OPERATIONS RESEARCH METHODS:
RELATED PRODUCTION, DISTRIBUTION, AND
INVENTORY MANAGEMENT APPLICATIONS
B. D. CRAVEN
University of Melbourne
S. M. N. ISLAM
Victoria University
March 2005
ii
PREFACE
Business managements make daily decisions on many issues, such
as how much and where to produce, for which market, what prices to
set, and how much stock to keep. Mathematical models can help to
make the best decisions, among the possible alternatives.
The objective of this book is to present a survey of selected
Operations Research methods, and some of their applications to
managerial decision making, concerning production, inventory,
distribution, and integrated supply chain modelling. Also included are
listing of some computer programs to compute operations research
models.1
Operations Research (O.R.) has been termed The Science of
Better. A problem in the real world is modelled, usually in mathematical
terms, then mathematical techniques, together with data analysis and
computational algorithms, are applied, in order to find ways to do the
job better. The word Operations derives from the many successful
applications of O.R. to military operations in the 1940s. But, since
then, most O.R. applications have been to peaceful activities, especially
to business management, of which planning industrial production, and
scheduling airlines, and other transportation, have been prominent. The
name Management Science denotes the same discipline, with some
emphasis on business management. Practitioners of Operations
Management will find many of these techniques relevant. The areas of
Logistics, Supply Chain Management, Decision Sciences, and
Manufacturing Management deal with similar applications.2
This book is concerned with O.R. methods. The commencing chapter
(numbered chapter 0 - computers count up from zero) gives a general
discussion of model building, various examples of their applications, and
some discussion of the limitations of some models. It is suggested that
an O.R. practitioner should understand the models and the techniques,
and, while using computers and computer packages extensively, should
not depend on them to decide a model (possibly inappropriate) to use.
Chapter 1 discusses linear programming . As well as the
mathematics, section 1.16 on cost data discusses when such models
are appropriate, and what sort of data must be sought.
Chapter 2 discusses dynamic programming, and several of its many
applications, to capital budgetting and to a rental problem. The effect
of random elements is also considered.
Chapter 3 discusses the critical path method.
Chapter 4 discusses planning over time. This includes discussion of
interest rates and present value, effect of inflation, risk-averse utility,
decision trees and planning over time, and forecasting. A concise
introduction to the use of spreadsheets for such planning and
calculations is given in chapter 9.
Chapter 5 discusses inventory, presenting various versions of
1
Some programs written in FORTRAN AND GAMS are included.
2
See Ragsdale (2001), and Gass and Harris (2001), for various such applications.
iii
economic order quantity, with applications including seasonal demand,
discounting, and the issues that arise when inventory and production
must be managed together, as especially in supply chain management.
Chapter 6 outlines the topic of networks, which often arise in
planning models.
Chapter 7 discusses various methods for nonlinear optimization ,
and lists some of their applications. Although many models are
described by linear equations, nonlinear models are also often needed.
Chapter 8 discusses simulation techniques, with applications to
queue, inventory and storage models, and also to networks models.
Often simulation can obtain results, when no exact formulas are
possible. The discussion of simulation methods, both those based on
events and those planned on a time scale, should enable a user to
understand what a computer package for simulation should be doing,
and perhaps write his own simulation program.
In chapter 9, a number of related computer programs are listed.
For these various methods and applications, the basic methods and
primary applications are presented, together with some reviews of
recent developments.
In comparison with some other books on the methods of Operations
Research, the present book gives a concise account of the methods and
applications. A moderate mathematical background is assumed (some
calculus and matrix algebra). A reader is assumed to want to
understand the methods, and the modelling assumptions made, rather
than leave it all to a computer package. Some aspects that may be
innovative include the following:
• A detailed discussion (in chapter 0) of O.R. modelling, including the
difficulties involved, and various areas of application to real problems.
• A discussion (section 1.16) of cost data, replacing the common
(inadequate) assumption that everything is linear.
• Various topics (in chapter 4) concerning planning over time, i
including discussion of riak-averse utility, and stability analysis.
• Some critical discussion (sections 5.11 and 5.12) of supply chain
models.
• Some critical discussion (sections 7.19 and 7.20) of multiobjective
optimization, and of optimization when the usual convexity
assumptions fail.
This book presents phases and aspects of applied operations
research studies with the emphasis on business, logistics, and operation
management applications. This book includes many examples of
applications of operations research methods in those areas. It can be
used as a text or a reference for a course on operations management
and logistics/supply chain management at the Masters or Doctoral level
and a text for undergraduate students in operations research. It will also
be of importance to corporate executives, academics, practitioners,
and business consultants.
iv
TABLE OF CONTENTS
Chapter 0
Introduction to Operations Research and Applications
0.1 Mathematical models for management
0.2 Some reference books
0.3 Initial example
0.4 Check-list, when setting up an Operations Research model
0.5 Some other examples of Operations Research models
in business
0.6 Importance of model building, and limitations of computers
0.7 Why the name?
0.8 The usefulness of spreadsheets
0.9 Presentation of conclusions and recommendations
6
11
12
12
12
Chapter 1 Linear Programming
1.1 Introduction - management models
1.2 Standard form for a linear program
1.3 Idea of simplex method
1.4 Simplex tableaus
1.5 Information from tableau
1.6 Two-phase method
1.7 Unsigned variables
1.8 Revised simplex method
1.9 Dual linear program
1.10 Dual properties
1.11 Sensitivity
1.12 Dual simplex method
1.13 Jumps in shadow costs
1.14 Integer linear programming
1.15 Transportation problem
1.16 Cost data (piecewise linear cost function)
1.17 Some exercises on linear programming
1.18 Computer input/output for linear programming
1.19 A note on interior-point methods
Appendix: Some theory of linear programming
14
16
16
17
18
19
20
20
22
23
25
25
26
27
29
33
36
40
43
44
Chapter 2
Dynamic Programming
2.1 Introduction
2.2 A routing problem
2.3 Why does this work?
2.4 Forward analysis
2.5 An investment (or capital budgetting) problem
2.6 A model for inventory and production
2.7 A rental problem
2.8 Stochastic inventory
2.9 Some exercises and examples
46
46
47
48
48
49
51
52
53
v
1
2
3
3
Chapter 3
Critical Path Method
3.1 Introduction
3.2 Example of CPM
3.3 Method of calculation
3.4 Example of computer output
3.5 Alternative presentation
3.6 Random variation in time durations
3.7 Exercises
55
55
56
57
58
59
59
Chapter 4
Planning over Time, Uncertainty and Forecasting
4.1 Present value
4.2 Annual cost and rate of return
4.3 Allowing for inflation?
4.4 Different interest rates
4.5 Decision trees
4.6 Risk-averse utility
4.7 Decision tree examples and exercises
4.8 Stability over time
4.9 Indifference curves
4.10 Forecasting and exponential smoothing
61
62
62
63
64
68
69
70
71
72
Chapter 5
Inventory Management
5.1 What is inventory?
5.2 Economic order quantity
5.3 Inventory model for constant sales rates
5.4 Inventory model for variable demand
5.5 Sequencing a number of products
5.6 What stock to carry of a critical component?
5.7 Managing inventory and production together
5.8 Seasonal demand
5.9 Discounted and floor prices
5.10 The newsboy problem
5.11 Supply chain management
5.12 Examples of supply chain models
5.13 Some exercises
74
74
75
76
77
78
78
80
80
81
81
83
85
Chapter 6
Network Planning
6.1 Why networks?
6.2 Idea of out-of-kilter algorithm
6.3 Idea of a shortest-path algorithm
6.4 Idea of Ford-Fulkerson transportation method
6.5 Branch and bound
6.6 Remarks on mixed-integer linear programming
6.7 Job-shop scheduling
6.8 Examples
86
87
88
88
89
92
92
94
Chapter 7
Nonlinear Optimization Methods
7.1 Introduction
95
vi
7.2
7.3
7.4
7.5
7.6
7.7
7.8
7.9
7.10
7.11
7.12
7.13
7.14
7.15
7.16
7.17
7.18
7.19
7.20
Separable programming
Convergence and convergence rate
Lagrange multipliers
Sensitivity
Computing an iterative algorithm
Descent methods
Fletcher-Reeves algorithm
Davidon-Fletcher-Powell (DFP) algorithm
Example
Linesearch
Descent methods and stability
Constrained minimization
Sequential unconstrained minimization technique (SUMT)
Projected gr5dient
Quadratic programming by Wolfe's method
Further comments on constrained mimimization
Multiobjective optimization
Invexity
Some applications of nonlinear programming
95
96
97
99
99
100
101
101
102
103
105
106
107
107
108
109
112
113
114
Chapter 8 Simulation
8.1 Idea of simulation
8.2 Queue and storage models
8.3 Examples of simulation
8.4 Getting pseudorandom numbers
8.5 Some simulation models
8.6 Some other methods and models
115
115
118
120
122
122
Chapter 9 Some Computer Programs
9.1 Remarks on computer programs for O. R. problems
9.2 Using spreadsheets
9.3 Using Excel Solver
9.4 An example of a GAMS package for capital budgetting
125
126
129
131
9.5
9.6
9.7
9.8
9.9
9.10
Introduction to the FORTRAN programming language
A linear programming package lpZSq
Documentation for the linear programming package lpZSq
A FORTRAN program for critical path
A program for one application of dynamic programming
FORTRAN programs to illustrate methods for unconstrained
optimization
Bibliography
133
138
144
147
151
153
160
vii
Chapter 0
INTRODUCTION TO OPERATIONS RESEARCH
AND APPLICATIONS
0.1
Mathematical models for management
A business firm must make decisions on many issues relating to
production, distribution, consumers, shareholders, information
processing, employees, society, pricing etc..These decisions typically
concern production and inventory planning, sales forecasting, capital
budgeting, investment planning, materials requirement planning,
locational decisions, personnel management and planning, pricing,
distribution, management and planning, integrated supply chain or
logistics management and planning, as they affect the several parts of
the firm. These questions are sometimes considered under the heading
of Operations Management.
Business decision making requires the choice of the best decision
among alternatives, or at any rate a decision that gives a substantial
improvement. The objectives may be revenue maximisation, cost
minimisation, satisfactory performance regarding social responsibility,
shareholder value maximisation, etc., or sometimes the survival of the
firm in adverse circumstances. The decisions are constrained by
requirements such as budget and resource constraints. Methods of
Operations Research (O.R.) are well adapted to such decision making in
business. Applications of OR to business management have been
discussed in different disciplines such as Management Science,
Operations Management, Logistics Management, Supply Chain
Management, and Decision Sciences.
In Operations Research, we set up and use mathematical models,
usually related to questions of planning in business, industry, or
management. Any model of a real-life situation must simplify it greatly,
by picking out those factors we think important for our purpose, and
neglecting the rest. Otherwise, we can't calculate, or predict. (Newton
approximated the sun and earth by masses concentrated at points.
This approximation succeeded). We must consider whether we have
we made a right selection. What check or validation is possible?
To put our model in mathematical terms makes us formulate our
ideas precisely - what exactly are we assuming? - and so we are less
likely to make hidden assumptions without knowing it. Moreover, once
we have a model in mathematical terms, we can manipulate it
effectively using concise mathematical language. Also calculations,
using a computer, are more readily set up. The model results provide
useful information for making business decisions.
The factors we recognize, as affecting the situation, consist of (i)
external factors ("exogenous factors"), which we regard as inputs to
the system, and whose effects we shall study, (ii) factors the model
1
tries to explain ("output", "dependent", or "endogenous" factors), and
(iii) factors we shall neglect. For example, if we are hired by a firm to
determine the level of production that will maximize profits, then profit
becomes an "output" for the system, and market factors become the
"inputs". But, for an economist who wants to explain production levels,
profit could be an internal variable (internal to the system being
studied, neither an input nor an output). We should be clear what is the
aim of our investigation. It often falls to the O.R. investigator to clarify
the objective. This involves much discussion with the propounder of the
problem, to find out "where the shoe is really pinching". For example, it
could be irrelevant to optimize a production schedule, if the real
problem is labor relations.
We can fall on our face by neglecting some essential factors. A
century ago, some mathematicians proved that aeroplanes were
impossible; they had neglected the essential "boundary layer", where
viscosity of the air plays an essential role. Likewise, in a business
model, it is possible to omit an essential factor - for example there
may be an upper bound on certain resources, which must not be
forgotten. This course has to present examples from books; however, in
a real-world situation, always go and see for yourself (the factory floor,
or the airport, or whatever) if you possibly can - it is usually different
from what was supposed.
0.2 Some reference books (a few out of the many published.)
The codes <DG> etc. are here for cross-referencing.
Note that a book such as <DG>, which includes discussion of
building suitable models, is of more value to an O.R. practitioner than a
book that mainly presents the computational algorithms. In practice,
the calculations are done on computers; however, one needs some
understanding of the algorithms, otherwise one never knows whether
the computer is producing sense or nonsense.
<DG> H.C. Daellenbach, J.A. George, D.C.. McNickle, 1984 (1st. ed.)
or 1978 (2nd. ed.). Introduction to Operations Research (Allyn &
Bacon).
<a good balance between model building and the mathematics>
<HW> R. Hesse and E. Woolsey, 1980. Applied Management Science
(a quick and dirty approach), (Science Research Associates).
<good for practical " real-world" details of application>
<B+> K.H. Bradshaw and others, 1982. An Operations Research
Casebook (Longman Cheshire).
<the source of many miniproject topics for an O. R. course>
J.C. Ecker and M. Kupferschmid, 1988.Introduction to Operations
Research (Wiley).
<MD> M.J.C. Martin and R.A. Denison, 1971. Case Exercises in
Operations Research ,(Wiley-Interscience)
2
<a good source of many project topics>
E.A. Bender, 1970. An introduction to mathematical modelling
(Wiley/Interscience).
H.A. Taha, 1982. Operations Research (Macmillan).
<better on calculation examples than model building>
B. Srinavasan and C.L. Sandblom, 1989, Quantitative Analysis for
Business Decisions (McGraw-Hill).
<includes some useful examples of decision trees>
<SDK> M. Syslo, N. Deo, J. Kowalik,1983. Discrete Optimization
Algorithms with Pascal Programs, (Prentice-Hall).
<See especially chapter 3: Optimization on Networks>
<WS> R.E.D.Woolsey and H. Swanson, 1969, 1975.Operations
Research for Immediate Application A Quick and Dirty Manual, (Harper &
Row).
<see especially chapters 6 and 7>
<MW> S. Makridakis and S. C. Wheelwright, 1989. Forecasting
Methods for Management (Wiley).
This list is not meant to exclude other, and more recent, textbooks.
Reference may be made to Thompson and Thore (1992), Ragsdale
(2001), and Vollman et al. (2005), listed in the bibliography, for more
examples of O. R. models in management. But it is not recommended
to use a text that is mainly a manual for a particular computer package
(see section 0.6).
0.3 Initial example
Think of a factory, whose running cost (which is to be minimized)
is a function of various production variables (how much do we make of
which product on which machine for which market?), subject to
constraints (limits on material resources, labor, time available, market
requirements). This may (but need not always) reduce to a linear
programming model). Note that we must, later, be specific as to what
we mean by cost. The following check-list (in 0.4) applies specifically
to linear programming models, but also gives general guidance in
setting up other kinds of O.R. models.
0.4 Check-list, when setting up an Operations Research model
The stages of an applied operations research study have been listed
as follows (following Taha, 1992):
a. defining the research or the policy problem
b. developing the operations research model for the relevant
system
c. implementing/solving the model numerically by some suitable
algorithms and computer programs
d. undertaking validation tests of the model and its results
e. analysing the implications of the model results for actions or
decisions related to the issues under the study.
3
A more detailed check-list is as follows:
(1). Find out what the problem really is! It won't come to you packages
in mathematical terms. Ask simple, relevant questions. Go and see for
yourself. "Obvious" assumptions can be wrong.
(2) What are the variables of the problem? What, in fact, can you
vary? If you seek to optimize, what is the objective function? Or are
there several ? Check that the variables relate to the decisions that
must be made. Would they give us enough information to act on? Many
variables cannot be negative. What alternatives are possible?
(3) List the constraints. They often include: input constraints
(restrictions or requirements on raw materials), capacity constraints
(e.g. available machine time, pipeline capacity, storage available, rate
at which a product can be made), materials balance ("what goes in
must come out", e.g. old inventory + production - part used to make
new product = new inventory + deliveries to customers), output
constraints (requirements, or limits, on sales, production, or rate of
return). Check that units of measurement are consistent, or make
them so. (A factory once measured its storage tank capacities in cubic
feet, but its flow rates - inconsistently - in gallons per minute. For
mathematical modelling, the flow rates were converted to cubic feet
per hour - to avoid hopeless confusion. Today, SI units (metres,
kilograms,. etc.) would be preferred).
(4) What data are needed? You may need all your tact to get it! Exact
figures may be unobtainable, or make little sense; maybe get upper and
lower estimates instead. Are the cost figures you get relevant ? If you
are planning production schedules, then likely costs of raw materials
and energy enter; wages may or may not, depending on the time scale
of planning.
(5) Check out a simplified example! So, find any blunders in your
model, or your computer program, before you risk your reputation. And
get quickly a "ballpark estimate" to please the boss ! (Don't be
frightened of a rough "back of envelope" calculation for this limited
purpose - if you can do it quickly !) Do the results make practical
sense? Are the computations, perhaps, only telling you what everyone
in tbe industry has known for twenty years ?
(6) Compute the full model. Often it is better to start from present
operating conditions, and try to improve them, rather than start from
scratch. What can be recommended for action (expressed in practical
4
terms, not mathematical symbols) ? Note the importance of writing a
report, intelligible to management. Don't expect managers to cope with
a schedule of how much to make of which product, if it is expressed as:
x⁄ = 17.5
x‹ = 29.7
and so on !
Instead, tell the, in their language ! Something like:
Daily production schedule
Dinguses
17.5 tonnes
Whatsits
29.7 tonnes
and so on.
(7) Model Validation and Verification.
In Gass and Harris (2001, p. 865), model validation is defined as
the process of determining how well the outputs of a mathematical
model of real-world problem conform to reality. Model validation is
often synonymously used with other terms such as model verification
and model testing. Model validation is one of the important steps of an
O.R. study. In an applied O.R. study, the reliability of the O.R. model and
its results need to be tested. For this purpose, the usual model
validation criteria can be adopted. According to Hazell and Norton
(1986, p. 269) model validation and its purposes are as follows
“"Validation of model is a process that leads to (1) a numerical report
of the models fidelity to the historical data, (2) improvements of the
model as a consequence of imperfect validation, (3) a qualitative
judgment on how reliable the model is for its stated purposes, and (4) a
conclusion (preferably explicit) for the kinds of uses it should not be
used for"
Several criteria can be used to test the validation of a model (see
Gass and Harris, 2001; Labys, 1982; Taha, 1992; Hazell and Norton,
1986) at three levels of validation tests: descriptive, analytical and
experimental. Three different types of validation criteria are applied to
these three levels of validation tests (although 2 types of validations
are suggested in Gass and Harris (2001) which are face validity and
predictive validity) (see Kresge 1980 for further details):
(i) Descriptive validation criteria:
The attainment of the objectives of the model.
The appropriateness of the model structure and the plausibility of
results.
(ii) Analytical validation criteria:
The plausibility and characteristics of models and their results.
The robustness of the results.
(iii) Experimental validation criteria:
Methodological tests of model documentation.
Cost and efficiency in model, storage, transfer and extension.
Tests for accuracy and efficiency of implementation as well as cost
of and efficiency in the software transfer, storage and extension.
5
The above is a detailed set of model validation criteria for O.R.
models. In a real life O.R. study often some simple validation tests are
used (Hazell and Norton, 1986): plausibility of results, and comparisons
of model results with actual or historical results, and other model
results. Also, it is important to compare the model predictions with an
independent set of observations. Those may be observations from a
different time period, from the period whose data were used to
estimate the parameters of the model.
(8) What happens if something changes?
Murphy's law makes sure it does ! How sensitive are your
conclusions to small changes in the data ? (The jargon phrases are
"post-optimality analysis" for linear programming, or (more generally)
"sensitivity analysis".) If your recommendation is very sensitive to a
particular item of data, or a particular assumption, find this out before
it finds you out!
If you calculate a linear programming model, it is seldom good
enough to just present an optimum schedule! Consider also what
happens if some likely things change. For example, a sales target for
some product may have been set; but would it be profitable to increase
it by a few percent, even though the distribution costs would then also
increase ? Also, there may (quite often) be several other schedules,
alternative to the calculated optimum schedule, which are nearly as
good; and sometimes these alternatives should be presented to a
management. They may want to take additional factors into account such as qualitative factors, that could not be included in the
mathematical model.
(9) Try to persuade the management to put your recommendations into
effect. This can be the hardest part ! If they do, observe carefully
what happens. Get some feedback of information. Should your model
then be altered ?
0.5 Some other examples of Operations Research models in business
There will be many such in later chapters. Chapter 1 discusses
various linear programming examples. For the present, consider without going into details - the following additional examples of
Operations Research models in business, including supply chain
questions.
(a ) A management must choose between several plans for future
action. For example, do they start at once to produce their new product
line ? Or do they conduct a market survey first ? That gives them,
hopefully, better information, but runs the risk of letting a competitor
6
get in first. The various possibilities may be shown on a diagram called
a decision tree. In this diagram, a branch divides into two or more
branches whenever the management must take some decision (go
ahead, or not? how much to make?), or whether the external world
decides it for them (the new product sells like hot cakes, or only a few
people want to buy it?) Based on this diagram, some evaluation of
alternative courses of action can be made. Of course, one is
estimating uncertain things (how likely is it that the sales will go up by
50% for the new product?), and such estimates have to be rough. But
one is better off analysing even rough information in a systematic and
quantitative way, than just relying on intuition - or optimism - to put
the picture together. (See Chapter 4)
(b) Planning industrial operations over months or years, as well as
comparing alternatives as in a, needs to be done systematically. A
dollar now is worth more than a dollar at some future date. (Even
without inflation, money can still earn interest.) So one has to discount
future payments or receipts, to an extent depending on the time
interval, to allow for this. One may thus calculate a present value of
operations that will happen gradually over time. A variant of this is to
calculate a (percentage) rate of return - a measuring stick for how
profitable or productive the planned operations will be (see Chapter 4).
Such calculations are often, and usefully, done using a spreadsheet
program on a desktop computer, so that all the relevant figures are
handily displayed, say with years as columns and various kinds of
payments or receipts as rows. This makes it easy to vary some
parameter, such as some interest rate, and see what will happen. Such
a presentation also allows another important aspect, called cash flow,
to be checked. A business must have enough money coming in, from
somewhere, to meet its essential obligations in each year. (A firm may
go bankrupt now, for lack of cash flow this year, even though its
prospects for future years appear bright.)
(c) Network models are often useful. This may describe some actual
network, of pipelines for oil or water, or an electrical or telephone
network, or a network of routes for road transport, whose performance
(and bottlenecks !) must be analysed. However, scheduling problems
can often be described by networks. If a number of tasks have to be
done in a suitable order (you can't put the roof on the building until you
have built some walls !), then a network can describe the scheduling.
Thus, each line (or "pipe", or arc) of the network could represent a
task, and each dot (or "junction", or node) of the network represents a
time of completion. One may then calculate how soon the whole
project could be completed, and which are the "bottlenecks" (see
Chapters 3 and 6).
7
(d) Inventory means things kept in store - whether raw materials, or
products partly made ("in-process inventory"), or completed products.
If the firm has no inventory of raw materials, then an interruption to
supply will stop production - and starting and stopping is expensive
(even, sometimes, disastrous !). If there is no inventory of finished
products, then the customers can't get the products when they need
them (and then do they buy them from another firm ?), or the firm's
production is further disrupted by starting and stopping to match
varying demand for the products. But carrying inventory costs money;
not only do storage facilities, and wages of people to run them, cost
money, but (often, most importantly) the cost of the materials held in
stock has to be borrowed from the bank, and costs bank charges. How
should these conflicting requirements be balanced ? Guesswork usually
gets it wrong ! (Many firms have spent far too much in connection with
inventory, keeping far too much, or far too little.) However,
mathematical models are available, and these things can be calculated
(see Chapter 5).
(e) Rail trucks, carrying raw materials for a factory, come to a goods
yard to be unloaded. Their rate of arrival is somewhat variable; and the
rate at which they can be handled is also somewhat variable. The usual
result is that some queue builds up of trucks, waiting to be unloaded.
That costs money - and the situation is even worse, if the materials are
perishable ! Similar queues arise in many other situations, including
(for example):
Passengers, and baggage, arriving at airline counters;
Telephone calls coming into a telephone exchange;
Waiting to use specialized machines in a factory.
Mathematical models are available for various kinds of queues, if
something is known (or can be measured) about variabilities, as well as
average rates. It is these quantities that make the differences, not
whether the queue is of people or rail trucks or telephone calls. The
mathematics is very close to that for inventories (see Section 8.1).
(f) Many scheduling problems, including some questions of inventory
management, scheduling production and employment, budgetting for
capital expenditure between several projects, have the common (and
awkward) feature that the number of possibilities is very great, far too
many to make a list of them to compare. Suppose that the scheduling
can be put on a time scale (weeks or months, say), or at any rate put in
order (number off the alternative investment projects 1,2,3,...).
Suppose also that the "cost" (maybe actual cost, or some other
quantity such as waste product, to be minimized), adds up over the
different "times". Then a technique called dynamic programming can
8
often be used, to avoid listing an impossible number of possibilities.
Under the "adds up" assumption, one may calculate from each "time"
to the next, instead of listing all possible schedules. Roughly stated, the
production (or whatever) for "today" depends on past history, only to
the extent that it is summed up in the production (or whatever) for
"yesterday" (see Chapter 3).
(g ) The property just stated (not needing to look further back than
"yesterday") may be thought of as a "short memory" property.
(Technically, it is called the Markovian property). This property
happens, quite often, with stochastic systems - those including some
random elements - in particular with many queues, and inventory
systems where (as usual) there is some variability in demand for the
products. Of course, it does not always happen; but, when it does, it
usually makes some calculations possible.
(h) The order of scheduling tasks in a workshop - say, when various
jobs must be scheduled on successive machines - often leads to job
shop scheduling problems, involving integers (in which order should a
list of jobs be done), rather than continuous variables. For some of
these practical problems, there are good techniques; for some others,
there are only rough approximations (see Chapter 6).
(i)
Suppose the yield (ratio of useful output to raw material input)
for some industrial process must be maximized, subject to some
constraints (it is no use blowing up the factory, or making the reaction
vessel freeze solid !). Usually, the functions here are nonlinear, so that
linear programming does not apply. Sometimes, such a problem can be
reformulated, using different variables, to allow linear programming to
be used as an approximation. But, quite often, there is too much
nonlinearity in the problem to allow this simplification. There are a
diversity of methods for calculating such nonlinear problems. Questions
of sensitivity to changes in the data are also important (see Chapter 7).
(j) Optimization is not only about maximizing profit or minimizing cost,
but can, and often should, consider environmental variables as well. For
example, a model may include a constraint which sets an upper bound
to air or water pollution. Or an environmental variable may be
assigned a cost, and included in the cost function that is to be
minimized.
(k) A model for optimal waste management was given by Sahoo, Kim,
Kim, Kraas, and Popov (2005). If only vehicle travel time is minimized,
the model is a mixed integer linear program (MILP). Thus (see Chapter
1 and Section 6.6), the model is described by linear equations, but
9
some of them may only take integer values. A more complicated
version of the model also seeks to minimize the number of vehicles, and
balance the workload. Of course, there must be some compromise
between these different objectives. While there are various computer
programs for MILP, they may not be suitable for this specialized
problem. The authors use several heuristic (approximate) methods,
including a clustering algorithm to generate the initial routes, related to
the travelling salesman problem, and also a simulated annealing method
(see Section 8.4) to improve the result.
(l) Various models for supply chain have been given recently. One, by
de Kok, Janssen, van Donemalen, van Wachem, Clerkx and Peeters
(2005) is cited in section 5.12. The variability of demand increases as
one moves back through the stages of the production process. The
authors propose a model, with linear equations describing inventory.
However, their main concern is to reduce time lags in the planning
process.
Another supply chain model, by Troyer, Smith, Marshall, Yaniv,
Tayur, Barkman, Kaya and Liu (2005), seeks to improve the proportion
of deliveries that are on time, over a network of factories and dealers,
while reducing inventories. Here, inventories must be held at various
places in the network, with appropriate levels of safety stock (see
chapter 5 for what this means for a single inventory). But the demands
must be propagated down the network (see Section 5.11.)
(m) Metty, Harlan, Samelson, Moore, Schnweur, Raskina, Schneur,
Kanner, Potts and Robbins (2005) gave a model for a supplier
negotiation process in the telecommunications industry. This led again
to a MILP model.
Hicks, Madrid, Milligan, Pruneau, Kanaley, Dumas, Lacroix,
Desrisoers and Soumis (2005) gave a MILP model for scheduling the use
of aircraft which are shared between several firms. This model
contained many zero-one variables (taking value 1 when the aircraft is
used for a particular journey, and 0 otherwise). The (heavy)
computational requirement was reduced by partitioning the MILP into
smaller subprograms, having only a few variables in common. So the
problem was handled by optimizing the subprograms, then
improving the values of the common variables, then repeating the
process.
Kuchta, Newman and Topal (2005) gave a MILP model for
scheduling production at an iron mine, to meet demands from the mills
that process it. Here, zero-one variables occur, taking value 1 if a
particular block is mined at a particular time, and otherwise zero.
LeBlanc, Hill, Greenwell and Czesnat (2005) developed a
distribution model, involving different warehouses and shipping
distances, to minimize the total shipping, holding and handling costs,
using a linear programming model.
10
0.6 Importance of model building, and limitations of computers
Any course on Operations Research must detail various standard
models, and methods to analyse and calculate with them. But one must
not impose a standard model on a real-world situation, if it does not fit.
It may well happen that no standard model is of any relevance, to a
particular real situation. One must learn, and be prepared, to set up a
model, to fit the real situation. To practice this, one should also
undertake project work on specific problems.
Most Operations Research investigations involve serious
computation, and many computer packages are available, e.g. for linear
programming. But it is important not to allow an available package to
determine the model to be used. The computer output may be very
misleading if you do not know what model, or assumptions, are built
into the package, or if you do not know, even in outline, what
computational method is being used. Indeed, if a very rough, quick,
pencil-and-paper calculation can be done, it may provide a useful check
(are we out by a factor of ten?), as well as giving some answer to an
impatient boss or client.
One instance of misuse of a computer package was cited by
Greenberg (1987). A factory had a two-year supply of unsaleable red
widgets in its warehouse, and more being made, because the
computer-based production and inventory control system said they
should be made. The system was not documented, to they did not
understand it. Eventually, it was found that an order for blue widgets
had been received, but it was entered by mistake as a large order for
red widgets. This caused a special production order to replenish the
stock of red widgets, and a spurious forecast demand for red widgets,
so even more were called for. The mistaken order was corrected, but
the system did not recover from the mistake.
One source of projects (of moderate size) is An Operations
Research Casebook, by K. H. Bradshaw, D. N. Foster, B. R. Smith, and G.
A. Vignaux (Longman Cheshire, Melbourne, 1982). Many actual
applications of operations research may be found in journals, especially
Journal of the Operational Research Society and Interfaces. However
those in Interfaces are often large models, involving large sets of data
and heavy computation. A student is advised to start on some smaller
models, found in the older literature. A few examples are the following:
K. Kolesar, 1975, Determining the relation between fire engine travel times and
travel distances in New York city, Operations Research 23 (4), 614.
W. T. Ziemba, C. Parkan and R. B rooks-Hill, 1974. Calculation of investment
portfolio with risk-free borrowing and lending, Management Science 21, 209-222.
R. Juseret, Long term optimization of electrical system generation by convex
programming, 1978. Mathematical Programming Study 9, 186-195.
A. Prékopa and T. Szántai, Flood control reservoir system design using stochastic
programming, 1978. Mathematical Programming Study 9, 138-151.
B. D. Craven, Mathematical Programming and Control Theory, 1978. Chapman & Hall,
11
London.
S. K. Singh, A function for size distribution of incomes, 1976. Econometrica 44, 963970.
R. P. O'Neill, M. Williard, B. Wilkins and R. Pike, 1979. A mathematical
programming model for allocation of natural gas, Operations Research 27, 857.
0.7 Why the name ?
Operations Research takes its name from the application of
scientific methods to the planning of military operations, by Britain and
U.S.A., during the 1939-1945 war. However, the ideas and methods
have long since been applied to the arts of peace.
Other names, such as Management Science and Industrial
Engineering, describe very similar disciplines.
0.8 The usefulness of spreadsheets
A spreadsheet program is often useful for data input (e.g. to a
linear programming package, see section 1.18 for an example), or for
data output for presentation, from the computation of an O. R. model.
The basic principles of a spreadsheet are quite simple (see section 9.2),
and the complications (especially graphical presentations) can be
learned later, as and when required.
However, the computational features of spreadsheet programs have
serious limitations, and often more specialized computer programs are
preferable (e.g. programs for linear programming or nonlinear
optimization.)
Moreover, a model that is easy to set up on a spreadsheet may be an
inadequate model. Nothing substitutes for the pencil-and-paper
modelling, before going near the computer.
0.9 Presentation of conclusions and recommendations
It is the responsibility of an Operations Research consultant
(whatever label he/she wears) to present the results intelligibly to the
client, without assuming the client is already familiar with the
technicalities. (It is always the job of a mathematician to explain
himself/herself to the client, rather than the other way about.)
Usually, both a written report and an oral presentation are required.
The following notes relate to a written report, what it should contain,
and how it should be arranged, so as to be acceptable to a business
management.
The layout of a report should be something like the following:
SUMMARY (or OVERVIEW (on a single page; no math
symbols!)
Introduction and objectives
Summary of findings and conclusions
Summary of recommendations (perhaps cross-referenced to
main report)
12
MAIN REPORT
(maximum nine pages)
Problem definition
Method (but relegate math. technicalities to an appendix)
Assumptions made (list assumptions - so that your work is
checkable)
Analysis of data
Findings and conclusions
(some graphical, or
spreadsheet, presentation may be useful)
Recommendations
(some specific things to do)
APPENDICES
(as the problem requires; put here such
things as extensive tables of data, computer output,
mathematical formulation, e.g. of a linear programming
model.)
The stated limitation on the size of the Summary (one page, for
the big boss), and the Main Report (maximum 9 pages), are important.
Do not expect managers to read masses of material! Don't expect them
to read mathematical symbols, when these are not essential! For
example, do not present the results of a linear programming optimization
as x1 = 123.45
, x2 = 17.4 , etc. .
Instead present some table of recommended quantities, with names that
the management will be familiar with, for example:
PRODUCTION SCHEDULE
123.5 tons of Wotsernames,
17.4 tons of Gizmos,
etc (putting in whatever the names of the products really are).
Do not present spurious precision in the conclusions - consider how
accurate the data are! Also, a spreadsheet presentation is often
persuasive.
The names given for headings are not set in concrete; vary them to
suit the topic.
There must be (usually in an appendix) enough relevant mathematical
detail to enable the calculations to be checked, or repeated with some
changes. For example, for a linear programming model, there must be a
list of variables, and the problem description in some form - a data
matrix, maybe on a spreadsheet, can be better than pages of equations.
Note carefully that the data for a real-world problem is often
inadequate (then, often, you must make assumptions, and state clearly
what you have assumed), and may contain items that are irrelevant, or
misleading (you must use some judgement !)
A report must be typed, using any available word processor. You don't
have to use a computer program to do diagrams; graphs can be pasted
in the report. However, spreadsheet programs, such as Excel, have some
useful graphics capability. And the added neatness of graphs done with a
computer program is a selling point, as well as a requirement by journal
editors.
13
Chapter 1
LINEAR PROGRAMMING
1.1 Introduction : management models
To introduce linear programming, consider the following two models.
Model A Allocation of production to machines Six products can be
made on each of two machines.The following tables show, for each
combination of product and machine, the production rate (expressed as
hours per ton) and the cost ($ per ton); also the demand in tons for
each product. Also shown are the variables, x⁄, x¤, ..., chosen to
denote the amount (tons) of each product to be made on each
machine. The hours available are 780 hours on machine 1, and 3077
hours on machine 2. A schedule is required, by which the requirements
can be met at minimum total cost.
Production rate
Product
Product
Product
Product
Product
Product
1
2
3
4
5
6
Cost
Demand
($/ton)
(tons)
Mach.1 Mach.2 Mach.1
.228 .119
36.0
.228 .119
42.4
.231 .132
34.5
.231 .132
41.0
.182 .112
22.1
.182 .112
28.5
Variables (hours/ton)
Mach.2
44.2
36.2
38.4
30.2
21.4
13.4
2929
9420
1655
8978
2989
2977
x⁄
x¤
x‹
x›
xfi
xfl
x‡
x°
x·
x⁄‚
x⁄⁄
x⁄¤
There is a constraint for each product (x⁄ + x‡ = 2929 for the first
product), and a constraint for each machine
(.228x⁄+.228x¤+.231x‹+.231x›+.182xfi+.182xflˆ780 for machine 1);
and all the variables xj are nonnegative. The constraints need to be laid
out tidily in rows and columns (thus as a matrix) - or there will be
confusion, especially when entering data into a computer program.
The use of dots for zeros is unconventional, but helps the eye
considerably. The objective function to be minimized is
36.0x⁄ + 42.4x¤ + ... + 13.4x⁄¤ .
14
x⁄ x¤
1 1
2 .
3 .
4 .
5 .
6 .
7
8
x‹
.
1
.
.
.
.
x›
.
.
1
.
.
.
xfi
.
.
.
1
.
.
xfl
.
.
.
.
1
.
x‡
.
.
.
.
.
1
.228 .228 .231 .231 .182 .182
.
.
.
.
.
.
x°
1
.
.
.
.
.
x· x⁄‚ x⁄⁄ x⁄¤
.
.
.
.
.
1 .
.
.
.
.
1
.
.
.
.
.
1
.
.
.
.
.
1
.
.
.
.
.
1
=2929
=9420
=1655
=8978
=2989
=2977
.
.
.
.
.
.
ˆ 780
.119 .119.132 .132 .112 .112 ˆ3077
Model B The Novelty Company makes Thingos, Dinguses and Whatsits.
Each Thingo takes 1.5 hours to make, and requires 4.5 kg of raw
material I and 950 g of raw material II. Each Dingus takes 4.6 hours to
make, and requires 2.0 kg of raw material I and 100 g of raw material
II. Each Whatsit takes 2.1 hours to make, and requires 11.0 kg of raw
material I and 240 g of raw material II. The net profit (= sale price
minus costs) for each Thingo is $220, for each Dingus is $150, for
each Whatsit is $270. The Company has available 25 hours, 130 kg of
raw material I, and 26 kg of raw material II. What is the maximum
profit they can make, and how do they do it ?
Suppose they make x⁄ Thingos, x¤ Dinguses, and x‹ Thingos.
Then x⁄ ˜ 0, x¤ ˜ 0 and x‹ ˜ 0, since negative amounts have no meaning
here. The restrictions on time, raw material I, and raw material II,
require the following constraints to be satisfied:
1.5x⁄ + 4.6x¤ + 2.1x‹
ˆ 25,
4.5x⁄ + 2.0x¤ + 11.0x‹ ˆ 130,
950x⁄ +100x¤ +240x‹
ˆ26000.
Subject to these constraints, the objective function
220x⁄ + 150x¤ + 270x‹
is to be maximized. So there is a linear programming problem:
Maximize 220x⁄ + 150x¤ + 270x‹
subject to x⁄ ˜ 0, x¤ ˜ 0, x‹ ˜ 0,
1.5x⁄ + 4.6x¤ + 2.1x‹
ˆ 25 ,
.45x⁄ + .20x¤ + 1.10x‹ ˆ 13.0 ,
.95x⁄ +.10x¤ + .24x‹ ˆ 26.0
.
Observe that, in a linear program, the objective function and the
constraint functions are linear, and the veariables are nonnegative .
The constraints may be equalities, ˆ inequalities, or ˜ inequalities.
[Note also , in Model B, that the second and third constraints have been
scaled (multiplying by 0.1 and 0.001 respectively), so as to make the
largest coefficient in each constraint not too far from 1 is absolute
value; this helps the accuracy of computation.]
Linear programming applies to many other situations; other
examples are given in later sections.
15
1.2
Standard form for a linear program
The inequalities are converted to equalities by putting in extra
variables. Thus, in Model B, the inequality1.5x⁄ + 4.6x¤ + 2.1x‹ ˆ 25
becomes 1.5x⁄ + 4.6x¤ + 2.1x‹ + x› = 25,
where x› is a nonnegative slack variable . Thus Model B becomes:
Minimize 220x⁄ + 150x¤ + 270x‹ + 0x› + 0xfi + 0xfl
subject to x⁄, x¤, x‹, x›, xfi, xfl ˜ 0,
1.5x⁄ + 4.6x¤ + 2.1x‹
+ x›
= 25.0 ,
.45x⁄+ .20x¤ + 1.10x‹
+ xfi
= 13.0 ,
.95x⁄ + .10x¤ + .24x‹
+ xfl = 26.0 .
Note that each constraint now has a variable with coefficient 1, that
occurs nowhere else (thus, x› in the first constraint). Thus there is an
initial feasible solution (feasible means satisfying the constraints) x› =
25.0, x› = 13.0, xfl = 26.0, all other xj = 0. (Notice that there are here
just as many nonzero xj (three) as there are constraints.) If the
coefficients in these constraInts are laid out as a matrix, thus
x⁄
1.5
.45
.95
x¤
4.6
.20
.10
x‹
2.1
1.10
.24
x›
1
.
.
xfi
.
1
.
xfl
.
.
1
25
13
26
then this matrix contains a unit matrix (here, columns 4,5,6). This is
necessary in order to start the simplex method - see Section 1.3).
If a model has a ˜ constraint, for example 3x⁄ + 4x¤ ˜ 12, then a
negative slack variable is introduced, to convert to an equality, thus
3x⁄ + 4x¤ - xnegslack = 12. But this is not enough, since a term
with coefficient +1 is needed, to provide the unit matrix. So an
artificial variable is added; thus,
3x⁄ + 4x¤ - xnegslack + xartificial=12.
But xartificial is not part of the original problem; so a penalty cost term
+Mxartificial must be added to the objective function (to be minimized).
Here, M is a large enough positive number, so that minimizing the
objective function will get rid of the artificial variable. Consider now an
equality constraint, say 3x⁄ + 4x¤ = 12. This still needs an artificial
variable (though not a slack variable), thus 3x⁄+4x¤ + xartificial= 12,
with +Mxartificial in the objective function.
This account standardizes on Minimization problems. Maximization
(as in many textbooks) is equally good. Note that MAX 2x⁄ - 3x¤ + 5x‹
is equivalent to MIN -2x⁄ + 2x¤ -+ 5x‹ .Thus, the profit coefficients
-2, 2, -5 are the negatives of the cost coefficients 2, -3, 5 ; and the
z j-cj (see Section 1.4) have opposite sign; and -Mxartificial is added to the
objective being maximized. [Also some books have a different
definition of standard form.]
1.3 Idea of simplex method
The simplex method is the most widely used algorithm for
computing linear programs. (See also Section 1.19.)
16
Consider the linear program:
C: Minimize 3x⁄ + 4x¤ subject to x⁄˜0, x¤˜0, 2x⁄ + x¤ ˆ 7, x⁄ + x¤ ˜ 5.
Putting this into standard form gives the (initial) tableau :
3
x⁄
2
1
7
Cost coeffts
zj -cj
4
x¤
1
1
6
0
x‹
.
-1
-10
0
x›
1
.
0
10
xfi
.
1
0
Requirement
7
5
50
Multiplier
-.5
-3.5
Note that the cost coefficients (c⁄=3, c¤=4, etc.) are not part of
the tableau; but we need to refer to them. Multiplier and zj-cj are
explained below. Here, x› is a slack variable, x‹ is a negative slack, and
xfi is an artificial variable; M is taken here as 10, for simplicity. (Usually
a much larger number is required, like 10000, to be large enough,
compared to all the other costs in the problem). Columns 4 and 5 give
the required unit matrix.
The constraints solve immediately to give x›=7-2x⁄-2x¤-0x‹ and
xfi=5-x⁄-x¤+x‹. (The variables x› and xfi form the initial basis. Initially
x›=7 and xfi=5, so the objective function f = 0(7)+10(5)=50. ) Now
the objective function equals
f = 3x⁄+4x¤+0x‹+0x›+10xfi = 3x⁄+4x¤+0x‹+10(5-x⁄-x¤+x‹)
= 50 - [ 10(1) -3]x⁄ - [10(1) - 4]x¤ - [10(-1)-0]x‹
= 50 - 7x⁄ - 6x¤ + 10x‹.
[The numbers 7, 6, -10 are zj -cj in the tableau.]
Initially x⁄=x¤=x‹=0 (nonbasic variables). Consider increasing one of
them from zero. The fastest rate of decrease of f is obtained by
increasing x⁄ from zero. How far can x⁄ increase ? From x›=7-2x⁄
˜0and xfi=5-x⁄˜0, there follows x⁄ˆ3.5 and x⁄ˆ5, hence x⁄ is taken as
3.5. The new basis variables are x⁄ (replacing x›) and xfi. [The order
of the basis variables matters.] [Here, the pivot is the element in
column 1 (new basis variable) and row 1 (the first constraint here tells
which variable leaves the basis. Row 1 is here the pivot row.]
It is now required to solve for x⁄ and xfi. Add 0.5 (equation 1) to
equation 2; add -3,5(equation 1) to the zj-cj row); divide the pivot row
(here, row 1) by the 2 (the pivot element). This gives
x⁄ +.5x¤ +.5x› = 3.5 and .5x¤-x‹-.5x›+xfi=1.5.
Solving for the new basis variables, x⁄=3.5-.5x¤-.5x› and
xfi=1.5-.5x¤+x‹+.5x›.
Therefore f = 3x⁄+4x¤+10xfi
=25.5 -[ 3(.5)+10(.5)-4]x¤ -[3(0)+10(-1)-0]x‹ - [3(.5)+10(-.5)-0]x›
=25.5 - 2.5x¤ +10x‹ + 3.5x› ;
so x¤ enters the basis. Then x⁄=3.5-.5x¤˜0 and xfi=1.5-.5x¤˜0, require
that x¤ˆ3.5/.5=7.0 and x¤ˆ1.5/.5=3; so take x¤=3; the pivot is the
second element in column 2. And so on.
1.4 Simplex tableaus
The above calculations are done systematically, using arrays of
numbers, called tableaus. The step from each tableau to the next is
17
called an iteration.
3
x⁄
Tabl.1 2
1
zj -cj 7
Cost coeffts
Tabl.2 1
.
0
Tabl.3 1
.
0
4
x¤
1
1
6
0
x‹
.
-1
-10
0
x›
1
.
0
.5
.5
2.5
.
-1
-10
.5
-.5
-3.5
.
1
0
1
-2
-5
1
-1
-1
10
xfi
.
1
0
Requirement
5
50
.
1
0
3.5
1.5
25.5
-1
2
-5
2
3
18 .
Multiplier
7
-.5
-3.5
-1
-5
In tableau 1, the basis columns are 4 and 5; the cost coefficients are
[c› cfi] = [0 10]. Calculate z⁄-c⁄ =[0 10] ã2õ-3=7;
z¤-c¤=[0 10] ã1õ-4=6;
and so on; note basic zj-cj=0.
Ã1Õ
The largest positive zj-cj is z⁄-c⁄=7; so x⁄ enters the basis. Choose
the smallest nonnegative ratio among 7/2 and 5/1 [look at
requirements column and pivot column; ignore negative ratios, if
present, since they don't stop the variable going positive]. The first
ratio is the smallest, so the first basis variable (x›) is replaced by x⁄.
The pivot is shown in bold.
For rows other than the pivot row, the
multiplier = - (element in pivot column)/(pivot element).
[Thus the multiplier for second row is -(1)/(2). ]
To update the tableau, for rows other than the pivot row:
new element = old element +(row multiplier)(element in same
column and pivot row)
Thus, for row 2 and column 1, 1 ü 1 + (-.5)(1) = .6. The zj-cj row is
updated like any other row. Finally, the pivot row is divided by the
pivot.Thus tableau 2 is obtained. The largest positive zj-cj is z¤-c¤=2.5;
so x¤ enters the basis. The second basis element (xfi) leaves the basis
(consider ratios 3.5/.5 and 1.5/.5).
In the third tableau, all zj-cj are ˆ 0; so an optimum has been
reached. (The numbers zj-cj are called reduced costs.)
1.5 Information from tableau
The optimal solution is x⁄ = 2, x¤ = 3, other xj = 0. Observe that
the unit matrix in the initial tableau consists of columns 4 and 5 (in that
order). The optimal inverse basis matrix then consists of columns 4
and 5 in the optimal tableau. From z›-c›=-1 and zfi-cfi=-5, with c›=0
and cfi=10, obtain z›=-1 and zfi=5. In the optimal tableau, columns 1
and 2 form the unit matrix; columns 1 and 2 in the initial tableau from
18
the optimal basis.
Check that: ã 1 -1õã2 1õ =ã1
0õ and [3 4]ã -1õ = [-1 5]
Ã-1 2ÕÃ1 1Õ Ã0
1Õ
Ã-1 2Õ
i.b.m.
b.m.
Here, [3 4] is the vector of cost coefficients for the optimal basic
variables; and the shadow costs are [z›, zfi] = [-1 5] (see below).
What happens if some requirement is perturbed a little? If the
second requirement 5 is changed to 5+h, then (using the inverse basis
matrix)
ãx⁄ õ =
ã1
-1õã7 õ =ã2- h õ ˜ ã0õ provided hˆ2
Ãx¤ Õ
Ã-1
2ÕÃ5+hÕ Ã3+2hÕ
Ã0Õ
h˜-1.5
So the basic variables x⁄, x¤ stay basic when h varies between -1.5 and
2, although the values change. Within that range, the optimal objective
function equals 3(2-h)+4(3+2h)=18+5h. The number 18 is already
given in the optimal tableau; and 5 is the second shadow cost.
1.6 Two-phase method
So far, a numerical value was given for M. If, instead, M is left as a
symbol, then the zj-cj for the initial tableau for example C (in 1.3) are
M-3, M-4, -M, 0,0, and the initial objective is 5M (instead of 50), with
multiplier -(M-3)/2. The second tableau obtains 0, (M-5)/2,-M, -(M3)/2,0; objective =(3M+21)/2, and multiplier = -(M-5). The third
tableau has
zj-cj as 0, 0, -5, -1, -M+5, and objective 18. This is the
big M method, sometimes used for hand calculation.
An alternative, often used in computer programs for LP, is the
two-phase method. In phase I, M is taken as the unit of cost, and other
numbers considered negligible, so that the cost coefficients become 0,
0, 0, 0, 1, and the initial objective value is 5.
Phase I
Cost coeffts 0
0
0
0
1
x⁄
x¤
x‹
x›
xfi
Requirement Multiplier
Tabl.1 2
1
.
1
.
7
1
1
-1
.
1
5 -.5
zj -cj
1
1
-1
0
0
5 -.5
Tabl.2 1
.
0
.5
.5
.5
.
-1
-1
.5
-.5
- .5
Tabl.3 1
.
z j - cj
0
New zj - cj 0
.
1
0
0
1
-2
0
-5
1
-1
0
-1
.
1
0
3.5
1.5
1.5
-1
2
-1
5
2
3
0
-1
-1
At the end of phase I, the zj-cj are all zero, except artificial columns.
They must be recalculated, using the original costs, for example
19
z‹-c‹= [3
4] ã 1 õ -0 = -5 (using costs 3, 4 for basic x⁄,x¤
Ã-2 Õ
and zero now instead of M )
Here, by coincidence and unusually, all the (new) zj-cj happen to be ˆ 0,
so the optimum has been reached. But usually some (new) zj-cj is > 0,
so a phase II must follow, consisting of one or more simplex iterations.
If the objective function in example C (in 1.3) is changed to x⁄ - x¤,
then the tableaus become
x⁄
x¤ x‹
x› xfi
Requirement
2
1
.
1
.
7
1
1 -1
.
1
5
1
1
-1
.
.
5
1
1
0
-2
.
1
0
.
1
-1
0
1
1
.
0
.
-1
1
-1
X
2
5
0
-5 öRecalculated
After the end of Phase I, the zj - cj row and objective value are
recalculated as shown above, for example
z⁄-c⁄ = [c‹ c¤]
[1] - c⁄ =
[0 -1
[1] -1 = -3 .
[2]
[2]
[Here X is no longer relevant, since the artificial column will ot be put back into the basis.
For comparison, X would be -M-1 for the big M method.] An iteration of Phase II
gives:
1
2
-3
.
1
.
1
.
.
1
1
-1
-1
.
X
2
7
-7
which happens to be optimal, with x⁄=0, x¤=7, and (slack) x‹=2. (Again,
X is not relevant here - although its matrix column is part if the inverse
basis natrix.)
1.7 Unsigned variables
A variable (z say) that is not required to be nonnegative can occur
in a linear program - usually representing the change (up or down) of
a variable from its usual level. To make the simplex method work, z
must be written as z = z⁄ - z¤ with z⁄˜0 and z¤˜0. But z⁄ and z¤ must
not both occur in a basis (3 = 3-0 and -3 = 0-3 okay, not 3 = 4-1).
1.8 Revised simplex method
Usually (see model A for an example) the matrix of coefficients in a
linear program is sparse, meaning that there is a high proportion of
zeros. It then becomes computationally useful to not work with the
whole simplex tableau, but only with part of it - the inverse basis
matrix, with an extra row and column. The revised simplex method thus
saves both data storage and computing time (multiplying uselessly by
zero). But it is then necessary to record separately which variables are
20
in the basis; the whole simplex tableau is no longer available to show it.
Req
Matrix
Col
Mult
Req
50 | 0 10 | 7 | -3.5
7 | 1 . |2 |
5 | .
1 |1 | -.5
Basis x›,xfi
zj - cj =7,6,-10,0,0 ; col 1
Matrix
Col
Mult
25.5 | -3.5 10 | 2.5 | -5
3.5 | 0.5 . | .5 |-1
1.5 | -.5 1 | .5 |
Basis x⁄,xfi
zj - cj =0,2.5,-10,-3.5,0 ; col 2
Req
18
2
3
Matrix
.
| -1 5|
| 1 -1|
| -1 2|
Basis x⁄,x¤
zj - cj =0,0,-5,-1,[-5] ;optimal.
In this calculation, M is taken as 10. The initial objective value,
50, is obtained as in the simplex method; it is convenient here to put
the requirements column at the left. The matrix tabulated is the
inverse basis matrix , so starts with a unit matrix. The top row starts
with the cost coefficients for the initial basis columns.
The revised simplex (unlike the simplex) regularly refers back to
the original data matrix:
x⁄
x¤
x‹
x›
xfi
Requirement
2
1
.
1
.
7
1
1
-1
.
1
5
The cost coefficients are 3,4,0,0,10. To obtain the zj-cj for the first
revised-simplex tableau, use the top row with the columns of the
data matrix. Thus z⁄-c⁄ = [0 10]
ã2 õ - 3 = 7; others are
Ã1 Õ
similar.This shows that column 1 should enter the basis. To
reconstruct column 1 of the simplex, use the inverse basis matrixwith
column 1 of the data matrix :
ã1 0 õã2õ = ã2õ
Ã0 1ÕÃ1Õ Ã1Õ
This column is marked Cols in the above revised-simplex tableau; the 7
at the top is z⁄-c⁄. The pivot is chosen, in this column, as for the
simplex; and the update (of the whole matrix) is done, just as for the
simplex (except that it is a smaller matrix). [For example, 0-(3.5)(1)=-3.5]
In the second revised-simplex tableau, z¤-c¤=[-3.5 10]ã1õ-4= 2.5;
others are similar. Thus x¤ enters the basis.
Ã1Õ
Column 2 of the simplex is reconstructed as ã.5 0õ ã1õ=ã.5õ;note that
the column is got from the data matrix
Ã-.5 1Õ Ã1Õ Ã.5Õ
Considering the ratios 3.5/.5 and 1.5/.5, the first basis variable (x›)
leaves the basis, being replaced by x¤. In the update (for example) the
matrix element 0.5 ü 0.5+(-1)(-.5)=1.
The third tableau is optimal, since all zj-cj are ˆ 0; zfi-cfi (for the
artificial column) is put in [ ], to remind that an artificial column is not
considered for putting back in the basis.
Computer programs for linear programming use the revised simplex
(or some development of it that saves space by storing, not the inverse
matrix itself, but other information from which it can be constructed, or
some other form that is more numerically stable for large LPs), not the
simplex, to save both storage space and time. (Only integer
21
programming needs the full simplex tableau).
Normally, one uses decimals, rather than vulgar fractions otherwise denominators soon get unmanageably large. Note that the
change from simplex to revised simplex, and the change from big M to
two-phase, are separate things; the revised simplex example here used
big M (with M=10 for convenient presentation), but could have used
two-phase instead. Observe also that the calculations work down
columns rather than across rows. For this reason, some computer
programs for linear programming (e.g. Craven's program - see Section
1.19, and the Minos package) enter the matrix data going down
columns (thus not across rows, like equations), and enter only nonzero
matrix elements (and which row they are in). Alternatively, some
programs accept the data in spreadsheet format (see section 4.10);
here also, only nonzero data need be entered.
1.9 Dual linear program
Consider the following linear programs:
Primal program
in matrix language
MIN 3x⁄+4x¤+Mxfi subject to
MIN cTx = [3 4 0 0 M] x
2x⁄+x¤+x› = 7,
(where x = [x⁄, x¤, x‹, x›, xfi]T )
x⁄+x¤-x‹+xfi=5,
ü
subject to
x ˜ 0,
x⁄,x¤,x‹,x›.xfi ˜0
Ax = b,
[Optimum x⁄=2, x¤=3]
where A = ã2 1 . 1 .õ, b=ã7õ .
ï
Ã1 1 -1 . 1Õ
Ã5Õ
ï (rearrange the pieces)
Dual program
ö MAX bTw subject to ATw ˆ cT,
MAX 7w⁄ + 5w¤ subject to
2w⁄+w¤ˆ3, w⁄+w¤ˆ4,
thus MAX [7 5]ãw⁄õ subject to
0w⁄ -1w¤ˆ 0 [so w¤˜0]
à w¤Õ
1w⁄+0w¤ ˆ M, [so w¤ˆM]
[w⁄ w¤]ã2 1 . 1 .õ ˆ [3 4 0 0 M]
ï [Optimum w⁄=-1,w¤=5
Ã1 1 -1 . 1Õ
set z⁄=-w⁄, z¤=w¤
Note that the dual does not
MAX -7z⁄+5z¤ subject to
include a constraint w ˜ 0.
-2z⁄+z¤ˆ3, -z⁄+z¤ˆ4, z⁄˜0, z¤˜0.
Thus, a dual problem can sometimes be usefully rearranged. Note that,
if M is very large, a constraint w¤ ˆ M has no effect.
1.10 Dual properties
I Weak duality If x is feasible for the primal, and w is feasible for the
dual, then cTx ˜ bTw.
Proof cTx - bTw = cTx - wTb = cTx - wTAb = (cT-wTA)x ˜ 0
since each vector has components ˜ 0.
Remark The range of values that the primal objective takes lies
entirely above the range of values that the dual objective takes
(assuming x and w feasible). Suppose that the primal is minimized
as x=x*, and the dual is maximized at w=w*; then cTx* ˜ bTw*.
22
II Zero Duality Gap (ZDG) cTx* = bTw* ; and w* is the vector of
shadow costs for the primal.
Proof In the optimal simplex tableau, denote by Q the inverse
basis matrix, u the vector of basic variables, q the vector of cost
coefficients for them. Then u = Qb; the matrix in the optimal tableau
is QA; the vector of the zj-cj is cTQA-cT, which is ˆ 0 by the
optimality condition. So the vector qTQ of shadow costs is feasible
for the dual. Since (qTQ)b = qTu = cTx*, and no w feasible for the
dual can (by I) have wTb>cTx*, it follows that w*=qTb is optimal for
the dual, and cTx* = bTw*.
III Karush-Kuhn-Tucker conditions (see chapter 7). Necessary
conditions for x* to minimize the primal are that, for Lagrange
multipliers r* ˜ 0 and w*, there hold cT=w*TA + r* and r*Tx* = 0.
Proof In the optimal tableau, xj>0 only for basic xj, and then r*j=cj-zj =0.
So r*j = cj-zj = 0; so r*Tx = 0.
1.11 Sensitivity
It is usually not enough to calculate an optimum to a linear
program; it is also necessary to consider how sensitive the optimum is
to small changes of the data. (Requirements are not all "set in
concrete"; and there can be last-minute changes to data.)
•a
In 1.5, a small change to one requirement is considered; over a
sufficiently small right hand side range (in the example, if the second
requirement lies between 5-1.5 and 5+2), then optimal objective =
original value+(shadow cost)(change in requirement); and the same
basic variables stay optimal. (If more than one requirement changes,
the shadow costs still apply, but the RHS ranges are reduced.)
•b
Consider a change to a cost coefficient.
(a) Change c‹ away from 0. Since x‹ is not in the optimal basis, the
z j are unchanged. So all the zj-cj remain ˆ 0 (still optimal), provided that
z‹-c‹ stays ˆ0. From c‹=0 and z‹-c‹=-5 follows z‹=-5, so now z‹-c‹=5-c‹ˆ0 provided c‹˜-5.
(b) Change c⁄ away from 3. Since x⁄ is basic, the zj are changed; it
is necessary to check all nonbasic zj-cj (except artificials). Thus,
z‹-c‹= [c⁄ 4] ã 1 õ-0=c⁄-8ˆ0 provided c⁄ˆ8. (Hereã 1õ is column 3 of
Ã-2Õ
Ã-2Õ
the optimal tableau in 1.4.) Similarly z›-c›=c⁄-4ˆ0 provided c⁄ˆ4. So
the tableau is still optimal when c⁄ˆ4. (If c⁄ increases above 4, a
further simplex iteration is needed.)
(c) Suppose that both c⁄ and c‹ are both functions of some
parameter r. For a simple example, suppose c⁄=3+r and c¤=4-2r. Then
z‹-c‹ = [3+r 4-2r]ã 1õ-0 = 5-4r ˆ 0 provided r˜5/4; z›-c› = ... .
Ã-2Õ
Such a situation may arise in a profit-maximizing problem, with profit
23
coefficients p⁄ = 30-20 (thus, revenue - variable cost) and p¤ = 85-45
(representing products manufactured) in the objective being
maximized. Suppose that a unit of the first product requires 0.1 kg of a
certain ingredient, and a unit of the second product requires 0.2 kg,
where this ingredient costs $q per kg (currently q = $50, but may
vary). Then
p⁄ = 30 - [.1q+15] and p¤ = 85 - [.2q + 35].
•c Consider the example of 1.4, with requirements changed to 7 and
3.5. (The latter is on a boundary of its RHS range.). The optimal tableau
now becomes:
Tabl.3
1
.
0
.
1
0
1
-2
-5
1
-1
-1
-1
2
-5
3.5
0
25.5
(Note one basic xj is zero.) But there is also another optimal tableau
(got using dual simplex, see 1.12) :
Tabl.3 1
(new) .
0
1
-1
0
-1
2
-3
.
1
0
1
-2
-7
3.5
0
25.5
The shadow costs are different: -1 and 5 for the first tableau, 0 and 3
for the second.
But also the RHS ranges are different, when the
second requirement (b¤) varies: from 3.5 to 5.0 for the first tableau,
from 0 to 3.5 for the second.
If the optimal objective value is
considered as a function of b¤, then this function is piecewise linear
(thus, made up of a finite number of linear pieces), and the slope
changes when b¤=3.5 .
•d Consider adding a new column to the problem . Suppose the
example of 1.3 is modified to:
Minimize 3x⁄+4x¤ (+10xfi)-7xfl subject to all xj˜0, and
2x⁄+x¤+x› +xfl = 7, x⁄+x¤-x‹+xfi -xfl = 5.
(Including slacks x3, x4 and artificial x5).
What would zfl-cfl be for tableau 3 in 1.4 (which was optimal without xfl) ?
This is calculated using the shadow costs, as
zfl-cfl= [-1 5]ã 1õ-(-7) =+1, so no longer optimal. Then the new
Ã-1Õ
column 6 must be calculated, applying the inverse basis matrix,to the
new column, as
ã1 -1õ ã 1õ = ã 2õ . A simplex iteration would then follow.
Ã-1 2Õ Ã-1Õ
Ã-3Õ
•e Consider a change in (one or more) matrix coefficients. The
constraints Ax=b, x˜0 may be written as Bu + Kv = b, where B = matrix
of basic columns, K = matrix of nonbasic columns, u = vector of basic
24
variables, v = vector of nonbasic variables. So the optimum (u*,v*)
satisfies Bu*=b, u*˜0, v*=0. Now perturb B to B+E; this will change u to
u+z, say, satisfying
(B + E)(u + z) = b. If the elements of the
matrix E are "small enough", so that z is "small", and the product Ez
can be neglected, then Eu + Bz Ú 0; so z Ú -B-1Eu.
In the example of
1.4, suppose the coefficient of x⁄ in the first constraint changes from 2
to 2.1 Then
B =
ã2 1õ
Ã1 1Õ
; E =
ã0.1 0õ;
à 0 0Õ
z
Ú
ã1 -1õã.1 0õã2õ .
Ã-1 2Õ Ã 0 0ÕÃ3Õ
1.12 Dual simplex method
In the simplex, requirements must be ˜0; feasibility is kept at all
times, but the optimality conditions are not satisfied until the last
tableau. In the dual simplex, which in fact solves the dual problem,
but using a tableau suitable for the primal, requirements may be
negative; the optimality conditions are satisfied always; but feasibility is
not fulfilled until the last tableau. Consider the example:
D:
MIN x⁄+3x¤ subject to x⁄,x¤˜0, x⁄+4x¤˜8, x⁄+2x¤˜6,2x⁄+x¤˜6.
Instead of introducing artificial variables then using the simplex method,
the dual simplex may be used, writing the constraints in the form
-x⁄ -4x¤ + xslack = -8 (similarly for the others). This gives the first
two tableaus as follows:
-1 -4
1
.
.
| -8 ö
.25 1 -.25 . . | 2
ü
-1 -2
.
1 .
| -6
-.5 . -.5
1 . | -2
-2
-1
-1
-3
.
0
.
0
1
0
| -6
| 0
-1.75 .
-.25 .
-.25 . 1 | -4 ö
-.75 0 0 | 6
In the first tableau, x‹,x›,xfi form an infeasible basis. The zj-cj are
calculated as for the simplex; they are all ˆ0 (optimality conditions satisfied),
so this basis is dual feasible. Choose now the row with most negative
requirement as the pivot row (shown ö). Look at the pivot row and the
z j-cj row, considering the ratios -1/-1 , -3/-4, 0/1 . The smallest
positive ratio picks out column 2 to enter the basis; the pivot (-4) is
shown bold in the tableau. [In a maximizing problem, choose instead the
negative ratio nearest zero. ] An ordinary simplex iteration, with this
pivot, gives the second tableau shown.
Consider now the parametric programming question from 1.11•c,
where the second requirement (h say) is the parameter, and the
optimum is required, as this parameter varies. The tableau obtained is:
25
Tabl.3 1
.
0
.
1
0
1
-2
-5
1
-1
-1
-1
2
-5
|
7-h
| (optimal
| -7+2h ö |3.5ˆhˆ7)
| 21-3h
(Note that -7+2h=0 when h=3.5). If h decreases below 3.5, then x¤
becomes negative. Therefore a dual simplex iteration is done, with the
second row as the pivot row. This leads to the tableau:
1
1
-1
.
1
|
h
(optimal
.
-1
2
1
-2
| 7-2h
0ˆhˆ3.5)
0
0
-3
0
-7
| 3h
(Note that h˜0 and 7-2h˜0 provided that 0ˆhˆ3.5).
optimization could be continued, for h˜7 or for hˆ0.
simplex is also used in integer linear programming.
This parametric
The dual
1.13 Jumps in shadow costs
Consider the following initial and optimal tableaus for a linear
program. (This example is modified from an example in <DG>, page
125, by scaling the rows, so that the largest coefficients in each row
are comparable in size. Such scaling is recommended for
computational accuracy.)
Costs
-24
-20
0
0
0
0
100
12
20
24
0
0
.5
1
1.5
1
1
1
1.2
144
1
.
.
-.8
-60
.
1
.
0
.
.
1
.
0
.
.
.
.
.
.
.
-1
1
-100 0
.
.
.
-1
-6
-.5
.
-.5
.
1.5
1
.
1
0
-14
6
2
9.6
12
-408
.
.
.
1
0
1
.
.
1,5
-.5
-2.4
.
0
0
.
0
.
.
-1
-100
.
Suppose that the first requirement is perturbed from 12 to 12+h. When
h=-6, the basic xj are no longer all ˜ 0; multiplying the optimal inverse
basis matrix by the column vector (12+h, 20, 24, 0) gives
(x¤, x›, xfl, x⁄0 = (-3, 5, 24, 18). The dual simplex can be used here;
two iterations proceed from the tableau on the left to the optimal
tableau on the right, (If the simplex is run again, starting with first
requirement = 6, the same optimum is obtained, but with the rows in a
different order).
26
.
.
.
1
0
1 1.5 . -.5
.
-.5 1 -.5
. -2.4 . 1.6
. -1 . 1
0 -6 0 -14
. .
. .
1 -1
. .
0
-3
. -2 -3
5
. -1 -2
24 ü . 3.2 2.4 .
18
1
2 2
-372
0 -28 -48
. 1 . .
6
1 . . .
8
. 1 -1
14.4
. .
. .
12
0 0 0 0 -288
Now consider what happens when the first requirement is
changed to 19. The dual simplex gives an optimum, x2=12, x4=0, x3=3,
x1=8; with shadow costs (got from columns 3,4,5,7 ) 0, 0, -18, -2.5.
But now, for comparison, let us solve this problem (with requirements
19,20,24,0) using the simplex. This gives x2=12, x3=3,x6=0,x1=8, and
shadow costs 0, -12, -8, 0, thus a different optimum, though with the
same optimal total cost of -432. Why the difference? Observe that
each optimum has some basic xj at zero. Using the inverse basis matrix
in each case, a calculation with the optimal inverse basis matrix, and
using here a RHS parameter p, gives:
x2
. .
1/2 -5/8
19
12+p/2
0
x4 = . 1 -5/6 5/24
20 = 0-5p/8 ! 0
if -24"p"0
x3
1
. -2/3
5/12
24+p
3-2p/3
0
x1
. .
1/3 5/12
0
8+p/3
0
x2
1 -2
1
.
19
3+p
0
x3 = .
3 -2
.
20 =
12-2p ! 0
if 0 " p "6
x6
. -4.8 4
-1
24+p
0+4p
x1
. -2
2
.
0
8+2p
Thus p=0 is a boundary between the first case (where optimal cost =
-432-18p, -24ˆpˆ0), and the second case (optimal cost = -432-8p,
0ˆp<6). Here -18 and -8 are the respective shadow costs. As usual
with parametric linear programming, the optimal total cost is a
piecewise linear function of a RHS parameter p.
1.14 Integer linear programming
Often some, or all, variables in a linear program must be
restricted to be integers (e.g. rolls of paper, or lengths of steel, to be
cut; number of people in a work schedule. There are also zero-one
linear programming problems, where some variables must be 0 or 1
only, describing things that are either on or off (present or absent).
Except for transportation problems (see 1.15) the usual linear
programming algorithm does not generally give integer solutions, and
rounding a non-integer solution can give a very poor approximation to
an optimum integer solution, when small integers are involved. (If all
the numbers are large, one would certainly round to integers, and
expect to get a usable answer.) The following is an example of
Gomory's cutting-plane algorithm for integer linear programming.
Unfortunately it, and also the rival branch-and-bound algorithm (see
section 6.5) are much slower that the usual (revised) simplex method,
27
and are restricted in practice to integer problems of modest size.
Consider the following example:
Maximize x⁄ + x¤ subject to x⁄, x¤ nonnegative integers, and to
2x⁄ + x¤ ˆ 6, 4x⁄ + 5x¤ ˆ 20. The initial and final tableaus for the usual
simplex method are then:
2
4
-1
1
5
-1
1 .
. 1
.
.
6
20
1
.
.
. 5/6
1 -2/3
. 1/6
-1/6
1/3
1/6
5/3
8/3
13/3
Since this problem is maximizing, the zj-cj in the optimal simplex tableau
are ˜ 0.
Consider one of the equations involving a noninteger variable, say
the second equation. (Usually the equation having the largest fractional
part on the right hand side is chosen; but here, there is a choice
between 2/3 and 2/3.) The equation may be rewritten as:
0.x⁄ + 0.x¤ + (1/3)x‹ + (1/3)x› = (2/3) + [2 - x¤ + x‹].
Here, the left side coefficients are fractions, thus 0 ˆ coefficient < 1;
this has been arranged by shifting integer multiples (like 1.x‹) to the
right hand side. An optimum integer solution must satisfy
(1/3)x‹ + (1/3)x› ˜ 0, hence the right hand side of the rewritten
equation is ˜ 0, hence the integer [2 - x¤ + x‹] ˜ 0, hence
(1/3)x‹ + (1/3)x› ˜ (2/3).
The problem is modified by adding this additional constraint; it may be
simplified (multiplying by the denominator 3) to x‹ + x› ˜ 2. This
additional constraint is called a cutting plane, because it cuts off a part
of the feasible region of the original linear program.
The previous optimum now no longer satisfies all the constraints. A
new optimum may be found, using the dual simplex method. For this,
the cutting plane is rewritten as -x‹ - x› + xfi = -2, in which xfi ˜ 0 is a
new slack variable. The dual simplex tableau, and the result of an
iteration on it, are shown as follows; the pivot element chosen is shown
in bold.
1
0
5/6 -1/6 .
5/3
1 .
1
. -1/6
2
.
1 -2/3
1/3 .
8/3 ü
. 1 -1
.
1/3
2
.
.
-1
-1 1
-2
.
.
1 1
-1
2
.
.
1/6
1/6 .
13/3
.
.
0 .
1/6
4
There is here a choice of pivots; if row 3, column 2 is chosen instead, a
different optimum : (x⁄,x¤,x‹,x›) = (0,4,2,0) , is obtained, in place of
(2,2,0,4). The integer-optimum objective function has the same value,
4, for each optimum. (When a problem involves small integers, it often
happens that there are more than one optimum point.)
This example is unusually fortunate, in that a single cutting plane
leads at once to an integer optimum. More typically, a number of
iterations are required, each introducing a cutting plane - so the tableau
grows more rows and columns. Note also that, for integer programming
(though not usually for ordinary linear programming), calculations in
28
exact vulgar fractions may be required (as done here).
To see what is happening geometrically, note that
x‹ + x› ˜ 2 Ä (6=2x⁄-x¤) + (20-4x⁄-2x¤) ˜ 2 Ä x⁄ + x¤ ˆ 4.
The original (non-integer) optimum (x⁄,x¤) = (5/3. 8/3) does not satisfy
this constraint; the new optimum (2,2) satisfies it.
With the present example, if constraint 1 had been chosen, instead
of constraint 2, to generate a cutting plane, then it happens that two
cutting planes would have been needed, to find an optimum.
[ A (less recommended) alternative to using the dual simplex here
is to solve the linear program
(a) Minimize (1/6)x1 + (1/6)x2 + Mx6 subject to the original
constraints, and also x3 + x4 -x5 + x6 = 2. (Here x5 is a slack variable,
x6 is an artificial variable, and 1/6, 1/6 come from the previous zj-cj
row, changing sign because we are switching from a maximum to a
minimum problem.) The calculated numbers are the same as for the
dual simplex, but there is more calculation. ]
1.15 Transportation problem
As an example, suppose that 3, 8 and 2 buses are required at three
destinations A, B, and C respectively. They must come from garages P,
Q, R, which can supply respectively 2, 6 and 5 buses. Note that the
numbers match: 2+6+5 = 3+8+2. The number of garages need not
equal the number of destinations; it just happens to in this example. The
buses are to be scheduled so as to minimize the total of the delivery
times. The delivery times, for each combination of garage and
destination, are given (in units of 10 minutes) in the following table:
From
P
Q
R
To
A
4
10
6
B
3
5
8
C
7
9
7
Suppose that xij buses are sent from source i (thus from P, Q, or
R) to destination j ( thus to A, B, or C). All the xij must be nonnegative,
and in fact integers, and must satisfy the constraints:
x⁄⁄ + x⁄¤ + x⁄‹ = 2,
x⁄⁄ + x¤⁄ + x‹⁄ = 3,
x¤⁄ + x¤¤ + x¤‹ = 6,
x⁄¤ + x¤¤ + x⁄‹ = 8,
x‹⁄ + x‹¤ + x‹‹ = 5,
x⁄‹ + x¤‹ + x‹‹ = 2.
The "cost" f = 4x⁄⁄ + 3x⁄¤ + 7x⁄‹ + 10x¤⁄ + 5x¤¤ + 9x¤‹ + 6x‹⁄ + 8x‹¤
+ 7x‹‹
is to be minimized, subject to these constraints.
This problem is a linear program, except for the requirement that
all xij are integers; however, it will turn out, for a linear program of this
special form, that if the data are all integers, then the resulting xij will all
be integers. The problem could be solved by the simplex method, but
the tableau would be needlessly large, and would consist mostly of
zeros. More efficient computational methods are therefore required.
Dantzig's stepping stone method (an adaptation of the simplex method)
29
is described here;. The alternative labelling method of Ford and
Fulkerson is much more efficient on a computer (see Chapter 6).
If there are m sources and n destinations, then there are m+n
constraints (3+3=6) in the example above). However, these equations
are linearly dependent; in the example, the left set sum to ÛiÛjxij = 13,
and so also do the right set. So one constraint is superfluous; there
should be m+n+1 equations. The xij will be set out in a table with m
rows and n columns. The supplies are put at the left, and the demands
at the top.
Table of
A
P
4
Q
10
R
6
"costs"
B
3
5
8
Initial
C
7
9
7
allocation
3
2+´ .
6+´ .
5+´ 3
6
8
2+´
6-´
.
3
2+3´
. 0
2´
2+´
7
2
0
For the present, ignore the ´ terms.
The minimum cost method for obtaining an initial feasible solution
begins as follows.The cheapest cell in the first row is in column 2. Put as
much as possible in the cell at row 1, column 2, thus MIN(2,8)=2. Thus
the first supply is reduced to 0, and the second demand to 8-2=6. Since
there is no more room in row 1, look for the cheapest remaining cell in
column 2; this is row2, column 2, and put as much as possible in that
cell, thus:
MIN(remaining supply in row 2, remaining demand in column 2)
= MIN(2,6) = 6.
But now row 2 and column 2 are filled up, and it is not clear where to go
next. This difficulty arises because the supplies for rows 1 and 2 sum to
6, the demand for column 2. To overcome it, perturb the supply for
each row by adding ´ to it, where ´ is considered as positive and very
small. To keep the total demand equal to the total supply, a total of 3´
(3 being the number of rows) must be added to the demands
somewhere. As shown, the 3´ is added to just one of the demands.
Then 2+´=MIN(2+´,8) goes in row 1, column 2, leaving 8-(2+´)=6-´
demand in column 2. Then 6-´=MIN(6+´,6-´) goes in row 2, column 2,
so that a supply of 2´ remains in row 2. This goes in the cheapest
available cell in row 2, namely column 3, leaving 2+´ demand in column
3. That goes in the only available cell, namely row 3, column 2, leaving a
demand of 3 remaining in row 3. That goes in the only available cell at
row 1, column 1. So all demands and supplies are accounted for, so a
feasible solution is obtained. Note that there are just 5 stepping stones
("occupied cells", to which some xij has been allocated, corresponding to
basic variables in the simplex method), and that
5 = number of rows + number of columns - 1.
Is it optimal ? Suppose that an amount t is added to the cell in row 1,
column 1. To keep the row and column totals correct, -t must be added
30
to row 1, column 2, +t to row 2, column 2, -t to row 2, column 3, +t to
row 3, column 3, and -t to row 3, column 1. This path is following (some
or all of the) stepping stones previously obtained. The change in total
"cost" is Î = c⁄⁄t-c⁄¤t+c¤¤t-c¤‹t+c‹‹t-c‹⁄t = 4t-3t+5t-9t+7t-6t=-2t, so
the total "cost" would be reduced by increasing t from 0. The largest
possible t is chosen, subject to 2+´-t˜0´, 2´-t˜0, ´3-t˜0, hence t=2´.
Table of
A
P
4
Q
10
R
6
"costs" cij
B
C
3
7
5
9
8
7
Adding t to one unoccupied cell ?
3
8
2+3´
ui
2+´ t
2+´-t
.
0
6+´ .
6-´+t
2´-t
2
5+´
3-t
.
2+´+t 0
vj
6
3
7
cij-ui-vj
-2 . 0
2 . .
. 5 .
But is row 1, column 1 the best cell to choose? To find this, dual
variables (sometimes called"fictitious costs") are calculated ; since each
row and each column of the table represents a constraint, there are dual
variables ui for the rows and vj for the columns. They must satisfy ui +
vj = cij for each occupied square. They are calculated by stepping
stones; let u⁄=0; u⁄+v¤=c¤¤ gives v¤=3; u¤+v¤=c¤¤ gives u¤=2;
u¤+v‹=c¤‹ gives v‹=7; u‹+v‹=c‹‹ gives u‹=0; u‹+v⁄=c‹⁄ gives v⁄=6.
Then
Î = [c⁄⁄-(u⁄+v¤)+(u¤+v¤)-(u¤+v‹)+(u‹+v‹)-(u‹+v⁄)]t
=[c⁄⁄ - u⁄ - v⁄]t= -2t <0.
For an unoccupied cell in row i and column j, cij - ui - vj would be
calculated. In this example, row 1 and column 1 gives the only negative
value, so this cell is the only choice. If cij - ui - vj ˜ 0 for all unoccupied
cells, then a minimum has been reached. (Note that the arbitrary
choice of u⁄=0 can be made, because one constraint is dependent on
the others.)
How large can the change t be ? Since no xij can be negative, t must
satisfy all of 2+´-t ˜0, 2´-t˜0, 3-t˜0. Hence, giving t the largest
possible value, t=2´. The new table is shown below left; note that row 2,
column 3 has now 2´-t=0, and becomes "unoccupied". The total "cost"
has decreased from 68+23´ to 68+19´, this decrease equalling the
product of t=2´ by -[c⁄⁄-u⁄-v⁄]=2. The new ui and vj are also shown;
the consequent cij-ui-vj for the three unoccupied squares are 2,4,3, all
>0, so an optimum has been reached - shown below at right (now
dropping the ´s.)
With different data, more iterations may be needed; and values of t
larger than "´ terms" may be found.
cij-ui-vj
Optimal table
3
8
2+3´ ui
2+´ 2´ 2-´ .
0
.
. 2
0 2 .
6+´ .
6+´ .
2
4 . 2
.
6 .
5+´ 3-2´
.2+3´ 2
.
3 .
3 . 2
vj
4
3
5
31
Integer solutions ? The supplies and demands were given as integers; all
the calculations added and subtracted, but never divided; so an integer
optimum schedule was obtained. This happens with transportation
problems, though not with general linear programs.
Unbalanced demands ? In the above example, suppose now that the
third source can supply 7 buses instead of 5; then total supplies =15,
total demands =13, no longer balanced. To make the method work,
balance must be restored by introducing an additional column (D) for a
dummy destination, having zero "costs" (because those buses do not, in
fact. go anywhere), and a demand of 2 (the extra supply available). The
previous optimum can be used, as shown, to give an initial allocation.
Table of "costs" cij , and an initial allocation, for unbalanced
A
B
C
D
3
8
2+3´
P
4
3
7
0
2+´ 2´
2-´
.
Q
10 5
9
0
6+´ .
6+2´
.
R
6
8
7
0
5+´
3-2´
.
2+3´
problem
2
.
.
2
Connection with linear programming A transportation problem can be
written as a linear program: MIN cTx subject to x˜0, Ax=b, where x is a
vector of the elements xij. For the above example, bT=[2 6 5 3 8 2], and
the matrix A has the "structured" form:
1
.
.
1
.
.
1
.
.
.
1
.
1
.
.
.
.
1
.
1
.
1
.
.
.
1
.
.
1
.
.
1
.
.
.
1
.
.
1
1
.
.
.
.
1
.
1
.
.
.
1
.
.
1
The dual variables are v⁄,v¤,v‹,u⁄,u¤,u‹. The vector wTA-cT in the duality
theory for linear programming has as elements the quantities ui+vj-cij for
the transportation problem. Thus, in fact, the u and vj are shadow costs.
However, one would not use the ordinary simplex method for a
transportation problem, because it would be very wasteful in time and
numbers stored.
A special case of production allocation Consider the allocation of
production for m products and n machines. Such a linear programming
model could be:
MIN Ûi,j cijxij subject to all xij ˜ 0,
Ûj xij = di (i=1,2,...,m), Ûi rijxij ˆ hj (j=1,2,...,n),
in which a quantity xij (tons or kg or whatever) of product i is to be made
on machine j, and total production cost is to be minimized, subject to
satisfying demands di for product i (i=1,2,...,m), and staying within the
available time of hj hours on machine j (j=1,2,...,n), where the rate of
production for product i on machine j is expressed by rij tons (or kg, etc.)
32
per hour. In general, this problem required the simplex method. However,
if rij = åi!j holds, for all i and j, then the problem can be scaled into
transportation form. The constraints can be rewritten as:
Ûj (åixij) = (åidi) (i=1,2,...,m); Ûi (åixij) ˆ dj/!j (j=1,2,...,n).
Thus a transportation problem is obtained, in the new variables åixij. The ˆ
constraints are dealt with by introducing an additional "slack row" in the
transportation table, with zero costs attached.
Note that transportation problems are commonly computed using a
faster network method by Ford and Fulkerson (see Section 6.4)
1.16 Cost data (piecewise linear cost function)
Every textbook of linear programming includes examples where cost
figures like "cost of processing material X is $20 per ton". This assumes
that (i) the cost is known, and (ii) is exactly proportional to the quantity
(tons) of the material. Life, whether or not meant to be easy, is rarely so
simple! Let us look first at (ii). The following idealized diagram gives the
picture. (Here only one product is considered, in order to get a plane
diagram, whereas there are usually many products to be considered.)
In the diagram, A shows the initial cost (you have to have a factory
and a machine before you can make anything); AB is a linear part of the
cost curve (suitable for applying LP - note that this line does not pass
through the origin); BC is a steeper part of the curve (extra production,
at overtime rates, or using more expensive raw material); the step CD
shows the cost of a new and better machine; DE represents normal
production (cheaper with the new machine), and EF overtime production.
A linear programming model can only apply to a straight-line part of
a cost curve, not to the whole curve. A cost coefficient for the LP model
is then the slope of the linear part of the cost curve. Note also that,
when the same machine is changed over from making one product to
making another, there is always a setup cost incurred, in addition to a
cost proportional to the amount made of the new product. Oftern, the
setup cost includes a substantial contribution from production time lost,
in getting the machine properly adjusted to the new product.
Now, what do we mean by cost ? There are practical difficulties in
33
getting told (maybe no one knows, or the man who knows sees the O.R.
type as a threat, and won't say). But, beyond these, there is the
question of what kind of cost. Accountants classify cost components in
several ways; as direct costs (directly related to the goods being made)
versus indirect costs (not directly related), or as fixed costs (not
varying with level of production) and variable costs (varying with level of
production, often taken as proportional to it). Some examples are as
follows.
Direct costs
Indirect costs
Wages of production workers
Purchase of food for factory canteen
Cost of testing output
Rates, heating, etc., for the factory
Purchase of special tools
Wages of cleaners
Fixed costs
Rental of factory building
Manager's salary
Variable costs
Cost of raw materials
Cost of electric power to run
production machines
Wages of production workers (?)
Note that wages of production workers are not always a variable
cost over a short time scale (the wages may well be fixed by a union
contract, so that they will be paid, even if there is not much work to do);
but wages become a variable cost over a longer time scale (over which
the number of men employed must move up or down with the general
level of production.)
Some examples of cost components are shown in the following
diagrams. The abscissa is always level of production or output; the
ordinates are not always on the same scale. For an optimization model
(such as LP) , some kind of variable costs are required. However, we
must consider what actually varies, if the changes considered in the
model are made. (For example, if we reallocated production between
different machines, would this change the wages of the production
workers, or would they be paid the same anyway? If it would not change
them, then they would not enter the cost coefficients of the LP model.)
However, after optimizing the LP model, we would have to also calculate
the remaining fixed costs, for the alternative plans considered; they
could turn out to dominate. (For example, is it useful to buy very low
variable costs - low slope in the cost curves - by incurring enormous
capital costs, of installations, machines, etc. ?) The term marginal cost
is also used for the slope of a linear part of a cost curve.
34
Cost curves
Remember that accountants use cost data for quite different purposes
- including what to charge for the product, a means of stopping anyone
pinching the cash, and for tax purposes, all essential. Often, the
absorption costing method is used. The system is divided into cost
centres (e.g. departments, or machines). To each cost centre is assigned
the direct costs applicable, and also a proportion of overheads (= indirect
manufacturing costs); the proportioning may be according to hours of
labor, or e.g. machine time. But these proportions are likely to change if
we replan the production, e.g. using a LP model; we don't want to include
the overheads in the costs for the LP. Sometimes variable costing is used,
in which fixed costs (which do not depend on level of production) are
separated out. The following simple example shows the differences.
(This does not look like "mathematics" ! But an O.R. person must be
able to get needed data from accountants' records.)
Data
Direct material
$18000
Number of units made 10000
Direct labor
$12000
Number of units sold
6000
Overheads:variable
$15000
Sale price per unit
$12
Overheads: fixed
$15000
Absorption method Total manufacturing cost $60000.
Manufacturing cost per unit $6.
Cost of goods sold (6000 x $6)
$72000
Inventories at cost (4000 x $6)
$24000
Sales revenue (6000 x $12)
$72000
Contribution to profit (=72000-36000).
$36000
Variable costing method Variable costs $45000. So unit cost is $4.50.
Cost of goods sold (6000 x $4.50)
$27000
Inventories at cost(4000 x $4.50)
$18000
Fixed costs
$15000
Contribution to profit (=72000-27000-15000)
$30000.
35
1.17 Some exercises on linear programming
Formulate each model as a linear programming problem, with a
function to be maximized or minimized, subject to some constraints
(equations or inequalities).Then convert the problem into standard form
by introducing slack variables and artificial variables as required. (Of
course, inequalities like xj ˜ 0 are unchanged.)
LP1 A firm can make four different products. each requiring different
amounts of the inputs, which are labor (measured in man-weeks), energy
(KWh), and material (tons). The data are tabulated as follows:
Product
Unit profit
Production level
INPUTS
Labor (man-weeks)
Energy (KWh)
Material (tons)
I
4
x1
II
5
x2
1
7
3
1
5
5
III
9
x3
1
3
10
IV
11 (in some suitable unit)
x4 units per week
TOTAL AVAILABILITIES
1
ˆ 15
2
ˆ120
15
ˆ100.
The firm wants the maximum profit it can get. Formulate as a LP
problem. What happens to the formulation if there are also minimum
quantities of certain of the products which have to be supplied? (Remark:
This product-mix problem is nearly formulated already. It is included to
stress that variables need to be listed, with units of measurement
specified (otherwise one will get lost with a real-world problem).
LP2 Two products are produces successively on two machines. The
manufacturing times in hours per unit of each product are tabulated
below for the two machines.
Time per unit (hours)
Machine
Product 1
Product 2
Product 3
Product 4
1
2
3
4
2
2
3
2
1
2
The total cost of producing 1 unit of each product is based directly on
the machine times. Assume that the cost per hour for machines 1 and 2
is $10 and $15 respectively. The total hours budgeted for all the
products on machines 1 and 2 are 500 and 380. The sales price per unit
for products 1,2,3 and 4 are $65, $70, $55, $45.
LP3 For model A in Section 1.1, use a computer program (e.g. lpZ7ap) to
compute an optimum. (Note that the data are needed in a tidy input
form, as in the second table in 1.1, before entering data into the
program.) Then tabulate the optimum obtained in the style:
36
machine 1
machine 2
shadow cost
product 1
.
2929
100.31
product 2
product 3
product 4
product 5
product 6
shadow cost
Some such tabulation is essential, in order that a non-mathematician may
make sense of the output !
What then happens if the requirement for product 1 is increased, say
by 2% ? Use a computer program to find out; and see if the change
relates to shadow costs. (If it doesn't, then why not ?).
LP4 Multi-period scheduling A firm makes both thingos and whatsits.
They want to plan for a year ahead, to meet the following sales targets
for the four quarter
quarter 1
2 !!!!
3
4
thingos
4000
1800
4500
3000
whatsits
2000
2500
1500
2400
They keep inventory of each product (thus, they make some in advance
of the demand). In quarter t, they make at thingos and bt whatsits; and ht
labor hours are available; ht must not vary by more than 10% (up or
down) from each quarter to the next. At the end of quarter t, after
making the deliveries for that quarter, they have an inventory of pt
thingos and qt whatsits. Each thingo requires 2 working hours to make;
each whatsit requires 3 working hours to make. The initial conditions
(where they start) are given as h‚ = 5000 hours, and given values of p‚
and q‚ (initial inventoroes). If the variables are listed by quarters, thus
a⁄, b⁄, p⁄, q⁄, h⁄, then a¤, b¤, p¤, p¤, q¤, h¤, then ..., then the matrix of
coefficients has a "block structure" :
quarter
quarter
quarter
quarter
1
2
3
4
’
’
0
0
0
’
’
0
0
0
’
’
0
0
0
’
where ’ is a submatrix of
coefficients (not all zero),
and 0 is a zero submatrix.
Write down those constraints which affect the second quarter; and thus
verify the stated block structure. Note here that
initial inventory + production - demand = final inventory
(for each product and each quarter); and that ht is linked by inequalities
to ht-1. (Such "block structures" are typical for linear programming
models for multi-period scheduling.)
LP5 The following example was given by Trustrum (1971). Solve it both
by the simplex method, and by the revised simplex method, and write
37
down the dual problem.
Minimize x1 + 6x2 + 2x3 -x4 + x5 -3x6 subject to all xj˜ 0, and
x1 + 2x2 + x3 + 5x6 = 3,
-3x2 + 2x3 + x4 + x6 = 1,
5x2 + 3x3 + x5 -2x6 = 2.
(Remark: Is it necessary in this example to introduce more variables in
order to have an initial basis ? This example is cited to show this
feature.)
LP6 Solve the following transportation problem. (Note that there may,
very likely, be more than one optimum. And epsilons may be essential!)
An efficiency expert (?) has rated four candidates Ci for four jobs Jj as
follows. Find the optimal assignment of candidates to jobs (so as to
maximize the sum of the ratings).
J1
6
7
6
1
C1
C2
C3
C4
J2
9
1
8
3
J3
5
2
2
9
J4
6
4
5
3
LP7 Three products are processed through three different operations.
Time per unit (minutes)
Operation capacity
Product 1 Product 2 Product 3
Operation
(minutes per day)
1
1
2
1
430
2
3
0
2
460
3
1
4
0
420
Net profit/unit($)
3
2
5
Let x1, x2, x3 units of products 1,2,3 be produced each day. The LP
model to maximize profit is:
MAX 3x1+2x2+5x3 s.t. xj˜0,
x1+2x2+x3ˆ430, 3x1+2x3ˆ460, x1+4x2ˆ420.
(This assumes profit proportional to amount made - not always true !).
The optimal tableau is
100
230
20
1350
-.25
1.5
2
4
1
.
.
.
.
1
.
.
.5
0
-2
1
-.25
.5
1
2
.
.
1
.
(i) Suppose the second "requirement" of 460 minutes per day is
decreased, by some accident to the machine. How far can we decrease it
without upsetting the optimal schedule, and what does that cost us?
(ii) Suppose a fourth product is scheduled, through the same operations,
with net profit $9 per unit, and times 3,2,4 minutes, through the same
operations. How much of the new product should we make? What would
the net total profit become?
38
LP8 A firm can make four products. The data for daily production
scheduling are:
product
1
2
3
4
availability
floor area (m™)
10
30
80
40
900 m™
labor (persons)
2
1
1
3
80 persons
variable cost ($/unit) 20
30
45
58
sales revenue ($/unit) 30
50
85
90
The optimal tableau (using 10 m™ and $10 as units) to maximize profit is:
10
20
74
|
|
|
1
.
.
-1
-4
1
3
0.2 1.6
.
1
.
-0.6
0.4
0.68
+0.8
-0.2
0.16
(i) A raw material used in making products 1 and 3 is variable in price. At
present it costs $100 per ton (included in the above variable costs).
Product 1 uses 0.1 ton, and product 3 uses 0.2 tons. Suppose the cost of
this raw material changes to $p per ton. How far can p vary, without
changing the optimal schedule ? (Hint: Compare 1.11c. The first profit
coefficient is 30 - (0.1–100 + 10) when p = 100; it changes to
30 - (0.1p+10), so gives 2.0 - 0.01p in $10 units.)
(ii) The firm can increase its effective floor space to 1000 m™ by
installing some new equipment, that would cost $50 per day. Should
they ? If so, what would be the new production schedule ?
LP9 A furniture maker wants to maximize his weekly profit. His data are
as follows:
Product number :
Available
Machine hours
Wood (square metres)
Profit($) per item made
1
2
3
4
1
4
20
3
2
20
4
6
50
3
8
40
1000 hours
2500 sq. m
A linear program (in scaled variables) to maximize profit is
MAX 2x1+2x2+5x3+4x4
s.t. all xj˜0, x1+3x2+4x3+3x4ˆ10,
4x1+2x2+6x3+8x4ˆ25. The optimal tableau is
x1=4.0
x3=1.5
15.5
|
|
|
1
.
.
-1
1
1
.
1
.
1.4
.4
.8
-.6
.4
.8
.4
-.1
.3
(i) What happens to the optimum schedule, and to the optimum profit, as
the unit profit for product 3 (at present 5) varies from 0 to Ç ?
(ii) A major customer insists that 100 units of product 2 be delivered in
as particular week. For that week, what are the optimum schedule and
profit?
(iii) A shipping delay reduces the wood supply to 2000 sq. m. What is the
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optimal production schedule and profit under that restriction? However,
another source can supply up to 1000 sq. m at $6 per sq. m instead of
the normal price of $4 per sq. m). How much (if any) should he buy at
that increased price?
(iv) He is considering another product, which uses 4 hours of machine
time, and only 2 sq. m of wood. It would give a profit of $36 per item.
Should he make any?
LP10
Solve the integer programming problem:
MAX 4x1 + 6x2 + 2x3 subject to
4x1 - 4x2 ˆ 5, -x1 + 6x2 ˆ 5, -x1 + x2 + x3 ˆ5,
where x1, x2 , x3 are nonnegative integers .
The optimal simplex tableau is
1
.
.
.3
.2
.
.
1
.
.05 .2
.
.
.
1
.25 .
1
.
.
.
2
2
2
| 2.5
| 1.25
| 6.25
| 30
LP11 In the example of 1.5, the basic variables x⁄ and x¤ remained
basic when the second requirement was varied between 5.0-1.5 and
5.0+2.0. Outside this range, some x jruns negative. Use the dual simplex
method to calculate the optimum value of this linear program as a
function of the second requirement, as it varies from 3.5 down to 0.0.
Also, what happens if the second requirement is increased above 7.0 ?
1.18 Computer input/output for a linear program
Consider the (trivial) linear program:
Minimize -3.4x⁄ -5.6x¤ +6.0x‹ subject to x⁄˜0, x¤˜0. x‹˜0, and
2.0x⁄ + 3.1x¤
ˆ 3.0,
-4,5x⁄ + 6.0x¤ + 4.0x‹ ˆ 4.0 .
For Craven's LP program (lpZ - see chapter 9), the output file is as
follows. Note that the nonzero matrix elements are given sequence
numbers 0,1,2,..., to enable changes to data. Only the basic optimal
variables xj are listed, since the nonbasic variables are zero. The optimal
dual variables, and the objective function value, are listed. The optimal
inverse basis matrix is given, with an extra "zero row" for optimal dual
variables, and an extra "zero column" for optimal xj variables; thus the
inverse basis matrix is:
ã.2312
-.1195õ
Ã.1734
.0721Õ
(to 4 decimal, places).
The zj-cj (reduced costs) are not tabulated; however this program shows
them on the screen as : 0.0000, 0.0000, -2.1017, -1.7572, -0.0254.
Sample output file
SeqNo Row Col Entry
0 -1 1 2.000000 1 2 1 -4.500000 2 -1 2 3.100000
3 2 2 6.000000 4 -2 3 4.000000 5 -1 4 1.000000
6 -2 5 1.000000 7 0 0
.000000
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C( 1 )= -3.4
C( 2 )= -5.6
C( 3 )= 2
C( 4 )= 0
C( 5 )= 0
B( 1 )= 3 -1
B( 2 )= 4 -1
Optimum
X 1 = .215799614643
X 2 = .828516377651
Dual 1 = -1.75722543353
Dual 2 = -.0254335260086
Objective = -5.37341040462
W( 0 0 ) -5.3734
W( 1 0 )
.2158
W( 2 0 )
.8285
W( 0 1 ) -1.7572
W( 1 1 ) .2312
W( 2 1 ) .1734
W( 0 2 ) -.0254
W( 1 2 ) -.1195
W( 2 2 ) .0771
With this program, the matrix can also be saved for future use, in the format:
2,5,3,5,7
0 ,-1 , 1 , 2
1 , 2 , 1 ,-4.5
2 ,-1 , 2 , 3.1
3,2,2,6
4 ,-2 , 3 , 4
(The first line records the number of different kinds of data, for setting up arrays.)
The following page shows the format for data input. This can be done
either by keyboard entry (nonzero matrix elements only), or by
spreadsheet input.
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1.19 A note on interior-point methods
Consider a linear program, in standard form, with m constraints and n
variables; typically, n is considerably larger than m. The simplex method
moves from vertex to vertex, thus along the boundary of the feasible region,
which is a polyhedron in n-dimensional space. The number of vertices is
large, but usually only a few need to be visited. While an example exists
where all the vertices are visited, the number of iterations of the simplex
method, for typical industrial planning models, typically averages about 2m,
though there is much variation about this average.
Would it be quicker to take a more direct path through the interior of the
feasible region? There are more recent interior-path methods which do this,
and they are often found useful for large linear programs, which in any case
require heavy computation on big computers. Usually they are implemented
as central-path methods. The basic idea is discussed below. However, their
efficiency depends on the details of how the computation is arranged.
When the dimension n of the space is large, most of the( n-dimensional)
volume is close to the boundary. Consider, for example, a hypercube (the ndimensional analog of a cube), defined by:
C = {x = (x1, x2, ..., xn) : 0 ˆ xj ˆ 1, j=1,2,...,n},
and another hypercube inside it:
Cå = {x = (x1, x2, ..., xn) : å/2 ˆ xj ˆ 1 - å/2, j=1,2,...,n},
where å is small positive. The n-dimensional volume of the part of C not in
Cå is 1 - (1-å)n , which approaches 1 as n becomes large. Thus, in large
dimension, keeping close to the boundary may in fact be passing through a
large part of the volume.
Consider now a linear program, in standard form:
MIN cTx subject to Ax = b, x ˜ 0,
where A is a matrix with m rows and n columns. If the KKT conditions for a
minimum are added, this gives:
Ax = b, x ˜ 0, ATy + z = c, z ˜ 0, xi zi = 0 (i=1,2,...,n).
(PD)
(This is called a primal-dual formulation, sin c e y and z are the variables of
the dual linear program, and the added constraints are those of the dual
program.) The system (PD) is linear except for the nonlinear constraints
xi zi = 0., and these divide the computation into separate branches,
depending on which subsets of the xi are nonzero.
Consider then a modification of (PD):
Ax = b, x > 0, ATy + z = c, z > 0, xi zi = µ (i=1,2,...,n),
(PDµ)
in which µ is a positive parameter. The system (PDµ) can be solved quickly
by Newton's method (see Section 7.3), giving a solution (x*(µ),y*(µ)). The
central path is the graph of x*(µ) against µ. Of course, x*(µ) is only
computes for a chosen set of values of µ = µ0 > µ1 > µ2 > ... tending to 0.
Efficient computation requires a balance between the steps in µ (say, how tg
choose µj+1/µj) and the precision of computing each x*(µj).
Newton's method, to find a zero of a vector function F(w), generates a
sequence w0, w1, w2, ...
F(wj=1) Ú F(wj) + G(wj) (wj+1 - wj) = 0
to determine the step wj+1 - wj . Here G(wj) is the gradient of F at wj .
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