")  R Ð!ß 'Ñ œ "!Þ""
R Ð!ß )Þ$$$Ñ œ "Þ$"& † )Þ$$$ œ "!Þ*&)
B# œ #&  R Ð!ß )Þ$$$Ñ œ "%Þ!%#
This solution is feasible.
^8 œ 0 ÐBÑ œ "!(ß %'(
Since Ð^8  ^- ÑÎX œ *Þ!#%(, the probability of accepting this solution as the next trial
solution is PÖacceptance× œ /*Þ!#%( œ !Þ!!!"#. From Table 20.3, the next random
number is !Þ*''&(  !Þ!!!"#, so we reject Ð"!Þ""ß "%Þ!%#Ñ as the next trial solution.
(b) Because of the randomness in the algorithm, the output will vary.
14-11
14.4-1.
(a)
P1:
P2:
010011 and
100101
Only the last digits agree, the children then become:
C1:
C2:
xxxxx1 and
xxxxx1,
where x represents the unknown digits. Random numbers are used to identify these
unknown digits and let random numbers:
0.00000 - 0.49999 correspond to x = 0,
0.50000 - 0.99999 correspond to x = 1.
Starting from the front of the top row of Table 20.3, the first 10 random numbers are:
0.09656, 0.96657, 0.64842, 0.49222, 0.49506, 0.10145, 0.48455, 0.23505, 0.90430,
0.04180. The corresponding digits are: 0,1,1,0,0,0,0,0,1,0. The children then become:
C1:
C2:
011001 and
000101.
Next, we consider the possibility of mutations. The probability of a mutation in any
generation is set at 0.1, and let random numbers
0.00000 - 0.09999 correspond to a mutation,
0.10000 - 0.99999 correspond to no mutation.
Starting from the second row of Table 20.3, we obtain the next 12 random numbers.
Accordingly, the 8th and 11th ones correspond to a mutation, so the final conclusion is
that the two children are
(b)
C1:
C2:
011001 and
010111.
P1:
P2:
000010 and
001101
The first and second digits agree, the children then become:
C1:
C2:
00xxxx and
00xxxx,
where x represents the unknown digits. Random numbers are used to identify these
unknown digits and let random numbers:
0.00000 - 0.49999 correspond to x = 0,
0.50000 - 0.99999 correspond to x = 1.
Starting from the front of the top row of Table 20.3, the first 8 random numbers
correspond to digits: 0,1,1,0,0,0,0,0. The children then become:
C1:
C2:
000110 and
000000.
14-12
Next, we consider the possibility of mutations. The probability of a mutation in any
generation is set at 0.1, and let random numbers
0.00000 - 0.09999 correspond to a mutation,
0.10000 - 0.99999 correspond to no mutation.
Use Table 20.3 to obtain the next 12 random numbers. Accordingly, the 2nd and 10th ones
correspond to a mutation, so the final conclusion is that the two children are
(c)
C1:
C2:
010110 and
000100.
P1:
P2:
100000 and
101000
All but the third digits agree, the children then become:
C1:
C2:
10x000 and
10x000,
where x represents the unknown digits. Random numbers are used to identify these
unknown digits and let random numbers:
0.00000 - 0.49999 correspond to x = 0,
0.50000 - 0.99999 correspond to x = 1.
Starting from the front of the top row of Table 20.3, the first 2 random numbers
correspond to digits: 0,1. The children then become:
C1:
C2:
100000 and
101000.
Next, we consider the possibility of mutations. The probability of a mutation in any
generation is set at 0.1, and let random numbers
0.00000 - 0.09999 correspond to a mutation,
0.10000 - 0.99999 correspond to no mutation.
Use Table 20.3 to obtain the next 12 random numbers. Accordingly, only the 8th one
corresponds to a mutation, so the final conclusion is that the two children are
C1:
C2:
100000 and
111000.
14-13
14.4-2.
(a)
P1:
P2:
1-2-3-4-7-6-5-8-1 and
1-5-3-6-7-8-2-4-1
Start from city 1.
Possible links: 1-2, 1-8, 1-5, 1-4
Random numbers:
0.09656
choose 1-2
0.96657
no mutation
Start from city 2. Current tour: 1-2
Possible links: 2-3, 2-8, 2-4
Random numbers:
0.64842
choose 2-8
0.49222
no mutation
Start from city 8. Current tour: 1-2-8
Possible links: 8-5, 8-7
Random numbers:
0.49506
choose 8-5
0.10145
no mutation
Start from city 5. Current tour: 1-2-8-5
Possible links: 5-6, 5-3
Random numbers:
0.48455
choose 5-6
0.23505
no mutation
Start from city 6. Current tour: 1-2-8-5-6
Possible links: 6-7, 6-7, 6-3
Random numbers:
0.90430
choose 6-3
0.04180
mutation
Reject 6-3 and consider all other possible links: 6-4, 6-7
Random numbers:
0.24712
choose 6-4
Start from city 4. Current tour: 1-2-8-5-6-4
Possible links: 4-3, 4-7
Random numbers:
0.55799
choose 4-7
0.60857
no mutation
The only remaining city is 3. Hence, C1 = 1-2-8-5-6-4-7-3-1.
(b)
P1:
P2:
1-6-4-7-3-8-2-5-1 and
1-2-5-3-6-8-4-7-1
Start from city 1.
Possible links: 1-6, 1-5, 1-2, 1-7
Random numbers:
0.09656
choose 1-6
0.96657
no mutation
Start from city 6. Current tour: 1-6
Possible links: 6-4, 6-3, 6-8
Random numbers:
0.64842
choose 6-3
0.49222
no mutation
Start from city 3. Current tour: 1-6-3
Possible links: 3-7, 3-8, 3-5
Random numbers:
0.49506
choose 3-8
0.10145
no mutation
14-14
Start from city 8. Current tour: 1-6-3-8
Possible links: 8-2, 8-4
Random numbers:
0.48455
choose 8-2
0.23505
no mutation
Start from city 2. Current tour: 1-6-3-8-2
Possible links: 2-5
Random numbers:
0.04180
mutation
Reject 2-5 and consider all other possible links: 2-4, 2-7
Random numbers:
0.24712
choose 2-4
Start from city 4. Current tour: 1-6-3-8-2-4
Possible links: 4-7
Random numbers:
0.60857
no mutation
The only remaining city is 5. Hence, C1 = 1-6-3-8-2-4-7-5-1.
(c)
P1:
P2:
1-5-7-4-6-2-3-8-1 and
1-3-7-2-5-6-8-4-1
Start from city 1.
Possible links: 1-5, 1-8, 1-3, 1-4
Random numbers:
0.09656
choose 1-5
0.96657
no mutation
Start from city 5. Current tour: 1-5
Possible links: 5-7, 5-2, 5-6
Random numbers:
0.64842
choose 5-2
0.49222
no mutation
Start from city 2. Current tour: 1-5-2
Possible links: 2-6, 2-3, 2-7
Random numbers:
0.49506
choose 2-3
0.10145
no mutation
Start from city 3. Current tour: 1-5-2-3
Possible links: 3-8, 3-7
Random numbers:
0.48455
choose 3-8
0.23505
no mutation
Start from city 8. Current tour: 1-5-2-3-8
Possible links: 8-6, 8-4
Random numbers:
0.90430
choose 8-4
0.04189
mutation
Reject 8-4 and consider all other possible links: 8-6, 8-7
Random numbers:
0.24712
choose 8-6
Start from city 6. Current tour: 1-5-2-3-8-6
Possible links: 6-4
Random numbers:
0.55799
choose 6-4
0.60857
no mutation
The only remaining city is 7. Hence, C1 = 1-5-2-3-8-6-4-7-1.
14-15
14.4-3.
(a) Because of the randomness in the algorithm, the output will vary.
(b) Because of the randomness in the algorithm, the output will vary.
14.4-4.
Integer nonlinear programming:
maximize 0 ÐBÑ œ B$  '!B#  *!!
subject to ! Ÿ B Ÿ $"
(a)
(b) Because of the randomness in the algorithm, the output will vary.
14.4-5.
Because of the randomness in the algorithm, the output will vary.
14.4-6.
Because of the randomness in the algorithm, the output will vary.
14.4-7.
(a) Because of the randomness in the algorithm, the output will vary.
(b) Because of the randomness in the algorithm, the output will vary.
14-16
14.4-8.
(a) Genetic Algorithm
(b) Because of the randomness in the algorithm, the output will vary.
14.4-9.
Because of the randomness in the algorithm, the output will vary.
14.4-10.
Because of the randomness in the algorithm, the output will vary.
14.4-11.
Answers will vary.
14.5-1.
See the solution for Problem 14.2-6(a) for the output from the basic tabu search
algorithm. Because of the randomness in the basic simulated annealing and genetic
algorithms, their outputs will vary.
14.5-2.
See the solution for Problem 14.2-7(a) for the output from the basic tabu search
algorithm. Because of the randomness in the basic simulated annealing and genetic
algorithms, their outputs will vary.
14-17
CHAPTER 15: GAME THEORY
15.1-1.
Let player 1 be the labor union with strategy 3 being to decrease the wage demand by
"!Ð3  "Ñ¢ and player 2 be the management with strategy 3 being to increase the offer by
"!Ð3  "Ñ¢. The payoff matrix is:
"
#
$
%
&
'
"
"Þ$&
"Þ&
"Þ%
"Þ$
"Þ#
"Þ"
#
"Þ#
"Þ$&
"Þ%
"Þ$
"Þ#
"Þ"
$
"Þ$
"Þ$
"Þ$&
"Þ$
"Þ#
"Þ"
%
"Þ%
"Þ%
"Þ%
"Þ$&
"Þ#
"Þ"
&
"Þ&
"Þ&
"Þ&
"Þ&
"Þ$&
"Þ"
'
"Þ'
"Þ'
"Þ'
"Þ'
"Þ'
"Þ$&
where the rows represent the strategy of player 1 and the columns the strategy of player 2.
15.1-2.
Label the products as A and B respectively. The strategies for each manufacturer are:
1- Normal development of both products
2- Crash development of product A
3- Crash development of product B.
Let :34 œ "# [(% increase to manufacturer 1 from A)  (% increase to manufacturer 1 from B)] when
manufacturer 1 uses strategy 3 and manufacturer 2 uses strategy 4. The payoff matrix is:
"
#
$
"
)
%
%
#
"!
%
"$
$
"!
"$
%
row min
)
%
%
col max
)
"$
"$
)
The rows correspond to the strategy of manufacturer 1 and the columns to the strategy of
manufacturer 2. The minimum of the column maxima and the maximum of the row
minima is ), so both manufacturers should use strategy 1, namely choose normal development
of both products. Consequently, manufacturer 1 will increase its share by )%.
15.1-3.
Each player has the same strategy set. A strategy must specify the first chip chosen, the
second and third chips chosen for every choice first chip by the opponent. Denote the
white, red and blue chips by W, R and B respectively. Then a strategy is of the form:
Choose 3 − ÖW, R, B× as first chip, if the opponent chooses 4 − ÖW, R, B×, then choose
54 − ÖW, R, B×\Ö3×, and let 64 − ÖW, R, B×\Ö3, 5×. There are three choices of 3 and for
each 3, eight choices of second and third chips, so 24 strategies in total. Player 1 can
either win all three games, or win one and get a draw in another one, or lose all three.
15-1
Hence, the payoff to player 1 can be either 120, 0, or -120. The payoff to player 1 in each
possible scenario is given in the table below, where the rows and the columns represent
the order of chips played by player 1 and 2 respectively.
WRB
WBR
RWB
RBW
BWR
BRW
WRB
0
0
0
120
-120
0
WBR
0
0
120
0
0
-120
RWB
0
-120
0
0
0
120
RBW
-120
0
0
0
120
0
BWR
120
0
0
-120
0
0
BRW
0
120
-120
0
0
0
15.2-1.
(a) Strategies 4, 5, and 6 of each player are dominated by their strategy 3. Then strategy 1
can be eliminated, since it is dominated by strategy 3 for each player. Once these are
eliminated, strategy 2 of each is dominated by strategy 3. Thus, the best strategy of the
labor union is to decrease its demand by #!¢ and the best for the management if to
increase its offer by #!¢. The resulting wage is $"Þ$&.
(b)
"
#
$
%
&
'
col max
"
"Þ$&
"Þ&
"Þ%
"Þ$
"Þ#
"Þ"
#
"Þ#
"Þ$&
"Þ%
"Þ$
"Þ#
"Þ"
"Þ&
"Þ%
$
"Þ$
"Þ$
"Þ$&
"Þ$
"Þ#
"Þ"
%
"Þ%
"Þ%
"Þ%
"Þ$&
"Þ#
"Þ"
&
"Þ&
"Þ&
"Þ&
"Þ&
"Þ$&
"Þ"
'
"Þ'
"Þ'
"Þ'
"Þ'
"Þ'
"Þ$&
row min
"Þ#
"Þ$
"Þ$&
"Þ$
"Þ#
"Þ"
"Þ$&
"Þ%
"Þ&
"Þ'
"Þ$&
15.2-2.
Strategy 3 of player 1 is dominated by strategy 2.
Strategy 3 of player 2 is dominated by strategy 1.
Strategy 1 of player 1 is dominated by strategy 2.
Strategy 2 of player 2 is dominated by strategy 1.
Therefore, the optimal strategy is strategy 2 for player 1 and strategy 1 for player 2 and
the resulting payoff is 1 to player 1.
15.2-3.
Strategy 1 of player 2 is dominated by strategy 3.
Strategy 4 of player 2 is dominated by strategy 2.
Strategies 1 and 2 of player 1 are dominated by strategy 3.
Strategy 2 of player 2 is dominated by strategy 3.
Therefore, the optimal strategy is strategy 3 for each player and the resulting payoff is 1
to player 2.
15-2
15.2-4.
"
#
$
col max
"
1
2
$
$
#
"
0
"
"
$
1
3
#
$
row min
"
#
"
"
The best strategy is strategy 3 for player 1 and strategy 2 for player 2, the resulting payoff
is " to player 1. The game is stable with a saddle point Ð$ß #Ñ, since the minimax value
equals the maximin value.
15.2-5.
"
#
$
col max
"
#
$
%
$ $ # %
% # "
"
" "
#
!
" "
#
"
row min
%
%
"
"
The best strategy is strategy 3 for player 1 and strategy 2 for player 2, the resulting payoff
is 1 to player 2. The game is stable with a saddle point Ð$ß #Ñ.
15.2-6.
(a)
"
#
$
col max
"
#
"
$
$
#
$
%
#
%
$
"
!
"
"
row min
"
!
#
"
The best strategy is strategy 1 for player 1 and strategy 3 for player 2, the resulting payoff
is 1 to player 1. The game is stable with a saddle point Ð"ß $Ñ.
(b)
Strategy 1 of player 2 is dominated by strategy 3.
Strategy 3 of player 1 is dominated by strategies 1 and 2.
Strategy 2 of player 2 is dominated by strategy 3.
Strategy 2 of player 1 is dominated by strategy 1.
The optimal strategy is strategy 1 for player 1 and strategy 3 for player 2, with a payoff of
1 to player 1.
15.2-7.
(a)
"
#
$
col max
"
(
"
&
(
#
"
!
$
!
$
$
#
"
"
row min
"
!
&
!
The best strategy is to use issue 2 for each politician, with zero payoff to each.
15-3
(b) Let :34 be the probability that politician 1 wins the election or the election results in a
tie when politician 1 chooses issue 3 and politician 2 issue 4. Then the new payoff matrix
is:
"
#
$
"
"
"Î&
!
#
!
!
!
$
$Î&
#Î&
"Î&
Strategies 2 and 3 of politician 1 are dominated by strategy 2.
Strategies 1 and 3 of politician 2 are dominated by strategy 2.
Hence, by eliminating dominated strategies, one gets issue 1 as the best strategy for
politician 1 and issue 2 for politician 2, the payoff is zero. Thus, politician 2 can prevent
politician 1 from winning or getting a tie.
(c) Let :34 œ 
"
!
if politician 1 will win or tie
if politician 2 will win
Then the payoff matrix becomes:
"
#
$
"
"
!
!
#
!
!
!
$
!
!
!
where the minimax of the columns and the maximin of the rows both equal zero, i.e.,
politician 1 cannot win. Politician 1 can use any use, politician 2 can choose issue 2 or 3;
however, since issue 1 offers politician 1 his only chance of winning, he should use that
one and hope that politician 2 chooses issue 1 by mistake.
15.2-8.
Advantages: It provides the best possible guarantee on what the worst outcome can be,
regardless of how skillfully the opponent plays the game and hence, reduces the
possibility of undesirable outcomes to a minimum.
Disadvantages: Since it aims at eliminating worst cases, it is conservative and may yield
payoffs that are far from the best ones.
15.3-1.
(a)
"
#
col max
"
"
"
"
#
"
"
"
row min
"
"
The minimax payoff is not the same as the maximin payoff, so the game does not have a
saddle point.
15-4
(b)
Expected payoff for player 1: ÐB" C"  B# C# Ñ  ÐB" C#  B# C" Ñ
B"  B# œ C"  C# œ "
(i)
C" œ "ß C# œ !:
B"  B# œ B"  Ð"  B" Ñ œ #B"  "
(ii)
C" œ !ß C# œ ":
B#  B" œ Ð"  B" Ñ  B" œ "  #B"
"
"
(ii)
C" œ # ß C# œ # :
!
(c)
Expected payoff for player 1: ÐB" C#  B# C" Ñ  ÐB" C"  B# C# Ñ
B"  B# œ C"  C# œ "
(i)
C" œ "ß C# œ !:
B#  B" œ Ð"  B" Ñ  B" œ "  #B"
(ii)
C" œ !ß C# œ ":
B"  B# œ B"  Ð"  B" Ñ œ #B"  "
(ii)
C" œ "# ß C# œ "# :
!
15.3-2.
(a)
Strategies for player1: 1- Pass on heads or tails
2- Bet on heads or tails
3- Pass on heads, bet on tails
4- Bet on heads, pass on tails
Strategies for player 2:1- If player 1 bets, call.
2- If player 1 bets, pass.
(b)
"
#
$
%
"
&
!
(Þ&
#Þ&
#
&
&
!
!
Strategies 1 and 3 of player 1 are dominated bye strategy 2. Upon eliminating them, the
table is reduced to:
#
%
(c)
"
#
$
%
col max
"
!
#Þ&
#
&
!
"
#
&
&
!
&
(Þ&
!
#Þ&
!
#Þ&
&
row min
&
!
(Þ&
!
The minimum of the column maxima is not equal to the maximum of the row minima,
there is no saddle point. If either player chooses a pure strategy, the other one can choose
a strategy to cause the first player to change his strategy. One needs mixed strategies to
find an equilibrium.
15-5
(d) The dominated strategies will not be chosen. Let B# and B% be the probabilities that
player 1 uses strategy 2 and 4 respectively, C" and C# be the probabilities that player 2
uses strategy 1 and 2 respectively. Hence, B#  B% œ " and C"  C# œ " and the expected
payoff can be expressed as :#" B# C"  :## B# C#  :%" B% C"  :%# B% C# .
Case (i): C" œ ", C# œ !
Ê #Þ&B% œ #Þ&Ð"  B# Ñ
Case (ii): C" œ !, C# œ "
Ê &B# œ &Ð"  B% Ñ
Case (iii): C" œ C# œ !Þ&
Ê &B#  "#   #Þ&B%  "#  œ !Þ#&B#  "Þ#&
15.4-1.
Expected payoff for player 1: (i) C" œ "ß C# œ !: #B"  "
(ii) C" œ !ß C# œ ": "  #B"
Expected payoff for player 2: (i) B" œ "ß B# œ !: "  #C"
(ii) B" œ !ß B# œ ": #C"  "
The corresponding value of the game is zero.
15-6
15.4-2.
ÐC" ß C# Ñ
Ð"ß !Ñ
Ð!ß "Ñ
Expected Payoff
#Þ&Ð"  B# Ñ
&B#
#Þ&Ð"  B# Ñ œ &B# Ê ÐB‡" ß B‡# ß B‡$ ß B‡% Ñ œ Ð!ß "Î$ß !ß #Î$Ñ and @ œ &Î$.
#Þ&C"‡ Ð"  B# Ñ  &C#‡ B# œ &Î$ for ! Ÿ B# Ÿ " Ê #Þ&C"‡ œ &Î$ and &C#‡ œ &Î$
Ê ÐC"‡ ß C#‡ Ñ œ Ð#Î$ß "Î$Ñ.
15.4-3.
ÐC" ß C# Ñ
Ð"ß !Ñ
Ð!ß "Ñ
Expected Payoff
$B"  "Ð"  B" Ñ œ %B"  "
#B"  #Ð"  B" Ñ œ %B"  #
%B"  " œ %B"  # Ê ÐB‡" ß B‡# Ñ œ Ð $) ß &) Ñ and @ œ %Ð $) Ñ  " œ "# .
$C"‡  #C#‡ œ "# and C"‡  #C#‡ œ "# Ê ÐC"‡ ß C#‡ Ñ œ Ð "# ß "# Ñ.
The payoff matrix for player 2 is:
"
#
"
$
"
15-7
#
#
#
ÐB" ß B# Ñ
Ð"ß !Ñ
Ð!ß "Ñ
Expected Payoff
$C"  #Ð"  C" Ñ œ &C"  #
C"  #Ð"  C" Ñ œ $C"  #
&C"  # œ $C"  # Ê C"‡ œ C#‡ œ "# Þ
15.4-4.
ÐC" ß C# ß C$ Ñ
Ð"ß !ß !Ñ
Ð!ß "ß !Ñ
Ð!ß !ß "Ñ
Expected Payoff
%B"
$B"  Ð"  B" Ñ œ #B"  "
B"  #Ð"  B" Ñ œ B"  #
%B" œ B"  # Ê ÐB‡" ß B‡# Ñ œ Ð#Î&ß $Î&Ñ and @ œ )Î&.
C"‡ Ð%B" Ñ  C$‡ ÐB"  #Ñ œ )Î& for ! Ÿ B" Ÿ " Ê #C$‡ œ )Î& and %C"‡  C$‡ œ $Î&
Ê ÐC"‡ ß C#‡ ß C$‡ Ñ œ Ð"Î&ß !ß %Î&Ñ.
15.4-5.
(a)
Strategies for A.J. Team:
1- John does not swim butterfly.
2- John does not swim backstroke.
3- John does not swim breaststroke.
Strategies for G.N. Team:
1- Mark does not swim butterfly.
2- Mark does not swim backstroke.
3- Mark does not swim breaststroke.
Let the payoff entries be the total points earned in all three events by A.J. Team when a
given pair of strategies are chosen by the teams. Then the payoff matrix becomes:
15-8
"
#
$
"
"%
"$
"#
#
"$
"#
"#
$
"#
"#
"$
Strategy 2 of A.J. Team is dominated by strategy 1 and strategy 1 of G.N. Team is dominated by strategy 2. When we eliminate these strategies we obtain the table:
"
$
#
"$
"#
$
"#
"$
ÐC" ß C# Ñ
Ð"ß !Ñ
Ð!ß "Ñ
Expected Payoff
"$B"  "#Ð"  B" Ñ œ B"  "#
"#B"  "$Ð"  B" Ñ œ B"  "$
B"  "# œ B"  "$ Ê ÐB‡" ß B‡# ß B‡$ Ñ œ Ð!Þ&ß !ß !Þ&Ñ and @ œ "#Þ&.
C#‡ ÐB"  "#Ñ  C$‡ ÐB"  "$Ñ œ "#Þ& for ! Ÿ B" Ÿ " Ê "#C#‡  "$C$‡ œ "#Þ& and
"$C#‡  "#C$‡ œ "#Þ& Ê ÐC"‡ ß C#‡ ß C$‡ Ñ œ Ð!ß !Þ&ß !Þ&Ñ.
Hence, John should always swim backstroke and should swim butterfly and breaststroke
each with probability "Î#. Also, Mark should always swim butterfly and should swim
backstroke and breaststroke each with probability "Î#. Consequently, A.J. Team can
expect to get "#Þ& points on average in three events.
(b) The strategies for the two teams are the same as in (a). If :34 denotes the total points
w
earned by A.J. Team, let :34
be the new payoff that is defined as:
w
:34
œ
"Î#
"Î#
if :34 "$, i.e., if A.J. Team wins
.
if :34  "$, i.e., if A.J. Team loses
Then, the new payoff matrix becomes:
"
#
$
"
"Î#
"Î#
"Î#
#
"Î#
"Î#
"Î#
$
"Î#
"Î#
"Î#
where strategy 2 of A.J. Team is dominated by strategy 1 and strategy 1 of G.N. Team is
dominated by strategy 2. After eliminating these, the reduced payoff matrix is:
"
$
#
"Î#
"Î#
$
"Î#
"Î#
Adding the constant "#Þ& to every entry does not change the optimal strategies.
Furthermore, the payoff matrix in (a) is obtained by doing so. Hence, the best strategies
found in (a) are still optimal, the new payoff is @w œ "#Þ&  "#Þ& œ !.
(c) Since John and Mark are the best swimmers of their teams, they will always swim in
two events. Their teams cannot do better if they do not swim or if they swim in only one
event. Hence, if either one of them does not swim in the first event, namely butterfly, he
will surely swim the last two events. Accordingly, the strategies for A.J. Team are:
15-9
1- John swims butterfly and then backstroke regardless of whether Mark swims butterfly.
2- John swims butterfly and then backstroke if Mark swims butterfly, breaststroke else.
3- John swims butterfly and then breaststroke if Mark swims butterfly, backstroke else.
4- John swims butterfly and then breaststroke regardless of whether Mark swims butterfly.
5- John does not swim butterfly, swims both backstroke and breaststroke.
The strategies for G.N. Team are the same but with the roles of John and Mark are
reversed. The associated payoff matrix is:
"
#
$
%
&
"
"Î#
"Î#
"Î#
"Î#
"Î#
#
"Î#
"Î#
"Î#
"Î#
"Î#
$
"Î#
"Î#
"Î#
"Î#
"Î#
%
"Î#
"Î#
"Î#
"Î#
"Î#
&
"Î#
"Î#
"Î#
"Î#
"Î#
"
#
$
%
&
$
"Î#
"Î#
"Î#
"Î#
"Î#
Strategy 3 of G.N. Team dominates all others, by eliminating them, we obtain the payoff
matrix on the right. It shows that if G.N. Team uses strategy 3, it will win regardless of
what strategy is employed by A.J. Team.
(d) Strategy 2 of A.J. Team dominates strategies 1, 3, and 4. Thus, if the coach of G.N.
Team may choose any of their strategies at random, the coach of A.J. Team should
choose either strategy 2 or 5. After eliminating the dominated strategies, the payoff
matrix becomes:
#
&
"
"Î#
"Î#
#
"Î#
"Î#
$
"Î#
"Î#
%
"Î#
"Î#
&
"Î#
"Î#
The two rows are identical except for columns 1 and 4. Thus, if the coach of A.J. team
knows that the other coach has a tendency to enter Mark in butterfly and backstroke more
often than breaststroke, that means column 1 is more likely to be chosen than column 4,
so the coach of A.J. team should choose strategy 2.
15.5-1.
(a)
Player 1:
maximize
subject to
Player 2:
minimize
subject to
B$
B"  B#  B$ !
B"  B#  B$ !
B"  B# œ "
B" ß B# !
C$
C"  C#  C$ Ÿ !
C"  C#  C$ Ÿ !
C"  C# œ "
C" ß C# !
(b) Optimal Solution: B" œ B# œ C" œ C# œ !Þ&ß B$ œ C$ œ !
15-10
15.5-2.
After adding 3 to the entries of Table 15.6, the payoff table becomes:
"
#
"
$
)
#
"
(
$
&
!
The new linear programming problem for player 1 is:
maximize
subject to
B$
$B"  )B#  B$ !
B"  (B#  B$ !
&B"  B$ !
B"  B# œ "
B" ß B# ß B$ !
The new linear programming problem for player 2 is:
maximize
subject to
C%
$C"  C#  &C$  C% Ÿ !
)C"  (C#  C% Ÿ !
C"  C#  C$ œ "
C" ß C# ß C$ ß C%
!
Based on the information given in Section 15.5, the optimal solutions for these new
models are:
ÐB‡" ß B‡# ß B‡$ Ñ
œ Ð(Î""ß %Î""ß $&Î""Ñ
ÐC"‡ ß C#‡ ß C$‡ ß C%‡ Ñ œ Ð!ß &Î""ß 'Î""ß $&Î""Ñ.
Note that B‡$ œ C%‡ œ @  $ where @ is the value for the original version of the game.
15.5-3.
(a)
maximize
B%
subject to
&B"  #B#  $B$  B% !
%B#  #B$  B% !
$B"  $B#  B% !
B"  #B#  %B$  B% !
B"  B#  B$ œ "
B" ß B# ß B$ ß B% !
(b)
15-11
15.5-4.
(a) To insure B%
!, add $ to each entry of the payoff table.
maximize
B%
subject to
(B"  #B#  &B$  B%
&B"  $B#  'B$  B%
'B#  B$  B% !
B"  B#  B$ œ "
B" ß B# ß B$ ß B% !
!
!
(b)
15.5-5.
(a) To insure B&
!, add 4 to each entry of the payoff table.
maximize B&
subject to &B"  'B#  %B$  B& !
B"  (B#  )B$  %B%  B& !
'B"  %B#  $B$  #B%  B& !
#B"  (B#  B$  'B%  B& !
&B"  #B#  'B$  $B%  B& !
B"  B#  B$  B% œ "
B" ß B# ß B$ ß B% ß B& !
(b)
15-12
15.5-6.
Following Table 6.14, the dual of player 1's problem is:
C8"
minimize
subject to
:"" C"w  :"# C#w  â  :"8 C8w  C8"
!
w
w
w
:#" C"  :## C#  â  :#8 C8  C8"
!
ã
:7" C"w  :7# C#w  â  :78 C8w  C8"
!
w
w
w
C" 
C#  â 
C8
œ "
C3w Ÿ !, 3 œ "ß #ß á ß 8; (C8" free).
Now, let C3 œ C3w for 3 œ "ß #ß á ß 8 to get the linear program for player 2.
15.5-7.
Taking the dual of player 1's problem gives:
C%
minimize
subject to
#C#w  #C$w  C%
!
w
w
w
&C"  %C#  $C$  C%
!
w
w
w
C"  C#  C$
œ "
C"w ß C#w ß C$ Ÿ !; (C% free).
Now, let C3 œ C3w for 3 œ "ß #ß $ to get the linear program for player 2.
15.5-8.
The feasible region may be algebraically described
$Î& Ÿ B" Ÿ #Î$. The restrictions may be rewritten as:
B$ Ÿ &B"  &
B$ Ÿ 'B"  %
B$ Ÿ &B"  $
$Î& Ÿ B" Ÿ #Î$
$Î& Ÿ B" Ÿ #Î$
$Î& Ÿ B" Ÿ #Î$
15-13
by:
B# œ "  B"
and
'B"  % œ &B"  $ Ê B" œ (Î"".
Therefore, the algebraic expression for the maximizing value of B$ for any point in the
feasible region is:
B$ œ 
&B"  $
'B"  %
for $Î& Ÿ B" Ÿ (Î""
for (Î"" Ÿ B" Ÿ #Î$
Hence, the optimal solution is:
ÐB‡" ß B‡# ß B‡$ Ñ œ Ð(Î""ß "  (Î""ß &Ð(Î""Ñ  $Ñ œ Ð(Î""ß %Î""ß #Î""Ñ.
15.5-9.
Optimal primal solution: ÐB" ß B# Ñ œ Ð!Þ'$'ß !Þ$'%Ñ with a payoff of !Þ")#
Optimal dual solution: ÐC" ß C# ß C$ Ñ œ Ð!ß !Þ%&&ß !ß &%&Ñ
15-14
15.5-10.
(a) Since the saddle points can be found by linear programming, (a) follows from (b).
(b) Consider the linear programming formulation of the problem for player 2. The 3th and
5th constraints are:
:3" C"  :3# C#  â  :38 C8 Ÿ C8"
:5" C"  :5# C#  â  :58 C8 Ÿ C8"
If row 5 weakly dominates row 3, then
:3" C"  :3# C#  â  :38 C8 Ÿ :5" C"  :5# C#  â  :58 C8
for every C" ß á ß C8 . In that case, the 3th constraint is redundant, as it is implied by the
5th constraint. Hence, eliminating dominated pure strategies for player 1 corresponds to
eliminating redundant constraints from the linear program for player 2. Similarly,
eliminating dominated strategies of player 2 is equivalent to eliminating redundant
constraints of player 1's linear program. Since this process cannot eliminate any feasible
solutions or create new ones, all optimal strategies are preserved and no new ones are
added.
15-15
CHAPTER 1': DECISION ANALYSIS
16.2-1.
Phillips Petroleum Company developed a decision analysis tool named DISCOVERY to
evaluate available investment opportunities and decide on the participation levels. The
need for a systematic decision analysis tool arose from the uncertainty associated with
various alternatives, the lack of a consistent risk measure across the organization and the
scarcity of capital resources. The notion of risk is incorporated in the model with the use
of risk-averse exponential utility function. The objective is to maximize expected utility
rather than expected return. DISCOVERY provides a decision-tree display of available
alternatives at various participation levels. A simple version of the problem is one where
Phillips needs to decide first on the participation level and second on whether to drill or
not. The exploration of petroleum when drilled is uncertain. The analysis is performed for
different levels of risk-aversion and the sensitivity of the decisions to the risk-aversion
level is observed. When additional seismic information is available at a cost, the value of
information is computed.
This study "has increased management's awareness of risk and risk tolerance, provided
insight into the financial risks associated with its set of investment opportunities, and
provided the company a formalized decision model for allocating scarce capital" [p. 55].
The software package developed has been a valuable aid in decision making. It provided
a systematic treatment of risk and uncertainty. Other petroleum exploration firms started
to use DISCOVERY in analyzing decisions, too.
16.2-2.
(a)
Alternative
Build Computers
Sell Rights
State of Nature
Sell "!ß !!! Sell "!!ß !!!
!
&%
"&
"&
(b)
1'-1
(c) Let : be the prior probability of selling 10,000 computers.
Build: EP œ :Ð!Ñ  Ð"  :ÑÐ&%Ñ œ &%:  &%
EP œ :Ð"&Ñ  Ð"  :ÑÐ"&Ñ œ "&
Sell:
The expected profit for Build and Sell is the same when &%:  &% œ "& Ê : œ !Þ(##.
They should build when : Ÿ !Þ(## and sell if : !Þ(##.
(d)
(e)
50%
Sell 10,000
0
Build Computers
-6
27
6
0
50%
Sell 100,000
54
1
60
54
27
Sell Rights
15
15
15
Building computers should be chosen, since it has an expected payoff of $27 million.
16.2-3.
(a)
Alternative
Buy 12 Cases
Buy 13 Cases
Buy 14 Cases
Buy 15 Cases
Prior Probability
Sell 12 Cases
1$#
"#&
"")
"""
!Þ"
State of Nature
Sell 13 Cases Sell 14 Cases
"$#
"$#
"%$
"%$
"$'
"&%
"#*
"%(
!Þ$
!Þ%
1'-2
Sell 15 Cases
"$#
"%$
"&%
"'&
!Þ#
(b) According to the maximin payoff criterion, Jean should purchase 12 cases.
Alternative
Buy 12 Cases
Buy 13 Cases
Buy 14 Cases
Buy 15 Cases
Prior Probability
Sell 12 Cases
1$#
"#&
"")
"""
!Þ"
State of Nature
Sell 13 Cases Sell 14 Cases
"$#
"$#
"%$
"%$
"$'
"&%
"#*
"%(
!Þ$
!Þ%
Sell 15 Cases
"$#
"%$
"&%
"'&
!Þ#
Min
"$#
"#&
"")
"""
(c) She will be able to sell 14 cases with highest probability and the maximum possible
profit from selling 14 cases is earned when she buys 1% cases. Hence, according to the
maximum likelihood criterion, Jean should purchase 1% cases.
(d) According to Bayes' decision rule, Jean should purchase 1% cases.
Alternative
Buy 12 Cases
Buy 13 Cases
Buy 14 Cases
Buy 15 Cases
Prior Probability
(e)
Sell 12 Cases
1$#
"#&
"")
"""
!Þ"
State of Nature
Sell 13 Cases Sell 14 Cases
"$#
"$#
"%$
"%$
"$'
"&%
"#*
"%(
!Þ$
!Þ%
Sell 15 Cases
"$#
"%$
"&%
"'&
!Þ#
Exp.
Profit
"$#
"%"Þ#
"%&
"%"Þ'
!Þ# and !Þ&: Jean should purchase 1% cases.
Alternative
Buy 12 Cases
Buy 13 Cases
Buy 14 Cases
Buy 15 Cases
Prior Probability
Sell 12 Cases
1$#
"#&
"")
"""
!Þ"
State of Nature
Sell 13 Cases Sell 14 Cases
"$#
"$#
"%$
"%$
"$'
"&%
"#*
"%(
!Þ#
!Þ&
Sell 15 Cases
"$#
"%$
"&%
"'&
!Þ#
Exp.
Profit
"$#
"%"Þ#
"%'Þ)
"%$Þ%
!Þ% and !Þ$: Jean should purchase 1% cases.
Alternative
Buy 12 Cases
Buy 13 Cases
Buy 14 Cases
Buy 15 Cases
Prior Probability
Sell 12 Cases
1$#
"#&
"")
"""
!Þ"
State of Nature
Sell 13 Cases Sell 14 Cases
"$#
"$#
"%$
"%$
"$'
"&%
"#*
"%(
!Þ%
!Þ$
Sell 15 Cases
"$#
"%$
"&%
"'&
!Þ#
Exp.
Profit
"$#
"%"Þ#
"%$Þ#
"$*Þ)
!Þ& and !Þ#: Jean should purchase 1% cases.
Alternative
Buy 12 Cases
Buy 13 Cases
Buy 14 Cases
Buy 15 Cases
Prior Probability
Sell 12 Cases
1$#
"#&
"")
"""
!Þ"
State of Nature
Sell 13 Cases Sell 14 Cases
"$#
"$#
"%$
"%$
"$'
"&%
"#*
"%(
!Þ&
!Þ#
1'-3
Sell 15 Cases
"$#
"%$
"&%
"'&
!Þ#
Exp.
Profit
"$#
"%"Þ#
"%"Þ%
"$)
16.2-4.
(a) The optimal (maximin) actions are conservative and countercyclical investments, both
incur a loss of $"! million in the worst case.
(b) The economy is most likely to be stable and the alternative with the highest profit in
this state of nature is to make a speculative investment. According to the maximum
likelihood criterion, Warren should choose speculative investment.
(c) To maximize his expected payoff, Warren should make a countercyclical investment.
Alternative
Conservative
Speculative
Countercyclical
Prior Probability
State of Nature
Improving Stable Worsening
$!
&
"!
%!
"!
$!
"!
!
"&
!Þ"
!Þ&
!Þ%
Exp.
Profit
"Þ&
$
&
16.2-5.
(a) Warren should make a countercyclical investment.
Alternative
Conservative
Speculative
Countercyclical
Prior Probability
State of Nature
Improving Stable Worsening
$!
&
"!
%!
"!
$!
"!
!
"&
!Þ"
!Þ$
!Þ'
Exp.
Profit
"Þ&
""
)
(b) Warren should make a speculative investment.
Alternative
Conservative
Speculative
Countercyclical
Prior Probability
State of Nature
Improving Stable Worsening
$!
&
"!
%!
"!
$!
"!
!
"&
!Þ"
!Þ(
!Þ#
1'-4
Exp.
Profit
%Þ&
&
#
(c) The expected profit from countercyclical and conservative investments is the same
when : ¸ !Þ'#. The expected profit lines for conservative and speculative investments
cross at : ¸ !Þ'). Those for countercyclical and speculative investments cross at
: ¸ !Þ'&; however, this crossover point does not result in a decision shift.
(d) Let : be the prior probability of stable economy.
Conservative:
Speculative:
Countercyclical:
EP œ Ð!Þ"ÑÐ$!Ñ  :Ð&Ñ  Ð"  !Þ"  :ÑÐ"!Ñ œ "&:  '
EP œ Ð!Þ"ÑÐ%!Ñ  :Ð"!Ñ  Ð"  !Þ"  :ÑÐ$!Ñ œ %!:  #$
EP œ Ð!Þ"ÑÐ"!Ñ  :Ð!Ñ  Ð"  !Þ"  :ÑÐ"&Ñ œ "&:  "#Þ&
Countercyclical and conservative cross when "&:  "#Þ& œ "&:  ' Ê : œ !Þ'"(.
Conservative and speculative cross when "&:  ' œ %!:  #$ Ê : œ !Þ').
Accordingly, Warren should choose countercyclical investment when :  !Þ'"(,
conservative investment when !Þ'"( Ÿ :  !Þ') and speculative investment when
: !Þ').
(e)
1'-5
16.2-6.
(a) A2 should be chosen.
Alternative
A"
A2
Prior Probability
State of Nature
W"
W#
W$
##! "(! ""!
#!! ")! "&!
!Þ' !Þ$ !Þ"
Min
""!
"&!
(b) The most likely state of nature is W" and the alternative with highest profit in this state
is E" .
(c) A" should be chosen.
Alternative
A"
A2
Prior Probability
State of Nature
W"
W#
W$
##! "(! ""!
#!! ")! "&!
!Þ' !Þ$ !Þ"
Exp.
Payoff
"*%
")*
(d)
Let : be the prior probability of W".
A1 :
A2 :
EP œ :Ð##!Ñ  Ð"  !Þ"  :ÑÐ"(!Ñ  Ð!Þ"ÑÐ""!Ñ œ &!:  "'%
EP œ :Ð#!!Ñ  Ð"  !Þ"  :ÑÐ")!Ñ  Ð!Þ"ÑÐ"&!Ñ œ #!:  "((
A1 and A2 cross when &!:  "'% œ #!:  "(( Ê : œ !Þ%$$. They should choose A2
when : Ÿ !Þ%$$ and A1 if :  !Þ%$$.
1'-6
(e)
Let : be the prior probability of W".
A1 :
A2 :
EP œ :Ð##!Ñ  Ð!Þ$ÑÐ"(!Ñ  Ð"  !Þ$  :ÑÐ""!Ñ œ ""!:  "#)
EP œ :Ð#!!Ñ  Ð!Þ$ÑÐ")!Ñ  Ð"  !Þ$  :ÑÐ"&!Ñ œ &!:  "&*
A1 and A2 cross when ""!:  "#) œ &!:  "&* Ê : œ !Þ&"(. They should choose A2
when : Ÿ !Þ&"( and A1 if :  !Þ&"(.
(f)
Let : be the prior probability of W#.
A1 :
A2 :
EP œ Ð!Þ'ÑÐ##!Ñ  :Ð"(!Ñ  Ð"  !Þ'  :ÑÐ""!Ñ œ '!:  "('
EP œ Ð!Þ'ÑÐ#!!Ñ  :Ð")!Ñ  Ð"  !Þ'  :ÑÐ"&!Ñ œ $!:  ")!
A1 and A2 cross when '!:  "(' œ $!:  ")! Ê : œ !Þ"$$. They should choose A2
when : Ÿ !Þ"$$ and A1 if :  !Þ"$$.
(g) A1 should be chosen.
1'-7
16.2-7.
(a)
Alternative
Crop 1
Crop 2
Crop 3
Crop 4
Prior Probability
State of Nature
Dry Moderate Damp
#!
$&
%!
##Þ&
$!
%&
$!
#&
#&
#!
#!
#!
!Þ$
!Þ&
!Þ#
(b) Grow Crop 1.
Alternative
Crop 1
Crop 2
Crop 3
Crop 4
Prior Probability
(c)
Dry
#!
##Þ&
$!
#!
!Þ$
State of Nature
Moderate Damp
$&
%!
$!
%&
#&
#&
#!
#!
!Þ&
!Þ#
Exp.
Payoff
$"Þ&
$!Þ(&
#'Þ&
#!
Prior probability of moderate weather œ !Þ#: Grow Crop 2.
State of Nature
Exp.
Alternative
Dry Moderate Damp Payoff
Crop 1
#!
$&
%!
$$
Crop 2
##Þ&
$!
%&
$'Þ#&
Crop 3
$!
#&
#&
#'Þ&
Crop 4
#!
#!
#!
#!
Prior Probability !Þ$
!Þ#
!Þ&
Prior probability of moderate weather œ !Þ$: Grow Crop 2.
State of Nature
Exp.
Alternative
Dry Moderate Damp Payoff
Crop 1
#!
$&
%!
$#Þ&
Crop 2
##Þ&
$!
%&
$$Þ(&
Crop 3
$!
#&
#&
#'Þ&
Crop 4
#!
#!
#!
#!
Prior Probability !Þ$
!Þ$
!Þ%
Prior probability of moderate weather œ !Þ%: Grow Crop #.
State of Nature
Exp.
Alternative
Dry Moderate Damp Payoff
Crop 1
#!
$&
%!
$#
Crop 2
##Þ&
$!
%&
$#Þ#&
Crop 3
$!
#&
#&
#'Þ&
Crop 4
#!
#!
#!
#!
Prior Probability !Þ$
!Þ%
!Þ$
1'-8
Prior probability of moderate weather œ !Þ': Grow Crop 1.
State of Nature
Exp.
Alternative
Dry Moderate Damp Payoff
Crop 1
#!
$&
%!
$"
Crop 2
##Þ&
$!
%&
#*Þ#&
Crop 3
$!
#&
#&
#'Þ&
Crop 4
#!
#!
#!
#!
Prior Probability !Þ$
!Þ'
!Þ"
16.2-8.
The prior distribution is PÖ) œ )" × œ #Î$, PÖ) œ )# × œ "Î$.
Order 15: EP œ #Î$Ð"Þ"&& † "!( Ñ  "Î$Ð"Þ%"% † "!( Ñ œ "Þ#%" † "!(
Order 20: EP œ #Î$Ð"Þ!"# † "!( Ñ  "Î$Ð"Þ#!( † "!( Ñ œ "Þ!(( † "!(
Order 25: EP œ #Î$Ð"Þ!%( † "!( Ñ  "Î$Ð"Þ"$& † "!( Ñ œ "Þ!(' † "!(
The maximum expected profit, or equivalently the minimum expected cost is that of
ordering 25, so the optimal decision under Bayes' decision rule is to order 25.
16.3-1.
This article describes the use of decision analysis at the Workers' Compensation Board of
British Columbia (WCB), which is "responsible for the occupational health and safety,
rehabilitation, and compensation interests of British Columbia's workers and employers"
[p. 15]. The focus of the study is on the short-term disability claims that can later turn
into long-term disability claims and can be very costly for the WCB. First, logistic
regression is employed to estimate the probability of conversion for each claim. Then
using decision analysis, a threshold is determined to classify the claims as high- and lowrisk claims. For any fixed conversion probability, the problem consists of a simple
decision tree. First the WCB chooses between classifying the claim as high risk or low
risk and then whether the claim converts or not determines the actual cost. If the claim is
identified as a high-risk claim, the WCB intervenes. The early intervention lowers the
costs and ensures faster rehabilitation. The expected total cost is computed for various
cutoff points and the point with minimum expected cost is identified as the optimal
threshold.
The new policy offers accurate predictions of high-risk claims. As a result, future costs
are reduced and injured workers start working sooner. This study is expected to save the
WCB $4.7 per year. The scorecard system developed to implement the new policy
improved the efficiency of claim management and the productivity of staff. Overall, the
benefits accrued from this study paved the way for the WCB's adoption of operations
research in other aspects of the organization.
1'-9
16.3-2.
(a)
Alternative
Build Computers
Sell Rights
Prior Probability
Maximum Payoff
State of Nature
Sell "!ß !!! Sell "!!ß !!!
!
&%
"&
"&
!Þ&
!Þ&
"&
&%
Expected Payoff with Perfect Information: !Þ&Ð"&Ñ  !Þ&Ð&%Ñ œ $%Þ&
Expected Payoff without Information: !Þ&Ð!Ñ  !Þ&Ð&%Ñ œ #(
EVPI œ $%Þ&  #( œ $(Þ& million
(b) Since the market research will cost $" million, it might be worthwhile to perform it.
(c)
1'-10
(d)
Data:
State of
Nature
Sell 10,000
Sell 100,000
Posterior
Probabilities:
Finding
Predict Sell 10,000
Predict Sell 100,000
Prior
Probability
0.5
0.5
Predict Sell 10,000
0.667
0.333
P(Finding)
0.5
0.5
Sell 10,000
0.667
0.333
P(Finding | State)
Finding
Predict Sell 100,000
0.333
0.667
P(State | Finding)
State of Nature
Sell 100,000
0.333
0.667
(e) EVE œ Ò!Þ&Ð")!!Ñ  !Þ&Ð$'!!ÑÓ  #(!! œ !, so performing the market research is
not worthwhile.
16.3-3.
(a) Choose A# with expected payoff $"!!!.
Alternative
A"
A2
A$
Prior Probability
State of Nature
W"
W # W$
%
!
!
!
#
!
$
!
"
!Þ# !Þ& !Þ$
Exp.
Payoff
!Þ)
"Þ!
!Þ*
(b)
Alternative
A"
A2
A$
Prior Probability
Maximum Payoff
State of Nature
W"
W # W$
%
!
!
!
#
!
$
!
"
!Þ# !Þ& !Þ$
%
#
"
Expected Payoff with Perfect Information: !Þ#Ð%Ñ  !Þ&Ð#Ñ  !Þ$Ð"Ñ œ #Þ"
Expected Payoff without Information: "Þ!
EVPI œ #Þ"  "Þ! œ $"Þ" thousandÞ
(c) Since the information will cost $"!!! and the value is $""!!, it might be worthwhile
to spend the money.
16.3-4.
(a) Choose A1 with expected payoff $$&.
Alternative
A"
A2
A$
Prior Probability
State of Nature
W"
W#
W$
&! "!! "!!
!
"! "!
#!
%! %!
!Þ& !Þ$
!Þ#
1'-11
Exp.
Payoff
$&
"
"%
(b)
Alternative
A"
A2
A$
Prior Probability
Maximum Payoff
State of Nature
W"
W#
W$
&! "!! "!!
!
"! "!
#!
%! %!
!Þ& !Þ$
!Þ#
&! "!! "!
Expected Payoff with Perfect Information: !Þ&Ð&!Ñ  !Þ$Ð"!!Ñ  !Þ#Ð"!Ñ œ &$
Expected Payoff without Information: $&
EVPI œ &$  $& œ $")
(c) Betsy should consider spending up to $") to obtain more information.
16.3-5.
(a) Choose A3 with expected payoff $$&ß !!!.
Alternative
A"
A2
A$
Prior Probability
State of Nature
W"
W # W$
"!! "! "!!
"! #! &!
"! "! '!
!Þ#
!Þ$ !Þ&
Exp.
Payoff
$$
#*
$&
(b) If S1 occurs for certain, then choose A3 with expected payoff $"!ß !!!. If S1 does not
occur for certain, then the probability that S2 will occur is $) and the probability that S3
will occur is &) .
A1 :
A2 :
A3 :
Ð $) ÑÐ"!Ñ  Ð &) ÑÐ"!!Ñ œ ''Þ#&
Ð $) ÑÐ#!Ñ  Ð &) ÑÐ&!Ñ œ $)Þ(&
Ð $) ÑÐ"!Ñ  Ð &) ÑÐ'!Ñ œ %"Þ#&
Hence, choose A1 which offers an expected payoff of $''ß #&!.
Expected Payoff with Information: !Þ#Ð"!Ñ  !Þ)Ð''Þ#&Ñ œ &&
Expected Payoff without Information: $&
EVI œ &&  $& œ $#! thousand
The maximum amount that should be paid for the information is $#!ß !!!. The decision
with this information will be to choose A$ if the state of nature is W" and E" otherwise.
(c) If S2 occurs for certain, then choose A2 with expected payoff $#!ß !!!. If S2 does not
occur for certain, then the probability that S1 will occur is #( and the probability that S3
will occur is &( .
A1 :
A2 :
A3 :
Ð #( ÑÐ"!!Ñ  Ð &( ÑÐ"!!Ñ œ %#Þ)&(
Ð #( ÑÐ"!Ñ  Ð &( ÑÐ&!Ñ œ $#Þ)&(
Ð #( ÑÐ"!Ñ  Ð &( ÑÐ'!Ñ
œ %&Þ("%
Hence, choose A$ which offers an expected payoff of $%&ß ("%.
1'-12
Expected Payoff with Information: !Þ$Ð#!Ñ  !Þ(Ð%&Þ("%Ñ œ $)
Expected Payoff without Information: $&
EVI œ $)  $& œ $$ thousand
The maximum amount that should be paid for the information is $$!!!. The decision
with this information will be to choose A# if the state of nature is W# and E$ otherwise.
(d) If S3 occurs for certain, then choose A1 with expected payoff $"!!ß !!!. If S3 does
not occur for certain, then the probability that S1 will occur is #& and the probability that
S2 will occur is $& .
A1 :
A2 :
A3 :
Ð #& ÑÐ"!!Ñ  Ð $& ÑÐ"!!Ñ œ $%
Ð #& ÑÐ"!Ñ  Ð $& ÑÐ#!Ñ œ )
Ð #& ÑÐ"!Ñ  Ð $& ÑÐ"!Ñ
œ "!
Hence, choose A$ which offers an expected payoff of $"!ß !!!.
Expected Payoff with Information: !Þ&Ð"!!Ñ  !Þ&Ð"!Ñ œ &&
Expected Payoff without Information: $&
EVI œ &&  $& œ $#! thousand
The maximum amount that should be paid for the information is $%ß !!!. The decision
with this information will be to choose A" if the state of nature is W$ and E$ otherwise.
(e)
Expected Payoff with Perfect Information: !Þ#Ð"!Ñ  !Þ$Ð#!Ñ  !Þ&Ð"!!Ñ œ &)
Expected Payoff without Information: $&
EVPI œ &)  $& œ $#$ thousand
The maximum amount that should be paid for the information is $#$ß !!!. The decision
with this information will be to choose A$ if the state of nature is W" , A# if the state of
nature is W# and E" otherwise.
(f) The maximum amount that should be paid for testing is $#$ß !!!, since any additional
information cannot add more value than perfect information.
1'-13
16.3-6.
(a)
(b)
Prior
Probability
0.25
0.75
P(Finding | State)
Finding
FSS
USS
0.8
0.2
0.4
0.6
Posterior
Probabilities:
Finding
P(Finding)
FSS
0.5
USS
0.5
P(State | Finding)
State of Nature
Oil
Dry
0.4
0.6
0.1
0.9
Data:
State of
Nature
Oil
Dry
(c) The optimal policy is to do a seismic survey, to drill if favorable seismic surroundings
are obtained, and to sell if unfavorable surroundings are obtained.
16.3-7.
(a) Choose A1 with expected payoff $"!!.
Alternative
A"
A2
Prior Probability
State of Nature
W"
W#
%!!
"!!
!
"!!
!Þ%
!Þ'
Exp.
Payoff
"!!
'!
1'-14
(b)
Alternative
A"
A2
Prior Probability
Maximum Payoff
State of Nature
W"
W#
%!!
"!!
!
"!!
!Þ%
!Þ'
%!!
"!!
Expected Payoff with Perfect Information: !Þ%Ð%!!Ñ  !Þ'Ð"!!Ñ œ ##!
Expected Payoff without Information: "!!
EVPI œ ##!  "!! œ $"#!, so it might be worthwhile to do the research.
(c) Let \ denote the state of nature and ] denote the prediction. From Bayes' Rule,
P(\ œ B and ] œ C) œ P(\ œ B)P(] œ Cl\ œ B).
(i)
(ii)
(iii)
(iv)
P(\
P(\
P(\
P(\
œ S1 and ]
œ S1 and ]
œ S2 and ]
œ S2 and ]
œ S1 ) œ Ð!Þ%ÑÐ!Þ'Ñ œ !Þ#%
œ S2 ) œ Ð!Þ%ÑÐ!Þ%Ñ œ !Þ"'
œ S1 ) œ Ð!Þ'ÑÐ!Þ#Ñ œ !Þ"#
œ S2 ) œ Ð!Þ'ÑÐ!Þ)Ñ œ !Þ%)
(d) P(S1 ) œ !Þ#%  !Þ"# œ !Þ$', P(S2 ) œ !Þ"'  !Þ%) œ !Þ'%
(e) Bayes' Rule: P(\ œ Bl] œ C) œ
P(\œB and ] œC)
P(\œBÑ
P(S1 lS1 ) œ !Þ#%Î!Þ$' œ !Þ''(
P(S1 lS2 ) œ !Þ"'Î!Þ'% œ !Þ#&
P(S2 lS1 ) œ !Þ"#Î!Þ$' œ !Þ$$$
P(S2 lS2 ) œ !Þ%)Î!Þ'% œ !Þ(&
(f)
Prior
Probability
0.4
0.6
P(Finding | State)
Finding
Predict S1 Predict S2
0.6
0.4
0.2
0.8
Posterior
Probabilities:
Finding
P(Finding)
Predict S1
0.36
Predict S2
0.64
P(State | Finding)
State of Nature
Actual S1 Actual S2
0.667
0.333
0.250
0.750
Data:
State of
Nature
Actual S1
Actual S2
(g) If S1 is predicted, then choose A1 with expected payoff $#$$Þ$$.
Alternative
A"
A2
Prior Probability
State of Nature
W"
W#
%!!
"!!
!
"!!
!Þ''( !Þ$$$
Exp.
Payoff
#$$Þ&
$$Þ$
1'-15
(h) If S2 is predicted, then choose A2 with expected payoff $(&.
Alternative
A"
A2
Prior Probability
State of Nature
W"
W#
%!!
"!!
!
"!!
!Þ#&
!Þ(&
Exp.
Payoff
#&
(&
(i) Given that the research is done, the expected payoff is
Ð!Þ$'ÑÐ#$$Þ$$Ñ  Ð!Þ'%ÑÐ(&Ñ  "!! œ $$#.
(j) The optimal policy is to not do research and to choose A1 .
16.3-8.
(a) EVPI œ ÒÐ#Î$ÑÐ"Þ!"# † "!( Ñ  Ð"Î$ÑÐ"Þ"$& † "!( ÑÓ  Ð"Þ!(' † "!( Ñ
œ $#$!ß !!!.
(b)
P() œ #"l 30 spares required) œ
œ
P(30 spares required l) œ#")P() œ#")
P(30 spares required l) œ#")P() œ#")P(30 spares required l) œ#%)P() œ#%)
Ð!Þ!"$ÑÐ#Î$Ñ
Ð!Þ!"$ÑÐ#Î$ÑÐ!Þ!$'ÑÐ"Î$Ñ
œ !Þ%"*
P() œ #%l 30 spares required) œ "  !Þ%"* œ !Þ&)"
Order 15:
Order 20:
Order 25:
EP œ !Þ%"*Ð"Þ"&& † "!( Ñ  !Þ&)"Ð"Þ%"% † "!( Ñ œ "Þ$!& † "!(
EP œ !Þ%"*Ð"Þ!"# † "!( Ñ  !Þ&)"Ð"Þ#!( † "!( Ñ œ "Þ"#& † "!(
EP œ !Þ%"*Ð"Þ!%( † "!( Ñ  !Þ&)"Ð"Þ"$& † "!( Ñ œ "Þ!*) † "!(
The optimal alternative is to order 25.
16.3-9.
(a)
Alternative
Extend Credit
Not Extend Credit
Prior Probability
State of Nature
Poor Risk Average Risk
"&ß !!!
"!ß !!!
!
!
!Þ#
!Þ&
Good Risk
#!ß !!!
!
!Þ$
(b) Choose to extend credit with expected payoff $)ß !!!.
Alternative
Extend Credit
Not Extend Credit
Prior Probability
Poor Risk
"&ß !!!
!
!Þ#
State of Nature
Average Risk
"!ß !!!
!
!Þ&
1'-16
Good Risk
#!ß !!!
!
!Þ$
Exp.
Payoff
)ß !!!
!
(c)
Alternative
Extend Credit
Not Extend Credit
Prior Probability
Maximum Payoff
State of Nature
Poor Risk Average Risk
"&ß !!!
"!ß !!!
!
!
!Þ#
!Þ&
!
"!ß !!!
Expected Payoff with Perfect Information:
!Þ#Ð!Ñ  !Þ$Ð"!ß !!!Ñ  !Þ%Ð#!ß !!!Ñ œ ""ß !!!
Expected Payoff without Information: )ß !!!
EVPI œ ""ß !!!  )ß !!! œ $$ß !!!
Hence, the credit-rating organization should not be used.
(d)
1'-17
Good Risk
#!ß !!!
!
!Þ$
#!ß !!!
(e)
Data:
State of
Nature
Poor
Average
Good
Prior
Probability
0.2
0.5
0.3
Posterior
Probabilities:
Finding
P(Finding)
Poor
0.360
Average
0.450
Good
0.190
Poor
0.5
0.4
0.2
P(Finding | State)
Finding
Average
Good
0.4
0.1
0.5
0.1
0.4
0.4
Poor
0.278
0.178
0.105
P(State | Finding)
State of Nature
Average
Good
0.556
0.167
0.556
0.267
0.263
0.632
(f) Vincent should not get the credit rating and extend credit.
16.3-10.
(a) Given that the test is positive, the athlete is a drug user with probability !Þ'()'.
(b) Given that the test is positive, the athlete is not a drug user with probability !Þ$#"%.
(c) Given that the test is negative, the athlete is a drug user with probability !Þ!!&).
(d) Given that the test is negative, the athlete is not a drug user with probability !Þ**%#.
1'-18
(e) The answers in Excel agree with those found in parts (a), (b), (c), and (d).
Prior
Probability
0.1
0.9
P(Finding | State)
Finding
Positive Negative
0.95
0.05
0.05
0.95
Posterior
Probabilities:
Finding
P(Finding)
Positive
0.14
Negative
0.86
P(State | Finding)
State of Nature
User
Nonuser
0.6786
0.3214
0.0058
0.9942
Data:
State of
Nature
User
Nonuser
1'Þ3-11.
(a)
Alternative
Develop New Product
Not Develop New Product
Prior Probability
State of Nature
Successful Unsuccessful
"ß &!!ß !!! "ß )!!ß !!!
!
!
!Þ''(
!Þ$$$
(b) Choose to develop new product with expected payoff $%!!ß !!!.
Alternative
Develop New Product
Not Develop New Product
Prior Probability
State of Nature
Successful Unsuccessful
"ß &!!ß !!! "ß )!!ß !!!
!
!
!Þ''(
!Þ$$$
Alternative
Develop New Product
Not Develop New Product
Prior Probability
Maximum Payoff
State of Nature
Successful Unsuccessful
"ß &!!ß !!! "ß )!!ß !!!
!
!
!Þ''(
!Þ$$$
"ß &!!ß !!!
!
Exp.
Payoff
%!!ß !!!
!
(c)
Expected Payoff with Perfect Information: !Þ''(Ð"ß &!!ß !!!Ñ  !Þ$$$Ð!Ñ œ "ß !!!ß !!!
Expected Payoff without Information: %!!ß !!!
EVPI œ "ß !!!ß !!!  %!!ß !!! œ $'!!ß !!!
This indicates that consideration should be given to conducting the market survey.
1'-19
(d)
Data:
State of
Nature
Successful
Unsuccessful
Prior
Probability
0.667
0.333
Posterior
Probabilities:
Finding
Predict Successful
Predict Unsuccessful
P(Finding)
0.633
0.367
P(Finding | State)
Finding
Predict Successful Predict Unsuccessful
0.8
0.2
0.3
0.7
Successful
0.8421
0.3636
P(State | Finding)
State of Nature
Unsuccessful
0.1579
0.6364
(e)
Action
Develop product
Not develop product
Develop product
Not develop product
Prediction
Successful
Successful
Unsuccessful
Unsuccessful
Expected Payoff
Ò!Þ)%#"Ð"Þ&Ñ  !Þ"&(*Ð"Þ)ÑÓ † "!' œ $*(*ß !!!
!
Ò!Þ$'$'Ð"Þ&Ñ  !Þ'$'%Ð"Þ)ÑÓ † "!' œ $'!!ß !!!
!
It is optimal to develop the product if it is predicted to be successful and to not develop
otherwise. Let S be the event that the product is predicted to be successful. Then,
P(S) œ P(Sl)" )P()" )  P(Sl)# )P()# ) œ !Þ)Ð#Î$Ñ  !Þ#Ð"Î$Ñ œ !Þ'.
The expected payoff given the information is !Þ'Ð*(*ß !!!Ñ  !Þ%Ð!Ñ œ $&)(ß !!!, so
EVE œ &)(ß !!!  %!!ß !!! œ $")(ß !!!  $$!!ß !!! œ Cost of survey.
Hence, the optimal strategy is to not conduct the market survey, and to market the
product.
16.3-12.
(a)
Alternative
Screen
Not Screen
Prior Probability
State of Nature
: œ !Þ!& : œ !Þ#&
"ß &!! "ß &!!
(&! $ß (&!
!Þ)
!Þ#
(b) Choose to not screen with expected loss $"ß $&!.
Alternative
Screen
Not Screen
Prior Probability
State of Nature
: œ !Þ!& : œ !Þ#&
"ß &!! "ß &!!
(&! $ß (&!
!Þ)
!Þ#
1'-20
Exp.
Payoff
"ß &!!
"ß $&!
(c)
Alternative
Screen
Not Screen
Prior Probability
Maximum Payoff
State of Nature
: œ !Þ!& : œ !Þ#&
"ß &!! "ß &!!
(&! $ß (&!
!Þ)
!Þ#
(&! "ß &!!
Expected Payoff with Perfect Information: !Þ)Ð(&!Ñ  !Þ#Ð"ß &!!Ñ œ *!!
Expected Payoff without Information: "ß $&!
EVPI œ *!!  Ð"ß $&!Ñ œ $%&!
This indicates that consideration should be given to inspecting the single item.
(d)
Data:
State of
Nature
p = 0.05
p = 0.25
Posterior
Probabilities:
Finding
Defective
Nondefective
(e)
Prior
Probability
0.8
0.2
P(Finding)
0.09
0.91
Defective
0.05
0.25
P(Finding | State)
Finding
Nondefective
0.95
0.75
p = 0.05
0.4444
0.8352
P(State | Finding)
State of Nature
p = 0.25
0.5556
0.1648
P(defective) œ Ð!Þ!&ÑÐ!Þ)Ñ  Ð!Þ#&ÑÐ!à #Ñ œ !Þ!* and P(nondefective) œ !Þ*"
EVE œ ÒÐ!Þ!*ÑÐ"&!!Ñ  Ð!Þ*"ÑÐ"#%&ÑÓ  Ð"$&!Ñ œ )#Þ!&
Since the cost of the inspection is $"#&  $)#Þ!&, inspecting the single item is not worthwhile.
(f) If defective:
EP(screen, )ldefective) œ !Þ%%%Ð"&!!Ñ  !Þ&&'Ð"&!!Ñ œ "&!!
EP(no screen, )ldefective) œ !Þ%%%Ð(&!Ñ  !Þ&&'Ð$(&!Ñ œ #%")
If nondefective:
EP(screen, )ldefective) œ "&!!
EP(no screen, )ldefective) œ !Þ)$&Ð(&!Ñ  !Þ"'&Ð$(&!Ñ œ "#%&
Hence, the optimal policy with experimentation is to screen if defective is found and not
screen if nondefective is found. On the other hand, from part (e), inspecting a single item,
in other words experimenting is not worthwhile. Using part (b), the overall optimal
policy is to not inspect the single items, to not screen each item in the lot, instead, rework
each item that is ultimately found to be defective.
1'-21
16.3-13.
(a)
Say coin 1 tossed:
Say coin 2 tossed:
EP œ !Þ'Ð!Ñ  !Þ%Ð"Ñ œ !Þ%
EP œ !Þ'Ð"Ñ  !Þ%Ð!Ñ œ !Þ'
The optimal alternative is to say coin 1 is tossed.
(b) If the outcome is heads (H):
P(coin 1lH) œ
P(Hlcoin 1)P(coin 1)
P(Hlcoin 1)P(coin 1)P(Hlcoin 2)P(coin 2)
P(coin 2lH) œ
%
(
Say coin 1:
EP œ $( Ð!Ñ  %( Ð"Ñ œ  %(
Say coin 2:
EP œ $( Ð"Ñ  %( Ð!Ñ œ  $(
œ
!Þ$Ð!Þ'Ñ
!Þ$Ð!Þ'Ñ!Þ'Ð!Þ%Ñ
œ
$
(
The optimal alternative is to say coin 2.
If the outcome is tails (T):
P(coin 1lT) œ
P(Tlcoin 1)P(coin 1)
P(Tlcoin 1)P(coin 1)P(Tlcoin 2)P(coin 2)
œ
!Þ(Ð!Þ'Ñ
!Þ(Ð!Þ'Ñ!Þ%Ð!Þ%Ñ
œ !Þ(#%"
P(coin 2lT) œ !Þ#(&*
Say coin 1:
EP œ !Þ(#%"Ð!Ñ  !Þ#(&*Ð"Ñ œ !Þ#(&*
Say coin 2:
EP œ !Þ(#%"Ð"Ñ  !Þ#(&*Ð!Ñ œ !Þ(#%"
The optimal alternative is to say coin 1.
16.3-14.
(a)
Alternative
Predict 0 H
Predict 1 H
Predict 2 H
Prior probabilities
Predict 0 H:
Predict 1 H:
Predict 2 H:
State of Nature
Coin 1 Coin 2
4
36
32
48
64
16
0.5
0.5
EP œ !Þ&Ð4Ñ  !Þ&Ð36Ñ œ 20
EP œ !Þ&Ð32Ñ  !Þ&Ð48Ñ œ 40
EP œ !Þ&Ð64Ñ  !Þ&Ð16Ñ œ 40
The optimal alternative is to predict one or two heads with expected payoff of $40.
(b)
Prior
Probability
0.5
0.5
P(Finding | State)
Finding
Heads
Tails
0.8
0.2
0.4
0.6
Posterior
Probabilities:
Finding
P(Finding)
Heads
0.6
Tails
0.4
P(State | Finding)
State of Nature
Coin 1
Coin 2
0.6667
0.3333
0.2500
0.7500
Data:
State of
Nature
Coin 1
Coin 2
1'-22
(c) If the outcome is heads (H):
Predict 0 H:
Predict 1 H:
Predict 2 H:
EP œ !Þ66(Ð4Ñ  !Þ$33Ð36Ñ œ 14.67
EP œ !Þ66(Ð32Ñ  !Þ33$Ð48Ñ œ 37.33
EP œ !Þ66(Ð64Ñ  !Þ33$Ð16Ñ œ 48
The optimal alternative is to predict two heads.
If the outcome is tails (T):
Predict 0 H:
Predict 1 H:
Predict 2 H:
EP œ !Þ25Ð4Ñ  !Þ(5Ð36Ñ
EP œ !Þ25Ð32Ñ  !Þ(5(48Ñ
EP œ !Þ25Ð64Ñ  !Þ(5Ð16Ñ
œ 28
œ 44
œ 28
The optimal alternative is to predict one heads.
The expected payoff œ !Þ'Ð$%)Ñ  !Þ%Ð$%%Ñ œ $%'Þ%!.
(d) EVE œ $46.40  $%! œ $'Þ%!  $$!, so it is better to not pay for the experiment and
choose to predict either one or two heads.
16.4-1.
Driven by "the pressure to reduce costs and deliver high-impact technology quickly while
justifying investments" [p. 57], Westinghouse initiated this study to evaluate R and D
efforts effectively. At any point in time, the firm chooses between launching, delaying
and abandoning an innovation. When the launch is delayed, there is a chance of losing
the opportunity. R and D is hence treated as a call option with flexibility. The value of
the innovation and the optimal decision rule in subsequent stages are found by using
dynamic programming. This value is then used in the analysis of the decision tree
constructed to find the present value of the project. In this tree, decisions consist of
whether to fund or not at different stages and each decision node is followed by a chance
node that represents either a technical milestone or strategic fit. Sensitivity analysis is
performed to ensure robustness of the model.
As a result of this study, explicit decision rules for funding R and D projects are obtained.
Including flexibility in the model yields a more realistic model. The new system helps
identifying cost-effective research portfolios with simplified data acquisition and easy
implementation.
1'-23
16.4-2.
67%
Sell 10,000
Build Computers
-6
50%
Predict Sell 10,000
0
17
-1
6
17
-1
33%
Sell 100,000
1
60
53
53
Sell Rights
15
14
14
Market Research
-1
33%
Sell 10,000
26
Build Computers
50%
Predict Sell 100,000
1
0
35
27
2
-6
6
35
-1
67%
Sell 100,000
60
14
14
50%
Sell 10,000
Build Computers
-6
6
27
No Market Research
1
0
27
53
53
Sell Rights
15
-1
0
0
50%
Sell 100,000
60
54
54
Sell Rights
15
15
15
The optimal policy is to build the computers without doing market research.
16.4-3.
40%
2500
2500 2500
0 580
20%
2
60%
-700 -700
-700
0 900
0 820
900 900
900
80%
1
800
820
800 800
750
750 750
1'-24
16.4-4.
(a)
Alternative
Hold Campaign
Not Hold Campaign
Prior Probability
State of Nature
[
P
$
#
!
!
!Þ'
!Þ%
(b) Choose to hold the campaign with expected payoff $" million.
Alternative
Hold Campaign
Not Hold Campaign
Prior Probability
State of Nature
[
P
$
#
!
!
!Þ'
!Þ%
Alternative
Hold Campaign
Not Hold Campaign
Prior Probability
Maximum Payoff
State of Nature
[
P
$
#
!
!
!Þ'
!Þ%
$
!
(c)
Exp.
Payoff
"
!
Expected Payoff with Perfect Information: !Þ'Ð$Ñ  !Þ%Ð!Ñ œ "Þ)
Expected Payoff without Information: "
EVPI œ "Þ)  " œ $!Þ) million
(d)
1'-25
(e)
Data:
State of
Nature
Winning Season
Losing Season
Posterior
Probabilities:
Finding
Predict W
Predict L
Prior
Probability
0.6
0.4
P(Finding)
0.55
0.45
P(Finding | State)
Finding
Predict L
0.25
0.75
Predict W
0.75
0.25
P(State | Finding)
State of Nature
Winning Season Losing Season
0.818
0.182
0.333
0.667
(f) Leland University should hire William. If he predicts a winning season, then they
should hold the campaign and if he predicts a losing season, then they should not hold the
campaign.
82%
Win
Hold Campagin
0 1.991
55%
Predict Win
0
3
1.991
1
2.9
18%
Lose
-2 -2.1
Don't Hold Campaign
0
-0.1
33%
Win
1.05
Hold Campaign
0
45%
Predict Lose
0
1.05
3
-0.43
2
67%
Lose
-2 -2.1
Don't Hold Campagin
0
-0.1
3
3
-2
-2
60%
Win
Hold Campaign
1
Don't Hire William
1
0
1
Don't Hold Campaign
0
2.9
2.9
-2.1
-0.1
1
0
-2.1
-0.1
-0.1
Hire William
2.9
0
1'-26
40%
Lose
-0.1
3
-2
0
16.4-5.
(a)
70%
Growth
70%
Growth
20
133.2
Stocks Y2
24
0
30%
Recession
133.2
144
-12
1
108
70%
Growth
Bonds Y2
0
6
127.8
126
30%
Recession
12
Stocks Y1
100
132
20%
Growth
122.94
18
70%
Recession
Stocks Y2
0
108
82.8
144
108
126
132
108
81
-9
81
10%
Depression
30%
Recession
-10
99
-45
2
0
94.5
70%
Recession
Bonds Y2
99
9
99
10%
Depression
1
18
1'-27
45
20%
Growth
4.5
122.94
45
108
94.5
99
108
70%
Growth
70%
Growth
Stocks Y2
21
0
30%
Recession
116.55
126
-10.5
1
5 116.55
94.5
70%
Growth
Bonds Y2
5.25 110.25
0 111.825
30%
Recession
10.5
Bonds Y1
110.25
115.5
132
22
132
Stocks Y2
70%
Recession
0
-11
101.2
94.5
115.5
20%
Growth
100 117.885
126
99
99
10%
Depression
30%
Recession
10
121
-55
2
55
20%
Growth
5.5
70%
Recession
Bonds Y2
0
115.5
121
11
121
10%
Depression
22
132
55
115.5
121
132
(b) The comptroller should invest in stocks the first year. If growth occurs in the first
year, then she should invest in stocks again the second year. If recession occurs in the
first year, then she should invest in bonds the second year.
1'-28
16.4-6.
The optimal policy is to wait until Wednesday to buy if the price is $9 on Tuesday. If the
price is $10 or $11 on Tuesday, then buying on Tuesday is optimal.
Buy
900
30%
Close at $9
0
900
900
40%
10% Lower
2
891
810
30%
Unchanged
Wait
0
810
810
891
900
900
30%
10% Higher
990
990
Buy
900
990
1000
1000 1000
30%
Close at $10
1007.3
0
1000
20%
10% Lower
1
900
Wait
0 1040
900
20%
Unchanged
1000
1000
900
1000
60%
10% Higher
1100
1100
Buy
1100
1100 1100
40%
Close at $11
0
10%
10% Lower
1
1100
990
Wait
0 1166
990
990
20%
Unchanged
1100
1100
70%
10% Higher
1210
1'-29
1100
1210
1100
1210
16.4-7.
(a)
(b)
Prior Distribution:
P) Ð5Ñ
B
\"
\#
\$
)"
!Þ#
)#
!Þ&
)$
!Þ$
U\l)œ5 ÐBÑ
)" )# ) $
!Þ& !Þ% !Þ#
!Þ% !Þ& !Þ%
!Þ" !Þ" !Þ%
Posterior Distribution:
B
\"
\#
\$
2)l\œB Ð5Ñ
)"
)#
!Þ#() !Þ&&'
!Þ"() !Þ&&'
!Þ"!& !Þ#'$
)$
!Þ"'(
!Þ#'(
!Þ'$#
(c) It is optimal to not use credit rating, but to extend credit, see part (a).
1'-30
16.4-8.
(a)
(b)
Prior Distribution:
P) Ð5Ñ
B
\"
\#
)"
!Þ''(
)#
!Þ$$$
U\l)œ5 ÐBÑ
)"
)#
!Þ) !Þ$
!Þ# !Þ(
Posterior Distribution:
B
\"
\#
2)l\œB Ð5Ñ
)"
)#
!Þ)%# !Þ"&)
!Þ$'% !Þ'$'
(c) It is optimal to not conduct a survey, but to market the new product, see part (a).
1'-31
16.4-9.
(a)
(b)
Prior Distribution:
P) Ð5Ñ
B
\"
\#
)"
!Þ)
)#
!Þ#
U\l)œ5 ÐBÑ
)"
)#
!Þ*& !Þ(&
!Þ!& !Þ#&
Posterior Distribution:
B
\"
\#
2)l\œB Ð5Ñ
)"
)#
!Þ)$& !Þ"'&
!Þ%%% !Þ&&'
(c) It is optimal to not test and to not screen, see part (a).
1'-32
16.4-10.
(a)
(b)
Prior Distribution:
P) Ð5Ñ
B
\"
\#
)"
!Þ'
)#
!Þ%
U\l)œ5 ÐBÑ
)"
)#
!Þ$ !Þ'
!Þ( !Þ%
Posterior Distribution:
B
\"
\#
2)l\œB Ð5Ñ
)"
)#
!Þ%#* !Þ&("
!Þ(#% !Þ#('
(c) It is optimal to choose coin 1 if the outcome is tails and coin 2 if the outcome is
heads, see part (a).
1'-33
16.4-11.
(a)
95%
Successful
1980
Hire
2000 1980
0
66%
Pass
1870.9
5%
Unsuccessful
-420
1
0
-400 -420
1870.9
Don't Hire
-20
0
-20
Test
21%
Successful
-20 1260
1980
Hire
2000 1980
0
74.118
34%
Fail
79%
Unsuccessful
-420
1
0
-400 -420
74.118
2
Don't Hire
1280
-20
0
-20
70%
Successful
Hire
2000
0
1280
2000
2000
30%
Unsuccessful
Don’t Test
-400
1
-400
-400
0 1280
Don't Hire
0
0
0
(b)
Data:
State of
Nature
Successful
Not Successful
Posterior
Probabilities:
Finding
Pass Test
Fail Test
Prior
Probability
0.7
0.3
P(Finding | State)
Finding
Pass Test
Fail Test
0.9
0.1
0.1
0.9
P(Finding)
0.6600
0.3400
P(State | Finding)
State of Nature
Successful Not Successful
0.9545
0.0455
0.2059
0.7941
(c) The optimal policy is to not pay for testing and to hire Matthew.
(d) Even if the fee is zero, hiring Matthew without any further investigation is optimal, so
Western Bank should not pay anything for the detailed report.
1'-34
16.5-1.
(a)
Alternative
Build Computers
Sell Rights
State of Nature
Sell "!ß !!! Sell "!!ß !!!
!
&%
"&
"&
Sell 10,000
Build Computers
-6
27
27
1
6
50%
Sell 100,000
60
Sell Rights
15
0
0
54
54
15
15
They should build computers with an expected payoff of $#( million.
(b)
Prob(Sell 10,000)
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Decision
Build
Build
Build
Build
Build
Build
Build
Build
Build
Sell
Sell
Sell
Expected Payoff ($million)
27
54
48.6
43.2
37.8
32.4
27
21.6
16.2
15
15
15
1'-35
16.5-2.
(a) The optimal policy is to not do market research and build the computers. The
expected payoff is $#( million.
67%
Sell 10,000
Build Computers
50%
Predict Sell 10,000
0
17
-6
17
1
6
-1
33%
Sell 100,000
60
53
Sell Rights
15
Market Research
-1
33%
Sell 10,000
Build Computers
50%
Predict Sell 100,000
1
0
35
27
2
-6
35
6
-1
67%
Sell 100,000
60
53
Sell Rights
15
14
50%
Sell 10,000
Build Computers
-6
6
27
No Market Research
1
0
27
0
50%
Sell 100,000
60
54
Sell Rights
15
53
14
14
26
-1
15
-1
53
14
0
54
15
(b)
Prior(Sell 10,000) Action 1
No Market Research
0
No Market Research
0.1
No Market Research
0.2
No Market Research
0.3
No Market Research
0.4
No Market Research
0.5
No Market Research
0.6
No Market Research
0.7
Market Research
0.8
Market Research
0.9
No Market Research
1
No Market Research
Action 2
Expected Payoff
Build
27
Build
54
Build
48.6
Build
43.2
Build
37.8
Build
32.4
Build
27
Build
21.6
Build if Predict Sell 100,000
18.3
Build if Predict Sell 100,000
15.2
Sell
15
Sell
15
(c) If the rights can be sold for $"'Þ& or $"$Þ& million, the optimal policy is still to build the
computers with an expected payoff of $#( million. If the cost of setting up the assembly line
is $&Þ% million or $'Þ' million, the optimal policy is still to build the computers with an
expected payoff of $#(Þ' or $#'Þ% million respectively. If the difference between the selling
price and the variable cost of each computer is $&%! or $''!, the optimal policy is still to
build the computers with an expected payoff of $#$Þ( or $$$Þ$ million respectively. For
each combination of financial data, the expected payoff is as shown in the following table.
In all cases, the optimal policy is to build the computers without doing market research.
1'-36
Sell Rights
$13.5 million
$13.5 million
$13.5 million
$13.5 million
$16.5 million
$16.5 million
$16.5 million
$16.5 million
Cost of Assembly Line
$5.4 million
$5.4 million
$6.6 million
$6.6 million
$5.4 million
$5.4 million
$6.6 million
$6.6 million
Selling Price  Variable Cost
$540
$660
$540
$660
$540
$660
$540
$660
Expected Payoff
$23.4 million
$30.9 million
$23.1 million
$29.7 million
$24.3 million
$30.9 million
$23.1 million
$29.7 million
16.5-3.
40%
2500
2500 2500
0 580
20%
0 900
2
0 820
60%
-700 -700
900 900
-700
900
80%
820
1
800 800
800
750
750 750
1'-37
16.5-4.
70%
Growth
70%
Growth
20
133.2
Stocks Y2
24
0
30%
Recession
133.2
144
-12
1
108
70%
Growth
Bonds Y2
0
6
127.8
126
30%
Recession
12
Stocks Y1
100
132
20%
Growth
122.94
18
70%
Recession
Stocks Y2
0
108
82.8
144
108
126
132
108
81
-9
81
10%
Depression
30%
Recession
-10
99
-45
2
0
94.5
70%
Recession
Bonds Y2
99
9
99
10%
Depression
1
18
1'-38
45
20%
Growth
4.5
122.94
45
108
94.5
99
108
70%
Growth
70%
Growth
Stocks Y2
21
0
30%
Recession
116.55
126
-10.5
1
5 116.55
94.5
70%
Growth
Bonds Y2
5.25 110.25
0 111.825
30%
Recession
10.5
Bonds Y1
110.25
115.5
132
22
132
Stocks Y2
70%
Recession
0
-11
101.2
94.5
115.5
20%
Growth
100 117.885
126
99
99
10%
Depression
30%
Recession
10
121
-55
2
55
20%
Growth
5.5
70%
Recession
Bonds Y2
0
115.5
121
11
121
10%
Depression
22
1'-39
132
55
115.5
121
132
16.5-5.
Buy
900
30%
Close at $9
0
891
900
900
40%
10% Lower
2
810
30%
Unchanged
Wait
0
810
891
900
900
30%
10% Higher
990
990
1000
1000 1000
1007.3
0
1000
20%
10% Lower
1
900
Wait
0 1040
900
990
Buy
30%
Close at $10
810
900
20%
Unchanged
1000
1000
60%
10% Higher
1100
1100
Buy
900
1000
1100
1100
1100 1100
40%
Close at $11
0
1100
10%
10% Lower
1
990
Wait
0 1166
990
990
20%
Unchanged
1100
1100
70%
10% Higher
1210
1210
1100
1210
The optimal policy is to wait until Wednesday to buy if the price is $* on Tuesday. If the
price is $"! or $"" on Tuesday, then buy on Tuesday.
1'-40
16.5-6.
P(Finding | State)
Finding
Not Excellent
0.2
0.7
Data:
State of
Nature
Satisfactory Box
Unsatisfactory Box
Posterior
Probabilities:
Finding
Excellent
Not Excellent
Prior
Probability
0.9
0.1
Excellent
0.8
0.3
P(State | Finding)
State of Nature
Satisfactory Box Unsatisfactory Box
0.960
0.040
0.720
0.280
P(Finding)
0.75
0.25
96%
Satisfactory
Buy
200
0
75%
Excellent
0
152
152
1
200
4%
Unsatisfactory
-1000
200
-1000
-1000
Reject
0
Sample
0
0
0
72%
Satisfactory
114
Buy
200
0
25%
Not Excellent
0
-136
2
200
200
28%
Unsatisfactory
-1000
-1000
-1000
0
1
Reject
114
0
0
90%
Satisfactory
Buy
200
0
80
Don't Sample
1
0
80
200
10%
Unsatisfactory
-1000
Reject
0
200
0
-1000
-1000
0
0
The optimal policy is to sample the fruit and buy if it is excellent and reject if it is
unsatisfactory.
1'-41
16.5-7.
(a) Choose to introduce the new product with expected payoff of $"#Þ& million.
Alternative
Introduce New Product
Don't Introduce New Product
Prior Probabilities
State of Nature
Successful Unsuccessful
$%! million $"& million
!
!
!Þ&
!Þ&
Exp.
Payoff
$"#Þ& million
!
(b) With perfect information, Morton Ward should introduce the product if it will be
successful and not introduce it if it will not be successful.
Expected Payoff with Perfect Information: !Þ&Ð%!Ñ  !Þ&Ð!Ñ œ #!
Expected Payoff without Information: "#Þ&
EVPI œ #!  "#Þ& œ $(Þ& million
(c) The optimal policy is to not test but to introduce the new product, with expected
payoff $"#Þ& million.
P(Finding | State)
Data:
Finding
State of
Prior
Nature
Probability Approved Not Approved
Successful
0.5
0.8
0.2
Unsuccessful
0.5
0.25
0.75
P(State | Finding)
Posterior
State of Nature
Probabilities:
Finding
P(Finding) Successful Unsuccessful
Approved
0.525
0.762
0.238
Not Approved
0.475
0.211
0.789
1'-42
76%
Successful
Introduce Product
0
53%
Approved
40
24.9048
1
38
38
24%
Unsuccessful
-15 -17
-17
0 24.9048
Don't Introduce Product
0
-2
-2
Test
21%
Successful
-2 12.125
Introduce Product
0 -5.42105
48%
Not Approved
0
40
-2
2
2
38
38
79%
Unsuccessful
-17
-15 -17
Don't Introduce Product
12.5
0
-2
50%
Successful
Introduce Product
0
12.5
Don’t Test
0
12.5
1
40
40
50%
Unsuccessful
-15
40
-15
-15
Don't Introduce Product
0
-2
0
0
(d)
Prior(Successful) Action 1
Don't Test
0
Don't Test
0.1
Don't Test
0.2
Test
0.3
Test
0.4
Test
0.5
Don't Test
0.6
Don't Test
0.7
Don't Test
0.8
Don't Test
0.9
Don't Test
1
Don't Test
Action 2
Expected Payoff
Introduce
12.5
Don't Introduce
0
Don't Introduce
0
Introduce if Approved
1.4
Introduce if Approved
4.975
Introduce if Approved
8.55
Introduce
12.5
Introduce
18
Introduce
23.5
Introduce
29
Introduce
34.5
Introduce
40
(e) If the net profit if successful is only $$! million, then the optimal policy is to test and
to introduce the product only if the test market approves. The expected payoff is $)Þ"#&
million. If the net profit if successful is $&! million, then the optimal policy is to skip the
test and to introduce the product, with an expected payoff of $"(Þ& million. If the net loss
if unsuccessful is only $""Þ#& million, then the optimal policy is to skip the test and to
introduce the product, with an expected payoff of $"%Þ$(& million. If the net loss if
unsuccessful is $")Þ(& million, then the optimal policy is to conduct the test and to
1'-43
introduce the product only if the test market approves. The expected payoff is $""Þ'&'
million. For each combination of financial data, the expected payoff and the optimal
policy are as shown below.
Successful
$30 million
$30 million
$50 million
$50 million
Unsuccessful
-$11.25 million
-$18.75 million
-$11.25 million
-$18.75 million
Optimal Policy
Skip Test, Introduce Product
Test, Introduce Product if Approved
Skip Test, Introduce Product
Test, Introduce if Approved
Expected Profit
$9.375 million
$7.656 million
$19.375 million
$15.656 million
16.5-8.
(a) Chelsea should run in the NH primary. If she does well, then she should run in the ST
primaries. If she does poorly in the NH primary, then should not run the ST primaries.
The expected payoff is $'''ß ''(.
57%
Do Well in ST
14.4
Run in ST
16
0 3.25714
47%
Do Well in NH
14.4
43%
Do Poorly in ST
-11.6
1
-10 -11.6
0 3.257143
Don't Run in ST
0
-1.6
-1.6
Run in NH
25%
Do Well in ST
-1.6 0.666667
14.4
Run in ST
0
16
-5.1
53%
Do Poorly in NH
0.6667
75%
Do Poorly in ST
-11.6
2
0
14.4
-10 -11.6
-1.6
1
Don't Run in ST
0
-1.6
-1.6
40%
Do Well in ST
16
16
Run in ST
0
0.4
16
60%
Do Poorly in ST
Don't Run in NH
-10
1
0
-10
-10
0.4
Don't Run in ST
0
0
1'-44
0
(b)
Prior(Well in NH)
0.0000
0.0667
0.1333
0.2000
0.2667
0.3333
0.4000
0.4667
0.5333
0.6000
0.6667
0.7333
0.8000
0.8667
0.9333
1.0000
Action 1
Run in NH
Don't Run
Don't Run
Don't Run
Don't Run
Don't Run
Don't Run
Don't Run
Run in NH
Run in NH
Run in NH
Run in NH
Run in NH
Run in NH
Run in NH
Run in NH
Run in NH
in
in
in
in
in
in
in
NH
NH
NH
NH
NH
NH
NH
Action 2
Run in ST
Run in ST
Run in ST
Run in ST
Run in ST
Run in ST
Run in ST
Run in ST
Run in ST
Run in ST
Run in ST
Run in ST
Run in ST
Run in ST
Run in ST
Run in ST
Run in ST
if do well in NH
if
if
if
if
if
if
if
if
if
do
do
do
do
do
do
do
do
do
well
well
well
well
well
well
well
well
well
in
in
in
in
in
in
in
in
in
NH
NH
NH
NH
NH
NH
NH
NH
NH
Expected Payoff
0.6667
0.4000
0.4000
0.4000
0.4000
0.4000
0.4000
0.4000
0.6667
0.9905
1.3143
1.6381
1.9619
2.2857
2.6095
2.9333
3.2571
(c) If the payoff for doing well in ST is only $"# million, Chelsea should not run in either
NH or ST, with expected payoff of $!. If the payoff for doing well in ST is $#! million,
Chelsea should not run in NH, but run in ST, with expected payoff of $# million. If the
loss for doing poorly in ST is $(Þ& million, Chelsea should not run in NH, but run in ST,
with expected payoff of $"Þ* million. If the loss for doing poorly in ST is only $"#Þ&
million, Chelsea should run in NH and run in ST if she does well in NH, with expected
payoff of $"''ß ''(. For each combination of financial data, the expected payoff and the
optimal policy is as shown below.
Well in ST
$12 million
$12 million
$20 million
$20 million
Poorly in ST
-$7.5 million
-$12.5 million
-$7.5 million
-$12.5 million
Optimal Policy
Run in ST Only
Don't Run in Either
Run in ST Only
Run in NH, Run in ST if Well
1'-45
Expected Funds
$300,000
$0
$3.5 million
$1.233 million
16.6-1.
(a) - (b) The optimal policy is to not conduct a survey and to sell the land.
14%
Successful
Drill
0 0.1414
70%
Unfavorable
0
0.99 0.99
0.4
0.99
86%
Unsuccessful
2
0
0
0
Sell
0.4
Do Seismic Survey
0
0.4
0.4
50%
Successful
0.4285
Drill
0
30%
Favorable
0
0.45
0.495
2
0.99 0.99
0.495
50%
Unsuccessful
1
0
Sell
0.4
25%
Oil
1
0 0.3025
No Seismic Survey
0
0.45
2
1
1
75%
Dry
0.07
0.07
0.07
Sell
0.45
0.45
0.45
16.6-2.
(a) Choose to not buy insurance with expected payoff $#%*ß )%!.
Alternative
Buy Insurance
Not Buy Insurance
Prior Probability
(b)
0
0
0.4
0.4
Drill
0.99
State of Nature
Earthquake No Earthquake
#%*ß )#!
#%*ß )#!
*!ß !!!
#&!ß !!!
!Þ!!"
!Þ***
Exp.
Payoff
#%*ß )#!
#%*ß )%!
Y (insurance) œ Y Ð#&!ß !!!  ")!Ñ œ #%*ß )#! œ %**Þ)#
Y (no insurance) œ !Þ***Y Ð#&!ß !!!Ñ  !Þ!!"Y Ð*!ß !!!Ñ œ %**Þ)
The optimal policy is to buy insurance.
1'-46
16.6-3.
Expected utility of $"*ß !!!: Y Ð"*Ñ œ #& œ &
Expected utility of investment: !Þ$Y Ð"!Ñ  !Þ(Y Ð$!Ñ œ !Þ$"'  !Þ($' œ &Þ%
Choose the investment to maximize expected utility.
16.6-4.
Expected utility of A1 œ Expected utility of A2
:Y Ð"!Ñ  Ð"  :ÑY Ð$!Ñ œ Y Ð"*Ñ
!Þ$Y Ð"!Ñ  !Þ(Ð#!Ñ œ "'Þ( Ê Y Ð"!Ñ œ *
16.6-5.
(a) Expected utility of A1 œ Expected utility of A2
:Y Ð"!Ñ  Ð"  :ÑY Ð!Ñ œ Y Ð"Ñ
!Þ"#&Y Ð"!Ñ  !Þ)(&Ð!Ñ œ " Ê Y Ð"!Ñ œ )
(b) Expected utility of A1 œ Expected utility of A2
:Y Ð"!Ñ  Ð"  :ÑY Ð!Ñ œ Y Ð&Ñ
!Þ&'#&Ð)Ñ  !Þ%$(&Ð!Ñ œ Y Ð&Ñ Ê Y Ð&Ñ œ %Þ&
(c) Answers will vary.
16.6-6.
(a)
Expected utility of A1 œ :Y Ð25Ñ  Ð"  :ÑY Ð36Ñ œ 5:  6Ð"  :Ñ œ 6  :
Expected utility of A2 œ :Y Ð"00Ñ  Ð"  :ÑY Ð!Ñ œ "0:  ! œ "0:
Expected utility of A3 œ :Y Ð!Ñ  Ð"  :ÑY Ð49Ñ œ !  7Ð"  :Ñ œ 7  7:
A1 and A3 cross when 6  : œ (  (: Ê : œ "' .
A1 and A# cross when '  : œ "!: Ê : œ #$ .
Thus, A3 is best when : Ÿ "' , A1 is best when
"
'
1'-47
Ÿ : Ÿ #$ , and A2 is best when :
#
$.
(b)
Y ÐQ Ñ œ &!Ð"  /Q Î&! Ñ
: œ #&%, alternative 1 is optimal
25%
State 1
25
Alternative 1
25
0 33.015
0.4833
25
0.39347
75%
State 2
36
36
36
0.51325
25%
State 1
100
Alternative 2
100
1
33.0149
0.4833
0 12.178
0.2162
100
0.86466
75%
State 2
0
0
0
0
25%
State 1
0
Alternative 3
0
0 31.604
0.4685
0
0
75%
State 2
49
49
49
0.62469
: œ &!%, alternative 1 is optimal
50%
State 1
Alternative 1
25
0 30.198
0.4534
25
25
0.39347
50%
State 2
36
36
50%
State 1
Alternative 2
100
1
30.1981
0.45336
0 28.311
0.4323
36
0.51325
100
0.86466
50%
State 2
0
100
0
0
0
50%
State 1
0
Alternative 3
0
0 18.723
0.3123
50%
State 2
49
1'-48
0
0
49
0.62469
49
: œ (&%, alternative 2 is optimal
75%
State 1
Alternative 1
25
0 27.532
0.4234
25%
State 2
36
75%
State 1
52.2771
0.6485
2
Alternative 2
100
0 52.277
0.6485
25%
State 2
0
75%
State 1
Alternative 3
0
0 8.4903
0.1562
25
36
36
0.51325
100
100
0.86466
0
0
0
0
0
0
25%
State 2
49
1'-49
25
0.39347
49
0.62469
49
16.6-7.
The optimal policy is to not test for disease A, but to treat disease A.
50%
Poor Health
80%
Have A
10
0 17
7
7
50%
Good Health
Treat A
27
-1
30
13
27
20%
Have B / Die
-3
0 -3
50%
Positive
20%
Die
1
0
13
80%
Have A
0
0
6
-2
80%
Poor Health
Don't Treat A
0 5.4
-2
10
8
0
-2
8
50%
Die
20%
Have B
-2
0
3
Test for A
50%
Poor Health
10
-2 8.3
8
8
50%
Poor Health
20%
Have A
10
0 17
7
7
50%
Good Health
Treat A
27
-1
30
1
27
80%
Have B / Die
-3
0 -3
50%
Negative
0 3.6
2
20%
Die
20%
Have A
-2
0
0
6
-2
80%
Poor Health
Don't Treat A
10
8
0
-2
8
2
9
0 3.6
50%
Die
80%
Have B
-2
0
3
50%
Poor Health
8
10
1'-50
8
50%
Poor Health
50%
Have A
10
0
19
9
50%
Good Health
Treat A
-1
29
30 29
9
50%
Have B / Die
0
-1
-1
Test for B
20%
Die
1
0
9
9
50%
Have A
0
0
8
0
0
80%
Poor Health
10
Don't Treat A
10 10
0 6.5
50%
Die
50%
Have B
0
0
5
0
0
50%
Poor Health
10
10 10
16.6-8.
(a)
At : œ !Þ#&, "!&:  & œ #"Þ#& and max Ð##:  #ß #Ñ œ max Ð$Þ&ß #Ñ œ $Þ&, so A1 is optimal.
1'-51
(b)
As can be seen on the graph, A1 stays optimal for "Î"& Ÿ : Ÿ !Þ&.
CASE 16.1 Brainy Business
(a) The decision alternatives are to price the product high ($50), medium ($40), or low
($30), or don't market the product at all. The possible states of nature are the demand
could be high (50,000), medium (30,000), or low (20,000), which in turn depends upon
the price, and whether the competition is severe, moderate, or weak. The various data are
summarized in the spreadsheet below. The payoff table can be generated based on the
results in the far right column.
High
Medium
Low
Price
$50
$40
$30
Sales
(thousands)
High
50
Medium
30
Low
20
Prior Probability
Severe
0.2
Moderate
0.7
Weak
0.1
High Price
Sales High
Sales Medium
Sales Low
Severe
0.20
0.25
0.55
Moderate
0.25
0.30
0.45
Weak
0.30
0.35
0.35
Medium Price
Sales High
Sales Medium
Sales Low
Severe
0.25
0.35
0.40
Moderate
0.30
0.40
0.30
Weak
0.40
0.50
0.10
0.3
0.4
0.3
2,000
1,200
800
Low Price
Sales High
Sales Medium
Sales Low
Severe
0.35
0.40
0.25
Moderate
0.40
0.50
0.10
Weak
0.50
0.45
0.05
0.4
0.475
0.125
1,500
900
600
1'-52
Prior
Revenue
Probability ($thousands)
0.245
2,500
0.295
1,500
0.46
1,000
The decision tree for this probem follows (over three pages):
20%
Sales High
2,500
2,500
20%
Severe Competition
2,500
25%
Sales Medium
1,500
0 1425
1,500
1,500
55%
Sales Low
1,000
1,000
1,000
25%
Sales High
2,500
2,500
70%
Moderate Competition
Price High
2,500
30%
Sales Medium
1,500
0
1515
0 1525
1,500
1,500
45%
Sales Low
1,000
1,000
1,000
30%
Sales High
2,500
2,500
10%
Weak Competition
2,500
35%
Sales Medium
1,500
0 1625
1,500
1,500
35%
Sales Low
1,000
1,000
1'-53
1,000
25%
Sales High
2,000
2,000
20%
Severe Competition
2,000
35%
Sales Medium
1,200
0 1240
1,200
1,200
40%
Sales Low
800
800
800
2,000
2,000
30%
Sales High
2,000
70%
Moderate Competition
Price Medium
40%
Sales Medium
1
1515
1,200
0
1320
0 1320
1,200
1,200
30%
Sales Low
800
800
800
2,000
2,000
40%
Sales High
2,000
10%
Weak Competition
50%
Sales Medium
1,200
0 1480
1,200
1,200
10%
Sales Low
800
800
1'-54
800
35%
Sales High
1,500
1,500
20%
Severe Competition
1,500
40%
Sales Medium
900
0 1035
900
900
600
600
1,500
1,500
25%
Sales Low
600
40%
Sales High
1,500
70%
Moderate Competition
Price Low
50%
Sales Medium
900
0 1102.5
0 1110
900
900
600
600
1,500
1,500
10%
Sales Low
600
50%
Sales High
1,500
10%
Weak Competition
45%
Sales Medium
900
0 1185
900
900
600
600
5%
Sales Low
600
(b) The scenario "moderate competition, sales of 30,000 units at a unit price of $30" has
the largest total probability. Therefore, under the maximum likelihood criterion,
Charlotte should price the product at $30.
To find out best maximin alternative, note that for a price of
$30: 20,000 units at a unit price $30 is the worst case,
$40: 20,000 units at a unit price $40 is the worst case,
$50: 20,000 units at a unit price $50 is the worst case.
The maximum of these three is for the price of $50, so it is optimal under the maximin
criterion.
1'-55
(c) As shown in the decision tree for part a (recall that decision trees assume Bayes'
decision rule), Charlotte should charge the high price ($50), since this maximizes the
expected revenue ($1.515 million). Alternatively, the expected revenues for each possible
decision
can
be
calculated
directly
as
shown
in
the
following
spreadsheet.
High
Medium
Low
Price
$50
$40
$30
Sales
(thousands)
High
50
Medium
30
Low
20
Prior Probability
Severe
0.2
Moderate
0.7
Weak
0.1
Expected
Prior
Revenue
Revenue
Probability ($thousands) ($thousands)
0.245
2,500
0.295
1,500
1,515
0.46
1,000
High Price
Sales High
Sales Medium
Sales Low
Severe
0.20
0.25
0.55
Moderate
0.25
0.30
0.45
Weak
0.30
0.35
0.35
Medium Price
Sales High
Sales Medium
Sales Low
Severe
0.25
0.35
0.40
Moderate
0.30
0.40
0.30
Weak
0.40
0.50
0.10
0.3
0.4
0.3
2,000
1,200
800
1,320
Low Price
Sales High
Sales Medium
Sales Low
Severe
0.35
0.40
0.25
Moderate
0.40
0.50
0.10
Weak
0.50
0.45
0.05
0.4
0.475
0.125
1,500
900
600
1,102.5
(d) With more information from the marketing research company, the posterior
probabilities for the state of competition can be found using the template for posterior
probabilities as follows.
Data:
State of
Nature
Severe
Moderate
Weak
Posterior
Probabilities:
Finding
Predict Severe
Predict Moderate
Predict Weak
Prior
Probability
0.2
0.7
0.1
P(Finding)
0.268
0.597
0.135
P(Finding | State)
Finding
Predict Severe Predict Moderate Predict Weak
0.8
0.15
0.05
0.15
0.8
0.05
0.03
0.07
0.9
Severe
0.597
0.050
0.074
P(State | Finding)
State of Nature
Moderate
0.392
0.938
0.259
Weak
0.011
0.012
0.667
To keep the decision tree from becoming too unwieldy, we will break it into parts. The
first three parts consider the situation after each possible prediction by the marketing
research company. The decision tree from part a is reused with the only change being the
prior probabilities of severe, moderate and weak competition used in part a are replaced
by the appropriate posterior probabilities calculated above, depending upon the
prediction of the marketing research company. For example, if the marketing research
company predicts the competition will be severe, the probability of severe, moderate, and
weak competition are 0.597, 0.392, and 0.011, respectively.
1'-56
The optimal decision if the marketing research company predicts severe is to price high
($50), with expected revenue of $1.466 million.
High
Medium
Low
Price
$50
$40
$30
Sales
(thousands)
High
50
Medium
30
Low
20
Prior Probability
Severe
0.597
Moderate
0.392
Weak
0.011
High Price
Sales High
Sales Medium
Sales Low
Severe
0.20
0.25
0.55
Moderate
0.25
0.30
0.45
Weak
0.30
0.35
0.35
Prior
Revenue
Probability ($thousands)
0.220709
2,500
0.270709
1,500
0.5085821
1,000
Medium Price
Sales High
Sales Medium
Sales Low
Severe
0.25
0.35
0.40
Moderate
0.30
0.40
0.30
Weak
0.40
0.50
0.10
0.2712687
0.3712687
0.3574627
2,000
1,200
800
Low Price
Sales High
Sales Medium
Sales Low
Severe
0.35
0.40
0.25
Moderate
0.40
0.50
0.10
Weak
0.50
0.45
0.05
0.3712687
0.4397388
0.1889925
1,500
900
600
Optimal Decision Price High
Expected Revenue 1466.42
($thousands)
The optimal decision if the marketing research company predicts moderate competition is
to price high ($50), with expected revenue of $1.521 million.
High
Medium
Low
Price
$50
$40
$30
Sales
(thousands)
High
50
Medium
30
Low
20
Prior Probability
Severe
0.050
Moderate
0.938
Weak
0.012
High Price
Sales High
Sales Medium
Sales Low
Severe
0.20
0.25
0.55
Moderate
0.25
0.30
0.45
Weak
0.30
0.35
0.35
Prior
Revenue
Probability ($thousands)
0.2480737
2,500
0.2980737
1,500
0.4538526
1,000
Medium Price
Sales High
Sales Medium
Sales Low
Severe
0.25
0.35
0.40
Moderate
0.30
0.40
0.30
Weak
0.40
0.50
0.10
0.29866
0.39866
0.3026801
2,000
1,200
800
Low Price
Sales High
Sales Medium
Sales Low
Severe
0.35
0.40
0.25
Moderate
0.40
0.50
0.10
Weak
0.50
0.45
0.05
0.39866
0.4943886
0.1069514
1,500
900
600
Optimal Decision Price High
Expected Revenue 1521.15
($thousands)
The optimal decision if the marketing research company predicts weak competition is to
price high ($50), with expected revenue of $1.584 million.
1'-57
High
Medium
Low
Price
$50
$40
$30
Sales
(thousands)
High
50
Medium
30
Low
20
Prior Probability
Severe
0.074
Moderate
0.259
Weak
0.667
High Price
Sales High
Sales Medium
Sales Low
Severe
0.20
0.25
0.55
Moderate
0.25
0.30
0.45
Weak
0.30
0.35
0.35
Prior
Revenue
Probability ($thousands)
0.2796296
2,500
0.3296296
1,500
0.3907407
1,000
Medium Price
Sales High
Sales Medium
Sales Low
Severe
0.25
0.35
0.40
Moderate
0.30
0.40
0.30
Weak
0.40
0.50
0.10
0.362963
0.462963
0.1740741
2,000
1,200
800
Low Price
Sales High
Sales Medium
Sales Low
Severe
0.35
0.40
0.25
Moderate
0.40
0.50
0.10
Weak
0.50
0.45
0.05
0.462963
0.4592593
0.0777778
1,500
900
600
Optimal Decision Price High
Expected Revenue 1584.26
($thousands)
Then, incorporating the expected payoff with each possible prediction by the marketing
company, along with the expected revenue without information from part a, we combine
the whole problem into the following decision tree.
Proceed with No Info
1515
1515 1515
27%
Predict Severe
1
1456.41791
1515
1466.41791
1456.41791
60%
Predict Moderate
Employ Marketing Research
1511.147404
-10 1505
1521.147404 1511.147404
14%
Predict Weak
1574.259259
1584.259259 1574.259259
Charlotte should not purchase the services of the market research company. The
information is not worth anything since it does not affect the decision. Regardless of the
prediction, the optimal policy is to set the price at $50.
1'-58
CASE 16.2 Smart Steering Support
(a) The available data are summarized in the table.
Research
Development
Marketing
Costs
($million)
0.3
0.8
0.2
Probability
of Success
0.8
0.65
Revenues from R&D
($million)
Sell Product Rights
1
Sell Research Results
0.2
Sell Current DSS
2
High
Medium
Low
Sales
Revenue
($million)
8
4
2.2
Probability
0.3
0.5
0.2
(b) The basic decision tree is shown below.
No Success (Sell Current DSS)
Do Research
Don't Develop (Sell Current DSS & Research Results)
Success
No Success (Sell Current DSS & Research Results)
Develop
Don't Market (Sell Current DSS & Rights)
Success
High
Market
Medium
Low
Don't do Research (Sell Current DSS)
1'-59
(c) The decision tree displays all the expected payoffs and probabilities.
20%
No Success (Sell Current DSS)
1.7
2
1.7
Do Research
Don't Develop (Sell Current DSS & Research Results)
1.9
-0.3 2.4888
2.2
1.9
80%
Success
35%
No Success (Sell Current DSS & Research Results)
2
0%
1.1
2.69
2.2
1.1
Don't Market (Sell Current DSS & Rights)
Develop
1.9
-0.8
2.686
3
1.9
65%
Success
30%
High
1
2
2.4888
0
6.7
3.54
8
6.7
50%
Medium
Market
2.7
-0.2
3.54
4
2.7
20%
Low
0.9
2.2
0.9
Don't Do Research (Sell Current DSS)
2
2
2
(d) The best course of action is to do the research project. The expected payoff is $#Þ%)9
million.
(e) The decision tree with perfect information on research is displayed. The expected
value in this case equals $#Þ&%9 million. The difference between the expected values with
and without information is $'!ß !!!, which is the value of perfect information on
research.
No Research (Sell Current DSS)
2
2
2
80%
Research will be Successful
2
0 2.686
Don't Develop (Sell Current DSS & Research Results)
1.9
2.2
1.9
35%
No Success (Sell Current DSS & Research Results)
Do Research
2
1.1
-0.3 2.686
2.2
1.1
Develop
Don't Market (Sell Current DSS & Rights)
1.9
3 1.9
-0.8 2.686
65%
Success
30%
High
2
2.5488
6.7
0 3.54
8 6.7
Market
50%
Medium
2.7
-0.2 3.54
4 2.7
20%
Low
0.9
2.2 0.9
20%
Research won't be Successful (Sell Current DSS)
2
2
2
1'-60
(f) The decision tree with perfect information on development is displayed. The expected
value in this case equals $#Þ('2 million. The difference between the expected values with
and without information is $#($ß !!!, which is the value of perfect information on
development
No Research (Sell Current DSS)
2
2
2
65%
Development Successful
20%
No Success (Sell Current DSS)
2
0
1.7
3.172
2
1.7
Do Research
Don't Develop (Sell Current DSS & Research Results)
1.9
-0.3 3.172
2.2
1.9
80%
Success
Don't Market (Sell Current DSS & Product Rights)
2
1.9
0 3.54
3
1.9
30%
High
Develop
2
2.7618
6.7
-0.8 3.54
8 6.7
50%
Medium
Market
2.7
-0.2
3.54
4 2.7
20%
Low
0.9
2.2 0.9
35%
Development would not be Successful (Sell Current DSS)
2
2
2
(g) - (h) - (i) The decision tree with expected utilities is displayed. The expected utilities
are calculated in the following way: for each of the outcome branches of the decision tree
(e.g., profit of $'ß (!!ß !!!), the corresponding utility is computed (e.g., "#Þ%&**#). Once
this is done, the expected utilities are calculated. The best course of action is to not do
research (expected utility of "!Þ"%%'* vs. *Þ)%'#'( in the case of doing research).
1'-61
(j) The expected utility for perfect information on research equals *Þ*$*$*(, which is still
less than the expected utility of not doing research ("!Þ"%%'*). Therefore, the best course
of action is to not do research, implying a value of zero for perfect information on the
outcome of the research effort.
(k) The expected utility for perfect information on development equals "!Þ$#"$%(, which
is more than the expected utility without information ("!Þ"%%'*). The value of perfect
information on development is the difference between the inverses of these two utility
values, U" Ð"!Þ$#"$%(Ñ  U" Ð"!Þ"%%'*Ñ œ #!Þ*$#(%  #! œ !Þ*$#(%. The value of
perfect information on the outcome of the development effort is $*$Þ#(%.
1'-62
CASE 16.3 Who Wants to be a Millionaire
(a) The course of action that maximizes the expected payoff is to answer $500,000
question alone. If you get the question correct, then use the phone-a-friend lifeline to help
answer $1 million question. The expected payoff is $440,980.
$1 million (alone)
$1 million (Phone‐a‐Friend)
$500k (alone)
$500k (Phone‐a‐Friend)
Prob(Correct)
50%
65%
65%
80%
50%
Correct
$1,000,000
Answer Alone
$1,000,000
0
516000
80%
Correct
$1,000,000
50%
Incorrect
$32,000
1
0
$32,000
$32,000
$516,000
Answer with Phone‐a‐Friend
Don't Answer
$500,000
0
$419,200
$500,000
$500,000
20%
Incorrect
$32,000
$32,000
$32,000
65%
Correct
$1,000,000
Answer with Phone‐a‐Friend
0
$661,200
65%
Correct
2
$1,000,000
$1,000,000
35%
Incorrect
$32,000
$440,980
1
0
$32,000
$32,000
$661,200
Answer Alone
Don't Answer
$500,000
0
$440,980
$500,000
$500,000
35%
Incorrect
$32,000
$32,000
$32,000
Don't Answer
$250,000
$250,000
$250,000
(b) Answers will vary depending on your level of risk aversion. One possible solution is
obtained by setting
Y ÐMaximumÑ œ Y Ð$1 millionÑ œ " and Y ÐMinimumÑ œ Y Ð$32,000Ñ œ !.
If getting $250,000 for sure is equivalent to a 60% chance of getting $1 million vs. a 40%
chance of getting $32,000, then Y Ð$250,000Ñ œ : œ !Þ'.
If getting $500,000 for sure is equivalent to a 90% chance of getting $1 million vs. a 10%
chance of getting $32,000, then Y Ð$500,000Ñ œ : œ !Þ*.
1'-63
(c) With the utilities derived in part (b), the decision changes to using the phone-a-friend
lifeline to help answer the $500,000 question, and then walk away.
$1 million (alone)
$1 million (Phone‐a‐Friend)
$500k (alone)
$500k (Phone‐a‐Friend)
Prob(Correct)
50%
65%
65%
80%
50%
Correct
Answer Alone
1
0
U($1 million) =
U($500k) =
U($250k) =
U($32k) =
1
0.9
0.6
0
1
$1,000,000
50%
Incorrect
2
0
0
0
$32,000
0 0.9
Answer with Phone‐a‐Friend
0
0.5
80%
Correct
1
Don't Answer
0.72
0.9
0.9
$500,000
0.9
20%
Incorrect
0
0
$32,000
0
65%
Correct
Answer with Phone‐a‐Friend
0
0.65
65%
Correct
1
0.72
1
1
1
$1,000,000
0
0
0
$32,000
35%
Incorrect
2
0 0.9
Answer Alone
Don't Answer
0
0.585
0.9
0.9
0.9
$500,000
35%
Incorrect
0
0
0
$32,000
Don't Answer
0.6
0.6
$250,000
0.6
CASE "'Þ4 University Toys and the Business Professor Action Figures
(a)
Prior
Probability
0.5
0.5
P(Finding | State)
Finding
Well in Test
Poor in Test
0.8
0.2
0.4
0.6
P(Finding)
0.6
0.4
P(State | Finding)
State of Nature
Well in Full Market Poor in Full Market
0.6667
0.3333
0.25
0.75
Data:
State of
Nature
Well in Full Market
Poor in Full Market
Posterior
Probabilities:
Finding
Well in Test
Poor in Test
1'-64
(b) The best course of action is to skip the test market, and immediately market the
product fully. The expected payoff is $1750.
67%
Do Well
Probabilities:
P("Well in Full Market")
P("Well in Test")
P("Well in Full Market | Well in Test"
P("Well in Full Market | Poor in Test")
P("LSPAF")
$1,300
0.5
0.6
0.6667
0.25
0.2
Fully Market
($1,000)
$2,000
$700
20%
LSPAF Enters Market
0
$2,000
$5,000
$200
$500
$400
$40
$1,000
$100
33%
Do Poorly
($500)
1
Cost & Revenue Data:
Do Well, LSPAF
Do Well, No LSPAF
Do Poorly, LSPAF
Do Poorly, No LSPAF
Do Well in Test Market
Do Poorly in Test Market
Full Market Cost
Test Market Cost
$1,300
$200
($500)
$700
Don't
$300
60%
Do Well
0
$400
$300
67%
Do Well
$2,380
$4,300
Fully Market
$5,000
($1,000) $2,800
80%
No LSPAF
$4,300
33%
Do Poorly
($200)
1
0
$500
($200)
$2,800
Don't
$300
0
$300
Test Market
($100)
25%
Do Well
$1,604
$940
Fully Market
$2,000
($1,000) ($410)
20%
LSPAF Enters Market
75%
Do Poorly
($860)
2
0
$940
$200
($860)
($60)
Don't
($60)
40%
Do Poorly
0
$40
($60)
25%
Do Well
$440.00
$3,940
Fully Market
$5,000
$3,940
2
$1,750
($1,000)
565
80%
No LSPAF
75%
Do Poorly
($560)
1
0
$500
($560)
$565
Don't
($60)
0
($60)
50%
Do Well
$4,000
Fully Market
($1,000)
$5,000
1750
$4,000
50%
Do Poorly
No Test Market
($500)
1
0
$500
($500)
$1,750
Don't
0
0
$0
(c) If the probability that the LSPAFs enter the market before the test marketing would be
completed increases this would make the test market even less desirable, so it would still
not be worthwhile to do. However, if the probability decreases, this would make the test
market more desirable. It might reach the point where the test market is worthwhile.
1'-65
(d) Let : denote the probability that the LSPAFs will enter and EP the expected payoff.
:
!Þ!
!Þ"
!Þ#
!Þ$
!Þ%
!Þ&
!Þ'
!Þ(
!Þ)
!Þ*
"Þ!
EP
$1,750
$1,906
$1,755
$1,750
$1,750
$1,750
$1,750
$1,750
$1,750
$1,750
$1,750
$1,750
Test Market?
No
Yes
Yes
No
No
No
No
No
No
No
No
No
(e) It is better to perform the test market if the probability that the LSPAFs will enter the
market is 10% or less. It is better to skip the test market if this probability is greater than
10%.
1'-66
CHAPTER 17: QUEUEING THEORY
17.2-1.
A typical barber shop is a queueing system with input source being the population having
hair, customers being the people who want haircut and servers being the barbers. The
queue forms as customers wait for a barber to serve them. The customers are served
usually with the first-come-first-served discipline. The service mechanism involves the
barbers and equipment.
17.2-2.
(a) Average number of customers in the shop, including those getting their haircut:
(b)
# in queue
probability
product
Average number of customers waiting in the shop:
(c)
Expected number of customers being served:
(d)
hours
minutes
hours
minutes
Hence, each customer will be in the shop for half an hour on the average. This includes
the time to get a haircut. The average waiting time for a customer before getting a haircut
is
minutes.
(e)
hours
minutes
17.2-3.
(a) A parking lot is a queueing system for providing parking. The customers are the cars
and the servers are the parking spaces. The service time is the amount of time a car stays
parked in a space and the queue capacity is zero.
(b)
2
3
3
2
5 cars
cars
5
7 hours
hours
(c) A car spends an average of 45 minutes in a parking space.
17-1
17.2-4.
(a) FALSE. The queue is where customers wait before being served.
(b) FALSE. Queueing models conventionally assume infinite capacity.
(c) TRUE. The most common is first-come-first-served.
17.2-5.
(a) A bank is a queueing system with people as the customers and tellers as the servers.
(b)
minute
minutes
customers
customers
17.2-6.
The utilization factor represents the fraction of time that the server is busy. The server
is busy except when there is nobody in the system.
is the probability of having zero
customers in the system, so
.
17.2-7.
17.2-8.
(a)
when nobody is in the system
otherwise
(b)
(c)
17.2-9.
17-2
17.3-1.
Part
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
Customers
Customers waiting for checkout
Fires
Cars
Broken bicycles
Ships to be loaded or unloaded
Machines needing operator
Materials to be handled
Calls for plumbers
Custom orders
Typing requests
Servers
Checkers
Firefighting units
Toll collectors
Bicycle repairpersons
Longshoremen & equipment
Operator
Handling equipment
Plumbers
Customized process
Typists
17.4-1.
for
(a)
for
for
and
next arrival before 1:00
next arrival between 1:00 and 2:00
next arrival after 2:00
(b) Probability that the next arrival will occur between 1:00 and 2:00 given no arrivals
between 12:00 and 1:00 is
.
(c)
no arrivals between 1:00 and 2:00
one arrival between 1:00 and 2:00
two or more arrivals between 1:00 and 2:00
(d)
none served by 2:00
none served by 1:10
none served by 1:01
17.4-2.
for
arrivals in an hour
(a)
arrivals in an hour
(b)
arrivals in an hour
(c)
or more arrivals in an hour
arrivals in an hour
17-3
17.4-3.
Expected pay:
old
special
Expected increase in pay:
old
special
17.4-4.
Given the memoryless property, the system becomes a two-server after the first
completion occurs. Let
be the amount of time after
until the next service
completion occurs.
min
By Property 3,
has an exponential distribution with mean
.
17.4-5.
By memoryless property,
min
Exp
. By Property 3
Exp
, where
Exp
Then, the expected waiting time is
Exp
,
Exp
, and
.
minutes.
17.4-6.
(a) From aggregation property of Poisson process, the arrival process does still have a
Poisson distribution with mean rate
per hour, so the distribution of the time between
consecutive arrivals is exponential with a mean of
hours
minutes.
(b) The waiting time of this type 2 customer is the minimum of two exponential random
variables, so by Property 3, it is exponentially distributed with a mean of minutes.
17.4-7.
(a) This customer's waiting time is exponentially distributed with a mean of
(b) The total waiting time of the customer in the system is
are independent from each other.
minutes
var
(c)
var
var
minutes, var
17-4
minutes.
, where
hour
and
17.4-8.
(a) FALSE.
and var
, p.775.
(b) FALSE. "The exponential distribution clearly does not provide a close approximation
to the service-time distribution for this type of situation," second paragraph, p.776.
(c) FALSE. A new arrival would have an expected waiting time, before entering service
of
, second last paragraph, p.777.
17.4-9.
Let
min
.
for all
17.5-1.
(a)
(b)
(c)
17-5
17.5-2.
(a)
(b)
(c)
(d)
17.5-3.
(a)
(b)
(1)
(2)
(3)
(4)
(5)
(6)
17-6
(c)
(1)
(2)
(3)
(4)
(5)
(d)
17.5-4.
(a)
17-7
(b)
(c) The mean arrival rate to the system and the mean service rate for each server when it
is busy serving customers are both .
17.5-5.
(a)
(b)
(1)
(2)
(3)
(4)
(5)
(c)
(1)
(2)
(3)
(5)
The same equations can be obtained as follows:
,
,
(d)
hours
17.5-6.
(a) Let the state represent the number of machines that are broken down.
17-8
.
(b)
(c)
hours
hours
(d)
(e)
17.5-7.
(a)
(b)
17.5-8.
(a)
17-9
(b)
for
(c)
,
17.5-9.
(a) Let the state represent the number of documents received, but not completed.
(b)
below corresponds to the steady-state probability that
not completed.
(c)
17-10
documents are received but
17.5-10.
for
17.5-11.
(a)
(b)
(c)
17.5-12.
(a) Let
be the number of customers in the system.
Balance equations:
17-11
(b) Let the state
be the number of customers in service and in queue respectively.
Balance equations:
1
17.5-13.
(a) Let the state
be the number of type 1 and type 2 customers in the systems.
(b) Balance equations:
(c)
(d) Type 1 customers are blocked when the system is in state
fraction of type 1 customers who cannot enter the system is
customers are blocked when the system is in state
,
or
type 2 arrivals that are blocked is
.
or
, so the
. Type 2
, so the fraction of
17.6-1.
KeyCorp deploys queueing theory as part of its Service Excellence Management System
(SEMS) to improve productivity and service in its branches. The main objective of this
study is to enhance customer satisfaction by reducing wait times without increasing the
staffing costs. To do this, first a system that collects data about various phases of
customer transactions is developed. Then, a preliminary analysis is conducted to
determine the number of tellers required for at most 10% of customers to wait more than
five minutes. The underlying model is an M/M/k queue with an average service time of
246 seconds. The arrival and service rates, and are estimated from the data. By using
steady state equations, measures such as average queue length, average waiting time, and
17-12
probability of having zero customers waiting are computed. The analysis revealed that
with the current service time, the bank needed over 500 new employees. Hiring so many
new tellers was too costly and physically impossible. Alternatively, the bank could
achieve its goal by reducing the average service time. The investigation of the collected
data helped to identify potential improvements in service. Accordingly, customer
processing is reengineered, proficiency of tellers is improved and efficient schedules are
obtained. Heuristic algorithms are incorporated in the model to make it more realistic.
The model allowed KeyCorp to reduce the processing time by 53%. As a result of this,
the customer wait time has decreased and the percentage of customers who wait more
than five minutes is reduced to 4%. In addition to increased customer satisfaction, the
new system resulted in the reduction of operating costs. Savings from personnel expenses
is estimated to be $98 million over five years whereas the cost of the new system was
only half a million dollars. The reports generated from the data are used in obtaining
better schedules and identifying service components that are open to improvement.
Efficient scheduling and reduced personnel released 15% of the capacity, which can now
be used for more profitable investments. KeyCorp also gained more credibility by using a
systematic approach in making decisions. KeyCorp management, customers, employees
and shareholders all benefit from this study.
17.6-2.
(a)
M/M/1 queue with
and
Proportion of the time the storage space will be adequate:
(b)
n
Pn
0
1
2
3
4
0.5
0.25
0.125
0.0625
0.03125
Total
17.6-3.
(proportion of time no one is waiting)
17.6-4.
(a)
Exp
(b)
if
if
17-13
17.6-5.
Use the equalities
and
.
17.6-6.
The system without the storage restriction is an M/M/1 queue with
and
. The
proportion of the time that
square feet floor space is adequate for waiting jobs is
. Hence, the goal is to find
such that
for
and
,
,
.
ln
ln
ln
ln
Part
(a)
(b)
(c)
ln
Floor space required
ln
17.6-7.
(a) TRUE. A customer does not wait before the service begins if and only if there is no
one in the system, so the long-run probability that the customer does not wait is
.
(b) FALSE. The expected number of customers in the system is
not proportional to .
(c) FALSE. When is increased from
to
increased from
to
, increases from
17.6-8.
, increases from
to
.
, so it is
to
. When it is
(a) FALSE. A temporary return to the state where no customers are present is possible.
(b) TRUE. Since
(c) TRUE. Since
, the queue grows without bound.
, the system can reach steady-state conditions.
17.6-9.
(a) TRUE. "
(b) FALSE. "
p.787.
has an exponential distribution with parameter
does not quite have an exponential distribution, because
17-14
," p.787.
,"
(c) TRUE. "
represents the conditional waiting time given customers already in
the system. As discussed in Sec. 17.7,
is known to have an Erlang distribution,"
p.787.
17.6-10.
(a)
customers,
hours
hours
customers
,
There is a
,
% chance of having more than 2 customers at the checkout stand.
(b)
s=
Data
30
40
1
Results
(mean arrival rate)
(mean service rate)
(# servers)
Pr(W > t) = 3.98E-31
when t =
L=
Lq =
3
2.25
W=
Wq =
0.1
0.075
7
0.75
Prob(W q > t) = 1.45E-22
when t =
5
Pn
n
0
1
2
(c)
cumulative
0.25
0.1875
0.140625
customer,
hrs
hrs,
,
There is a
0.25
0.4375
0.578125
customers
,
% chance of having more than 2 customers at the checkout stand.
(d)
s=
Data
20
40
1
Results
(mean arrival rate)
(mean service rate)
(# servers)
Pr(W > t) = 1.58E-61
when t =
L=
Lq =
1
0.5
W=
Wq =
0.025
0.05
7
0.5
Prob(W q > t) = 1.86E-44
when t =
5
n
0
1
2
Pn
cumulative
0.5
0.25
0.125
0.5
0.75
0.875
(e) The manager should hire another person to help the cashier by bagging the groceries.
17-15
17.6-11.
(a) All the criteria are currently satisfied.
s=
Data
10
20
1
Results
(mean arrival rate)
(mean service rate)
(# servers)
Pr(W > t) = 0.006738
when t =
L=
Lq =
1
0.5
W=
Wq =
0.05
0.1
0.5
0.5
Prob(W q > t) = 0.003369
when t =
0.5
n
0
1
2
3
4
5
Pn
cumulative
0.5
0.25
0.125
0.0625
0.03125
0.015625
0.5
0.75
0.875
0.9375
0.96875
0.984375
(b) None of the criteria are satisfied.
s=
Data
15
20
1
Results
(mean arrival rate)
(mean service rate)
(# servers)
Pr(W > t) = 0.082085
when t =
L=
Lq =
3
2.25
W=
Wq =
0.15
0.2
0.5
0.75
Prob(W q > t) = 0.061564
when t =
0.5
n
Pn
cumulative
0
0.25
0.25
1
0.1875
0.4375
2
0.140625
0.578125
3 0.10546875 0.68359375
4 0.079101563 0.76269531
5 0.059326172 0.82202148
(c) The first and third criteria are satisfied, but the second is not.
s=
Data
25
20
2
Results
L = 2.051282051
Lq = 0.801282051
(mean arrival rate)
(mean service rate)
(# servers)
W = 0.082051282
W q = 0.032051282
Pr(W > t) = 0.001022
when t =
0.5
0.625
Prob(W q > t) = 0.000266
when t =
0.5
17-16
n
Pn
0
1
2
3
4
5
0.230769231
0.288461538
0.180288462
0.112680288
0.07042518
0.044015738
cumulative
0.23076923
0.51923077
0.69951923
0.81219952
0.8826247
0.92664044
17.6-12.
(a) All the guidelines are currently met.
s=
Data
2
1
4
(mean arrival rate)
(mean service rate)
(# servers)
W = 1.086956522
W q = 0.086956522
Pr(W > t) = 0.007902
when t =
Results
L = 2.173913043
Lq = 0.173913043
5
0.5
Prob(W q > t) =
when t =
7.9E-06
5
n
Pn
0
1
2
3
4
5
6
7
8
9
0.130434783
0.260869565
0.260869565
0.173913043
0.086956522
0.043478261
0.02173913
0.010869565
0.005434783
0.002717391
cumulative
0.1304
0.3913
0.6522
0.8261
0.9130
0.9565
0.9783
0.9891
0.9946
0.9973
(b) The first two guidelines will not be satisfied in a year, but the third will be.
s=
Data
3
1
4
(mean arrival rate)
(mean service rate)
(# servers)
W = 1.509433962
W q = 0.509433962
Pr(W > t) = 0.023901
when t =
Results
L = 4.528301887
Lq = 1.528301887
5
0.75
Prob(W q > t) = 0.003433
when t =
5
(c) Five tellers are needed in a year.
17-17
n
Pn
0
1
2
3
4
5
6
7
8
9
0.037735849
0.113207547
0.169811321
0.169811321
0.127358491
0.095518868
0.071639151
0.053729363
0.040297022
0.030222767
cumulative
0.0377
0.1509
0.3208
0.4906
0.6179
0.7134
0.7851
0.8388
0.8791
0.9093
17.6-13.
(a)
(b)
17.6-14.
s=
Data
10
12
1
(mean arrival rate)
(mean service rate)
(# servers)
W=
0.5
W q = 0.416666667
Pr(W > t) = 2.06E-09
when t =
Results
L=
5
Lq = 4.166666667
10
0.833333333
Prob(W q > t) = 0.833333
when t =
0
n
Pn
0 0.166666667
1 0.138888889
s=
Data
10
12
2
(mean arrival rate)
(mean service rate)
(# servers)
0.1667
0.3056
Results
L = 1.008403361
Lq = 0.175070028
W = 0.100840336
W q = 0.017507003
Pr(W > t) = 1.89E-52
when t =
cumulative
10
0.416666667
Prob(W q > t) = 0.245098
when t =
0
n
Pn
0 0.411764706
1 0.343137255
2 0.142973856
17-18
cumulative
0.4118
0.7549
0.8979
s=
Data
10
12
3
Results
L = 0.855529512
Lq = 0.022196179
(mean arrival rate)
(mean service rate)
(# servers)
W = 0.085552951
W q = 0.002219618
Pr(W > t) = 8.05E-53
when t =
10
0.277777778
Prob(W q > t) =
when t =
s=
0.05771
0
Data
10
12
4
Pn
0
1
2
3
0.432132964
0.360110803
0.150046168
0.041679491
cumulative
0.4321
0.7922
0.9423
0.9840
Results
L = 0.836234411
Lq = 0.002901077
(mean arrival rate)
(mean service rate)
(# servers)
W = 0.083623441
W q = 0.000290108
Pr(W > t) = 7.71E-53
when t =
n
10
0.208333333
Prob(W q > t) = 0.011024
when t =
s=
0
10
12
5
(mean arrival rate)
(mean service rate)
(# servers)
Pn
0
1
2
3
4
0.434331675
0.361943063
0.150809609
0.041891558
0.008727408
cumulative
0.4343
0.7963
0.9471
0.9890
0.9977
L = 0.833682622
Lq = 0.000349289
W = 0.083368262
W q = 3.49289E-05
Pr(W > t) = 7.67E-53
when t =
n
10
0.166666667
Prob(W q > t) = 0.001746
when t =
0
Part
Number of servers
(a)
(b)
(c)
(d)
17-19
(e)
(f)
n
Pn
0
1
2
3
4
5
0.434571213
0.362142678
0.150892782
0.041914662
0.008732221
0.00145537
(g)
cumulative
0.4346
0.7967
0.9476
0.9895
0.9983
0.9997
17.6-15.
M/M/1 queue with
An arriving customer does not have to wait before service
Expected price of gasoline per gallon:
$
17.6-16.
Expected cost per customer:
17.6-17.
Let
and
.
Differentiate both sides of the equation.
Integrate to get
.
17.6-18.
(a) Let
and
.
Differentiate both sides of the equation.
(b) Let
and
.
17-20
for
Differentiate both sides of the equation.
17.6-19.
Mean rate at which service completion occurs during the periods when no customers are
waiting in the queue:
17.6-20.
s=
Data
4
6
2
Pr(W > t) =
1
when t =
0
Results
L=
0.75
Lq = 0.083333333
(mean arrival rate)
(mean service rate)
(# servers)
W=
0.1875
W q = 0.020833333
0.333333333
Prob(W q > t) = 0.003053
when t =
0.5
n
Pn
0
0.5
1 0.333333333
number of customers
number of customers
number of customers
17.6-21.
(a)
hours
Clara
hours
Clarence
total
Clara
Clara
minutes
minutes
minutes
Clarence
Clarence
hours
17-21
(b) It is an M/M/2 queue,
hours.
,
, and
. OR Courseware gives
(c)
An expected processing time of
minutes results in the same expected waiting time.
17.6-22.
(a)
Current system:
Next year's system:
The next year's system yields smaller , but larger
,
and
.
(b)
(c)
17.6-23.
(a) The future evolution of the queueing system is affected by whether the parameter of
the service time distribution for the customer currently in service is
or . Therefore,
the current state of the system needs to include this information from the history of the
process. Let the state
be the number of customers in the system and the index of
the current service rate. Note that the state
does not need an index of service rate.
if the current parameter is
if the current parameter is
,
.
17-22
(b)
for
for
(c) Truncate the balance equations at a very large and then solve the resulting finite
system of equations numerically. The resulting approximation of the stationary
distribution should be good if the steady-state probability that the number of customers in
the original system exceeds is negligible.
(d)
(e) Because the input is Poisson, the distribution of the state of the system is the same just
before an arrival and at an arbitrary point in time.
A new arrival finds the system in state
A new arrival finds the system in state
A new arrival finds the system in state
The three conditional distributions of
and Erlang
, (3) Erlang
are (1) Exp
, (2) a convolution of Exp
respectively.
17.6-24.
(a)
(0)
(1)
( )
The solution given in Sec. 17.6 is:
the balance equations.
for
(0)
( )
Hence, the solution satisfies the balance equations.
17-23
. Substitute this in
(b)
The solution given in Sec. 17.6 is:
for
balance equations.
Hence, the solution satisfies the balance equations.
(c)
The solution given in Sec. 17.6 is:
for
.
Substitute this in the balance equations.
Hence, the solution satisfies the balance equations.
17-24
. Substitute this in the
17.6-25.
(a)
s=
Data
6
4
3
Results
L = 1.736842105
Lq = 0.236842105
(mean arrival rate)
(mean service rate)
(# servers)
W = 0.289473684
W q = 0.039473684
Pr(W > t) = 0.025817
when t =
1
0.5
Prob(W q > t) = 0.236842
when t =
0
Pn
n
0 0.210526316
1 0.315789474
2 0.236842105
(b)
A phone is answered immediately
Or
A phone is answered immediately
if
if
(d) Finite Queue Variation
s=
K=
Data
6
4
3
3
Results
L = 1.29851
Lq =
0
(mean arrival rate)
(mean service rate)
(# servers)
(max customers)
W = 0.2500
Wq =
0
0.5
An arriving call is lost
n
Pn
0
1
2
3
0.23881
0.35821
0.26866
0.13433
All three servers are busy
17-25
0.2105
0.5263
0.7632
At least one server is free
(c)
calls on hold
cumulative
17.6-26.
These form M/M/1/K queues with
, and
and the fraction of customers lost is
respectively,
and
, so
.
(a) Zero spaces:
(b) Two spaces:
(c) Four spaces:
17.6-27.
M/M/s/K model
17.6-28.
and
represent the waiting times of arriving customers who enter the system. The
probability that such a customer finds customers in the system already is:
for
for
customers in system system not full
(a)
(b)
17-26
.
17.6-29.
(a) - (b)
s=
K=
Data
20
30
1
2
Results
L = 0.73684
Lq = 0.21053
(mean arrival rate)
(mean service rate)
(# servers)
(max customers)
W = 0.0467
W q = 0.01333
0.66667
s=
K=
Data
20
30
1
3
Results
L = 1.01538
Lq = 0.43077
(mean arrival rate)
(mean service rate)
(# servers)
(max customers)
W = 0.0579
W q = 0.02456
0.66667
s=
K=
Data
20
30
1
4
Results
L = 1.24171
Lq = 0.62559
(mean arrival rate)
(mean service rate)
(# servers)
(max customers)
W = 0.0672
W q = 0.03385
0.66667
s=
K=
Data
20
30
1
5
Results
L = 1.42256
Lq = 0.78797
(mean arrival rate)
(mean service rate)
(# servers)
(max customers)
W = 0.0747
W q = 0.04139
0.66667
(c)
Spaces
Rate
at which
customers are lost
Change in
Profit / hour
$
$
$
$
$
Change in
Profit / hour
$
$
$
(d) Since it costs $
per month per car length rented, each additional space must bring
at least $
per month (or $ per hour) in additional profit. Five spaces still bring more
than that, so five should be provided.
17-27
17.6-30.
(a) The M/M/s model with finite calling population fits this queueing system.
(b) The probabilities that there are 0, 1, 2, or 3 machines not running are
respectively. The mean of this distribution is
.
Data
0.111111
0.5
s=
1
N=
3
(exponential parameter)
(mean service rate)
(# servers)
(size of population)
,
,
, and
Results
L = 0.71805274
Lq = 0.21095335
W=
Wq =
2.832
0.832
0.66666667
-bar =
(c)
0.2535497
n
Pn
0
1
2
3
0.49290061
0.32860041
0.14604462
0.03245436
hours
(d) The expected fraction of time that the repair technician will be busy is the system
utilization, which is
.
(e) M/M/s model
Data
0.333333 (mean arrival rate)
0.5
(mean service rate)
s=
1
(# servers)
Pr(W > t) =
1
when t =
0
Results
L=
2
Lq = 1.333333333
W=
Wq =
6
4
0.666666667
M/M/s/K model
Data
0.333333
0.5
s=
1
K=
3
(mean arrival rate)
(mean service rate)
(# servers)
(max customers)
Results
L = 1.01538
Lq = 0.43077
W = 3.4737
W q = 1.47368
0.66667
17-28
(f)
Data
0.111111
0.5
s=
2
N=
3
(exponential parameter)
(mean service rate)
(# servers)
(size of population)
Results
L = 0.55280899
Lq = 0.00898876
W = 2.03305785
W q = 0.03305785
0.33333333
-bar = 0.27191011
n
Pn
0
1
2
3
0.54606742
0.36404494
0.08089888
0.00898876
The probabilities that there are 0, 1, 2, or 3 machines not running are , , , and
respectively. The mean of this distribution is
. The expected fraction of time
that the repair technician will be busy is the system utilization,
.
17.6-31.
(a) This is an M/M/s model with a finite calling population, with
and
.
,
(b)
s=
N=
Data
1
2
1
3
(exponential parameter)
(mean service rate)
(# servers)
(size of population)
Results
L = 1.42105263
Lq = 0.63157895
W=
Wq =
0.9
0.4
1.5
-bar = 1.57894737
17-29
n
Pn
0
1
2
3
0.21052632
0.31578947
0.31578947
0.15789474
,
,
17.6-32.
(a) Alternative 1:
s=
N=
Data
0.4
4
1
3
(exponential parameter)
(mean service rate)
(# servers)
(size of population)
Results
L = 0.32064422
Lq = 0.05270864
W = 0.29918033
W q = 0.04918033
Three machines are the maximum that can be assigned to an operator while still
achieving the required production rate. The average number of machines not running is
, so
% of the machines are running on the average. The
utilization of servers is
.
(b) Alternative 2:
s=
N=
Data
0.4
4
3
12
(exponential parameter)
(mean service rate)
(# servers)
(size of population)
Results
L = 1.1246521
Lq = 0.03711731
W = 0.25853244
W q = 0.00853244
Three operators are needed to achieve the required production rate. The average number
of machines not running is
, so
% of the machines are
running on the average. The utilization of servers is
.
(c) Alternative 3:
s=
N=
Data
0.4
8
1
12
(exponential parameter)
(mean service rate)
(# servers)
(size of population)
Results
L = 1.03577079
Lq = 0.48755933
W = 0.23617045
W q = 0.11117045
Two operators are needed to achieve the required production rate. The average number of
machines not running is
, so
% of the machines are
running on the average. The utilization of servers is
.
17.6-33.
(a) Let the state
be the number of failed machines (
and the stage of
service for the machine under repair (
if all machines are running properly or
otherwise).
17-30
(b)
(c)
State
Balance Equation
17.7-1.
(a)
(i)
Exponential:
(ii)
Constant:
(iii)
Erlang:
Exp
C
Erlang
Exp
C
(b) Let
respectively. Now,
Erlang
when the distribution is exponential, constant and Erlang
and
.
Hence, the expected waiting time is reduced by
mained the same.
17-31
% and the expected queue length re-
17.7-2.
(a)
s=
Data
0.2
4
4
1
s=
Data
0.2
4
3
1
s=
Data
0.2
4
2
1
s=
Data
0.2
4
1
1
s=
Data
0.2
4
0
1
(mean arrival rate)
(expected service time)
(standard deviation)
(# servers)
(mean arrival rate)
(expected service time)
(standard deviation)
(# servers)
(mean arrival rate)
(expected service time)
(standard deviation)
(# servers)
(mean arrival rate)
(expected service time)
(standard deviation)
(# servers)
(mean arrival rate)
(expected service time)
(standard deviation)
(# servers)
(b) If
,
is half of the value with
ability of the service times.
(c)
L=
Lq =
Results
4.000
3.200
W=
Wq =
20.000
16.000
L=
Lq =
Results
3.300
2.500
W=
Wq =
16.500
12.500
L=
Lq =
Results
2.800
2.000
W=
Wq =
14.000
10.000
L=
Lq =
Results
2.500
1.700
W=
Wq =
12.500
8.500
L=
Lq =
Results
2.400
1.600
W=
Wq =
12.000
8.000
, so it is quite important to reduce the vari-
Change
largest reduction
smallest reduction
(d)
needs to be increased by
to achieve the same
17-32
.
17.7-3.
M/G/1 with
:
(a) FALSE. When
fixed.
and
(b) FALSE. Smaller
increase, both
and
and
increase too provided that
do not necessarily imply a smaller
. Even though
is
. For example, let
and
,
.
(c) TRUE. If the service time is exponential,
constant,
and
so that
.
(d) FALSE. It is possible to find a distribution with
.
17.7-4.
(a)
hours
hours
(b)
hours
hours
(c)
in (b) is half of
in (a).
(d) Marsha needs to reduce her service time to approximately 61 seconds.
17-33
. If it is
17.7-5.
(a)
(b) Poisson input with
and Erlang service times with
,
.
(c)
(d)
(e)
k=
s=
Data
1
2
2
1
(mean arrival rate)
(mean service rate)
(shape parameter)
(# servers)
L=
Lq =
Results
0.875
0.375
W=
Wq =
0.875
0.375
17.7-6.
(a)
Current Policy:
Results
s=
Data
1
2
1
(mean arrival rate)
(mean service rate)
(# servers)
L=
Lq =
1
0.5
W=
Wq =
0.5
1
Proposal:
k=
s=
Data
0.25
0.5
4
1
(mean arrival rate)
(mean service rate)
(shape parameter)
(# servers)
L=
Lq =
Results
0.8125
0.3125
W=
Wq =
3.25
1.25
Under the current policy, an airplane loses one day of flying time as opposed to 3.25 days
under the proposed policy.
17-34
(b) Under the current policy, one airplane is losing flying time each day as opposed to
0.8125 airplanes under the proposed policy.
(c) The comparison in (b) is the appropriate one for making the decision, since it takes
into account that airplanes will not have to come in for service as often.
17.7-7.
(a) Let the state
be the number of airplanes at the base and the stage of service of
the airplane being overhauled.
(b)
17.7-8.
For the current arrangement,
3 and
, so
8.
Model
Fig. 17.6
Fig. 17.8
Fig. 17.10
Fig. 17.11
24 and
3 , so
Current
at each crib Total
4
2.4
48
3.2
6.4
2.2
4.4
17-35
8. For the proposal,
Proposal
167
098
133
99
4 4
3.1
3.7
2.8
093
.064
078
058
48,
17.7-9.
(a) Let the state
served at the
denote calling units in the system with the calling unit being
th stage of its service. Then, the state space is
.
Note that this analysis is possible because an Erlang distribution with parameters
and
is equivalent to the distribution of the sum of two independent
exponential random variables each with parameter
. The steady-state
equations are:
.
(b) The solution of the steady-state equations:
(c) If the service time is exponential, then the system is an M/M/1 queue with capacity
,
and
.
17-36
17.7-10.
Let the state
represent the number of customers in the system
number of completed arrival stages for currently arriving customer
and the
.
17.7-11.
(a) Let
be the repair time.
minor repair needed
major repair needed
hours
Now let
be a Bernoulli random variable with
and
,
be an exponential random variable with mean
.
,
where
are independent.
var
var
var
var
17-37
for
, where
and
var
var
var
Observe that
has a much larger variance than
exponential random variable with the same mean.
, the variance of an
(b) M/G/1 queue with
(c)
major repair needed
minor repair needed
major repair machines
minor repair machines
(d) Let the state
denote the number of failed machines and the type of repair being
done on the machine under repair (
represents minor repair and
represents
major repair).
(e)
17-38
17.7-12.
(a)
if
if
where
is the number of arrivals in
and
,
minutes.
and
(b) Using the OR Courseware:
(c)
M/D/1 model:
17.8-1.
(a) This system is an example of a nonpreemptive priority queueing system.
(b)
n=
s=
Data
2
20
1
(# of priority classes)
(mean service rate)
(# servers)
Results
Priority
Priority
Priority
Priority
Priority
Class
Class
Class
Class
Class
1
2
3
4
5
i
2
10
1
1
1
L
Lq
W
Wq
0.1666667
1.3333333
0.2642857
0.3357143
0.45
0.0666667
0.8333333
0.2142857
0.2857143
0.4
0.0833333
0.1333333
0.2642857
0.3357143
0.45
0.0333333
0.0833333
0.2142857
0.2857143
0.4
12
0.6
(c)
(d)
17.8-2.
If
hours
hours
is the primary concern, one should choose the alternative with one fast server. If
is the primary concern, one should choose the alternative with two slow servers.
17-39
17.8-3.
(a)
(b)
The approximation is not good for
and
.
17.8-4.
(a) First-come-first-served:
days
(b) Nonpreemptive priority:
days
days
days
(c) Preemptive priority:
days
days
days
17-40
17.8-5.
Preemptive
Nonpreemptive
17.8-6.
(a) The expected number of customers would not change since customers of both types
have exactly the same arrival pattern and service times. The change of the priority would
not affect the total service rate from the server's view and thus, the total queue size stays
the same.
(b) Using the template for M/M/s nonpreemptive priorities queueing model:
n=
s=
Data
2
6
2
(# of priority classes)
(mean service rate)
(# servers)
Results
Priority
Priority
Priority
Priority
Priority
Class
Class
Class
Class
Class
1
2
3
4
5
i
5
5
1
1
1
L
Lq
W
Wq
1.3744589 0.5411255 0.2748918 0.1082251
4.0800866 3.2467532 0.8160173 0.6493506
4.7121212 4.5454545 4.7121212 4.5454545
#DIV/0!
#DIV/0!
#DIV/0!
#DIV/0!
#DIV/0!
#DIV/0!
#DIV/0!
#DIV/0!
10
0.8333333
17-41
Using the template for M/M/s queueing model:
Results
L = 5.454545455
Lq = 3.787878788
s=
Hence,
Data
10
6
2
(mean arrival rate)
(mean service rate)
(# servers)
W = 0.545454545
W q = 0.378787879
0.833333333
.
17.8-7.
Let the state
denote jobs of high priority and jobs of low priority.
State
Balance Equation
for
for
for
17.9-1.
GM launched this project to improve the throughput of its production lines. A sequence
of stations through which parts move sequentially until completion is called a production
line. These stations are separated by finite-capacity buffers. Since machines may have
unequal speeds and fail randomly, analyzing even simple production lines is not easy. To
overcome the difficulties in measuring throughput and identifying bottlenecks, GM
developed a throughput-analysis tool named C-MORE. The analysis assumes unreliable
stations with deterministic speeds, exponential failure and repair times. Analytic
decomposition and simulation methods are deployed. Analytic decomposition is based on
first solving the two-station problem and then extending the results to multiple stations.
Each station is modeled as a single-server queueing system with constant interarrival and
service times. The server at each station can fail randomly. The first station is blocked
17-42
and shuts down if its buffer is full and the second station is starved and shuts down if
there are no jobs completed by the first station. The state of the system includes
information about blocked and starved stations, downtimes, and buffer contents. Closedform expressions for the steady-state distribution of buffer contents when both stations
are up are obtained. The output includes throughput, system-time and work-in-process
averages, average state of the system, bottleneck and sensitivity analysis.
The results of this study include enhanced throughput, lowered overtime and increased
sales of high-demand products. These improvements translated into savings of more than
$2.1 billion. The use of a systematic approach enabled GM to make reliable decisions
about equipment purchases, product launch times and maintenance schedules while
meeting its production targets. Consequently, unprofitable investments and unfruitful
improvement efforts are avoided. Alternatives are evaluated efficiently and questions are
answered accurately. Continuous improvement of productivity is made possible. Overall,
this study provided GM a competitive advantage in the industry. Following this study,
OR has been widely adopted throughout the organization.
17.9-2.
(a) Let the state
(b) Let the state
be the number of type 1 customers in the system.
be the number of customers in the system.
17-43
(c) Let the state
respectively
be the number of type 1 and type 2 customers in the system
17.9-3.
(a)
(b)
(c)
hour
minutes
17.9-4.
In a system of infinite queues in series, customers are served at
service facilities in a
fixed order. Each facility has an infinite queue capacity. The arrivals from outside the
system to the first facility form a Poisson process with rate
. There are no arrivals
from outside the system to other facilities, so
for
, this is a Poisson process
with parameter . From the equivalence property, under steady-state conditions, the
arrivals to each facility have a Poisson distribution with rate . Facility has servers
whose service time is exponentially distributed with rate . A customer leaving facility
is routed to facility
with probability if
and leaves the system if
, so for
,
if
else,
and
. It is assumed that
so that the queue does not grow without bound.
17-44
17.9-5.
(a)
(b)
for
for
for
for facility 1
for facility 2
for facility 3
(c)
(d)
(e)
17.10-1.
(a) The optimal number of servers is one.
s=
Data
8
10
1
Results
(mean arrival rate)
(mean service rate)
(# servers)
Pr(W > t) = 0.904837
when t =
L=
Lq =
4
3.2
W=
Wq =
0.4
0.5
0.05
0.8
Prob(W q > t) =
when t =
0.72387
0.05
n
Economic Analysis:
Cs = $100.00 (cost / server / unit time)
Cw = $10.00 (waiting cost / unit time)
Cost of Service $100.00
Cost of Waiting $40.00
Total Cost $140.00
17-45
0
1
2
3
4
5
6
7
Pn
0.2
0.16
0.128
0.1024
0.08192
0.065536
0.0524288
0.04194304
(b) The optimal number of servers is two.
s=
Data
8
10
2
(mean arrival rate)
(mean service rate)
(# servers)
W = 0.119047619
W q = 0.019047619
Pr(W > t) = 0.672495
when t =
Results
L = 0.952380952
Lq = 0.152380952
0.05
0.4
Prob(W q > t) = 0.125443
when t =
0.05
Economic Analysis:
Cs = $100.00 (cost / server / unit time)
Cw = $100.00 (waiting cost / unit time)
Cost of Service $200.00
Cost of Waiting $95.24
Total Cost $295.24
n
Pn
0
1
2
3
4
5
6
7
0.428571429
0.342857143
0.137142857
0.054857143
0.021942857
0.008777143
0.003510857
0.001404343
(c) The optimal number of servers is three.
s=
Data
8
10
3
(mean arrival rate)
(mean service rate)
(# servers)
W = 0.102365115
W q = 0.002365115
Pr(W > t) = 0.618397
when t =
Results
L = 0.818920916
Lq = 0.018920916
0.05
0.266666667
Prob(W q > t) =
when t =
0.01732
0.05
Economic Analysis:
Cs = $10.00 (cost / server / unit time)
Cw = $100.00 (waiting cost / unit time)
Cost of Service $30.00
Cost of Waiting $81.89
Total Cost $111.89
17-46
n
Pn
0
1
2
3
4
5
6
7
0.447154472
0.357723577
0.143089431
0.038157182
0.010175248
0.0027134
0.000723573
0.000192953
17.10-2.
Jim should operate four cash registers during the lunch hour.
s=
Data
66
30
4
(mean arrival rate)
(mean service rate)
(# servers)
W = 0.037533312
W q = 0.004199979
Pr(W > t) = 0.267335
when t =
Results
L = 2.477198599
Lq = 0.277198599
0.05
0.55
Prob(W q > t) = 0.015242
when t =
0.05
Economic Analysis:
Cs =
$9.00
Cw = $18.00
Cost of Service
Cost of Waiting
Total Cost
(cost / server / unit time)
(waiting cost / unit time)
$36.00
$44.59
$80.59
n
Pn
0
1
2
3
4
5
6
7
0.104562001
0.230036403
0.253040043
0.185562698
0.102059484
0.056132716
0.030872994
0.016980147
17.10-3.
The company needs a total of six machines to minimize its expected total cost per hour.
s=
Data
30
12
6
(mean arrival rate)
(mean service rate)
(# servers)
W = 0.084462972
W q = 0.001129638
Pr(W > t) = 0.556903
when t =
Results
L = 2.533889152
Lq = 0.033889152
0.05
0.416666667
Prob(W q > t) =
when t =
0.00581
0.05
Economic Analysis:
Cs =
$1.50
Cw = $25.00
Cost of Service
Cost of Waiting
Total Cost
(cost / server / unit time)
(waiting cost / unit time)
$9.00
$63.35
$72.35
17.11-1.
Answers will vary.
17.11-2.
Answers will vary.
17-47
n
Pn
0
1
2
3
4
5
6
7
0.081620259
0.204050648
0.25506331
0.212552759
0.132845474
0.066422737
0.02767614
0.011531725
Case%17.1%
!
a)! Status!quo!at!the!presses!–!7.52!sheets!of!in6process!inventory.!
A
1
2
3
4
5
6
B
C
D
E
G
H
Template for the M/M/s Queueing Model
λ=
µ=
s=
Data
7
1
10
(mean arrival rate)
(mean service rate)
(# servers)
L=
Lq =
Results
7.517372837
0.517372837
!
Status!quo!at!the!inspection!station!–!3.94!wing!sections!of!in6process!inventory.!
A
1
2
3
4
5
6
B
C
D
E
F
G
Template for M/D/1 Queueing Model
λ=
µ=
s=
Data
7
8
1
(mean arrival rate)
(mean service rate)
(# servers)
L=
Lq =
Results
3.9375
3.0625
!
Inventory!cost!=!(7.52!+!3.94)($8/hour)!=!$91.68!/!hour!
Machine!cost!=!(10)($7/hour)!=!$70!/!hour!
Inspector!cost!=!$17!/!hour!
!
Total!cost!=!$178.68!/!hour!
!
b)! Proposal!1!will!increase!the!in6process!inventory!at!the!presses!to!11.05!sheets!
since!the!mean!service!rate!has!decreased.!
A
1
2
3
4
5
6
B
C
D
E
G
H
Template for the M/M/s Queueing Model
Data
λ=
7
(mean arrival rate)
µ = 0.83333333 (mean service rate)
s=
10
(# servers)
L=
Lq =
Results
11.04740664
2.647406638
!
The!in6process!inventory!at!the!inspection!station!will!not!change.!
!
Inventory!cost!=!(11.05!+!3.94)($8/hour)!=!$119.92!/!hour!
Machine!cost!=!(10)($6.50)!=!$65!/!hour!
Inspector!cost!=!$17!/!hour!
!
Total!cost!=!$201.92!/!hour!
!
This!total!cost!is!higher!than!for!the!status!quo!so!should!not!be!adopted.!!The!
main!reason!for!the!higher!cost!is!that!slowing!down!the!machines!won’t!change!
in6process!inventory!for!the!inspection!station.!
17-48
!
c)! Proposal!2!will!increase!the!in6process!inventory!at!the!inspection!station!to!
4.15!wing!sections!since!the!variability!of!the!service!rate!has!increased.!
!
The!in6process!inventory!at!the!presses!will!not!change.!
!
Inventory!cost!=!(7.52!+!4.15)($8/hour)!=!$93.36!/!hour!
Machine!cost!=!(10)($7/hour)!=!$70!/!hour!
Inspector!cost!=!$17!/!hour!
!
Total!cost!=!$180.36!/!hour!
!
This!total!cost!is!higher!than!for!the!status!quo!so!should!not!be!adopted.!The!
main!reason!for!the!higher!cost!is!the!increase!in!the!service!rate!variability!and!
the!resulting!increase!in!the!in6process!inventory.!
!
!
!
d)! They!should!consider!increasing!power!to!the!presses!(increasing!there!cost!to!
$7.50!per!hour!but!reducing!their!average!time!to!form!a!wing!section!to!0.8!
hours).!This!would!decrease!the!in6process!inventory!at!the!presses!to!5.69.!
A
1
2
3
4
5
6
B
C
D
E
G
H
Template for the M/M/s Queueing Model
λ=
µ=
s=
Data
7
1.25
10
(mean arrival rate)
(mean service rate)
(# servers)
!
!
Inventory!cost!=!(5.69!+!3.94)($8/hour)!=!$77.04!/!hour!
Machine!cost!=!(10)($7.50/hour)!=!$75!/!hour!
Inspector!cost!=!$17!/!hour!
!
Total!cost!=!$169.04!/!hour!
!
This!total!cost!is!lower!than!the!status!quo!and!both!proposals.!
%
17-49
L=
Lq =
Results
5.688419945
0.088419945
Case%17.2%
The!operations!of!the!records!and!benefits!call!center!can!be!modeled!as!an!
M/M/s!queueing!system.!We,!therefore,!use!the!template!for!the!M/M/s!queueing!
model!throughout!this!case.!The!mean!arrival!rate!equals!70!per!hour,!and!the!
mean!service!rate!of!every!representative!equals!6!per!hour.!Mark!needs!at!least!
s=12!representatives!answering!phone!calls!to!ensure!that!the!queue!does!not!
grow!indefinitely.!
!
a)! In!order!to!solve!this!problem!we!have!to!determine!the!number!of!servers!by!
"trial!and!error"!until!we!find!a!number!s!such!that!the!probability!of!waiting!
more!than!4!minutes!in!the!queue!is!above!35%.!
!
For!13!servers,!the!probability!that!a!customer!has!to!wait!more!than!4!minutes!
equals!36.3%.!It!appears!that!Mark!currently!employs!13!servers.!
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
B
C
D
E
G
H
Template for the M/M/s Queueing Model
Data
70
6
13
λ=
µ=
s=
(mean arrival rate)
(mean service rate)
(# servers)
Pr(W > t) = 0.8256082
when t = 0.06666667
Prob(Wq > t) = 0.36291401
when t = 0.06666667
0.08
0.07
0.06
0.05
Probability 0.04
0.03
0.02
0.01
0
0
1
2
3
4
5
6
7
8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
Number of Customers in System
17-50
L=
Lq =
Results
17.07963527
5.4129686
W=
Wq =
0.24399479
0.077328123
ρ=
0.897435897
n
0
1
2
3
4
5
6
7
8
9
10
5.32592E-06
6.21358E-05
0.000362459
0.001409561
0.004111221
0.009592849
0.018652761
0.031087935
0.045336573
0.058769631
0.06856457
Pn
!
!
b)! Using!the!same!procedure!as!in!part!a!we!find!that!for!s!=!18!servers!the!
probability!of!waiting!more!than!1!minute!drops!below!5%:!
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
B
C
D
E
G
H
Template for the M/M/s Queueing Model
Data
70
6
18
λ=
µ=
s=
(mean arrival rate)
(mean service rate)
(# servers)
Pr(W > t) = 0.90907539
when t = 0.01666667
Prob(Wq > t) = 0.03207826
when t = 0.01666667
L=
Lq =
Results
11.77798802
0.111321353
W=
Wq =
0.168256972
0.001590305
ρ=
0.648148148
n
0
1
2
3
4
5
6
7
8
9
10
0.14
0.12
0.1
Probability
0.08
0.06
0.04
0.02
0
0
1
2
3
4
5
6
7
8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
Number of Customers in System
Pn
8.49029E-06
9.90534E-05
0.000577812
0.002247045
0.006553882
0.015292391
0.029735204
0.049558673
0.072273065
0.093687307
0.109301858 !
!
!
!
c)! Using!the!same!"trial!and!error"!method!as!before,!we!find!the!minimal!number!
of!servers!necessary!to!ensure!that!80%!of!customers!wait!one!minute!or!less!to!
be!s!=!15.!
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
B
C
D
E
G
H
Template for the M/M/s Queueing Model
Data
70
6
15
λ=
µ=
s=
(mean arrival rate)
(mean service rate)
(# servers)
Pr(W > t) = 0.92671158
when t = 0.01666667
Prob(Wq > t) = 0.19421332
when t = 0.01666667
0.12
0.1
0.08
Probability 0.06
0.04
0.02
0
0
1
2
3
4
5
6
7
8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
Number of Customers in System
17-51
L=
Lq =
Results
12.61532951
0.948662841
W=
Wq =
0.180218993
0.013552326
ρ=
0.777777778
n
0
1
2
3
4
5
6
7
8
9
10
7.80062E-06
9.10072E-05
0.000530875
0.002064516
0.006021504
0.014050177
0.027319789
0.045532982
0.066402265
0.08607701
0.100423178
Pn
!
!
!
The!minimal!number!of!servers!to!ensure!that!95%!of!customers!wait!90!
seconds!or!less!is!s!=!17.!
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
B
C
D
E
G
H
Template for the M/M/s Queueing Model
Data
70
6
17
λ=
µ=
s=
Pr(W > t) =
when t =
(mean arrival rate)
(mean service rate)
(# servers)
0.8705238
0.025
Prob(Wq > t) = 0.04645911
when t =
0.025
0.14
0.12
0.1
Probability
0.08
0.06
0.04
0.02
0
0
1
2
3
4
5
6
7
8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
Number of Customers in System
L=
Lq =
Results
11.89284685
0.226180186
W=
Wq =
0.169897812
0.003231146
ρ=
0.68627451
n
0
1
2
3
4
5
6
7
8
9
10
8.39517E-06
9.79436E-05
0.000571338
0.002221869
0.006480452
0.015121055
0.029402052
0.049003419
0.07146332
0.092637637
0.108077243
Pn
!
!
When!an!employee!of!Cutting!Edge!calls!the!benefits!center!from!work!and!has!
to!wait!on!the!phone,!the!company!loses!valuable!work!time!for!this!customer.!
Mark!should!try!to!estimate!the!amount!of!work!time!employees!lose!when!they!
have!to!wait!on!the!phone.!Then!he!could!determine!the!cost!of!this!waiting!time!
and!try!to!choose!the!number!of!representatives!in!such!a!fashion!that!he!
reaches!a!reasonable!trade6off!between!the!cost!of!employees!waiting!on!the!
phone!and!the!cost!of!adding!new!representatives.!
!
Clearly,!Mark's!criteria!would!be!different!if!he!were!dealing!with!external!
customers.!While!the!internal!customers!might!become!disgruntled!when!they!
have!to!wait!on!the!phone,!they!cannot!call!somewhere!else.!Effectively,!the!
benefits!center!holds!monopolistic!power.!On!the!contrary,!if!Mark!were!running!
a!call!center!dealing!with!external!customers,!these!customers!could!decide!to!do!
business!with!a!competitor!if!they!become!angry!from!waiting!on!the!phone.!
17-52
!
d)! If!the!representatives!can!only!handle!6!calls!per!hour,!then!Mark!needs!to!
employ!18!representatives!(see!part!b).!If!a!representative!can!handle!8!calls!per!
hour,!then!the!minimal!number!of!representatives!equals!14.!
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
B
C
D
E
G
H
Template for the M/M/s Queueing Model
Data
70
8
14
λ=
µ=
s=
(mean arrival rate)
(mean service rate)
(# servers)
Pr(W > t) = 0.88174766
when t = 0.01666667
Prob(Wq > t) =
0.0366495
when t = 0.01666667
0.16
0.14
0.12
0.1
Probability 0.08
0.06
0.04
0.02
0
0
1
2
3
4
5
6
7
8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
Number of Customers in System
L=
Lq =
Results
8.873005049
0.123005049
W=
Wq =
0.126757215
0.001757215
ρ=
0.625
n
0
1
2
3
4
5
6
7
8
9
10
0.000156459
0.001369018
0.005989453
0.017469238
0.038213959
0.066874429
0.097525208
0.12190651
0.133335246
0.129631489
0.113427553
Pn
!
The!cost!of!training!14!employees!equals!(14)($2,500)!=!$35,000!and!saves!
Mark!(4)($30,000)!=!$120,000!in!annual!salary.!In!the!first!year!alone!Mark!
would!save!$85,000!if!he!chose!to!train!all!his!employees!so!that!they!can!handle!
8!instead!of!6!phone!calls!per!hour.!
!
e)! Mark!needs!to!carefully!check!the!number!of!calls!arriving!at!the!call!center!per!
hour.!In!this!case!we!have!made!the!simplifying!assumption!that!the!arrival!rate!
is!constant.!That!assumption!is!unrealistic;!clearly!we!would!expect!more!calls!
during!certain!times!of!the!day,!during!certain!days!of!the!week,!and!during!
certain!weeks!of!the!year.!We!might!want!to!collect!data!on!the!number!of!calls!
received!depending!on!the!time.!This!data!could!then!be!used!to!forecast!the!
number!of!calls!the!center!will!receive!in!the!near!future,!which!in!turn!would!
help!to!forecast!the!number!of!representatives!needed.!
!
Also,!Mark!should!carefully!check!the!number!of!phone!calls!a!
representative!can!answer!per!hour.!Clearly,!the!length!of!a!call!will!depend!on!
the!issue!the!caller!wants!to!discuss.!We!might!want!to!consider!training!
representatives!for!special!issues.!These!representatives!could!then!always!
answer!those!particular!calls.!Using!specialized!representatives!might!increase!
the!number!of!phone!calls!the!entire!center!can!handle.!
!
Finally,!using!an!M/M/s!model!is!clearly!a!great!simplification.!We!need!to!
evaluate!whether!the!assumptions!for!an!M/M/s!model!are!at!least!
approximately!satisfied.!If!this!is!not!the!case,!we!should!consider!more!general!
models!such!as!M/G/s!or!G/G/s.!
17-53
CHAPTER 18: INVENTORY THEORY
18.3-1.
(a)
O œ "&, 2 œ !Þ$!, . œ $! Ê U‡ œ  Ð#ÑÐ$!ÑÐ"&Ñ
œ &%Þ((
!Þ$!
>‡ œ U‡ Î. œ "Þ)$ months
(b)
 $!Þ$!
: œ $ Ê U‡ œ  #Ð$!ÑÐ"&Ñ
œ &(Þ%&
!Þ$!
$
$
W ‡ œ  #Ð$!ÑÐ"&Ñ
!Þ$!  $!Þ$! œ &#Þ##
>‡ œ U‡ Î. œ "Þ*" months
18.3-2.
(a)
O œ #&, 2 œ !Þ!&, . œ '!! Ê U‡ œ  #Ð'!!ÑÐ#&Ñ
œ ((%Þ'
!Þ!&
>‡ œ U‡ Î. œ "Þ#* weeks
(b)
 #!Þ!&
: œ # Ê U‡ œ  #Ð'!!ÑÐ#&Ñ
œ ()%Þ##
!Þ!&
#
#
 #!Þ!&
W ‡ œ  #Ð'!!ÑÐ#&Ñ
œ ('&Þ!*
!Þ!&
>‡ œ U‡ Î. œ "Þ$" weeks
18.3-3.
(a)
d=
K=
h=
L=
WD =
Data
676
$75
$600.00
3.5
365
Q=
Decision
5
(demand/year)
(setup cost)
(unit holding cost)
(lead time in days)
(working days/year)
18-1
Results
Reorder Point 6.482191781
Annual Setup Cost
Annual Holding Cost
Total Variable Cost
$10,140.00
$1,500.00
$11,640.00
(b)
(c)
d=
K=
h=
L=
WD =
Data
676
$75
$600.00
3.5
365
Q=
Decision
13
(demand/year)
(setup cost)
(unit holding cost)
(lead time in days)
(working days/year)
Results
Reorder Point 6.482191781
Annual Setup Cost
Annual Holding Cost
Total Variable Cost
$3,900.00
$3,900.00
$7,800.00
(d)
d=
K=
h=
L=
WD =
Data
676
$75
$600.00
3.5
365
Q=
Decision
13
Results
Reorder Point 6.482191781
(demand/year)
(setup cost)
(unit holding cost)
(lead time in days)
(working days/year)
Annual Setup Cost
Annual Holding Cost
Total Variable Cost
$3,900.00
$3,900.00
$7,800.00
(optimal order quantity)
The results are the same as those obtained in (c).
(e)
 #Ð(&ÑÐ'('Ñ
U‡ œ  #OH
2 œ
!Þ#Ð$!!!Ñ œ "$ computers purchased with each order
(f)
Number of order per year:
H
U
œ
'('
"$
œ &#
VST œ HÐPX Ñ œ Ð"$Ñ "#  œ 'Þ& inventory level when each order is placed
(g) The optimal policy reduces the total variable inventory cost by $$ß )%! per year,
which is a 33% reduction.
18-2
18.3-4.
(a)
d=
K=
h=
L=
WD =
Data
120000
$2,000
$0.48
0
365
Q=
Decision
10000
d=
K=
h=
L=
WD =
Data
120000
$2,000
$0.48
0
365
Q=
Decision
31622.78
(demand/year)
(setup cost)
(unit holding cost)
(lead time in days)
(working days/year)
Reorder Point
Annual Setup Cost
Annual Holding Cost
Total Variable Cost
Results
0
$24,000.00
$2,400.00
$26,400.00
(b)
(c)
(demand/year)
(setup cost)
(unit holding cost)
(lead time in days)
(working days/year)
Reorder Point
Annual Setup Cost
Annual Holding Cost
Total Variable Cost
Results
0
$7,589.47
$7,589.47
$15,178.93
If U is required to be integer:
d=
K=
h=
L=
WD =
Data
120000
$2,000
$0.48
0
365
Q=
Decision
31623
(demand/year)
(setup cost)
(unit holding cost)
(lead time in days)
(working days/year)
18-3
Reorder Point
Annual Setup Cost
Annual Holding Cost
Total Variable Cost
Results
0
$7,589.41
$7,589.52
$15,178.93
(d)
d=
K=
h=
L=
WD =
Q=
Data
120000
$2,000
$0.48
0
365
Decision
31622.78
(demand/year)
(setup cost)
(unit holding cost)
(lead time in days)
(working days/year)
Reorder Point
Results
0
Annual Setup Cost $7,589.47
Annual Holding Cost $7,589.47
Total Variable Cost $15,178.93
(optimal order quantity)
The results are the same as those in (c).
(e)
 #Ð#ß!!!ÑÐ"!ß!!!Ñ
U‡ œ  #OH
œ $"ß '##Þ() gallons purchased with each order
2 œ
!Þ!%
18.3-5.
(a) U‡ will decrease by half.
(b) U‡ will double.
(c) U‡ remains the same.
(d) U‡ will double.
(e) U‡ remains the same.
18.3-6.
(a)
 #Ð(&ÑÐ&!Ñ
U‡ œ  #OH
Ê 2 œ $$ per month,
2 Ê &! œ
2
which is 15% of the acquisition cost.
(b)
d=
K=
h=
L=
WD =
Data
600
$75
$36.00
0
365
Q=
Decision
50
(demand/year)
(setup cost)
(unit holding cost)
(lead time in days)
(working days/year)
Reorder Point
Annual Setup Cost
Annual Holding Cost
Total Variable Cost
Results
0
$900.00
$900.00
$1,800.00
Optimal Order Quantity
d=
K=
h=
L=
WD =
Data
600
$75
$48.00
0
365
Q=
Decision
43.30
(demand/year)
(setup cost)
(unit holding cost)
(lead time in days)
(working days/year)
18-4
Reorder Point
Annual Setup Cost
Annual Holding Cost
Total Variable Cost
Results
0
$1,039.23
$1,039.23
$2,078.46
Current Order Quantity
d=
K=
h=
L=
WD =
Data
600
$75
$48.00
0
365
Q=
Decision
50.00
d=
K=
h=
L=
WD =
Data
600
$75
$48.00
5
300
Q=
Decision
43.30
(demand/year)
(setup cost)
(unit holding cost)
(lead time in days)
(working days/year)
Reorder Point
Annual Setup Cost
Annual Holding Cost
Total Variable Cost
Results
0
$900.00
$1,200.00
$2,100.00
(c)
(demand/year)
(setup cost)
(unit holding cost)
(lead time in days)
(working days/year)
Reorder Point
Annual Setup Cost
Annual Holding Cost
Total Variable Cost
Results
10
$1,039.23
$1,039.23
$2,078.46
(d) VST œ &  Ð&!ÑÐ&Î#&Ñ œ "& hammers, which adds & ‚ $% œ $#! to TVC every
month, $#%! per year.
18.3-7.
O œ "#ß !!!, 2 œ !Þ$!, . œ )ß !!!, : œ &
 &!Þ$!
U‡ œ  #Ð)!!!ÑÐ"#!!!Ñ
œ #'ß !%'
!Þ$!
&
&
W ‡ œ  #Ð)!!!ÑÐ"#!!!Ñ
 &!Þ$!
œ #%ß &(#
!Þ$!
>‡ œ U‡ Î. œ $Þ#' months
18.3-8.
(a)
d=
K=
h=
L=
WD =
Data
6000
$1,000
$100.00
0
365
Q=
Decision
346.41
(demand/year)
(setup cost)
(unit holding cost)
(lead time in days)
(working days/year)
Reorder Point
Results
0
Annual Setup Cost $17,320.51
Annual Holding Cost $17,320.51
Total Variable Cost $34,641.02
(optimal order quantity)
18-5
(b)
Data
d = 6000 (demand/year)
K = $1,000 (setup cost)
h = $100.00 (unit holding cost)
p = $150.00 (unit shortage cost)
Max Inventory Level
Decision
Q = 447.214 (optimal order quantity)
S = 178.885 (optimal maximum shortage)
Results
268.33
Annual Setup Cost $13,416.41
Annual Holding Cost $8,049.84
Annual Shortage Cost $5,366.56
Total Variable Cost $26,832.82
18.3-9.
(a)
d=
K=
h=
p=
Data
676
$75
$600.00
$200.00
(demand/year)
(setup cost)
(unit holding cost)
(unit shortage cost)
Q=
S=
Decision
26
20
(order quantity)
(maximum shortage)
Max Inventory Level
Annual Setup Cost
Annual Holding Cost
Annual Shortage Cost
Total Variable Cost
This TVC is almost half of the optimal value found for Problem 18.3-3.
(b)
18-6
Results
6.00
$1,950.00
$415.38
$1,538.46
$3,903.85
(c)
18.3-10.
:
2
1/3
1
2
3
5
10
 #OH
U‡ œ  2:
:
2
Maximum Inventory Level
2,000
1,414
1,225
1,155
1,095
1,049
500
707
816
866
913
953
18.3-11.
(a)
Maximum inventory: Ð,+ÑU
,
Length of interval M : U,
Average inventory in interval M : Ð,+ÑU
#,
Length of interval MM : U+  U,
Average inventory in interval MM : Ð,+ÑU
#,
Average inventory per cycle:
Holding cost per cycle:
Ê X œ  +O
U 
Ð,+ÑU
#,
Ð,+ÑU
#+2
Ð,+Ñ2U
#,
 +-
18-7
Maximum Shortage
1,500
707
408
289
183
95
(b)
.X
.U
œ  +O
U# 
Ð,+Ñ2
#,
#+,5
œ ! Ê U‡ œ  Ð,+Ñ2
18.3-12.
(a)
D=
K=
I=
N=
Data
5200
$50
0.2
3
(demand/year)
(setup cost)
(inventory holding cost rate)
(number of discount categories)
Range of order quantities
Category Price
Lower Limit Upper Limit
1
$100.00
0
99
2
$95.00
100
499
3
$90.00
500
100000000
Optimal Q
Results
500
Total Variable Cost
$473,020
EOQ
161
165
170
Q*
99
165
500
Annual
Purchase
Cost
$520,000
$494,000
$468,000
Annual
Setup
Cost
$2,626
$1,572
$520
Annual
Holding
Cost
$990
$1,572
$4,500
Total
Variable
Cost
$523,616
$497,143
$473,020
(b)
&#!!
&!! œ "!Þ%
U
&!!
orders: H
œ &#!!
œ
Orders placed per year:
Time interval between
H
U
œ
!Þ!*' years ¸ & weeks
18.3-13.
(a)
D=
K=
I=
N=
Category
1
2
3
Data
365
$5
0.2
3
(demand/year)
(setup cost)
(inventory holding cost rate)
(number of discount categories)
Price
$4.00
$3.90
$3.80
Range of order quantities
Lower Limit Upper Limit
0
49
50
99
100
100000000
EOQ
68
68
69
Q*
49
68
100
Annual
Purchase
Cost
$1,460
$1,424
$1,387
Annual
Setup
Cost
$37
$27
$18
Annual
Holding
Cost
$20
$27
$38
Results
100
Optimal Q
Total Variable Cost
$1,443
(b)
$'&
"!!
U
orders: H
Orders placed per year:
Time interval between
H
U
œ
œ $Þ'&
œ
"!!
$'&
18-8
œ !Þ#(% years ¸ "%Þ#& weeks
Total
Variable
Cost
$1,517
$1,477
$1,443
18.3-14.
(a)
Discount Category
1
2
3
U
X Z G œ -H  O H
U 2#
U
X Z G œ Ð)Þ&!ÑÐ%!!Ñ  Ð)!Ñ %!!
U   Ð!Þ#ÑÐ)Þ&!Ñ # 
U
X Z G œ Ð)Þ!!ÑÐ%!!Ñ  Ð)!Ñ %!!
U   Ð!Þ#ÑÐ)Þ!!Ñ # 
U
X Z G œ Ð(Þ&!ÑÐ%!!Ñ  Ð)!Ñ %!!
U   Ð!Þ#ÑÐ(Þ&!Ñ # 
(b)
Discount Category
1
2
3
U‡ œ  #OH
2
U‡ œ  #Ð)!ÑÐ%!!Ñ
!Þ#Ð)Þ&!Ñ œ "*%
U‡ œ  #Ð)!ÑÐ%!!Ñ
!Þ#Ð)Þ!!Ñ œ #!!
!ÑÐ%!!Ñ
U‡ œ  #Ð)
!Þ#Ð(Þ&!Ñ œ #!(
(c)
Discount Category
1
2
3
Feasible U
**
#!!
"!!!
(d)
18-9
U
X Z G œ -H  O H
U 2#
$$ß )!(Þ$)
$$ß &#!Þ!!
$$ß ()#Þ!!
(e) U‡ œ #!! with a TVC of $$ß &#!
(f)
D=
K=
I=
N=
Category
1
2
3
Data
400
$80
0.2
3
(demand/year)
(setup cost)
(inventory holding cost rate)
(number of discount categories)
Price
$8.50
$8.00
$7.50
Range of order quantities
Lower Limit Upper Limit
0
99
100
999
1000
100000000
Optimal Q
Total Variable Cost
EOQ
194
200
207
Q*
99
200
1000
Annual
Purchase
Cost
$3,400
$3,200
$3,000
Annual
Setup
Cost
$323
$160
$32
Annual
Holding
Cost
$84
$160
$750
Total
Variable
Cost
$3,807
$3,520
$3,782
Results
200
$3,520
(g) Since the value of U that minimizes TVC for discount category 2 is feasible, this
order quantity minimizes the annual setup and holding costs. Then, category 1 cannot
have lower annual setup and holding costs. Furthermore, since the purchase price per
case is higher for category 1, it cannot have lower purchasing costs. Hence, category 1
can be eliminated as a candidate for providing the optimal order quantity.
(h)
%!!
#!!
U
orders: H
Orders placed per year:
Time interval between
H
U
œ
œ#
œ
#!!
%!!
œ !Þ& years œ ' months
18.3-15.
(a)
Discount Category
1
2
3
U
X Z G œ -H  O H
U 2#
U
X Z G œ Ð"Þ!!ÑÐ#%!!Ñ  Ð%Ñ #%!!
U   Ð!Þ"(ÑÐ"Þ!!Ñ # 
U
X Z G œ Ð!Þ*&ÑÐ#%!!Ñ  Ð%Ñ #%!!
U   Ð!Þ"(ÑÐ!Þ*&Ñ # 
U
X Z G œ Ð!Þ*!ÑÐ#%!!Ñ  Ð%Ñ #%!!
U   Ð!Þ"(ÑÐ!Þ*!Ñ # 
(b)
Discount Category
1
2
3
U‡ œ  #OH
2
U‡ œ  #Ð%ÑÐ#%!!Ñ
!Þ"(Ð"Þ!!Ñ œ $$'
U‡ œ  #Ð%ÑÐ#%!!Ñ
!Þ"(Ð!Þ*&Ñ œ $%&
Ð%ÑÐ#%!!Ñ
œ $&%
U‡ œ  #!Þ"(Ð!Þ*!Ñ
18-10
(c)
Discount Category
1
2
3
Feasible U
"**
$%&
&!!
U
X Z G œ -H  O H
U 2#
$#ß %'&Þ"'
$#ß $$&Þ')
$#ß #"(Þ%&
(d)
(e) U‡ œ &!! with a TVC of $#ß #"(Þ%&
(f)
D=
K=
I=
N=
Category
1
2
3
Data
2400
$4
0.17
3
(demand/year)
(setup cost)
(inventory holding cost rate)
(number of discount categories)
Price
$1.00
$0.95
$0.90
Range of order quantities
Lower Limit Upper Limit
0
199
200
499
500
100000000
Optimal Q
Total Variable Cost
EOQ
336
345
354
Q*
199
345
500
Results
500
$2,217
18-11
Annual
Purchase
Cost
$2,400
$2,280
$2,160
Annual
Setup
Cost
$48
$28
$19
Annual
Holding
Cost
$17
$28
$38
Total
Variable
Cost
$2,465
$2,336
$2,217
(g) Since the value of U that minimizes TVC for discount category 2 is feasible, this
order quantity minimizes the annual setup and holding costs. Then, category 1 cannot
have lower annual setup and holding costs. Furthermore, since the purchase price per bag
is higher for category 1, it cannot have lower purchasing costs. Hence, category 1 can be
eliminated as a candidate for providing the optimal order quantity.
(h)
#%!!
&!! œ %Þ)
U
&!!
orders: H
œ #%!!
Orders placed per year:
Time interval between
H
U
œ
œ !Þ#" years ¸ #Þ& months
18.4-1.
G& œ %  $ œ (
Ð%Ñ
Ð&Ñ
G% œ (  %  # œ "$à G% œ %  &  !Þ$Ð$Ñ œ *Þ*à G% œ min Ö"$ß *Þ*× œ *Þ*
Ð$Ñ
Ð%Ñ
G$ œ *Þ*  %  # œ "&Þ*à G$ œ (  %  %  !Þ$Ð#Ñ œ "&Þ'à
Ð&Ñ
G$ œ %  (  !Þ$Ð#  'Ñ œ "$Þ%à G$ œ min Ö"&Þ*ß "&Þ'ß "$Þ%× œ "$Þ%
Ð#Ñ
Ð$Ñ
Ð"Ñ
Ð#Ñ
G# œ "$Þ&  %  % œ #"Þ&à G# œ *Þ*  %  '  !Þ$Ð#Ñ œ #!Þ&à
Ð%Ñ
Ð&Ñ
G# œ (  %  )  !Þ$Ð#  %Ñ œ #!Þ)à G# œ %  ""  !Þ$Ð#  %  *Ñ œ "*Þ&à
G# œ min Ö#"Þ&ß #!Þ&ß #!Þ)ß "*Þ&× œ "*Þ&
G" œ "*Þ&  %  # œ #&Þ&à G" œ "$Þ&  %  '  !Þ$Ð%Ñ œ #%Þ(à
Ð$Ñ
Ð%Ñ
G" œ *Þ*  %  )  !Þ$Ð%  %Ñ œ #%Þ$à G" œ (  %  "!  !Þ$Ð%  %  'Ñ œ #&Þ#à
Ð&Ñ
G" œ %  "$  !Þ$Ð%  %  '  "#Ñ œ #%Þ)à
G" œ min Ö#&Þ&ß #%Þ(ß #%Þ$ß #&Þ#ß #%Þ)× œ #%Þ$
The optimal production schedule is to produce ) in the first month, and 5 in the fourth at
a cost of $24.30.
18.4-2.
G % œ G&  # œ #
Ð$Ñ
G $ œ G%  # œ #  # œ %
Ð%Ñ
G$ œ G&  #  !Þ#Ð<% Ñ œ !  #  !Þ#Ð$Ñ œ #Þ'
G$ œ min Ö%ß #Þ'× œ #Þ'
Ð#Ñ
G# œ G$  # œ #Þ'  # œ %Þ'
Ð$Ñ
G# œ G%  #  !Þ#Ð<$ Ñ œ #  #  !Þ#Ð%Ñ œ %Þ)
Ð%Ñ
G# œ G&  #  !Þ#Ð<$  #<% Ñ œ !  #  !Þ#Ð%  'Ñ œ %
G# œ min Ö%Þ'ß %Þ)ß %× œ %
Ð"Ñ
G " œ G#  # œ %  # œ '
Ð#Ñ
G" œ G$  #  !Þ#Ð<# Ñ œ #Þ'  #  !Þ#Ð$Ñ œ &Þ#
Ð$Ñ
G" œ G%  #  !Þ#Ð<#  #<$ Ñ œ #  #  !Þ#Ð$  )Ñ œ 'Þ#
18-12
Ð%Ñ
G" œ G&  #  !Þ#Ð<#  #<$  $<% Ñ œ !  #  !Þ#Ð$  )  *Ñ œ '
G" œ min Ö'ß &Þ#ß 'Þ#ß '× œ &Þ#
The optimal production schedule is to produce ( units in the first and third periods at a
total variable cost of $&Þ# million.
18.4-3.
B%
!
"
#
B#
!
"
#
$
%
&
'
(
B"
!
D%
#
"
!
!


*Þ%
)Þ#
(Þ!
%Þ'
%Þ!
"Þ%
$
"'Þ)
G%‡ ÐB% Ñ
%
$
!
"

"#Þ%
""Þ#
""Þ!
(Þ'
)Þ!
%Þ%

D%‡
#
"
!
B$
!
"
#
$
%
&
G# ÐB# ß D# Ñ
#
$
"$Þ% "$Þ#
"#Þ# "$Þ!
"#Þ!
*Þ'
)Þ' "!Þ!
*Þ!
'Þ%
&Þ%





G" ÐB" ß D" Ñ
%
&
'
"(Þ# "(Þ) ")Þ%
!



%Þ!
$Þ#
!Þ%
%
"%Þ!
"!Þ'
""Þ!
(Þ%




(
"*Þ!
G$ ÐB$ ß D$ Ñ
"
#
$

 "!Þ#
 )Þ)
*Þ%
(Þ% )Þ!
'Þ'
'Þ' &Þ#

$Þ) 




&
""Þ'
"#Þ!
)Þ%





'
"$Þ!
*Þ%






)
")Þ)
*
"*Þ)
(
"!Þ%







"!
")Þ)
%
"!Þ)
)Þ!




G$‡ ÐB$ Ñ
*Þ%
)Þ!
'Þ'
%Þ!
$Þ#
!Þ%
&
*Þ%





G#‡ ÐB# Ñ
"!Þ%
*Þ%
)Þ%
(Þ%
'Þ%
%Þ'
%Þ!
"Þ%
G"‡ ÐB" Ñ
"'Þ)
D$‡
&
%
$
!
!
!
D#‡
(
'
&
%
$
!
!
!
D"‡
$
The optimal production schedule is to produce $ units in period 1 and ( units in period 2,
with a cost of $"'Þ) million.
18-13
18.4-4.
2œ#
FÐB8 ß D8 Ñ œ 
B$
!
"
#
$
%
B"
!
D$
%
$
#
"
!
!
)(
58  -8 D8  #maxÖ!ß D8  $×  2ÐB8  D8  <8 Ñ for !  D8 Ÿ %
2ÐB8  D8 Ñ
for D8 œ !
G$‡ ÐB$ Ñ
%(
$'
#(
")
%
"
*#
D$‡
%
$
#
"
!
G" ÐB" ß D" Ñ
#
$
%
*# )# )&
B#
!
"
#
$
%
!



%(
$)
G"‡ ÐB" Ñ
)#
"


'(
&)
&"
G# ÐB# ß D# Ñ
#
$

)(
(( ()
') ("
'" '%
&% &#
%
*!
)$
('
'%

G#‡ ÐB# Ñ
)(
((
'(
%(
$)
D#‡
$
#
"
!
!
D"‡
$
The optimal production schedule is to produce $ units in period 1 and % units in period 3,
with a cost of $)# thousand.
18.5-1.
Deere & Company uses inventory theory to determine optimal inventory levels ensuring
product availability, on-time delivery, and customer satisfaction. In doing this, the
multistage inventory planning and optimization (MIPO) tool developed by SmartOps is
deployed. The underlying model is a stochastic, capacitated, multiechelon, multiproduct
production and inventory model. In MIPO, the material flow in the supply chain is
represented as an acyclic-directed graph. The recommended stock levels are found by
minimizing the inventory costs among periodic-review replenishment policies with a
certain service level. The demand is stochastic and its probability distribution is
nonstationary over time. The latter allows to model seasonality of demand. The capacities
and supply paths can be nonstationary. Lower bounds on service levels and other
constraints can be encapsulated in the model. The main decision variables are safety
stocks. Once the optimal stock levels are found, what-if analyses are performed to
evaluate the impact of changes.
After the implementation of the results, on-time deliveries have increased from 63% to
92% with a 90% customer service level. The reduction in inventory provided a savings of
$890 million between 2001 and 2003 and a $107 million increase in annual shareholder
value added. Estimated savings by the end of 2004 exceed $1 billion. The new system
also allows Deere to reduce the amount of aged inventory and to offer customers newer
models. This, in turn, avoids discounts and saves Deere over $10 million per year. Other
benefits from this study include enhanced manufacturing flexibility, improved service
levels, accurate predictions, ability to respond to changes quickly and trust in the supply
chain.
18-14
18.5-2.
O" œ $1&ß !!!ß O# œ $&!!ß 2" œ $2!ß 2# œ $22ß . œ 5!!!
Optimizing separately:
#
U‡# œ  #.O
2# œ 477
G#‡ œ #.O# 2# œ $10,488
8
" 2#
8‡ œ  O
O# 2" œ 5.74ß Ò8‡ Ó Ÿ
‡
Ò8‡ Ó"
8‡
Ê8œ6
U‡" œ 8U‡# œ 2860
G"‡ œ
.O"
8U#

2" Ð8"ÑU#
#
œ $50,055
G ‡ œ G"‡  G#‡ œ $60,543
Optimizing simultaneously:
/" œ 2" œ 2!ß /# œ 2#  2" œ 2
8
" /#
8‡ œ  O
O# /" œ "Þ73ß Ò8‡ Ó 
‡
U‡#
œ
#.
O"
8 O# 
8/" /#
Ò8‡ Ó"
8‡
Ê8œ#
œ "$80
U‡" œ 8U‡# œ #760
G ‡ œ #.  O8"  O# Ð8/"  /# Ñ œ $57,966
Quantity
U‡#
8‡
8
U‡"
G‡
Separate Optimization
477
5.74
6
#860
$60,543
Simultaneous Optimization
"$80
"Þ73
#
#760
$57ß 966
The increase in the total variable cost per unit time if the results from separate
optimization were to be used instead of the ones from simultaneous optimization is 3%.
18.5-3.
(a) 2" œ $#&ß 2# œ $#&!ß . œ #ß &!!
Quantity
U‡#
8‡
8
U‡"
Ð$#&!!!ß $"!!!Ñ
"%*
"&
"&
##$'
Ð$"!!!!ß $#&!!Ñ
#$'
'
'
"%"%
18-15
Ð$&!!!ß $&!!!Ñ
$$$
$
$
"!!!
(b) O" œ $"!ß !!!ß O# œ $#&!!ß . œ #ß &!!
Quantity
U‡#
8‡
8
U‡"
Ð$"!ß $&!!Ñ
"'!
"%
"%
##$'
Ð$#&ß $#&!Ñ
#$'
'
'
"%"%
Ð$&!ß $"!!Ñ
&!!
#
#
"!!!
(c) O" œ $"!ß !!!ß O# œ $#&!!ß 2" œ $#&ß 2# œ $#&!
Quantity
U‡#
8‡
8
U‡"
"!!!
"%*
'
'
)*%
#&!!
#$'
'
'
"%"%
&!!!
$$$
'
'
#!!!
18.5-4.
O" œ $&ß !!!ß O# œ $#!!ß 2" œ $"!ß 2# œ $""ß . œ "!!
Quantity
U‡#
8‡
8
U‡"
G‡
Separate Optimization (a)
'!
&Þ#%
&
$!#
$&#)
Simultaneous Optimization (b)
"'!
"Þ&)
#
$#"
$$'(
(c) The decrease in the total variable cost per unit time G ‡ by using the approach in (b)
rather than the one in (a) is &%.
18.5-5.
O" œ $&!ß !!!ß O# œ $&!!ß 2" œ $&!ß 2# œ $'!ß . œ &!!
Quantity
U‡#
G#‡
8‡
8
U‡"
G"‡
G‡
Separate Optimization (a)
*"
&%((
"!Þ*&
""
"!!%
%((")
&$"*&
Simultaneous Optimization (b)
#%*
)%'*
%Þ%(
%
**&
%$()!
&##%*
(c) The assembly plant will lose money ($#ß **#) by using the joint inventory policy
obtained in (b) whereas the supplier will make money ($$ß *$)) by doing so. One
possible financial agreement between the supplier and the assembly plant is that the
supplier will compensate for the loss of the plant so that the plant agrees to a supply
contract inducing the inventory policy in (b). By using this policy instead of separately
optimal ones, the total saving is $#ß **#  $$ß *$) œ $*%'.
18-16
18.5-6.
O" œ $&!ß !!!ß O# œ $#ß !!!ß O$ œ $$'!ß 2" œ $"ß 2# œ $#ß 2$ œ $"!ß . œ &ß !!!
The cost G is about !Þ#*% above the optimal cost G of the relaxed problem. Since the
latter is a lower bound on the optimal cost G ‡ of the original problem, the optimal cost G
of the revised problem can exceed G ‡ at most by !Þ#*%.
18.5-7.
O" œ $"#&ß !!!ß O# œ $#!ß !!!ß O$ œ $'ß !!!ß O% œ $"!ß !!!ß O& œ $#&!
2" œ $#ß 2# œ $"!ß 2$ œ $"&ß 2% œ $#!ß 2& œ $$!ß . œ "ß !!!
/" œ $#ß /# œ $)ß /$ œ $&ß /% œ $&ß /& œ $"!
Since ÐO$ Î/$ Ñ œ "#!!  #!!! œ ÐO% Î/% Ñ, we need to merge the installation 3 and 4 as
a new installation 3' with O$w œ $"'ß !!! and /$w œ $"!.
The cost G is about "Þ)%% above the optimal cost G of the relaxed problem. Since the
latter is a lower bound on the optimal cost G ‡ of the original problem, the optimal cost G
of the revised problem can exceed G ‡ at most by "Þ)%%.
18-17
18.5-8.
O" œ $"ß !!!ß O# œ $&ß O$ œ $(&ß O% œ $)!
2" œ $!Þ&ß 2# œ $!Þ&&ß 2$ œ $$Þ&&ß 2% œ $(Þ&&ß . œ %ß !!!
The cost G is about "Þ")% above the optimal cost G of the relaxed problem. Since the
latter is a lower bound on the optimal cost G ‡ of the original problem, the optimal cost G
of the revised problem can exceed G ‡ at most by "Þ")%.
18.5-9.
O" œ $'!ß !!!ß O# œ $'ß !!!ß O$ œ $%!!ß 2" œ $$ß 2# œ $(ß 2$ œ $*ß . œ "!ß !!!
The cost G is about !Þ#"% above the optimal cost G of the relaxed problem. Since the
latter is a lower bound on the optimal cost G ‡ of the original problem, the optimal cost G
of the revised problem can exceed G ‡ at most by !Þ#"%.
18.6-1.
(a)
 #Ð"&!!ÑÐ*!!Ñ
 #OH
 $!!!"!!!
U œ  2:
œ '!
:
2 œ
"!!!
$!!!
(b) V œ .  OP 5 œ &!  !Þ'(&Ð"&Ñ œ '!
(c)
d=
K=
h=
p=
L=
Data
900
$1,500
$3,000.00
$1,000
0.75
(average demand/unit time)
(setup cost)
(unit holding cost)
(unit shortage cost)
(service level)
Demand During Lead Time
Distribution
Normal
mean =
50
stand. dev. =
15
18-18
Q=
R=
Results
60
60
(d) Safety Stock: V  mean œ '!  &! œ "!
(e) If demand during the delivery time exceeds the order quantity '!, then the reorder
point will be hit again before the order arrives, triggering another order.
18.6-2.
(a)
2:
 #Ð408ÑÐ40Ñ  81 1 œ 60
U œ  #HO
2  : œ
V œ +  PÐ,  +Ñ œ 5  !Þ)Ð15  5Ñ œ 13
(b)
d=
K=
h=
p=
L=
Data
40
$40
8
1
0.8
(average demand/unit time)
(setup cost)
(unit holding cost)
(unit shortage cost)
(service level)
Q=
R=
Results
60
13
Demand During Lead Time
Distribution
Uniform
a=
5
(lower endpoint)
b=
15
(upper endpoint)
Average number of orders per year: Ð4!ÑÐ"#ÑÎ6! œ 8
Probability of a stock-out before the order is received: "  !Þ) œ !Þ#
Average number of stock-outs per year: 8Ð!Þ#Ñ œ 1.6
(c)
18.6-3.
(a)
P
!Þ&
!Þ(&
!Þ*
!Þ*&
!Þ**
!Þ***
Case 1
2 œ $"ß 5 œ "
!
!Þ'(&
"Þ#)#
"Þ'%&
#Þ$#(
$Þ!*)
Case 2
2 œ $"!!ß 5 œ "
!
'(Þ&
"#)Þ#
"'%Þ&
#$#Þ(
$!*Þ)
Case 3
2 œ $"ß 5 œ "!!
!
'(Þ&
"#)Þ#
"'%Þ&
#$#Þ(
$!*Þ)
Case 4
2 œ $"!!ß 5 œ "!!
!
'(&!
"#ß )#!
"'ß %&!
#$ß #(!
$!ß *)!
?P
!Þ&
!Þ"&
!Þ!&
!Þ!%
!Þ!!*
Case 1
2 œ $"ß 5 œ "
!Þ'(&
!Þ'!(
!Þ$'$
!Þ')#
!Þ(("
Case 2
2 œ $"!!ß 5 œ "
'(Þ&
'!Þ(
$'Þ$
')Þ#
((Þ"
Case 3
2 œ $"ß 5 œ "!!
'(Þ&
'!Þ(
$'Þ$
')Þ#
((Þ"
Case 4
2 œ $"!!ß 5 œ "!!
'(&!
'!(!
$'$!
')#!
(("!
(b)
18-19
(c) As the service level gets higher, increasing the service level further costs more for
smaller increases. Thus, there will be diminishing returns when raising the service level
further and further. A manager should balance the cost of the safety stock with the cost of
stock-outs to determine the best service level.
18.6-4.
(a) G œ 2OP 5 œ Ð"!!ÑÐ"Þ#)#ÑÐ"!!Ñ œ $"#ß )#!
(b) 5 œ . 5" Ê "!! œ %5" Ê 5" œ &!
If the lead time were one day: G œ 2OP 5" œ Ð"!!ÑÐ"Þ#)#ÑÐ&!Ñ œ $'ß %"!. This is a
&!% reduction in the cost of the safety stock.
(c) 5 œ . 5" œ )Ð&!Ñ œ "%"Þ%, G œ 2OP 5" œ Ð"!!ÑÐ"Þ#)#ÑÐ"%"Þ%Ñ œ $")ß "#(
This is a %"% increase in the cost of the safety stock.
(d) The lead time would need to quadruple to "' days.
18.6-5.
(a) The safety stock drops to zero.
(b) The safety stock decreases.
(c) The safety stock remains the same for a given service level. However, with higher
shortage costs, there will be an incentive to increase the service level, which induces a
higher level of safety stock.
(d) The safety stock increases.
(e) The safety stock doubles.
(f) The safety stock doubles.
18.6-6.
(a)
Ground Chuck
d=
K=
h=
p=
L=
Data
26000
$25
$0.30
$3
0.95
(average demand/unit time)
(setup cost)
(unit holding cost)
(unit shortage cost)
(service level)
Demand During Lead Time
Distribution
Uniform
a=
50
(lower endpoint)
b=
150
(upper endpoint)
18-20
Q=
R=
Results
2,183
145
Chuck Wagon
d=
K=
h=
p=
L=
Data
26000
$200
$0.30
$3
0.95
(average demand/unit time)
(setup cost)
(unit holding cost)
(unit shortage cost)
(service level)
Q=
R=
Results
6,175
829
Demand During Lead Time
Distribution
Normal
mean =
500
stand. dev. =
200
(b)
Ground Chuck: V œ +  PÐ,  +Ñ œ &!  !Þ*&Ð"&!  &!Ñ œ "%&
Chuck Wagon: V œ .  OP 5 œ &!!  "Þ'%&Ð#!!Ñ œ )#*
(c)
Ground Chuck: safety stock V  mean œ "%&  "!! œ %&
Chuck Wagon: safety stock V  mean œ )#*  &!! œ $#*
(d)
Ground Chuck:
Annual average holding cost: Ð!Þ$!Ñ %&Ð#")$%&Ñ
 œ $$%!Þ*&
#
Chuck Wagon:
Annual average holding cost: Ð!Þ$!Ñ $#*Ð'"(&$#*Ñ
 œ $$ß %"'Þ&!
#
(e)
Ground Chuck:
#'ß!!!
Annual shipping cost: O  H
U  œ #& #")$  œ $#*(Þ('
Annual purchasing cost: Ð#'ß !!!ÑÐ"Þ%*Ñ œ $$)ß (%!
Average annual acquisition cost: $#*(Þ('  $$)ß (%! œ $$*ß !$(Þ('
Chuck Wagon:
#'ß!!!
Annual shipping cost: O  H
U   !Þ"!H œ #!! '"(&   !Þ"!Ð#'ß !!!Ñ
œ $$%%#Þ""
Annual purchasing cost: Ð#'ß !!!ÑÐ"Þ$&Ñ œ $$&ß "!!
Average annual acquisition cost: $$%%#Þ""  $$&ß "!! œ $$)ß &%#Þ""
(f)
Ground Chuck: $$%!Þ*&  $$*ß !$(Þ(' œ $$*ß $()Þ("
Chuck Wagon: $$ß %"'Þ&!  $$)ß &%#Þ"" œ $%"ß *&)Þ'"
Jed should choose Ground Chuck as their supplier.
(g) If Jed would like to use the beef within a month of receiving it, then Ground Chuck is
the best choice. The order quantity with Ground Chuck is roughly one month's supply
whereas with Chuck Wagon, it is roughly three months' supply.
18-21
18.7-1.
In this study, inventory theory is applied to the three-echelon distribution problem faced
by Time Inc., the largest magazine publisher in the US. For each issue of each magazine,
Time Inc. needs to solve three subproblems. The first is to determine the total number H
of copies to be printed and shipped. The second is to find an allocation H" ß á ß HR of
these H copies among R wholesalers. The third subproblem is to decide on the
distribution .34 of H4 copies among 84 retailers of wholesaler 4 for every 4. Complicated
cost and revenue structures, timing and constraints on available information complicate
these problems. The overall objective is to maximize the expected total profit. The
problem is solved backwards by using readily available results from the literature of
newsvendor problem under ideal conditions. The solution found is then adjusted to
incorporate deviations from the ideal.
To solve the store-level allocation problem, first the distribution of demand is estimated
using statistical analysis. If J Ð5l.34 Ñ is the probability that the demand in store 3 of
wholesaler 4 is at least 5 , then the optimal allocation to this retailer is determined as
.34 œ J " Ð-l.34 Ñ or the best approximation to this. With this allocation, the probability
of selling out is "  - for each store and the solution satisfies 3 .34 œ H3 . Similarly, the
wholesaler-level allocation is found from the equation 74 ÐH4 Ñ œ 7, where 74 Ð † Ñ is the
probability that wholesaler 4 will sell the last copy shipped and 7 is chosen such that
4 H4 œ H. Finally, a lower bound on the national print order is determined from
Q ÐH! Ñ œ -Î<, where Q Ð † Ñ is the probability of selling the last copy printed and
shipped, - and < are the marginal cost and revenue respectively. Because of the
complications in identifying - and <, Time Inc. aims at producing more than H! .
The new system increased Time Inc.'s annual profits by over $3.5 million. The benefits
include improvement of wholesaler and retailer allocations, and increase of sales
stimulation effect by over 1%.
18.7-2.
FÐW ‡ Ñ œ
18.7-3.
W ‡ #!!
100
œ
::2
œ
#&")
#&!Þ"
Ê W ‡ œ "!!# 
( 
#&Þ"
¸ ##)
(a) Freddie's most profitable alternative is to order "' copies.
Alternative
Order 15 copies
Order 16 copies
Order 17 copies
Order 18 copies
Prior Probability
State of Nature
15 16 17 18
15 15 15 15
14 16 16 16
13 15 17 17
12 14 16 18
0.4 0.2 0.3 0.1
18-22
Expected
Payoff
$15.00
$15.20 Maximum
$15.00
$14.20
(b) Freddie's most profitable alternative is to order "' copies.
Alternative
Order 15 copies
Order 16 copies
Order 17 copies
Order 18 copies
Prior Probability
State of Nature
15 16 17 18
0
1
2
3
1
0
1
2
2
1
0
1
3
2
1
0
0.4 0.2 0.3 0.1
Expected
Cost
$1.10
$0.90 Minimum
$1.10
$1.90
(c)
Alternative
Order "& copies
Order "' copies
Order "( copies
Order ") copies
Gunder
Optimal service level: Gunder
Gover œ
"
""
Service Level
!Þ%
!Þ'
!Þ*
"
œ !Þ&
Freddie should order "' copies.
(d)
18-23
18.7-4.
(a) Gunder œ $$  $" œ $#, Gover œ $"  $!Þ&! œ $!Þ&!
(b) Prepare % doughnuts everyday to minimize the costs.
(c)
Alternative
Make !
Make "
Make #
Make $
Make %
Make &
Gunder
Optimal service level: Gunder
Gover œ
#
#!Þ&
Service Level
!Þ"
!Þ#&
!Þ%&
!Þ(&
!Þ*
"
œ !Þ)
Prepare % doughnuts everyday.
(d) The probability of running short is "  !Þ* œ "!%.
(e) Before & doughnuts are prepared, the optimal service level needs to exceed !Þ*. Let 1
be the cost of lost customer goodwill. Then Gunder œ #  1.
Gunder
Gunder Gover
 !Þ* Í
#1
#1!Þ&
 !Þ* Í 1  #Þ&!
The goodwill cost should be at least $#Þ&! before & doughnuts are prepared.
18-24
18.7-5.
(a) Optimal service level:
Gunder
Gunder Gover
œ
"
"!Þ&
œ !Þ''(
(b)
(c) U‡ œ $!!  !Þ''(Ð'!!  $!!Ñ œ &!!
(d) The probability of running short is "  !Þ''( œ $$Þ$%.
(e) Optimal service level:
Gunder
Gunder Gover
œ
""Þ&
""Þ&!Þ&
œ !Þ)$$
U‡ œ $!!  !Þ)$$Ð'!!  $!!Ñ œ &&!
The probability of running short is "  !Þ)$$ œ "'Þ(%.
18.7-6.
(a)
Revenue (with shortages): &!!Ð$Ñ œ $"ß &!!
(b)
Average number of loaves sold (without shortages): $!! 
&!!$!!
#
œ %!!
Average daily revenue (without shortages): %!!Ð$Þ!!Ñ œ $"ß #!!
(c)
With shortages: "ß &!! ‚ !Þ$$$ œ $&!!
Without shortages: "ß #!! ‚ !Þ''( œ $)!!
Average daily revenue over all days: $&!!  $)!! œ $"ß $!!
(d)
Average number of loves not sold:
#!!!
#
œ "!!
Average number of day-old loaves obtained over all days: "!! ‚ !Þ''( œ ''Þ(
Average daily revenue from day-old loaves: ''Þ(Ð"Þ&!Ñ œ $"!!
(e)
Average total daily revenue: $"ß $!!  $"!! œ $"ß %!!
Average daily profit: $"ß %!!  $#Ð&!!Ñ œ $%!!
(f)
Average daily profit with '!! loaves: $Ð%&!Ñ  #Ð'!!Ñ  "Þ&!Ð"&!Ñ œ $$(&
18-25
(g)
Average daily profit with &&! loaves: $(& 
(h)
Average size of shortage with &&! loaves:
%!!$(&
#
'!!&&!
#
œ $$)(Þ&!
œ #& loaves
Average daily shortage over all days: #& ‚ !Þ"'( œ %Þ"'(
Average daily cost of lost goodwill: %Þ"'( ‚ "Þ&! œ $'Þ#&
Average daily profit with &&! loaves and lost goodwill: $$)(Þ&!  $'Þ#& œ $$)"Þ#&
(i)
Average size of shortage with &!! loaves:
"!!!
#
œ &! loaves
Average daily shortage over all days: &! ‚ !Þ$$$ œ "'Þ'(
Average daily cost of lost goodwill: "'Þ'( ‚ "Þ&! œ $#&
Average daily profit with &!! loaves and lost goodwill: $%!!  $#& œ $$(&
18.7-7.
(a) U‡ œ +  Ðservice levelÑÐ,  +Ñ œ +  Ð!Þ''(ÑÐ(&Ñ œ +  &!
(b) Probability of incurring shortage: "  !Þ''( œ $$Þ$% (same as in 18.7-4)
(c)
Maximum shortage: ,  Ð+  &!Ñ œ #&
Maximum number of loaves that will not be sold: &!
The corresponding numbers for 18.7-5 are "!! and #!! respectively, which are four times
the amounts in this problem.
(d) The average daily costs of underordering and overordering for the new plan are #&%
of the original costs, so it is quite valuable to obtain as much information as possible
about the demand before placing the final order for a perishable product.
(e)
U‡ œ +  Ðservice levelÑÐ,  +Ñ œ +  Ð!Þ)$$ÑÐ(&Ñ œ +  '#Þ&
Probability of incurring shortage: "  !Þ)$$ œ "'Þ'(%
Maximum shortage: ,  Ð+  '#Þ&Ñ œ "#Þ&
Maximum number of loaves that will not be sold: '#Þ&
18.7-8.
(a)
-2
"!!!$!!
W ‡ œ -ln :2
 œ &!ln "!!!!$!!
 ¸ "!$
(b) GÐCÑ œ -ÐC  MÑ  PÐCÑ œ -C  -M  PÐCÑ
Taking the derivative with respect to C, the term involving the initial inventory M
vanishes, so the optimal policy is the same as in (a), i.e., to order up to "!$ or
equivalently to order "!$  #$ œ )! parts.
(c)
W
PÖH Ÿ W× œ FÐWÑ œ "  / &! œ !Þ* Ê W œ &!lnÐ!Þ"Ñ ¸ ""&
(d)
::2
œ !Þ* Ê
:"!!!
:$!!
œ !Þ* Ê : œ $"#ß (!!
18-26
18.7-9.
(a) Optimal service level:
Gunder
Gunder Gover
œ
$!!!
$!!!"!!!
œ !Þ(&
(b) U œ .  OP 5 œ &!  Ð!Þ'(&ÑÐ"&Ñ œ '!
18.7-10.
PÐCÑ œ
" C
#!  ! ÐC
 BÑ.B  $C ÐB  CÑ.B œ
-C  PÐCÑ œ #C 
#!
C#
"!
 $C  $! œ
C#
"!
C#
"!
 $C  $!
 C  $!
Taking the derivative with respect to C: C&  " œ ! Ê W œ &. We could have used the
result PÖH Ÿ W× œ Ð:  -ÑÎÐ:  2Ñ directly:
PÖH Ÿ W× œ WÎ#! œ Ð:  -ÑÎÐ:  2Ñ œ Ð$  #ÑÎÐ$  "Ñ œ !Þ#& Ê W œ &.
GÐ=Ñ œ O  -W  PÐWÑ Ê -=  PÐ=Ñ œ O  -W  PÐWÑ
Ê
=#
"!
 =  $! œ "Þ&! 
Ê = œ &  "& ¸ "Þ"$
&#
"!
 &  $! Ê
=#
"!
="œ!
The Ð=ß WÑ œ Ð"Þ"$ß &Ñ policy is optimal.
18.7-11.
Single-period model with a setup cost:
Demand density is exponential with - œ #&. Per unit production/purchasing cost is
- œ ". Per unit inventory holding cost is 2 œ !Þ% and per unit shortage cost is : œ "Þ&.
The setup cost is O œ "!. The optimal policy is an Ð=ß WÑ policy with = œ ""Þ/$ and
W œ (Þ'$%&%.
18.8-1.
In each case, P œ #!!ß :" œ $"!!! and H has a normal distribution with mean '! and
standard deviation #!.
:# œ $$!! Ê J ÐB‡ Ñ œ " 
:#
:"
œ !Þ( Ê B‡ œ '!  O!Þ$ Ð#!Ñ œ '!  !Þ&#Ð#!Ñ œ (!Þ%
When the discount fare is $$!!, (! seats should be reserved for class 1 customers and the
request to make a sale to the class 2 customer should be accepted if there are (" or more
seats remaining.
:# œ $%!! Ê J ÐB‡ Ñ œ " 
:#
:"
œ !Þ' Ê B‡ œ '!  O!Þ% Ð#!Ñ œ '!  !Þ#&Ð#!Ñ œ '&
:# œ $&!! Ê J ÐB‡ Ñ œ " 
:#
:"
œ !Þ& Ê B‡ œ '!  O!Þ& Ð#!Ñ œ '!  !Ð#!Ñ œ '!
:# œ $'!! Ê J ÐB‡ Ñ œ " 
:#
:"
œ !Þ% Ê B‡ œ '!  O!Þ' Ð#!Ñ
œ '!  O!Þ% Ð#!Ñ œ '!  !Þ#&Ð#!Ñ œ &&
As the discount fare increases, the optimal number B‡ of reservation slots for class 1
customers decreases.
18-27
18.8-2.
The capacity P is "!!!ß the price :" paid by luxury-seeking customers is $#!ß !!! and the
discount fare is :# œ $"#ß !!!. The demand H by luxury-seeking customers has a normal
distribution with mean %!! and standard deviation "!!.
J ÐB‡ Ñ œ "  ::#" œ !Þ%
Ê B‡ œ %!!  O!Þ' Ð"!!Ñ œ %!!  O!Þ% Ð"!!Ñ œ %!!  !Þ#&Ð"!!Ñ œ "&!
Ê P  B‡ œ "!!!  "&! œ )&!
Hence, the maximum number of cabins that should be sold at the discount fare is )&!.
18.8-3.
P œ "!!ß :" œ $!!ß :# œ "!!
The demand H for full-fare tickets has a uniform distribution on integers between $" and
&!.
‡
"
:# Ÿ :" T ÐH B‡ Ñ Í ::#" œ "$ Ÿ &!B
Í B‡ Ÿ &"  #!
#!
$ œ %%Þ$$
B‡  "Ñ Í
:#  :" T ÐH
:#
:"
œ
"
$

&!B‡
#!
Í B‡  &! 
#!
$
œ %$Þ$$
Thus B‡ œ %% slots should be reserved to full-fare customers.
18.8-4.
P œ "&!ß : œ !Þ)ß < œ $$!!ß = œ $"&!!
T ÖHÐ8‡ Ñ
"&!× œ
<
=:
œ !Þ#&
HÐ8Ñ is normally distributed with mean !Þ)8 and standard deviation !Þ%8.
O!Þ#& œ
"&!!Þ)8
!Þ%8
Ê 8 œ
Ê !Þ'( œ
"&!!Þ)8
!Þ%8
Ê !Þ)8  !Þ#')8  "&! œ !
!Þ#')Ð!Þ#')Ñ# %Ð!Þ)ÑÐ"&!Ñ
"Þ'
œ "$Þ&#( Ê 8‡ œ Ð"$Þ&#(Ñ# œ ")$
We chose the smallest integer that is greater than Ð"$Þ&#(Ñ# to determine 8‡ . Hence, the
number of reservations to accept for this flight is ")$.
18.8-5.
P œ "#&ß < œ $#&!ß = œ $!!  $!! œ $'!!
Finding the optimal overbooking requires finding the smallest integer 8 with ?IÐT Ð8ÑÑ
nonpositive.
œ #&!  '!!  Ð.  "#&ÑÒT ÖHÐ8  "Ñ œ .×  T ÖHÐ8Ñ œ .×Ó
8
?IÐT Ð8ÑÑ
.œ"#'
Let \ denote the random variable associated with no-shows.
œ #&!  '!!  Ð8  5  "#&ÑÒT Ö\ œ 5  "×  T Ö\ œ 5×Ó
8"#'
?IÐT Ð8ÑÑ
œ #&!  '!!  T Ö\ œ 5  "× œ #&!  '!!T Ö\ Ÿ 8  "#&×
5œ!
8"#'
5œ!
18-28
Then the problem is to find the smallest 8 such that
T Ö\ Ÿ 8  "#&× #&!
'!! œ !Þ%"(.
B
T Ö\ Ÿ B×
!
!
"
!Þ!&
#
!Þ"&
$
!Þ#&
%
!Þ%
&
!Þ'
'
!Þ(&
(
!Þ)&
)
!Þ*&
*
"
From the cumulative distribution of \ , 8‡ is found to be "#&  & œ "$!, so &
reservations can be accepted in addition to the capacity.
18.8-6.
P œ $ß : œ !Þ&ß < œ $"!!!ß = œ $&!!!
To determine the optimal number of reservations to accept, we need to find the smallest
integer 8 such that
=:T ÖHÐ8Ñ
$×
< Í T ÖHÐ8Ñ
$×
!Þ% Í T ÖHÐ8Ñ Ÿ #× Ÿ !Þ'
8
8
8
Í        !Þ&8 Ÿ !Þ'
!
"
#
Í Ð8#  8  #Ñ!Þ&8" Ÿ !Þ'
A first guess can be 8 œ ', since then the average number of customers with reservation
and who actually show up is P œ $. It satisfies
Ð'#  '  #Ñ!Þ&(  !Þ',
so =:T ÖHÐ'Ñ $× <. This suggests 8‡ Ÿ '. Now consider 8 œ &.
Ð&#  &  #Ñ!Þ&'  !Þ',
so =:T ÖHÐ&Ñ $× <. Then 8‡ Ÿ &. For 8 œ %,
Ð%#  %  #Ñ!Þ&&  !Þ',
so =:T ÖHÐ%Ñ
$×  <. Hence the optimal number of reservations to accept is &.
18.8-7.
P œ "!!ß : œ !Þ*ß < œ $$!!!ß = œ $#!!!!
T ÖHÐ8‡ Ñ
"!!× œ
<
=:
œ
"
'
¸ !Þ"'(
HÐ8‡ Ñ is normally distributed with mean !Þ*8 and standard deviation !Þ$8.
O!Þ"'( œ
"!!!Þ*8
!Þ$8
Ê 8 œ
Ê !Þ*( œ
"!!!Þ*8
!Þ$8
!Þ#*"Ð!Þ#*"Ñ# %Ð!Þ*ÑÐ"!!Ñ
"Þ)
¸ "!Þ$) Ê 8‡ œ Ð"!Þ$)Ñ# œ "!)
The number of reservations to accept is "!), so ) reservations should be overbooked.
18.8-8.
Answers will vary.
18.9-1.
Answers will vary.
18.9-2.
Answers will vary.
18-29
CASES
CASE 18.1 Brushing Up on Inventory Control
(a) Robert's problem can be solved using the basic EOQ model, with the data:
H œ "#Ð#&!Ñ œ $ß !!!, O œ ")Þ(&Î$ œ 'Þ#&,
2 œ !Þ"#Ð"Þ#&Ñ œ !Þ"&, P œ !, [ H œ "#Ð$!Ñ œ $'!
D=
K=
h=
L=
WD =
Q=
Data
3,000
$6.25
$0.15
0
360
(demand/year)
(setup cost)
(unit holding cost)
(lead time in days)
(working days/year)
Reorder Point
Annual Setup Cost
Annual Holding Cost
Total Variable Cost
Results
0
$37.50
$37.50
$75.00
Decision
500
(optimal order quantity)
Robert should order 500 toothbrushes 6 times per year.
(b) EOQ model with P œ ' days
D=
K=
h=
L=
WD =
Q=
Data
3,000
$6.25
$0.15
5
360
(demand/year)
(setup cost)
(unit holding cost)
(lead time in days)
(working days/year)
Reorder Point
Annual Setup Cost
Annual Holding Cost
Total Variable Cost
Results
41.7
$37.50
$37.50
$75.00
Decision
500
(optimal order quantity)
Whenever the inventory drops down to &!, Robert should place an order for &!! toothbrushes. He needs to place ' orders per year.
(c) Planned shortages with : œ $"Þ&!/unit
D=
K=
h=
p=
Q=
S=
Data
3,000
$6.25
$0.15
$1.50
(demand/year)
(setup cost)
(unit holding cost)
(unit shortage cost)
Max Inventory Level
Decision
524
(optimal order quantity)
48
(optimal maximum shortage)
Annual Setup Cost
Annual Holding Cost
Annual Shortage Cost
Total Variable Cost
Results
476.7
$35.75
$32.50
$3.25
$71.51
Robert should order about &#% toothbrushes. Since the lead time is ' days, the reorder
point is %(Þ'(  'Ð$!!!Î$'!Ñ œ #Þ$$. The maximum shortage size is approximately
%).
18-30
(d) Two extreme cases: : œ $!Þ)&/unit and : œ $#&/unit
D=
K=
h=
p=
Q=
S=
Data
3,000
$6.25
$0.15
$0.85
(demand/year)
(setup cost)
(unit holding cost)
(unit shortage cost)
Max Inventory Level
Annual Setup Cost
Annual Holding Cost
Annual Shortage Cost
Total Variable Cost
Decision
542
(optimal order quantity)
81
(optimal maximum shortage)
Results
461.0
$34.57
$29.39
$5.19
$69.15
The reorder point when : œ $!Þ)&/unit is )"  'Ð$!!!Î$'!Ñ œ $".
D=
K=
h=
p=
Q=
S=
Data
3,000
$6.25
$0.15
$25.00
(demand/year)
(setup cost)
(unit holding cost)
(unit shortage cost)
Max Inventory Level
Annual Setup Cost
Annual Holding Cost
Annual Shortage Cost
Total Variable Cost
Decision
501
(optimal order quantity)
3
(optimal maximum shortage)
Results
498.5
$37.39
$37.17
$0.22
$74.78
The reorder point when : œ $#&/unit is 3  'Ð$!!!Î$'!Ñ œ %(. This suggests that as
the shortage cost increases, the reorder point increases.
(e) EOQ model with quantity discounts, with three prices $"Þ#&, $"Þ"& and $"Þ!! and
holding cost rate M œ !Þ"#.
D=
K=
I=
N=
Category
1
2
3
Data
3,000
$6.25
0.12
3
Price
$1.25
$1.15
$1.00
(demand/year)
(setup cost)
(inventory holding cost rate)
(number of discount categories)
Range of order quantities
Lower Limit Upper Limit
0
499
500
999
1,000
10,000,000
Optimal Q
Total Variable Cost
EOQ
500
521
559
Annual
Purchase
Q*
Cost
499
$3,750
521
$3,450
1,000
$3,000
Annual
Setup
Cost
$38
$36
$19
Annual
Holding
Cost
$37
$36
$60
Total
Variable
Cost
$3,825
$3,522
$3,079
Results
1,000
$3,079
The optimal order quantity is U œ "ß !!! and Robert should order $ times a year.
18-31
CASE 18.2 TNT: Tackling Newsboy's Teaching
For the analysis of this case, we use the template for perishable products.
(a) First we need to determine the optimal service level for Howie. The unit sale price is
$&, the unit purchase cost is $$, and the unit salvage value is !Þ& ‚ $$ œ $".
Unit Sales Price
Unit Purchase Cost
Unit Salvage Value
Data
$5
$3
$1
Cost of Overordering
Cost of Underordering
Optimal Service Level
Results
$2
$2
0.5
Since Talia assumes that the demand is uniformly distributed between "#! and %#! sets,
Howie should order "#!  !Þ& ‚ $!! œ #(! sets.
(b) If Leisure Limited refunds 75% of the purchase cost, then the unit salvage value for a
returned set becomes !Þ(& ‚ $$  $!Þ& œ $"Þ(&. We determine the new optimal service
level.
Unit Sales Price
Unit Purchase Cost
Unit Salvage Value
Data
$5
$3
$1.75
Cost of Overordering
Cost of Underordering
Optimal Service Level
Results
$1
$2
0.615
The order quantity is now "#!  !Þ'"& ‚ $!! œ $!%Þ5. Note that Howie can now order
more sets at one time than he could under the scenario of part (a) because he is not
punished as severely as before when he fails to sell all sets.
When the refund is 25%, the unit salvage cost is $!Þ#&.
Unit Sales Price
Unit Purchase Cost
Unit Salvage Value
Data
$5
$3
$0.25
Cost of Overordering
Cost of Underordering
Optimal Service Level
Results
$3
$2
0.421
Consequently, the order quantity is reduced to "#!  !Þ%#" ‚ $!! œ #%'Þ$. In this case,
Howie should purchase fewer sets at one time (compared to previous scenarios), since he
is punished more severely for failing to sell all the sets.
(c) The unit sale price is now $' and there is a 50% refund on returned firecracker sets.
Unit Sales Price
Unit Purchase Cost
Unit Salvage Value
Data
$6
$3
$1
Cost of Overordering
Cost of Underordering
Optimal Service Level
Results
$2
$3
0.6
However, if Howie raises the price of a firecracker set, one would expect a decrease in
the demand for his sets, so Talia should not use the same uniform demand distribution
that she used for her previous calculations of the optimal order quantity.
18-32
(d) Talia's strategy for estimating the demand is overly simplistic. She makes the very
simplifying assumption that the demand is uniformly distributed between "#! and %#!
sets. However, she does not take into account that the demand depends on the price of a
firecracker set. She should expect that stands charging less than the average price of $&
per set typically sell more sets than stands charging more. Talia should call Buddy again
to try to obtain more detailed information such as the range of sales and the average sale
of stands charging $& or $' per set.
Talia should also reevaluate her assumption that the demand is uniformly distributed. She
should check how her forecasts change if she uses other demand distribution like normal
distribution.
CASE 18.3 Jettisoning Surplus Stock
(a) We can use Excel to compute the sample mean and variance.
25
31
18
22
40
Observations
19 38 21
25
36
34
28
27
Mean
28
Std. Dev.
7.29154
Hence, the sample mean is 28 and the sample variance is 7.291542 ¸ 53.1667.
(b) Based on the findings of Scarlett Windermere, American Aerospace can use an
ÐVß UÑ policy for the inventory of part 10003487. The assumptions of the model are
satisfied.
1- The part is a stable product.
2- Its inventory level is under continuous review.
3- While the production of the part itself has no lead time, it is typically delayed by the
lead time of "Þ& months of the little steel part. Assume the lead time is "Þ& months.
4- The demand for the part is the same as for the jet engine MX332, since it is used only
for this particular engine. Hence, assume that the demand is approximately normally
distributed with mean #) and variance &$Þ"''(.
5- Excess demand is backlogged.
6- There is a fixed setup cost O œ $&ß )!!, a holding cost 2 œ $(&! and a shortage cost
: œ $$ß #&!.
Note that the average demand per year is "# ‚ #) œ $$', the average demand during the
lead time is "Þ& ‚ #) œ %# and it has a standard deviation of "Þ& ‚ (Þ#*"&% œ "!Þ*$($#.
18-33
D=
K=
h=
p=
L=
Data
336
$5,800
$750
$3,250
0.85
(average demand/unit time)
(setup cost)
(unit holding cost)
(unit shortage cost)
(service level)
Q=
R=
Results
80
53
Demand During Lead Time
Distribution
Normal
mean =
42
stand. dev. =
10.9
American Aerospace should implement the ÐVß UÑ policy with V œ &$ and U œ )!.
(c) The average inventory just before an order arrives is &$  %# œ "" and the one just
after an order has arrived is ""  )! œ *". Then, the average inventory is
Ð""  *"ÑÎ# œ &", with an average holding cost of &"Ð(&!Ñ œ $$)ß #&! per year. The
average number of setups in a year is $$'Î)! œ %Þ#, with a resulting average setup cost
of %Þ#Ð&ß )!!Ñ œ $#%ß $'! per year.
(d) The new service level is P œ !Þ*&.
D=
K=
h=
p=
L=
Data
336
$5,800
$750
$3,250
0.95
(average demand/unit time)
(setup cost)
(unit holding cost)
(unit shortage cost)
(service level)
Q=
R=
Results
80
60
Demand During Lead Time
Distribution
Normal
mean =
42
stand. dev. =
10.9
V œ '! and U œ )!. The average inventory just before an order arrives is )!  '! œ #!
and just after an order has arrived is #!  )! œ "!!, so the average inventory is '! and
the resulting average inventory holding cost is '!Ð(&!Ñ œ $%&ß !!! per year. Note that
the average holding cost has increased substantially. This is a consequence of increasing
the safety stock to #! from "". The average number of setups per year is still %Þ# and the
average setup cost is $#%ß $'! per year.
(e) Scarlett's independent analysis of the stationary part 10003487 can be justified since
there is only one jet engine that needs this part and this part appears to be the bottleneck
in the production process. However, in general, a stationary part is used for several jet
engines, so the demand for stationary parts depends on the demand for several jet engines
and a stock-out in one stationary part affects the demand for other parts. These
interdependencies cannot be captured by an independent analysis of each part; therefore,
Scarlett's approach is most likely to result in rather inaccurate inventory policies for
many other stationary parts.
(f) Scarlett could try to forecast the demand for jet engines based on sales data from
previous years.
18-34
SUPPLEMENT 1 TO CHAPTER 18
DERIVATION OF THE OPTIMAL POLICY FOR THE STOCHASTIC
SINGLE-PERIOD MODEL FOR PERISHABLE PRODUCTS
18S1-1.
GÐWÑ œ -W  2! ÐW  BÑ0 ÐBÑ.B  :W ÐB  WÑ0 ÐBÑ.B  5T ÖH
W
∞
H is uniformly distributed on Ò+ß ,Ó, so T ÖH
GÐWÑ œ -W  5 ,W
,+  PÐWÑ Ê
Ê J ÐWÑ œ
.GÐWÑ
.W
œ-
W× œ
5
,+
W×
,W
,+ .
 2J ÐWÑ  :Ö"  J ÐWÑÓ œ !
5
: ,+
:2
Let : œ -  #ß 5 œ "%ß 2 œ Ð-  "Ñß + œ %!ß , œ '!.
Ê J ÐWÑ œ
W%!
#!
œ
#Þ(
$
œ !Þ* Ê W œ &)
18S1-2.
(a)
GÐMß WÑ œ -ÐW  MÑ  :T ÖH  W× œ -ÐW  MÑ  :/W
Ê
`GÐMßWÑ
`W
œ -  :/W œ ! Ê W œ ln Ð-Î:Ñ
Order up to W if M  W , do not order otherwise.
(b)
GÐMß WÑ œ 
O  -ÐW  MÑ  :/W
:/M
if M  W
if M œ W
An Ð=ß WÑ policy is optimal with W œ ln Ð-Î:Ñ and = being the smallest value such that
-=  :/= œ O  - ln Ð-Î:Ñ  - .
18S1-1
SUPPLEMENT 2 TO CHAPTER 18
STOCHASTIC PERIODIC-REVIEW MODELS
18S2-1.
(a) Single-period model with no setup cost:
Demand density is exponential with - œ #&. Per unit production/purchasing cost is
- œ "!. Per unit inventory holding cost is 2 œ ' and per unit shortage cost is : œ "&. The
optimal one-period inventory level is WÐ!Ñ œ 'Þ(*)$%.
(b) Two-period model with no setup cost:
Demand density is exponential with - œ #&. Per unit production/purchasing cost is
- œ "!. Per unit inventory holding cost is 2 œ ' and per unit shortage cost is : œ "&. The
optimal two-period policy consists of the inventory levels W" Ð!Ñ œ #$Þ#*$# and
W# Ð!Ñ œ 'Þ(*)$%.
18S2-2.
(a) Single-period model with no setup cost:
Demand density is uniform on Ò!ß &!Ó. Per unit production/purchasing cost is - œ "!. Per
unit inventory holding cost is 2 œ ) and per unit shortage cost is : œ "&. The optimal
one-period inventory level is W ‡ œ "!Þ)'*'. It is optimal to order up to W ‡ if the initial
inventory is below W ‡ and not to order otherwise.
(b) Two-period model with no setup cost:
Demand density is uniform on Ò!ß &!Ó. Per unit production/purchasing cost is - œ "!. Per
unit inventory holding cost is 2 œ ) and per unit shortage cost is : œ "&. The optimal
two-period policy consists of the inventory levels W"‡ œ *Þ#'"&' and W#‡ œ "!Þ)'*'. It is
optimal to order up to W3‡ if the initial inventory is below W3‡ in period 3 and not to order
otherwise.
18S2-3.
Two-period model with no setup cost:
Demand density is exponential with - œ #&. Per unit production/purchasing cost is
- œ ". Per unit inventory holding cost is 2 œ !Þ#& and per unit shortage cost is : œ #.
The discount factor is !Þ*. The optimal two-period policy is the same as the one for the
infinite-period model, so consists of the inventory level WÐ!Ñ œ %'Þ&")).
18S2-4.
Two-period model with no setup cost:
Demand density is exponential with - œ #&. Per unit production/purchasing cost is
- œ ". Per unit inventory holding cost is 2 œ !Þ#& and per unit shortage cost is : œ #.
The optimal two-period policy consists of the inventory levels W" Ð!Ñ œ $'Þ&#" and
W# Ð!Ñ œ "%Þ'*%(.
18S2-1
18S2-5.
Infinite-period model with no setup cost:
Demand density is exponential with - œ #&. Per unit production/purchasing cost is
- œ ". Per unit inventory holding cost is 2 œ !Þ#& and per unit shortage cost is : œ #.
The discount factor is !Þ*. The optimal policy consists of the inventory level
WÐ!Ñ œ %'Þ&")).
18S2-6.
Infinite-period model with no setup cost:
Demand density is exponential with - œ ". Per unit production/purchasing cost is - œ #.
Per unit inventory holding cost is 2 œ " and per unit shortage cost is : œ &. The discount
factor is !Þ*&. The optimal policy consists of the inventory level WÐ!Ñ œ "Þ'*'%&.
18S2-7.
12-period model with no setup cost:
The answer is the same as in 18S2-6, so the optimal policy consists of the inventory level
WÐ!Ñ œ "Þ'*'%&.
18S2-8.
Infinite-period model with no setup cost:
Demand density is uniform on Ò#!!!ß $!!!Ó. Per unit production/purchasing cost is
- œ "&!. Per unit inventory holding cost is 2 œ # and per unit shortage cost is : œ $!.
The discount factor is !Þ*. The optimal policy consists of the inventory level
WÐ!Ñ œ #ß %')Þ(&.
18S2-9.
Infinite-period model with no setup cost:
Demand density is exponential with - œ "!!!. Per unit production/purchasing cost is
- œ )!. Per unit inventory holding cost is 2 œ !Þ(! and per unit shortage cost is : œ #.
The discount factor is !Þ**). The optimal policy consists of the inventory level
WÐ!Ñ œ %*(.
18S2-10.
2 œ !Þ$ß : œ #Þ&
BÎ#&
BÎ#&
KÐWÑ œ !Þ$! ÐWBÑ
.B  #Þ&W ÐBWÑ
.B œ !Þ$W  (!/WÎ#&  (Þ&
#& /
#& /
Kw ÐWÑ œ !Þ$  #Þ)/WÎ#& œ ! Ê W œ &&Þ)%
WÎ#&
Kww ÐWÑ œ #Þ)
 ! Ê W œ &&Þ)% minimizes KÐWÑ.
#& /
W
∞
KÐ5Ñ œ KÐ5  "!!Ñ Í !Þ$5  (!/5Î#& œ !Þ$Ð5  "!!Ñ  (!/Ð5"!!ÑÎ#&
Í (!/5Î#& Ð"  /% Ñ œ $! Í 5 œ #!Þ(# ¸ #"
5 œ #"  W œ &&Þ)%  "#" œ 5  "!! and KÐ#"Ñ ¸ KÐ"#"Ñ
Hence, the optimal policy is a Ð5ß UÑ œ Ð#"ß "!!Ñ policy.
18S2-2
18S2-11.
Since - œ !, the answer is identical to that for 18.S2-10, viz., Ð5ß UÑ œ Ð#"ß "!!Ñ is
optimal.
18S2-12.
PÐWÑ œ ! 2ÐW  BÑ0 ÐBÑ.B  W :ÐB  WÑ0 ÐBÑ.B
W
∞
.PÐWÑ
.W
œ ! 20 ÐBÑ.B  W :0 ÐBÑ.B œ 2J ÐWÑ  :Ò"  J ÐWÑÓ
.PÐWÑ
.W
 -Ð"  αÑ œ ! Ê :  :J ÐWÑ  2J ÐWÑ  -Ð"  αÑ œ !
W
Ê J ÐWÑ œ
∞
:-Ð"αÑ
:2
18S2-3
CHAPTER 19: MARKOV DECISION PROCESSES
19.2-1.
Bank One, one of the major credit card issuers in the United States has developed the
portfolio control and optimization (PORTICO) system to manage APR and credit-line
changes of its card holders. Customers prefer low APR and high credit lines, which can
reduce the bank's profitability and increase the risk. Consequently, the bank faces the
need to find a balance between revenue growth and risk. PORTICO formulates the
problem as a Markov decision process. The state variables are chosen in a way to satisfy
Markovian assumption as closely as possible while keeping the dimension of the state
space at a tractable level. The resulting variables are ÐBß CÑ, where B corresponds to the
credit line and APR level and C represents the behavior variables. The transition
probabilities are estimated from the available data. The objective is to maximize the
expected net present value of the cash flows over a 36-month horizon. The dynamic
programming equation for the decision periods of the problem is
Z> ÐBß CÑ œ max <ÐB „ +ß CÑ  "  :ÐB „ +ß Cà 4ÑZ>" ÐB „ +ß 4Ñ,
+−EÐBßCÑ
4−f
where <Ð † Ñ denotes the immediate net cash flow and " is the discount factor. The
solution obtained is then adjusted to conform to business rules.
Benchmark tests are performed to evaluate the output policy. These tests suggest that the
new policy improves profitability. By adopting this policy, Bank One is expected to
increase its annual profit by more than $75 million.
19-1
19.2-2.
(a) Let the states 3 œ !ß "ß # be the number of customers at the facility. There are two
possible actions when the facility has one or two customers. Let decision 1 be to use the
slow configuration and decision 2 be to use the fast configuration. Also let G34 denote the
expected net immediate cost of using decision 4 in state 3. Then,
G""
G"#
G!"
G!#
œ G#" œ $ 
œ G## œ * 
œ$
œ*
$
&
%
&
‚ &! œ #(
‚ &! œ $"
(b) In state !, the configuration chosen does not affect the transition probabilities, so it is
best to choose the slow configuration when there are no customers in line. Consequently,
the number of stationary policies is four.
3
"
#
.3 ÐV" Ñ
"
"
Policy
.3 ÐV# Ñ
"
#
.3 ÐV$ Ñ
#
"
V"
Transition Matrix
"
"
 # # !
 $ " "
V#


V$
#
#
V%
#
#
Policy
V"
V#
V$
V%
1!
!Þ$"!$
!Þ$#%$
!Þ%!')
!Þ%"'
!
"!
"
#
$
"!
!
"
!
&
"
!
&
#
$
&
"
#
"
#
%
&
"
#
"
#
$
&
"
#
"
#
%
&
.3 ÐV% Ñ
#
#
Expected Average Cost
&
#
&

G" œ $1! #(1"  #(1#
!
" 
&
"
&

G# œ $1! #(1"  $"1#
!
" 
"!
#
&

G$ œ $1! $"1"  #(1#
!
" 
"!
"
&

G% œ $1! $"1"  $"1#
(c)
1"
!Þ&"(#
!Þ&%!&
!Þ&!)&
!Þ&"*
1#
!Þ"(#%
!Þ"$&"
!Þ!)%(
!Þ!'&
Average Cost
G" œ "(Þ'*
G# œ "(Þ)"
G$ œ "'Þ)$
G% œ "'Þ)(
G# is the minimum, so the optimal policy is V# , i.e., to use slow configuration when no
customer or only one customer is present and fast configuration when there are two
customers.
19-2
19.2-3.
(a) Let the states represent whether the student's car is dented, 3 œ ", or not, 3 œ !.
Decision
"
#
$
%
&
Action
Park on street in one space
Park on street in two spaces
Park in lot
Have it repaired
Drive dented
State
!
!
!
"
"
Immediate Cost
G!" œ !
G!# œ %Þ&
G!$ œ &
G"% œ &!
G"& œ *
(b) Assuming the student's car has no dent initially, once she decides to park in lot, state
" will never be entered. In that case, the decision chosen in state " does not affect the
expected average cost. Hence, it is enough to consider five stationary deterministic
policies.
3
!
"
.3 ÐV" Ñ
"
%
Policy
V"
V#
V$
V%
V&
.3 ÐV# Ñ
"
&
.3 ÐV$ Ñ
#
%
Transition Matrix
!Þ* !Þ"
 "
! 
!Þ* !Þ"
 !
" 
!Þ*) !Þ!#
 "
! 
!Þ*) !Þ!#
 !
" 
" !
 
.3 ÐV% Ñ
#
&
.3 ÐV& Ñ
$

Expected Average Cost
G" œ !10  &!1"
G# œ !10  *1"
G$ œ %Þ&10  &!1"
G% œ %Þ&10  *1"
G & œ &1 0
(c)
Policy
V"
V#
V$
V%
V&
1!
!Þ*!*
!
!Þ*)
!
"
1"
!Þ!*"
"
!Þ!#
"
!
Average Cost
%Þ&&
*
&Þ%"
*
& (if initially not dented)
The policy V" has the minimum cost, so it is optimal to park on the street in one space if
not dented and to have it repaired if dented.
19-3
19.2-4.
(a) Let states ! and " denote the good and the bad mood respectively. The decision in
each state is between providing refreshments or not.
Decision
"
#
"
#
Action
Provide refreshments
Not provide refreshments
Provide refreshments
Not provide refreshments
State
!
!
"
"
Immediate Cost
G!" œ "%
G!# œ !
G"" œ "%
G"# œ (&
(b) There are four possible stationary policies.
3
!
"
.3 ÐV" Ñ
"
"
Policy
V"
V#
V$
V%
.3 ÐV# Ñ
"
#
.3 ÐV$ Ñ
#
"
Transition Matrix
!Þ)(& !Þ"#&
 !Þ)(& !Þ"#& 
!Þ)(& !Þ"#&
 !Þ"#& !Þ)(& 
!Þ"#& !Þ)(&
 !Þ)(& !Þ"#& 
!Þ"#& !Þ)(&
 !Þ"#& !Þ)(& 
.3 ÐV% Ñ
#
#
Expected Average Cost
G" œ "%1!  "%1"
G# œ "%1!  (&1"
G$ œ "%1"
G% œ (&1"
(c)
Policy
V"
V#
V$
V%
1!
!Þ)(&
!Þ&
!Þ&
!Þ"#&
1"
!Þ"#&
!Þ&
!Þ&
!Þ)(&
Average Cost
G" œ "%
G# œ %%Þ&
G$ œ (
G% œ '&Þ'#&
The optimal policy is V$ , i.e., to provide refreshments only if the group begins the night
in a bad mood.
19-4
19.2-5.
(a) Let state ! denote point over, two serves to go on next point and state " denote one
serve left. The decision in each state is to attempt an ace or a lob.
Decision
Action
State
"
Attempt ace
!
#
Attempt lob
!
"
Attempt ace
"
#
Attempt lob
"
Immediate Cost
G!" œ $)  #$ Ð"Ñ  "$ Ð"Ñ œ  ")
G!# œ ()  "$ Ð"Ñ  #$ Ð"Ñ œ
G"" œ $)  #$ Ð"Ñ  "$ Ð"Ñ  &) Ð"Ñ œ
G"# œ ()  "$ Ð"Ñ  #$ Ð"Ñ  ") Ð"Ñ œ
(b) There are four possible stationary deterministic policies.
3
!
"
.3 ÐV" Ñ
"
"
Policy
V"
V#
V$
V%
.3 ÐV# Ñ
"
#
.3 ÐV$ Ñ
#
"
Transition Matrix
$Î) &Î)
 "
! 
$Î) &Î)
 "
! 
(Î) "Î)
 "
! 
(Î) "Î)
 "
! 
.3 ÐV% Ñ
#
#
Expected Average Cost
G" œ Ð"Î)Ñ1!  Ð"Î#Ñ1"
G# œ Ð"Î)Ñ1!  Ð&Î"#Ñ1"
G$ œ Ð(Î#%Ñ1!  Ð"Î#Ñ1"
G% œ Ð(Î#%Ñ1!  Ð&Î"#Ñ1"
(c)
Policy
V"
V#
V$
V%
1!
!Þ'"&
!Þ'"&
!Þ))*
!Þ))*
1"
!Þ$)&
!Þ$)&
!Þ"""
!Þ"""
(
#%
Average Cost
G" œ !Þ#(!
G# œ !Þ#$(
G$ œ !Þ$"&
G% œ !Þ$!'
The optimal policy is V$ , i.e., to attempt lob in state ! and ace in state ".
19-5
"
#
&
"#
19.2-6.
(a) Let states 3 œ !ß "ß # represent the state of the market, ""ß !!!, "#ß !!! and "$ß !!!
respectively. The decision is between two funds, namely the Go-Go Fund and the GoSlow Mutual Fund. All the costs are expressed in thousand dollars.
Decision
"
#
"
#
"
#
Action
Invest in the Go-Go
Invest in the Go-Slow
Invest in the Go-Go
Invest in the Go-Slow
Invest in the Go-Go
Invest in the Go-Slow
State
!
!
"
"
#
#
Immediate Cost
G!" œ !Þ&Ð#!Ñ  !Þ#Ð&!Ñ œ #!
G!# œ !Þ&Ð"!Ñ  !Þ#Ð#!Ñ œ *
G"" œ !Þ"Ð#!Ñ  !Þ%Ð#!Ñ œ '
G"# œ !Þ"Ð"!Ñ  !Þ%Ð"!Ñ œ $
G#" œ !Þ#Ð&!Ñ  !Þ%Ð#!Ñ œ ")
G## œ !Þ#Ð#!Ñ  !Þ%Ð"!Ñ œ )
(b) There are eight possible stationary policies.
3
!
"
#
.3 ÐV" Ñ
"
"
"
.3 ÐV# Ñ
"
"
#
.3 ÐV$ Ñ
"
#
#
.3 ÐV% Ñ
"
#
"
All V3 's have the same transition matrix:
Policy
V"
V#
V$
V%
V&
V'
V(
V)
 !Þ$
!Þ"
 !Þ#
.3 ÐV& Ñ
#
#
"
!Þ&
!Þ&
!Þ%
.3 ÐV' Ñ
#
"
#
!Þ# 
!Þ% .
!Þ" 
.3 ÐV( Ñ
#
"
"
.3 ÐV) Ñ
#
#
#
Expected Average Cost
G" œ #!1!  '1"  ")1#
G# œ #!1!  '1"  )1#
G$ œ #!1!  $1"  )1#
G% œ #!1!  $1"  ")1#
G& œ *1!  $1"  ")1#
G' œ *1!  '1"  )1#
G( œ *1!  '1"  ")1#
G) œ *1!  $1"  )1#
(c) 1 œ Ð!Þ"("ß !Þ%'$ß !Þ$''Ñ
Policy
V"
V#
V$
V%
V&
V'
V(
V)
Average Cost
!Þ$*
$Þ#(
"Þ))"
"Þ((*
$Þ''
"Þ$)*
#Þ#("
!
The optimal policy is V# , i.e. to invest in the Go-Go Fund in states ! and ", in the GoSlow Fund in state #.
19-6
19.2-7.
(a) Let states ! and " represent whether the machine is broken down or is running
respectively. The decision is between Buck and Bill.
Decision
"
#
"
#
Action
Buck
Bill
Buck
Bill
State
!
!
"
"
Immediate Cost
G!" œ !
G!# œ !
G"" œ "#!!
G"# œ "#!!
(b) There are four possible stationary deterministic policies.
3
!
"
.3 ÐV" Ñ
"
"
Policy
V"
V#
V$
V%
.3 ÐV# Ñ
"
#
.3 ÐV$ Ñ
#
"
Transition Matrix
!Þ% !Þ'
 !Þ' !Þ% 
!Þ% !Þ'
 !Þ% !Þ' 
!Þ& !Þ&
 !Þ' !Þ% 
!Þ& !Þ&
 !Þ% !Þ' 
.3 ÐV% Ñ
#
#
Expected Average Cost
G" œ "#!!1"
G# œ "#!!1"
G$ œ "#!!1"
G% œ "#!!1"
(c)
Policy
V"
V#
V$
V%
1!
!Þ&
!Þ%
!Þ&%&
!Þ%%%
1"
!Þ&
!Þ'
!Þ%&&
!Þ&&'
Average Cost
G" œ '!!
G# œ (#!
G$ œ &%'
G% œ ''(Þ#
The largest expected average profit is given by V# .
19-7
19.2-8.
(a) Let the states be the number of items in inventory at the beginning of the period and
the decision be the number of items ordered. To conform to the software package, one
needs to relabel the decisions as "ß #ß $ respectively. The cost matrix is:
-35
!
"
#
"
%!Î$
%
%
#
&'Î$
"*

$
#%


Let V$ denote the policy to order # items when the inventory level is initially ! and not to
order when the inventory level is initially either ! or ". In other words, .! ÐV$ Ñ œ $ and
." ÐV$ Ñ œ .# ÐV$ Ñ œ ".
 "Î$
T ÐV$ Ñ œ #Î$
 "Î$
"Î$
"Î$
"Î$
"Î$ 
!
Ê 1 œ Ð%Î*ß $Î*ß #Î*Ñ
"Î$ 
Expected average cost: Ð%Î*ÑG!$  Ð$Î*ÑG""  Ð#Î*ÑG#" œ ""'Î* ¸ $"#Þ)*/period
(b) There are $$ œ #( stationary policies, since one can order !ß " or # items in each state.
However, only six of these are feasible. The remaining #" policies are infeasible and the
decision at least in one of the states leads to over capacity.
3
!
"
#
.3 ÐV" Ñ
"
"
"
.3 ÐV# Ñ
#
"
"
.3 ÐV$ Ñ
$
"
"
.3 ÐV% Ñ
"
#
"
.3 ÐV& Ñ
#
#
"
.3 ÐV' Ñ
$
#
"
19.3-1.
(a)
minimize
$C!"  *C!#  $C""  *C"#  #)C#"  $%C##
subject to
C!"  C!#  C""  C"#  C#"  C## œ "
C!"  C!#   "# C!"  "# C!# 
$
"! C""
 #& C"#  œ !
C""  C"#   "# C!"  "# C!#  "# C""  "# C"#  $& C#"  %& C##  œ !
#
C#"  C##   "!
C"" 
C35
"
"! C"#
 &# C#"  &" C##  œ !
! for 3 œ !ß "ß # and 5 œ "ß #
(b) Using the simplex method, we find C!" œ !Þ$#%$#ß C"" œ !Þ&%!&%ß C## œ !Þ"$&"% and
the remaining C35 's are zero. Hence, the optimal policy uses decision " in states ! and ",
decision # in state #.
19-8
19.3-2.
(a)
minimize
%Þ&C!#  &C!$  &!C"%  *C"&
subject to
C!"  C!#  C!$  C"%  C"& œ "
*
C!"  C!#  C!$   "!
C!" 
"
C"%  C"&   "!
C!" 
"
&! C!#
C!" ß C!# ß C!$ ß C"% ß C"&
%*
&! C!#
 C!$  C"%  œ !
 C"&  œ !
!
(b) Using the simplex method, all C35 's turn out to be zero except that C!" œ !Þ*!*!* and
C"% œ !Þ!*!*", so the policy that uses decision " in state ! and decision % in state " is
optimal.
19.3-3.
(a)
minimize
"%C!"  "%C""  (&C"#
subject to
C!"  C!#  C""  C"# œ "
C!"  C!#   () C!"  ") C!#  () C""  ") C"#  œ !
C""  C"#   ") C!"  () C!#  ") C""  () C"#  œ !
C35
! for 3 œ !ß " and 5 œ "ß #
(b) Using the simplex method, we find C!# œ C"" œ !Þ&ß C!" œ C"# œ !, so the optimal
policy is to use decision # in state ! and decision " in state ".
19.3-4.
(a)
minimize
 ") C!" 
(
#% C!#
 "# C"" 
&
"# C"#
subject to
C!"  C!#  C""  C"# œ "
C!"  C!#   $) C!"  () C!#  C""  C"#  œ !
C""  C"#   &) C!"  ") C!#  œ !
C35
! for 3 œ !ß " and 5 œ "ß #
(b) Using the simplex method, we find C!# œ !Þ)))*ß C"" œ !Þ""""ß C!" œ C"# œ !, so the
optimal policy is to use decision # (lob) in state ! and decision " (ace) in state ".
19-9
19.3-5.
(a) minimize #!C!"  *C!#  'C""  $C"#  ")C#"  )C##
subject to C!"  C!#  C""  C"#  C#"  C## œ "
$
C!"  C!#   "!
C!" 
&
C""  C"#   "!
C!" 
#
C#"  C##   "!
C!" 
C35
$
"! C!#

"
"! C""

"
"! C"#

#
"! C#"

&
"! C!#

&
"! C""

&
"! C"#

%
"! C#"

#
"! C!#

%
"! C""

%
"! C"#

%
"! C#"

#
"! C## 
%
"! C## 
%
"! C## 
œ!
œ!
œ!
! for 3 œ !ß "ß # and 5 œ "ß #
(b) Using the simplex method, we find C!" œ !Þ"("ß C"" œ !Þ%'$ß C## œ !Þ$'' and the
remaining C35 's are zero. Hence, the optimal policy uses decision " (the Go-Go Fund) in
states ! and ", decision # in state # (the Go-Slow Fund).
19.3-6.
(a)
minimize
"#!!C"" "#!!C"#
subject to
C!"  C!#  C""  C"# œ "
C!"  C!#  !Þ%C!"  !Þ&C!#  !Þ'C""  !Þ%C"#  œ !
C""  C"#  !Þ'C!"  !Þ&C!#  !Þ%C""  !Þ'C"#  œ !
C35
! for 3 œ !ß " and 5 œ "ß #
(b) Using the simplex method, we find C!" œ !Þ%ß C"# œ !Þ'ß C!# œ C"" œ !, so the
optimal policy is to use decision " (Buck) in state ! and decision # (Bill) in state ".
19.3-7.
(a) minimize
%!
$ C!"

&'
$ C!#
 #%C!$  %C""  "*C"#  %C#"
subject to C!"  C!#  C!$  C""  C"#  C#" œ "
C!"  C!#  C!"  #$ C!#  "$ C!$  #$ C""  "$ C"#  "$ C#"  œ !
C""  C"#   "$ C!#  "$ C!$  "$ C""  "$ C"#  "$ C#"  œ !
C#"   "$ C!$  "$ C""  "$ C"#  "$ C#"  œ !
C35
! for 3 œ !ß "ß # and 5 œ "ß #ß $
(b) Using the simplex method, we find C!$ œ !Þ%%%%ß C"" œ !Þ$$$$ß C#" œ !Þ#### and the
remaining C35 's are zero. Hence, the optimal policy is to order # items in state ! and not to
order in states " and #.
19-10
SUPPLEMENT 1 TO CHAPTER 19
A POLICY IMPROVEMENT ALGORITHM FOR
FINDING OPTIMAL POLICIES
19S1-1.
19S1-1
19S1-2.
19S1-2
19S1-3.
19S1-3
19S1-4.
19S1-4
19S1-5.
19S1-5
19S1-6
19S1-6.
19S1-7
19S1-7.
19S1-8
19S1-8.
When the number of pints of blood delivered can be specified at the time of delivery, the
starting number of pints including the delivery will never exceed the largest possible
demand in a period, so we can restrict our attention to states 3 œ !ß "ß #ß $. The admissible
actions in state 3 are to order ! Ÿ 5 Ÿ $  3. Given a decision 5 , the transition
probabilities and the immediate cost are computed as follows:
:34 Ð5Ñ œ T ÖH œ 3  5  4× if 4
:3! Ð5Ñ œ T ÖH
"
3  5×
G35 œ &!5  IÒ"!!Ð3  5  HÑ Ó.
Initialization: .3 ÐV" Ñ œ " for 3 œ !ß "ß # and .$ ÐV" Ñ œ !
! 
 !Þ' !Þ% !
 *! 
 !Þ$ !Þ$ !Þ% ! 
 '! 
PÐV" Ñ œ 
 GÐV" Ñ œ  
!Þ" !Þ# !Þ$ !Þ%
&!
 !Þ" !Þ# !Þ$ !Þ% 
 ! 
Iteration 1:
Step 1: Value determination:
1ÐV" Ñ œ *!  !Þ'@! ÐV" Ñ  !Þ%@" ÐV" Ñ  @! ÐV" Ñ
1ÐV" Ñ œ '!  !Þ$@! ÐV" Ñ  !Þ$@" ÐV" Ñ  !Þ%@# ÐV" Ñ  @" ÐV" Ñ
1ÐV" Ñ œ &!  !Þ"@! ÐV" Ñ  !Þ#@" ÐV" Ñ  !Þ$@# ÐV" Ñ  !Þ%@$ ÐV" Ñ  @# ÐV" Ñ
1ÐV" Ñ œ !  !Þ"@! ÐV" Ñ  !Þ#@" ÐV" Ñ  !Þ$@# ÐV" Ñ  !Þ%@$ ÐV" Ñ  @$ ÐV" Ñ
@$ ÐV" Ñ œ !
Ê 1ÐV" Ñ œ &(Þ)ß @! ÐV" Ñ œ "*'Þ$ß @" ÐV" Ñ œ ""&Þ*ß @# ÐV" Ñ œ &!ß @$ ÐV" Ñ œ !
Step 2: Policy improvement:


minimize 
"!!  @! ÐV" Ñ  @! ÐV" Ñ œ "!!

*!  !Þ'@! ÐV" Ñ  !Þ%@" ÐV" Ñ  @! ÐV" Ñ œ &(Þ)


""!  !Þ$@! ÐV" Ñ  !Þ$@" ÐV" Ñ  !Þ%@# ÐV" Ñ  @! ÐV" Ñ œ #(Þ$'
 "&!  !Þ"@! ÐV" Ñ  !Þ#@" ÐV" Ñ  !Þ$@# ÐV" Ñ  !Þ%@$ ÐV" Ñ  @! ÐV" Ñ œ ""Þ&" 
Ê .! ÐV# Ñ œ $
minimize

%!  !Þ'@! ÐV" Ñ  !Þ%@" ÐV" Ñ  @" ÐV" Ñ œ ))Þ#%

'!  !Þ$@! ÐV" Ñ  !Þ$@" ÐV" Ñ  !Þ%@# ÐV" Ñ  @" ÐV" Ñ œ &(Þ)
 "!!  !Þ"@! ÐV" Ñ  !Þ#@" ÐV" Ñ  !Þ$@# ÐV" Ñ  !Þ%@$ ÐV" Ñ  @" ÐV" Ñ œ %"Þ*" 
Ê ." ÐV# Ñ œ #
minimize 
"!  !Þ$@! ÐV" Ñ  !Þ$@" ÐV" Ñ  !Þ%@# ÐV" Ñ  @# ÐV" Ñ œ ($Þ''
&!  !Þ"@! ÐV" Ñ  !Þ#@" ÐV" Ñ  !Þ$@# ÐV" Ñ  !Þ%@$ ÐV" Ñ  @# ÐV" Ñ œ &(Þ) 
Ê .# ÐV# Ñ œ "
V# is not identical to V" , so optimality test fails.
19S1-9
Iteration #:
Step 1: Value determination:
1ÐV# Ñ œ "&!  !Þ"@! ÐV# Ñ  !Þ#@" ÐV# Ñ  !Þ$@# ÐV# Ñ  !Þ%@$ ÐV# Ñ  @! ÐV# Ñ
1ÐV# Ñ œ "!!  !Þ"@! ÐV# Ñ  !Þ#@" ÐV# Ñ  !Þ$@# ÐV# Ñ  !Þ%@$ ÐV# Ñ  @" ÐV# Ñ
1ÐV# Ñ œ &!  !Þ"@! ÐV# Ñ  !Þ#@" ÐV# Ñ  !Þ$@# ÐV# Ñ  !Þ%@$ ÐV# Ñ  @# ÐV# Ñ
1ÐV# Ñ œ !  !Þ"@! ÐV# Ñ  !Þ#@" ÐV# Ñ  !Þ$@# ÐV# Ñ  !Þ%@$ ÐV# Ñ  @$ ÐV# Ñ
@$ ÐV# Ñ œ !
Ê 1ÐV# Ñ œ &!ß @! ÐV# Ñ œ "&!ß @" ÐV# Ñ œ "!!ß @# ÐV# Ñ œ &!ß @$ ÐV# Ñ œ !
Step 2: Policy improvement:


minimize 
"!!  @! ÐV# Ñ  @! ÐV# Ñ œ "!!

*!  !Þ'@! ÐV# Ñ  !Þ%@" ÐV# Ñ  @! ÐV# Ñ œ (!


""!  !Þ$@! ÐV# Ñ  !Þ$@" ÐV# Ñ  !Þ%@# ÐV# Ñ  @! ÐV# Ñ œ &&
 "&!  !Þ"@! ÐV# Ñ  !Þ#@" ÐV# Ñ  !Þ$@# ÐV# Ñ  !Þ%@$ ÐV# Ñ  @! ÐV# Ñ œ &! 
Ê .! ÐV$ Ñ œ $
minimize

%!  !Þ'@! ÐV# Ñ  !Þ%@" ÐV# Ñ  @" ÐV# Ñ œ (!

'!  !Þ$@! ÐV# Ñ  !Þ$@" ÐV# Ñ  !Þ%@# ÐV# Ñ  @" ÐV# Ñ œ &&
 "!!  !Þ"@! ÐV# Ñ  !Þ#@" ÐV# Ñ  !Þ$@# ÐV# Ñ  !Þ%@$ ÐV# Ñ  @" ÐV# Ñ œ &! 
Ê ." ÐV$ Ñ œ #
minimize 
"!  !Þ$@! ÐV" Ñ  !Þ$@" ÐV" Ñ  !Þ%@# ÐV" Ñ  @# ÐV" Ñ œ &&
&!  !Þ"@! ÐV" Ñ  !Þ#@" ÐV" Ñ  !Þ$@# ÐV" Ñ  !Þ%@$ ÐV" Ñ  @# ÐV" Ñ œ &! 
Ê .# ÐV$ Ñ œ "
V$ is identical to V# , so it is optimal to start every period with $ pints of blood after
delivery of the order.
19S1-10
SUPPLEMENT 2 TO CHAPTER 19
A DISCOUNTED COST CRITERION
19S2-1.
Let states !, " and # denote $'!!, $)!! and $"!!! offers respectively and let state $
designate the case that the car has already been sold (state ∞ of the hint). Let decisions "
and # be to reject and to accept the offer respectively.
G!" œ G"" œ G#" œ '!, G!# œ '!!, G"# œ )!! and G## œ "!!!
 &Î)
 &Î)
T Ð"Ñ œ 
&Î)
 !
"Î%
"Î%
"Î%
!
!
!
!
!
ß T Ð#Ñ œ 
!
!
!
"
"Î)
"Î)
"Î)
!
!
!
!
!
!
!
!
!
"
"

"
"
Start with the policy to reject only the $'!! offer. The relevant equations are:
Z! œ '!  !Þ*& &) Z!  "% Z"  ") Z# 
Z" œ )!!  !Þ*&Z$
Z# œ "!!!  !Þ*&Z$
Z$ œ !Þ*&Z$ ,
which admit the unique solution ÐZ! ß Z" ß Z# ß Z$ Ñ œ Ð(*'!Î"$ß )!!ß "!!!ß !Ñ.
Policy improvement:
State ! with decision #: '!!  !Þ*&Z$ œ '!!  Z!
State " with decision ": '!  !Þ*&ÒÐ&Î)ÑZ!  Ð"Î%ÑZ"  Ð"Î)ÑZ# Ó œ (*'!Î"$  Z"
State # with decision ": '!  !Þ*&ÒÐ&Î)ÑZ!  Ð"Î%ÑZ"  Ð"Î)ÑZ# Ó œ (*'!Î"$  Z#
Hence, the policy to reject the $'!! offer and to accept $)!! and $"!!! offers is optimal.
19S2-2.
(a) minimize '!C!"  '!!C!#  '!C""  )!!C"#  '!C#"  "!!!C##
subject to
C!" 
C"" 
C#" 
C35
C!#  !Þ*& &) C!"  C""  C#"  œ
C"#  !Þ*& "% C!"  C""  C#"  œ
C##  !Þ*& ") C!"  C""  C#"  œ
"
$
"
$
"
$
! for 3 œ !ß "ß # and 5 œ "ß #
(b) Using the simplex method, we find C!" œ !Þ)"*(*ß C"# œ !Þ&#((ß C## œ !Þ%$!&' and
the remaining C35 's are zero. Hence, the optimal policy is to reject the $'!! offer and to
accept the $)!! and $"!!! offers.
19S2-1
19S2-3.
Z38 œ minÖ'!  !Þ*&ÐÐ&Î)ÑZ!8"  Ð"Î%ÑZ"8"  Ð"Î)ÑZ#8" Ñß ÐofferÑ× for 3 œ !ß "ß #
Z3! œ ! for 3 œ !ß "ß #
Iteration 1:
Z3" œ minÖ'!ß ÐofferÑ× œ ÐofferÑ for 3 œ !ß "ß # Ê Accept
Iteration 2:
Z!# œ minÖ'!&ß '!!× œ '!& Ê Reject
Z"# œ minÖ'!&ß )!!× œ )!! Ê Accept
Z## œ minÖ'!&ß "!!!× œ "!!! Ê Accept
Iteration 3:
Z!$ œ minÖ'!(Þ*(ß '!!× œ '!(Þ*( Ê Reject
Z"$ œ minÖ'!(Þ*(ß )!!× œ )!! Ê Accept
Z#$ œ minÖ'!(Þ*(ß "!!!× œ "!!! Ê Accept
The approximate optimal solution is to reject the $'!! offer and to accept the $)!! and
$"!!! offers. This policy is indeed optimal, as found in Problem 19S2-1 and 19S2-2.
19S2-4.
Let states !, " and # denote the selling price of $"!, $#! and $$! respectively and let state
$ designate the case that the stock has already been sold. Let decisions " and # be to hold
and to sell the stock respectively.
G!" œ G"" œ G#" œ !, G!# œ "!, G"# œ #! and G## œ $!
 %Î&
 "Î%
T Ð"Ñ œ 
!
 !
"Î&
"Î%
$Î%
!
!
"Î#
"Î%
!
!
!
!
!
ß T Ð#Ñ œ 
!
!


"
!
!
!
!
!
!
!
!
!
"
"

"
"
Start with the policy to sell only when the price is $$!. The relevant equations are:
Z! œ !  !Þ* %& Z!  "& Z" 
Z" œ !  !Þ* "% Z!  "% Z"  "# Z# 
Z# œ $!  !Þ*Z$
Z$ œ !  !Þ*Z$ ,
which admit the unique solution ÐZ! ß Z" ß Z# ß Z$ Ñ œ Ð%)'!Î$&$ß (&'!Î$&$ß $!ß !Ñ.
Policy improvement:
State ! with decision #: "!  !Þ*Z$ œ "!  Z!
State " with decision #: #!  !Þ*Z$ œ #!  Z"
State # with decision ": !  !Þ*ÒÐ$Î%ÑZ"  Ð"Î%ÑZ# Ó œ #"Þ#"  Z#
Hence, the policy to hold the stock when the price is $"! and $#!, and to sell it when the
price is $$!.
19S2-2
19S2-5.
"!C!#  #!C"#  $!C##
(a) minimize
subject to C!"  C!#  !Þ* %& C!"  "% C"" 
C""  C"#  !Þ* "& C!"  "% C""  $% C#" 
C#"  C##  !Þ* "# C""  "% C#" 
C35
œ
"
$
œ
"
$
œ
"
$
! for 3 œ !ß "ß # and 5 œ "ß #
(b) Using the simplex method, we find C!" œ "Þ*'!&*ß C"" œ !Þ*&)&"ß C## œ !Þ('%'$ and
the remaining C35 's are zero. Hence, the optimal policy is to hold the stock at the prices
$"! and $#! and to sell it at the price $$!.
19S2-6.
Z!8 œ minÖ!Þ*ÐÐ%Î&ÑZ!8"  Ð"Î&ÑZ"8" Ñß "!×
Z"8 œ minÖ!Þ*ÐÐ"Î%ÑZ!8"  Ð"Î%ÑZ"8"  Ð"Î#ÑZ#8" Ñß #!×
Z#8 œ minÖ!Þ*ÐÐ$Î%ÑZ"8"  Ð"Î%ÑZ#8" Ñß $!×
Z3! œ ! for 3 œ !ß "ß #
Iteration 1:
Z!" œ minÖ!ß "!× œ "! Ê Sell
Z"" œ minÖ!ß #!× œ #! Ê Sell
Z#" œ minÖ!ß $!× œ $! Ê Sell
Iteration 2:
Z!# œ minÖ"!Þ)ß "!× œ "!Þ) Ê Hold
Z"# œ minÖ#!Þ#&ß #!× œ #!Þ#& Ê Hold
Z## œ minÖ#!Þ#&ß $!× œ $! Ê Sell
Iteration 3:
Z!$ œ minÖ""Þ%#ß "!× œ ""Þ%# Ê Hold
Z"$ œ minÖ#!Þ%*ß #!× œ #!Þ%* Ê Hold
Z#$ œ minÖ#!Þ%#ß $!× œ $! Ê Sell
The approximate optimal solution is to sell if the price is $$! and to hold otherwise. This
policy is indeed optimal, as found in Problem 19S2-3 and 19S2-4.
19S2-3
19S2-7.
(a) Let states ! and " be the chemical produced this month, G" and G# respectively, and
decisions " and # refer to the process to be used next month, E and F respectively. There
are four stationary deterministic policies.
3
!
"
.3 ÐV" Ñ
"
"
.3 ÐV# Ñ
"
#
.3 ÐV$ Ñ
#
"
.3 ÐV% Ñ
#
#
The transition matrix is the same for every decision, viz.
T œ
!Þ$
!Þ%
!Þ(
.
!Þ' 
The costs G35 correspond to the expected amount of pollution using the process 5 in the
next period.
G!" œ !Þ$Ð"&Ñ  !Þ(Ð#Ñ œ &Þ*ß
G!# œ !Þ$Ð$Ñ  !Þ(Ð)Ñ œ 'Þ&ß
G"" œ !Þ%Ð"&Ñ  !Þ'Ð#Ñ œ (Þ#ß
G"# œ !Þ%Ð$Ñ  !Þ'Ð)Ñ œ '.
(b)
19S2-4
19S2-8.
(a) minimize &Þ*C!"  'Þ&C!#  (Þ#C""  'C"#
subject to
C!" 
C"" 
C35
$
C!#  "#  "!
C!" 
(
C"#  "#  "!
C!" 
%
"! C""

$
"! C!#

'
"! C""

(
"! C!#

%
"! C"# 
'
"! C"# 
œ
"
#
œ
"
#
! for 3 œ !ß " and 5 œ "ß #
(b) Using the simplex method, we find C!" œ !Þ)&(ß C"# œ "Þ"%$ and C!# œ C"" œ !.
Hence, the optimal policy is to use process E if G" is produced and F if G# is produced
this month.
19S2-5
19S2-9.
19S2-10.
The three iterations of successive approximations in Problem 19S2-9 gives the optimal
policy for the three-period problem. The optimal policy is, therefore, to use the process E
if G" is produced and F if G# is produced in all periods.
19S2-6
19S2-11.
Z!8 œ minÖ!  !Þ*!ÐÐ(Î)ÑZ"8"  Ð"Î"'ÑZ#8"  Ð"Î"'ÑZ$8" Ñß %!!!  !Þ*!Z"8" ß '!!!  !Þ*!Z!8" ×
Z"8 œ minÖ"!!!  !Þ*!ÐÐ$Î%ÑZ"8"  Ð"Î)ÑZ#8"  Ð"Î)ÑZ$8" Ñß %!!!  !Þ*!Z"8" ß '!!!  !Þ*!Z!8" ×
Z#8 œ minÖ$!!!  !Þ*!ÐÐ"Î#ÑZ#8"  Ð"Î#ÑZ$8" Ñß %!!!  !Þ*!Z"8" ß '!!!  !Þ*!Z!8" ×
Z$8 œ '!!!  !Þ*!Z!8"
Z3! œ ! for 3 œ !ß "ß #ß $
Iteration 1:
Z!" œ minÖ!ß %!!!ß '!!!× œ ! Ê Do nothing
Z"" œ minÖ"!!!ß %!!!ß '!!!× œ "!!! Ê Do nothing
Z#" œ minÖ$!!!ß %!!!ß '!!!× œ $!!! Ê Do nothing
Z$" œ '!!! Ê Replace
Iteration 2:
Z!# œ minÖ"#*$Þ(&ß %*!!ß '!!!× œ "#*$Þ(& Ê Do nothing
Z"# œ minÖ#')(Þ&ß %*!!ß '!!!× œ #')(Þ& Ê Do nothing
Z## œ minÖ(!&!ß %*!!ß '!!!× œ %*!! Ê Overhaul
Z$# œ '!!! Ê Replace
Iteration 3:
Z!$ œ minÖ#(#*Þ&$ß '%")Þ(&ß ("'%Þ$)× œ #(#*Þ&$ Ê Do nothing
Z"$ œ minÖ%!%!Þ$"ß '%")Þ(&ß ("'%Þ$)× œ %!%!Þ$" Ê Do nothing
Z#$ œ minÖ(*!&ß '%")Þ(&ß ("'%Þ$)× œ '%")Þ(& Ê Overhaul
Z$$ œ ("'%Þ$) Ê Replace
Iteration 4:
Z!% œ minÖ$*%&Þ)!ß ('$'Þ#)ß )%&'Þ&)× œ $*%&Þ)! Ê Do nothing
Z"% œ minÖ&#&&Þ$"ß ('$'Þ#)ß )%&'Þ&)× œ &#&&Þ$" Ê Do nothing
Z#% œ minÖ*""#Þ%"ß ('$'Þ#)ß )%&'Þ&)× œ ('$'Þ#) Ê Overhaul
Z$% œ )%&'Þ&) Ê Replace
The optimal policy is to do nothing in states !ß " and to replace in state $ in all periods.
When in state #, it is best to overhaul in periods "ß #ß $ and to do nothing in period %.
19S2-7
CHAPTER 20: SIMULATION
20.1-1.
(a)
to
to
Random observations:
heads,
(b)
to
to
Random observations:
ball,
(c)
to
to
to
correspond to tails.
correspond to heads.
heads,
tails,
tails,
heads
correspond to strikes.
correspond to balls.
ball,
strike,
strike,
strike,
green,
green,
ball
correspond to green lights.
correspond to yellow lights.
correspond to red lights.
Random observations:
red,
red
red,
green,
20.1-2.
(a)
If it is raining:
to
to
correspond to rain next day,
correspond to clear next day.
If it is clear:
to
to
correspond to clear next day,
correspond to rain next day.
Day
tails,
Random Number
Weather
Clear
Rain
Clear
Clear
Clear
Clear
Clear
Clear
Clear
Rain
20-1
(b)
If Clear, Prob(Stays Clear) =
If Rain, Prob(Stays Rain) =
Day
Random
Number
1
2
3
4
5
6
7
8
9
10
0.8815
0.0252
0.8081
0.5692
0.0277
0.9160
0.2733
0.0558
0.4683
0.8070
0.8
0.6
Weather
Clear
Rain
Rain
Clear
Clear
Clear
Rain
Rain
Rain
Rain
Clear
20.1-3.
(a)
(b)
Mean:
(c)
(d)
stoves
to
to
to
to
to
stoves,
The average of these is
correspond to stoves being sold.
correspond to stoves being sold.
correspond to stoves being sold.
correspond to stoves being sold.
correspond to stoves being sold.
stoves,
stoves
, which exceeds the mean in (b) by
20-2
.
(e) Answers will vary. The following
.
Day
1
2
3
4
5
6
7
297
298
299
300
Random
Number
0.7167
0.3367
0.1763
0.9230
0.6635
0.4588
0.2529
0.8098
0.4217
0.4709
0.0008
Demand
4
3
3
5
4
4
3
5
3
4
2
Average =
3.723
-day simulation yielded an average demand of
Distribution of
Demand
Probability
0.16
0.28
0.32
0.20
0.04
Cumulative
0
0.16
0.44
0.76
0.96
Demand
2
3
4
5
6
20.1-4.
(a)
(b)
Est
Est
Est
Est
Est
customers
Est
customers
Customers
Est
Est
Arrival Time
Service Time
sum of observed system times
number of observed system times
sum of observed waiting times
number of observed waiting times
Departure Time
minutes
minutes
20-3
System Time
Wait Time
(c)
(d)
Est
Est
Est
Est
customers
Est
customers
Customers
Est
Est
Arrival Time
Service Time
Departure Time
sum of observed system times
number of observed system times
sum of observed waiting times
number of observed waiting times
System Time
Wait Time
minutes
minutes
20.1-5.
(a) Interarrival Time ~Exp
per minute , Service Time ~Exp
Next interarrival time:
Next service time:
per minute
ln
ln
Let and
denote the time in minutes and the number of customers in the system at
time respectively. In the table below, N.I.T. stands for Next Interarrival Time and
N.S.T. for Next Service Time.
N.I.T.
N.S.T.
20-4
Next Arriv.
Next Dep.
Next Event
Arrival
Departure
Arrival
Departure
Arrival
(b)
arrival in two-minute period
departure in two-minute period
arrival occurred,
arrival did not occur.
departure occurred,
departure did not occur.
Let and
denote the time in minutes and the number of customers in the system at
time respectively.
12
18
24
30
36
42
48
0
(c) Interarrival Time ~Exp
569
764
492
950
610
145
484
350
Arrival?
Yes
No
No
No
No
No
Yes
No
Yes
, Service Time ~Exp
20-5
Departure?
0.665
842
224
No
No
Yes
0.552
Yes
Average number waiting to begin service:
Average number waiting for or in service:
Average waiting time excluding service:
Average waiting time including service:
(d)
Number of Servers =
Interarrival Times
Distribution =
Mean =
Data
1
L=
Lq =
W=
Wq =
Exponential
0.2
5
Mean =
P0 = 0.50138152
0.487114273
0.51564877
P1 = 0.25139558
0.244202927
0.258588225
0.1
P2 = 0.12327761
0.117377792
0.129177429
25
P3 = 0.06076612
0.055624158
0.065908089
Length of Simulation Run
Number of Arrivals =
Results
95% Confidence Interval
Low
High
0.924695993 1.110539606
0.435985367 0.602013275
0.187680035 0.221688189
0.088497477 0.120286046
Exponential
Service Times
Distribution =
Point
Estimate
1.0176178
0.51899932
0.20468411
0.10439176
10,000
Run Simulation
20-6
P4 = 0.02844012
0.02469865
0.032181586
P5 = 0.01636786
0.012907523
0.019828193
P6 = 0.00918318
0.006044617
0.012321737
P7 = 0.00363525
0.002194291
0.005076203
P8 = 0.00176598
0.000781557
0.002750408
P9 = 0.00085645
0.000299473
0.001413424
P10 = 0.00086804
-1.54083E-05
0.001751489
(e)
Results
s=
Every measure is inside the
Data
5
10
1
(mean arrival rate)
(mean service rate)
(# servers)
L=
Lq =
1
0.5
W=
Wq =
0.2
0.1
0.5
% confidence level.
20.1-6.
(a) The system is a single-server queueing system with the crew being servers and the
machines being customers. The service time has a uniform distribution between and
twice the mean. The interarrival time is exponentially distributed with mean being
hours. A simulation clock records the amount of simulated time that elapses. The state
of the system at time is the number of machines that need repair at time . The
breakdowns and repairs that occur over time are randomly generated by generating
random observations from the distributions of interarrival and service times. The state of
the system needs to be adjusted when a breakdown or repair occurs:
if a breakdown occurs at time ,
if a repair occurs at time .
Reset
The time on the simulation clock is adjusted by using the next-event time advance procedure. The time is in hours.
(b) The random numbers
and
are obtained from Table 20.3 starting from the front
of the first row. N.I.T. stands for Next Interarrival Time and N.S.T. for Next Service
Time. Interarrival times are computed as
ln
and service times correspond to
.
Initially there is one broken machine in the system.
N.I.T.
N.S.T.
20-7
Next Arriv.
Next Dep.
Next Event
Departure
Arrival
Arrival
Arrival
Departure
Arrival
Arrival
Departure
Departure
(c)
arrival in one-hour period
departure in one-hour period
arrival occurred,
arrival did not occur.
departure occurred,
departure did not occur.
Let and
denote the time in hours and the number of broken machines in the
system at time respectively.
and
are obtained from Table 20.3 starting from the
front of the first row.
Arrival?
Yes
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
Yes
No
No
Departure?
No
No
No
No
No
No
No
Yes
No
No
No
No
No
Yes
No
No
20-8
(d)
Crew size
Average waiting time excluding service:
Average waiting time including service:
Average number waiting to begin service:
Average number waiting or in service:
20-9
hours
hours
Crew size
Average waiting time excluding service:
Average waiting time including service:
Average number waiting to begin service:
Average number waiting or in service:
20-10
hours
hours
Crew size
Average waiting time excluding service:
Average waiting time including service:
Average number waiting to begin service:
Average number waiting or in service:
20-11
hours
hours
(e)
Crew size
Number of Servers =
Interarrival Times
Distribution =
Mean =
Data
1
L=
Lq =
W=
Wq =
Exponential
5
5
P0 = 0.20754019
Service Times
Distribution =
Point
Estimate
2.79160748
1.99914767
14.1265258
10.1163976
Results
95% Confidence Interval
Low
High
2.487589825 3.095625137
1.708379282 2.289916051
12.76418957 15.48886212
8.769419342 11.46337583
0.189943343
0.225137028
0.217466416
Uniform
P1 = 0.20310594
0.18874546
Minimum Value =
0
P 2 = 0.16447728
0.153931932
0.175022627
Maximum Value =
8
P3 = 0.1219194
0.113228192
0.130610613
P4 = 0.08985019
0.081414391
0.098285997
P5 = 0.06348721
0.055611754
0.071362667
P6 = 0.0471849
0.038913859
0.055455941
Length of Simulation Run
Number of Arrivals =
10,000
Run Simulation
P7 = 0.03473004
0.027372469
0.042087618
P8 = 0.02360601
0.017520132
0.029691895
P9 = 0.01630948
0.010667279
0.02195168
P10 = 0.00959473
0.005360122
0.013829346
Crew size
Number of Servers =
Interarrival Times
Distribution =
Mean =
Data
1
L=
Lq =
W=
Wq =
Exponential
5
5
Point
Estimate
1.24431993
0.64592258
6.29742625
3.26897425
Results
95% Confidence Interval
Low
High
1.142543237 1.346096621
0.554073233 0.737771925
5.859692708 6.735159796
2.842349379 3.695599122
P0 = 0.40160265
0.387532006
0.415673294
Uniform
P1 = 0.28107892
0.272769162
0.289388675
Minimum Value =
0
P2 = 0.15597768
0.14921628
0.162739075
Maximum Value =
6
P3 = 0.0793332
0.073170906
0.085495493
P4 = 0.04275587
0.037208336
0.048303406
Service Times
Distribution =
Length of Simulation Run
Number of Arrivals =
10,000
Run Simulation
20-12
P5 = 0.01981074
0.016184441
0.023437041
P6 = 0.00908873
0.006525147
0.011652319
P7 = 0.00422182
0.002175617
0.006268018
P8 = 0.00178977
0.000376454
0.003203086
P9 = 0.00161276
-0.000465062
0.003690573
P10 = 0.00097323
-0.00055135
0.002497807
Crew size
Number of Servers =
Interarrival Times
Distribution =
Mean =
Data
1
L=
Lq =
W=
Wq =
Exponential
5
5
Point
Estimate
0.57461519
0.17109501
2.84672424
0.84762866
Results
95% Confidence Interval
Low
High
0.554018788 0.595211587
0.157890875 0.184299136
2.779360906 2.914087583
0.79143139 0.903825926
P0 = 0.59647982
0.587464562
0.605495074
Uniform
P1 = 0.27673768
0.27132101
0.282154358
Minimum Value =
0
P 2 = 0.09234343
0.087802456
0.096884413
Maximum Value =
4
P 3 = 0.02626569
0.0235337
0.028997676
P4 = 0.00677101
0.005226946
0.008315078
P5 = 0.00118034
0.000591788
0.001768898
P6 = 0.00017176
6.54659E-06
0.000336978
P7 = 2.4834E-05
-1.98852E-05
6.95524E-05
P8 = 2.5425E-05
-2.43593E-05
7.52092E-05
Service Times
Distribution =
Length of Simulation Run
Number of Arrivals =
10,000
Run Simulation
P9 =
0
0
0
P10 =
0
0
0
According to these simulation runs, a crew size of
time before repair below hours.
is enough to get the average waiting
(f) ,
, , and denote the mean breakdown rate, the expected repair time, the
variance of the repair time, and the number of servers respectively. The variance of a
random variable uniformly distributed between and is
.
Crew size
:
,
,
Crew size
:
,
,
Crew size
:
,
,
A crew size of
than hours.
is enough to have the average waiting time before repair begins no more
20-13
20.1-7.
(a)
(b)
Est
Est
Est
Est
(c)
Est
customers
Est
customers
(d)
Est
Est
sum of observed system times
number of observed system times
sum of observed waiting times
number of observed waiting times
20-14
minutes
minutes
20.1-8.
(a)
Average number waiting to begin service:
Average number waiting for or in service:
Average waiting time excluding service:
Average waiting time including service:
20-15
(b)
Two Tellers
Number of Servers =
Data
2
Interarrival Times
Distribution = Translated Exponential
Minimum Value =
0.5
Mean =
1
Mean =
k=
P0 = 0.07904235
0.070993855
0.08709085
P1 = 0.33535638
0.317913662
0.352799095
1.5
P2 = 0.35727575
0.344441715
0.370109776
4
P3 = 0.15966204
0.146452148
0.17287193
Length of Simulation Run
Number of Arrivals =
5,000
Run Simulation
(c)
Results
95% Confidence Interval
Low
High
1.751958482 1.910622557
0.267225477 0.382237728
1.75482672
1.88644754
0.267716503 0.377968501
Erlang
Service Times
Distribution =
L=
Lq =
W=
Wq =
Point
Estimate
1.83129052
0.3247316
1.82063713
0.3228425
P4 = 0.04853863
0.03897433
0.058102924
P5 = 0.01520428
0.008218196
0.022190355
P6 = 0.0029803
0.000526793
0.005433813
P7 = 0.00124712
-0.000891002
0.003385241
P8 = 0.00062945
-0.000595242
0.001854135
P9 = 6.3713E-05
-6.02511E-05
0.000187678
0
0
P10 =
0
L=
Lq =
W=
Wq =
Point
Estimate
1.50182365
0.01101285
1.50567682
0.01104111
Three Tellers
Number of Servers =
Data
3
Interarrival Times
Distribution = Translated Exponential
Minimum Value =
0.5
Mean =
1
P0 = 0.11036737
0.102191363
0.118543369
Erlang
P1 = 0.4043504
0.393136309
0.415564497
1.5
P2 = 0.3693863
0.35881755
0.379955043
4
P3 = 0.10544811
0.097351282
0.113544943
P4 = 0.00991044
0.007638312
0.012182566
P5 = 0.00050974
5.55269E-05
0.000963948
P6 = 2.7646E-05
-2.6457E-05
8.17495E-05
Service Times
Distribution =
Mean =
k=
Length of Simulation Run
Number of Arrivals =
Results
95% Confidence Interval
Low
High
1.471908962 1.531738345
0.008142707 0.013882997
1.484221201 1.527132443
0.008201326 0.013880889
5,000
Run Simulation
20-16
P7 =
0
0
0
P8 =
0
0
0
P9 =
0
0
0
P10 =
0
0
0
(d)
Two Tellers
Number of Servers =
Data
2
Interarrival Times
Distribution = Translated Exponential
Minimum Value =
0.5
Mean =
0.9
Mean =
k=
P0 = 0.04013233
0.033488277
0.046776385
P 1 = 0.22981231
0.204481462
0.255143148
1.5
P2 = 0.31263989
0.285380805
0.339898978
4
P3 = 0.20077625
0.183373045
0.218179462
P4 = 0.10913211
0.092724399
0.125539826
P5 = 0.0496635
0.036931841
0.062395156
Length of Simulation Run
Number of Arrivals =
Results
95% Confidence Interval
Low
High
2.245999547 2.951254527
0.583387757 1.234020251
2.026140761 2.645193244
0.526819275 1.106681746
Erlang
Service Times
Distribution =
L=
Lq =
W=
Wq =
Point
Estimate
2.59862704
0.908704
2.335667
0.81675051
5,000
Run Simulation
P 6 = 0.02257618
0.0131468
0.032005558
P7 = 0.00911677
0.002426257
0.015807287
P8 = 0.00704568
-0.000481236
0.014572602
P9 = 0.00471191
-0.0012258
0.01064961
P10 = 0.00538407
-0.002296853
0.013064986
Three Tellers
Number of Servers =
Data
3
Interarrival Times
Distribution = Translated Exponential
Minimum Value =
0.5
Mean =
0.9
Mean =
k=
0.064654631
0.077395038
Erlang
P 1 = 0.34936892
0.337158282
0.361579559
1.5
P2 = 0.40664103
0.396507503
0.416774558
4
P3 = 0.14889467
0.139047346
0.158742002
P4 = 0.02199304
0.017683072
0.026303015
P5 = 0.00200283
0.000918837
0.003086824
P 6 = 7.4667E-05
-5.7164E-05
0.000206498
Length of Simulation Run
Number of Arrivals =
Results
95% Confidence Interval
Low
High
1.672660236 1.742878424
0.020528952 0.031916459
1.521784195 1.567952832
0.018683478 0.028759266
P0 = 0.07102483
Service Times
Distribution =
L=
Lq =
W=
Wq =
Point
Estimate
1.70776933
0.02622271
1.54486851
0.02372137
5,000
Run Simulation
20-17
P7 =
0
0
0
P8 =
0
0
0
P9 =
0
0
0
P10 =
0
0
0
(e) Let
denote the average time between customer arrivals. Some performance
measures are given for two-teller and three-teller systems in the following tables.
Two Tellers
Idle
Three Tellers
Two Tellers
Three Tellers
Idle
The last row corresponds to the probability that at least one of the tellers is idle. For the
two-teller system it is
and for the three-teller system it is
. There is
a big difference between the idle-time ratios of the two-teller and three-teller systems for
both values. For this reason, it may be better to hire two tellers. Two tellers also
provide reasonable wait times,
minutes for
and
minutes
for
. A thorough analysis would also incorporate the cost of hiring and the profit
from the completion of each job.
20.1-9.
20-18
Class 1 Customers:
Average number waiting to begin service:
Average number waiting for or in service:
Average waiting time excluding service:
Average waiting time including service:
Class 2 Customers:
Average number waiting to begin service:
Average number waiting for or in service:
Average waiting time excluding service:
Average waiting time including service:
20.1-10.
(a) For parts (a) through (f), each type of car corresponds to an M/M/1 system and they
are independent of each other. For parts (g) through (i), the system is an M/M/2 system.
Both interarrival and service times are exponentially distributed. A simulation clock
records the amount of simulated time that elapses. The state of the system at time
consists of the number
of Japanese cars that need to be repaired at time and the
number
of German cars that need to be repaired at time . The breakdowns and
repairs that occur over time are generated by random observations with exponential
distributions. The state of the system follows the dynamics:
if a Japanese car arrives to the shop,
if a Japanese car is repaired,
if a German car arrives to the shop,
if a German car is repaired.
The time is advanced using the next-event time advance procedure.
20-19
(b)
(c)
German Cars
Number of Servers =
Interarrival Times
Distribution =
Mean =
Data
1
L=
Lq =
W=
Wq =
Exponential
0.25
0.9
Mean =
0.168354591
0.215864553
Exponential
P1 = 0.14288828
0.126749608
0.159026958
0.2
P2 = 0.11977427
0.106778075
0.132770458
4
P3 = 0.10506269
0.093858332
0.116267048
P4 = 0.08706215
0.078233744
0.095890555
P5 = 0.07384061
0.064676221
0.083005004
P6 = 0.05661734
0.049562873
0.063671813
P7 = 0.04646493
0.039897523
0.053032328
P8 = 0.03867518
0.03211558
0.045234783
P9 = 0.02872709
0.022373996
0.03508018
P10 = 0.02231773
0.016062246
0.028573221
Length of Simulation Run
Number of Arrivals =
Results
95% Confidence Interval
Low
High
3.234015207 5.569448935
2.443475896
4.74420739
0.823761633 1.370728015
0.623425709 1.168289042
P0 = 0.19210957
Service Times
Distribution =
Point
Estimate
4.40173207
3.59384164
1.09724482
0.89585738
10,000
Run Simulation
20-20
(d)
Japanese Cars
Number of Servers =
Interarrival Times
Distribution =
Mean =
Data
1
L=
Lq =
W=
Wq =
Exponential
0.5
0.9
Point
Estimate
0.67784073
0.27524319
0.33354389
0.13543843
P0 = 0.59740246
Service Times
Distribution =
Mean =
0.586266118
0.608538804
Exponential
P1 = 0.2405621
0.234555839
0.246568367
0.2
P2 = 0.09556771
0.090486659
0.100648771
4
P3 = 0.03852888
0.034728947
0.04232882
P4 = 0.01624329
0.013664229
0.018822344
P5 = 0.00723714
0.005338826
0.009135444
P6 = 0.00266085
0.001585541
0.003736161
Length of Simulation Run
Number of Arrivals =
Results
95% Confidence Interval
Low
High
0.641699572 0.713981894
0.247874873 0.302611514
0.318934855 0.348152933
0.123198616
0.14767825
10,000
Run Simulation
(e)
20-21
P7 = 0.00113441
0.00035521
0.001913604
P8 = 0.00048975
-2.19148E-05
0.001001412
P9 = 0.00016031
-7.0534E-05
0.000391154
P10 = 1.3099E-05
-1.25453E-05
3.87434E-05
(f)
German Cars
Number of Servers =
Interarrival Times
Distribution =
Mean =
Data
2
L=
Lq =
W=
Wq =
Exponential
0.25
0.9
Mean =
0.410429578
Exponential
P1 = 0.34257354
0.334821362
0.35032571
0.2
P2 = 0.13990835
0.134017407
0.145799301
4
P3 = 0.05765995
0.053234966
0.062084937
P4 = 0.02285889
0.019882112
0.025835667
P5 = 0.00909262
0.00719708
0.010988165
P6 = 0.00355787
0.002290188
0.004825549
Length of Simulation Run
Number of Arrivals =
Results
95% Confidence Interval
Low
High
0.929376886 1.005304084
0.135159564 0.175031414
0.229378389 0.243625338
0.033338445 0.042499117
P0 = 0.42259073
Service Times
Distribution =
Point
Estimate
0.96734048
0.15509549
0.23650186
0.03791878
10,000
Run Simulation
0.43475189
P7 = 0.00084521
0.000306354
0.001384058
P8 = 0.00055155
3.71251E-05
0.001065979
P9 = 0.00026383
-0.000120885
0.000648547
P10 = 5.0835E-05
-4.87318E-05
0.000150401
(g) This option significantly decreases the waiting time for German cars without the
added cost of an additional mechanic.
Number of Servers =
Interarrival Times
Distribution =
Mean =
Data
2
L=
Lq =
W=
Wq =
Exponential
0.166666667
0.9
Mean =
P0 = 0.20555772
0.195123634
0.215991801
P1 = 0.26310835
0.253142531
0.273074175
0.22
P2 = 0.17488354
0.168697104
0.181069974
4
P3 = 0.12221465
0.116798427
0.12763088
P4 = 0.07857001
0.073827865
0.083312156
Length of Simulation Run
Number of Arrivals =
Results
95% Confidence Interval
Low
High
2.228936755 2.550158169
0.9248612 1.202681301
0.373871382 0.421598812
0.155175541 0.198949387
Exponential
Service Times
Distribution =
Point
Estimate
2.38954746
1.06377125
0.3977351
0.17706246
20,000
Run Simulation
20-22
P5 = 0.05118593
0.046911158
0.055460707
P6 = 0.0346834
0.030632654
0.038734137
P7 = 0.02396612
0.020211346
0.027720893
P8 = 0.01535973
0.012326221
0.018393236
P9 = 0.00995903
0.007331075
0.012586987
P10 = 0.00600944
0.004038714
0.00798016
(h)
Part
(c)
(d)
(f)
(g)
Est
The results of the simulation were quite accurate.
(i) Answers will vary. The option of training the two current mechanics significantly decreases the waiting time for German cars, without a significant impact on the wait for
German cars, and does so without the added cost of a third mechanic. Adding a third
mechanic reduces the average wait for German cars even more, but comes with the added
cost of a third mechanic.
20.1-11.
(a) There are two independent G/M/1 systems: printers and monitors. For printers, the
arrival stream is deterministic; for monitors, the arrival process is uniformly distributed
between
and . The inspection time is exponentially distributed with a mean of
minutes. A simulation clock records the amount of simulated time that elapses. The state
of the system at time consists of the number
of monitors in the inspection station
at time and the number
of printers in the inspection station at time . The arrivals
to the stations and the inspection times are generated by sampling distributions according
to interarrival and service time distributions. The system evolves according to the law:
if a monitor arrives to the inspection station,
if a monitor is repaired,
if a printer arrives to the inspection station,
if a printer is repaired.
The time is advanced using the next-event time advance procedure.
(b)
20-23
(c)
(d)
Monitors
Number of Servers =
Interarrival Times
Distribution =
Minimum Value =
Maximum Value =
Data
1
L=
Lq =
W=
Wq =
Uniform
10
20
Mean =
0.327224738
0.353060676
Exponential
P1 = 0.38164142
0.372075535
0.391207303
10
P2 = 0.16352709
0.155778216
0.171275956
4
P3 = 0.06811831
0.060998944
0.075237683
P4 = 0.02888903
0.023293304
0.034484764
P5 = 0.01168415
0.007821024
0.015547284
P6 = 0.00424548
0.002158582
0.006332375
P7 = 0.00128359
0.000174338
0.002392845
P8 = 0.00038837
-0.000179162
0.000955894
P9 = 7.9852E-05
-6.35832E-05
0.000223288
0
0
Length of Simulation Run
Number of Arrivals =
Results
95% Confidence Interval
Low
High
1.060901038 1.189721042
0.411484076 0.519423417
15.94273751
17.8635494
6.183798912 7.799235411
P0 = 0.34014271
Service Times
Distribution =
Point
Estimate
1.12531104
0.46545375
16.9031435
6.99151716
10,000
Run Simulation
P10 =
20-24
0
Printers
Number of Servers =
Interarrival Times
Distribution =
Value =
Data
1
L=
Lq =
W=
Wq =
Constant
15
20
Mean =
0.316172923
0.340971954
Exponential
P1 = 0.39246319
0.382956315
0.401970063
10
P2 = 0.16410908
0.156167074
0.172051091
4
P3 = 0.06829616
0.061331286
0.075261033
P4 = 0.02874872
0.023039412
0.034458025
P5 = 0.01227622
0.008565564
0.01598688
P6 = 0.00479104
0.002763062
0.006819019
P7 = 0.00070382
0.000113351
0.001294289
P8 = 3.933E-05
-3.76108E-05
0.00011627
Length of Simulation Run
Number of Arrivals =
10,000
Run Simulation
(e)
Results
95% Confidence Interval
Low
High
1.077190283
1.19467659
0.415782342 0.513229408
16.15785425 17.92014885
6.236735131 7.698441118
P0 = 0.32857244
Service Times
Distribution =
Point
Estimate
1.13593344
0.46450587
17.0390016
6.96758812
P9 =
0
0
0
P10 =
0
0
0
L=
Lq =
W=
Wq =
Point
Estimate
0.75155745
0.08784252
11.3286273
1.32409728
Monitors
Number of Servers =
Interarrival Times
Distribution =
Minimum Value =
Maximum Value =
Data
1
Uniform
10
20
P0 = 0.33628507
0.329400438
0.343169707
Erlang
P1 = 0.58164576
0.576199748
0.587091777
10
P2 = 0.07650568
0.07048566
0.082525703
4
P3 = 0.00535361
0.003131199
0.007576026
P4 = 0.00020987
Service Times
Distribution =
Mean =
k=
Length of Simulation Run
Number of Arrivals =
Results
95% Confidence Interval
Low
High
0.736774593 0.766340302
0.078522615 0.097162425
11.11532534 11.54192936
1.184672596 1.463521969
10,000
Run Simulation
20-25
-2.40367E-05
0.000443779
P5 =
0
0
0
P6 =
0
0
0
P7 =
0
0
0
P8 =
0
0
0
P9 =
0
0
0
P10 =
0
0
0
Printers
Number of Servers =
Interarrival Times
Distribution =
Value =
Data
1
L=
Lq =
W=
Wq =
Constant
15
20
Mean =
k=
0.326911198
Erlang
P1 = 0.6005944
0.595376773
0.605812026
10
P2 = 0.06184121
0.05625764
0.067424777
4
P3 = 0.00364284
0.001938538
0.00534714
P4 = 0.00025637
-0.000124119
0.000636866
P5 = 2.0194E-05
-1.93445E-05
5.97317E-05
0
0
0
Length of Simulation Run
Number of Arrivals =
Results
95% Confidence Interval
Low
High
0.722583868 0.750079721
0.06156358 0.078389982
10.83875803 11.25119582
0.923453699 1.175849737
P0 = 0.33364499
Service Times
Distribution =
Point
Estimate
0.73633179
0.06997678
11.0449769
1.04965172
10,000
P6 =
Run Simulation
0.340378775
P7 =
0
0
0
P8 =
0
0
0
P9 =
0
0
0
P10 =
0
0
0
The new inspection equipment would drastically reduce the average waiting time for both
monitors from minutes to
minutes and printers from minutes to minute .
20.2-1.
Merrill Lynch launched the Management Science Group to deal with the issues raised by
the rise of electronic trading in the late 1990s. The group studied various product
structure and pricing alternatives. They focused on two main pricing options, viz., an
asset-based pricing option and a direct online pricing option. Monte Carlo simulation is
applied to simulate the behavior of the clients who choose between the two product and
pricing options in the light of economic and qualitative factors. In the simulation model,
"the observed system data consist of every revenue-generating component of every
account of every client at Merrill Lynch. The output measures are the resulting revenue at
the firm level, the compensation impact on each FA, and the percentage of clients
considered adverse selectors" [p. 13]. Sensitivity analysis is performed to evaluate
various scenarios.
"The benefits were significant and fell into four areas: seizing the marketplace initiative,
finding the pricing sweet spot, improving financial performance, and adopting the
approach in other strategic initiatives in other strategic initiatives" [p. 15]. As a result of
this study, Merrill Lynch also acquired new clients.
20.2-2.
Answers will vary.
20-26
20.3-1.
(a)
(b)
(c)
20.3-2.
(a)
,
(b)
,
(c)
,
20.3-3.
20-27
20.3-4.
20.3-5.
(a)
(b)
,
20.3-6.
(a)
(b) Each integer appears only once in part (a).
(c)
will repeat the cycle
with length
20-28
.
20.4-1.
(a) Answers will vary.
(b) The formula in cell D10 is
Required Difference
Cash At End of Game
VLOOKUP C10 $J$8:$K$9 2 .
3
$8
Distribution of Coin Flips
Probability Cumulative Result
0.5
0
Heads
0.5
0.5
Tails
Summary of Game
Number of Flips
7
Winnings
$1
Flip
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
Random
Number
0.9683
0.8270
0.2837
0.1236
0.8999
0.7532
0.5228
0.3227
0.4547
0.2282
0.0403
0.6744
0.0852
0.9229
0.9497
0.4296
Result
Tails
Tails
Heads
Heads
Tails
Tails
Tails
Heads
Heads
Heads
Heads
Tails
Heads
Tails
Tails
Heads
Total
Heads
0
0
1
2
2
2
2
3
4
5
6
6
7
7
7
8
20-29
Total
Tails
1
2
2
2
3
4
5
5
5
5
5
6
6
7
8
8
Stop?
Stop
NA
NA
NA
NA
NA
NA
NA
NA
NA
(c) A simulation with 14 replications:
(d) A simulation with 1000 replications:
Play
1
2
3
4
5
6
7
8
9
10
11
12
13
14
Number
of Flips
7
31
3
7
29
3
17
5
21
3
9
13
7
3
7
Winnings
1
-23
5
1
-21
5
-9
3
-13
5
-1
-5
1
5
1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
997
998
999
1000
Average:
11.286
-3.286
Average:
Play
Number
of Flips
7
13
11
17
13
17
3
5
7
5
3
3
11
11
15
9
9
5
3
11
3
5
Winnings
1
-5
-3
-9
-5
-9
5
3
1
3
5
5
-3
-3
-7
-1
-1
3
5
-3
5
3
8.972
-0.972
20.4-2.
(a)
(b)
(c) If cell A1 contains the uniform random number, then the Excel function is "
20-30
A1
."
20.4-3.
(a)
¼
(b)
(c)
20.4-4.
(a) To determine whether
or is distributed uniformly between
at a three-digit random number from Table 20.3.
and
, look
.
is uniformly distributed.
If
, nothing else need to be done. Otherwise, use the next three-digit random
number as a decimal to generate .
~
~
Hence, the sequence is
.
20-31
(b)
,
For
For ½
½,
.
,
.
(c) Let
be a Bernoulli random variable with
, i.e.,
and
. Then,
is a random variable denoting the number of trials until the
Bernoulli random variable takes the value .
.
.
Hence, the sequence is
.
20.4-5.
(a) Answers will vary.
(b)
to
to
correspond to heads.
correspond to tails.
Group 1: HHH, Group 2: THH, Group 3: HTT, Group 4: THT,
Group 5: THH, Group 6: HHT, Group 7: THT, Group 8: TTH
Number of groups with 0 heads: 0
Number of groups with 1 heads: 4
Number of groups with 2 heads: 3
Number of groups with 3 heads: 1
20-32
(c)
Random
Number
0.6459
0.3080
0.0353
Flip
1
2
3
Result
Tails
Heads
Heads
Total Number of Heads =
2
(d) Answers will vary. The following eight replications have no replications with no
heads, two replications with one head
, six replication with two heads
, and no
replication with three heads. This is not very close to the expected probability
distribution.
Replication
1
2
3
4
5
6
7
8
Number of
Heads
2
1
2
2
1
2
2
2
2
(e) Answers will vary. Among the following
,
have one head
,
have three heads
distribution.
Replication
1
2
3
4
5
6
7
8
9
10
798
799
800
Number with 0 heads =
Number with 1 head =
Number with 2 heads =
Number with 3 heads =
replications, 6 have no heads
have two heads
, and
. This is quite close to the expected probability
Number of
Heads
2
0
1
3
1
2
1
1
3
1
0
1
2
1
96
302
286
116
20-33
20.4-6.
(a)
(b) Answers will vary. Below is the results from a 25-replication simulation.
Game
Win?
(c) 9 wins and 16 loses
win
and
lose
(d)
~
20.4-7.
We can use
directly instead of
, since both have uniform distribution. The
following values
are obtained in Excel using the function NORMINV
.
Average:
20-34
20.4-8.
(a)
(b)
20.4-9.
(a)
Let
denote the chi-square observations, for
and
.
(b)
(c)
From (a),
From (b),
and
and
.
.
20.4-10.
(a)
(b)
. Then
ln
ln
20-35
(c)
20.4-11.
(a)
Uniform Random Number
Random Observation
(b) If cell C4 contains the uniform random number, then the Excel function would be:
IF(C4<0.2, 7+(2/0.2)*C4, IF(C4<0.8, 9+(2/0.6)*(C4-0.2), 11+(2/0.2)*(C4-0.8))).
20.4-12.
ln
Hence, the Erlang observation is
.
20.4-13.
(a) TRUE. Both
and
are uniformly distributed.
(b) FALSE. Numerically,
.
(c) TRUE. The sum of independent exponential random variables each with the same
mean has Erlang distribution.
20-36
20.4-14.
(a) It is not valid, since
Modify it as
and
(b) It is valid. When
,
wouldn't reach .
.
(c) It is not valid, since
,
,
,
,
,
,
, and
so this method does not cover the number . Instead, let
then it is a valid method.
,
20.4-15.
Accept?
No
No
No
No
Yes
Yes
Yes
The three samples from the triangular distribution are
20.4-16. Let
, and
.
.
Accept?
No
No
No
No
No
No
No
Yes
No
No
No
No
No
Yes
No
No
No
No
No
Yes
The three samples from the given distribution are
20-37
, and
.
,
,
20.4-17.
if
if
if
size of risk
if
if
size of loss
Run 1
size
Total loss:
½
½
Run 2
size
size
size
Two simulation runs give
and
. Actually,
runs give
.
20.4-18.
Since the number
with
and
of employees incurring medical expenses has a binomial distribution
:
,
,
,
.
Let
.
if
Total amount
Only
$
.
if
if
causes an actual payment from the insurance company and the total payment is
20-38
20.5-1.
Answers will vary.
20.5-2.
Answers will vary.
20.6-1.
(a) Answers will vary. A typical set of 5 runs:
(b) Answers will vary. A typical set of 5 runs:
(c) The mean profits in part (b) seem to be more consistent.
20.6-2.
Land Purchase
Construction Cost
Operating Profit
Selling Price
Fixed
Triangular(min,likely,max)
Normal(mean,s.dev.)
Uniform(min,max)
Now
-1
-2.4
0.7
4
-2
0.7
8
-1.6
Year 1
Year 3
Year 4
Year 5
1.353
1.440
0.992
0.225
5.199
-1.867
Total Cash Flow
-1
Discount Factor
10%
Net Present Value ($million)
3.549
Mean
2.925
Minimum Annual Operating Profit ($million in y2-y5)
0.225
-0.007
(a) The mean NPV is approximately $
Year 2
-1.867 1.35327 1.44046 0.99231 5.42381
million.
(b) The probability that the NPV will be at least $ million is approximately
20-39
%.
(c) The mean value of the minimum annual operating profit is approximately zero.
(d) The probability that the minimum annual operating profit will be at least zero in all
four years of operation is approximately
%.
20.6-3.
The expected cost with the proposed system of replacing all relays whenever any one of them
fails is approximately $
per hour. This is cheaper than the current system of replacing
each relay as it fails. Therefore, they should replace all four relays with the first failure.
1
2
3
4
Time to
Failure
(hours)
1,759
1,354
1,597
1,605
Time to First Failure
Time to End of Shutdown
Total Cost
Cost per Hour
Mean Cost per Hour
1,354
1,356
$2,800
$2.07
$2.37
Relay
Relay
Relay
Relay
Uniform
Uniform
Uniform
Uniform
20-40
Min
1,000
1,000
1,000
1,000
Max
2,000
2,000
2,000
2,000
20.6-4.
The chance of negative clearance is approximately
%.
Shaft Radius 1.00122 Triangular(min,likely,max)
Bushing Radius 1.00317 Normal(mean,st.dev.)
Clearance 0.00196
Mean Clearance 0.00100
20-41
1.000
1.002
1.001
0.001
1.002
20.6-5.
Toss
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Die 1
6
5
1
6
3
3
4
2
5
4
6
3
1
4
3
2
3
2
3
6
6
4
3
6
6
5
2
1
6
2
Die 2
4
3
2
6
3
2
1
2
5
3
6
3
1
6
5
5
6
1
2
5
4
1
3
5
4
2
6
5
4
6
Sum
10
8
3
12
6
5
5
4
10
7
12
6
2
10
8
7
9
3
5
11
10
5
6
11
10
7
8
6
10
8
Win?
No
No
No
No
No
No
No
No
Yes
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
Lose?
No
No
No
No
No
No
No
No
No
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
Continue?
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
No
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
Win
Game?
(1=yes,0=no)
1
Mean (Win Game?)
0.500
(a) Answers will vary. The standard error is approximately
should be between
and
.
, so the typical values
(b) Answers will vary. The standard error is approximately
should be between
and
.
, so the typical values
(c) Answers will vary. The standard error is approximately
should be between
and
.
, so the typical values
(d) Answers will vary. There is a fair amount of variability in the number of wins, so a
large number of iterations, say
, is necessary to predict the true probability. With
iterations, the standard error is
.
20-42
20.6-6.
The order quantity that maximizes the mean profit is approximately
.
Order Quantity Mean Profit
50
$46.45
51
$46.74
52
$46.97
53
$47.13
54
$47.22
55
$47.26
56
$47.22
57
$47.13
58
$46.97
59
$46.74
60
$46.45
20.6- .
(a) and (b) The expected value of the college fund at year 5 is approximately $36
thousand. The standard deviation of the college fund at year 5 is just over $1700.
Stock Fund
Bond Fund
Initial
$3,000
$3,000
Annual
$2,000
$2,000
Stock Fund Investment
Stock Fund Start
Stock Fund Return (%)
Stock Fund End
Year 0
$5,000
$5,000
10%
$5,519
Year 1
$2,000
$7,519
6%
$7,962
Year 2
$2,000
$9,962
8%
$10,804
Year 3
$2,000
$12,804
8%
$13,794
Year 4
$2,000
$15,794
12%
$17,738
Year 5
$2,000
$19,738
Bond Fund Investment
Bond Fund Start
Bond Fund Return (%)
Bond Fund End
$5,000
$5,000
6%
$5,298
$2,000
$7,298
5%
$7,649
$2,000
$9,649
-1%
$9,516
$2,000
$11,516
2%
$11,722
$2,000
$13,722
8%
$14,862
$2,000
$16,862
Normal
Total $36,600
Mean $35,993
St. Deviation $1,729
20-43
Normal
Mean St. Dev.
8%
6%
4%
3%
(c) The probability that the college fund at year 5 will be at least $35,000 is
approximately 69.5%.
(d) The probability that the college fund at year 5 will be at least $40,000 is
approximately 1.4%.
20-44
20.6-8.
(a) The mean profit is approximately $107. There is an approximately 96.3% change of
making at least $0 profit.
Purchase Price
Selling Price
$0.75
$1.25
Order Quantity
350
Demand 298.7126
Rounded Demand
299
Revenue
Purchase Cost
Total Profit
Mean Total Profit
Normal
Mean
300
St. Dev.
50
$373.75
$262.50
$111.25
$107.29
(b) An order quantity of 275 maximizes the mean profit. An order quantity of 300 is also
very close to maximizing the mean profit. The order quantity that actually maximizes the
mean profit is probably somewhere between these two quantities.
Order Quantity Mean Profit
250
$119.80
275
$125.14
300
$125.07
325
$118.89
350
$107.30
20-45
(c)
(d) An order quantity of approximately 287 maximizes Michael's mean profit.
Purchase Price
Selling Price
$0.75
$1.25
Order Quantity
287
Demand 348.3337
Rounded Demand
348
Revenue
Purchase Cost
Total Profit
Mean Total Profit
Normal
Mean
300
$358.75
$215.25
$143.50
$125.85
20-46
St. Dev.
50
20.6-9.
(a) The mean profit is approximately $0.3 million. The probability of winning the bid is
approximately 52.6%.
Data
Our Project Cost ($million)
Our Bid Cost ($million)
5.000
0.050
Competitor Bids
Bid ($million)
Competitor 1
5.824
Competitor 2
5.959
Competitor 3
6.114
Competitor 4
6.136
Triangular
Triangular
Triangular
Triangular
Distribution
Competitor Distribution Parameters (Proportion of Our Project Cost)
Minimum
105%
105%
105%
Most Likely
120%
120%
120%
Maximum
140%
140%
140%
105%
120%
140%
Competitor Distribution Parameters ($millions)
Minimum
5.250
Most Likely
6.000
Maximum
7.000
5.250
6.000
7.000
Minimum Competitor
Bid ($million)
5.824
Our Bid ($million)
5.700
Win Bid?
Profit ($million)
Mean Profit ($million)
1
5.250
6.000
7.000
(1=yes, 0=no)
0.650
0.303
20-47
5.250
6.000
7.000
(b) A bid of approximately $5.5 million maximizes RPI's mean profit.
OurBid
5.3
5.4
5.5
5.6
5.7
5.8
5.9
6.0
Mean Profit ($million)
0.248
0.323
0.364
0.356
0.313
0.234
0.140
0.061
(c)
20-48
(d) The optimal bid is approximately $5.57 million, as found by Solver.
Data
Our Project Cost ($million)
Our Bid Cost ($million)
5.000
0.050
Competitor Bids
Bid ($million)
Competitor 1
6.133
Competitor 2
5.864
Competitor 3
6.312
Competitor 4
6.341
Triangular
Triangular
Triangular
Triangular
Distribution
Competitor Distribution Parameters (Proportion of Our Project Cost)
Minimum
105%
105%
105%
Most Likely
120%
120%
120%
Maximum
140%
140%
140%
105%
120%
140%
Competitor Distribution Parameters ($millions)
Minimum
5.250
Most Likely
6.000
Maximum
7.000
5.250
6.000
7.000
Minimum Competitor
Bid ($million)
5.864
Our Bid ($million)
5.569
Win Bid?
Profit ($million)
Mean Profit ($million)
1
5.250
6.000
7.000
(1=yes, 0=no)
0.519
0.366
20-49
5.250
6.000
7.000
20.6-10.
(a)
Airline Overbooking
Data
Seats Available
Fixed Cost
Discount Fare
Full Coach Fare
Cost of Bumping
112
$10,000
$150
$400
$600
Discount
Reservations
to Accept
50
Total
Reservations
to Accept
112
Discount Ticket Demand (Triangular)
Minimum
50
Most Likely
90
Maximum
150
Probability to Show Up
95%
Discount-Fare Demand
Rounded
Tickets Purchased
Number that Show
77.01
77
50
47
Full-Coach Ticket Demand (Uniform)
Minimum
30
Maximum
70
Probability to Show Up
85%
Full Coach Demand
Rounded
Tickets Purchased
Number that Show
52.32
52
52
47
Number Denied Boarding
Number of Filled Seats
0
94
Mean
0.00
89.33
Revenue (Discount Fare)
Revenue (Full Coach)
Bumping Cost
Fixed Cost
Profit
$7,500
$18,800
$0
$10,000
$16,300
Mean
$14,231
20-50
(b)
DiscountReservationsToAccept TotalReservationsToAccept
112
117
122
$14,220 $14,453 $14,492
50
$14,617 $15,271 $15,657
60
$14,183 $15,232 $15,925
70
$13,100 $14,479 $15,289
80
$11,830 $13,347 $14,132
90
20-51
127
$14,492
$15,722
$16,024
$15,220
$13,912
132
$14,492
$15,713
$15,860
$14,879
$13,406
(c) They should accept approximately 68 discount reservations and up to approximately
125 total in order to maximize mean profit, as found by Solver.
Airline Overbooking
Data
Seats Available
Fixed Cost
Discount Fare
Full Coach Fare
Cost of Bumping
112
$10,000
$150
$400
$600
Discount
Reservations
to Accept
68
Total
Reservations
to Accept
125
Discount Ticket Demand (Triangular)
Minimum
50
Most Likely
90
Maximum
150
Probability to Show Up
95%
Discount-Fare Demand
Rounded
Tickets Purchased
Number that Show
73.05
73
68
63
Full-Coach Ticket Demand (Uniform)
Minimum
30
Maximum
70
Probability to Show Up
85%
Full Coach Demand
Rounded
Tickets Purchased
Number that Show
31.78
32
32
30
Number Denied Boarding
Number of Filled Seats
0
93
Mean
0.69
104.28
Revenue (Discount Fare)
Revenue (Full Coach)
Bumping Cost
Fixed Cost
Profit
$10,200
$12,000
$0
$10,000
$12,200
Mean
$16,045
20.7-1.
Answers will vary.
20.7-2.
Answers will vary.
20-52
Case 20.1 Reducing In-Process Inventory (Revisited)
a) Status quo at the presses – 7.5 sheets of in-process inventory.
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
B
C
D
E
F
G
H
Template for Queueing Simulation
Number of Servers =
Interarrival Times
Distribution =
Mean =
Service Times
Distribution =
Mean =
Length of Simulation Run
Number of Arrivals =
Data
10
Exponential
0.142857143
5
Exponential
1
25
10,000
Run Simulation
L=
Lq =
W=
Wq =
Point
Estimate
7.48596004
0.55020043
1.0770836
0.07916311
P0 =
P1 =
P2 =
P3 =
P4 =
P5 =
P6 =
P7 =
P8 =
P9 =
P10 =
0.00110924
0.00582387
0.02306409
0.05166684
0.0866959
0.12118604
0.14062225
0.14294653
0.12452751
0.08806336
0.06192446
Results
95% Confidence Interval
Low
High
7.122474949
7.849445126
0.347368991
0.753031867
1.036422591
1.117744603
0.050901621
0.107424593
0.000312762
0.003292739
0.018701971
0.043052172
0.077527167
0.112124348
0.13442836
0.134902634
0.11900339
0.084082813
0.055935883
0.001905718
0.008355008
0.027426208
0.060281501
0.09586463
0.130247735
0.14681614
0.150990419
0.130051626
0.092043901
0.067913035
Status quo at the inspection station – 3.6 wing sections of in-process inventory.
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
B
C
D
E
F
G
H
Template for Queueing Simulation
Number of Servers =
Interarrival Times
Distribution =
Mean =
Service Times
Distribution =
Value =
Length of Simulation Run
Number of Arrivals =
Data
1
Exponential
0.142857143
5
Constant
0.125
25
10,000
Run Simulation
L=
Lq =
W=
Wq =
Point
Estimate
3.57765981
2.71234549
0.51681506
0.39181506
P0 =
P1 =
P2 =
P3 =
P4 =
P5 =
P6 =
P7 =
P8 =
P9 =
P10 =
0.13468567
0.18444199
0.16054199
0.12577666
0.09279878
0.07546784
0.0548405
0.04326737
0.03643173
0.02983638
0.02245891
Inventory cost = (7.5 + 3.6)($8/hour) = $88.80 / hour
Machine cost = (10)($7/hour) = $70 / hour
Inspector cost = $17 / hour
Total cost = $175.80 / hour
20-53
Results
95% Confidence Interval
Low
High
3.096884037
4.058435589
2.244158962
3.180532014
0.454627294
0.57900283
0.329627294
0.45400283
0.118076335
0.164766618
0.145653686
0.114607169
0.083029162
0.065828646
0.045754492
0.033313657
0.026094365
0.020206033
0.014710033
0.151295015
0.204117359
0.175430299
0.136946159
0.102568391
0.085107034
0.063926513
0.053221074
0.046769093
0.039466733
0.030207788
b) Proposal 1 will increase the in-process inventory at the presses to 10.6 sheets since
the mean service rate has decreased.
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
B
C
D
E
F
G
H
Template for Queueing Simulation
Number of Servers =
Interarrival Times
Distribution =
Mean =
Service Times
Distribution =
Mean =
Length of Simulation Run
Number of Arrivals =
Data
10
Exponential
0.142857143
5
Exponential
1.2
25
10,000
Run Simulation
L=
Lq =
W=
Wq =
Point
Estimate
10.6124208
2.34034351
1.5192496
0.33503816
P0 =
P1 =
P2 =
P3 =
P4 =
P5 =
P6 =
P7 =
P8 =
P9 =
P10 =
0.00034416
0.00330079
0.00683624
0.0225304
0.0437143
0.06530488
0.08305601
0.09066307
0.09495054
0.09944674
0.08672109
Results
95% Confidence Interval
Low
High
10.07045277
11.15438883
1.812410733
2.868276277
1.422904809
1.615594383
0.255248897
0.41482742
-0.000135983
0.002146705
0.005191338
0.017788623
0.041059108
0.058747044
0.074729232
0.081970997
0.09393376
0.090813615
0.077951648
0.000824295
0.004454878
0.008481139
0.027272181
0.0463695
0.071862716
0.091382794
0.099355138
0.095967318
0.108079863
0.095490525
The in-process inventory at the inspection station will not change.
Inventory cost = (10.6 + 3.6)($8/hour) = $113.60 / hour
Machine cost = (10)($6.50) = $65 / hour
Inspector cost = $17 / hour
Total cost = $195.60 / hour
This total cost is higher than for the status quo so should not be adopted. The main
reason for the higher cost is that slowing down the machines won’t change in-process
inventory for the inspection station.
20-54
c) Proposal 2 will increase the in-process inventory at the inspection station to 4.2 wing
sections since the variability of the service rate has increased.
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
B
C
D
E
F
G
H
Template for Queueing Simulation
Number of Servers =
Interarrival Times
Distribution =
Mean =
Data
1
Exponential
0.142857143
5
Service Times
Distribution =
Mean =
k=
Erlang
0.12
2
Length of Simulation Run
Number of Arrivals =
10,000
Run Simulation
L=
Lq =
W=
Wq =
Point
Estimate
4.15349196
3.31066782
0.58953022
0.46990309
P0 =
P1 =
P2 =
P3 =
P4 =
P5 =
P6 =
P7 =
P8 =
P9 =
P10 =
0.15717586
0.16164362
0.1417251
0.11157869
0.08340382
0.0729656
0.05422094
0.04033746
0.03068653
0.02468793
0.02288346
Results
95% Confidence Interval
Low
High
3.51945922
4.787524705
2.691612222
3.929723426
0.506614288
0.672446148
0.387653637
0.552152552
0.13797724
0.143938659
0.127306603
0.100074725
0.074497166
0.064546969
0.04655526
0.033104015
0.023437928
0.018553583
0.016278465
0.176374483
0.179348578
0.156143599
0.123082653
0.092310469
0.081384232
0.061886616
0.047570898
0.037935133
0.030822285
0.029488445
The in-process inventory at the presses will not change.
Inventory cost = (7.5 + 4.2)($8/hour) = $93.60 / hour
Machine cost = (10)($7/hour) = $70 / hour
Inspector cost = $17 / hour
Total cost = $180.60 / hour
This total cost is higher than for the status quo so should not be adopted. The main
reason for the higher cost is the increase in the service rate variability (Erlang rather
than constant) and the resulting increase in the in-process inventory.
20-55
d) They should consider increasing power to the presses (increasing there cost to $7.50
per hour but reducing their average time to form a wing section to 0.8 hours). This
would decrease the in-process inventory at the presses to 5.7.
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
B
C
D
E
F
G
H
Template for Queueing Simulation
Number of Servers =
Interarrival Times
Distribution =
Mean =
Service Times
Distribution =
Mean =
Length of Simulation Run
Number of Arrivals =
Data
10
Exponential
0.142857143
5
Exponential
0.8
2
10,000
Run Simulation
L=
Lq =
W=
Wq =
Point
Estimate
5.74237458
0.11624317
0.81487258
0.01649551
P0 =
P1 =
P2 =
P3 =
P4 =
P5 =
P6 =
P7 =
P8 =
P9 =
P10 =
0.00445475
0.0241519
0.06075455
0.10828334
0.14577459
0.1580859
0.14882682
0.12347465
0.0909915
0.05514285
0.03360049
Inventory cost = (5.7 + 3.6)($8/hour) = $74.40 / hour
Machine cost = (10)($7.50/hour) = $75 / hour
Inspector cost = $17 / hour
Total cost = $166.40 / hour
This total cost is lower than the status quo and both proposals.
20-56
Results
95% Confidence Interval
Low
High
5.581608211
5.903140941
0.076181593
0.156304743
0.801697429
0.828047729
0.011001805
0.021989206
0.002433487
0.019051394
0.0522877
0.096234
0.138731319
0.148929657
0.137378613
0.116102784
0.084900257
0.050413495
0.029185971
0.00647602
0.0292524
0.069221409
0.120332681
0.152817867
0.167242144
0.160275035
0.13084652
0.097082738
0.059872212
0.038015016
Case 20.2 Action Adventures
a) The spreadsheet model is spread over the next two pages:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
A
B
C
Cost & Revenue Data
Selling Price
$10
Replacement Part Cost $5,000
Monthly Fixed Cost $15,000
Minimum Balance $20,000
Starting Balance $25,000
Sales
D
E
F
G
H
Interest Rate Data
Initial Prime Rate
5%
Loan Rate Prime Gap
2%
Loan Rate Maximum
9%
Savings Rate Prime Gap
-2%
Savings Rate Minimum
2%
I
Seasonality Index
Base Sales
Actual Sales
Fraction Cash Customers
Dec
1.18
6,000
7,080
42%
Jan
0.79
5,250
4,148
42%
Feb
0.88
5,534
4,870
40%
Mar
0.95
4,937
4,690
45%
Apr
1.05
5,562
5,840
41%
May
1.09
5,706
6,219
32%
June
0.84
5,647
4,743
38%
July
0.74
5,137
3,801
39%
Interest Rates
Prime Rate Change
Prime Rate
Loan Interest Rate
Savings Interest Rate
5.00%
7.00%
3.00%
0.00%
5.00%
7.00%
3.00%
-0.25%
4.75%
6.75%
2.75%
0.50%
5.25%
7.25%
3.25%
0.25%
5.50%
7.50%
3.50%
0.00%
5.50%
7.50%
3.50%
0.00%
5.50%
7.50%
3.50%
0.00%
5.50%
7.50%
3.50%
3
0
0
1
0
0
0
$6.25
$7.96
$6.87
$6.85
$7.24
$6.39
$7.63
Minimum Balance
$25,000
$17,517
$41,064
-$15,000
-$25,916
-$15,000
$0
$0
$750
$28,415
$0
$28,415
>=
$20,000
$28,415
$19,344
$23,960
-$15,000
-$38,784
$0
$0
$0
$852
$18,788
$1,212
$20,000
>=
$20,000
$20,000
$21,251
$29,351
-$15,000
-$32,231
$0
-$1,212
-$82
$550
$22,627
$0
$22,627
>=
$20,000
$22,627
$24,195
$25,653
-$15,000
-$39,993
-$5,000
$0
$0
$735
$13,217
$6,783
$20,000
>=
$20,000
$20,000
$19,934
$34,203
-$15,000
-$45,055
$0
-$6,783
-$509
$700
$7,491
$12,509
$20,000
>=
$20,000
$20,000
$18,177
$42,257
-$15,000
-$30,309
$0
-$12,509
-$938
$700
$22,377
$0
$22,377
>=
$20,000
$22,377
$15,007
$29,254
-$15,000
-$28,994
$0
$0
$0
$783
$23,427
$0
$23,427
>=
$20,000
Simulated
Value
Ending Net Worth $14,935
Mean
$54,298
Manufacturing Costs
Replacement Parts Needed
Variable Cost
Cash Flows
Beginning Balance
Cash Receipts
30-Day Credit Receipts
Fixed Cost
Total Variable Cost
Repair Cost
Loan Payoff
Loan Interest
Savings Interest
Balance Before Loan
New Loan
Ending Balance
Maximum Loan
$25,000
$36,775
$25,990
20-57
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
A
J
K
L
M
N
O
P
Q
Seasonality Index
Base Sales
Actual Sales
Fraction Cash Customers
August
0.98
5,614
5,501
37%
Sept
1.06
5,652
5,991
33%
Interest Rates
Prime Rate Change
Prime Rate
Loan Interest Rate
Savings Interest Rate
0.00%
4.75%
6.75%
2.75%
0.00%
4.75%
6.75%
2.75%
0.00%
4.75%
6.75%
2.75%
0.00%
4.75%
6.75%
2.75%
0.00%
4.75%
6.75%
2.75%
Custom
Discrete
-0.50%
-0.25%
0%
0.25%
0.50%
0.05
0.1
0.7
0.1
0.05
Manufacturing Costs
Replacement Parts Needed
1
1
1
0
1
Binomial
10%
8
$7.18
$7.49
$7.05
$7.15
$6.63
Uniform
$6
$8
$20,000
$19,687
$34,523
-$15,000
-$44,866
-$5,000
-$4,776
-$322
$550
$4,796
$15,204
$20,000
>=
$20,000
$20,000
$22,385
$40,226
-$15,000
-$41,494
-$5,000
-$15,204
-$1,026
$550
$5,437
$14,563
$20,000
>=
$20,000
$20,000
$28,906
$36,509
-$15,000
-$47,529
$0
-$14,563
-$983
$550
$7,889
$12,111
$20,000
>=
$20,000
$20,000
$25,395
$37,548
-$15,000
-$43,428
-$5,000
-$12,111
-$818
$550
$7,137
$12,863
$20,000
>=
$20,000
$20,000
R
Selling Price
Replacement Part Cost
Monthly Fixed Cost
Minimum Balance
Starting Balance
Sales
Variable Cost
Cash Flows
Beginning Balance
Cash Receipts
30-Day Credit Receipts
Fixed Cost
Total Variable Cost
Repair Cost
Loan Payoff
Loan Interest
Savings Interest
Balance Before Loan
New Loan
Ending Balance
$27,676
$20,491
$25,782
-$15,000
-$39,486
-$5,000
$0
$0
$761
$15,224
$4,776
$20,000
>=
Minimum Balance $20,000
October November December January
1.1
1.16
1.18
5,354
5,729
5,549
Normal prev mo.
5,889
6,645
6,548
38%
43%
39%
Triangular
28%
20-58
$40,081
-$12,863
-$868
$550
$46,899
500
40%
48%
b) As seen on the spreadsheet in part a, the mean ending net worth is approximately
$54.4 thousand. The probability that it will be greater than $0 is approximately
84.1%.
20-59
c) The maximum short-term loan is shown in row 45 of the spreadsheet. The maximum
short-term loan averages just over $25 thousand. However, to be fairly sure that the
credit limit is high enough, it should probably be set quite a bit higher. The
cumulative chart shows the probability that any given credit limit will be large
enough. For example, a $70 thousand credit limit has about a 95% chance of being
sufficient.
20-60
Case 20.3 Planning Planers
Current Situation: A simulation run (shown below) indicates that the average number of
jobs in the system is 2.0. Of these, half will be platen castings (1) and half will be
housing castings (1). The waiting cost is therefore ($200)(1) + ($100)(1) = $300 / hour.
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
B
C
D
E
F
G
H
Template for Queueing Simulation
Number of Servers =
Interarrival Times
Distribution =
Mean =
Data
2
Exponential
15
5
Service Times
Distribution = Translated Exponential
Minimum Value =
10
Mean =
20
Length of Simulation Run
Number of Arrivals =
10,000
Run Simulation
L=
Lq =
W=
Wq =
Point
Estimate
1.98365641
0.66628639
30.0811805
10.1039076
P0 =
P1 =
P2 =
P3 =
P4 =
P5 =
P6 =
P7 =
P8 =
P9 =
P10 =
0.19988054
0.2828689
0.21948306
0.13257277
0.0722497
0.04178641
0.02261418
0.0129863
0.00771744
0.003861
0.00185903
Results
95% Confidence Interval
Low
High
1.870700578
2.096612244
0.575783306
0.756789465
28.73655618
31.4258049
8.845870432
11.36194473
0.188799674
0.271597628
0.211376682
0.125756108
0.0660523
0.036150923
0.018147395
0.009197547
0.004659116
0.001884639
0.000591296
0.210961405
0.294140162
0.227589435
0.13938943
0.078447105
0.047421901
0.027080961
0.016775062
0.010775773
0.005837354
0.003126755
Proposal 1: A simulation run (shown below) indicates that the average number of jobs in
the system with three planers is approximately 1.4. Of these, half will be platen castings
(0.7) and half will be housing castings (0.7). The waiting cost is therefore ($200)(0.7) +
($100)(0.7) = $210 / hour. The savings ($90 / hour) is substantially more than the added
cost of the third planer ($30 / hour), so this looks to be worthwhile. The net savings
would be $60 / hour.
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
B
C
D
E
F
G
H
Template for Queueing Simulation
Number of Servers =
Interarrival Times
Distribution =
Mean =
Data
3
Exponential
15
5
Service Times
Distribution = Translated Exponential
Minimum Value =
10
Mean =
20
Length of Simulation Run
Number of Arrivals =
10,000
Run Simulation
20-61
L=
Lq =
W=
Wq =
Point
Estimate
1.42409865
0.09771456
21.4712624
1.47325117
P0 =
P1 =
P2 =
P3 =
P4 =
P5 =
P6 =
P7 =
P8 =
P9 =
P10 =
0.25534157
0.33796761
0.231656
0.11158027
0.04233244
0.01406273
0.00409905
0.00133435
0.00080672
0.00036429
0.00025729
Results
95% Confidence Interval
Low
High
1.380256124
1.467941167
0.076609893
0.11881923
21.07385924
21.86866564
1.168148546
1.778353785
0.245577077
0.329893149
0.224819899
0.106409547
0.038546771
0.011836939
0.002819395
0.000425396
-0.000131529
-0.000147459
-0.000188128
0.265106061
0.346042064
0.238492092
0.116750986
0.046118111
0.016288531
0.005378708
0.002243309
0.001744969
0.000876045
0.000702701
Proposal 2: A simulation run (shown below) indicates that the average number of jobs in the
system with constant interarrival times is approximately 1.4. Of these, half will be platen
castings (0.7) and half will be housing castings (0.7). The waiting cost is therefore
($200)(0.7) + ($100)(0.7) = $210 / hour. The savings ($90 / hour) is somewhat more than
the added cost of changing the preceding production cost ($60 / hour). The net savings
($30) is less than for proposal 1, so this option is less worthwhile.
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
B
C
D
E
F
G
H
Template for Queueing Simulation
Number of Servers =
Interarrival Times
Distribution =
Value =
Data
2
Constant
15
5
Service Times
Distribution = Translated Exponential
Minimum Value =
10
Mean =
20
Length of Simulation Run
Number of Arrivals =
10,000
Run Simulation
L=
Lq =
W=
Wq =
Point
Estimate
1.40396168
0.06455913
21.0594253
0.96838689
P0 =
P1 =
P2 =
P3 =
P4 =
P5 =
P6 =
P7 =
P8 =
P9 =
P10 =
0.05142644
0.55774455
0.33345643
0.05060063
0.00635733
0.00041461
0
0
0
0
0
Results
95% Confidence Interval
Low
High
1.383412164
1.424511205
0.055139154
0.073979097
20.75118246
21.36766807
0.82708731
1.109686461
0.049243293
0.547877528
0.325678768
0.045262513
0.004037203
2.30036E-06
0
0
0
0
0
0.053609594
0.56761158
0.341234095
0.055938746
0.008677449
0.000826928
0
0
0
0
0
Proposal 1 and 2: A simulation run (shown below) indicates that the average number of
jobs in the system with both three planers and constant interarrival times is approximately
1.33. Of these, half will be platen castings (0.665) and half will be housing castings
(0.665). The waiting cost is therefore ($200)(0.665) + ($100)(0.665) = $200 / hour. The
savings ($85 / hour) is less than the combined cost of adding a third planer and changing
the preceding production cost ($90 / hour), so this combined option does not appear to be
worthwhile.
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
B
C
D
E
F
G
H
Template for Queueing Simulation
Number of Servers =
Interarrival Times
Distribution =
Value =
Data
3
Constant
15
5
Service Times
Distribution = Translated Exponential
Minimum Value =
10
Mean =
20
Length of Simulation Run
Number of Arrivals =
10,000
Run Simulation
20-62
L=
Lq =
W=
Wq =
Point
Estimate
1.32985569
0.00052554
19.9478354
0.00788307
P0 =
P1 =
P2 =
P3 =
P4 =
P5 =
P6 =
P7 =
P8 =
P9 =
P10 =
0.05754771
0.58946474
0.31909725
0.03336476
0.00052554
0
0
0
0
0
0
Results
95% Confidence Interval
Low
High
1.316690172
1.343021211
0.000184596
0.00086648
19.75035259
20.14531816
0.002768944
0.012997201
0.055474824
0.581401676
0.311531281
0.03022801
0.000184596
0
0
0
0
0
0
0.059620587
0.5975278
0.326663225
0.036501519
0.00086648
0
0
0
0
0
0
Overall recommendation: Proposal 1 appears to be the most worthwhile, with a net
savings of about $36 / hour over the current situation. Other proposals that may be worth
looking into should include giving priority to platen castings, because of the higher
waiting cost for that type of job.
Case 20.4 Pricing Under Pressure
a) Before we begin the formal problem, we must first calculate the mean µ and standard
deviation σ of the normally distributed random variable N. We are told that the
annual interest rate will be used to estimate µ and the historical annual volatility will
be used to estimate σ. Because the case is simulating weekly – not yearly – change,
we must convert these yearly values to weekly values.
We first convert the annual interest rate r = 8% to a weekly interest rate w with the
following formula:
w = (1 + r)(1/52) – 1
= (1 + 0.08)(1/52) – 1
= (1.08)(1/52) – 1
= 0.00148
We next convert the annual volatility Va = 0.30 to a weekly volatility Vw with the
following formula:
Vw = Va / √52
= 0.30 / √52
= 0.0416
Once we have the weekly interest rate and volatility, we can calculate µ and σ.
µ = w – 0.5(Vw)2
= 0.00148 – 0.5(0.0416)2
= 0.0006
σ = Vw
= 0.0416
1. One component appears in this system: the stock price. The stock price in the
previous week is used to calculate the stock price in the next week. The relationship
between the stock price in the previous week and the stock price in the next week is
given by sn = eN sc.
2. State of the system: P(t) = price of the stock at time t.
20-63
3. This simulation requires generating a series of random observations from the
normal distribution. Each random observation is a normally distributed random
variable that determines the increase or decrease of the stock price at the end of next
week. The random variable is substituted for N in the following equation:
sn = eN sc
To generate a series of random variables, we define an uncertain variable cell with
normal distribution, where µ = 0.0006 and σ = 0.0416.
4. The formula sn = eN sc gives us a procedure for changing the price (the state of the
system) when an event occurs.
5. In this simulation, the time periods are fixed. We have a twelve-week period, and
we need to calculate the change in the stock price each week. We have a formula sn
= eN sc that relates the stock price at the end of the next week to the stock price at the
end of the previous week. Thus, we do not have to worry about advancing the clock.
We simply have to generate N for each of the twelve weeks.
6. We need to build a spreadsheet using RSPE. We start with the current stock price
of $42.00. We then use the formula sn = eN sc to calculate the stock price at the end
of each of the twelve weeks. We substitute a RSPE uncertain variable cell with
normal distribution (with mean µ = 0.0006, and standard deviation σ = 0.0416) for N.
We then use the stock price at the end of the twelfth week to calculate the value of the
option at the end of the twelfth week. If the stock price at the end of the twelfth week
is greater than the exercise price of $44.00, the value of the option is the difference
between the value of the stock at the end of the twelfth week and the exercise price.
If the stock price at the end of the twelfth week is less than or equal to the exercise
price of $44.00, the value of the option is $0.
Finally, we need to discount the value of the option at the end of the twelfth week to
the value of the option in today’s dollars using the following formula:
(Value of the option at the end of the twelfth week) / (1.00148)12
The spreadsheet model is shown below. The uncertain variable cells are the N values
(B8:B19), the result cell is the price of the option today (C22), and the statistic cell is
the mean price of the option today (C23).
20-64
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
Week
1
2
3
4
5
6
7
8
9
10
11
12
C
D
Current Stock Price
Exercise Price
$42.00
$44.00
N
0.002944088
-0.041470376
0.051283439
0.012944376
-0.026906539
-0.079994242
0.006708864
0.00707491
-0.003553399
0.093240243
-0.086581855
-0.017522862
Stock Price at
End of Week
$42.12
$40.41
$42.54
$43.09
$41.95
$38.72
$38.99
$39.26
$39.12
$42.95
$39.38
$38.70
Price of Option at end of Week 12
Price of Option Today
Mean(Price of Option Today)
A
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
B
E
F
Simulation Model to Estimate Option Value
Annual Interest Rate
Weekly Interest Rate
8%
0.148%
Annual Volatility
Weekly Volatility
30%
4.160%
µ=
σ=
0.0006
0.0416
$0.00
$0.00
$1.91
B
C
Current Stock Price 42
Exercise Price 44
Week
1
2
3
4
5
6
7
8
9
10
11
12
N
=PsiNormal(Mean,StandardDeviation)
=PsiNormal(Mean,StandardDeviation)
=PsiNormal(Mean,StandardDeviation)
=PsiNormal(Mean,StandardDeviation)
=PsiNormal(Mean,StandardDeviation)
=PsiNormal(Mean,StandardDeviation)
=PsiNormal(Mean,StandardDeviation)
=PsiNormal(Mean,StandardDeviation)
=PsiNormal(Mean,StandardDeviation)
=PsiNormal(Mean,StandardDeviation)
=PsiNormal(Mean,StandardDeviation)
=PsiNormal(Mean,StandardDeviation)
Stock Price at
End of Week
=CurrentStockPrice*EXP(B8)
=EXP(B9)*C8
=EXP(B10)*C9
=EXP(B11)*C10
=EXP(B12)*C11
=EXP(B13)*C12
=EXP(B14)*C13
=EXP(B15)*C14
=EXP(B16)*C15
=EXP(B17)*C16
=EXP(B18)*C17
=EXP(B19)*C18
Price of Option at end of Week 12 =IF(C19>ExercisePrice,C19-ExercisePrice,0)
Price of Option Today =C21/(1+WeeklyInterestRate)^12 + PsiOutput()
Mean(Price of Option Today) =PsiMean(C22)
20-65
Range Name
AnnualInterestRate
AnnualVolatility
CurrentStockPrice
ExercisePrice
Mean
PriceOfOption
StandardDeviation
WeeklyInterestRate
WeeklyVolatility
Cells
F3
F6
C3
C4
F9
C22
F10
F4
F7
3
4
5
6
7
8
9
10
E
F
Annual Interest Rate 0.08
Weekly Interest Rate =((1+AnnualInterestRate)^(1/52))-1
Annual Volatility 0.3
Weekly Volatility =AnnualVolatility/SQRT(52)
m= =WeeklyInterestRate-0.5*(WeeklyVolatility^2)
s = =WeeklyVolatility
The mean of the “Price of Option Today” is the price of the option in today’s dollars.
The simulation results after 100, 1,000, and 10,000 trials will vary. Typical mean
values might be $1.70, $1.91, and $1.87. The variation is significantly reduced with
more trials (the mean standard error drops from $0.26 to $0.11 to $0.03 at 100, 1,000,
and 10,000 trials, respectively).
20-66
b) Using the Black-Scholes Formula, the price of the option is $1.88. The spreadsheet
used to calculate the Black-Scholes Formula in Excel follows:
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
B
C
D
E
F
Black-Scholes Calculation of Option Value
Current Stock Price
$42.00
Weeks to exercise date
Exercise Price
Exercise Price Present Value
12
$44.00
$43.23
Annual Interest Rate
Weekly Interest Rate
8%
0.148%
Annual Volatility
Weekly Volatility
30%
4.160%
µ=
σ=
0.0006
0.0416
Black-Scholes
d1 = -0.127503153
d2 = -0.271618491
N[d1] =
N[d2] =
0.449271051
0.39295775
Value =
$1.88
E
F
3 Black-Scholes
4
d1 = =LN(CurrentStockPrice/ExercisePricePV)/(StandardDeviation*SQRT(WeeksToExerciseDate))+StandardDeviation*SQRT(Weeks
5
d2 = =d_1-StandardDeviation*SQRT(WeeksToExerciseDate)
6
7
N[d1] = =NORMSDIST(d_1)
8
N[d2] = =NORMSDIST(d_2)
9
10
Value = =Nd1*CurrentStockPrice-Nd2*ExercisePricePV
Range Name
AnnualInterestRate
AnnualVolatility
CurrentStockPrice
d_1
d_2
ExercisePrice
ExercisePricePV
Mean
Nd1
Nd2
StandardDeviation
Value
WeeklyInterestRate
WeeklyVolatility
WeeksToExerciseDate
Cells
C9
C12
C3
F4
F5
C6
C7
C15
F7
F8
C16
F10
C10
C13
C5
The price of the option obtained by simulation and the price of the option obtained by
the Black-Scholes formula are fairly close. The 1,000-iteration simulation price is off
by just thirteen cents.
20-67
c) No, a random walk does not completely describe the price movement of the stock
because the random walk assumes a consistent lognormal increase or decrease in the
price of the stock. The price of the stock could change according to a different
distribution, however, especially if an event occurs to trigger a dramatic increase or
decrease in the stock. In this case, the European Space Agency may award Fellare
the International Space Station contract. The award notice would most likely trigger
a dramatic movement in the stock. The random walk does not take into account this
dramatic event.
20-68
SUPPLEMENT 1 TO CHAPTER 20
VARIANCE-REDUCING TECHNIQUES
20S1-1.
(a)
< œ T Ö\ Ÿ B× œ "
B .>
>#
<
!Þ!*'
!Þ&'*
!Þ''&
!Þ('%
!Þ)%#
!Þ%*#
!Þ##%
!Þ*&!
!Þ'"!
!Þ"%&
.
sœ
(b)
œ"
"
B
ÊBœ
"
"<
B œ "ÎÐ"  <Ñ
"Þ"!'
#Þ$#!
#Þ*)&
%Þ#$(
'Þ$#*
"Þ*'*
"Þ#)*
#!Þ!!!
#Þ&'%
"Þ"(!
%$Þ*'*
"!
œ %Þ$*'*
Stratum 1: <w œ !Þ!  !Þ'<
Stratum 2: <w œ !Þ'  !Þ$<
Stratum 3: <w œ !Þ*  !Þ"<
Let A denote the sampling weight.
Stratum
"
"
"
#
#
#
$
$
$
$
.
sœ
)(Þ''"
"!
<
!Þ!*'
!Þ&'*
!Þ''&
!Þ('%
!Þ)%#
!Þ%*#
!Þ##%
!Þ*&!
!Þ'"!
!Þ"%&
<w
!Þ!&)
!Þ$%"
!Þ$**
!Þ)#*
!Þ)&$
!Þ(%)
!Þ*##
!Þ**&
!Þ*'"
!Þ*"&
B œ "ÎÐ"  <w Ñ
"Þ!'#
"Þ&"(
"Þ''%
&Þ)%)
'Þ)!$
$Þ*')
"#Þ)#"
#!!Þ!!!
#&Þ'%"
""Þ('&
œ )Þ(''"
20S1-1
A
½
½
½
"
"
"
%
%
%
%
BÎA
#Þ"#%
$Þ!$%
$Þ$#)
&Þ)%)
'Þ)!$
$Þ*')
$Þ#!&
&!Þ!!!
'Þ%"!
#Þ*%"
(c)
<w œ "  < Ê Bw œ
œ
B œ "ÎÐ"  <w Ñ
"Þ"!'
#Þ$#!
#Þ*)&
%Þ#$(
'Þ$#*
"Þ*'*
"Þ#)*
#!Þ!!!
#Þ&'%
"Þ"(!
%$Þ*'*
%Þ$*'*
<
!Þ!*'
!Þ&'*
!Þ''&
!Þ('%
!Þ)%#
!Þ%*#
!Þ##%
!Þ*&!
!Þ'"!
!Þ"%&
Sum
.
s
.
sœ
"
"<w
%Þ$*'*$Þ##'"
#
"
<
Bw œ "Î<
"!Þ%"(
"Þ(&(
"Þ&!%
"Þ$!*
"Þ"))
#Þ!$$
%Þ%'%
"Þ!&$
"Þ'$*
'Þ)*(
$#Þ#'"
$Þ##'"
œ $Þ)""&
20S1-2.
B
)
&
"
'
$
(
$
&
#
#
Stratum
"
"
"
"
"
"
#
#
#
$
.
sœ
'#!Î")
"!
B#
'%
#&
"
$'
*
%*
*
#&
%
%
A
")Î"!
")Î"!
")Î"!
")Î"!
")Î"!
")Î"!
*Î"!
*Î"!
*Î"!
$Î"!
œ $ *% ß IÒ\ # Ó œ
BÎA
)!Î")
&!Î")
"!Î")
'!Î")
$!Î")
(!Î")
'!Î")
"!!Î")
%!Î")
"#!Î")
#)%!Î")
"!
B# ÎA
'%!Î")
#&!Î")
"!Î")
$'!Î")
*!Î")
%*!Î")
")!Î")
&!!Î")
)!Î")
#%!Î")
œ "& *(
20S1-3.
(a)
\œ
<3
\
!
if !Þ" Ÿ <3  "
"!!<3  & if ! Ÿ <3  !Þ"
!Þ!*'
"%Þ&
!Þ''&
!
!Þ)%#
!
!Þ##%
!
!Þ'"!
!
.
s œ #Þ*
20S1-2
(b)
Stratum 1: <‡ œ !Þ!  !Þ*<
Stratum 2: <‡ œ !Þ*  !Þ"<ß B œ "!!Ð<‡  !Þ*Ñ  &
<
!Þ!*'
!Þ''&
!Þ)%#
!Þ##%
!Þ'"!
Stratum
"
#
#
#
#
.
sœ
&Þ%$&
&
<‡
!Þ!)'%
!Þ*''&
!Þ*)%#
!Þ*##%
!Þ*'"!
B
!Þ!!
""Þ'&
"$Þ%#
(Þ#%
""Þ"!
A
#Î*
)
)
)
)
BÎA
!Þ!!
"Þ%'
"Þ')
!Þ*!&
"Þ$*
œ "Þ!)(
20S1-4.
(a)
!Þ!!!! to !Þ$*** correspond to a minor repair.
!Þ%!!! to !Þ**** correspond to a major repair.
Random observations: !Þ(#&' œ major, !Þ!)"( œ minor, !Þ%$*# œ major
Using random numbers, generate length of each repair: "Þ##%$ hoursß !Þ*&!$ hours,
"Þ'"!% hours. Then the average repair time is
Ð"Þ##%$  !Þ*&!$  "Þ'"!%ÑÎ$ œ "Þ#' hours.
(b)
(c)
J ÐBÑ œ 
!Þ%B
!Þ%  !Þ'ÐB  "Ñ
if ! Ÿ B Ÿ "
if B "
J ÐBÑ œ !Þ##%$ Ê B œ !Þ&'" hours
J ÐBÑ œ !Þ*&!$ Ê B œ "Þ*"( hours
J ÐBÑ œ !Þ'"!% Ê B œ "Þ$&" hours
Average repair time: Ð!Þ&'"  "Þ*"(  "Þ$&"ÑÎ$ œ "Þ#) hours
20S1-3
(d)
J ÐBÑ œ !Þ((&( Ê B œ "Þ'#' hours
J ÐBÑ œ !Þ!%*( Ê B œ !Þ"#% hours
J ÐBÑ œ !Þ$)*' Ê B œ !Þ*(% hours
Average repair time: Ð"Þ'#'  !Þ"#%  !Þ*(%ÑÎ$ œ !Þ*" hours
(e) Average repair time:
Ð!Þ&'"  "Þ*"(  "Þ$&"  "Þ'#'  !Þ"#%  !Þ*(%ÑÎ' œ "Þ!* hours
(f) The method of complementary random numbers in (e) gave the closest estimate. It
performs well because using complements helps counteract rather extreme random
numbers such as !Þ*&!$.
(g) Results will vary. The following 300-day simulation using the method of
complementary random numbers yielded an overall average service time of "Þ!*&
minutes. This is very close to the true mean, which is "Þ" minutes.
(h) We get !Þ(#&'ß !Þ#(%%ß !Þ!)"(ß !Þ*")$ß !Þ%$)#ß !Þ&'!) for minor repair times and
"Þ##%$ß "Þ((&(ß "Þ*&!$ß "Þ!%*(ß "Þ'"!%ß "Þ$)*' for major repair times. The weight for
minor repair times is Ð'Î"#ÑÎ!Þ% œ "Þ#& and the weight for major repair times is
Ð'Î"#ÑÎ!Þ' œ "Þ!)*. By dividing each sample by its corresponding weight, we obtain "Þ"
minutes as the estimate of the mean of the overall distribution of repair times.
20S1-4
20S1-5.
(a)
!Þ!!!! to !Þ$*** correspond to no claims filled.
!Þ%!!! to !Þ(*** correspond to small claims filled.
!Þ)!!! to !Þ**** correspond to large claims filled.
Random observations: !Þ(#&' œ small, !Þ!)"( œ no, !Þ%$*# œ small
Using random numbers, generate size of each claim:
!Þ##%$ † #ß !!! œ $%%)Þ'!ß $!, !Þ'"!% † #ß !!! œ $"ß ##!Þ)!.
Then the average claim size is
Ð%%)Þ'!  !  "ß ##!Þ)!ÑÎ$ œ $&&'Þ%(.
(b)
(c)

!
B
J ÐBÑ œ  !Þ%  !Þ% #ß!!!

ÐB#ß!!!Ñ
 !Þ)  !Þ# ")ß!!!
if B  !
if ! Ÿ B Ÿ #ß !!!
if #ß !!! Ÿ B Ÿ #!ß !!!
J ÐBÑ œ !Þ##%$ Ê B œ $!
J ÐBÑ œ !Þ*&!$ Ê B œ $"&ß &#(
J ÐBÑ œ !Þ'"!% Ê B œ $"ß !&#
Average claim size: Ð$!  $"&ß &#(  $"ß !&#ÑÎ$ œ $&ß &#'Þ$$
(d)
J ÐBÑ œ !Þ((&( Ê B œ $"ß ))!
J ÐBÑ œ !Þ!%*( Ê B œ $!
J ÐBÑ œ !Þ$)*' Ê B œ $!
Average claim size: Ð$"ß ))!  $!  $!ÑÎ$ œ $'#'Þ'(
(e) Average claim size: Ð$!  $"&ß &#(  $"ß !&#  $"ß ))!  $!  $!ÑÎ' œ $$ß !('Þ&!
(f) The method of complementary random numbers in (e) gave the closest estimate. It
performs well because using complements helps counteract rather extreme random
numbers such as !Þ*&!$.
20S1-5
(g) Results will vary. The following 300-day simulation using the method of
complementary random numbers yielded an overall average claim size of $#ß &%(Þ"&.
This is very close to the true mean, which is $#ß '!!.
(h) We get "%&"Þ#, "'$Þ%, )()Þ%, &%)Þ), ")$'Þ', ""#"Þ' for small claims and '!$(Þ%,
"*"!&Þ%, "#*)(Þ#, "&*'#Þ', #)*%Þ', *!"#Þ) for large claims. The weight for small claims
is Ð'Î"#ÑÎ!Þ% œ "Þ#& and the weight for large claims is Ð'Î"#ÑÎ!Þ# œ #Þ&. By dividing
each sample by its corresponding weight, we obtain $#ß '!! as the estimate of the mean
of the overall distribution of claim sizes.

!

"
ÐB  "Ñ#
B
J ÐBÑ œ ∞ 0 ÐCÑ.C œ  # "
#

 "  # Ð"  BÑ
"
20S1-6.
ÊBœ
<
!Þ!*'
!Þ&'*
#<  "
"  #Ð"  <Ñ
B
!Þ&'")
!Þ!("'
"<
!Þ*!%
!Þ%$"
if ! Ÿ <  "#
if "# Ÿ <  "
B
!Þ&'")
!Þ!("'
Ê sample mean: !
20S1-6
if B  "
if " Ÿ B  !
if ! Ÿ B  "
if B "
!
20S1-7.
J ÐBÑ œ
B
∞
0 ÐCÑ.C
$
ÊBœ
#<  "
<
!Þ!*'
!Þ&'*
B
!Þ*$"%
!Þ&"')
œ
"
B$ "
#
"<
!Þ*!%
!Þ%$"
if B  "
if " Ÿ B  "
if B "
B
!Þ*$"%
!Þ&"')
Ê sample mean: !
20S1-8.
(a)


 !Þ"#&
!Þ$(&
T Ö\ œ 5× œ 

 !Þ$(&
 !Þ"#&
<
!Þ!*'
!Þ&'*
!Þ''&
B
!
#
#
"<
!Þ*!%
!Þ%$"
!Þ$$&
if 5
if 5
if 5
if 5
œ!
œ"
œ#
œ$


!
"
\œ

#
$
B
$
"
"
Ê sample mean: Ð!  #  #ÑÎ$ œ "Þ$$
(b) Sample mean: Ð%  &ÑÎ' œ "Þ&
(c)
head
\œ
tail
if ! Ÿ <  "#
if "# Ÿ <  "
<" œ Ö!Þ!*'ß !Þ&'*ß !Þ''&× Ê \" œ "
<# œ Ö!Þ('%ß !Þ)%#ß !Þ%*#× Ê \# œ "
<$ œ Ö!Þ##%ß !Þ*&!ß !Þ'"!× Ê \$ œ "
sample mean: $Î$ œ "
(d)
<"‡ œ Ö!Þ*!%ß !Þ%$"ß !Þ$$&× Ê \"‡ œ #
<#‡ œ Ö!Þ#$'ß !Þ"&)ß !Þ&!)× Ê \#‡ œ #
<$‡ œ Ö!Þ(('ß !Þ!&!ß !Þ$*!× Ê \$‡ œ #
sample mean: Ð$  'ÑÎ$ œ $
20S1-7
if ! Ÿ <  !Þ"#&
if !Þ"#& Ÿ <  !Þ&
if !Þ& Ÿ <  !Þ)(&
if !Þ)(& Ÿ <  "
20S1-9.
(a)
Shaft radius: <= œ " %!!/%!!Ð>"Ñ .> œ "  /%!!Ð="Ñ Ê = œ " 
=
Bushing radius: <, œ " "!!.> œ "!!Ð,  "Ñ Ê , œ " 
,
<=
!Þ!*'
!Þ''&
!Þ)%#
!Þ##%
!Þ'"!
!Þ%)%
!Þ$&!
!Þ%$!
!Þ)!#
!Þ#&&
=
"Þ!!!#&#
"Þ!!#($%
"Þ!!%'"$
"Þ!!!'$%
"Þ!!#$&%
"Þ!!"'&%
"Þ!!"!((
"Þ!!"%!&
"Þ!!"%!&
"Þ!!!($'
<,
!Þ&'*
!Þ('%
!Þ%*#
!Þ*&!
!Þ"%&
!Þ&&#
!Þ&*!
!Þ!%"
!Þ%("
!Þ(**
,
"Þ!!&'*
"Þ!!('%
"Þ!!%*#
"Þ!!*&!
"Þ!!"%&
"Þ!!&&#
"Þ!!&*!
"Þ!!!%"
"Þ!!%("
"Þ!!(**
lnÐ"<= Ñ
%!!
<,
"!!
=  ,?
No
No
No
No
Yes
No
No
Yes
No
No
When =  , , interference occurs, so the probability of interference is estimated as
#Î"! œ #!%.
(b)
Stratum
"
#
$
Portion of Distribution
!Þ! Ÿ J Ð,Ñ Ÿ !Þ#
!Þ# Ÿ J Ð,Ñ Ÿ !Þ'
!Þ' Ÿ J Ð,Ñ Ÿ "Þ!
Stratum
"
"
"
"
"
"
#
#
$
$
<=
!Þ!*'
!Þ''&
!Þ)%#
!Þ##%
!Þ'"!
!Þ%)%
!Þ$&!
!Þ%$!
!Þ)!#
!Þ#&&
=
"Þ!!!#&#
"Þ!!#($%
"Þ!!%'"$
"Þ!!!'$%
"Þ!!#$&%
"Þ!!"'&%
"Þ!!"!((
"Þ!!"%!&
"Þ!!"%!&
"Þ!!!($'
Stratum Random Number
<,w œ !Þ#<,
<,w œ !Þ#  !Þ%<,
<,w œ !Þ'  !Þ%<,
<,
!Þ&'*
!Þ('%
!Þ%*#
!Þ*&!
!Þ"%&
!Þ&&#
!Þ&*!
!Þ!%"
!Þ%("
!Þ(**
<,w
!Þ""%
!Þ"&$
!Þ!*)
!Þ"*!
!Þ!#*
!Þ""!
!Þ%$'
!Þ#"'
!Þ())
!Þ*#!
Estimated probability of interference: %Î$! œ #Î"&
20S1-8
,
"Þ!!""%
"Þ!!"&$
"Þ!!!*)
"Þ!!"*!
"Þ!!!#*
"Þ!!""!
"Þ!!%$'
"Þ!!#"'
"Þ!!())
"Þ!!*#!
Size
'
#
#
Weight
"Î$
"Î#
"Î#
Interference Weight
!
"Î$
"Î$
!
"Î$
"Î$
!
!
!
!
(c)
<=
!Þ!*'
!Þ''&
!Þ)%#
!Þ##%
!Þ'"!
!Þ%)%
!Þ$&!
!Þ%$!
!Þ)!#
!Þ#&&
=
"Þ!!!#&#
"Þ!!#($%
"Þ!!%'"$
"Þ!!!'$%
"Þ!!#$&%
"Þ!!"'&%
"Þ!!"!((
"Þ!!"%!&
"Þ!!%!%)
"Þ!!!($'
<,
!Þ&'*
!Þ('%
!Þ%*#
!Þ*&!
!Þ"%&
!Þ&&#
!Þ&*!
!Þ!%"
!Þ%("
!Þ(**
,
"Þ!!&'*
"Þ!!('%
"Þ!!%*#
"Þ!!*&!
"Þ!!"%&
"Þ!!&&#
"Þ!!&*!
"Þ!!!%"
"Þ!!%("
"Þ!!(**
=  ,?
No
No
No
No
Yes
No
No
Yes
No
No
Estimated probability of interference: "#  "&  #&  œ $!%
=w
"Þ!!&)&*
"Þ!!"!#!
"Þ!!!%$!
"Þ!!$(%!
"Þ!!"#$'
"Þ!!")"%
"Þ!!#'#&
"Þ!!#""!
"Þ!!!&&#
"Þ!!$%"'
,w
"Þ!!%$"
"Þ!!#$'
"Þ!!&!)
"Þ!!!&!
"Þ!!)&&
"Þ!!%%)
"Þ!!%"!
"Þ!!*&*
"Þ!!&#*
"Þ!!#!"
=w  , w
Yes
No
No
Yes
No
No
No
No
Yes
Yes
Summary:
Method:
Interference Probability:
Monte Carlo
"Î&
Stratified Sampling
#Î"&
20S1-9
Complementary RNs
$Î"!
SUPPLEMENT 2 TO CHAPTER 20
REGENERATIVE METHOD OF STATISTICAL ANALYSIS
20S2-1.
(a)
C" œ !  &  %
C# œ !  #
C$ œ !  $  "  '
œ *à D" œ $
œ #à D# œ #
œ "!à D$ œ %
C œ #"Î$ œ (à
EstÖ[; × œ
(
$
D œ *Î$ œ $
œ # "$
=#"" œ Ð)"  %  "!!ÑÎ#  Ð*  #  "!Ñ# Î' œ "*
=### œ Ð*  %  "'ÑÎ#  Ð$  #  %Ñ# Î' œ "
=#"# œ Ð#(  %  %!ÑÎ#  Ð#"ÑÐ*ÑÎ' œ %
=# œ "*  Ð#ÑÐ(Î$ÑÐ%Ñ  Ð(Î$Ñ# œ &Þ(() Ê = œ #Þ%!%
"  #α œ !Þ*! Ê α œ !Þ!& Ê Oα œ "Þ'%&
T Ö"Þ&(# Ÿ [; Ÿ $Þ!*%× œ !Þ*!
(b)
C"
C#
C$
C%
C&
œ!$#
œ!$"&
œ!
œ!#%
œ!$&#
œ &à
œ *à
œ !à
œ 'à
œ "!à
D"
D#
D$
D%
D&
œ$
œ%
œ"
œ$
œ%
C œ $!Î& œ 'à
EstÖ[; × œ
'
$
D œ "&Î& œ $
œ#
=#"" œ Ð#&  )"  $'  "!!ÑÎ%  Ð"!  '  !  *  &Ñ# Î#! œ "& "#
=### œ Ð*  "'  "  *  "'ÑÎ%  Ð$  %  "  $  %Ñ# Î#! œ " "#
=#"# œ Ð"&  $'  !  ")  %!ÑÎ%  Ð$!ÑÐ"&ÑÎ#! œ % $%
=# œ "& "#  Ð#ÑÐ#Ñ% $%   Ð#Ñ# " "#  œ # "# Ê = œ "Þ&)"
"  #α œ !Þ*! Ê α œ !Þ!& Ê Oα œ "Þ'%&
T Ö"Þ'"# Ÿ [; Ÿ #Þ$))× œ !Þ*!
20S2-2.
When a service completion occurs, > minutes have passed since the last arrival, where
! Ÿ > Ÿ #&. The time until the next arrival is uniformly distributed between > and #&  >,
where > œ maxÐ!ß &  >Ñ. Thus, the probabilistic structure of when future arrivals will
occur depends on the history, so this cannot be a regeneration point.
20S2-1
20S2-3.
(a) For any new tube, the time of the next failure is given by "current
time  "!!!  "!!!<," where < is a random number from Table 20.3. At each shutdown,
one hour is added to the time of the next failure for all tubes when simulating the status
quo and two hours are added when simulating the proposal.
Simulation of the status quo:
Time
!
"!*'
"&(!
"''(
"('(
#)*$
#*%$
$!'(
$(#"
%!*"
%&!)
%&&&
&!!!
<"
!Þ!*'
!Þ)%#




!Þ"%&


!Þ$&!



<#
!Þ&'*

!Þ%*#




!Þ%)%



!Þ%$!

<$
!Þ''&


!Þ##%

!Þ'"!




!Þ&*!


<%
!Þ('%



!Þ*&!



!Þ&&#




Time of Failure of
Tube 1 Tube 2 Tube 3
"!*'
"&'*
"''&
#*$*
"&(!
"'''
#*%!
$!'$
"''(
#*%"
$!'%
#)*#
#*%#
$!'&
#)*$
#*%$
$!''
%&!%
%!)*
$!'(
%&!&
%!*!
%&&#
%&!'
%!*"
%&&$
%&!(
&%%#
%&&%
%&!)
&%%$
%&&&
'!**
&%%%
&*)'
'"!!
&%%%
&*)'
'"!!
Tube 4
"('%
"('&
"(''
"('(
$(")
$("*
$(#!
$(#"
&#(%
&#(&
&#('
&#((
&#((
Estimated cost of the status quo: "" ‚ $"ß #!! œ $"$ß #!!
Simulation of the proposal:
Time
!
"!*'
#$##
$%'*
%&"#
<"
!Þ!*'
!Þ)%#
!Þ'"!
!Þ$&!
!Þ)!#
<#
!Þ&'*
!Þ%*#
!Þ"%&
!Þ&*!
!Þ%("
<$
!Þ''&
!Þ##%
!Þ%)%
!Þ%$!
!Þ#&&
<%
!Þ('%
!Þ*&!
!Þ&&#
!Þ!%"
!Þ(**
First Tube to Fail
Tube 1
Tube 3
Tube 2
Tube 4
Tube 3
Time of Failure
"!*'
#$##
$%'*
%&"#
&('*
Estimated cost of the proposal: % ‚ $#ß )!! œ $""ß #!!
(b) Based on the simulation results in part (a), the proposal should be accepted.
(c) For the proposed policy, each shutdown is a regeneration point because all tubes are
replaced and the process begins a new. For the status quo, the process never repeats itself
because each tube is replaced when it fails.
20S2-2
(d)
Cycle
"
#
$
%
Cycle Cost
$#ß )!!
$#ß )!!
$#ß )!!
$#ß )!!
Cycle Length
"!*'
"##'
""%(
"!%$
C œ $#ß )!!ß D œ ""#)
EstÖcost/hour× œ #)!!Î""#) œ $#Þ%)#
=#"" œ
Ð%‚#)!!# Ñ
$
=### œ
Ð"!)'# "##'# ""%(# "!%$# Ñ
$
=#"# œ
Ð#)!!ÑÐ"!)'"##'""%("!%$Ñ
$

Ð%‚#)!!Ñ#
"#
œ!

Ð"!)'"##'""%("!%$Ñ#
"#

œ '!(" $"
Ð%ÑÐ#)!!ÑÐ"!)'"##'""%("!%$Ñ
"#
œ!
=# œ !  Ð#Þ%)#ÑÐ!ÑÐ#Ñ  Ð#Þ%)#Ñ# '!(" "$  œ $(%"! Ê = œ "*$Þ%
"  #α œ !Þ*& Ê α œ !Þ!#& Ê Oα œ "Þ*'
T Ö#Þ$"% Ÿ cost/hour Ÿ #Þ'&!× œ !Þ*&
20S2-4.
(a)
(i)
Number of Servers =
Interarrival Times
Distribution =
Mean =
Data
1
L=
Lq =
W=
Wq =
Exponential
1.25
5
Mean =
P0 = 0.1899996
0.16425264
0.215746552
P1 = 0.15101797
0.132489844
0.169546091
1
P2 = 0.12530975
0.111337121
0.139282371
25
P3 = 0.09541037
0.084720833
0.106099913
P4 = 0.07620596
0.066901918
0.085509995
P5 = 0.06224509
0.054038618
0.070451571
Length of Simulation Run
Number of Arrivals =
Results
95% Confidence Interval
Low
High
3.344529192 6.413546353
2.552881224 5.585193514
4.231161152 7.942948994
3.233440493
6.91956742
Exponential
Service Times
Distribution =
Point
Estimate
4.87903777
4.06903737
6.08705507
5.07650396
10,000
Run Simulation
20S2-3
P6 = 0.05620591
0.047527927
0.064883901
P7 = 0.0420322
0.035157883
0.048906526
0.037328273
P8 = 0.03118735
0.025046424
P9 = 0.02715091
0.020825618
0.033476206
P10 = 0.02295245
0.016668188
0.029236719
(ii)
Data
1
Number of Servers =
Interarrival Times
Distribution =
Mean =
L=
Lq =
W=
Wq =
Exponential
1.25
5
Point
Estimate
2.72338571
1.92941897
3.42557184
2.4268921
Results
95% Confidence Interval
Low
High
2.426711315 3.020060108
1.646668271 2.212169662
3.099322189 3.751821491
2.104137696 2.749646512
P0 = 0.20603325
0.188347397
0.223719112
Erlang
P1 = 0.21238852
0.196737339
0.228039706
Mean =
1
P2 = 0.16915551
0.157922441
0.180388584
k=
4
P3 = 0.12039424
0.111814695
0.128973778
P4 = 0.0820109
0.074401677
0.089620118
P5 = 0.06040587
0.053059888
0.067751862
P6 = 0.04648171
0.038778844
0.054184579
P7 = 0.03450976
0.02699144
0.042028079
P8 = 0.02377114
0.017106209
0.030436065
P9 = 0.01509947
0.009781903
0.020417036
P10 = 0.01074926
0.005500231
0.015998293
Service Times
Distribution =
Length of Simulation Run
Number of Arrivals =
10,000
Run Simulation
(iii)
Data
1
Number of Servers =
Interarrival Times
Distribution =
Mean =
L=
Lq =
W=
Wq =
Exponential
1.25
5
Value =
P0 = 0.20135878
0.18542182
0.217295736
P1 = 0.24659089
0.230719649
0.262462131
1
P2 = 0.18551928
0.175252934
0.195785632
4
P3 = 0.1236632
0.115231596
0.1320948
P4 = 0.08618065
0.078092827
0.094268471
0.063597968
Length of Simulation Run
Number of Arrivals =
Results
95% Confidence Interval
Low
High
2.133069489 2.588975073
1.346964384 1.777797733
2.715448043 3.197150042
1.715448043 2.197150042
Constant
Service Times
Distribution =
Point
Estimate
2.36102228
1.56238106
2.95629904
1.95629904
10,000
Run Simulation
P5 = 0.05600588
0.048413787
P6 = 0.03835197
0.030657389
0.04604656
P7 = 0.02567582
0.018972941
0.032378704
P8 = 0.0158616
0.010224825
0.021498383
P9 = 0.00962322
0.005462589
0.013783841
P10 = 0.00542071
0.002088955
0.00875247
P; # ÎP; " œ "Þ*$Î%Þ!( œ !Þ%(ß P; $ ÎP; " œ "Þ&'Î%Þ!( œ !Þ$)
(b)
P; œ
-# 5# 3#
#Ð"3Ñ ß P
(i)
P; " œ
!Þ'%!Þ'%
#‚!Þ#
(ii)
P; # œ
!Þ'%‚!Þ#&!Þ'%
#‚!Þ#
(iii)
P; $ œ
!Þ'%
#‚!Þ#
œ 3  P; ß [; œ
P;
- ß[
œ [; 
"
.
œ $Þ#ß P" œ %ß [; " œ %ß [" œ &
œ #ß P# œ #Þ)ß [; # œ #Þ&ß [# œ $Þ&
œ "Þ'ß P$ œ #Þ%ß [; $ œ #ß [$ œ $
P; # ÎP; " œ !Þ'(&ß P; $ ÎP; " œ "Þ'Î$Þ# œ !Þ&
They all fall into *&% confidence intervals in (a).
20S2-4
20S2-5.
(i)
Number of Servers =
Interarrival Times
Distribution =
Mean =
Data
2
L=
Lq =
W=
Wq =
Exponential
0.625
5
Mean =
0.106050845
0.133310916
Exponential
P1 = 0.18241551
0.165942487
0.198888524
1
P2 = 0.14682206
0.135064103
0.158580013
4
P3 = 0.1174116
0.108084485
0.126738716
P4 = 0.09455108
0.086683682
0.102418474
P5 = 0.07588536
0.068940487
0.082830237
Length of Simulation Run
Number of Arrivals =
Results
95% Confidence Interval
Low
High
3.55223048 4.535979034
2.004050365 2.927713682
2.243162514 2.820872423
1.266283183 1.821498531
P0 = 0.11968088
Service Times
Distribution =
Point
Estimate
4.04410476
2.46588202
2.53201747
1.54389086
10,000
Run Simulation
P6 = 0.06080669
0.053342123
0.06827125
P7 = 0.04646041
0.039609107
0.053311723
P8 = 0.03437865
0.0287365
0.0400208
P9 = 0.02643309
0.021235661
0.031630518
P10 = 0.02198272
0.016342638
0.027622795
(ii)
Number of Servers =
Interarrival Times
Distribution =
Mean =
k=
Data
2
L=
Lq =
W=
Wq =
Erlang
0.625
4
Mean =
0.079559951
Exponential
P1 = 0.23998572
0.223898835
0.25607261
1
P2 = 0.21905849
0.206494297
0.231622688
4
P3 = 0.15304732
0.143823719
0.162270921
P4 = 0.10218281
0.094322966
0.110042646
P5 = 0.06743454
0.059859123
0.075009951
Length of Simulation Run
Number of Arrivals =
Results
95% Confidence Interval
Low
High
2.655401428
3.08340411
1.094469272 1.475828087
1.667426546 1.929761643
0.687401521 0.923707836
P0 = 0.08788009
Service Times
Distribution =
Point
Estimate
2.86940277
1.28514868
1.79859409
0.80555468
10,000
Run Simulation
20S2-5
0.096200237
P6 = 0.04697817
0.03996655
0.053989791
P7 = 0.02872811
0.022787465
0.034668759
P8 = 0.02146079
0.01479521
0.028126376
P9 = 0.01418783
0.00823392
0.020141735
P10 = 0.00984255
0.004906822
0.014778274
(iii)
Number of Servers =
Interarrival Times
Distribution =
Value =
Data
2
L=
Lq =
W=
Wq =
Constant
0.625
4
Mean =
0.067303152
0.082360281
Exponential
P1 = 0.26131418
0.243064069
0.279564296
1
P2 = 0.25789822
0.244238516
0.271557916
4
P3 = 0.15619313
0.146337505
0.166048745
P4 = 0.09746869
0.08773456
0.107202822
P5 = 0.06257259
0.052995794
0.072149386
Length of Simulation Run
Number of Arrivals =
Results
95% Confidence Interval
Low
High
2.409533768 2.796065206
0.845468642 1.182085562
1.505958605 1.747540754
0.528417901 0.738803476
P0 = 0.07483172
Service Times
Distribution =
Point
Estimate
2.60279949
1.0137771
1.62674968
0.63361069
10,000
Run Simulation
P6 = 0.0388295
0.030093467
0.047565538
P7 = 0.02255516
0.015853677
0.029256644
P8 = 0.01252675
0.008005617
0.017047883
P9 = 0.00585078
0.002756411
0.008945158
P10 = 0.00430886
0.001317909
0.007299815
P; # ÎP; " œ "Þ29Î#Þ47 œ !Þ&2ß P; $ ÎP; " œ "Þ01Î#Þ47 œ !Þ41
20S2-6.
Number of Servers =
Interarrival Times
Distribution =
Mean =
Data
1
L=
Lq =
W=
Wq =
Exponential
1
4
Mean =
P0 = 0.19081784
0.166501011
0.215134664
P1 = 0.15255487
0.134895841
0.170213906
0.8
P2 = 0.11905166
0.106286404
0.13181692
4
P3 = 0.10175524
0.092077494
0.111432992
P4 = 0.08389876
0.075564479
0.092233044
0.068495604
Length of Simulation Run
Number of Arrivals =
Results
95% Confidence Interval
Low
High
3.580199095 5.002042509
2.790979562 4.172897717
3.619225209 4.929792236
2.822990545 4.113927697
Exponential
Service Times
Distribution =
Point
Estimate
4.2911208
3.48193864
4.27450872
3.46845912
10,000
Run Simulation
20S2-6
P5 = 0.06123564
0.053975668
P6 = 0.05137418
0.044309177
0.058439176
P7 = 0.04114546
0.034871539
0.047419377
0.043317632
P8 = 0.03639616
0.029474684
P9 = 0.03166299
0.023347529
0.039978447
P10 = 0.02653407
0.018696268
0.034371877
Number of Servers =
Interarrival Times
Distribution =
Mean =
k=
Data
1
L=
Lq =
W=
Wq =
Erlang
1
4
Mean =
0.181119032
0.215863063
Exponential
P1 = 0.23984341
0.222687222
0.256999605
0.8
P2 = 0.17199717
0.160956734
0.18303761
4
P3 = 0.11861122
0.109457495
0.127764955
P4 = 0.08105923
0.073189799
0.088928655
P5 = 0.0537207
0.046173793
0.061267614
P6 = 0.04044456
0.032967741
0.047921373
P7 = 0.03226234
0.024660756
0.03986393
P8 = 0.02285132
0.016014453
0.029688193
Length of Simulation Run
Number of Arrivals =
10,000
Run Simulation
Number of Servers =
Interarrival Times
Distribution =
Value =
Mean =
0.009235805
0.018709353
0.004580435
0.011960443
L=
Lq =
W=
Wq =
Constant
1
4
Point
Estimate
2.12966982
1.32953423
2.12966982
1.32953423
Results
95% Confidence Interval
Low
High
1.879312107 2.380027537
1.091852419
1.56721604
1.879312107 2.380027537
1.091852419
1.56721604
P0 = 0.19986441
0.183544775
0.21618404
Exponential
P1 = 0.29281298
0.273527473
0.312098488
0.8
P2 = 0.18827746
0.177005913
0.199549014
4
P3 = 0.12222875
0.111746308
0.132711198
P4 = 0.0780542
0.068346215
0.087762184
P5 = 0.04689815
0.037890832
0.05590546
Length of Simulation Run
Number of Arrivals =
P9 = 0.01397258
P10 = 0.00827044
Data
1
Service Times
Distribution =
Results
95% Confidence Interval
Low
High
2.316932956 2.923228649
1.529344973 2.107798727
2.322735858 2.914249992
1.533437067 2.101502373
P0 = 0.19849105
Service Times
Distribution =
Point
Estimate
2.6200808
1.81857185
2.61849292
1.81746972
10,000
Run Simulation
P6 = 0.02863435
0.020994488
0.036274217
P7 = 0.01633121
0.010115182
0.022547243
P8 = 0.01019452
0.00485594
0.015533105
P9 = 0.00647183
0.001373016
0.011570642
P10 = 0.00323684
-4.77602E-05
0.006521442
P; # ÎP; " œ "Þ8#Î3.48 œ !Þ52ß P; $ ÎP; " œ "Þ$2Î3.48 œ !Þ$8
20S2-7
CHAPTER 21: THE ART OF MODELING WITH SPREADSHEETS
21.1.
LT Rate
ST Rate
Savings Interest
5%
7%
3%
Start Balance
Minimum Cash
1
0.5
Year
2014
2015
2016
2017
2018
2019
2020
2021
2022
2023
2024
Cash
Flow
-8
-2
-4
3
6
3
-4
7
-2
10
(all cash figures in millions of dollars)
LT
Loan
7.50
ST
Loan
0.00
2.36
6.89
4.73
0.00
0
1.02
0
0
0
LT
Interest
ST
Interest
-0.38
-0.38
-0.38
-0.38
-0.38
-0.38
-0.38
-0.38
-0.38
-0.38
0.00
-0.17
-0.48
-0.33
0.00
0
-0.07
0
0
0
LT
ST
Payback Payback
-7.50
0.00
-2.36
-6.89
-4.73
0.00
0
-1.02
0
0
0
Savings
Interest
0.02
0.01
0.02
0.01
0.03
0.11
0.01
0.18
0.12
0.41
Balance
0.50
0.50
0.50
0.50
1.08
3.74
0.50
6.04
3.85
13.59
6.12
>=
>=
>=
>=
>=
>=
>=
>=
>=
>=
>=
Minimum
Balance
0.50
0.50
0.50
0.50
0.50
0.50
0.50
0.50
0.50
0.50
0.50
21.2.
(a) The COO will need to know how many of each product to produce. Thus, the
decisions are how many end tables, how many coffee tables, and how many dining room
tables to produce. The objective is to maximize total profit.
(b)
Pine wood used œ (3 end tables)(8 pounds/end table)
 (3 dining room tables)(80 pounds/dining room table)
œ 264 pounds
Labor used
œ (3 end tables)(1 hour/end table)
 (3 dining room tables)(4 hours/dining room table)
œ 15 hours
(c)
End Tables Coffee Tables
Dining Room Tables
Unit Profit
Resource Used per unit Produced
Pine Wood
Labor
Total Used
Available
<=
<=
End Tables Coffee Tables
Dining Room Tables
Units Produced
21-1
Total Profit
(d)
End Tables Coffee Tables Dining Room Tables
Unit Profit
$50
$100
$220
Resource Used per unit Produced
8
15
80
1
2
4
Pine Wood
Labor
Units Produced
Total Used
3000
<=
200
<=
End Tables Coffee Tables Dining Room Tables
0
40
30
Available
3000
200
Total Profit
$10,600
21.3.
(a) Top management will need to know how much to produce in each quarter. Thus, the
decisions are the production levels in quarters 1, 2, 3, and 4. The objective is to maximize
the net profit.
(b)
Ending Inventory(Q1)
œ Starting Inventory(Q1)  Production(Q1)  Sales(Q1)
œ "ß !!!  &ß !!!  $ß !!! œ $ß !!!
Ending Inventory(Q2)
œ Starting Inventory(Q2)  Production(Q2)  Sales(Q2)
œ $ß !!!  &ß !!!  %ß !!! œ %ß !!!
Profit from Sales(Q1)
œ Sales(Q1) ‚ Ð$#!Ñ œ $ß !!! ‚ Ð$#!Ñ œ $'!ß !!!
Profit from Sales(Q2)
œ Sales(Q2) ‚ Ð$#!Ñ œ %ß !!! ‚ Ð$#!Ñ œ $)!ß !!!
Inventory Cost(Q1) œ Ending Inventory(Q1) ‚ Ð$)Ñ œ $ß !!! ‚ Ð$)Ñ œ $#%ß !!!
Inventory Cost(Q2) œ Ending Inventory(Q2) ‚ Ð$)Ñ œ %ß !!! ‚ Ð$)Ñ œ $$#ß !!!
(c)
Inventory Holding Cost
Gross Profit from Sales
Starting
Inventory
Quarter 1
Quarter 2
Quarter 3
Quarter 4
Maximum
Production
Production
Demand/ Ending
Sales Inventory
<=
<=
<=
<=
Inventory Gross Profit
Cost
from Sales
>=
>=
>=
>=
Net Profit
21-2
(d)
Inventory Holding Cost
Gross Profit from Sales
Quarter 1
Quarter 2
Starting
Inventory
1,000
0
$8
$20
Maximum
Production
Production
2,000
<= 6,000
4,000
<= 6,000
Demand/ Ending
Sales Inventory
3,000
0
>= 0
4,000
0
>= 0
Inventory Gross Profit
Cost
from Sales
$0
$60,000
$0
$80,000
Totals
$0
$140,000
Net Profit
$140,000
(e)
Inventory Holding Cost
Gross Profit from Sales
Quarter 1
Quarter 2
Quarter 3
Quarter 4
Starting
Inventory
1,000
1,000
3,000
1,000
$8
$20
Production
3,000
6,000
6,000
6,000
<=
<=
<=
<=
Maximum
Production
6,000
6,000
6,000
6,000
Demand/ Ending
Sales Inventory
3,000
1,000 >= 0
4,000
3,000 >= 0
8,000
1,000 >= 0
7,000
0
>= 0
Inventory Gross Profit
Cost
from Sales
$8,000
$60,000
$24,000
$80,000
$8,000
$160,000
$0
$140,000
Totals $40,000
$440,000
Net Profit
$400,000
21.4.
(a) Fairwinds needs to know how much to participate in each of the three projects and
what their ending balances will be. The decisions to be made are how much to participate
in each of the three projects. The objective is to maximize the ending balance at the end
of six years.
(b)
Ending Balance(Y1) œ Starting Balance  Project A  Project C  Other Projects
œ "!  Ð"!!%ÑÐ%Ñ  Ð&!%ÑÐ"!Ñ  ' œ $( million
Ending Balance(Y2) œ Starting Balance  Project A  Project C  Other Projects
œ (  Ð"!!%ÑÐ'Ñ  Ð&!%ÑÐ(Ñ  ' œ $$Þ& million
(c)
Starting Cash
Year
1
2
3
4
5
6
Cash Flow (at full participation, $million)
Project A
Project B
Project C
Total
Cash Flow
From ABC
Other
Projects
Ending
Balance
Minimum
Balance
>=
>=
>=
>=
>=
>=
Participation
<=
100%
<=
100%
<=
100%
21-3
(d)
Starting Cash
Year
1
2
Participation
10
all cash numbers are in $millions
Cash Flow (at full participation, $million)
Project A
Project B
Project C
-4
-8
-10
-6
-8
-7
0%
<=
100%
0%
<=
100%
Total
Cash Flow
From ABC
0
0
Other
Projects
6
6
Ending
Balance
16
22
Minimum
Balance
>=
1
>=
1
Ending
Balance
5.25
3.125
1
6.5
9.5
59.5
Minimum
Balance
1
1
1
1
1
1
0%
<=
100%
(e)
Starting Cash
Year
1
2
3
4
5
6
Participation
10
all cash numbers are in $millions
Cash Flow (at full participation, $million)
Project A
Project B
Project C
-4
-8
-10
-6
-8
-7
-6
-4
-7
24
-4
-5
0
30
-3
0
0
44
18.75%
<=
100%
0%
<=
100%
Total
Cash Flow
From ABC
-10.75
-8.125
-8.125
-0.5
-3
44
Other
Projects
6
6
6
6
6
6
>=
>=
>=
>=
>=
>=
100%
<=
100%
21.5.
(a) Web Mercantile needs to know each month how many square feet to lease and for
how long. The decisions therefore are for each month how many square feet to lease for
one month, two months, three months, etc. The objective is to minimize the overall
leasing cost.
(b) Total Cost œ (30,000 sq feet)($190/sq foot)  (20,000 sq feet)($100/sq foot) œ $(Þ( million
21-4
(c)
Month Covered by Lease?
Month of Lease: 1 1 1 1 1 2 2 2 2 3 3 3
Length of Lease: 1 2 3 4 5 1 2 3 4 1 2 3
Month 1
Month 2
Month 3
Month 4
Month 5
4
1
4
2
Total
Leased
(sq. ft.)
5
1
Space
Required
(sq. ft.)
>=
>=
>=
>=
>=
Cost of Lease
(per sq. ft.)
Total Cost
Lease (sq. ft.)
(d)
Month Covered by Lease? Total
Month of Lease:
1
1
2
Leased
Length of Lease:
1
2
1
(sq. ft.)
Month 1
1
1
30,000 >=
Month 2
1
1
20,000 >=
Cost of Lease
(per sq. ft.)
$65
$100
Space
Required
(sq. ft.)
30,000
20,000
$65
Lease (sq. ft.) 10,000 20,000
Total Cost
$2,650,000
0
(e)
Month of Lease:
Length of Lease:
Month 1
Month 2
Month 3
Month 4
Month 5
1
1
1
1
2
1
1
1
3
1
1
1
1
4
1
1
1
1
1
5
1
1
1
1
1
Cost of Lease $65 $100 $135 $160 $190
(per sq. ft.)
Month Covered by Lease?
2
2
2
2
3
1
2
3
4
1
1
1
1
1
1
1
1
1
1
1
$65 $100 $135 $160
1
$65
3
2
1
1
3
3
1
1
1
4
1
1
4
2
1
1
$100 $135 $65 $100
5
1
1
Total
Leased
(sq. ft.)
30,000
30,000
40,000
30,000
50,000
Space
Required
(sq. ft.)
30,000
20,000
40,000
10,000
50,000
>=
>=
>=
>=
>=
$65
Total Cost
Lease (sq. ft.)
0
0
0
0
30,000
0
0
0
0
10,000
0
0
0
0
20,000
$7,650,000
21.6.
(a) Larry needs to know how many employees should work each possible shift. Therefore, the decision variables are the number of employees that work each shift. The
objective is to minimize the total cost of the employees.
Working 8 A.M.-noon:
3 FT morning  3 PT œ '
Working Noon-4 P.M.:
3 FT morning  2 FT afternoon  3 PT œ )
Working 4 P.M-8 P.M:
2 FT afternoon  4 FT evening  3 PT œ *
Working 8 P.M-midnight:
4 FT evening  3 PT œ (
Total cost per day œ Ð* FTÑÐ) hrsÑÐ$%!/hrÑ  Ð"# PTÑÐ% hrsÑÐ$$!/hrÑ œ $%ß $#!
(b)
21-5
(c)
Full Time
8am-4pm
Full Time
Full Time
Part Time
noon-8pm 4pm-midnight 8am-noon
Part Time
noon-4pm
Part Time
Part Time
4pm-8pm 8pm-midnight
Cost per Shift
Total
Working
Shift Covers Time of Day? (1=yes, 0=no)
8am-noon
noon-4pm
4pm-8pm
8pm-midnight
Total
Needed
>=
>=
>=
>=
Workers per Shift
Total
Time of Day Full Time
8am-noon
noon-4pm
4pm-8pm
8pm-midnight
Times Total
Part Time
Total
Cost
>=
>=
>=
>=
(d)
Full Time
8am-4pm
Cost per Shift
$320
8am-noon
noon-4pm
4pm-8pm
8pm-midnight
1
1
Workers per Shift
3
Time of Day
8am-noon
noon-4pm
4pm-8pm
8pm-midnight
Total
Full Time
3
6
7
4
Full Time
Full Time
noon-8pm 4pm-midnight
$320
$320
1
1
Part Time Part Time Part Time
Part Time
8am-noon noon-4pm 4pm-8pm 8pm-midnight
$120
$120
$120
$120
Shift Covers Time of Day? (1=yes, 0=no)
1
1
1
1
3
4
>=
>=
>=
>=
2
Times Total
Part Time
2
4
6
4
1
2
1
1
3
Total
Working
4
8
10
6
>=
>=
>=
>=
Total
Needed
4
8
10
6
2
Total
Cost
$4,160
21.7.
(a) Al will need to know how much to invest in each possible investment each year.
Thus, the decisions are how much to invest in investment A in year 1, 2, 3, and 4; how
much to invest in B in year 1, 2, and 3; how much to invest in C in year 2; and how much
to invest in D in year 5. The objective is to accumulate the maximum amount of money
by the beginning of year 6.
(b)
Ending Cash(Y1) œ ($60,000)(Starting Balance)-($20,000)(A in Y1) œ $40,000
Ending Cash(Y2) œ ($40,000)(Starting Balance)-($20,000)(B in Y2)-($20,000)(C in Y2) œ $0
Ending Cash(Y3) œ ($0)(Starting Balance)+($20,000)(1.4)(investment A) œ $28,000
Ending Cash(Y4) œ ($28,000)(Starting Balance)
Ending Cash(Y5) œ ($28,000)(Starting Balance)+($20,000)(1.7)(investment B) œ $62,000
Ending Cash(Y6) œ ($62,000)(Starting Balance)+($20,000)(1.9)(investment C) œ $100,000
21-6
(c)
Beginning Balance
Minimum Balance
Investment
Year
Year 1
Year 2
Year 3
Year 4
Year 5
Year 6
A
1
A
2
A
3
A
4
B
1
B
2
B
3
C
2
D
5
Ending
Balance
Minimum
Balance
>=
>=
>=
>=
>=
>=
Dollars Invested
(d)
Beginning Balance $60,000
Minimum Balance
$0
Investment
Year
Year 1
Year 2
Year 3
A
1
-1
A
2
A
3
B
1
-1
B
2
-1
1.4
B
3
-1
$0
$0
Ending
Minimum
Balance
Balance
$0
>=
$0
$0
>=
$0
$84,000 >=
$0
-1
-1
Dollars Invested $60,000
C
2
-1
$0
$0
$0
$0
(e)
Beginning Balance $60,000
Minimum Balance
$0
Investment
Year
Year 1
Year 2
Year 3
Year 4
Year 5
Year 6
A
1
-1
A
2
A
3
A
4
B
1
-1
-1
1.4
Dollars Invested $60,000
B
2
B
3
-1
-1
1.4
D
5
-1
-1
-1
1.7
1.4
1.7
1.4
$0 $84,000
C
2
$0
$0
$0
1.7
1.9
-1
1.3
$0
$0
$117,600
Ending
Balance
$0
$0
$0
$0
$0
$152,880
>=
>=
>=
>=
>=
>=
Minimum
Balance
$0
$0
$0
$0
$0
$0
21.8.
In the poor formulation, the data are not separated from the formula - they are buried
inside the equations in column C. In contrast, the spreadsheet in Figure 21.6 separates all
of the data in their own cells, and then the formulas for hours used and total profit refer to
these data cells.
In the poor formulation, no range names are used. The spreadsheet in Figure 21.6 uses
range names for UnitProfit, HoursUsed, TotalProfit, etc.
21-7
The poor formulation uses no borders, shading, or colors to distinguish between cell
types. The spreadsheet in Figure 21.6 uses borders and shading to distinguish the data
cells, changing cells, and target cell.
The poor formulation does not show the entire model on the spreadsheet. There is no
indication of the constraints on the spreadsheet (they are only displayed in the Solver
dialogue box). Furthermore, the right-hand-sides of the constraints are not on the
spreadsheet, but buried in the Solver dialogue box. The spreadsheet in Figure 21.6 shows
all of the constraints of the model in three adjacent cells on the spreadsheet.
21.9.
Cell F16 has -0.23 for LT Interest, rather than -LTRate*LTLoan.
Cell G14 for the 2017 ST Interest uses the LT Loan amount rather than the ST Loan
amount.
Cell H21 for the LT Payback refers to the 2017 ST Loan rather than the LT Loan to
determine the payback amount.
21.10.
Cell G21 for the 2024 ST Interest uses LTRate instead of STRate.
Cell H21 for the LT Payback in 2024 has -4.65 instead of -LTLoan.
Cell I15 for ST Payback in 2018 has -LTLoan instead of -E14 (STLoan for 2017).
21-8
CASES
CASE 21.1 Prudent Provisions for Pensions
(a) PFS needs to know how many units of each of the four bonds to purchase, how much
to invest in the money market, and their ending balance in the money market fund each
year after paying the pensions. The decisions are how many units of each bond to
purchase, as well as the initial investment in 2014 in the money market. The objective is
to minimize the overall initial investment necessary in 2014 in order to meet the pension
payments through 2023.
(b)
Payment received from Bond 1 (2015) œ (10,000 units)($1,000 face value)
 (10,000 units)($1,000 face value)(0.04)
œ $10.4 million
Payment received from Bond 1 (2016) œ $0
Payment received from Bond 2 (2015) œ (10,000 units)($1,000 face value)(0.02)
œ $0.2 million
Payment received from Bond 2 (2016) œ (10,000 units)($1,000 face value)(0.02)
œ $0.2 million
Balance in money market fund (2014) œ $28 million (initial investment)
 $8 million (pension payment)
œ $20 million
Balance in money market fund (2015) œ $20 million (starting balance)
 $10.4 million (payment from Bond 1)
 $0.2 million (payment from Bond 2)
 $12 million (pension payment)
 $0.4 million (money market interest)
œ $19 million
Balance in money market fund (2016) œ $19 million (starting balance)
 $0.2 million (payment from Bond 2)
 $13 million (pension payment)
 $0.38 million (money market interest)
œ $6.58 million
21-9
(c) PFS will need to track the flow of cash from bond investments, the initial investment,
the required pension payments, interest from the money market, and the money market
balance. The decisions are the number of units to purchase of each bond. Data for the
problem include the yearly cash flows from the bonds (per unit purchased), the money
market rate, and the minimum required balance in the money market fund at the end of
each year. A sketch of a spreadsheet model might appear as follows.
Money Market Rate
Minimum Required Balance
Bond Cash Flows (per unit)
Bond 1
Bond 2
Bond 3
Bond 4
Bond
Initial
Flow Investment
Required
Pension
Flow
Money
Market
Interest
2014
2015
2016
2017
2018
2019
2020
2021
2022
2023
Money
Market
Balance
>=
>=
>=
>=
>=
>=
>=
>=
>=
>=
0
0
0
0
0
0
0
0
0
0
Units Purchased
(d) The bond cash flows (per unit) are calculated in B7:E9. For example, one unit of
Bond 1 costs $0.98 in 2014, and returns the face value ($1) plus the coupon rate ($0.04)
in 2015. The total cash flow from bonds is then calculated in column F. The Initial
Investment (G7) is both a decision variable and the target cell. It includes all money
invested on January 1, 2014 (including enough to pay for the bonds and pension payment
in 2014, as well as any initial investment in the money market).
If just years 2014 through 2016 are considered, then 23.79 thousand units of Bond 1
should be purchased at a cost of $23.32 million, along with an initial $8 million
investment in the money market fund on January 1, 2014.
Money Market Rate
Minimum Required Balance
2014
2015
2016
Bond Cash Flows (per unit)
Bond 1
Bond 2
Bond 3
Bond 4
-0.98
-0.92
-0.75
-0.80
1.04
0.02
0.03
0.02
0.03
Units Purchased
(thousands)
23.79
0
0
0
Cost of Bonds
0.98
0.92
0.75
0.8
Bond
Initial
Flow Investment
-23.32
31.32
24.75
0.00
Required
Pension
Flow
-8
-12
-13
2%
0
Money
Market
Interest
0.00
0.25
all cash figures in $millions
21-10
Money
Market
Balance
0.00
>= 0
12.75 >= 0
0.00
>= 0
(e) Expanded to consider all years through 2023, the spreadsheet is as shown below. PFS
should purchase 20.26 thousand units of Bond 1, 26.53 thousand units of Bond 2, 52.89
thousand units of Bond 3, and 44.20 thousand units of Bond 4 (at a cost of $119.29
million), and invest an additional $8 million in the money market on January 1, 2014.
Money Market Rate
Minimum Required Balance
2014
2015
2016
2017
2018
2019
2020
2021
2022
2023
Units Purchased
(thousands)
Bond Cash Flows (per unit)
Bond 1
Bond 2
Bond 3
Bond 4
-0.98
-0.92
-0.75
-0.80
1.04
0.02
0.03
0.02
0.03
1.02
0.03
0.03
1.00
0.03
0.03
0.03
1.03
20.26
26.53
52.89
Bond
Initial
Flow Investment
-119.29 127.29
22.92
1.86
28.39
1.33
54.22
1.33
1.33
45.53
0.00
44.20
21-11
Required
Pension
Flow
-8
-12
-13
-14
-16
-17
-20
-21
-22
-24
2%
0
Money
Market
Interest
0.00
0.22
0.00
0.29
0.00
0.74
0.39
0.00
0.47
all cash figures in $millions
Money
Market
Balance
0.00
10.92
0.00
14.39
0.00
37.22
19.29
0.00
23.53
0.00
>=
>=
>=
>=
>=
>=
>=
>=
>=
>=
0
0
0
0
0
0
0
0
0
0
CHAPTER 22: PROJECT MANAGEMENT WITH PERT/CPM
22.2-1.
22.2-2.
22-1
22.2-3.
22-2
22.3-1.
(a)
(b)
Start Ä E Ä F Ä N Ä P Ä Finish
Start Ä G Ä H Ä N Ä P Ä Finish
Start Ä I Ä J Ä N Ä P Ä Finish
Start Ä K Ä L Ä M Ä N Ä P Ä Finish
Start Ä O Ä P Ä Finish
Hence, Start Ä E Ä F Ä N Ä P Ä Finish is the critical path.
22-3
Length œ (& minutes
Length œ %& minutes
Length œ (# minutes
Length œ '( minutes
Length œ %& minutes
(c) - (d) - (e)
Critical Path: Start Ä E Ä F Ä N Ä P Ä Finish
(f) Dinner will be delayed three minutes because of the phone call. If the food processor
is used, dinner will not be delayed, since there was a slack of three minutes, five minutes
of cutting time is saved and the call used only six minutes of these eight minutes.
22.3-2.
(a)
Start Ä E Ä G Ä Finish
Start Ä E Ä H Ä I Ä Finish
Start Ä F Ä G Ä Finish
Start Ä F Ä H Ä I Ä Finish
Length œ % weeks
Length œ ( weeks
Length œ & weeks
Length œ ) weeks
Hence, Start Ä F Ä H Ä I Ä Finish is the critical path.
(b)
Critical Path: Start Ä F Ä H Ä I Ä Finish
(c) No, this will not shorten the length of the project because the activity is not on the
critical path.
22-4
22.3-3.
(a)
Start Ä E Ä H Ä Finish
Start Ä E Ä I Ä Finish
Start Ä E Ä J Ä O Ä Finish
Start Ä E Ä K Ä L Ä M Ä N Ä Finish
Start Ä F Ä H Ä Finish
Start Ä F Ä G Ä I Ä Finish
Start Ä F Ä G Ä L Ä M Ä N Ä Finish
Start Ä F Ä G Ä O Ä Finish
Length œ % weeks
Length œ & weeks
Length œ ) weeks
Length œ ) weeks
Length œ $ weeks
Length œ ' weeks
Length œ ) weeks
Length œ ( weeks
Critical Paths: Start Ä E Ä J Ä O Ä Finish
Start Ä E Ä K Ä L Ä M Ä N Ä Finish
Start Ä F Ä G Ä L Ä M Ä N Ä Finish
(b)
Critical Paths: Start Ä E Ä J Ä O Ä Finish
Start Ä E Ä K Ä L Ä M Ä N Ä Finish
Start Ä F Ä G Ä L Ä M Ä N Ä Finish
(c) No, this will not shorten the length of the project because E is not on all of the critical
paths.
22.3-4.
(a)
Start Ä E Ä H Ä L Ä Q Ä Finish
Start Ä F Ä I Ä N Ä Q Ä Finish
Start Ä G Ä J Ä O Ä R Ä Finish
Start Ä E Ä M Ä Q Ä Finish
Start Ä G Ä K Ä P Ä R Ä Finish
Critical Paths: Start Ä F Ä I Ä N Ä Q Ä Finish
Start Ä G Ä K Ä P Ä R Ä Finish
22-5
Length œ "* weeks
Length œ #! weeks
Length œ "' weeks
Length œ "( weeks
Length œ #! weeks
(b)
Ken will be able to meet his deadline.
(c) Critical Paths:
Start Ä F Ä I Ä N Ä Q Ä Finish
Start Ä G Ä K Ä P Ä R Ä Finish
Focus attention on activities with no slack.
(d) If activity M takes two more weeks, there will be no delay because its slack is three. If
activity L takes two extra weeks, then there will be a delay of one week because its slack
is only one week. If activity N takes two more weeks, there will be a delay of two weeks,
since it has no slack.
22-6
22.3-5.
(a)
(b)
Critical Path: Start Ä E Ä I Ä J Ä Finish
(c) 6 months
22-7
22.3-6.
Critical Path: Start Ä E Ä F Ä G Ä H Ä K Ä L Ä Q Ä Finish
Total duration: (! weeks
22.3-7.
Critical Path: Start Ä E Ä F Ä G Ä I Ä J Ä N Ä O Ä R Ä Finish
Total duration: #' weeks
22-8
22.3-8.
Critical Path: Start Ä E Ä F Ä G Ä I Ä J Ä N Ä O Ä R Ä Finish
Start Ä E Ä F Ä G Ä I Ä J Ä N Ä P Ä R Ä Finish
Total duration: #) weeks
22.4-1.
.œ
9%7:
'
œ
$!%Ð$'Ñ%)
'
œ $(
%)$!
5# œ  :9
'  œ  '  œ *
#
#
22.4-2.
(a)
Start Ä E Ä I Ä M Ä Finish
Start Ä E Ä G Ä J Ä M Ä Finish
Start Ä F Ä H Ä K Ä N Ä Finish
Start Ä F Ä L Ä N Ä Finish
Length œ "( months
Length œ "( months
Length œ "( months
Length œ ") months
Critical Path: Start Ä F Ä L Ä N Ä Finish
(b)
..:
5:
œ
##")
$"
œ !Þ(") Ê PÖX Ÿ ##× ¸ !Þ((
(c) Start Ä E Ä I Ä M Ä Finish:
..:
5:
œ
Start Ä E Ä G Ä J Ä M Ä Finish:
Start Ä F Ä H Ä K Ä N Ä Finish:
##"(
#&
..:
5:
..:
5:
œ
œ
œ " Ê PÖX Ÿ ##× ¸ !Þ)%
##"(
#(
##"(
#)
œ !Þ*'# Ê PÖX Ÿ ##× ¸ !Þ)%
œ !Þ*%& Ê PÖX Ÿ ##× ¸ !Þ)%
(d) There is approximately a ((% chance that the drug will be ready in ## weeks.
22-9
22.4-3.
Start Ä B Ä H Ä J Ä Finish
Start Ä A Ä E Ä I Ä Finish
Mean Critical
Path
18.417

31.201
 
Mean Critical
Path
18.417

31.201
 
P(T<=d) =
where
d=
0.7394
P(T<=d) =
where
d=
22
Start Ä A Ä C Ä F Ä I Ä Finish
Mean Critical
Path
17.583

27.368
 
P(T<=d) =
where
d=
0.7394
22
Start Ä B Ä D Ä G Ä J Ä Finish
Mean Critical
Path
17.833

28.042
 
0.8007
P(T<=d) =
where
d=
22
0.7843
22
There is approximately a ($% chance that the drug will be ready in ## weeks.
22.4-4.
(a)
22-10
(b)
.
Activity
A
B
C
D
E
(c)
4
2
4.83
3
3.17
5#
0.111
0
0.25
0.444
0.25
Start Ä E Ä F Ä G Ä Finish
Start Ä E Ä F Ä I Ä Finish
Start Ä E Ä H Ä I Ä Finish
Length œ "!Þ)$ weeks
Length œ *Þ"( weeks
Length œ "!Þ"( weeks
Critical Path: Start Ä E Ä F Ä G Ä Finish
(d)
..:
5:
œ
"""!Þ)$
!Þ$'"
œ !Þ!#) Ê PÖX Ÿ ""× œ !Þ'
(e) Make the bid, since there is approximately a '!% chance that the project will be
completed in "" weeks or less.
22.4-5.
(a)
Activity
A
B
C
D
E
F
(b)
.
12
23
15
27
18
6
5#
0
16
1
9
4
4
Start Ä E Ä G Ä I Ä J Ä Finish
Start Ä F Ä H Ä Finish
Length œ &" days
Length œ &! days
Critical Path: Start Ä E Ä G Ä I Ä J Ä Finish
(c)
..:
5:
œ
&(&"
*
œ # Ê PÖX Ÿ &(× œ !Þ*((# (Normal Distribution table)
(d)
..:
5:
œ
&(&!
#&
œ "Þ% Ê PÖX Ÿ &(× œ !Þ*"*# (Normal Distribution table)
(e) Ð!Þ*((#ÑÐ!Þ*"*#Ñ œ !Þ)*)#, so the procedure used in (c) overestimates the
probability of completing the project within &( days.
22-11
22.4-6.
(a)
Activity
A
B
C
D
E
F
G
H
I
J
.
32
27.7
36
16
32
53.7
16.7
20.3
34
17.7
5#
1.78
2.78
11.1
0.444
0
32.1
4
2.78
7.11
9
Start Ä E Ä G Ä N Ä Finish
Start Ä F Ä J Ä N Ä Finish
Start Ä F Ä I Ä L Ä Finish
Start Ä F Ä I Ä M Ä Finish
Start Ä F Ä H Ä K Ä L Ä Finish
Start Ä F Ä H Ä K Ä M Ä Finish
(b)
Length œ )&Þ( weeks
Length œ **Þ" weeks
Length œ )! weeks
Length œ *$Þ( weeks
Length œ )!Þ( weeks
Length œ *%Þ% weeks
Critical Path: Start Ä F Ä J Ä N Ä Finish
(c)
..:
5:
œ
"!!**Þ"
%$Þ)*
œ !Þ"$' Ê PÖX Ÿ "!!× œ !Þ%%%$ (Normal Distribution table)
(d) Higher
22.4-7.
(a) TRUE. The optimistic and pessimistic estimates lie at the extremes of what is
possible, p.33.
(b) FALSE. The probability distribution is a Beta distribution, p.33.
(c) FALSE. The mean critical path will turn out to be the longest path in the project
network.
22.5-1.
Activity to Crash
Crash Cost
F
F
H
G
H
G
H
$&ß !!!
$&ß !!!
$'ß !!!
$%ß !!!
$'ß !!!
$%ß !!!
$'ß !!!
22-12
Length of Path
EG FH
"%
"'
"%
"&
"%
"&
"%
"%
"$
"%
"$
"$
"#
"$
"#
"#
22.5-2.
(a) Let BE and BG be the reduction in E and G respectively, due to crashing.
minimize
G œ &!!!BE  %!!!BG
subject to
BE Ÿ $
BG Ÿ #
BE  BG #
BE ß BG !
and
Optimal Solution: ÐBE ß BG Ñ œ Ð!ß #Ñ and G ‡ œ )ß !!!.
(b) Let BF and BH be the reduction in F and H respectively, due to crashing.
minimize
G œ &!!!BF  '!!!BH
subject to
BF Ÿ #
BH Ÿ $
BF  BH %
BF ß BH !
and
Optimal Solution: ÐBF ß BH Ñ œ Ð#ß #Ñ and G ‡ œ ##ß !!!.
22-13
(c) Let BE , BF , BG , and BH be the reduction in the duration of E, F , G , and H
respectively, due to crashing.
minimize
subject to
and
G œ &!!!BE  &!!!BF  %!!!BG  '!!!BH
BE Ÿ $
BF Ÿ #
BG Ÿ #
BH Ÿ $
BE  BG #
BF  BH %
BE ß BF ß BG ß BH !
Optimal Solution: ÐBE ß BF ß BG ß BH Ñ œ Ð!ß #ß #ß #Ñ and G ‡ œ $!ß !!!.
(d) Let B4 be the reduction in the duration of activity 4 due to crashing for
4 œ Eß Fß Gß H. Also let C4 denote the start time of activity 4 for 4 œ Gß H and CFINISH
the project duration.
minimize
subject to
and
G œ &!!!BE  &!!!BF  %!!!BG  '!!!BH
BE Ÿ $ß BF Ÿ #ß BG Ÿ #ß BH Ÿ $
CG !  )  BE
CH !  *  BF
CFINISH CG  '  BG
CFINISH CH  (  BH
CFINISH Ÿ "#
BE ß BF ß BG ß BH ß CG ß CH ß CFINISH !
(e)
Normal
Crash
Time
Time
Activity (months) (months)
A
8
5
B
9
7
C
6
4
D
7
4
Normal
Cost
$25,000
$20,000
$16,000
$27,000
Crash
Cost
$40,000
$30,000
$24,000
$45,000
Maximum Time
Reduction
(months)
3
2
2
3
Crash Cost
per Month
Saved
$5,000
$5,000
$4,000
$6,000
Project Completion Time (months)
Start
Time
0
0
8
7
Time
Reduction
0
2
2
2
Finish
Time
8
7
12
12
12
<=
Max Time
12
Total Cost $118,000
(f) The solution found using LINGO agrees with the solution in (e), i.e., it is optimal to
reduce the duration of activities F, G , and H by two months. Then the entire project
takes 12 months and costs #&  $!  #%  Ð#(  "#Ñ œ "") thousand dollars.
22-14
(g)
Deadline of "" months
Activity
A
B
C
D
Normal
Time
(months)
8
9
6
7
Crash
Time
(months)
5
7
4
4
Normal
Cost
$25,000
$20,000
$16,000
$27,000
Crash
Cost
$40,000
$30,000
$24,000
$45,000
Maximum Time Crash Cost
Reduction
per Month
(months)
Saved
3
$5,000
2
$5,000
2
$4,000
3
$6,000
Project Completion Time (months)
Start
Time
0
0
7
7
Time
Reduction
1
2
2
3
Finish
Time
7
7
11
11
11
<=
Max Time
11
Start
Time
0
0
8
7
Time
Reduction
0
2
1
1
Finish
Time
8
7
13
13
13
<=
Max Time
13
Total Cost $129,000
Deadline of "$ months
Activity
A
B
C
D
Normal
Time
(months)
8
9
6
7
Crash
Time
(months)
5
7
4
4
Normal
Cost
$25,000
$20,000
$16,000
$27,000
Crash
Cost
$40,000
$30,000
$24,000
$45,000
Maximum Time Crash Cost
Reduction
per Month
(months)
Saved
3
$5,000
2
$5,000
2
$4,000
3
$6,000
Project Completion Time (months)
Total Cost $108,000
22.5-3.
(a)
Activity to Crash
B
B
B
Crash Cost
$10,000
$10,000
$10,000
Length of Path B-D
50
49
48
47
(b)
Activity to Crash
C
C
C
E
Crash Cost
$10,000
$10,000
$10,000
$15,000
Length of Path A-C-E-F
51
50
49
48
47
22-15
(c)
Activity
A
B
C
D
E
F
Normal
Time
(days)
12
23
15
27
18
6
Crash
Time
(days)
9
18
12
21
14
4
Normal
Cost
$210,000
$410,000
$290,000
$440,000
$350,000
$160,000
Crash
Cost
$270,000
$460,000
$320,000
$500,000
$410,000
$210,000
Maximum Time Crash Cost
Reduction
per Day
(days)
Saved
3
$20,000
5
$10,000
3
$10,000
6
$10,000
4
$15,000
2
$25,000
Project Completion Time (days)
Total Cost
Start
Time
0
0
12
20
24
41
Time
Reduction
0
3
3
0
1
0
Finish
Time
12
20
24
47
41
47
47
<=
Max Time
47
$1,935,000
22.5-4.
(a)
Activity
Start
A
B
C
D
E
Finish
ES
0
0
3
3
7
8
12
EF
0
3
7
8
10
12
12
LS
0
0
4
3
9
8
12
LF
0
3
8
8
12
12
12
Slack
0
0
1
0
2
0
0
Critical Path
Yes
Yes
No
Yes
No
Yes
Yes
Critical Path: Start Ä E Ä G Ä I Ä Finish
Total Duration: "# weeks
(b) $7,834 is saved by the new plan given below.
Activity to Crash
G
I
H&I
F&G
Activity
E
F
G
H
I
Crash Cost
$"ß $$$
$#ß &!!
$%ß !!!
$%ß $$$
Duration
$ weeks
$ weeks
$ weeks
# weeks
# weeks
Length of Path
EFH EFI EG I
"!
""
"#
"!
""
""
"!
"!
"!
*
*
*
)
)
)
Cost
$&%ß !!!
$'&ß !!!
$&)ß '''
$%"ß &!!
$)!ß !!!
22-16
(c)
Activity
A
B
C
D
E
Normal
Time
(weeks)
3
4
5
3
4
Crash
Time
(weeks)
2
3
2
1
2
Normal
Cost
$54,000
$62,000
$66,000
$40,000
$75,000
Crash
Cost
$60,000
$65,000
$70,000
$43,000
$80,000
Maximum Time Crash Cost
Reduction
per Week
(weeks)
Saved
1
$6,000
1
$3,000
3
$1,333
2
$1,500
2
$2,500
Project Completion Time (weeks)
Start
Time
0
4
3
9
8
Time
Reduction
0
0
0
0
0
Finish
Time
3
8
8
12
12
12
<=
Max Time
12
Start
Time
0
3
3
8
7
Time
Reduction
0
0
1
0
0
Finish
Time
3
7
7
11
11
11
<=
Max Time
11
Start
Time
0
3
3
7
7
Time
Reduction
0
0
1
0
1
Finish
Time
3
7
7
10
10
10
<=
Max Time
10
Start
Time
0
3
3
7
7
Time
Reduction
0
0
1
1
2
Finish
Time
3
7
7
9
9
9
<=
Max Time
9
Total Cost $297,000
Activity
A
B
C
D
E
Normal
Time
(weeks)
3
4
5
3
4
Crash
Time
(weeks)
2
3
2
1
2
Normal
Cost
$54,000
$62,000
$66,000
$40,000
$75,000
Crash
Cost
$60,000
$65,000
$70,000
$43,000
$80,000
Maximum Time Crash Cost
Reduction
per Week
(weeks)
Saved
1
$6,000
1
$3,000
3
$1,333
2
$1,500
2
$2,500
Project Completion Time (weeks)
Total Cost $298,333
Activity
A
B
C
D
E
Normal
Time
(weeks)
3
4
5
3
4
Crash
Time
(weeks)
2
3
2
1
2
Normal
Cost
$54,000
$62,000
$66,000
$40,000
$75,000
Crash
Cost
$60,000
$65,000
$70,000
$43,000
$80,000
Maximum Time Crash Cost
Reduction
per Week
(weeks)
Saved
1
$6,000
1
$3,000
3
$1,333
2
$1,500
2
$2,500
Project Completion Time (weeks)
Total Cost $300,833
Activity
A
B
C
D
E
Normal
Time
(weeks)
3
4
5
3
4
Crash
Time
(weeks)
2
3
2
1
2
Normal
Cost
$54,000
$62,000
$66,000
$40,000
$75,000
Crash
Cost
$60,000
$65,000
$70,000
$43,000
$80,000
Maximum Time Crash Cost
Reduction
per Week
(weeks)
Saved
1
$6,000
1
$3,000
3
$1,333
2
$1,500
2
$2,500
Project Completion Time (weeks)
Total Cost $304,833
22-17
Activity
A
B
C
D
E
Normal
Time
(weeks)
3
4
5
3
4
Crash
Time
(weeks)
2
3
2
1
2
Normal
Cost
$54,000
$62,000
$66,000
$40,000
$75,000
Crash
Cost
$60,000
$65,000
$70,000
$43,000
$80,000
Maximum Time Crash Cost
Reduction
per Week
(weeks)
Saved
1
$6,000
1
$3,000
3
$1,333
2
$1,500
2
$2,500
Project Completion Time (weeks)
Start
Time
0
3
3
6
6
Time
Reduction
0
1
2
1
2
Finish
Time
3
6
6
8
8
8
<=
Max Time
8
Total Cost $309,167
Crash to ) weeks.
22.5-5.
(a) Let B4 be the reduction in the duration of activity 4 and C4 be the start time of activity
4.
minimize
G œ 'BE  "#BF  %BG  'Þ'(BH  "!BI  (Þ$$BJ  &Þ(&BK  )BL
subject to
! Ÿ BE Ÿ # ! Ÿ BF Ÿ "
! Ÿ BI Ÿ " ! Ÿ BJ Ÿ $
CE  &  BE Ÿ CG
CF  $  BF Ÿ CI
CG  %  BG Ÿ CK
CI  &  BI Ÿ CK
CK  *  BK Ÿ CFINISH
! Ÿ CFINISH Ÿ "&
C4 !
! Ÿ BG Ÿ # ! Ÿ BH Ÿ $
! Ÿ BK Ÿ % ! Ÿ BL Ÿ #
CE  &  BE Ÿ CH
CF  $  BF Ÿ CJ
CH  '  BH Ÿ CL
CJ  (  BJ Ÿ CL
CL  )  BL Ÿ CFINISH
(b) Finish Time: "& weeks, total crashing cost: $%&Þ(& million, total cost: $#&*Þ(& million.
22-18
22.5-6.
(a) Let B4 be the reduction in the duration of activity 4 and C4 be the start time of activity 4.
minimize
G œ &BE  (BF  )BG  %BH  &BI  'BJ  $BK  %BL  *BM  #BN
subject to
! Ÿ BE Ÿ % ! Ÿ BF Ÿ $
! Ÿ BJ Ÿ ( ! Ÿ BK Ÿ #
CE  $#  BE Ÿ CG
CF  #)  BF Ÿ CI
CG  $'  BG Ÿ CN
CI  $#  BI Ÿ CL
CJ  &%  BJ Ÿ CN
CK  "(  BK Ÿ CM
CM  $%  BM Ÿ CFINISH
! Ÿ CFINISH Ÿ *#
C4 !
! Ÿ BG Ÿ & ! Ÿ BH Ÿ $
! Ÿ BL Ÿ $ ! Ÿ BM Ÿ %
CF  #)  BF Ÿ CH
CF  #)  BF Ÿ CJ
CH  "'  BH Ÿ CK
CI  $#  BI Ÿ CM
CK  "(  BK Ÿ CL
CL  #!  BL Ÿ CFINISH
CN  ")  BN Ÿ CFINISH
! Ÿ BI Ÿ &
! Ÿ BN Ÿ #
(b) Finish Time: *# weeks, total crashing cost: $%$ million, total cost: $"Þ$)) billion.
22.6-1.
(a)
Activity
Start
A
B
C
D
E
Finish
ES
0
0
3
3
6
6
8
EF
0
3
6
6
8
8
8
Total Duration: ) weeks
22-19
(b) - (c) - (d)
Activity
A
B
C
D
E
Estimated
Duration
(weeks)
3
3
3
2
2
Estimated
Cost
$54,000
$65,000
$68,667
$41,500
$80,000
Start
Time
0
3
3
6
6
Cost Per Week
of Its Duration
$18,000
$21,667
$22,889
$20,750
$40,000
Week
1
$18,000
$0
$0
$0
$0
Week
2
$18,000
$0
$0
$0
$0
Week
3
$18,000
$0
$0
$0
$0
Week
4
$0
$21,667
$22,889
$0
$0
Weekly Project Cost $18,000
Cumulative Project Cost $18,000
$18,000
$36,000
$18,000
$54,000
$44,556
$98,556
Week
5
$0
$21,667
$22,889
$0
$0
Week
6
$0
$21,667
$22,889
$0
$0
Week
7
$0
$0
$0
$20,750
$40,000
Week
8
$0
$0
$0
$20,750
$40,000
$44,556 $44,556 $60,750 $60,750
$143,111 $187,667 $248,417 $309,167
(e)
Michael should concentrate his efforts on activity G , since it is not yet completed.
22.6-2.
(a)
Activity
Start
A
B
C
D
E
F
Finish
ES
0
0
0
6
6
10
11
20
EF
0
6
2
10
11
17
20
20
LS
0
0
4
9
6
13
11
20
LF
0
6
6
13
11
20
20
20
Slack
0
0
4
3
0
3
0
0
The earliest finish time for this project is #! weeks.
(b)
Estimated Estimated
Duration
Cost
Start
Activity (weeks) ($thousands) Time
A
6
420
0
B
2
180
0
C
4
540
6
D
5
360
6
E
7
590
10
F
9
630
11
Cost Per Week
all costs in $thousands
of Its Duration Week Week Week Week Week Week Week Week
($thousands)
1
2
3
4
5
6
7
8
70
70
70
70
70
70
70
0
0
90
90
90
0
0
0
0
0
0
135
0
0
0
0
0
0
135
135
72
0
0
0
0
0
0
72
72
84.286
0
0
0
0
0
0
0
0
70
0
0
0
0
0
0
0
0
Weekly Project Cost ($thousands)
Cumulative Project Cost ($thousands)
160
160
22-20
160
320
70
390
70
460
70
530
70
600
207
807
207
1014
Week Week Week Week Week Week Week Week Week Week Week Week
9
10
11
12
13
14
15
16
17
18
19
20
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
135
135
0
0
0
0
0
0
0
0
0
0
72
72
72
0
0
0
0
0
0
0
0
0
0
0
84.29 84.29 84.29 84.29 84.29 84.29 84.29
0
0
0
0
0
0
70
70
70
70
70
70
70
70
70
207
1221
207
1428
156.3 154.3 154.3 154.3 154.3 154.3 154.3
1584 1739 1893 2047 2201 2356 2510
70
2580
70
2650
70
2720
(c)
Estimated Estimated
Cost Per Week
all costs in $thousands
Duration
Cost
Start of Its Duration Week Week Week Week Week Week Week Week
Activity (weeks) ($thousands) Time ($thousands)
1
2
3
4
5
6
7
8
A
6
420
0
70
70
70
70
70
70
70
0
0
B
2
180
4
90
0
0
0
0
90
90
0
0
C
4
540
9
135
0
0
0
0
0
0
0
0
D
5
360
6
72
0
0
0
0
0
0
72
72
E
7
590
13
84.286
0
0
0
0
0
0
0
0
F
9
630
11
70
0
0
0
0
0
0
0
0
Weekly Project Cost ($thousands)
Cumulative Project Cost ($thousands)
70
70
70
140
70
210
70
280
160
440
160
600
Week Week Week Week Week Week Week Week Week Week Week Week
9
10
11
12
13
14
15
16
17
18
19
20
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
135
135
135
135
0
0
0
0
0
0
0
72
72
72
0
0
0
0
0
0
0
0
0
0
0
0
0
0
84.29 84.29 84.29 84.29 84.29 84.29 84.29
0
0
0
70
70
70
70
70
70
70
70
70
72
816
207
207
205
205 154.3 154.3 154.3 154.3 154.3 154.3 154.3
1023 1230 1435 1640 1794 1949 2103 2257 2411 2566 2720
(d)
3000
Cumulative Project Cost ($thousands)
2500
2000
1500
Early Start
Late Start
1000
500
0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Week
22-21
72
672
72
744
(e)
The project manager should focus attention on activity H, since it is not yet finished and
they are running over budget.
22.6-3.
(a)
Estimated Estimated
Duration
Cost
Start
Activity (weeks) ($thousands) Time
A
6
180
0
B
3
75
0
C
4
120
0
D
4
140
6
E
7
175
3
F
4
80
4
G
6
210
4
H
3
45
10
I
5
125
6
J
4
100
10
K
3
60
8
L
5
50
10
M
6
90
14
N
5
150
15
Cost Per Week
all costs in $thousands
of Its Duration Week Week Week Week Week Week Week Week
($thousands)
1
2
3
4
5
6
7
8
30
30
30
30
30
30
30
0
0
25
25
25
25
0
0
0
0
0
30
30
30
30
30
0
0
0
0
35
0
0
0
0
0
0
35
35
25
0
0
0
25
25
25
25
25
20
0
0
0
0
20
20
20
20
35
0
0
0
0
35
35
35
35
15
0
0
0
0
0
0
0
0
25
0
0
0
0
0
0
25
25
25
0
0
0
0
0
0
0
0
20
0
0
0
0
0
0
0
0
10
0
0
0
0
0
0
0
0
15
0
0
0
0
0
0
0
0
30
0
0
0
0
0
0
0
0
Weekly Project Cost ($thousands)
Cumulative Project Cost ($thousands)
85
85
85
170
85
255
85
340
110
450
110
560
Week Week Week Week Week Week Week Week Week Week Week Week
9
10
11
12
13
14
15
16
17
18
19
20
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
35
35
0
0
0
0
0
0
0
0
0
0
25
25
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
35
35
0
0
0
0
0
0
0
0
0
0
0
0
15
15
15
0
0
0
0
0
0
0
25
25
25
0
0
0
0
0
0
0
0
0
0
0
25
25
25
25
0
0
0
0
0
0
20
20
20
0
0
0
0
0
0
0
0
0
0
0
10
10
10
10
10
0
0
0
0
0
0
0
0
0
0
0
15
15
15
15
15
15
0
0
0
0
0
0
0
30
30
30
30
30
140
980
140
95
50
50
35
25
45
45
45
45
45
1120 1215 1265 1315 1350 1375 1420 1465 1510 1555 1600
22-22
140
700
140
840
(b)
Estimated Estimated
Duration
Cost
Start
Activity (weeks) ($thousands) Time
A
6
180
1
B
3
75
0
C
4
120
0
D
4
140
7
E
7
175
3
F
4
80
8
G
6
210
4
H
3
45
11
I
5
125
9
J
4
100
10
K
3
60
12
L
5
50
10
M
6
90
14
N
5
150
15
Cost Per Week
of Its Duration
($thousands)
30
25
30
35
25
20
35
15
25
25
20
10
15
30
all costs in $thousands
Week Week Week Week Week Week Week Week
1
2
3
4
5
6
7
8
0
30
30
30
30
30
30
0
25
25
25
0
0
0
0
0
30
30
30
30
0
0
0
0
0
0
0
0
0
0
0
35
0
0
0
25
25
25
25
25
0
0
0
0
0
0
0
0
0
0
0
0
35
35
35
35
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
Weekly Project Cost ($thousands)
Cumulative Project Cost ($thousands)
55
55
85
140
85
225
85
310
90
400
Week Week Week Week Week Week Week Week Week Week Week Week
9
10
11
12
13
14
15
16
17
18
19
20
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
35
35
35
0
0
0
0
0
0
0
0
0
25
25
0
0
0
0
0
0
0
0
0
0
20
20
20
20
0
0
0
0
0
0
0
0
35
35
0
0
0
0
0
0
0
0
0
0
0
0
0
15
15
15
0
0
0
0
0
0
0
25
25
25
25
25
0
0
0
0
0
0
0
0
25
25
25
25
0
0
0
0
0
0
0
0
0
0
20
20
20
0
0
0
0
0
0
0
10
10
10
10
10
0
0
0
0
0
0
0
0
0
0
0
15
15
15
15
15
15
0
0
0
0
0
0
0
30
30
30
30
30
115
790
140
930
115
95
95
95
45
45
45
45
45
45
1045 1140 1235 1330 1375 1420 1465 1510 1555 1600
(c)
1800
1600
Cumulative Project Cost ($thousands)
1400
1200
1000
Early Start
800
Late Start
600
400
200
0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Week
22-23
90
490
90
580
95
675
(d)
The project manager should investigate activities H, I and M , since they are not yet
finished and they are running over budget.
22-24
CASE 22.1 "School's Out Forever ..." Alice Cooper
(a)
The estimated project duration equals the length of the longest path in the project
network. To calculate this length, we use the layout of the Excel spreadsheets for
Reliable's project in this chapter. We need to modify the spreadsheet to reflect the
network unique to this case.
22-25
Activity
A
B
C
D
E
F
G
H
I
J
K
L
M
N
O
P
Q
R
S
Time
Description
(days)
Register online
2
Attend orientation
5
Write initial resume
7
Search internet
10
Attend company sessions
25
Review industry, etc.
7
Attend mock interview
4
Submit initial resume
2
Meet resume expert
1
Revise resume
4
Attend career fair
1
Search jobs
5
Decide jobs
3
Bid
3
Write cover letters
10
Submit cover letters
4
Revise cover letters
4
Mail
6
Drop
2
ES
0
0
0
0
0
5
7
7
9
10
14
12
17
20
25
35
39
43
43
Project Duration (days)
Week
EF
LS
2
8
5
5
7
7
10
12
25
0
12
10
11
45
9
14
10
16
14
17
15
21
17
17
20
22
23
46
35
25
39
35
43
39
49
43
45
47
LF
10
10
14
22
25
17
49
16
17
21
22
22
25
49
35
39
43
49
49
Slack
(days) Critical?
8
No
5
No
7
No
12
No
0
Yes
5
No
38
No
7
No
7
No
7
No
7
No
5
No
5
No
26
No
0
Yes
0
Yes
0
Yes
0
Yes
4
No
49
Brent can start the interviews in %* days. The critical steps in the process are:
Start Ä I Ä S Ä T Ä U Ä V Ä Finish.
22-26
(b) We substitute first the pessimistic, then the optimistic estimates for the time values
used in part (a).
Pessimistic Estimates:
Activity
A
B
C
D
E
F
G
H
I
J
K
L
M
N
O
P
Q
R
S
Description
Register online
Attend orientation
Write initial resume
Search internet
Attend company sessions
Review industry, etc.
Attend mock interview
Submit initial resume
Meet resume expert
Revise resume
Attend career fair
Search jobs
Decide jobs
Bid
Write cover letters
Submit cover letters
Revise cover letters
Mail
Drop
Time
(days)
4
10
14
12
30
12
8
6
1
6
1
10
4
8
12
7
9
10
3
Week
EF
LS
4
6
10
0
14
4
12
20
30
6
22
10
22
66
20
18
21
24
27
25
28
31
32
22
36
32
44
66
48
36
55
48
64
55
74
64
67
71
ES
0
0
0
0
0
10
14
14
20
21
27
22
32
36
36
48
55
64
64
Project Duration (days)
LF
10
10
18
32
36
22
74
24
25
31
32
32
36
74
48
55
64
74
74
Slack
(days)
6
0
4
20
6
0
52
4
4
4
4
0
0
30
0
0
0
0
7
Critical?
No
Yes
No
No
No
Yes
No
No
No
No
No
Yes
Yes
No
Yes
Yes
Yes
Yes
No
74
Under the worst-case scenario, Brent will require (% days before he is ready to start
interviewing. The critical path is:
Start Ä F Ä J Ä P Ä Q Ä S Ä T Ä U Ä V Ä Finish.
Optimistic Estimates:
Activity
A
B
C
D
E
F
G
H
I
J
K
L
M
N
O
P
Q
R
S
Description
Register online
Attend orientation
Write initial resume
Search internet
Attend company sessions
Review industry, etc.
Attend mock interview
Submit initial resume
Meet resume expert
Revise resume
Attend career fair
Search jobs
Decide jobs
Bid
Write cover letters
Submit cover letters
Revise cover letters
Mail
Drop
Time
(days)
1
3
5
7
20
5
3
1
1
3
1
3
2
2
3
2
3
4
1
Week
EF
LS
1
9
3
7
5
7
7
11
20
0
8
10
8
29
6
12
7
13
10
14
11
17
11
15
13
18
15
30
23
20
25
23
28
25
32
28
29
31
ES
0
0
0
0
0
3
5
5
6
7
10
8
11
13
20
23
25
28
28
Project Duration (days)
LF
10
10
12
18
20
15
32
13
14
17
18
18
20
32
23
25
28
32
32
Slack
(days)
9
7
7
11
0
7
24
7
7
7
7
7
7
17
0
0
0
0
3
Critical?
No
No
No
No
Yes
No
No
No
No
No
No
No
No
No
Yes
Yes
Yes
Yes
No
32
Under the best-case scenario, Brent will require $# days before he is ready to begin
interviewing. The critical path remains the same as in (a).
22-27
(c) The mean critical path is the path in the project network that would be critical path if
the duration of each activity equals its mean. To compute the mean duration of each
activity, we use the Excel spreadsheet named PERT.
Activity
A
B
C
D
E
F
G
H
I
J
K
L
M
N
O
P
Q
R
S
Time Estimates
o
m
p
1
2
4
3
5
10
5
7
14
7
10
12
20
25
30
5
7
12
3
4
8
1
2
6
1
1
1
3
4
6
1
1
1
3
5
10
2
3
4
2
3
8
3
10
12
2
4
7
3
4
9
4
6
10
1
2
3
On Mean
Critical Path
*
*
*
*
*

2.17
5.5
7.83
9.83
25
7.5
4.5
2.5
1
4.17
1
5.5
3
3.67
9.17
4.17
4.67
6.33
2

0.25
1.36
2.25
0.69
2.78
1.36
0.69
0.69
0
0.25
0
1.36
0.11
1
2.25
0.69
1
1
0.11
Mean Critical
Path
49.333

7.722
 
P(T≤d) = 0.99994
where
d=
60
Now, substitute the mean duration of each activity for the time values.
Activity
A
B
C
D
E
F
G
H
I
J
K
L
M
N
O
P
Q
R
S
Week
Time
Slack
Description
(days) ES
EF
LS
LF
(days) Critical?
Register online
2.167
0
2.1667 6.8333
9
6.8333
No
Attend orientation
5.5
0
5.5
3.5
9
3.5
No
Write initial resume
7.833
0
7.8333 5.5 13.333
5.5
No
Search internet
9.833
0
9.8333 12.167
22
12.167
No
Attend company sessions
25
0
25
0
25
0
Yes
Review industry, etc.
7.5
5.5
13
9
16.5
3.5
No
Attend mock interview
4.5 7.8333 12.333 44.833 49.333
37
No
Submit initial resume
2.5 7.8333 10.333 13.333 15.833
5.5
No
Meet resume expert
1
10.333 11.333 15.833 16.833
5.5
No
Revise resume
4.167 11.333 15.5 16.833
21
5.5
No
Attend career fair
1
15.5
16.5
21
22
5.5
No
Search jobs
5.5
13
18.5
16.5
22
3.5
No
Decide jobs
3
18.5
21.5
22
25
3.5
No
Bid
3.667 21.5 25.167 45.667 49.333 24.167
No
Write cover letters
9.167
25
34.167
25
34.167
0
Yes
Submit cover letters
4.167 34.167 38.333 34.167 38.333
0
Yes
Revise cover letters
4.667 38.333
43
38.333
43
0
Yes
Mail
6.333
43
49.333
43
49.333
0
Yes
Drop
2
43
45
47.333 49.333 4.3333
No
Project Duration (days) 49.333
The mean critical path is the same as in (a). To compute the variance of the project
duration, we use the PERT template again.
The mean and the variance of the mean critical path are . œ %*Þ$$$ and 5# œ (Þ(##.
22-28
(d) We use the PERT template as in part (c). Brent will be ready for his interviews within
'! days with probability **Þ**%%.
(e) The earliest start time for the career fair is day #% and the career fair itself still lasts
one day. To ensure that the earliest start time for the career fair is day #%, we add a
dummy node X with duration #% days to the project network, directly following the
START node and preceding the career fair node O .
22-29
(f) To obtain the mean critical path for the new network and the probability that Brent
will complete the project within '! days, we first use the PERT template to compute the
mean duration for each activity. We add the new node X to the list of activities.
Activity
A
B
C
D
E
F
G
H
I
J
K
L
M
N
O
P
Q
R
S
T
Time Estimates
o
m
p
1
2
4
3
5
10
5
7
14
7
10
12
20
25
30
5
7
12
3
4
8
1
2
6
1
1
1
3
4
6
1
1
1
3
5
10
2
3
4
2
3
8
3
10
12
2
4
7
3
4
9
4
6
10
1
2
3
24
24
24
On Mean
Critical Path
*
*
*
*
*
*
*

2.17
5.5
7.83
9.83
25
7.5
4.5
2.5
1
4.17
1
5.5
3
3.67
9.17
4.17
4.67
6.33
2
24

0.25
1.36
2.25
0.69
2.78
1.36
0.69
0.69
0
0.25
0
1.36
0.11
1
2.25
0.69
1
1
0.11
0
We next substitute these mean duration values for the time values to find the critical path.
We need to add node X to the spreadsheet used in (a).
Activity
A
B
C
D
E
F
G
H
I
J
K
L
M
N
O
P
Q
R
S
T
Week
Time
Slack
Description
(days) ES
EF
LS
LF
(days) Critical?
Register online
2.167
0
2.1667 9.8333
12
9.8333
No
Attend orientation
5.5
0
5.5
6.5
12
6.5
No
Write initial resume
7.833
0
7.8333 8.5 16.333
8.5
No
Search internet
9.833
0
9.8333 15.167
25
15.167
No
Attend company sessions
25
0
25
3
28
3
No
Review industry, etc.
7.5
5.5
13
12
19.5
6.5
No
Attend mock interview
4.5 7.8333 12.333 47.833 52.333
40
No
Submit initial resume
2.5 7.8333 10.333 16.333 18.833
8.5
No
Meet resume expert
1
10.333 11.333 18.833 19.833
8.5
No
Revise resume
4.167 11.333 15.5 19.833
24
8.5
No
Attend career fair
1
24
25
24
25
0
Yes
Search jobs
5.5
13
18.5
19.5
25
6.5
No
Decide jobs
3
25
28
25
28
0
Yes
Bid
3.667
28
31.667 48.667 52.333 20.667
No
Write cover letters
9.167
28
37.167
28
37.167
0
Yes
Submit cover letters
4.167 37.167 41.333 37.167 41.333
0
Yes
Revise cover letters
4.667 41.333
46
41.333
46
0
Yes
Mail
6.333
46
52.333
46
52.333
0
Yes
Drop
2
46
48
50.333 52.333 4.3333
No
Dummy
24
0
24
0
24
0
Yes
Project Duration (days) 52.333
The mean project duration is now &#Þ$$ days and the new mean critical path is:
Start Ä X Ä O Ä Q Ä S Ä T Ä U Ä V Ä Finish.
22-30
We specify this new critical path in the PERT spreadsheet to obtain the probability that
Brent will complete the project within '! days.
Mean Critical
Path
52.333


5.056
 
P(T≤d) = 0.99967
where
d=
60
Brent will be ready for his interviews within '! days with probability **Þ*'(%, which is
slightly less than the probability computed in part (d). This decrease is a result of the
increase in the mean project duration. However, since the variance of the project duration
is smaller than the one found in (d), the probability decreases only slightly.
22-31
CHAPTER 23: ADDITIONAL SPECIAL TYPES OF
LINEAR PROGRAMMING PROBLEMS
23.1-1.
(a) Locations 1, 2, 3 are supply centers and locations 4, 5, 6, 7 are receiving centers.
Shipments can be sent via intermediate points.
(b)
"
#
$
%
&
'
(
.4
"
!
#*
&!
'#
*$
((
M
#!!
#
#"
!
"(
&%
'(
M
%)
#!!
$
&!
"(
!
'!
*)
'(
#&
#!!
%
'#
&%
'!
!
#(
M
$)
#$!
&
*$
'(
*)
#(
!
%(
%#
#'!
'
((
M
'(
M
%(
!
$&
#&!
(
M
%)
#&
$)
%#
$&
!
#'!
=3
#(!
#)!
#&!
#!!
#!!
#!!
#!!
"
#!!
#
(!
"$!
$
%
&
'
(
=3
#(!
#)!
#&!
#!!
#!!
#!!
#!!
(c)
"
#
$
%
&
'
(
.4
#!!
#!!
"&!
&!
#!!
#!!
$!
#$!
"(!
*!
#'!
""!
"%!
#&!
'!
#!!
#'!
The shipping pattern obtained with the northwest corner rule forms a chain where
location 3 ships only to location 3  ".
23-1
(d)
Shipping pattern:
23.1-2.
(a) Let the supply center be year 0 with a supply of 1 and the receiving center be year 3
with a demand of 1. Years 1 and 2 are transshipment points. The parameter table is as
follows:
Years
!
"
#
$
Demand
!
!
Q
Q
Q
!
"
"$
!
Q
Q
!
#
#)
"(
!
Q
!
$
%)
$$
#!
!
"
Supply
"
!
!
!
(b) The transportation problem is the same as above except that all supplies and demands
are increased by one.
23-2
(c) Vogel's approximation
(d) Vogel's approximation prices out optimal.
23.1-3.
(a) Let -345 be the cost of buying a very old car Ð5 œ "Ñ or a moderately old car Ð5 œ #Ñ at
the beginning of year 3 and trading it in at the end of year 4. This cost is the difference of
the purchase price, operating and maintenance costs for years "ß #ß á ß 43" from the
trade in value after 43" years.
-34"
"
#
$
%
"
#%!!
M
M
M
-34#
#
%)!!
#%!!
M
M
$
(%!!
%)!!
#%!!
M
%
"!$!!
(%!!
%)!!
#%!!
"
#
$
%
"
$!!!
M
M
M
#
&!!!
$!!!
M
M
$
(#!!
&!!!
$!!!
M
%
"!(!!
(#!!
&!!!
$!!!
Let -3ß4" œ min Ö-34" ß -34# ×. Let the supply center be year 1 with unit supply and the
demand center be year 5 with unit demand. Years 2, 3, 4 are transshipment points.
-33 œ !, -3" œ M for 3  " and -&4 œ M for 4  &. The following is the parameter table of
this transshipment problem:
Year 3
"
#
$
%
&
Demand
"
!
M
M
M
M
!
#
#%!!
!
M
M
M
!
Year 4
$
%)!!
#%!!
!
M
M
!
%
(#!!
%)!!
#%!!
!
M
!
&
"!$!!
(#!!
%)!!
#%!!
!
"
23-3
Supply
"
!
!
!
!
(b) The cost and requirements table of the equivalent transportation problem is identical
to the one in (a) except that all supplies and demands need to be increased by one.
(c)
"
#
$
%
&
Demand
"
"
#
"
$
%
&
Supply
#
"
"
"
"
Cost: *ß '!!
"
"
"
"
"
"
"
"
#
The optimal solution is to purchase a very old car for year 1 and a moderately old one for
years 2, 3, and 4. The cost of this is $*ß '!!.
23.1-4.
Suppose there are 7 supply centers, 8 receiving centers and : transshipment points.

 -34 B34
78: 78:
minimize
3œ"
4œ"
=
 ÐB34  B43 Ñ œ  .3
4œ"
!
78:
subject to
B34
3
for 3 œ "ß #ß á ß 7
for 3 œ 7"ß á ß 78
for 3 œ 78"ß á ß 78:
!, for all 3 Á 4
This model has the special structure that each decision variable appears in exactly two
constraints, once with a coefficient of " and once with a coefficient of ". The table of
constraint coefficients is:
B"#
"
"
ã
!
B"$
"
!
ã
!
â
â
â
â
B"ß78:
"
!
ã
"
B#"
"
"
ã
!
B#$
!
"
ã
!
â
â
â
â
B#ß78:
!
"
ã
"
â
â
â
â
B78:ß"
"
!
ã
"
B78:ß#
!
"
ã
"
â
â
â
â
B78:ß78:"
!
!
ã
"
23.2-1.
(a)
#B"  %B#  $B$  #B%  &B&  $B'
&B"  #B#  $B$  %B%  #B&  B' Ÿ #!
#B"  %B#
 #B%
 $B' Ÿ '!
Subproblem 1
$B"  #B#  $B$
Ÿ $!
&B"
 B$
Ÿ $!
B"  #B#  B$
#!
Subproblem 2
B%
Ÿ "&
B%
$
Subproblem 3
#B&  B' Ÿ #!
#B&  $B' Ÿ %!
B4 !, for 4 œ "ß #ß á ß '
Maximize
Master Problem
23-4
inequalities to Ÿ inequalities, the coefficient table becomes:
(b) After converting
B"
&
#
$
&
"
B#
#
%
#
#
B$
$
B%
%
#
B&
#
B'
"
$
$
"
"
"
"
#
#
"
$
B"
%
!
!
"
#
B%
#
!
"
"
%
23.2-2.
(a)
Master Problem
Subproblem 1
Subproblem 2
Subproblem 3
Constraint
$
'
#
&
*
"
)
%
(
B#
$
&
B&
%
"
B(
"
%
"
#
"
"
"
$
B$
#
$
B'
!
#
#
!
%
"
(b) The first constraint of Subproblem 1 and the second constraint of Subproblem 3 are
the upper-bound constraints. The second constraint of Subproblem 1 and the first
constraint of Subproblem 2 are the GUB constraints.
23.2-3.
(a)
maximize
(B"  $B#  &B$  %B%  (B&  &B'
subject to
"'B"  (B#  "$B$  )B%  #!B&  "!B' Ÿ "&!
"!B"  $B#  (B$ Ÿ &!
%B"  #B#  &B$ Ÿ $!
'B%  "$B&  *B' Ÿ %&
$B%  )B&  #B' Ÿ #&
B4 !, for 4 œ "ß #ß á ß '
(b)
B"
"'
"!
%
B#
(
$
#
B$
"$
(
&
B%
)
B&
#!
B'
"!
'
$
"$
)
*
#
23-5
23.3-1.
Eœ
$
"
!
#
! , E" œ Ð$Ñ, E# œ Ð#Ñ, E$ œ Ð"Ñ, E% œ Ð#Ñ
#
Ä
Ä
-" œ Ð$Ñ, -# œ Ð&Ñ, B" œ ÐB" Ñ, B# œ ÐB# Ñ, , œ "), ," œ %, ,# œ "#
Subproblem 1: maximize
subject to
D" œ $B"
B" Ÿ %, B"
!
‡
B‡"" œ ! Ä 3"" , B"#
œ % Ä 3"#
Subproblem 2: maximize
subject to
D# œ &B#
#B# Ÿ "#, B#
!
‡
B‡#" œ ! Ä 3#" , B##
œ ' Ä 3##
Reformulate: maximize
subject to
(1) Start with BF œ
"#3"#  $!3##
"#3"#  "#3##  B& œ ")
3""  3"# œ "
3#"  3## œ "
3 !ß B& !
 B& 
 ") 
"
3"" , F œ M œ F " , F " , œ
 3#" 
 " 
4 œ ": minimize
subject to
A" œ $B"
B" Ÿ %, B" ! Ä B‡" œ % œ B‡"# , A‡" œ "#
4 œ #: minimize
subject to
A# œ &B#
#B# Ÿ "#, B#
! Ä B‡# œ ' œ B‡## , A‡# œ $!
Not optimal, A‡#  A‡" , so 3## enters the basis.
E5w œ
 "# 
 ") 
! , F " , œ
" , minimum ratio: "Î", so 3#" leaves the basis.
 " 
 " 
 B& 
(2) BF œ 3"" , -F œ  !
 3## 
!
 " " "# 
 " ! "# 
"
$! , F œ ! " ! , F œ ! "
!
! ! " 
! !
" 
A" œ $B" , B‡" œ % œ B‡"# , A‡" œ "#
A# œ &B#  $!, B‡# œ ' œ B‡## , A‡# œ !
Not optimal, A‡"  A‡# , so 3"# enters the basis.
E5w
 "# 
'
"
" , F , œ " , minimum ratio: 'Î"#, so B& leaves the basis.
œ
 ! 
"
23-6
(3) BF œ
F
"
 3"# 
3"" , -F œ  "#
 3## 
 "Î"#
œ "Î"#
 !
!
"
!
!
$! , F œ
" 
"
"
 "#
"
 !
! "# 
" ! ,
! " 
A" œ !B"  !
A# œ $B#  "), B‡# œ ' œ B‡## , A‡# œ !
-F F " œ "  !, so the solution is optimal, stop.
 3"# 
 "Î# 
"
BF œ 3"" œ F , œ "Î#
 3## 
 " 
Ê B" œ !Ð"Î#Ñ  %Ð"Î#Ñ œ #, B# œ !Ð!Ñ  'Ð"Ñ œ ', D œ $'
23.3-2.
(a) Reformulate:
!
&
&Î#
!
‡
Subproblem 1: B‡"" œ  , B‡"# œ  , B‡"$ œ 
, B"%
œ 

!
!
"&Î#
"!
!
&
"!Î$
!
Subproblem 2: B‡#" œ  , B‡## œ  , B‡#$ œ 
, B‡ œ
!
!
"!Î$  "%  & 
maximize
&!3"# 
"#&
# 3"$
 &!3"%  %!3##  &!3#$  $&3#%
subject to
$!3"# 
"!&
# 3"$
 &!3"%  #!3## 
"!!
$ 3#$
 $!3#%  B& œ %!
3""  3"#  3"$  3"% œ "
3#"  3##  3#$  3#% œ "
3
!ß B&
!
 B& 
 %! 
"
"
" ß -F œ !
(b) Start with BF œ 3"" , F œ M œ F , F , œ
 3#" 
 " 
4 œ ": minimize
subject to
A" œ "!B"  &B#
$B"  B# Ÿ "&, B"  B# Ÿ "!, B" ß B#
!
&Î#
B‡"$ œ 
is optimal, A‡" œ "#&Î#.
"&Î# 
4 œ #: minimize
subject to
B‡#$ œ 
A# œ )B$  (B%
B$  #B% Ÿ "!, #B$  B% Ÿ "!, B$ ß B%
"!Î$
is optimal, A‡# œ &!.
"!Î$ 
Not optimal, A‡"  A‡# , so 3"$ enters the basis.
23-7
!
E5w œ
 "!&Î# 
 %! 
"
" , minimum ratio: )!Î"!&, so B& leaves the basis.
, F " , œ
 ! 
 " 
 3"$ 
(2) BF œ 3"" , -F œ  "#&Î#
 3#" 
F
"
 #Î"!&
œ #Î"!&

!
A" œ  #!
( B" 
!
! !
" !
! "
#!
#" B# ,
 "!&Î#
! , F œ
"
 !
!
! ,
"
!
"
!
B‡"# is optimal, A‡" œ "%Þ#).
"
‡
‡
A# œ  ')
#" B$  ( B% , B## is optimal, A# œ "'Þ"*.
Not optimal, A‡#  A‡" , so 3## enters the basis.
E5w œ
 %!Î"!& 
 )!Î"!& 
%!Î"!& , F " , œ #&Î"!& , minimum ratio: "Î", so 3#" leaves the basis.


 " 
"
 3"$ 
(3) BF œ 3"" , -F œ  "#&Î#
 3## 
F " œ
 #Î"!&
#Î"!&

!
A" œ  #!
( B" 
%!Î"!& 
%!Î"!&

"
!
"
!
#!
#" B# ,
!
 "!&Î#
%! , F œ
"
 !
!
"
!
B‡"# is optimal, A‡" œ "%Þ#).
"
A# œ  ')
#" B$  ( B% 
&!!
#"
 %!, B‡## is optimal, A‡# œ !.
Not optimal, A‡"  A‡# , so 3"# enters the basis.
E5w
#! 
! ,
" 
 '!Î"!& 
 %!Î"!& 
"
œ &&Î"!& , F , œ '&Î"!& , minimum ratio: %!Î'!, so 3"$ leaves the basis.
 ! 
 " 
 3"# 
(4) BF œ 3"" , -F œ  &!
 3## 
F " œ
A" œ
 "Î$!
"Î$!
 !
"!
$ B# ,
!
"
!
!
#Î$ 
#Î$
" 
 $!
%! , F œ
"
 !
!
"
!
#! 
! ,
" 
B‡"" and B‡"# are both optimal, A‡" œ !.
A# œ  %$ B$  $B% 
"!!
$
 %!, B‡## is optimal, A‡# œ !.
-F F " œ &Î$  !, so optimality test holds, stop.
23-8
BF œ
 3"# 
 #Î$ 
3"" œ F " , œ "Î$
 3## 
 " 
Ä
Ä
!
&
"!Î$
&
&
Ê B" œ "$    #$   œ 
, B# œ "  œ  

!
!
!
!
!
Ê B" œ "!Î$, B# œ !, B$ œ &, B% œ !, D œ ##!Î$
23.3-3.
The problem has three subproblems and two linking constraints.
 B&" 
 B&# 


(1) Initial basis: BF œ  3"" , F œ F " œ M , -F œ !


3#"
 3$" 
4 œ ": minimize
subject to
B‡"5
#B"  %B#  $B$ Ÿ "!
(B"  $B#  'B$ Ÿ "&
&B"
 $B$ Ÿ "#
B" ß B# ß B$ !
 "&Î"" 
œ #!Î"" is optimal, A‡" œ #!.
 ! 
4 œ #: minimize
subject to
B‡#5
)B"  &B#  'B$
*B%  (B&  *B'
$B%  B&  #B' Ÿ (
#B%  %B&  $B' Ÿ *
B% ß B& ß B' !
 $Î& 
!
œ
is optimal, A‡# œ #)Þ).
 "$Î& 
4 œ $: minimize
subject to
B‡$5 œ 
'B(  &B)
)B(  &B) Ÿ #&
(B(  *B) Ÿ $!
'B(  %B) Ÿ #!
B( ß B) !
(&Î$(
is optimal, A‡# œ #!Þ*&.
'&Î$( 
A‡# is smallest, so 3#5 enters the basis.
23-9
‡
 E# B#5

E5w œ 

!
"
!
 #
 
$
 
œ

 
!
(

$Î&  
$ 
* 
 $! 
 
!




! 
*Î&
#! 
"$Î&   
 "
 
 œ  ! , F , œ  " 
 

 
!
"
"

"
 ! 
 " 

!
minimum ratio: "Î", so 3#" leaves.
 B&" 
 B&# 


(2) BF œ  3"" , -F œ  !


3#5
 3$" 
!
!
"%%Î&
! , F "
*
" ! !
 ! " ! *Î&

!
œ ! ! "

! ! !
"
! ! !
!
!
!

!

!
"
A" same, A‡" œ #!
Ä
A# œ  * ( * B#  "%%Î&, A‡# œ !
A$ same, A‡$ œ #!Þ*&
A‡$ is smallest, so 3$5 enters the basis.
‡
 %
 E$ B$5    "
 !  
E5w œ 
œ
!

 " 

'
(&Î$(   ")Þ'& 
 #" 
!  '&Î$(    #Þ!$ 
 *"Î& 
 
 "


 œ  ! , F , œ  " 
!
 



!
"
!
  " 
 " 
"
minimum ratio: "Î", so 3$" leaves.
 B&" 
 B&# 


Let BF œ  3""  and continue. This suggests that in the next iteration, 3"" will be


3#5
 3$5 
replaced by 3"5 .
23.4-1.
Constraint
"
#
$
%
&
'
(
B$
!
"
!
"
!
"
!
B'
!
!
!
"
!
"
!
B(
!
!
!
"
!
"
!
B"
$
"
B#
"
#
B%
"
#
!
B&
B)
B"!
&
"
"
"
#
23-10
B*
$
"
#
"
23.4-2.
(a) Let B34 denote the number of units of product 3 to be produced in year 4 for 3 œ "ß #
and 4 œ "ß #ß $. Let C34 denote the number of units of product 3 to be sold in year 4 for
3 œ "ß # and 4 œ "ß #ß $. Let D345 denote the number of units of product 3 to be produced
and stored in year 4 and sold in year 5 , for 3 œ "ß #, 4 œ "ß #ß $, and 5 œ 4"ß 4#ß á ß $.
maximize
$C""  &C#"  %C"#  %C##  &C"$  )C#$
 #D""#  #D#"#  %D""$  %D#"$  #D"#$  #D##$
subject to
B"" Ÿ %
#B#" Ÿ "#
$B""  #B#" Ÿ ")
B""  C""  D""#  D""$ œ !
B#"  C#"  D#"#  D#"$ œ !
B"# Ÿ '
#B## Ÿ "#
$B"#  #B## Ÿ #%
D""#  B"#  C"#  D"#$ œ !
D""#  C"# Ÿ !
D#"#  B##  C##  D##$ œ !
D#"#  C## Ÿ !
B"$ Ÿ $
#B#$ Ÿ "!
$B"$  #B#$ Ÿ "&
D""$  D"#$  B"$  C"$ œ !
D#"$  D##$  B#$  C#$ œ !
B34 !ß C34 !ß D345 !ß for all 3ß 4ß 5 .
(b) Table of constraint coefficients:
23-11
23.5-1.
Constraint
$
(
"
'
#
)
&
*
"!
%
B#
"
"
!
!
"
"
"
!
"
!
B)
!
"
"
!
#
"
#
!
!
"
B"
&
#
#
"
B%
"
$
$
"
B$
#
!
B(
$
!
"
#
#
"
B&
"
!
B*
!
"
B"!
%
!
#
"
%
&
#
"
$
"
&
B'
!
#
"
23.5-2.
(a) Let types 1 and 2 denote raw lumber and plywood respectively. Let B34 be the
thousand board feet of type 3 to be purchased in season 4, for 3 œ "ß # and 4 œ "ß #ß $ß %.
Let C34 be the thousand board feet of type 3 to be sold in season 4, for 3 œ "ß # and
4 œ "ß #ß $ß %. Let D345 be the thousand board feet of type 3 to be purchased and stored in
season 4 and sold in season 5 , for 3 œ "ß #, 4 œ "ß #ß $ß %, and 5 œ 4"ß 4#ß á ß %.
maximize
subject to
%"!B""  %#&C""  "(D""#  #(D""$  $(D""%
')!B#"  (!&C#"  #%D#"#  %#D#"$  '!D#"%
%$!B"#  %%!C"#  "(D"#$  #(D"#%
("&B##  ($!C##  #%D##$  %#D##%
%'!B"$  %'&C"$  "(D"$% ('!B#$  ((!C#$  #%D#$%
%&!B"%  %&&C"% (%!B#%  (&!C#%
B""  C""  D""#  D""$  D""% œ !
B#"  C#"  D#"#  D#"$  D#"% œ !
B""  B#" Ÿ #!!!
C"" Ÿ "!!!
C#" Ÿ )!!
D""#  B"#  C"#  D"#$  D"#% œ !
D""#  C"# Ÿ !
D#"#  B##  C##  D##$  D##% œ !
D#"#  C## Ÿ !
D""#  D""$  D""%  D#"#  D#"$  D#"%  B"#  B## Ÿ #!!!
C"# Ÿ "%!!
C## Ÿ "#!!
D""$  D"#$  B"$  C"$  D"$% œ !
D""$  D"#$  C"$ Ÿ !
D#"$  D##$  B#$  C#$  D#$% œ !
D#"$  D##$  C#$ Ÿ !
D""$  D""%  D"#$  D"#%  D#"$  D#"%  D##$  D##%  B"$  B#$ Ÿ #!!!
C"$ Ÿ #!!!
C#$ Ÿ "&!!
D""%  D"#%  D"$%  B"%  C"% œ !
D#"%  D##%  D#$%  B#%  C#% œ !
D""%  D"#%  D"$%  D#"%  D##%  D#$%  B"%  B#% Ÿ #!!!
C"% Ÿ "'!!
C#% Ÿ "!!
B34 !ß C34 !ß D345 !ß for all 3ß 4ß 5 .
23-12
(b)
23-13
CHAPTER 24: PROBABILITY THEORY
24.1.
(a) The six colored sides: red, white, blue, green, yellow, and violet.
(b) T Ö\ œ !× œ T Ö\ œ "× œ T Ö\ œ #× œ "Î$
(c) IÐ] Ñ œ IÐ\  "Ñ# œ  Ð5  "Ñ# T Ö\ œ 5× œ % #$
#
5œ!


 T ÖA" ∪ A# × œ T ÖA" ×  T ÖA# × œ "Î$  "Î& œ )Î"&
T ÖA$ × œ $Î"!
(a) T\" Ð3Ñ œ 

 T ÖA% × œ "Î'
!
24.2.
(b) IÐ\" Ñ œ " †
)
"&
%†
$
"!
&†
"
'
if 3 œ "
if 3 œ %
if 3 œ &
else
"(
œ # $!


 T ÖA" ∪ A# × œ T ÖA" ×  T ÖA# × œ "Î$  "Î& œ )Î"&
T ÖA$ × œ $Î"!
(c) T\" \# Ð3Ñ œ 

 T ÖA% × œ "Î'
!
(d)
IÐ\"  \# Ñ œ # †
IÐ\# Ñ œ " †  "$ 
)
"&
"
&
&†

$
"! 
$
"!
 "! †
&†
"
'
"
'
if 3 œ #
if 3 œ &
if 3 œ "!
else
(
œ % $!
œ " $#
IÐ\# Ñ œ IÐ\"  \# Ñ  IÐ\" Ñ

!
for ,"  " or ,#  "



 )Î"& for " Ÿ ,"  % and " Ÿ ,#  ∞
for % Ÿ ,"  & and " Ÿ ,#  ∞
(e) J\" \# Ð," ß ,# Ñ œ  &Î'


for % Ÿ ,"  ∞ and " Ÿ ,#  &

 &Î'
for & Ÿ ," and & Ÿ ,#
"
or
(f)
3œ
IÒ\" IÐ\" ÑÓÒ\# IÐ\# ÑÓ
IÒ\" IÐ\" ÑÓ# IÒ\# IÐ\# ÑÓ#
Since IÐ\" Ñ œ ((Î$!, IÐ\"# Ñ œ #)&Î$!, IÐ\# Ñ œ &!Î$!, IÐ\## Ñ œ "&!Î$! and
IÐ\" \# Ñ œ "((Î$!, 3 ¶ !Þ'%.
(g) IÐ#\"  $\# Ñ œ #IÐ\" Ñ  $IÐ\# Ñ œ #Î"&
24-1
24.3.
(a)
GG
GM
GB
MG
MM
MB
BG
BM
BB
(b)
%
$
#
$
#
"
#
"
!
(c)
"Î%
"Î'
"Î"#
"Î'
"Î*
"Î")
"Î"#
"Î")
"Î$'
(d) \ − Ö!ß "ß #ß $ß %×
T Ö\ œ !× œ "Î$',
T Ö\ œ "× œ "Î")  "Î") œ "Î*,
T Ö\ œ #× œ "Î"#  "Î*  "Î"# œ &Î"),
T Ö\ œ $× œ "Î'  "Î' œ "Î$,
T Ö\ œ %× œ "Î%,
T Ö\ œ 5× œ ! for 5  Ö!ß "ß #ß $ß %×.
(e) IÐ\Ñ œ ! † "Î$'  " † "Î*  # † &Î")  $ † "Î$  % † "Î% œ # #$
24.4.
(a) " œ ! 0\ ÐCÑ.C œ ! ).C  ) O.C œ )#  O  O ), so O œ
"
"
)
Ð") Ñ#
Ð") Ñ
œ ")

!


 ,
 ).C œ ),
J\ Ð,Ñ œ  !#
,

)  ) Ð"  )Ñ.C œ )#  Ð"  )Ñ,  Ð"  ) Ñ) œ ,  ) ,  )


"
(b)
(c) IÐ\Ñ œ ! C).C  ) CÐ"  )Ñ.C œ Ð"  )  )# ÑÎ#
)
if ,  !
if ! Ÿ ,  )
if ) Ÿ ,  "
if " Ÿ ,
"
(d) No, a counterexample is obtained by choosing ! Ÿ + Ÿ ) œ "Î$. In that case,
T Ö\  "Î$  +×
œ T Ö\  +  "Î$× œ J\ Ð+  "Î$Ñ
œ Ð+  "Î$Ñ  Ð"Î$ÑÐ+  "Î$Ñ  "Î$ œ Ð%Î$Ñ+  "Î*
T ÖÐ\  "Î$Ñ  +× œ T Ö\   +  "Î$× œ "  J\ Ð  +  "Î$Ñ
œ "  Ð"Î$ÑÐ  +  "Î$Ñ œ Ð"Î$Ñ+  )Î*,
so the equality does not hold.
24-2
24.5.
IÐ\Ñ œ "% B"  $% B# œ ! Ê B" œ $B#
(a)
varÐ\Ñ œ IÐ\ # Ñ  ÒIÐ\ÑÓ# œ IÐ\ # Ñ œ "% B#"  $% B## œ "!
Ê
"
#
% Ð$B# Ñ
B" œ $"!Î$ and B# œ "!Î$
 $% B## œ $B## œ "! Ê 
B" œ $"!Î$ and B# œ "!Î$
(b) Depending on B" and B# , the CDF can be represented as either one of the following
two graphs
24.6.
(a) T Ö\
#&!× œ "  T Ö\  #&!× œ "  ! 0\ ÐCÑ.C œ "  "!!
#&!
œ "   "!!
C 
(b) IÐ\Ñ œ ! C0\ ÐCÑ.C œ
∞
#&!
#&! "!!
C# .C
œ "  #Î&  " œ #Î&
"!!
∞ "!!
"!! .C
C
œ "!!Ðln ∞  ln "!!Ñ œ ∞
24.7.

T Ö"  \  #× œ T Ö\ œ !×  T Ö\ œ "× œ !Þ%



 T Ö\ œ !× œ !Þ$
(a)  T Öl\l Ÿ "× œ T Ö\ œ "×  T Ö\ œ !×  T Ö\ œ "× œ !Þ'



 T Ö\ #× œ T Ö\ œ #× œ T Ö\ œ "×  T Ö\ œ "×
 T Ö\ œ #×  T Ö\ œ "×  T Ö\ œ !×  T Ö\ œ "×  T Ö\ œ #× œ "
Solving this system of equations gives: 5
T Ö\ œ 5×
#
!Þ"
"
!Þ#
!
!Þ$
"
!Þ"
(b)
(c) IÐ\Ñ œ !Þ" † Ð#Ñ  !Þ# † Ð"Ñ  !Þ$ † Ð!Ñ  !Þ" † Ð"Ñ  !Þ$ † Ð#Ñ œ !Þ$
24-3
#
!Þ$
24.8.
(a) " OÐ"  C# Ñ.C œ O C 
"
"
C$
$ "
%O
$
œ
œ" Ê Oœ
$
%
(b)

!
,
J\ Ð,Ñ œ  "
OÐ"  C# Ñ.C œ $% C 

"
C$
$ "
,
if ,  "
œ $% Ð,  "Ñ  "% Ð,$  "Ñ if "
(c) IÐ#\  "Ñ œ #IÐ\Ñ  " œ #" C $% Ð"  C# Ñ.C  " œ $#  C# 
#
"
if "
"
C%
% "
,"
,
 " œ "
Note that IÐ\Ñ œ !.
(d) varÐ\Ñ œ IÐ\ # Ñ  ÒIÐ\ÑÓ# œ IÐ\ # Ñ œ " C# $% Ð"  C# Ñ.C œ "Î&
"
(e) From the Central Limit Theorem, \ is approximately normal with mean IÐ\Ñ and
variance varÐ\Ñ, equivalently \IÐ\Ñ
µ NÐ!ß "Ñ and hence
varÐ\ÑÎ8
T Ö\  !× œ T  \IÐ\Ñ

varÐ\ÑÎ8
24.9.
(a) " œ !
"!!! +
"!!! "
(b) IÐ\Ñ œ !
"!!!

C
"!!! .C
#
C "!!!
" 
œ
IÐ\Ñ
varÐ\ÑÎ8 
+
"!!! C
C
"!!! .C
œ

!
C
#
(c) J\ Ð,Ñ œ  !, "!!!
"  "!!!
.C œ

"
!
(d) J^ Ð,Ñ œ J\ Ð,Î$Ñ œ 
"
,
"&!!

,#
*†1!'

œ T ÖNÐ!ß "Ñ  !× œ !Þ&
"!!!
C#

#!!! !
C#
"
&!!  #
"
&!! C


œ
+
#
Ê +œ#
"!!!
C$

$!!! !
C
#!!! !
#
,
œ $$$ $"
if ,  !
œ
,
&!!

,#
1!'
if ! Ÿ ,  "!!!
if "!!! Ÿ ,
if ,  !
if ! Ÿ ,  $!!!
if $!!! Ÿ ,
24.10.
(a)
T Ö\
#&× œ "  T Ö\ Ÿ #%× œ "  !Þ%($ œ !Þ&#(
T Ö\ œ #!× œ T Ö\ Ÿ #!×  T Ö\ Ÿ "*× œ !Þ")&  !Þ"$% œ !Þ!&"
(b) T Öshortage× œ T Ö\  $&× œ "  T Ö\ Ÿ $&× œ "  !Þ*() œ !Þ!##
24-4
24.11.
(a) IÐ\Ñ œ  #8 Ð"Î#Ñ8 œ  " œ ∞
∞
∞
8œ"
8œ"
Hence, player B should pay ∞ to player A so that the game is fair. Otherwise, the game
can never be made fair.
(b) Since the mean is infinite and IÐ\ # Ñ
not well-defined.
ÒIÐ\ÑÓ# œ ∞, the variance is ∞  ∞, so
(c) T Ö\ Ÿ )× œ T Ö\ œ #×  T Ö\ œ %×  T Ö\ œ )× œ "Î#  "Î%  "Î) œ (Î)
24.12.
(a) " œ T ÖH œ "×  T ÖH œ !×  T ÖH œ "× œ "Î)  &Î)  -Î) œ 'Î)  -Î)
Solving this equation for - gives - œ #.
(b) I /H  œ
#
"
)
†/
&
)
†"
#
)
† / œ ") Ð&  $/Ñ
(c)
24.13.
(a) Let \3 denote the volume of bottle 3 for 3 œ "ß #ß $ and ^ œ \"  \#  \$ .
IÐ^Ñ
œ IÐ\" Ñ  IÐ\# Ñ  IÐ\$ Ñ œ $ † "& œ %&
varÐ^Ñ œ varÐ\" Ñ  varÐ\# Ñ  varÐ\$ Ñ œ $ † Ð!Þ!)Ñ# œ !Þ!"*#
5^
(b)
œ varÐ^Ñ œ !Þ"$*
^ µ NÐ%&ß !Þ!"*#Ñ
T Ö^
%&Þ#× œ T  ^%&
!Þ"$*
%&Þ#%&
!Þ"$* 
œ T ÖNÐ!ß "Ñ
"Þ%%× œ !Þ!(&
24.14.

!
#
(a) J\ Ð,Ñ œ  !, 'CÐ"  CÑ.C œ ' C# 

"
$
C
$
 œ $,#  #,$
,
!
24-5
if ,  !
if ! Ÿ ,  "
if " Ÿ ,
œ ! C'CÐ"  CÑ.C œ ' C$ 
"
C%
% !
$
"
IÐ\Ñ
(b)
œ !Þ&
varÐ\Ñ œ IÐ\ # Ñ  ÒIÐ\ÑÓ# œ ! C# 'CÐ"  CÑ.C  !Þ#&
"
œ ' C% 
%
"
C&

& !
 !Þ#& œ !Þ!&
(c) T Ö\  !Þ&× œ "  T Ö\ Ÿ !Þ&× œ "  Ð$ † !Þ&#  # † !Þ&$ Ñ œ !Þ&
% \& \'
(d) I  \" \# \$ \
œ
'
"
'
% \& \'
(e) var \" \# \$ \
œ
'
† ' † IÐ\" Ñ œ !Þ&
"
$'
† ' † varÐ\" Ñ œ "Î"#!
24.15.
(a) Let \" and \# be the voltage of battery 1 and 2 respectively, and ^ œ \"  \# .
Since
\" µ N" "# ß !Þ!'#&Ñ and \# µ N" "# ß !Þ!'#&Ñ, ^ µ NÐ$ß !Þ"#&Ñ.
T Öfailure× œ T Ö^  #Þ(&×  T Ö^  $Þ#&× œ # † T Ö^  $Þ#&×
œ # † T NÐ!ß "Ñ 
$Þ#&$
!Þ"#& 
œ # † T ÖNÐ!ß "Ñ  !Þ(!(× œ !Þ%)
The second equality is a result of the symmetry of normal distribution.
(b) Chebyshev's Inequality states T Öl\  .l O 5× Ÿ "ÎO # . Hence, the probability
T Ö^  #Þ(&×  T Ö^  $Þ#&× œ T Öl\  .l !Þ#&× Ÿ "ÎÐ!Þ#&Î5 Ñ# and since 5 ¶ !Þ$&%,
the upper bound is "ÎÐ!Þ(!'Ñ# . This value exceeds ", so it is not a useful bound on the
probability.
24.16.
T "!!! †
"
&!!!
† l\  .l Ÿ "& œ !Þ*! Í T Öl\  .l Ÿ (&× œ !Þ*!
Í T Öl\  .l  (&× œ !Þ"! Í T Ö\  .  (&× œ !Þ!&
.l
Í T  l\

5
\
Í
(&
5\
(&
5\ 
œ !Þ!& Í T NÐ!ß "Ñ 
(&
5\ 
œ !Þ!&
#
œ "Þ'%& Í 5\ œ %&Þ' or 5\
¶ #!(*
#
#
Since 5\
œ 5\
Î8, #!(* œ %!!!!Î8 Ê 8 œ "*Þ#%. Hence, choosing 8
sufficient.
24-6
#! is
24.17.
(a) 0\" Ð=Ñ œ ∞ 0\" ß\# Ð=ß >Ñ.>
∞
Let . œ
=.\"
5\"
0\" Ð=Ñ œ
œ
>.\#
5\ #
so that .> œ 5\# .@.
"
 ∞ exp " # Ð.#
#Ð"3 Ñ
#15\" "3# ∞
 #3./  / # Ñ.@
"
 ∞ exp " # Ð/ #
#Ð"3 Ñ
#15\" "3# ∞
Now let D œ
0\" Ð=Ñ œ
and / œ
 #3./  3# .#  3# .#  .# Ñ.@
so that .@ œ "  3# .D .
/ 3.
"3#
expÐ.# Î#Ñ  ∞
#
∞ expÐD Î#Ñ.D
#15\"
expÐ.# Î#Ñ
#15\"
œ
† #1 œ
"
" =.\"
#15\ exp #  5\" 
"
#
Hence, \" µ NÐ.\" ß 5\
Ñ and the same analysis leads to the conclusion
"
#
\# µ NÐ.\# ß 5\# Ñ.
IÒ\" IÐ\" ÑÓÒ\# IÐ\# ÑÓ
5\" 5\#
(b) CorrÐ\" ß \# Ñ œ
œ
Let . œ
=.\"
5\"
and / œ
CorrÐ\" ß \# Ñ œ
œ
Now let D œ
"
∞ ∞
5\" 5\# ∞ ∞ Ð=
œ
>.\#
5\ # .
"
 ∞  ∞ ./ exp " # Ð.#
#Ð"3 Ñ
#1"3# ∞ ∞
"
 ∞ ..
#1"3# ∞
/ 3.
"3#
CorrÐ\" ß \# Ñ œ
 .\" ÑÐ>  .\# Ñ0\" ß\# Ð=ß >Ñ.=.>
.
"
 ∞ ..
#1"3# ∞
3 ∞
#1 ∞ . .
 #3./  / # Ñ. .. /
"
#
./. Î# ∞ . / / exp #Ð"
3# Ñ Ð/  3.Ñ 
∞
#
./. Î# Ò!  3."  3# #1Ó
#
#
./. Î# œ 3
(c) See part (a).
(d) Let . œ
B" .\"
5\"
and / œ
0\" l\# ÐB" lB# Ñ œ
B# .\#
5\# .
0\" ß\# ÐB" ßB# Ñ
0\# ÐB# Ñ
œ

"
#15\ 5\ "3#
"
#
"
#
#
exp #Ð"3# Ñ Ð. #3./ / Ñ
 #15
"
\#
œ
"
"
#15\ "3# exp # 
"
24-7
#
// Î#
5\
"
B# .\" 3 5
ÐB# .\# Ñ
5\" "3#
\#

#

24.18.
(a) " œ "!! &! -.=.> œ #&!!- Ê - œ "Î#&!!
"&!
"!!

!



," ,#

&!
 "!!
(b)
for ,"  "!! or ,#  &!
"
"
#&!! .=.> œ #&!! Ð,"  "!!ÑÐ,#  &!Ñ for "!! Ÿ ,"  "&! and &! Ÿ ,#  "!!
"&! ,# "
Ð,# &!Ñ
&!
J\" \# Ð," ß ,# Ñ œ  "!!
for "&! Ÿ ," and &! Ÿ ,#  "!!
#&!! .=.> œ
&!


," "!! "
Ð,
"!!Ñ
"

 
for "!! Ÿ ,"  "&! and "!! Ÿ ,#

 "!! &! #&!! .=.> œ &!
"
for "&! Ÿ ," and "!! Ÿ ,#
!
for ,"  "!!
," "!! "
Ð," "!!Ñ


J\" Ð," Ñ œ  "!! &! #&!! .=.> œ #&!!
for "!! Ÿ ,"  "&!
"
for "&! Ÿ ,"
!
for ,#  &!
"&! ,# "
Ð,
&!Ñ
#
J\# Ð,# Ñ œ  "!! &! #&!! .=.> œ #&!!
for &! Ÿ ,#  "!!
"
for "!! Ÿ ,#
(c) 0\" Ð=Ñ œ "Î&! for "!! Ÿ =  "&!
0\# l\" œ= Ð>Ñ œ
24.19.
0\" ß\# Ð=ß>Ñ
0\" Ð=Ñ
œ
"Î#&!!
"Î&!
œ
"
&!
for "!! Ÿ =  "&! and 0\# l\" œ= Ð>Ñ œ ! else.
T\" Ð!Ñ œ  T\" ß\# Ð!ß 5Ñ œ "Î#
#
(a)
5œ!
T\" Ð"Ñ œ "  T\" Ð!Ñ œ "Î#
T\# Ð!Ñ œ  T\" ß\# Ð5ß !Ñ œ "Î)
"
5œ!
T\# Ð"Ñ œ  T\" ß\# Ð5ß "Ñ œ $Î)
"
5œ!
T\# Ð#Ñ œ "  T\# Ð!Ñ  T\# Ð"Ñ œ "Î#
(b)
T\" l\# œ" Ð!Ñ œ
T\" ß\# Ð!ß"Ñ
T\# Ð"Ñ
œ
"Î%
$Î)
œ
#
$
T\" l\# œ" Ð"Ñ œ
T\" ß\# Ð"ß"Ñ
T\# Ð"Ñ
œ
"Î)
$Î)
œ
"
$
(c) No, consider T\" l\# œ" Ð!Ñ œ #Î$ Á "Î# œ T\" Ð!Ñ.
(d)
IÐ\" Ñ œ "Î# and varÐ\" Ñ œ "Î%
IÐ\# Ñ œ ""Î) and varÐ\# Ñ œ $"Î'%
24-8
(e)
T\" \# Ð!Ñ œ "Î)
T\" \# Ð"Ñ œ "Î%  ! œ "Î%
T\" \# Ð#Ñ œ "Î)  "Î) œ "Î%
T\" \# Ð$Ñ œ $Î)
24.20.
7
(a) T ÖJ × œ T ÖJ ∩ H× œ T ÖJ ∩ ÐI" ∪ I# ∪ á ∪ I7 Ñ× œ T Ö ∪ ÐJ ∩ I3 Ñ×
œ  T ÖJ ∩ I3 × since T ÖI3 ∩ I4 × œ ! for 3 Á 4
7
3œ"
œ  T ÖJ l I3 ×T ÖI3 × since T ÖJ l I3 × œ
7
3œ"
(b) T ÖI3 l J × œ
T ÖI3 ∩J ×
T ÖJ ×
œ
T ÖI3 ∩J ×
 T ÖJ l I3 ×T ÖI3 ×
7
œ
3œ"
24-9
T ÖJ ∩I3 ×
T ÖI3 ×
T ÖJ l I3 ×T ÖI3 ×
 T ÖJ l I3 ×T ÖI3 ×
7
3œ"
3œ"
CHAPTER 25: RELIABILITY
25.1-1.
The minimal paths for the system are \" \# and \" \$ . Hence,
9Ð\" ß \# ß \$ Ñ œ maxÒ\" \# ß \" \$ Ó œ \" maxÒ\# ß \$ Ó
œ \" Ò"  Ð"  \# ÑÐ"  \$ ÑÓ.
25.1-2.
The minimal paths for the system are \" \# \$ and \" \# \% . Hence,
9Ð\" ß \# ß \$ ß \% Ñ œ maxÒ\" \# \$ ß \" \# \% Ó œ \" \# maxÒ\$ ß \% Ó
œ \" \# Ò"  Ð"  \$ ÑÐ"  \% ÑÓ.
25.2-1.
Note that throughout this chapter we assume that the component reliabilities are independent.
VÐ:" ß :# ß :$ Ñ œ EÒ9Ð\" ß \# ß \$ ÑÓ œ :" Ò"  Ð"  :# ÑÐ"  :$ ÑÓ
25.2-2.
VÐ:" ß :# ß :$ ß :% Ñ œ EÒ9Ð\" ß \# ß \$ ß \% ÑÓ œ :" :# Ò"  Ð"  :$ ÑÐ"  :% ÑÓ
25.3-1.
(a) Yes, 5 œ #, 8 œ $.
(b)
(c) 9Ð\" ß \# ß \$ Ñ œ "  Ð"  \" \# ÑÐ"  \" \$ ÑÐ"  \# \$ Ñ
œ \"# \# \$  \" \## \$  \" \# \$#  \" \#  \" \$  \"# \## \$#
(d) VÐ:" ß :# ß :$ Ñ œ "  Ð"  :" :# ÑÐ"  :" :$ ÑÐ"  :# :$ Ñ
25-1
25.3-2.
(a)
(b) 9Ð\" ß \# ß \$ ß \% ß \& Ñ œ "  Ð"  \" \% ÑÐ"  \# \& ÑÐ"  \# \$ \% Ñ
(c) VÐ>Ñ œ "  Ð"  V" Ð>ÑV% Ð>ÑÑÐ"  V# Ð>ÑV& Ð>ÑÑÐ"  V# Ð>ÑV$ Ð>ÑV% Ð>ÑÑ
25.3-3.
Let \3 and \3w denote the two units of type 3 œ "ß #ß $. Then, the two systems to be compared can be represented as follows.
System A
System B
9E Ð\" ß \# ß \$ ß \"w ß \#w ß \$w Ñ œ ÒmaxÐ\" ß \"w ÑÓÒmaxÐ\# ß \#w ÑÓÒmaxÐ\$ ß \$w ÑÓ
9F Ð\" ß \# ß \$ ß \"w ß \#w ß \$w Ñ œ maxÐ\" \# \$ ß \"w \#w \$w Ñ
ÒmaxÐ\" ß \"w ÑÓÒmaxÐ\# ß \#w ÑÓÒmaxÐ\$ ß \$w ÑÓ
\" \# \$
ÒmaxÐ\" ß \"w ÑÓÒmaxÐ\# ß \#w ÑÓÒmaxÐ\$ ß \$w ÑÓ
\"w \#w \$w
Hence, 9E Ð\ß \ w Ñ maxÐ\" \# \$ ß \"w \#w \$w Ñ œ 9F Ð\ß \ w w Ñ and system A is more
reliable than system B.
25-2
25.4-1.
(a)
Minimal paths: \" \$ and \# \%
Minimal cuts: \" \# , \" \% , \# \$ and \$ \%
(b) From the minimal path representation:
9Ð\" ß \# ß \$ ß \% Ñ œ maxÒ\" \$ ß \# \% Ó œ "  Ð"  \" \$ ÑÐ"  \# \% Ñ
VÐ:" ß :# ß :$ ß :% Ñ œ "  Ð"  :" :$ ÑÐ"  :# :% Ñ.
If :3 œ : œ !Þ*! for all 3, VÐ:Ñ œ !Þ*'$*.
(c)
Upper bound œ "  Ð"  :" :$ ÑÐ"  :# :% Ñ
Lower bound œ Ð"  ;" ;# ÑÐ"  ;" ;% ÑÐ"  ;# ;$ ÑÐ"  ;$ ;% Ñ
where ;3 œ "  :3 . If :3 œ : œ !Þ*! for all 3, then the upper bound is !Þ*'$* and the
lower bound is !Þ*'!'!.
25.4-2.
(a)
Minimal paths: \" \& , \" \$ \% , \# \$ \& and \# \%
Minimal cuts: \" \# , \" \$ \% , \# \$ \& and \% \&
(b) VÐ:" ß :# ß :$ ß :% ß :& Ñ
œ PÖÐ\" \& œ "Ñ ∪ Ð\" \$ \% œ "Ñ ∪ Ð\# \$ \& œ "Ñ ∪ Ð\# \% œ "Ñ×
œ PÐ\" \& œ "Ñ  PÐ\" \$ \% œ "Ñ  PÐ\# \$ \& œ "Ñ  PÐ\# \% œ "Ñ
 PÐ\" \$ \% \& œ "Ñ  PÐ\" \# \$ \& œ "Ñ  PÐ\" \# \% \& œ "Ñ
 PÐ\" \# \$ \% \& œ "Ñ  PÐ\" \# \$ \% œ "Ñ  PÐ\# \$ \% \& œ "Ñ
 PÐ\" \# \$ \% \& œ "Ñ  PÐ\" \# \$ \% \& œ "Ñ  PÐ\" \# \$ \% \& œ "Ñ
 PÐ\" \# \$ \% \& œ "Ñ  PÐ\" \# \$ \% \& œ "Ñ
œ :" :&  :" :$ :%  :# :$ :&  :# :%  :" :$ :% :&  :" :# :$ :&  :" :# :% :&  :" :# :$ :%
 :# :$ :% :&  #:" :# :$ :% :&
If :3 œ : œ !Þ*! for all 3, VÐ:Ñ œ !Þ*()%).
(c)
Upper bound œ "  Ð"  :" :& ÑÐ"  :" :$ :% ÑÐ"  :# :$ :& ÑÐ"  :# :% Ñ
Lower bound œ Ð"  ;" ;# ÑÐ"  ;" ;$ ;% ÑÐ"  ;# ;$ ;& ÑÐ"  ;% ;& Ñ
where ;3 œ "  :3 . If :3 œ : œ !Þ*! for all 3, then the upper bound is !Þ**($& and the
lower bound is !Þ*()"%.
25.4-3.
(a)
Minimal paths: \" \# and \# \$
Minimal cuts: \" \$ and \#
(b) From the minimal path representation:
9Ð\" ß \# ß \$ Ñ œ maxÒ\" \# ß \# \$ Ó œ \# Ò"  Ð"  \" ÑÐ"  \$ ÑÓ
VÐ:" ß :# ß :$ Ñ œ :# Ò"  Ð"  :" ÑÐ"  :$ ÑÓ œ :" :#  :# :$  :" :# :$ .
If :3 œ : œ !Þ*! for all 3, VÐ:Ñ œ !Þ)*".
25-3
(c)
Upper bound œ "  Ð"  :" :# ÑÐ"  :# :$ Ñ
Lower bound œ Ð"  ;" ;$ ÑÐ"  ;# Ñ
where ;3 œ "  :3 . If :3 œ : œ !Þ*! for all 3, then the upper bound is !Þ*'$* and the
lower bound is !Þ)*".
25.4-4.
(a)
Minimal paths: \" \& , \" \$ \' , \# \' and \# \% \&
Minimal cuts: \" \# , \" \% \' , \# \$ \& and \& \'
(b) VÐ:" ß :# ß :$ ß :% ß :& ß :' Ñ
œ PÖÐ\" \& œ "Ñ ∪ Ð\" \$ \' œ "Ñ ∪ Ð\# \' œ "Ñ ∪ Ð\# \% \& œ "Ñ×
œ :" :&  :" :$ :'  :# :'  :# :% :&  :" :$ :& :'  :" :# :& :'  :" :# :% :&  :" :# :$ :'
 :# :% :& :'  :" :# :$ :& :'  :" :# :% :& :'
If :3 œ : for all 3, VÐ:Ñ œ #:#  #:$  &:%  #:& and if : œ !Þ*, then VÐ:Ñ œ !Þ*()%)..
(c)
Upper bound œ "  Ð"  :" :& ÑÐ"  :" :$ :' ÑÐ"  :# :' ÑÐ"  :# :% :& Ñ
Lower bound œ Ð"  ;" ;# ÑÐ"  ;" ;% ;' ÑÐ"  ;# ;$ ;& ÑÐ"  ;& ;' Ñ
where ;3 œ "  :3 . If :3 œ : œ !Þ*! for all 3, then the upper bound is !Þ**($& and the
lower bound is !Þ*()"%.
25-5.1.
(a) VÐ>Ñ
/>Î. for > Ÿ . Ê VÐ"Î%Ñ
/Ð"Î%ÑÎ!Þ' ¸ !Þ'&*, so !Þ'&* Ÿ VÐ"Î%Ñ Ÿ ".
(b) VÐ>Ñ Ÿ /A> for >  . where "  .A œ /A> , so we need to find A such that /A œ
"  !Þ'A.
Hence, A ¸ *Î) and ! Ÿ VÐ>Ñ Ÿ /*Î) ¸ !Þ$#&.
25-5.2.
0 Ð>Ñ œ "( >"" />
"
Î(
and VÐ>Ñ œ />
"
Î(
, so <Ð>Ñ œ
0 Ð>Ñ
VÐ>Ñ
œ "( >"" ,
which is nondecreasing if " ", nonincreasing if " Ÿ ". Therefore, the Weibull
distribution is IFR for " " and DFR for " Ÿ ".
25-4
25-5.3.
VÐ>Ñ œ PÖX"  > and X#  >× œ /
 )>
"
/
 )>
#
œ/
> )"  )" 
"
#
,
so the failure rate of the system is exponentially distributed with parameter
Ð"Î)" ÑÐ"Î)# Ñ and as noted in Section 25.5, the exponential distribution is both IFR and
DFR.
25.5-4.
Let \3 denote the failure time of component 3 and \ the failure time of the system. Also
let -3 œ "Î.3 . Then
J Ð>Ñ œ PÖ\ Ÿ >× œ PÖ\" Ÿ >, \# Ÿ >× œ Ð"  /-" > ÑÐ"  /-# > Ñ,
<Ð>Ñ œ
0 Ð>Ñ
"J Ð>Ñ
œ
-" /-" > -# /-# > Ð-" -# Ñ/Ð-" -# Ñ>
.
/-" > /-# > /Ð-" -# Ñ>
Note that <Ð!Ñ œ !.
.<Ð>Ñ
.>
œ
œ
-"# /Ð-" #-# Ñ> -## /Ð#-" -# Ñ> Ð-" -# Ñ# /Ð-" -# Ñ>
Ò/-" > /-# > /Ð-" -# Ñ> Ó#
/Ð-" -# Ñ> Ò-"# /-# > -## /-" > Ð-" -# Ñ# Ó
Ò/-" > /-# > /Ð-" -# Ñ> Ó#
Let OÐ>Ñ œ -"# /-# >  -## /-" >  Ð-"  -# Ñ# and note that:
OÐ!Ñ œ #-" -#  !,
OÐ∞Ñ œ Ð-"  -# Ñ#  !, since -" Á -# and
.OÐ>Ñ
.>
œ -"# -# /-# >  -" -## /-" >  !.
Hence, OÐ>Ñ is a strictly decreasing function of >. It is positive at > œ ! and negative as >
tends to ∞. These together with the continuity imply that OÐ>Ñ œ ! has a unique
solution. Now, suppose OÐ>! Ñ œ ! for some !  >!  ∞.
  ! for >  >
  ! for >  >
OÐ>Ñ œ !
 !
!
for > œ >!
for >  >!
.<Ð>Ñ
.> 
!
œ ! for > œ >!
  ! for >  >!
Then, <Ð>Ñ is increasing for > Ÿ >! and decreasing for > >! . Thus, the system can be IFR
if and only if >! œ ∞. But since OÐ∞Ñ œ Ð-"  -# Ñ# , this can occur if and only if
-" œ -# , which contradicts the assumption that ." Á .# .
25.5-5.
Each component has an exponential failure time. The exponential distribution is IFR and
hence the time to failure distribution of each component is IFRA, so the system of
Problem 25.5-4 is composed of two independent IFRA components. The last paragraph
of Section 25.5 states the result that the time to failure distribution of the system is IFRA.
25-5
CHAPTER 26: THE APPLICATION OF QUEUEING THEORY
26.2-1.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
Service Costs
Salaries of checkers, cost of cash registers
Salaries of firemen, cost of fire trucks
Salaries of toll takers, cost of constructing toll lane
Salaries of repairpersons, cost of tools
Salaries of longshoremen, cost of equipment
Salary of an operator as a function of their experience
Salaries of operators, cost of equipment
Salaries of plumbers, cost of tools
Salaries of employees, cost of equipment
Salaries of typists, cost of typewriters
Waiting Costs
Lost profit from lost business
Cost of destruction due to waiting
Cost of waiting for commuters
Lost profit from lost business
Lost profit from ships not loaded or unloaded
Lost profit/productivity from unused machines
Lost profit/productivity from waiting materials
Lost profit from lost business
Lost profit from lost business
Lost profit from unfinished jobs
26.3-1.
= œ "ß - œ #ß . œ % Ê 3 œ !Þ& Ê T8 œ !Þ&8" and 08 Ð>Ñ œ #/#>
The answers in (a) and (b) are based on the following identities.
 8B8 œ
∞
(i)
B
Ð"BÑ#
8œ!
 8# B8 œ
∞
(ii)
8œ!
(iii)
(iv)
(a)
if lBl  "
#B#
Ð"BÑ$
!, B/αB .B œ

"
α# Ð"
 ∞ B$ /αB .B œ
B
Ð"BÑ#
if lBl  "
 /α,  α,/α, Ñ Ê ! B/αB .B œ
∞
"
α#
'
α%
!
E(WC) œ  Ð"!8  #8# ÑP8 œ "! 8!Þ&8"  # 8# !Þ&8"
∞
∞
∞
8œ!
8œ!
8œ!
#
œ & 8!Þ&8   8# !Þ&8 œ & !Þ& #    #†!Þ& $ 
"!Þ&
"!Þ&
(b)
∞
∞
8œ!
8œ!
∞
E(WC) œ -E[2Ðj Ñ] œ #! Ð#&A  A$ ÑÐ#/#A Ñ.A
œ "!!! A/#A .A  %! A$ /#A .A œ "!! †
∞
∞
"
##
%†
26.3-2.
The answers in (a) and (b) are based on the following identities.
 8B8 œ
∞
(i)
8œ!
 8# B8 œ
∞
(ii)
8œ!
 8$ B8 œ
∞
(iii)
8œ!
(iv)
B
Ð"BÑ#
if lBl  "
#B#
Ð"BÑ$

B
Ð"BÑ#
'B$
Ð"BÑ%

'B#
Ð"BÑ$
!, B/αB .B œ
"
α# Ð"
if lBl  "

B
Ð"BÑ#
if lBl  "
 /α,  α,/α, Ñ Ê ! B/αB .B œ
∞
26-1
"
α#
'
#%
!Þ&
"!Þ&
#
œ #'Þ&
 œ "'
(v)
(a)
,∞ B# /αB .B œ
 #α,  α# ,# Ñ/α,
E(WC) œ "! 8!Þ&8"   '8# !Þ&8"   8$ !Þ&8"
#
&
∞
8œ!
8œ$
8œ'
œ "! †
"
%
"
)
 #! †
%"*
"#)
œ #! 
(b)
"
α$ Ð#
 &% †
"
"'
 *' †
"
$#
 "&! †
"
'%
  8$ !Þ&8"
∞
8œ'
œ #$Þ#($
E(WC) œ #! A#/#A .A  #" A# #/#A .A
"
∞
œ % #"# Ð"  /#  #/# Ñ  % #"$ Ð#  %  %Ñ/# 
œ "  $/#  &/# œ "Þ#("
26.4-1.
- œ %, . œ &, GW œ #!
!
1ÐR Ñ œ  "#!
 "#!  ")!ÐR  "Ñ
for R œ !
for R œ "
for R #
E(WC) œ  1Ð8ÑP8 œ "#! P8  ")! 8P8  ")! P8
∞
∞
∞
∞
8œ!
8œ"
8œ#
8œ#
œ "#!Ð"  P! Ñ  ")!ÐL  P" Ñ  ")!Ð"  P!  P" Ñ œ '!P!  ")!L  '!
=
"
#
$
%
3 œ %Î&&
!Þ)
!Þ%
!Þ#'(
!Þ#
T!
!Þ#!
!Þ%$
!Þ%&
!Þ%%
P
%Þ!
!Þ*&
!Þ)#
!Þ)!
E(WC)
'(#Þ!!
"$'Þ)!
""%Þ'!
""!Þ%!
E(SC)
#!Þ!
%!Þ!
'!Þ!
)!Þ!
E(TC)
'*#Þ!!
"('Þ)!
"(%Þ'!
"*!Þ%!
Hence, s‡ œ $ and E(TC) œ $ "(%Þ'! per hour.
26.4-2.
(a) Model 2 with s œ " fixed, A œ Ö$!ß %!×, - œ #!,
0 Ð. Ñ œ 
%
"#
for . œ $!
for . œ %!
We need to choose between a slow server consisting of only the cashier and a fast one
consisting of the cashier and a box boy.
(b) E(WC) œ -E[2Ðj Ñ] œ -E[Ð!Þ!)Ñj ] œ -Ð!Þ!)ÑW œ !Þ!)L œ !Þ!) .- .
$!
%!
0 ( .)
%
"#
E(WC)
!Þ"'
!Þ!)
Hence, the status quo should be maintained.
26-2
E(TC)
%Þ"'
"#Þ!)
26.4-3.
(a)
P œ "Þ& Ê [ œ
P
-
œ
"Þ&
!Þ#
œ (Þ& Ê [; œ [ 
"
.
œ (Þ& 
"
!Þ"'(
œ "Þ&
Ê P; œ -[; œ !Þ#Ð"Þ&Ñ œ !Þ$
(b)
Template for M/D/1 Queueing Model
Data
0.2
(mean arrival rate)

 0.333333 (mean service rate)
s=
1
(# servers)
(c)
L=
Lq =
Results
1.05
0.45
W=
Wq =
5.25
2.25

0.6
P0 =
0.4
TC(Alternative 1) œ $(!  Ð$"!!ÑÐPÑ œ $##!
TC(Alternative 2) œ $"!!  Ð$"!!ÑÐPÑ œ $#!&
Alternative 2 should be chosen.
26.4-4.
(a)
Template for the M/G/1 Queueing Model



s=
Data
0.05
15
15
1
(mean arrival rate)
(expected service time)
(standard deviation)
(# servers)
L=
Lq =
Results
3.000
2.250
W=
Wq =
60.000
45.000

0.75
P0 =
0.25
L=
Lq =
Results
2.963
2.163
W=
Wq =
59.250
43.250
(b)
Data
0.05
(mean arrival rate)

16
(expected service time)

9.486833
(standard
deviation)

s=
1
(# servers)

0.8
P0 =
0.2
(c) The new proposal shows that they will be slightly better off if they switch to the new
queueing system.
(d)
TC(Status quo) œ $%!  ÐP; ÑÐ$#!Ñ œ $)&/hour
TC(Proposal) œ $%!  ÐP; ÑÐ$#!Ñ œ $)$/hour
26-3
26.4-5.
(a)
Pœ#Ê[ œ
P
-
œ
#
!Þ$
œ 'Þ'( Ê [; œ [ 
"
.
œ 'Þ'( 
"
!Þ#
œ "Þ'(
Ê P; œ -[; œ !Þ$Ð"Þ'(Ñ œ !Þ&
(b)
Template for the M/G/1 Queueing Model



s=
(c)
Data
0.3
3
1.19
1
(mean arrival rate)
(expected service time)
(standard deviation)
(# servers)
L=
Lq =
Results
5.587
4.687
W=
Wq =
18.624
15.624

0.9
P0 =
0.1
TC(Alternative 1) œ $$!!!  Ð$"&!ÑÐPÑ œ $$ß $!!
TC(Alternative 2) œ $#(&!  Ð$"&!ÑÐPÑ œ $$ß &)*
Alternative 1 should be chosen.
26.4-6.
For the status quo, the system has Poisson arrivals with - œ "&, exponential service time
with . œ "&, = œ " and the capacity of the waiting room is O œ %. There is a waiting
cost of '[; for each customer due to loss of good will and also a waiting cost of $%& per
hour when the system is full (i.e., when there are four cars in the system) due to loss of
potential customers.
E(TC) œ E(WC) œ -'[;  %&T% œ 'P;  %&T%
3 œ -Î. œ " Ê T8 œ
"
O"
œ
"
&
for 8 œ !ß "ß #ß $ß %
P œ  8T8 œ "& Ð"  #  $  %Ñ œ #
O
8œ"
P; œ P  Ð"  T! Ñ œ # 
E(TC) œ ' †
'
&
 %& †
"
&
%
&
œ
'
&
œ $"'Þ#! per hour
For Proposal 1, the system has Poisson arrivals with - œ "&, exponential service time
with . œ #! and = œ ". In addition to the waiting cost of 'P; due to loss of good will,
there is an expected waiting cost of $# per customer that waits longer than half an hour
before his car is ready. The expected value of this additional waiting cost is given by:
#-T Öj  !Þ&× œ #-/.Ð"3ÑÎ# œ $!/#Þ& œ #Þ%'.
P; œ
-#
.Ð.-Ñ
œ
##&
#!†&
œ #Þ#&
E(TC) œ $  ' † #Þ#&  #Þ%' œ $")Þ*' per hour,
where $$ is the capitalized cost of the new equipment.
26-4
For Proposal 2, the system has Poisson arrivals with - œ "&, Erlang service time with
. œ $!ß 5 œ # and = œ ". The only waiting cost is 'P; due to loss of good will.
P; œ  5"
#5  .Ð.-Ñ  œ
#
$
%
†
##&
$!†"&
œ !Þ$(&
E(TC) œ "!  #Þ#& œ $"#Þ#& per hour
Hence, Proposal 2 should be adopted.
26.4-7.
(a) The customers are trucks to be loaded or unloaded and the servers are crews. The
system currently has one server.
(b)
Template for the M/M/s Queueing Model
1
0
0
0
0
0
0


s=
Data
1
4
1
(mean arrival rate)
(mean service rate)
(# servers)
W = 0.333333333
W q = 0.083333333
Pr(W > t) = 0.049787
when t =
Results
L = 0.333333333
Lq = 0.083333333
1

0.25
0 Prob(W q > t) = 0.012447
0
when t =
1
n
0
Pn
0
0.75
(c)


s=
Data
1
3
1
Results
L=
0.5
Lq = 0.166666667
(mean arrival rate)
(mean service rate)
(# servers)
W=
0.5
W q = 0.166666667
Pr(W > t) = 0.135335
when t =
1
 0.333333333
Prob(W q > t) = 0.045112
when t =
1
n
Pn
0 0.666666667
(d)


s=
Data
1
2
1
Results
(mean arrival rate)
(mean service rate)
(# servers)
Pr(W > t) = 0.367879
when t =
Prob(W q > t) =
when t =
L=
Lq =
1
0.5
W=
Wq =
0.5

0.5
1
1
0.18394
1
n
0
Pn
0.5
(e) A one person team should not be considered since that would lead to a utilization
factor of 3 œ ", which is not permitted in this model.
26-5
(f) - (g)
TCÐ7Ñ œ Ð$#!ÑÐ7Ñ  Ð$$!ÑÐP; Ñ
TCÐ%Ñ œ Ð$#!ÑÐ%Ñ  Ð$$!ÑÐ!Þ!)$$Ñ œ $)#Þ&!/hour
TCÐ$Ñ œ Ð$#!ÑÐ$Ñ  Ð$$!ÑÐ!Þ"'(Ñ œ $'&/hour
TCÐ#Ñ œ Ð$#!ÑÐ#Ñ  Ð$$!ÑÐ!Þ&Ñ œ $&&/hour
A crew of 2 people will minimize the expected total cost per hour.
(h)
s
"
#
$
%
&
.s œ  s
"Þ!!!
"Þ%"%
"Þ($#
#Þ!!!
#Þ#$'
L œ .s-∞
#Þ%"%
"Þ$''
"Þ!!!
!Þ)!*
Since clearly E(SC)  &!Þ%* for s
E(WC) œ "&L
∞
$'Þ#"
#!Þ%*
"&Þ!!
"$Þ(&
E(SC) œ "!s
"!
#!
$!
%!
&!
E(TC)
∞
&'Þ#"
&!Þ%*
&&Þ!!
'$Þ(&
', it follows that s‡ œ $.
26.4-8.
- œ %, . œ '8, E(R ) œ -ÎÐ.  -Ñ œ %ÎÐ'8  %Ñ
Hourly cost -Ð8Ñ œ ")8  #!E(R ) œ ")8 
)!
'8%
One can easily check that -Ð8Ñ is convex in 8. When 8 is restricted to be integer, -Ð8Ñ
attains its minimum at 8 œ #, so two leaders would minimize the expected hourly cost.
26.4-9.
- œ $, E(T) œ Ð.  $Ñ"
Expected cost -Ð.Ñ œ &.  '!E(T) † - œ &.  ")!Ð.  $Ñ"
- w Ð.Ñ œ &  ")!Ð.  $Ñ#
The derivative is zero at . œ * and -Ð.Ñ is convex in ., so -Ð.Ñ attains its minimum at
. œ *. Equivalently, an hourly wage of $%& minimizes the expected total cost.
26.4-10.
(a) - œ !Þ&, = œ "
Recall: 3 œ
Lœ
=. ,
P! œ "  3, P8 œ Ð"  3Ñ38
.- ,
Lq œ
-#
.Ð.-Ñ
PÐj  >Ñ œ /.Ð"3Ñ> , PÐjq  >Ñ œ 3/.Ð"3Ñ>
Wœ
"
.- ,
Wq œ
.Ð.-Ñ
. œ #: 3 œ !Þ#&, P! œ !Þ(&, P8 œ !Þ(& † !Þ#&8
L œ "Î$, Lq œ !Þ!)$
PÐj  =Ñ œ !Þ!!!&&$, PÐjq  =Ñ œ !Þ!!!"$)
W œ !Þ'(, Wq œ !Þ"(
26-6
. œ ": 3 œ !Þ&, P! œ !Þ&, P8 œ !Þ&8"
L œ ", Lq œ !Þ&
PÐj  =Ñ œ !Þ!)#, PÐjq  =Ñ œ !Þ!%"
W œ #, Wq œ "
. œ #Î$: 3 œ !Þ(&, P! œ !Þ#&, P8 œ !Þ#& † !Þ(&8
L œ $, Lq œ #Þ#&
PÐj  =Ñ œ !Þ%$&, PÐjq  =Ñ œ !Þ$#'
W œ ', Wq œ %Þ&
(b) TC(mean œ !Þ&Ñ œ "Þ'!  !Þ)Ð"Î$Ñ œ "Þ)(
TC(mean œ "Ñ œ !Þ%!  !Þ)Ð"Ñ œ "Þ#!
TC(mean œ "Þ&Ñ œ !Þ#!  !Þ)Ð$Ñ œ #Þ'!
Hence, .‡ œ ".
26.4-11.
Given that = œ ", from the optimality of a single server result,
E(TC) œ G< .  GA L œ G< .  GA  .- - 
. E(TC)
..
.# E(TC)
..#
œ G<  GA  Ð.--Ñ#  œ ! Ê . œ -  -GA ÎG<
œ #GA  Ð.--Ñ$   ! for all .  -.
Assuming GA  ! and GA Á !, E(TC) is strictly convex in . and . œ -  -GA ÎG< is
the unique minimizer.
26.4-12.
E(TC) œ H. 
.E(TC)
..
.# E(TC)
..#
œH
œ
-G
Ð.-Ñ#
#-G
Ð.-Ñ$
'-G
Ð.-Ñ%
$
œ!Ê.œ-
#-GÎH
 ! for all G  !,
$
so E(TC) is strictly convex in . and . œ -  
#-GÎH is the unique minimizer.
26.4-13.
(a) The original design would give a smaller expected number of customers in the system
because of the pooling effect of multiple servers.
(b) The original design is an M/M/2 queue where - œ & and . œ '. Running ProMod, we
find L œ "Þ" from Figure 17.7. The alternative design consists of two M/M/1 queues with
L œ #-ÎÐ.  -Ñ œ "!. This result agrees with the claim in (a).
26.4-14.
(a) Part (a) of Problem 17.6-31 is a special case of Model 3, in which = œ " is fixed and
the goal is to determine the mean arrival rate -, or equivalently the number of machines
assigned to one operator.
26-7
(b)
(i) The resulting system is an M/M/s queue with finite calling population, whose
size equals the total number of machines. The associated decision problem fits Model 1,
with = being unknown.
(ii) The resulting system is a collection of independent M/M/1 queues with finite
calling populations. The appropriate decision model is a combination of Model 2 and
Model 3, since the goal is to determine ., depending on the number of operators
assigned, and -, depending on the number of machines assigned. In this case, = œ " is
fixed.
(iii) This system does not fit any of the models described in section 26.4.
Each of the proposed alternatives allows resource (operator) sharing to some extent in
contrast to the original proposal. Since in the original proposal, the operators would be
idle most of the time, it is reasonable to expect that allowing interaction will result in an
increase of the production rate obtained with the same number of operators. As a
consequence of this, the number of operators needed to achieve a given production rate
will decrease. Then, the question is what could prevent this from happening. In
alternatives (i) and (iii), the travel time, which is not considered in the preceding
argument, may pose a problem. The idle time could turn into travel time rather than
service time. Moreover, in alternative (iii), the service rate of a group of 8 workers can
be smaller than 8 times the individual service rate, since they will not be working
together regularly. This is not the case in alternative (ii), where the members of a crew do
work together regularly; even then, the service rate of a crew of 8 operators may be
strictly less than 8 times the individual rate.
26.4-15.
From Table 17.3:
W" 
W# 
W$ 
"
.
"
.
"
.
=œ"
!Þ!#%
=œ#
!Þ!!!$(
!Þ"&%
!Þ!!(*$
"Þ!$$
!Þ!'&%#
Note that -" œ !Þ#, -# œ !Þ' and -$ œ "Þ#.
s
"
#
critical
%)!Þ!!
(Þ%!
E(WC)
serious stable
*#Þ%!
"#Þ%!
%Þ('
!Þ(*
total
&)%Þ)!
"#Þ*&
E(SC)
E(TC)
%!Þ!!
)!Þ!!
'#%Þ)!
*#Þ*&
Hiring two doctors incurs less cost.
26.5-1.
+ œ , œ - œ . œ $!! and @ œ $ miles/hour œ #'% feet/min
E(T) œ
Ð$!!Ñ# Ð$!!Ñ#
"

#'%
Ð$!!$!!Ñ

Ð$!!Ñ# Ð$!!Ñ#
Ð$!!$!!Ñ 
œ #Þ#( minutes
26-8
26.5-2.
. œ $!, = œ ", -: œ #%, G0 œ #!, G= œ "&, G> œ #&
8 œ ": - œ -: Î8 œ #%, + œ , œ . œ &! and - œ "!!
2E(T) œ
Lœ
.-
"
&!# "!!#
&ß!!!  &!"!!

&!# &!#
&!&! 
œ !Þ!#'( hours
œ%
8 œ #: - œ -: Î8 œ "#, + œ , œ . œ &! and - œ #& by relabeling symmetric areas:
E(T) œ
Lœ
"
&!# #&#
&ß!!!  &!#&
.-
œ

&!# &!#
&!&! 
œ !Þ!")$ hours
#
$
E(TC) œ 8ÒÐG0  G= Ñ  G> L  -G> EÐTÑÓ
8
"
#
#%
"#
E(T)
!Þ!#'(
!Þ!")$
L
%
#Î$
G0  G=
$&
$&
G> L
"!!
&!Î$
So, there should be two facilities.
26-9
-G> EÐTÑ
"'
&Þ&
E(TC)
"&"
""%Þ$$
26.5-3.
The first step is to relabel Location 3 as the origin Ð!ß !Ñ for an ÐBß CÑ coordinate system
by subtracting 450 from all coordinates shown in the following figure.
The probability density function of \ is obtained by using the height of the area assigned
to the tool crib at Location 3 for each possible value of \ œ B and then dividing by the
size of the area, as given in figure 1-(a) below. This then yields the uniform distribution
of l\l shown in 1-(b).
Figure 1 - Probability density functions of (a) X and (b) l\l
Thus, EÐl\lÑ œ
"  "&!
"&! ! B.B
œ (&.
The probability density function of Y is obtained by using the width of the area assigned
to tool crib at Location 3 for each possible value of ] œ C and then dividing by the size
26-10
of the area, as given in figure 2-(a). This then leads to the probability density function of
l] l shown in 2-(b).
Figure 2 - Probability density functions of (a ) ] and (b) l] l
Thus, EÐl] lÑ œ
"  "&!
##& ! C.C

"  %&!
$!! "&! "
#
"&ß!!! (&
E(T) œ #@ ÒEÐl\lÑ  EÐl] lÑÓ œ

C
%&! C.C
œ "$$ $" .
 "$$ "$  œ !Þ!#() hr
26.5-4.
(a) Total area œ Ð#<Ñ#  Ð%<Ñ# œ #!<#
Probability density of \
EÐl\lÑ œ !
&< "
&< B.B
Probability density of l\l
œ #Þ&<
Probability density of ]
EÐl] lÑ œ !
< $
&< C.C
 <
#< #
&< C.C
E(T) œ #@ Ð#Þ&  !Þ*Ñ< œ
Probability density of l] l
œ !Þ*<
'Þ)<
@
26-11
(b) The area is symmetric about Ð!ß !Ñ, so EÐl\lÑ œ EÐl] lÑ and the total area is
&Ð#<Ñ# œ #!<# .
Probability density of \
EÐl\lÑ œ !
< $
&< B.B
 <
$< "
&< B.B
E(T) œ #@ Ð"Þ"  "Þ"Ñ< œ
Probability density of l\l
œ "Þ"<
%Þ%<
@
(c) Total area œ #Ð#<#  <#  !Þ&<# Ñ œ (<#
Probability density of \
EÐl\lÑ œ !
< &
(< B.B
 <
#< "
(<# Ð&<
Probability density of l\l
 #BÑB.B œ
Probability density of ]
EÐl] lÑ œ
"
<
(<#  ! Ð&<
"'
#" <
Probability density of l] l
 CÑC.C  < Ð%<  CÑC.C œ '& <
&
E(T) œ #@  "'
#"  ' < œ
#<
$Þ"*<
@
26-12
(d) Total area œ 'Ð%<# Ñ œ #%<#
Probability density of \
EÐl\lÑ œ
!#< % B.B
"#<

#<%< # B.B
"#<
Probability density of l\l
œ $& <
Probability density of ]
EÐl] lÑ œ !
< '
"#< C.C
Probability density of l] l
 <
E(T) œ #@  &$  &% < œ
$< $
"#< C.C
œ %& <
&Þ)$<
@
26.5-5.
Given G0 œ "!, G7 œ "&, G> œ %!, -: œ *!, @ œ #!ß !!! feet/hour, the expected
loading time is "Î#! hours. For unloading, .7 œ $!7 where 7 is the crew size.
8 œ ": + œ - œ $!!, , œ !, . œ '!!
E(T) œ
Lœ
Ð$!!Ñ# Ð$!!Ñ#
"
#!ß!!!  Ð$!!$!!Ñ
.7 -
œ

Ð'!!Ñ#
'!! 
œ !Þ!%& hours
$
7$
26-13
8 œ #: + œ - œ $!!, , œ !, . œ $!!
E(T) œ
Lœ
Ð$!!Ñ# Ð$!!Ñ#
"
#!ß!!!  Ð$!!$!!Ñ
.7 -
œ
$
#7$

since - œ
Ð$!!Ñ#
$!! 
-:
8
œ !Þ!$! hours
œ %&
8 œ $: The facilities would be located as follows:
Consider Locations 1 and 2, which are symmetric. Each can be labeled as:
with a total area of "$&ß !!!.
Probability density of \
EÐl\lÑ œ !
$!! "
%&! #

B
$!! B.B
Probability density of l\l
œ
%!!
$
26-14
Probability density of ] œ l] l
EÐl] lÑ œ EÐl\lÑ œ
EÐTÑ œ
Lœ
#
%!!
#!ß!!!  $
"$&Î%
$!7"$&Î%
œ
%!!
$

%!!
$ 
*
)7*
œ
%
"&!
œ !Þ!#'(
since - œ
"$&
%
Now consider Location 3. The area would be labeled as follows:
with a total area of *!ß !!!.
Probability density of \
EÐl\lÑ œ !
$!! "
"&! "

B
$!! B.B
Probability density of l\l
œ "!!
Probability density of ] œ l] l
EÐl] lÑ œ EÐl\lÑ œ "!!
E(T) œ
2
#!ß!!! Ð"!!
 "!!Ñ œ !Þ!#! hours
26-15
Lœ
%&Î#
$!7%&Î#
œ
$
%7$
8 œ %: The areas served by the four facilities would be identical to that of Location 3 for
8 œ $, so EÐTÑ œ %Î#!! œ !Þ!#! hours and L œ $ÎÐ%7  $Ñ.
8
"
#
$ L1,L2
L3
%
E(T) in hours
!Þ!%&
!Þ!$!
!Þ!#'(
!Þ!#!
!Þ!#!
L
$
7$
$
#7$
*
)7*
$
%7$
$
%7$
where L1, L2, and L3 represent Locations 1, 2 and 3 respectively.
If 8 œ ", E(TC) œ ÐG0  7G7 Ñ  G> L  -G> E(T)  -G> Î#! where - œ *!.
7
%
&
'
(
L
$
"Þ&
"
!Þ(&
E(T)
!Þ!%&
!Þ!%&
!Þ!%&
!Þ!%&
G0  7G7
(!
)&
"!!
""&
G> L
"#!
'!
%!
$!
-G> E(T)
"'#
"'#
"'#
"'#
-G> Î#!
")!
")!
")!
")!
E(TC)
&$#Þ!!
%)(Þ!!
%)#Þ!!
%)(Þ!!
For 8 œ ", the minimum cost per hour is $%)# with 7 œ '.
If 8 œ #, E(TC) œ #ÒÐG0  7G7 Ñ  G> L  -G> E(T)  -G> Î#!Ó where - œ %&.
7
#
$
%
&
L
$
"
!Þ'
$Î(
E(T)
!Þ!$!
!Þ!$!
!Þ!$!
!Þ!$!
G0  7G7
%!
&&
(!
)&
G> L
"#!
%!
#%
"(Þ"%
-G> E(T)
&%
&%
&%
&%
-G> Î#!
*!
*!
*!
*!
E(TC)
'!)Þ!!
%()Þ!!
%('Þ!!
%*#Þ#*
For 8 œ #, the minimum cost per hour is $%(' with 7 œ %.
If 8 œ $, at Locations 1 and 2 where - œ "$&Î%:
7
#
$
%
L
*Î(
$Î&
*Î#$
E(T)
!Þ!#'(
!Þ!#'(
!Þ!#'(
G0  7G7
%!
&&
(!
G> L
&"Þ%$
#%
"&Þ'&
-G> E(T)
$'
$'
$'
-G> Î#!
'(Þ&
'(Þ&
'(Þ&
E(TC)
"*%Þ*$
")#Þ&!
")*Þ"&
At Location 3 where - œ ##Þ&:
7
"
#
$
L
$
$Î&
"Î$
E(T)
!Þ!#!
!Þ!#!
!Þ!#!
G0  7G7
#&
%!
&&
G> L
"#!
#%
"$Þ$$
-G> E(T)
")
")
")
-G> Î#!
%&
%&
%&
E(TC)
#!)Þ!!
"#(Þ!!
"$"Þ$$
So, for 8 œ $, the minimum cost per hour is #Ð")#Þ&!Ñ  "#( œ %*# with 7 œ $ at
Locations 1 and 2, and 7 œ # at Location 3.
26-16
If 8 œ %, since all areas served are symmetric and each one is same as Location 3 of the
case with 8 œ $, the minimum cost per hour is %Ð"#(Ñ œ &!) with 7 œ #.
The following table summarizes these results.
8
"
#
$
%
7
'
% at both locations
$ at Locations 1 and 2
# at Location 3
# at all locations
E(TC)
%)#
%('
%*#
&!)
Therefore, the best is to have two facilities with a crew size of %.
26-17
CHAPTER 27: FORECASTING
27.4-1.
(a)
(b)
(c)
(d) The demand seems to be rising, so the average forecasting method may be
inappropriate, since it uses older, out of data.
27.4-2.
(a)
(b)
(c)
(d) The averaging method seems to be the best, since all five months of data are relevant
in determining the forecast of sales for the next month.
27.4-3.
27.4-4
27.4-5.
27.4-6.
27-1
27.4-7.
27.4-8.
The forecast remains equal to the best initial guess for the variable and never changes.
The forecast always equals the current value of the variable.
27.4-9.
(a)
Actual demand in April:
Actual demand in May:
(b)
Feb
Jan
March
Jan
Jan
Feb
,
Jan
March
Feb
Feb ,
Feb
Feb
Jan
Jan
Jan
Jan
March
Jan
Feb
March
April
May
June
Forecast
Actual
27.5-1.
(a)
Quarter
Call Volume
Seasonal Factor
(b)
Quarter
Seasonal Factor
Actual Call Volume
27-2
Seasonally Adjusted Call Volume
(c)
Quarter
Two-year Average
Seasonal Factor
(d)
Quarter
Seasonal Factor
Actual Call Volume
Seasonally Adjusted Call Volume
27.5-2.
(a)
Quarter
Unemployment Rate
Seasonal Factor
(b)
Quarter
Seasonal Factor
Act. Unemploy. Rate
Seasonally Adj. Unemploy. Rate
This progression indicates that the state's economy is improving with the unemployment
rate decreasing from 8% to 7% (seasonally adjusted) over the four quarters.
27.5-3.
(a)
Quarter
Three-year Average
(b) Seasonally adjusted value:
Seasonal Factor
forecast:
27-3
(c)
Quarter 1: seasonally adjusted value:
Quarter 2: seasonally adjusted value:
Quarter 3: seasonally adjusted value:
forecast:
forecast:
forecast:
0
0
(d)
Quarter
Seasonal Factor
Avg. House Sales
Seasonally Adjusted Forecast
27.5-4.
(a) - (b) - (c) - (d)
Year
2010
2011
2012
2013
2014
(e) There is a seasonal effect:
Quarter
1
2
3
4
1
2
3
4
1
2
3
4
1
2
3
4
1
2
3
4
down
Sales
6900
6700
7900
7100
8200
7000
7300
7500
9400
9200
9800
9900
11,400
10,000
9400
8400
8800
7600
7500
I
0.965
0.937
1.105
0.993
0.997
0.941
1.086
1.001
1.049
0.992
1.119
1.047
1.113
1.029
1.116
1.036
F
6900
6826
8069
7183
7245
7043
8372
7849
8442
8260
9513
8892
9402
8633
9256
8431
S
7150
7285
7303
7234
7266
7482
7711
7842
8047
8329
8505
8495
8448
8394
8293
8136
, and it is incorporated by the parameter .
(f) There is a substantial error in these estimates, the constant level assumption is not
good enough with
and
.
27.6-1.
27-4
27.6-2.
27.6-3.
27.6-4.
Time
Period
1
2
3
4
5
6
7
8
9
10
11
12
True
Value
15
21
24
32
37
41
40
47
51
53
Latest
Trend
5.00
5.20
4.79
5.39
5.38
5.15
3.93
4.35
4.19
3.66
Estimated
Trend
5.00
5.00
5.04
4.99
5.07
5.13
5.14
4.89
4.79
4.67
4.46
Forecast for next production yield:
Exponential
Smoothing
Forecast
15
20
25
30
35
41
46
50
54
58
62
Forecasting
Error
0
1
1
2
2
0
6
3
3
5
Smoothing Constants
0.2
0.2
Initial Estimates
Average =
Trend =
10
5
Mean Absolute Deviation
MAD =
2.3
Mean Square Error
MSE =
8.8
%
27.7-1.
Quarter
MAD
MSE
Forecast
True Value
Error
sum of forecasting errors
number of forecasts
sum of squares of forecasting errors
number of forecasts
27.7-2.
(a)
Method 1: MAD
Method 2: MAD
(b)
Method 1: MSE
Method 2: MSE
(c) She can use the older data to calculate more forecasting errors and compare MSE and
MAD for a longer time span. This may make her feel more comfortable with her
decision.
27-5
27.7-3.
(a)
(b) MAD
(c) MSE
(d)
27.7-4.
(a) Since sales are relatively stable, the averaging method would be appropriate for
forecasting future sales. This method uses a larger sample size than the last-value
method, which should make it more accurate and since the older data is still relevant, it
should not be excluded, as would be the case in the moving-average method.
(b) Last-Value Method
Time
Period
1
2
3
4
5
6
7
8
9
10
11
12
13
True
Value
23
24
22
28
22
27
20
26
21
29
23
28
Last-Value
Forecast
Forecasting
Error
23
24
22
28
22
27
20
26
21
29
23
28
1
2
6
6
5
7
6
5
8
6
5
Mean Absolute Deviation
MAD =
5.2
Mean Square Error
MSE =
30.6
(c) Averaging Method
Time
Period
1
2
3
4
5
6
7
8
9
10
11
12
13
True
Value
23
24
22
28
22
27
20
26
21
29
23
28
Averaging
Forecast
Forecasting
Error
23
24
23
24
24
24
24
24
24
24
24
24
1
2
5
2
3
4
2
3
5
1
4
27-6
Mean Absolute Deviation
MAD =
3.0
Mean Square Error
MSE =
11.1
(d) Moving-Average Method
Time
Period
1
2
3
4
5
6
7
8
9
10
11
12
13
Moving
Average
Forecast
True
Value
23
24
22
28
22
27
20
26
21
29
23
28
23
25
24
26
23
24
22
25
24
27
Forecasting
Error
5
3
3
6
3
3
7
2
4
(e) Considering the MAD values
Number of previous
periods to consider
n=
3
Mean Absolute Deviation
MAD =
3.9
Mean Square Error
MSE =
17.4
, the averaging method is the best.
(f) Considering the MSE values
, the averaging method is the best.
(g) Unless there is a reason to believe that sales will not continue to be relatively stable,
the averaging method should be the most accurate in the future as well.
27.7-5.
Ben Swanson should choose
for the smoothing constant.
Smoothing Constant
MAD
MSE
27.7-6.
(a) Answers will vary. The averaging or the moving-average methods seem to do a better
job than the last-value method.
(b) For the last-value method, a change in April affects only the forecast of May. For the
averaging method, it affects all forecasts after April and for the moving-average method,
it affects the forecasts for May, June and July.
(c) Answers will vary. The averaging and the moving-average methods seem to do
slightly better than the last-value method.
(d) Answers will vary. The averaging and the moving-average methods seem to do
slightly better than the last-value method.
27-7
27.7-7.
(a) Since the sales level is shifting significantly from month to month and there is no
consistent trend, the last-value method seems to be appropriate. The averaging method
will not do as well because it places too much weight on the old data. The movingaverage method will be better than the averaging method, but it will lag any short-term
trends. The exponential smoothing method will also lag trends by placing too much
weight on the old data. Exponential smoothing with trend will likely not to do well
because the trend is not consistent.
(b)
Last-Value Method
Time
Period
1
2
3
4
5
6
7
8
9
10
11
12
13
True
Value
126
137
142
150
153
154
148
145
147
151
159
166
Last-Value
Forecast
Forecasting
Error
126
137
142
150
153
154
148
145
147
151
159
166
11
5
8
3
1
6
3
2
4
8
7
Averaging
Forecast
Forecasting
Error
126
132
135
139
142
144
144
144
145
145
147
148
11
11
15
14
12
4
1
3
6
14
19
Mean Absolute Deviation
MAD =
5.3
Mean Square Error
MSE =
36.2
Averaging Method
Time
Period
1
2
3
4
5
6
7
8
9
10
11
12
13
True
Value
126
137
142
150
153
154
148
145
147
151
159
166
27-8
Mean Absolute Deviation
MAD =
10.0
Mean Square Error
MSE =
131.4
Moving-Average Method
Time
Period
1
2
3
4
5
6
7
8
9
10
11
12
13
True
Value
126
137
142
150
153
154
148
145
147
151
159
166
Comparing MAD values
value method is the best.
Moving
Average
Forecast
Forecasting
Error
135
143
148
152
152
149
147
148
152
159
15
10
6
4
7
2
4
11
14
Number of previous
periods to consider
n=
3
Mean Absolute Deviation
MAD =
8.1
Mean Square Error
MSE =
84.3
and MSE values
, the last-
(c) Using the template for exponential smoothing with an initial estimate of
, the
following forecast errors are obtained for various values of the smoothing constant .
MAD
MSE
Considering both MAD and MSE, a high value of the smoothing constant seems to be
appropriate.
(d) Using the template for exponential smoothing with trend using an initial estimate of
for the average value and
for the trend, the following forecast errors are obtained
for various values of the smoothing constants and .
MAD MSE
Considering both MAD and MSE, high values of the smoothing constants seem to be
appropriate.
27-9
(e) The management should use the last-value method to forecast sales. Using this
method, the forecast for January of the new year is
. Exponential smoothing with
trend using high smoothing constants, e.g.,
, also works well. With this
method, the forecast for January of the new year is
.
27.7-8.
(a) Answers will vary. The last-value method seems to be the best. Exponential smoothing
with trend is a close second.
(b) For the last-value method, a change in April affects only the forecast for May. For the
averaging method, exponential smoothing with or without trend, it affects all forecasts
after April. For the moving-average method, it affects the forecasts for May, June, and
July.
(c) Answers will vary. The last-value method and exponential smoothing seem to do better
than the others.
(d) Answers will vary. The last-value method and exponential smoothing seem to do better
than the others.
27.7-9.
(a)
MAD
Choose
(b)
.
MAD
Choose
(c)
.
MAD
Choose
.
27.7-10.
(a)
MAD
Choose
(b)
.
MAD
Choose
(c)
.
MAD
Choose
27.7-11.
(a) The time series is not stable enough for the moving-average method.
27-10
.
(b)
Time
Period
1
2
3
4
5
6
7
8
9
10
11
True
Value
382
405
398
421
426
415
443
451
446
464
Time
Period
1
2
3
4
5
6
7
8
9
10
11
True
Value
382
405
398
421
426
415
443
451
446
464
Time
Period
1
2
3
4
5
6
7
8
9
10
11
True
Value
382
405
398
421
426
415
443
451
446
464
Moving
Average
Forecast
Forecasting
Error
395
408
415
421
428
436
447
454
Number of previous
periods to consider
n=
3
26
18
0
22
23
10
17
Mean Absolute Deviation
MAD =
16.6
Mean Square Error
MSE =
346.0
(c)
Exponential
Smoothing
Forecast
380
381
393
396
408
417
416
430
440
443
454
Forecasting
Error
2
24
5
26
18
2
27
21
6
21
Smoothing Constant
0.5
Initial Estimate
Average =
380
Mean Absolute Deviation
MAD =
15
Mean Square Error
MSE =
323
(d)
Latest
Trend
10.50
13.72
9.21
12.32
10.82
5.33
9.94
9.53
5.87
8.20
Estimated
Trend
10.00
10.13
11.02
10.57
11.01
10.96
9.55
9.65
9.62
8.68
8.56
27-11
Exponential
Smoothing
Forecast
380
391
405
414
427
438
441
451
461
466
474
Forecasting
Error
2
14
7
7
1
23
2
0
15
2
Smoothing Constants
0.25
0.25
Initial Estimates
Average =
370
Trend =
10
Mean Absolute Deviation
MAD =
7.3
Mean Square Error
MSE =
105.1
(e) Exponential smoothing with a trend is recommended, since it offers the smallest MAD.
27.7-12.
27-12
Moving-Average Method: The forecasts typically lie below the demands.
Exponential Smoothing: The forecasts typically lie below the demands.
27-13
Exponential Smoothing with Trend: The forecasts are at about the same level as demands
(perhaps slightly above). This indicates that exponential smoothing with trend is the best
method to use hereafter.
27.7-13.
(a)
Year Quarter
1
1
1
2
1
3
1
4
2
1
2
2
2
3
2
4
3
1
3
2
3
3
3
4
True
Value
25
47
68
42
27
46
72
39
24
49
70
44
Type of Seasonality
Quarterly
Quarter
1
2
3
4
Estimate for
Seasonal Factor
0.5497
1.0271
1.5190
0.9042
27-14
(b) Forecast:
Year Quarter
1
1
1
2
1
3
1
4
2
1
2
2
2
3
2
4
3
1
3
2
3
3
3
4
4
1
4
2
4
3
4
4
5
1
5
2
5
3
5
4
6
1
(c)
acre-feet
True
Value
25
47
68
42
27
46
72
39
24
49
70
44
Seasonally Seasonally
Adjusted
Adjusted
Value
Forecast
45
46
45
45
46
46
45
49
46
45
49
47
45
43
47
44
43
48
44
46
48
49
46
#N/A
49
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
Winter:
Year Quarter
1
1
1
2
1
3
1
4
2
1
2
2
2
3
2
4
3
1
3
2
3
3
3
4
4
1
4
2
4
3
4
4
5
1
5
2
5
3
5
4
6
1
Forecasting
Error
47
70
40
26
50
68
43
24
45
72
42
27
0
2
2
1
4
4
4
0
4
2
2
Type of Seasonality
Quarterly
Quarter
1
2
3
4
Seasonal Factor
0.550
1.027
1.519
0.904
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
Mean Absolute Deviation
MAD =
2.4
Mean Square Error
MSE =
8
, Spring:
Summer:
(d) Forecast:
Actual
Forecast
,
, Fall:
acre-feet
True
Value
25
47
68
42
27
46
72
39
24
49
70
44
Seasonally Seasonally
Adjusted
Adjusted
Value
Forecast
45
46
45
45
46
46
45
49
46
45
46
47
46
43
46
44
46
48
46
46
46
49
46
#N/A
46
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
Actual
Forecast
Forecasting
Error
47
69
41
25
48
70
42
25
47
70
41
25
0
1
1
2
2
2
3
1
2
0
3
Type of Seasonality
Quarterly
Quarter
1
2
3
4
Seasonal Factor
0.550
1.027
1.519
0.904
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
Mean Absolute Deviation
MAD =
1.57
Mean Square Error
MSE =
3.07
27-15
(e) Forecast:
Year Quarter
1
1
1
2
1
3
1
4
2
1
2
2
2
3
2
4
3
1
3
2
3
3
3
4
4
1
4
2
4
3
4
4
5
1
5
2
5
3
5
4
6
1
6
2
6
3
6
4
(f) Forecast:
Year Quarter
1
1
1
2
1
3
1
4
2
1
2
2
2
3
2
4
3
1
3
2
3
3
3
4
4
1
4
2
4
3
4
4
5
1
5
2
5
3
5
4
6
1
6
2
6
3
6
4
7
1
7
2
acre-feet
True
Value
25
47
68
42
27
46
72
39
24
49
70
44
Seasonally Seasonally
Adjusted
Adjusted
Value
Forecast
45
46
#N/A
45
#N/A
46
#N/A
49
46
45
47
47
46
43
47
44
46
48
45
46
45
49
45
#N/A
47
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
Actual
Forecast
Forecasting
Error
Number of previous
periods to consider
n=
4
Type of Seasonality
Quarterly
25
48
70
42
25
46
69
41
26
2
2
2
3
1
3
1
3
Quarter
1
2
3
4
Seasonal Factor
0.550
1.027
1.519
0.904
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
Mean Absolute Deviation
MAD =
2.2
Mean Square Error
MSE =
5.5
acre-feet
True
Value
25
47
68
42
27
46
72
39
24
49
70
44
Seasonally Seasonally
Adjusted
Adjusted
Value
Forecast
45
46
46
46
45
46
46
46
49
46
45
46
47
46
43
46
44
46
48
46
46
46
49
46
#N/A
46
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
Actual
Forecast
25
47
70
41
25
47
70
42
25
47
70
41
25
Forecasting
Error
0
0
2
1
2
1
2
3
1
2
0
3
Smoothing Constant
0.1
Initial Estimate
Average =
46
Type of Seasonality
Quarterly
Quarter
1
2
3
4
Seasonal Factor
0.550
1.027
1.519
0.904
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
Mean Absolute Deviation
MAD =
1.4
Mean Square Error
MSE =
(g) Exponential smoothing results in the lowest MAD value,
(h) Exponential smoothing gives the lowest MSE value,
27-16
.
.
2.7
27.7-14.
(a)
Year Quarter
1
1
1
2
1
3
1
4
2
1
2
2
2
3
2
4
3
1
3
2
3
3
3
4
True
Value
23
22
31
26
19
21
27
24
21
26
32
28
Type of Seasonality
Quarterly
Quarter
1
2
3
4
Estimate for
Seasonal Factor
0.8400
0.9200
1.2000
1.0400
(b)
Year Quarter
1
1
1
2
1
3
1
4
2
1
2
2
2
3
2
4
3
1
3
2
3
3
3
4
4
1
4
2
4
3
4
4
5
1
5
2
5
3
5
4
6
1
True
Value
23
22
31
26
19
21
27
24
21
26
32
28
Seasonally Seasonally
Adjusted
Adjusted
Value
Forecast
27
24
27
26
24
25
26
23
25
23
23
23
23
23
23
25
23
28
25
27
28
27
27
#N/A
27
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
Actual
Forecast
Forecasting
Error
25
29
27
21
21
27
23
19
23
34
28
23
3
2
1
2
0
0
1
2
3
2
0
Type of Seasonality
Quarterly
Quarter
1
2
3
4
Seasonal Factor
0.840
0.920
1.200
1.040
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
Mean Absolute Deviation
MAD =
1.5
Mean Square Error
MSE =
3
27-17
(c)
Year Quarter
1
1
1
2
1
3
1
4
2
1
2
2
2
3
2
4
3
1
3
2
3
3
3
4
4
1
4
2
4
3
4
4
5
1
5
2
5
3
5
4
6
1
True
Value
23
22
31
26
19
21
27
24
21
26
32
28
Seasonally Seasonally
Adjusted
Adjusted
Value
Forecast
27
24
27
26
26
25
26
23
26
23
25
23
25
23
24
25
24
28
24
27
25
27
25
#N/A
25
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
Actual
Forecast
Forecasting
Error
25
31
27
21
23
30
25
20
22
30
26
21
3
0
1
2
2
3
1
1
4
2
2
Type of Seasonality
Quarterly
Quarter
1
2
3
4
Seasonal Factor
0.840
0.920
1.200
1.040
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
Mean Absolute Deviation
MAD =
1.94
Mean Square Error
MSE =
4.85
(d)
Year Quarter
1
1
1
2
1
3
1
4
2
1
2
2
2
3
2
4
3
1
3
2
3
3
3
4
4
1
4
2
4
3
4
4
5
1
5
2
5
3
5
4
6
1
6
2
6
3
6
4
True
Value
23
22
31
26
19
21
27
24
21
26
32
28
Seasonally Seasonally
Adjusted
Adjusted
Value
Forecast
27
24
#N/A
26
#N/A
25
#N/A
23
26
23
24
23
24
23
23
25
23
28
23
27
25
27
26
#N/A
27
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
Actual
Forecast
Forecasting
Error
Number of previous
periods to consider
n=
4
Type of Seasonality
Quarterly
21
22
29
24
19
21
30
27
22
2
1
2
0
2
5
2
1
Quarter
1
2
3
4
Seasonal Factor
0.840
0.920
1.200
1.040
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
Mean Absolute Deviation
MAD =
2.0
Mean Square Error
MSE =
5.3
27-18
(e)
Year Quarter
1
1
1
2
1
3
1
4
2
1
2
2
2
3
2
4
3
1
3
2
3
3
3
4
4
1
4
2
4
3
4
4
5
1
5
2
5
3
5
4
6
1
6
2
6
3
6
4
7
1
7
2
True
Value
23
22
31
26
19
21
27
24
21
26
32
28
Seasonally Seasonally
Adjusted
Adjusted
Value
Forecast
27
25
24
26
26
25
25
25
23
25
23
25
23
24
23
24
25
24
28
24
27
25
27
25
#N/A
26
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
Actual
Forecast
21
24
30
26
21
23
29
25
20
22
30
26
22
Forecasting
Error
2
2
1
0
2
2
2
1
1
4
2
2
Smoothing Constant
0.25
Initial Estimate
Average =
25
Type of Seasonality
Quarterly
Quarter
1
2
3
4
Seasonal Factor
0.840
0.920
1.200
1.040
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
Mean Absolute Deviation
MAD =
1.7
Mean Square Error
MSE =
3.6
(f)
Year Quarter
1
1
1
2
1
3
1
4
2
1
2
2
2
3
2
4
3
1
3
2
3
3
3
4
4
1
4
2
4
3
4
4
5
1
5
2
5
3
5
4
6
1
6
2
6
3
6
4
7
1
7
2
7
3
7
4
True
Value
23
22
31
26
19
21
27
24
21
26
32
28
Seasonally
Adjusted
Value
27
24
26
25
23
23
23
23
25
28
27
27
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
Latest
Trend
1
0
0
0
-1
-1
-1
0
0
1
1
1
Estimated
Trend
0
0
0
0
0
0
0
0
0
0
0
0
0
Seasonally
Adjusted
Actual Forecasting
Forecast Forecast
Error
25
21
2
26
24
2
25
30
1
26
27
1
25
21
2
25
23
2
24
29
2
23
24
0
23
19
2
23
21
5
25
29
3
25
26
2
26
22
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
(g) Using the last-value method with seasonality MAD
is houses.
(h) Quarter 2:
, Quarter 3:
27-19
Smoothing Constant
0.25
0.25
Initial Estimate
Average =
Trend =
25
0
Type of Seasonality
Quarterly
Quarter
1
2
3
4
Seasonal Factor
0.840
0.920
1.200
1.040
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
Mean Absolute Deviation
MAD =
2
Mean Square Error
MSE =
4
, the forecast for first quarter
, Quarter 4:
27.7-15.
(a)
Year
1
1
1
1
1
1
1
1
1
1
1
1
2
2
2
2
2
2
2
2
2
Last-Value Method with Seasonality
Month
Jan
Feb
Mar
Apr
May
June
July
Aug
Sep
Oct
Nov
Dec
Jan
Feb
Mar
Apr
May
June
July
Aug
Sep
True
Value
68
71
66
72
77
85
94
96
80
73
84
89
Seasonally Seasonally
Adjusted
Adjusted
Value
Forecast
76
81
76
73
81
77
73
80
77
78
80
80
78
83
80
82
83
80
82
80
80
82
80
#N/A
82
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
Actual
Forecast
Forecasting
Error
66
73
67
74
87
91
92
81
75
84
86
74
5
7
5
3
2
3
4
1
2
0
3
Type of Seasonality
Monthly
Month
Jan
Feb
Mar
Apr
May
June
July
Aug
Sep
Oct
Nov
Dec
Seasonal Factor
0.900
0.880
0.910
0.930
0.960
1.090
1.170
1.150
0.970
0.910
1.050
1.080
Mean Absolute Deviation
MAD =
3.07
Mean Square Error
MSE =
12.89
Averaging Method with Seasonality:
Year
1
1
1
1
1
1
1
1
1
1
1
1
2
2
2
2
2
2
2
2
2
Month
Jan
Feb
Mar
Apr
May
June
July
Aug
Sep
Oct
Nov
Dec
Jan
Feb
Mar
Apr
May
June
July
Aug
Sep
True
Value
68
71
66
72
77
85
94
96
80
73
84
89
Seasonally Seasonally
Adjusted
Adjusted
Value
Forecast
76
81
76
73
78
77
76
80
77
78
77
80
77
83
78
82
79
80
79
80
79
82
79
#N/A
79
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
Actual
Forecast
Forecasting
Error
66
71
71
73
84
91
89
76
72
83
86
71
5
5
1
4
1
3
7
4
1
1
3
Type of Seasonality
Monthly
Month
Jan
Feb
Mar
Apr
May
June
July
Aug
Sep
Oct
Nov
Dec
Seasonal Factor
0.900
0.880
0.910
0.930
0.960
1.090
1.170
1.150
0.970
0.910
1.050
1.080
Mean Absolute Deviation
MAD =
3.12
Mean Square Error
MSE =
13.07
27-20
Moving-Average Method with Seasonality
Year
1
1
1
1
1
1
1
1
1
1
1
1
2
2
2
2
2
2
2
2
2
2
2
2
Month
Jan
Feb
Mar
Apr
May
June
July
Aug
Sep
Oct
Nov
Dec
Jan
Feb
Mar
Apr
May
June
July
Aug
Sep
Oct
Nov
Dec
True
Value
68
71
66
72
77
85
94
96
80
73
84
89
Seasonally Seasonally
Adjusted
Adjusted
Value
Forecast
76
81
#N/A
73
#N/A
77
76
80
77
78
77
80
79
83
80
82
81
80
82
80
82
82
81
#N/A
81
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
Actual
Forecast
Forecasting
Error
71
74
84
92
91
78
75
86
87
73
1
3
1
2
5
2
2
2
2
Number of previous
periods to consider
n=
3
Type of Seasonality
Monthly
Month
Jan
Feb
Mar
Apr
May
June
July
Aug
Sep
Oct
Nov
Dec
Seasonal Factor
0.900
0.880
0.910
0.930
0.960
1.090
1.170
1.150
0.970
0.910
1.050
1.080
Mean Absolute Deviation
MAD =
2.18
Mean Square Error
MSE =
5.79
Exponential Smoothing Method with Seasonality
Year
1
1
1
1
1
1
1
1
1
1
1
1
2
2
2
2
2
2
2
2
2
2
2
2
3
3
Month
Jan
Feb
Mar
Apr
May
June
July
Aug
Sep
Oct
Nov
Dec
Jan
Feb
Mar
Apr
May
June
July
Aug
Sep
Oct
Nov
Dec
Jan
Feb
True
Value
68
71
66
72
77
85
94
96
80
73
84
89
Seasonally Seasonally
Adjusted
Adjusted
Value
Forecast
76
80
81
79
73
79
77
78
80
78
78
78
80
78
83
79
82
80
80
80
80
80
82
80
#N/A
81
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
Actual
Forecast
72
70
72
73
75
85
92
91
77
73
84
87
73
Forecasting
Error
4
1
6
1
2
0
2
5
3
0
0
2
Smoothing Constant
0.2
Initial Estimate
Average =
80
Type of Seasonality
Monthly
Month
Jan
Feb
Mar
Apr
May
June
July
Aug
Sep
Oct
Nov
Dec
Seasonal Factor
0.900
0.880
0.910
0.930
0.960
1.090
1.170
1.150
0.970
0.910
1.050
1.080
Mean Absolute Deviation
MAD =
2.34
Mean Square Error
MSE =
27-21
9.31
Method
Last-Value
Averaging
Moving-Average
Exponential Smoothing
MAD
MSE
(b) The moving-average method with seasonality has the lowest MAD value. With this
method, the forecast for January is
passengers.
27.7-16.
(a)
Method
Last-Value
Averaging
Moving-Average
Exp. Smoothing
MAD
MSE
(b) Forecast:
Year Month
1
Jan
1
Feb
1
Mar
1
Apr
1
May
1
June
1
July
1
Aug
1
Sep
1
Oct
1
Nov
1
Dec
2
Jan
2
Feb
2
Mar
2
Apr
2
May
2
June
2
July
2
Aug
2
Sep
2
Oct
2
Nov
2
Dec
3
Jan
3
Feb
3
Mar
3
Apr
True
Value
75
76
81
84
85
99
107
108
94
90
106
110
Seasonally
Adjusted
Value
83
86
89
90
89
91
91
94
97
99
101
102
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
Latest
Trend
2
2
3
2
2
2
1
1
2
2
2
2
Estimated
Trend
2
2
2
2
2
2
2
2
2
2
2
2
2
Seasonally
Adjusted
Actual Forecasting
Forecast Forecast
Error
82
74
1
84
74
2
87
79
2
90
83
1
92
88
3
93
102
3
95
111
4
96
110
2
97
95
1
99
90
0
101
106
0
103
111
1
104
94
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
MAD and MSE values are lower than those in (a).
27-22
Smoothing Constant
0.2
0.2
Initial Estimate
Average =
Trend =
80
2
Type of Seasonality
Monthly
Month
Jan
Feb
Mar
Apr
May
June
July
Aug
Sep
Oct
Nov
Dec
Seasonal Factor
0.900
0.880
0.910
0.930
0.960
1.090
1.170
1.150
0.970
0.910
1.050
1.080
Mean Absolute Deviation
MAD =
1.66
Mean Square Error
MSE =
4.21
(c) MAD and MSE values obtained are higher than the ones in (b).
Year Month
1
Jan
1
Feb
1
Mar
1
Apr
1
May
1
June
1
July
1
Aug
1
Sep
1
Oct
1
Nov
1
Dec
2
Jan
2
Feb
2
Mar
2
Apr
2
May
2
June
2
July
2
Aug
2
Sep
2
Oct
2
Nov
2
Dec
3
Jan
3
Feb
3
Mar
3
Apr
True
Value
68
71
66
72
77
85
94
96
80
73
84
89
75
76
81
84
85
99
107
108
94
90
106
110
Seasonally
Adjusted
Value
76
81
73
77
80
78
80
83
82
80
80
82
83
86
89
90
89
91
91
94
97
99
101
102
#N/A
#N/A
#N/A
#N/A
Latest
Trend
-1
0
-1
0
0
0
0
1
1
0
0
1
1
1
2
2
1
1
1
2
2
2
2
2
Estimated
Trend
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
1
1
1
1
1
1
1
2
2
2
Seasonally
Adjusted
Actual Forecasting
Forecast Forecast
Error
80
72
4
79
69
2
79
72
6
77
72
0
77
74
3
77
84
1
77
90
4
78
89
7
79
77
3
80
73
0
80
84
0
80
87
2
81
73
2
82
72
4
83
76
5
85
79
5
87
84
1
89
97
2
90
106
1
92
105
3
93
91
3
96
87
3
98
103
3
100
108
2
102
92
#N/A
#N/A
#N/A
Smoothing Constant
0.2
0.2
Initial Estimate
Average =
Trend =
80
0
Type of Seasonality
Monthly
Month
Jan
Feb
Mar
Apr
May
June
July
Aug
Sep
Oct
Nov
Dec
Seasonal Factor
0.900
0.880
0.910
0.930
0.960
1.090
1.170
1.150
0.970
0.910
1.050
1.080
Mean Absolute Deviation
MAD =
2.74
Mean Square Error
MSE =
10.44
(d) Exponential smoothing with seasonality and trend (with parameters as in (b) should be used.
27-23
27.7-17.
(a) Based on past sales:
Year
1
1
1
1
1
1
1
1
1
1
1
1
2
2
2
2
2
2
2
2
2
2
2
2
3
3
3
3
3
3
3
3
3
3
3
3
Month
Jan
Feb
Mar
Apr
May
June
July
Aug
Sep
Oct
Nov
Dec
Jan
Feb
Mar
Apr
May
June
July
Aug
Sep
Oct
Nov
Dec
Jan
Feb
Mar
Apr
May
June
July
Aug
Sep
Oct
Nov
Dec
True
Value
352
329
365
358
412
446
420
471
355
312
567
533
317
331
344
386
423
472
415
492
340
301
629
505
338
346
383
404
431
459
433
518
309
335
594
527
Type of Seasonality
Monthly
Month
Jan
Feb
Mar
Apr
May
June
July
Aug
Sep
Oct
Nov
Dec
Estimate for
Seasonal Factor
0.8082
0.8074
0.8764
0.9213
1.016
1.105
1.018
1.189
0.806
0.761
1.437
1.256
27-24
(b) Moving Average with Seasonality
Year
1
1
1
1
1
1
1
1
1
1
1
1
2
2
2
2
2
2
2
2
2
2
2
2
3
Month
Jan
Feb
Mar
Apr
May
June
July
Aug
Sep
Oct
Nov
Dec
Jan
Feb
Mar
Apr
May
June
July
Aug
Sep
Oct
Nov
Dec
Jan
True
Value
335
594
527
364
343
391
437
458
494
468
555
387
364
662
581
Seasonally Seasonally
Adjusted
Adjusted
Value
Forecast
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
440
#N/A
413
#N/A
420
#N/A
450
424
425
428
446
432
474
440
451
448
447
457
460
457
467
453
480
458
478
469
461
475
463
473
#N/A
467
Actual
Forecast
Forecasting
Error
Number of previous
periods to consider
n=
3
Type of Seasonality
Monthly
343
345
378
406
456
505
465
538
369
357
683
594
378
21
2
13
31
2
11
3
17
18
7
21
13
Month
Jan
Feb
Mar
Apr
May
June
July
Aug
Sep
Oct
Nov
Dec
Seasonal Factor
0.808
0.807
0.876
0.921
1.016
1.105
1.018
1.189
0.806
0.761
1.437
1.256
Mean Absolute Deviation
MAD =
13.30
Mean Square Error
MSE =
249.09
(c) Exponential Smoothing with Seasonality
Year
1
1
1
1
1
1
1
1
1
1
1
1
2
2
2
2
2
2
2
2
2
2
2
2
3
3
Month
Jan
Feb
Mar
Apr
May
June
July
Aug
Sep
Oct
Nov
Dec
Jan
Feb
Mar
Apr
May
June
July
Aug
Sep
Oct
Nov
Dec
Jan
Feb
True
Value
364
343
391
437
458
494
468
555
387
364
662
581
Seasonally Seasonally
Adjusted
Adjusted
Value
Forecast
450
420
425
426
446
426
474
430
451
439
447
441
460
442
467
446
480
450
478
456
461
461
463
461
#N/A
461
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
Actual
Forecast
339
344
373
396
446
488
450
530
363
347
662
579
373
Forecasting
Error
25
1
18
41
12
6
18
25
24
17
0
2
Smoothing Constant
0.2
Initial Estimate
Average =
420
Type of Seasonality
Monthly
Month
Jan
Feb
Mar
Apr
May
June
July
Aug
Sep
Oct
Nov
Dec
Seasonal Factor
0.808
0.807
0.876
0.921
1.016
1.105
1.018
1.189
0.806
0.761
1.437
1.256
Mean Absolute Deviation
MAD =
15.83
Mean Square Error
MSE =
384.99
27-25
(d) Exponential Smoothing with Seasonality and Trend
Year Month
1
Jan
1
Feb
1
Mar
1
Apr
1
May
1
June
1
July
1
Aug
1
Sep
1
Oct
1
Nov
1
Dec
2
Jan
2
Feb
2
Mar
2
Apr
2
May
2
June
2
July
2
Aug
2
Sep
2
Oct
2
Nov
2
Dec
3
Jan
3
Feb
3
Mar
3
Apr
True
Value
364
343
391
437
458
494
468
555
387
364
662
581
Seasonally
Adjusted
Value
450
425
446
474
451
447
460
467
480
478
461
463
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
Latest
Trend
6
1
5
10
5
3
5
6
7
6
2
1
Estimated
Trend
0
1
1
2
3
4
4
4
4
5
5
4
4
Seasonally
Adjusted
Actual Forecasting
Forecast Forecast
Error
420
339
25
427
345
2
428
375
16
433
399
38
445
452
6
450
497
3
453
461
7
458
545
10
464
374
13
472
359
5
479
688
26
479
602
21
480
388
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
#N/A
(e) The moving-average method results in the best MAD value
value
.
(f)
Month
Avg. Forecast Forecasting Error
January
February
March
April
May
June
July
August
September
October
November
December
Smoothing Constant
0.2
0.2
Initial Estimate
Average =
Trend =
420
0
Type of Seasonality
Monthly
Month
Jan
Feb
Mar
Apr
May
June
July
Aug
Sep
Oct
Nov
Dec
Seasonal Factor
0.808
0.807
0.876
0.921
1.016
1.105
1.018
1.189
0.806
0.761
1.437
1.256
Mean Absolute Deviation
MAD =
14.26
Mean Square Error
MSE =
314.71
and the best MSE
MAD
(g) The moving-average method performed better than the average of all three, so it
should be used next year.
27-26
27.7-18.
SMALLEST
27.7-19.
smallest
27.9-1.
(a)
(b)
27-27
(c)
(d)
(e)
(f) The average growth in monthly sales is
.
27.9-2.
(a)
27-28
(b)
(c)
(d)
Year 4
Year 5
Year 6
Year 7
Year 8
(e) It does not make sense to use the forecast obtained earlier,
. The relationship
between the variables has changed and thus the linear regression that was used is no
longer appropriate.
(f)
The linear regression line does not provide a close fit to the data. Consequently, the
forecast that it provides for year 8 is not likely to be accurate. It does not make sense to
continue to use a linear regression line when changing conditions cause a large shift in
the underlying trend in the data.
27-29
(g)
Time
Period
1
2
3
4
5
6
7
8
9
10
11
12
True
Value
4,600
5,300
6,000
6,300
6,200
5,600
5,200
Latest
Trend
700.00
700.00
700.00
500.00
150.00
-337.50
-546.88
Estimated
Trend
700.00
700.00
700.00
700.00
600.00
375.00
18.75
-264.06
Exponential
Smoothing
Forecast
4,600
5,300
6,000
6,700
7,100
7,025
6,331
5,502
Forecasting
Error
0
0
0
400
900
1,425
1,131
Smoothing Constants
0.5
0.5
Initial Estimates
Average =
3,900
Trend =
700
Mean Absolute Deviation
MAD =
550.9
Mean Square Error
MSE =
611,478.8
Casual forecasting takes all the data into account, even the data from before changing
conditions cause a shift. Exponential smoothing with trend adjusts to shifts in the
underlying trend.
27.9-3.
(a)
(b)
Time
Period
1
2
3
4
5
6
7
8
9
10
Independent
Variable
1
2
3
4
5
6
7
8
9
10
Dependent
Variable
382
405
398
421
426
415
443
451
446
464
Estimate
388
397
405
413
421
429
437
445
454
462
Estimation
Error
6.42
8.43
6.72
8.13
4.98
14.18
5.67
5.52
7.63
2.22
27-30
Square
of Error
41
71
45
66
25
201
32
30
58
5
Linear Regression Line
y = a + bx
a=
380.27
b=
8.15
Estimator
If x =
then y=
5,000
41,137.84
(c)
(d)
(e)
(f) The average growth per year is
tons.
27-31
(g)
27-32
27.9-4.
(a) The amount of advertising is the independent variable and sales is the dependent variable.
(b)
(c)
(d)
(e)
(f) An increase of
passengers
$
passengers can be attained.
27.9-5.
(a) If the sales increase from
to
when the amount of advertising is
, then the
linear regression line shifts below this point. The line actually shifts up, but not as much
as the data point has shifted up.
(b) If the sales increase from
to
when the amount of advertising is
, then the
linear regression line shifts below this point. The line actually shifts up, but not as much
as the data point has shifted up.
(c) If the sales increase from
to
when the amount of advertising is
, then the
linear regression line shifts below this point. The line actually shifts up, but not as much
as the data point has shifted up.
27-33
27.9-6.
(a) The number of flying hours is the independent variable and the number of wing flaps
needed is the dependent variable.
(b)
(c)
Time
Period
1
2
3
4
5
6
7
8
Independent
Variable
162
149
185
171
138
154
Dependent
Variable
12
9
13
14
10
11
Estimate
12
10
14
13
9
11
Estimation
Error
0.30
1.49
0.84
1.46
0.53
0.04
Square
of Error
0
2
1
2
0
0
Linear Regression Line
y = a + bx
a=
-3.382
b=
0.093
Estimator
If x =
then y=
(d)
(e)
(f)
27-34
150
10.584
27.9-7. Joe should use the linear regression line
forecast for jobs in the future.
Time Independent
Period
Variable
1
323
2
359
3
396
4
421
5
457
6
472
7
446
8
407
9
374
10
343
Dependent
Variable
24
23
28
32
34
37
33
30
27
22
to develop a
Estimation
Error
2.48
2.02
0.63
0.93
0.57
0.97
0.50
0.30
0.51
1.47
Estimate
22
25
29
31
35
36
34
30
26
23
Square
of Error
6
4
0
1
0
1
0
0
0
2
27.9-8.
(a)
(b)
The
% prediction interval is
.
(c) By interpolation:
The simultaneous
% prediction interval is
.
(d) By interpolation:
The simultaneous tolerance interval is
.
27.9-9.
(a)
,
,
,
(b)
The
% prediction interval is
.
(c)
The
% prediction interval is
.
(d) By interpolation:
The simultaneous tolerance interval is
.
27-35
Linear Regression Line
y = a + bx
a=
-9.954
b=
0.097
27.9-10.
(a)
log
log
log
log
log
The forecast for the distance traveled when log
approximately one million.
is then
(b)
log
log
log
log
log
(c)
Trend
27.9-11.
27-36
log
, which is
Cases
27#1% a)% We%need%to%forecast%the%call%volume%for%each%day%separately.%
%
1)%%To%obtain%the%seasonally%adjusted%call%volume%for%the%past%13%weeks,%we%first%
have%to%determine%the%seasonal%factors.%%Because%call%volumes%follow%seasonal%
patterns%within%the%week,%we%have%to%calculate%a%seasonal%factor%for%Monday,%
Tuesday,%Wednesday,%Thursday,%and%Friday.%%We%use%the%Template%for%Seasonal%
Factors.%The%0%values%for%holidays%should%not%factor%into%the%average.%Leaving%them%
blank%(rather%than%0)%accomplishes%this.%(A%blank%value%does%not%factor%into%the%
AVERAGE%function%in%Excel%that%is%used%to%calculate%the%seasonal%values.)%Using%
this%template%(shown%on%the%following%page,%the%seasonal%factors%for%Monday,%
Tuesday,%Wednesday,%Thursday,%and%Friday%are%1.238,%1.131,%0.999,%0.850,%and%
0.762,%respectively.%
27-37
%
A
1
2
3
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B
C
D
E
F
G
Template for Seasonal Factors
Week
44
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46
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47
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52/1
52/1
52/1
52/1
52/1
2
2
2
2
2
3
3
3
3
3
4
4
4
4
4
5
5
5
5
5
Day
Mon
Tue
Wed
Thur
Fri
Mon
Tue
Wed
Thur
Fri
Mon
Tue
Wed
Thur
Fri
Mon
Tue
Wed
Thur
Fri
Mon
Tue
Wed
Thur
Fri
Mon
Tue
Wed
Thur
Fri
Mon
Tue
Wed
Thur
Fri
Mon
Tue
Wed
Thur
Fri
Mon
Tue
Wed
Thur
Fri
Mon
Tue
Wed
Thur
Fri
Mon
Tue
Wed
Thur
Fri
Mon
Tue
Wed
Thur
Fri
Mon
Tue
Wed
Thur
Fri
True
Value
1,130
851
859
828
726
1,085
1,042
892
840
799
1,303
1,121
1,003
1,113
1,005
2,652
2,825
1,841
1,949
1,507
989
990
1,084
1,260
1,134
941
847
714
1,002
847
922
842
784
823
Type of Seasonality
Daily
Estimate for
Seasonal Factor
1.238
1.131
0.999
0.850
0.762
Day
Mon
Tue
Wed
Thur
Fri
Average Call Volume
1,025
401
429
1,209
830
1,082
841
1,362
1,174
967
930
853
924
954
1,346
904
758
886
878
802
945
610
910
754
705
729
772
%
27-38
%
%
2)%To%forecast%the%call%volume%for%the%next%week%using%the%last#value%forecasting%
method,%we%need%to%use%the%Last%Value%with%Seasonality%template.%%To%forecast%the%
next%week,%we%need%only%start%with%the%last%Friday%value%since%the%Last%Value%
method%only%looks%at%the%previous%day.%
%
A
1
2
3
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5
6
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8
9
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14
15
B
C
D
E
F
G
H
I
J
K
Template for Last-Value Forecasting Method with Seasonality
Week
5
5
5
5
5
6
6
6
6
6
Day
Mon
Tue
Wed
Thur
Fri
Mon
Tue
Wed
Thur
Fri
True
Value
772
1,254
1,146
1,012
860
771
Seasonally
Adjusted
Value
#N/A
#N/A
#N/A
#N/A
1,013
1,013
1,013
1,013
1,012
1,012
Seasonally
Adjusted
Forecast
#N/A
#N/A
#N/A
#N/A
1,013
1,013
1,013
1,013
1,012
Actual
Forecast
1,254
1,146
1,012
860
771
Forecasting
Error
0
0
0
0
0
Type of Seasonality
Daily
Day
Mon
Tue
Wed
Thur
Fri
Seasonal Factor
1.238
1.131
0.999
0.850
0.762
1.000
1.000
%
%
The%forecasted%call%volume%for%the%next%week%is%5,045%calls:%%1,254%calls%are%
received%on%Monday,%1,148%calls%are%received%on%Tuesday,%1,012%calls%are%received%
on%Wednesday,%860%calls%are%received%on%Thursday,%and%771%calls%are%received%on%
Friday.%
27-39
%
%
3)%To%forecast%the%call%volume%for%the%next%week%using%the%averaging%forecasting%
method,%we%need%to%use%the%Averaging%with%Seasonality%template.%
%
A
1
2
3
4
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7
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B
C
D
E
F
G
H
I
J
K
Template for Averaging Forecasting Method with Seasonality
Week
44
44
44
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44
45
45
45
45
45
46
46
46
46
46
47
47
47
47
47
48
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48
48
48
49
49
49
49
49
50
50
50
50
50
51
51
51
51
51
52/1
52/1
52/1
52/1
52/1
2
2
2
2
2
3
3
3
3
3
4
4
4
4
4
5
5
5
5
5
6
6
6
6
6
Day
Mon
Tue
Wed
Thur
Fri
Mon
Tue
Wed
Thur
Fri
Mon
Tue
Wed
Thur
Fri
Mon
Tue
Wed
Thur
Fri
Mon
Tue
Wed
Thur
Fri
Mon
Tue
Wed
Thur
Fri
Mon
Tue
Wed
Thur
Fri
Mon
Tue
Wed
Thur
Fri
Mon
Tue
Wed
Thur
Fri
Mon
Tue
Wed
Thur
Fri
Mon
Tue
Wed
Thur
Fri
Mon
Tue
Wed
Thur
Fri
Mon
Tue
Wed
Thur
Fri
Mon
Tue
Wed
Thur
Fri
True
Value
1,130
851
859
828
726
1,085
1,042
892
840
799
1,303
1,121
1,003
1,113
1,005
2,652
2,825
1,841
0
0
1,949
1,507
989
990
1,084
1,260
1,134
941
847
714
1,002
847
922
842
784
823
0
0
401
429
1,209
830
0
1,082
841
1,362
1,174
967
930
853
924
954
1,346
904
758
886
878
802
945
610
910
754
705
729
772
1171
1071
945
804
721
Seasonally
Adjusted
Value
913
752
860
975
953
877
921
893
989
1,048
1,053
991
1,004
1,310
1,319
2,143
2,497
1,842
0
0
1,575
1,332
990
1,165
1,422
1,018
1,002
942
997
937
810
749
923
991
1,029
665
0
0
472
563
977
734
0
1,274
1,103
1,100
1,038
968
1,095
1,119
746
843
1,347
1,064
995
716
776
803
1,112
800
735
666
706
858
1,013
946
947
946
946
946
Seasonally
Adjusted
Forecast
Actual
Forecast
Forecasting
Error
913
833
842
875
890
888
893
893
903
918
930
935
941
967
990
1,062
1,147
1,185
1,123
1,067
1,091
1,102
1,097
1,100
1,113
1,109
1,105
1,099
1,096
1,091
1,081
1,071
1,067
1,064
1,063
1,052
1,024
997
983
973
973
967
945
952
956
959
960
961
963
966
962
960
967
969
969
965
962
959
961
959
955
950
947
945
946
946
946
946
946
1,033
832
715
667
1,102
1,005
892
759
689
1,136
1,052
934
799
737
1,226
1,202
1,146
1,007
856
1,321
1,234
1,101
932
838
1,377
1,255
1,104
934
835
1,350
1,224
1,070
906
811
1,316
1,191
1,023
847
750
1,204
1,101
967
803
726
1,183
1,085
960
816
734
1,196
1,089
959
822
738
1,200
1,092
961
815
733
1,187
1,081
950
804
720
1,171
1,071
945
804
721
182
27
113
59
17
37
0
81
110
167
69
69
314
268
1,426
1,623
695
1,007
856
628
273
112
58
246
117
121
163
87
121
348
377
148
64
27
493
1,191
1,023
446
321
5
271
967
279
115
179
89
7
114
119
272
135
387
82
20
314
214
159
130
123
277
327
245
75
52
0
0
0
0
0
Type of Seasonality
Daily
Day
Mon
Tue
Wed
Thur
Fri
Seasonal Factor
1.238
1.131
0.999
0.850
0.762
1.000
1.000
1.000
1.000
1.000
1.000
1.000
Mean Absolute Deviation
MAD =
267.27
Mean Square Error
MSE =
187,916.17
%%
%
The%forecasted%call%volume%for%the%next%week%is%4,712%calls:%%1,171%calls%are%
received%on%Monday,%1,071%calls%are%received%on%Tuesday,%945%calls%are%received%
on%Wednesday,%804%calls%are%received%on%Thursday,%and%721%calls%are%received%on%
Friday.%
27-40
%
%
4)%To%forecast%the%call%volume%for%the%next%week%using%the%moving#average%
forecasting%method,%we%need%to%use%the%Moving%Averaging%with%Seasonality%
template.%%Since%only%the%past%5%days%are%used%in%the%forecast,%we%start%with%
Monday%of%the%last%week%to%forecast%through%Friday%of%the%next%week.%
%
A
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6
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10
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14
15
16
B
C
D
E
F
G
H
I
J
K
Template for Moving-Average Forecasting Method with Seasonality
Week
5
5
5
5
5
6
6
6
6
6
7
Day
Mon
Tue
Wed
Thur
Fri
Mon
Tue
Wed
Thur
Fri
Mon
True
Value
910
754
705
729
772
985
914
835
732
658
Seasonally
Adjusted
Value
735
666
706
858
1,013
796
808
836
862
863
#N/A
Seasonally
Adjusted
Forecast
#N/A
#N/A
#N/A
#N/A
796
808
836
862
863
833
Actual
Forecast
Forecasting
Error
Number of previous
periods to consider
n=
5
Type of Seasonality
Daily
985
914
835
732
658
1,031
0
0
0
0
0
Day
Mon
Tue
Wed
Thur
Fri
Seasonal Factor
1.238
1.131
0.999
0.850
0.762
%
%
The%forecasted%call%volume%for%the%next%week%is%4,124%calls:%%985%calls%are%received%
on%Monday,%914%calls%are%received%on%Tuesday,%835%calls%are%received%on%
Wednesday,%732%calls%are%received%on%Thursday,%and%658%calls%are%received%on%
Friday.%
%
5)%To%forecast%the%call%volume%for%the%next%week%using%the%exponential%smoothing%
forecasting%method,%we%need%to%use%the%Exponential%with%Seasonality%template.%We%
start%with%the%initial%estimate%of%1,125%calls%(the%average%number%of%calls%on%non#
holidays%during%the%previous%13%weeks).%
%
27-41
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70
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B
C
D
E
F
G
H
I
J
K
Template for Exponential Smoothing Forecasting Method with Seasonality
Week
44
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44
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45
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47
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52/1
52/1
52/1
52/1
52/1
2
2
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2
2
3
3
3
3
3
4
4
4
4
4
5
5
5
5
5
6
6
6
6
6
Day
Mon
Tue
Wed
Thur
Fri
Mon
Tue
Wed
Thur
Fri
Mon
Tue
Wed
Thur
Fri
Mon
Tue
Wed
Thur
Fri
Mon
Tue
Wed
Thur
Fri
Mon
Tue
Wed
Thur
Fri
Mon
Tue
Wed
Thur
Fri
Mon
Tue
Wed
Thur
Fri
Mon
Tue
Wed
Thur
Fri
Mon
Tue
Wed
Thur
Fri
Mon
Tue
Wed
Thur
Fri
Mon
Tue
Wed
Thur
Fri
Mon
Tue
Wed
Thur
Fri
Mon
Tue
Wed
Thur
Fri
True
Value
1,130
851
859
828
726
1,085
1,042
892
840
799
1,303
1,121
1,003
1,113
1,005
2,652
2,825
1,841
0
0
1,949
1,507
989
990
1,084
1,260
1,134
941
847
714
1,002
847
922
842
784
823
0
0
401
429
1,209
830
0
1,082
841
1,362
1,174
967
930
853
924
954
1,346
904
758
886
878
802
945
610
910
754
705
729
772
1074
982
867
737
661
Seasonally
Adjusted
Value
913
752
860
975
953
877
921
893
989
1,048
1,053
991
1,004
1,310
1,319
2,143
2,497
1,842
0
0
1,575
1,332
990
1,165
1,422
1,018
1,002
942
997
937
810
749
923
991
1,029
665
0
0
472
563
977
734
0
1,274
1,103
1,100
1,038
968
1,095
1,119
746
843
1,347
1,064
995
716
776
803
1,112
800
735
666
706
858
1,013
868
868
868
867
867
Seasonally
Adjusted
Forecast
1,025
1,014
988
975
975
973
963
959
952
956
965
974
976
978
1,012
1,042
1,152
1,287
1,342
1,208
1,087
1,136
1,156
1,139
1,142
1,170
1,155
1,139
1,120
1,107
1,090
1,062
1,031
1,020
1,017
1,018
983
885
796
764
744
767
764
687
746
782
814
836
849
874
898
883
879
926
940
945
922
908
897
919
907
890
867
851
852
868
868
868
868
868
Actual
Forecast
1,269
1,147
987
828
743
1,204
1,089
958
809
728
1,195
1,102
975
831
771
1,290
1,304
1,286
1,140
921
1,346
1,285
1,155
968
870
1,448
1,306
1,138
951
844
1,350
1,202
1,030
867
775
1,260
1,112
884
676
582
921
868
763
584
568
968
920
835
721
666
1,112
999
878
787
716
1,170
1,043
907
762
700
1,122
1,007
867
723
649
1,074
982
867
737
661
Forecasting
Error
139
296
128
0
17
119
47
66
31
71
108
19
28
282
234
1,362
1,521
555
1,140
921
603
222
166
22
214
188
172
197
104
130
348
355
108
25
9
437
1,112
884
275
153
288
38
763
498
273
394
254
132
209
187
188
45
468
117
42
284
165
105
183
90
212
253
162
6
123
0
0
0
0
0
Smoothing Constant
α=
Initial Estimate
Average =
0.1
1,025
Type of Seasonality
Daily
Day
Mon
Tue
Wed
Thur
Fri
Seasonal Factor
1.238
1.131
0.999
0.850
0.762
1.000
1.000
1.000
1.000
1.000
1.000
1.000
Mean Absolute Deviation
MAD =
261.3
Mean Square Error
MSE =
171,377.0
%
The%forecasted%call%volume%for%the%next%week%is%4,322%calls:%%1,074%calls%are%
received%on%Monday,%982%calls%are%received%on%Tuesday,%867%calls%are%received%on%
27-42
%
Wednesday,%737%calls%are%received%on%Thursday,%and%661%calls%are%received%on%
Friday.%
%
b)% To%obtain%the%mean%absolute%deviation%for%each%forecasting%method,%we%simply%
need%to%subtract%the%true%call%volume%from%the%forecasted%call%volume%for%each%day%
in%the%sixth%week.%%We%then%need%to%take%the%absolute%value%of%the%five%differences.%%
Finally,%we%need%to%take%the%average%of%these%five%absolute%values%to%obtain%the%
mean%absolute%deviation.%
%
1)%The%spreadsheet%for%the%calculation%of%the%mean%absolute%deviation%for%the%last#
value%forecasting%method%follows.%
%
A
1
2
3
4
5
6
7
8
9
B
C
D
E
Day
Monday
Tuesday
Wednesday
Thursday
Friday
True
Value
723
677
521
571
498
Actual
Forecast
1,254
1,146
1,012
860
771
Forecast
Error
531
469
491
289
273
F
G
H
Last Value
Week
6
6
6
6
6
Mean Absolute Deviation
MAD =
410.6
%
%
This%method%is%the%least%effective%of%the%four%methods%because%this%method%
depends%heavily%upon%the%average%seasonality%factors.%%If%the%average%seasonality%
factors%are%not%the%true%seasonality%factors%for%week%6,%a%large%error%will%appear%
because%the%average%seasonality%factors%are%used%to%transform%the%Friday%call%
volume%in%week%5%to%forecasts%for%all%call%volumes%in%week%6.%%We%calculated%in%part%
(a)%that%the%call%volume%for%Friday%is%0.762%times%lower%than%the%overall%average%
call%volume.%%In%week%6,%however,%the%call%volume%for%Friday%is%only%0.83%times%
lower%than%the%average%call%volume%over%the%week.%%Also,%we%calculated%that%the%call%
volume%for%Monday%is%1.34%times%higher%than%the%overall%average%call%volume.%%In%
Week%6,%however,%the%call%volume%for%Monday%is%only%1.21%times%higher%than%the%
average%call%volume%over%the%week.%%These%differences%introduce%a%large%error.%
%
27-43
%
%
2)%The%spreadsheet%for%the%calculation%of%the%mean%absolute%deviation%for%the%
averaging%forecasting%method%appears%below.%
%
A
1
2
3
4
5
6
7
8
9
B
C
D
E
Day
Monday
Tuesday
Wednesday
Thursday
Friday
True
Value
723
677
521
571
498
Actual
Forecast
1,171
1,071
945
804
721
Forecast
Error
448
394
424
233
223
F
G
H
Averaging
Week
6
6
6
6
6
Mean Absolute Deviation
MAD =
344.4
%
%
This%method%is%the%second#most%effective%of%the%four%methods.%%Again,%the%reason%
lies%in%the%average%seasonality%factors.%%Applying%the%average%seasonality%factors%to%
an%average%call%volume%yields%a%much%more%accurate%result%than%applying%average%
seasonality%factors%to%only%one%call%volume.%%This%method%is%not%the%most%effective%
method,%however,%because%the%centralized%call%center%experiences%not%only%daily%
seasonality,%but%also%weekly%seasonality.%%For%example,%the%call%volumes%in%weeks%
45%and%46%are%much%greater%than%the%call%volumes%in%week%6.%%Therefore,%these%
larger%call%volumes%inflate%the%average%call%volume,%which%in%turn%inflates%the%
forecasts%for%Week%6.%%%%
%
%
%
3)The%spreadsheet%for%the%calculation%of%the%mean%absolute%deviation%for%the%
moving#average%forecasting%method%appears%below.%
%
A
1
2
3
4
5
6
7
8
9
B
C
D
E
True
Value
723
677
521
571
498
Actual
Forecast
985
914
835
732
658
Forecast
Error
262
237
314
161
160
F
G
H
Moving Average
Week
6
6
6
6
6
Day
Monday
Tuesday
Wednesday
Thursday
Friday
Mean Absolute Deviation
MAD =
226.8
%
This%method%is%the%most%effective%of%the%four%methods%because%this%method%only%
uses%the%average%week%5%call%volume%to%forecast%the%call%volumes%for%week%6.%%
Again,%applying%the%average%seasonality%factors%to%an%average%call%volume%yields%a%
much%more%accurate%result%than%applying%average%seasonality%factors%to%only%one%
call%volume.%Also,%the%average%call%volume%used%in%this%method%is%not%overly%
inflated%since%it%is%an%average%of%the%week%5%call%volumes,%which%are%closer%to%the%
week%6%call%volumes%than%any%other%of%the%13%weeks.%
%
27-44
%
%
%
4)%The%spreadsheet%for%the%calculation%of%the%mean%absolute%deviation%for%
exponential%forecasting%method%follows.%
%
A
1
2
3
4
5
6
7
8
9
B
C
D
E
Actual
Forecast
1,074
982
867
737
661
Forecast
Error
351
305
346
166
163
F
G
H
Exponential Smoothing
Week
6
6
6
6
6
Day
Monday
Tuesday
Wednesday
Thursday
Friday
True
Value
723
677
521
571
498
Mean Absolute Deviation
MAD =
266.2
%
This%method%is%nearly%as%effective%as%the%moving%average.%This%method%is%a%little%
more%effective%than%the%averaging%forecasting%method%because%the%smoothing%
constant%causes%less%weight%to%be%placed%on%the%call%volumes%in%the%earlier%weeks.%
%
%
c)%This%problem%is%simply%a%linear%regression%problem.%
%
%1)%%To%find%a%mathematical%relationship,%we%use%the%Linear%Regression%template.%%
The%decentralized%case%volumes%are%the%independent%variables,%and%the%centralized%
case%volumes%are%the%dependent%variables.%%Substituting%the%case%volume%data,%we%
obtain%the%following%spreadsheet.%The%relationship%is%y%=%1576%+%0.756x,%where%x%is%
the%decentralized%case%volume,%and%y%is%the%estimated%centralized%case%volume.%
%
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
B
C
D
E
F
G
Estimate
2,038
2,121
2,099
1,984
2,623
2,012
1,882
1,909
2,071
2,008
1,935
1,976
2,025
Estimation
Error
13.84
49.45
679.61
350.27
109.26
298.70
45.32
741.89
521.66
118.08
402.41
60.17
72.69
Square
of Error
192
2,445
461,872
122,690
11,938
89,221
2,054
550,400
272,132
13,943
161,932
3,620
5,284
H
I
J
Template for Linear Regression
Week
44
45
46
47
48
49
50
51
52/1
2
3
4
5
Independent
Variable
612
721
693
540
1,386
577
405
441
655
572
475
530
595
Dependent
Variable
2,052
2,170
2,779
2,334
2,514
1,713
1,927
1,167
1,549
2,126
2,337
1,916
2,098
%
27-45
Linear Regression Line
y = a + bx
a=
1,576
b=
0.76
Estimator
If x =
613
then y=
2,038.9
%
%
%
2)%%To%forecast%the%week%6%call%volume%for%the%centralized%call%center,%we%simply%
input%the%week%6%decentralized%case%volume%for%the%value%of%x%in%the%Estimator%
section%of%the%Linear%Regression%Spreadsheet%(as%shown%in%part%1%above).%%The%
value%of%y%then%represents%the%week%6%centralized%case%volume.%%We%multiply%this%
value%of%y%by%1.5%to%obtain%the%week%6%centralized%call%volume.%%Thus,%the%forecasted%
number%of%calls%is%1.5%*%2,038.9%=%3,058.%
%
We%then%break%this%weekly%call%volume%into%daily%call%volume.%%We%do%this%
conversion%by%dividing%the%weekly%call%volume%by%the%sum%of%the%seasonal%factors%
calculated%in%part%(a)%and%then%multiplying%this%weekly%call%volume%by%the%
appropriate%seasonal%factor%to%find%the%call%volume%for%each%of%the%five%days%of%the%
week.%%The%spreadsheet%showing%these%calculations%follows:%
%
1
2
3
4
5
6
7
8
9
10
A
B
Week 6 Call Volume
Daily Call Volume
3058
611.6
C
Day
Monday
Tuesday
Wednesday
Thursday
Friday
Seasonal
Factor
1.238
1.131
0.999
0.850
0.762
Forecasted
Call Volume
757
692
611
520
466
%
%
The%forecasted%call%volume%for%week%6%is%3,046%calls:%%757%calls%are%received%on%
Monday,%692%calls%are%received%on%Tuesday,%611%calls%are%received%on%Wednesday,%
520%calls%are%received%on%Thursday,%and%466%calls%are%received%on%Friday.%
%
27-46
%
%
3)%To%calculate%the%mean%absolute%deviation,%we%need%to%subtract%the%true%call%
volume%from%the%forecasted%call%volume%for%each%day%in%the%sixth%week.%%We%then%
need%to%take%the%absolute%value%of%the%five%differences.%%Finally,%we%need%to%take%the%
average%of%these%five%absolute%values%to%obtain%the%mean%absolute%deviation.%
%
The%spreadsheet%for%the%calculation%of%the%mean%absolute%deviation%follows.%
%
A
1
2
3
4
5
6
7
8
9
B
C
D
E
True
Value
723
677
521
571
498
Actual
Forecast
757
692
611
520
466
Forecast
Error
34
15
90
51
32
F
G
H
Causal Forecasting
Week
6
6
6
6
6
Day
Monday
Tuesday
Wednesday
Thursday
Friday
Mean Absolute Deviation
MAD =
44.4
%
This%forecasting%method%is%by%far%the%most%effective%method.%%The%centralized%
center%performs%the%same%services%and%serves%the%same%population%as%the%
decentralized%center.%%Therefore,%the%call%volume%trends%are%the%same.%%Once%we%
have%a%factor%to%scale%the%decentralized%call%volumes%to%the%centralized%call%
volumes,%we%have%a%very%effective%forecasting%method.%
%
%
d)% We%would%definitely%recommend%using%the%causal%forecasting%method%
implemented%in%part%(c)%because%it%yields%the%lowest%error.%%The%causal%method%
shows%us%that%the%call%volume%trends%remain%relatively%the%same%year%after%year.%%
We%had%to%convert%between%case%volumes%and%call%volumes%in%part%(c),%however,%
and%such%a%conversion%introduces%error.%%For%example,%what%if%a%case%generates%a%
higher%or%lower%number%of%calls?%%We%therefore%recommend%that%call%volume%data%
be%meticulously%recorded%as%the%centralized%center%continues%its%operation.%%Once%
one%year’s%worth%of%call%volumes%have%been%collected,%the%causal%forecasting%model%
should%be%updated.%%The%model%should%be%updated%to%use%the%historical%centralized%
call%volume%data%instead%of%the%historical%decentralized%case%volume%data.%
27-47
CHAPTER 28: EXAMPLES OF PERFORMING SIMULATIONS ON
SPREADSHEETS WITH ANALYTIC SOLVER PLATFORM
28-­‐1. (a) Answers will vary. A typical set of 5 runs: 45.83, 46.26, 45.94, 45.98, and 46.89. (b) Answers will vary. A typical set of 5 runs: 46.49, 46.12, 46.38, 46.23, and 46.37. (c) The mean completion times in part b should be more consistent. 28-­‐2. (a) Error function (Scale = 0.0109, Shift = 460.94) (b) Normal Distribution (Mean = 460.94, Standard Deviation = 64.78). 28-­‐3. (a) Uniform Distribution (Min = 302, Max = 496). (b) Max Extreme Distribution (Mode = 62.01, Scale = 46.41, Shift = 301.99). 28-­‐4. A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
Activity
A Secure funding
B Design Building
C Site Preparation
D Foundation
E Framing
F Electrical
G Plumbing
H Walls and Roof
I Finish Work
J Landscaping
B
Predecessor
—
A
A
B, C
D
E
E
F, G
H
H
C
D
Distribution
Normal (mean, st. dev.)
Uniform (min, max)
Triangular (min, most likely, max)
Triangular (min, most likely, max)
Triangular (min, most likely, max)
Triangular (min, most likely, max)
Triangular (min, most likely, max)
Triangular (min, most likely, max)
Triangular (min, most likely, max)
Fixed (5)
E
Parameters
6
1
6
10
1.5
2
1.5
2
3
4
2
3
3
4
4
5
5
6
F
2.5
3
6
5
5
7
7
G
H
I
(all times in months)
Start
Activity
Finish
Time
Time
Time
0.0
6.257
6.3
6.3
9.188
15.4
6.3
1.661
7.9
15.4
2.219
17.7
17.7
4.100
21.8
21.8
2.322
24.1
21.8
3.514
25.3
25.3
4.862
30.1
30.1
6.000
36.1
30.1
5
35.1
Project Completion Time 35.140
Mean Project Completion Time 34.917
(a) The mean project completion time is approximately 35 months. 28-1
(b) The probability that the project completion time will be less than 36 months is approximately 72%. (c) Activity B and then Activity A have the largest impact on the project completion time. 28-2
28-­‐5. A
B
C
D
E
F
1 Size of Claim
Prob.
Distribution
Parameters
Claim (If Claim is This Size)
2
None
40%
Fixed
$0
$0
3
Small
40%
Uniform(Min,Max)
$0
$2,000
$54
4
Large
20%
Uniform(Min,Max)
$2,000
$20,000
$4,554
5
6 Size of Claim
2
7 (0=None,1=Small,2=Large)
Simulated Claim
$4,554
8
Mean Claim
$2,567
The mean claim is approximately $2,567. 28-3
28-­‐6. (a) Option 2 (Hotel Project only). The mean NPV is approximately $11.5 million, with an approximately 70% chance of being nonnegative. A
B
C
D
Project Simulated
3
Cash Flow
4
($millions)
5 Hotel Project:
6
Construction Costs:
Year 0
-80
7
Year 1
-76.763
8
Year 2
-74.188
9
Year 3
-98.612
10
Revenue per Share
Year 4
38.767
11
Year 5
69.990
12
Year 6
65.203
13
Selling Price per Share
Year 7
793.807
14
15 Shopping Center Project
16
Construction Costs:
Year 0
-90
17
Year 1
-49.723
18
Year 2
-26.847
19
Year 3
-57.320
20
Revenue per Share
Year 4
15.628
21
Year 5
9.743
22
Year 6
14.163
23
Selling Price per Share
Year 7
612.961
24
Think Big's
25
Simulated Cash Flow
26
($millions)
27
28
Year 0
-13.200
29
Year 1
-12.666
30
Year 2
-12.241
31
Year 3
-16.271
32
Year 4
6.397
33
Year 5
11.548
34
Year 6
10.759
35
Year 7
130.978
36
37
Net Present Value ($millions)
37.769
38
39
MeanNPV ($millions)
11.546
E
F
G
H
Normal
Normal
Normal
Normal
Normal
Normal
Uniform
-80
-80
-70
30
40
50
200
5
10
15
20
20
20
844
(mean, st. dev.)
(mean, st. dev.)
(mean, st. dev.)
(mean, st. dev.)
(mean, st. dev.)
(mean, st. dev.)
(lower, upper)
Normal
Normal
Normal
Normal
Normal
Normal
Uniform
-50
-20
-60
15
25
40
160
5
5
10
15
15
15
615
(mean, st. dev.)
(mean, st. dev.)
(mean, st. dev.)
(mean, st. dev.)
(mean, st. dev.)
(mean, st. dev.)
(lower, upper)
Share
16.50%
0.00%
Hotel
Shopping Center
Cost of Capital
10%
28-4
(b) Option 3 (Shopping Center Project only). The mean NPV is approximately $6.6 million, with an approximately 71% chance of being nonnegative. A
B
C
D
Project Simulated
3
Cash Flow
4
($millions)
5 Hotel Project:
6
Construction Costs:
Year 0
-80
7
Year 1
-79.864
8
Year 2
-56.819
9
Year 3
-69.994
10
Revenue per Share
Year 4
70.702
11
Year 5
57.821
12
Year 6
16.736
13
Selling Price per Share
Year 7
549.769
14
15 Shopping Center Project
16
Construction Costs:
Year 0
-90
17
Year 1
-56.321
18
Year 2
-26.445
19
Year 3
-87.406
20
Revenue per Share
Year 4
0.980
21
Year 5
38.831
22
Year 6
25.538
23
Selling Price per Share
Year 7
407.887
24
Think Big's
25
Simulated Cash Flow
26
($millions)
27
28
Year 0
-11.799
29
Year 1
-7.384
30
Year 2
-3.467
31
Year 3
-11.459
32
Year 4
0.128
33
Year 5
5.091
34
Year 6
3.348
35
Year 7
53.474
36
37
Net Present Value ($millions)
2.593
38
39
MeanNPV ($millions)
6.571
E
F
G
Normal
Normal
Normal
Normal
Normal
Normal
Uniform
-80
-80
-70
30
40
50
200
5
10
15
20
20
20
844
(mean, st. dev.)
(mean, st. dev.)
(mean, st. dev.)
(mean, st. dev.)
(mean, st. dev.)
(mean, st. dev.)
(lower, upper)
Normal
Normal
Normal
Normal
Normal
Normal
Uniform
-50
-20
-60
15
25
40
160
5
5
10
15
15
15
615
(mean, st. dev.)
(mean, st. dev.)
(mean, st. dev.)
(mean, st. dev.)
(mean, st. dev.)
(mean, st. dev.)
(lower, upper)
Hotel
Shopping Center
Cost of Capital
H
Share
0.00%
13.11%
10%
28-5
(c) Option 1 appears to be the best. It has the highest expected NPV ($18 million vs. less than $12 million vs. less than $7 million) and there is less chance of losing money (less than 20% vs. nearly 30% for options 2 and 3). 28-­‐7. (a) The mean profit is approximately $572. There is a 100% chance of making at least $0 profit. 1
2
3
4
5
6
7
8
9
10
11
12
13
14
A
Purchase Price
Selling Price
B
$100
$150
Order Quantity
14
Demand
14
Revenue
Purchase Cost
Total Profit
Mean Total Profit
C
Custom Discrete
$2,100
$1,400
$700
$572.50
D
E
Value
10
11
12
13
14
15
16
17
18
Probability
0.05
0.1
0.1
0.15
0.2
0.15
0.1
0.1
0.05
28-6
(b) Thirteen tickets maximizes Susan’s mean profit. (c) Order Quantity
10
11
12
13
14
15
16
17
18
Mean Total Profit
$500.00
$542.50
$570.00
$582.50
$572.50
$532.50
$470.00
$392.50
$300.00
28-7
(d) Thirteen tickets is the optimal order quantity found by Solver. 1
2
3
4
5
6
7
8
9
10
11
12
13
14
A
Purchase Price
Selling Price
B
$100
$150
Order Quantity
13
Demand
13
Revenue
Purchase Cost
Total Profit
Mean Total Profit
C
Custom Discrete
$1,950
$1,300
$650
$582.50
D
E
Value
10
11
12
13
14
15
16
17
18
Probability
0.05
0.1
0.1
0.15
0.2
0.15
0.1
0.1
0.05
28-­‐8. (a) A bid of approximately $5.3 million maximmizes the mean profit. OurBid
5.20
5.25
5.30
5.35
5.40
5.45
5.50
5.55
5.60
Mean Profit ($million)
0.469
0.481
0.485
0.482
0.476
0.467
0.405
0.315
0.253
28-8
(b) The optimal bid is approximately $5.302 million, as found by Solver. A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
B
C
D
E
Reliable Construction Co. Contract Bidding
Data
Our Project Cost ($million)
Our Bid Cost ($million)
Competitor Bids
Bid ($million)
Distribution
4.550
0.050
Competitor 1
4.753
Competitor 2
5.832
Competitor 3
5.648
Triangular
Triangular
Uniform
Competitor Distribution Parameters (Proportion of Our Project Cost)
Minimum
95%
110%
Most Likely
130%
125%
Maximum
160%
140%
Competitor Distribution Parameters ($millions)
Minimum
4.323
Most Likely
5.915
Maximum
7.280
Minimum Competitor
Bid ($million)
4.753
Our Bid ($million)
5.302
Win Bid?
Profit ($million)
Mean Profit ($million)
0
5.005
5.688
6.370
120%
130%
5.460
5.915
(1=yes, 0=no)
-0.050
0.496140938
28-­‐9. (a) A long-­‐term loan of approximately $5 million maximizes Everglades’s mean ending balance. LT Loan
0
5
10
15
20
Mean 2021 Ending Balance
8.95
9.17
8.21
6.13
3.67
28-9
(b) (c) The optimal long-­‐term loan is approximately $3.82 million, as found by Solver. A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
B
C
D
E
F
G
H
I
J
K
L
M
N
O
P
ST
Loan
4.41
7.46
11.79
11.81
6.67
4.30
9.12
0.99
0.00
0.00
Ending
Balance
0.50
0.50
0.50
0.50
0.50
0.50
0.50
0.50
0.78
8.57
4.56
>=
>=
>=
>=
>=
>=
>=
>=
>=
>=
>=
Minimum
Balance
0.50
0.50
0.50
0.50
0.50
0.50
0.50
0.50
0.50
0.50
0.50
Everglade Cash Flow Management Problem When Applying Simulation
LT Rate
ST Rate
5%
7%
Start Balance
Minimum Cash
1
0.5
(all cash figures in millions of dollars)
Cash Flow (Triangular Distribution)
Year
2011
2012
2013
2014
2015
2016
2017
2018
2019
2020
2021
Min
-9
-4
-7
0
3
1
-6
4
-5
5
Likely
-8
-2
-4
3
6
3
-4
7
-2
10
Max
-7
1
0
7
9
5
-2
12
4
18
Simulated
Cash
Flow
-8.73
-2.55
-3.62
0.99
6.16
3.03
-4.34
8.96
1.54
7.98
LT
Loan
3.82
LT
Interest
-0.19
-0.19
-0.19
-0.19
-0.19
-0.19
-0.19
-0.19
-0.19
-0.19
Balance
ST
LT
ST
Before
Interest Payback Payback ST Loan
-3.91
-0.31
-4.41
-6.96
-0.52
-7.46
-11.29
-0.83
-11.79
-11.31
-0.83
-11.81
-6.17
-0.47
-6.67
-3.80
-0.3008
-4.2972
-8.62
-0.64
-9.12
-0.49
-0.0696
-0.9938
0.78
0
0
8.57
0
-3.82
0
4.56
Mean 2021 Ending Balance
28-10
9.20
28-­‐10. (a) Accepting approximately 185 reservations maximizes the mean profit. (b) Reservations to Accept
180
181
182
183
184
185
186
187
188
189
190
Mean Profit
$11,612
$11,719
$11,806
$11,875
$11,918
$11,940
$11,936
$11,917
$11,875
$11,812
$11,732
28-11
(c) The optimal number of reservations to accept is approximately 185, as found by Solver. B
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
D
E
F
Available Seats
Fixed Cost
Avg. Fare / Seat
Cost of Bumping
C
Data
150
$30,000
$300
$450
Ticket Demand
Demand (rounded)
161.58
162
Normal
Mean
195
Standard Dev.
30
Binomial
Tickets
Purchased
162
Probability
to Show up
80%
Reservations to Accept
185
Number that Show
129
Number of Filled Seats
Number Denied Boarding
129
0
Mean Filled Seats
Mean Denied Boarding
Mean Profit
141.04
0.86
$11,925
28-12
Ticket Revenue
Bumping Cost
Fixed Cost
Profit
$38,700
$0
$30,000
$8,700
CHAPTER 29: MARKOV CHAINS
29.2-1.
(a) Since the probability of rain tomorrow is only dependent on the weather today,
Markovian property holds for the evolution of the weather.
(b) Let the two states be ! œ Rain and " œ No Rain. Then the transition matrix is
P œ PÐ"Ñ œ  !Þ&
!Þ"
!Þ& 
!Þ* .
29.2-2.
(a) Let " œ increased today and yesterday,
# œ increased today and decreased yesterday,
$ œ decreased today and increased yesterday,
% œ decreased today and yesterday.
P œ PÐ"Ñ
 α"
α
œ #
!
 !
!
!
α$
α%
"  α"
"  α#
!
!
! 
! 

"  α$
"  α% 
(b) The state space is properly defined to include information about changes yesterday
and today. This is the only information needed to determine the next state, namely
changes today and tomorrow.
29.2-3.
Yes, it can be formulated as a Markov chain with the following ) Ð œ #$ Ñ states.
State
1
2
3
4
5
6
7
8
Today
inc
inc
inc
inc
dec
dec
dec
dec
1 Day Ago
inc
inc
dec
dec
inc
inc
dec
dec
2 Days Ago
inc
dec
inc
dec
inc
dec
inc
dec
These states include all the information needed to predict the change in the stock
tomorrow whereas the states in Prob. 29.2-2 do not consider the day before yesterday, so
they do not contain all necessary information to predict the change tomorrow.
29-1
29.3-1.
(a)
!Þ$
PÐ#Ñ œ 
!Þ"%
!Þ"'(
PÐ"!Ñ œ 
!Þ"'(
!Þ(
!Þ)' 
!Þ)$$
!Þ)$$ 
!Þ"(& !Þ)#&
PÐ&Ñ œ 
!Þ"'& !Þ)$& 
!Þ"'(
PÐ#!Ñ œ 
!Þ"'(
!Þ)$$
!Þ)$$ 
(b)
Ð8Ñ
PÐRain 8 days from now | Rain todayÑ œ P""
Ð8Ñ
PÐRain 8 days from now | No rain todayÑ œ P#" .
If the probability it will rain today is !Þ&,
Ð8Ñ
Ð8Ñ
PÐRain 8 days from nowÑ œ :8 œ !Þ&P""  !Þ&P#" .
Hence, :# œ !Þ##, :& œ !Þ"(, :"! œ !Þ"'(, :#! œ !Þ"'(.
(c) We find 1" œ !Þ"'( and 1# œ !Þ)$$. As 8 grows large, PÐ8Ñ approaches
1"
1
"
1#
,
1# 
the stationary probabilities. Indeed,
PÐ"!Ñ œ PÐ#!Ñ œ 
1"
1"
1#
.
1# 
29.3-2.
(a) Let states ! and " denote that a ! and a " have been recorded respectively. Then the
transition matrix is
Pœ
";
;
;
ß
";
where ; œ !Þ!!&.
(b)
!Þ*&#
PÐ"!Ñ œ 
!Þ!%)
!Þ!%)
!Þ*&# 
The probability that a digit will be recorded accurately after the last transmission is
!Þ*&#.
(c)
PÐ"!Ñ œ 
!Þ*)
!Þ!#
!Þ!#
!Þ*) 
The probability that a digit will be recorded accurately after the last transmission is !Þ*).
29-2
29.3-3.
(a)
! !Þ& 
 ! !Þ& !
 !Þ& ! !Þ& !
! 


P œ  ! !Þ& ! !Þ& ! .


!
! !Þ& ! !Þ&
 !Þ& !
! !Þ& ! 
(b)
(c) 1" œ 1# œ 1$ œ 1% œ 1& œ !Þ#.
29.4-1.
(a) P has one recurrent communicating class: Ö!ß "ß #ß $×.
(b) P has 3 communicating classes: Ö!× absorbing, so recurrent; Ö"ß #× recurrent and Ö$×
transient.
29-3
29.4-2.
(a) P has one recurrent communicating class: Ö!ß "ß #ß $×.
(b) P has one recurrent communicating class: Ö!ß "ß #×.
29.4-3.
P has 3 communicating classes: Ö!ß "× recurrent, Ö#× transient and Ö$ß %× recurrent.
29.4-4.
P has one communicating class, so each state has the same period %.
29.4-5.
(a) P has two classes: Ö!ß "ß #ß %× transient and Ö$× recurrent.
(b) The period of Ö!ß "ß #ß %× is # and the period of Ö$× is ".
29.5-1.
Pœ
α
""
"α
" 
1P œ 1 Ê α1"  Ð"  " Ñ1# œ 1" and 1"  1# œ "
α
Ê 1 œ  #"α" " ß #"α
" .
29.5-2.
We need to show that 14 œ Q"" for 4 œ !ß "ß á ß Q satisfies the steady-state equations:
Q
Q
14 œ Q
3œ! 13 T34 and
3œ! 13 œ ". These are easily verified, using
3œ! T34 œ " for every
4. The chain is irreducible, aperiodic and positive recurrent , so this is the unique
solution.
29.5-3.
Q œ & Ê 1" œ 1# œ 1$ œ 1% œ 1& œ "Î& œ !Þ#
The steady-state probabilities do not change if the probabilities for moving steps change.
29.5-4.
1 œ Ð!Þ&""ß !Þ#)*ß !Þ#Ñ
The steady-state market share for A and B are !Þ&"" and !Þ#)* respectively.
29-4
29.5-5.
(a) Assuming demand occurs after delivery of orders:
!
!
!
! 
 !Þ' !Þ% !
 !Þ$ !Þ$ !Þ% !
!
!
! 


 !Þ" !Þ# !Þ$ !Þ% !
!
! 


! 
P œ  ! !Þ" !Þ# !Þ$ !Þ% !


! !Þ" !Þ# !Þ$ !Þ% ! 
 !


!
!
! !Þ" !Þ# !Þ$ !Þ%
 !
!
!
! !Þ" !Þ# !Þ( 
(b) 1P œ 1 and 4 14 œ " Ê 1 œ Ð!Þ"$* !Þ"$* !Þ"$* !Þ"$) !Þ"%" !Þ"$! !Þ"(%Ñ.
(c) The steady-state probability that a pint of blood is to be discarded is
PÐH œ !Ñ † PÐstate œ (Ñ œ !Þ% x !Þ"(% œ !Þ!'*'.
(d) PÐneed for emergency deliveryÑ
œ #3œ" PÐstate œ 3Ñ † PÐH  3Ñ
œ !Þ"$* x Ð!Þ#  !Þ"Ñ  !Þ"$* x !Þ"
œ !Þ!&&'
29.5-6.
For an Ð=ß WÑ policy with = œ # and W œ $:
-ÐB>" ß H> Ñ œ 
"!  #&Ð$  B>" Ñ  &!maxÐH>  $ß !Ñ
&!maxÐH>  B>" ß !Ñ
for B>"  #
for B>" #.
OÐ!Ñ œ IÒ-Ð!ß H> ÑÓ œ )&  &!Ò∞
4œ% Ð4  $Ñ † T ÐH> œ 4ÑÓ ¶ )'Þ#,
OÐ"Ñ œ IÒ-Ð"ß H> ÑÓ œ '!  &!Ò∞
4œ% Ð4  $Ñ † T ÐH> œ 4ÑÓ ¶ '"Þ#,
OÐ#Ñ œ IÒ-Ð#ß H> ÑÓ œ !  &!Ò∞
4œ% Ð4  #Ñ † T ÐH> œ 4ÑÓ ¶ &Þ#,
OÐ$Ñ œ IÒ-Ð$ß H> ÑÓ œ !  &!Ò∞
4œ% Ð4  #Ñ † T ÐH> œ 4ÑÓ ¶ "Þ#.
maxÐ$  H>" ß !Ñ
B>" œ 
maxÐB>  H>" ß !Ñ
for B>  #
for B> #
 !Þ!)!
 !Þ!)!
Pœ
!Þ#'%
 !Þ!)!
!Þ$') 
!Þ$') 

!
!Þ$') 
!Þ")%
!Þ")%
!Þ$')
!Þ")%
!Þ$')
!Þ$')
!Þ$')
!Þ$')
Solving the steady-state equations gives Ð1! ß 1" ß 1# ß 1$ Ñ œ Ð!Þ"%)ß !Þ#&#ß !Þ$')ß !Þ#$#Ñ.
Then the long-run average cost per week is $4œ! OÐ4Ñ † 14 œ $!Þ$(.
29-5
29.5-7.
(a)
maxÐB>  #  H>" ß !Ñ
B>" œ 
maxÐB>  H>" ß !Ñ
 !Þ#'%
 !Þ!)!
Pœ
!Þ#'%
 !Þ!)!
!Þ$')
!Þ")%
!Þ$')
!Þ")%
!Þ$')
!Þ$')
!Þ$')
!Þ$')
for B> Ÿ "
for B> #
! 
!Þ$') 

!
!Þ$') 
Solving the steady-state equations gives Ð1! ß 1" ß 1# ß 1$ Ñ œ Ð!Þ")#ß !Þ#)&ß !Þ$')ß !Þ"'&Ñ.
(b) lim8Ä∞ I  8" 8>œ" -ÐB> Ñ œ ! † 1!  # † 1"  ) † 1#  ") † 1$ œ 'Þ%).
29.5-8.
(a)
P"" œ PÐH8" œ !Ñ  PÐH8" œ #Ñ  PÐH8" œ %Ñ œ $Î&
P"# œ PÐH8" œ "Ñ  PÐH8" œ $Ñ œ #Î&
P#" œ PÐH8" œ "Ñ  PÐH8" œ $Ñ œ #Î&
P## œ PÐH8" œ !Ñ  PÐH8" œ #Ñ  PÐH8" œ %Ñ œ $Î&
Pœ
$Î&
#Î&
#Î&
$Î& 
(b) 1 œ 1P and 1"  1# œ " Ê 1" œ 1# œ "Î#.
(c) P is doubly stochastic and there are two states, so 1" œ 1# œ "Î#.
(d)
OÐ"Ñ
œ IÒ-Ð"ß H8 ÑÓ
œ Ð#Î&ÑÒ$  #Ð"ÑÓ  Ð#Î&ÑÒ$  #Ð#ÑÓ  Ð"Î&ÑÐ"Ñ  Ð%Î&ÑÒ"  #  $Ó
œ *Þ),
OÐ#Ñ
œ IÒ-Ð#ß H8 ÑÓ
œ Ð#Î&ÑÒ$  #Ð"ÑÓ  Ð"Î&ÑÒ$  #Ð#ÑÓ  Ð"Î&ÑÐ#  "Ñ  Ð%Î&ÑÒ"  #Ó
œ 'Þ%.
So the long-run average cost per unit time is *Þ)Ð"Î#Ñ  'Þ%Ð"Î#Ñ œ )Þ".
29-6
29.5-9.
Ð8Ñ
(a) PÐthe unit will be inoperable after 8 periodsÑ œ P!#
Ð8Ñ
Ð8Ñ
8 œ #: P!# œ !Þ!%; 8 œ &: P!# œ !Þ!$(;
Ð8Ñ
Ð8Ñ
8 œ "!: P!# œ !Þ!$*; 8 œ #!: P!# œ !Þ!$).
(b) 1! œ !Þ'"&, 1" œ !Þ"*#, 1# œ !Þ!$), and 1$ œ !Þ"&%.
(c) Long-run average cost per period is $!ß !!!1$ œ %ß '#!.
29.6-1.
(a)
Pœ
(b)
.!!
.!"
."!
.""
!Þ*&
!Þ&!
!Þ!&
!Þ&! 
œ "  !Þ!&."!
œ "  !Þ*&.!"
œ "  !Þ&!."!
œ "  !Þ&!.!"
Ê .!! œ "Þ"ß .!" œ #!ß ."! œ #ß ."" œ ""
29.6-2.
(a) States: ! œ Operational, " œ Down, # œ Repaired.
Pœ
 !Þ*
!
 !Þ*
!Þ" ! 
! "
!Þ" ! 
29-7
(b) We need to solve .34 œ "  5Á4 P35 .54 for every 3 and 4.
.!! œ "  !Þ"."!
."! œ "  .#!
.#! œ "  !Þ"."!
Ê .!! œ ""Î*ß ."! œ #!Î*ß .#! œ ""Î*
.!" œ "  !Þ*.!"
."" œ "  .#"
.#" œ "  !Þ*.!"
Ê .!" œ "!ß ."" œ ""ß .#" œ "!
.!# œ "  !Þ*.!#  !Þ"."#
."# œ "  !
.## œ "  !Þ*.!#  !Þ"."#
Ê .!# œ ""ß ."# œ "ß .## œ ""
The expected number of full days that the machine will remain operational before the
next breakdown after a repair is completed is .!" œ "!.
(c) It remains the same because of the Markovian property. The expected number of days
the machine will remain operational starting operational does not depend on how long the
machine remained operational in the past.
29.6-3.
(a) We order the states as Ð"ß "Ñ, Ð!ß "Ñ and Ð"ß !Ñ and write the transition matrix:
 !Þ*
P œ !Þ*
 !Þ*
!Þ" ! 
! !Þ" .
!Þ" ! 
(b) .$$ œ "Î1$ . From 1 œ 1P and 1 † " œ ", we get 1$ œ "Î""!, so the expected
recurrence time for the state Ð"ß !Ñ is .$$ œ ""!.
29.6-4.
(a)
 !Þ#& !Þ&
P œ !Þ(& !Þ#&
 !Þ#& !Þ&
!Þ#& 
!
!Þ#& 
29-8
(b)
PÐ#Ñ œ
Ð&Ñ
P
!Þ%
 !Þ%%*
œ !Þ%&" !Þ$**
 !Þ%%*
!Þ%
Ð"!Ñ
P
(c)
!Þ$(& !Þ"#& 
 !Þ&
!Þ$(& !Þ%$) !Þ"))
 !Þ&
!Þ$(& !Þ"#& 
 !Þ%&
œ !Þ%&
 !Þ%&
!Þ%
!Þ%
!Þ%
!Þ"& 
!Þ"%*
!Þ"& 
!Þ"& 
!Þ"&
!Þ"& 
.!! œ "  !Þ&."!  !Þ#&.#!
."! œ "  !Þ#&."!
.#! œ "  !Þ&."!  !Þ#&.#!
Ê .!! œ #!Î*ß ."! œ %Î$ß .#! œ #!Î*
.!" œ "  !Þ#&.!"  !Þ#&.#"
."" œ "  !Þ(&.!"
.#" œ "  !Þ#&.!"  !Þ#&.#"
Ê .!" œ #ß ."" œ # "# ß .#" œ #
.!# œ "  !Þ#&.!#  !Þ&."#
."# œ "  !Þ(&.!#  !Þ#&."#
.## œ "  !Þ#&.!#  !Þ&."#
Ê .!# œ #!Î$ß ."# œ )ß .## œ #!Î$
(d) The steady-state probability vector is Ð!Þ%& !Þ% !Þ"&Ñ.
(e) 1 † G œ !Ð!Þ%&Ñ  #Ð!Þ%Ñ  )Ð!Þ"&Ñ œ $ # Î week
29.6-5.
(a)
!
!
Pœ
!
"
!Þ)(& !Þ!'# !Þ!'# 
!Þ(& !Þ"#& !Þ"#& 

!
!Þ&
!Þ&
!
!
! 
1! œ !Þ"&%, 1" œ !Þ&$), 1# œ !Þ"&%, and 1$ œ !Þ"&%
(b) 1 † G œ "Ð!Þ&$)Ñ  $Ð!Þ"&%Ñ  'Ð!Þ"&%Ñ œ $ "*#$Þ!)
(c)
.!!
."!
.#!
.$!
œ "  !Þ)(&."!  !Þ!'#&.#!  !Þ!'#&.$!
œ "  !Þ(&."!  !Þ"#&.#!  !Þ"#&.$!
œ "  !Þ&.#!  !Þ&.$!
œ"!
So the expected recurrence time for state ! is .!! œ 'Þ&.
29-9
29.7-1.
(a) P!! œ PX X œ "; P3ß3" œ ; ; P3ß3" œ :; P3ß5 œ ! else.
" ! !
; ! :

ã

Pœ



(b)
!
!
ä





!

:
"
â
â
;
!
!
! :
; !
! !
Class 1: Ö!× absorbing
Class 2: ÖX × absorbing
Class 3: Ö"ß #ß á ß X  "× transient
(c) Let 03O œ PÐabsorption at O starting at 3Ñ. Then 0!! œ 0$$ œ ", 0$! œ 0!$ œ !.
Since P34 œ ! for l3  4l Á " and P3ß3" œ :, P3ß3" œ ; , we get:
0"!
0"$
0#!
0#$
œ ;  :0#!
œ "  0"!
œ ;0"!
œ "  0#!
Solving this system gives
0"! œ
;
":;
œ !Þ))', 0"$ œ !Þ""%, 0#! œ !Þ'#, 0#$ œ !Þ$).
(d) Plugging in : œ !Þ( in the formulas in part (c), we obtain
0"! œ !Þ$), 0"$ œ !Þ'#, 0#! œ !Þ""%, 0#$ œ !Þ))'.
Observe that when :  "Î#, the drift is towards X and when :  "Î#, it is towards !.
29.7-2.
(a)
! œ Have to honor warranty
" œ Reorder in 1st year
# œ Reorder in 2nd year
$ œ Reorder in 3rd year
!
 "
 !Þ!" !
Pœ
!Þ!& !
 !
!
!
! 
!Þ**
! 

!
!Þ*&
!
" 
(b) The probability that the manufacturer has to honor the warranty is 0"!.
0"! œ !Þ!"0!!  !0"!  !Þ**0#!  !0$!
0#! œ !Þ!&0!!  !0"!  !0#!  !Þ*&0$!
0!! œ " and 0$! œ !
Ê 0"! œ !Þ!"  !Þ**0#! and 0#! œ !Þ!&
Ê 0"! œ !Þ!&*& œ &Þ*&%.
29-10
29.8-1.
(a)
(b) Steady-state equations:
$1! œ #1"
%1" œ $1!  #1#
$1# œ #1"  #1$
#1$ œ 1#
1!  1 "  1 #  1 $ œ "
% ' ' $
(c) Solving the steady-state equations gives 1 œ  "*
ß "* ß "* ß "* .
29.8-2.
(a) Let the state be the number of jobs at the work center.
(b) Steady-state equations:
"
# 1!
œ 1"
$
# 1"
œ "# 1!  1#
1# œ "# 1"
1!  1 "  1 # œ "
(c) Solving the steady-state equations gives 1 œ  %( ß #( ß "( .
29-11