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Chapter 1 Solutions OSCM
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Chapter 01 - Operations and Supply Chain Management
CHAPTER 1
OPERATIONS AND SUPPLY CHAIN MANAGEMENT
Discussion Questions
1. Using Exhibit 1.2 as a model, describe the source-make-deliver-return relationships in the
following systems:
a. An airline
Source:
Make:
Deliver:
Return:
Aircraft manufacturer, in-flight food, repair parts, computer systems
Aircraft and flight crew scheduling, ground services provided at airports,
aircraft maintenance and repair
Outbound and arriving passenger service, baggage handling
Resolve any post-service issues such as lost or damaged luggage
b. An automobile manufacturer
Source: Suppliers of components and raw materials
Make: Manufacturing of vehicles and components or subassemblies to be sold as
spare parts
Deliver: Delivery to and sales from dealerships, delivery of spare parts to the
wholesale system
Return: Warranty and recall repairs, trade-ins
c. A hospital
Source:
Make:
Deliver:
Return:
Medical supplies, cleaning services, disposal services, food services, qualified
personnel
Inpatient rooms, outpatient clinics, emergency room, operating rooms
Scheduling patients, providing treatment, ambulance service, family
counseling
Billing errors, follow up visits
d. An insurance company
Source: Supplies needed for the office, underwriters, legal authority to operate
Make: Establish policy guidelines and pricing, field agent/representative and facility
network, develop Internet service capabilities, establish preferred vehicle
repair service network
Deliver: Meet with and advise clients, write policies, process and pay claims
Return: Refund of overpayments
2. Define the service package of your college or university. What is its strongest element? What is
its weakest one?
The categories with examples are:
Supporting facility - location, buildings, labs, parking
Facilitating goods – class schedules, computers, books, chalk
Explicit services – classes with qualified instructors, placement offices
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Chapter 01 - Operations and Supply Chain Management
Implicit services – status and reputation (e.g., Ivy League schools)
At Indiana University and the University of Southern California, among their strongest
elements are their business schools and their Operations Management programs (of course).
Both also have very dedicated alumni networks. A weak element of Indiana University is its
weak football program; for USC, weak elements are on-campus parking and housing.
3. What service industry has impressed you the most with its innovativeness?
Our vote goes to cruise lines which have introduced such onboard innovations as wave
machines for belly boarding and rock climbing walls, as well as all sorts of other amenities to
keep cruisers involved. The industry is doing record business as well.
Some of the standout companies in less innovative industries are Bank of America (has a
formalized research program to try out new customer services/amenities such as video screens
in next to teller lines), Intuit (e.g., putting Quicken money management software online), Ikea,
JetBlue Airlines, and Progressive Insurance (discussed later in the book).
4. What is product-service bundling and what are the benefits to customers?
Product-service bundling is adding value-added services to a firm’s product offerings to create
more value for the customer. This provides benefits in two areas. First, this differentiates the
organization from the competition. Secondly, these services tie customers to the organization
in a positive way. Alternatively, bundling can also involve adding products to a service, for
example, adding the sale of convenience items and snacks at a hotel.
5. What is the difference between a service and a good?
A service is an intangible process (you can’t hold it in your hands), while a good is the physical
output of a process. Some service businesses also provide a physical good as part of the
service, like a restaurant. Also, mots manufacturers of goods provide services for after-sales
support, like computer tech support or automobile warranty service. So while a service and a
good are definitely distinguishable, customers will often encounter both in their experiences
with a company.
6. Some people tend to use the terms effectiveness and efficiency interchangeably, though we’ve
seen they are different concepts. But is there any relationship at all between them? Can a
firm be effective but inefficient? Very efficient but essentially ineffective? Both? Neither?
Firms can be anywhere on these two dimensions. It is possible for a firm to be the best at what
they do in serving their market, but be very wasteful in doing so. Alternatively, a firm could
squeeze every last dollar out of their processes but fail to deliver what the market expects and
desires. Of course, the best firms will provide the goods and services that the market desires,
exactly as the market desires, and do so at a minimum cost. Firms that are both inefficient and
ineffective do not survive for long in any market.
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7. Two of the efficiency ratios mentioned in the chapter are the receivable turnover ratio and the
inventory turnover ratio. While they are two completely separate measures, they are very
similar in one way. What is the common thread between these two?
(There are a number of answers that students may come up with, from simplistic to more
thoughtful. Following is one of the latter.) Both are measuring the average amount of a
valuable asset that is not generating value for the company. Accounts receivable are an asset,
but they do not create value for the firm until the money is received. Reducing the average
amount of accounts receivable frees up that money for use by the company on a recurring
basis. Inventory is another asset, but while inventory is being held by the company it is not
making any money for the firm. Reducing inventory allows the firm to invest the money that
would otherwise be spent on the inventory.
8. Look at the job postings at http://www.apics.org and evaluate the opportunities for an OSCM
major with several years of experience.
There are pages and pages of these in the APICS Career Center. Here are some examples:
Nacelle Product Materials Leader
General Electric Corporation
US - Hattiesburg, Ohio
The Nacelle Product Materials Leader will demonstrate leadership in communicating business
goals, programs, and processes for an area or business segment. In this role, you will utilize
your experience or expertise to solve problems, develop and execute objectives for self and
others, and have the ability to effect short-term and some long-term business goals ...
May 12, 2015
Buyer – Supply Chain
Froedtert Health
US - Menomonee Falls, Wisconsin
As SUPPLY CHAIN BUYER, you will be responsible for the acquisition of supplies, equipment and
services in a timely manner, ensuring price, quality and delivery. This position serves as
resource to departments regarding procurement practices that meet customer needs ...
May 07, 2015
Demand Planning Manager
Cintas Corporation
US – Jackson, Mississippi
The Demand Planning Manager leads the development of strategy, processes, and tools for the
company’s key forecasting activities within the site. This position will represent the site
demand planning in key cross-functional decisions, including short- and long-term strategy
discussions, product initiatives, strategic business planning, and systems integration. ...
May 11, 2015
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Operations Team Leader - Supply Chain
Parker Aerospace
US – Dublin, Georgia
The Site Supply Chain leader role is responsible for S&OP, Planning, Production control,
purchasing, Warehousing and logistics functions at the Dublin, Georgia location. This role plays
a key role in the overall success of the Dublin Operation. The individual will lead a team of 1215 employees across different supply chain functions...
May 08, 2015
Production/Operations Planner
CG Industrial Specialties
US - Nationwide
Reporting to the Operations Manager or Branch Manager; this position is responsible for
preparing assembly schedules for shop technicians; coordinate material requirements with
purchasing as well as coordinate shipping / receiving activities with warehouse staff.
May 08, 2015
9. Recent outsourcing of parts and services that had previously been produced internally is
addressed by which current issue facing operations and supply management today?
The coordination of relationships between mutually supportive but separate organizations.
10. What factors account for the resurgence of interest in OSCM today?
With companies facing competition on a global scale, and ever-advancing manufacturing and
information technologies, firms realize the competitive advantage their OSCM functions can
provide if properly managed. Many have found that the same old way of doing business
leaves them unable to compete successfully. The 2011 tsunami in Japan and the 2015 LA ports
closure have also brought to the forefront how important supply chains are, as well as the
negative economic impact that disruptions in the supply chain can cause.
11. As the field of OSCM has advanced, new concepts have been applied to help companies
compete in a number of ways, including the advertisement of the firm’s products or services.
One recent concept to gain the attention of companies is promoting sustainability. Discuss
how you have seen the idea of sustainability used by companies to advertise their goods or
services.
There of course will be a number of examples that students will bring up, though they may
need some prodding to jog their memories. Some examples to start with might be IBM’s “I’m
an IBMer” campaign where they advertise how they are “building a smarter planet.” Bottled
water manufacturers have reduced the amount of plastic used in many of their products, thus
saving production and distribution costs, but also allowing them to advertise how the new
bottles are better for the environment because they result in less waste.
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Chapter 01 - Operations and Supply Chain Management
Objective Questions
1. What are the three elements that require integration to be successful in operations and supply
chain management?
Strategy, Processes, and Analytics
2. Operations and supply chain management is concerned with the design and management of
the entire system that has what function?
Produces a product or delivers a service
3. Consider the following financial data from the past year for Midwest Outdoor Equipment
Corporation.
Gross Income
Total Sales
Total Credit Sales
Net Income
Cost of Goods Sold
Total Assets
Average Inventory
Average Receivables
$25,240,000
24,324,000
18,785,000
2,975,000
12,600,000
10,550,000
2,875,000
3,445,000
a. Compute the receivable turnover ratio.
$ 18,785,000
=5.453
$ 3,445,000
b. Compute the inventory turnover ratio.
$ 12,600,000
=4.383
$ 2,875,000
c. Compute the asset turnover ratio.
$ 24,324,000
=2.306
$ 10,550,000
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Chapter 01 - Operations and Supply Chain Management
4. A manufacturing company has entered into a new contract with a major supplier of raw
materials used in the manufacturing process. Under the new arrangement, called vendor
managed inventory, the supplier manages their raw material inventory inside the
manufacturer’s plant, and only bills the manufacturer when the manufacturer consumes the
raw material. How is this likely to affect the manufacturer’s inventory turnover ratio?
This will reduce the average amount of money the firm has invested in raw material, so the
inventory turnover ratio should increase.
5. What is the name of the process in which one company studies the processes of another firm
in order to identify best practices?
Benchmarking
6. A company has recently implemented an automated online billing and payment processing
system for orders it ships to customers. As a result, it has reduced the average number of days
between billing a customer and receiving payment by 10 days. How will this affect the
receivables turnover ratio?
Quicker payments will reduce the average amount of accounts receivables, so the receivables
turnover ratio will increase.
7. Match the following OSCM job titles with the appropriate duties and responsibilities.
C
Plant manager
D
Supply chain manager
A
E
Project manager
Business process
improvement analyst
B
Logistics manager
A: Plans and coordinates staff activities such as new product
development and new facility location
B: Oversees the movement of goods throughout the supply
chain
C: Oversees the workforce and resources required to
produce the firm’s products
D: Negotiates contracts with vendors and coordinates the
flow of material inputs to the production process
E: Applies the tools of lean production to reduce cycle time
and eliminate waste in a process
8. What high-level OSCM position manager is responsible for working with the CEO and company
president to determine the company’s competitive strategy?
Chief Operating Officer
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Chapter 01 - Operations and Supply Chain Management
9. Order the following major concepts that have helped define the OSCM field on a time line.
Use 1 for the earliest to be introduced, and 5 for the most recent.
3
1
5
2
4
Supply chain management
Manufacturing strategy
Business analytics
Total quality management
Electronic commerce
10. Which major OSCM concept can be described as an integrated set of activities designed to
achieve high-volume production using minimal inventories of parts that arrive at workstations
exactly when they are needed?
Just-in-time (JIT) production
11. ___________________________ leverage the vast amount of data in enterprise resource
planning systems to make decisions related to managing resources.
Business analytics
12. Which current issue in OSCM relates to the ability of a firm to maintain balance in a system,
considering the ongoing economic, employee, and environmental viability of the firm?
Sustainability
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Chapter 01 - Operations and Supply Chain Management
Analytics Exercise: Comparing Companies Using Wall Street Efficiency Measures
Each student is asked to pick an industry and compare three companies within that industry based on
income per employee, revenue per employee, receivable turnover, inventory turnover, and asset
turnover. The following is typical of what you might obtain:
BP
Shell
ExxonMobil
Oil Industry
Net Income/Employee
315,300
343,533
414,328
289,320
Revenue/Employee
4.6 Mil
5.2 mil
4.7 mil
3 Mil
Receivable Turnover
9.38
6.29
13.17
13.5
Inventory Turnover
11.92
13.59
21.91
15.5
Asset Turnover
1.92
1.36
1.41
1.1
Management Efficiency
Students are then asked to identify which company appears to have the most productive employees.
With this data we see that ExxonMobil does very well in generating $414,328 net income per
employee. Comparing Shell to ExxonMobil we can observe that ExxonMobil appears to be more
efficient since it can generate more net income on lower revenue/employee, at least compared to
Shell. The inventory turnover is highest for ExxonMobil indicating that the company is the most
efficient from an operations and supply chain processes view. ExxonMobil also appears to do a good
job in collecting receivables as well, thus supporting the idea that the company is very efficient. BP
seems to do a little better in asset turnover, which relates to the use of its facility and equipment
assets. But ExxonMobil is very good especially in comparison to the oil industry average.
Overall, ExxonMobil appears to be the most efficient, so the other companies might find it valuable to
benchmark the company’s processes.
Of course, the data generated by each student will be different and an interesting interchange can be
developed with students each presenting what they found from their research. It is very interesting to
do comparisons across industries; retailers versus oil companies, and computer makes versus software
companies, for example.
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Financial Management (Lingnan University)
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Chapter 02 - Strategy
CHAPTER 2
STRATEGY
Discussion Questions
1. What is meant by a “triple-bottom-line” strategy? Give an example of a company that has
adopted this type of strategy.
A triple-bottom-line strategy places emphasis on a company’s environmental and social
responsibilities as well as the traditional bottom line of economic prosperity. It recognizes that
the long-term health of the firm is interdependent with the health of the environment and the
betterment of society. There are many examples – one if Kraft Foods. For details see their 2010
report:
http://www.kraftfoodscompany.com/SiteCollectionDocuments/pdf/kraftfoods_responsibility_rep
ort.pdf
2. Find examples where companies have used features related to environmental sustainability to
“win” new customers.
Car companies use environmental concerns in marketing ads. The development of hybrid and
flex-fuel cars is one way they have operationalized those concerns. Consumer goods companies
display the “made with recycled material” logo on the packaging. Bottled water manufacturers
are using and advertising bottles made with less plastic.
3. What are the major priorities associated with operations and supply chain strategy? How has
their relationship to each other changed over the years?
The four major imperatives are cost, quality, delivery, and flexibility. In the sixties, these four
imperatives were viewed from a tradeoffs perspective. For example, this meant that improving
quality would result in higher cost, and in many cases that was true. However, advances in
manufacturing and information technologies since then have reduced the size of those tradeoffs,
allowing firms to improve on several or all of these imperatives simultaneously, gaining greater
competitive advantage than was possible 50 years ago. The problem now becomes one of
prioritizing and managing towards orderly improvement.
4. Why does the “proper” operations and supply chain strategy keep changing for companies that
are world-class competitors?
The top three priorities have generally remained the same over time: make it good, make it fast,
and deliver it on time. Others have changed. Part of this may be explained by realizing that world
class organizations have achieved excellence in these three areas and are, therefore, focusing
attention on some of the more minor areas to gain competitive advantage. The changes in the
minor priorities may result from recognizing opportunities or from changes in customer desires or
expectations.
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5. What is meant by the expressions order winners and order qualifiers? What was the order
winner(s) for your last purchase of a product or service?
Order winners are dimensions that differentiate the product or service or services of one firm from
another. Order qualifiers are dimensions that are used to screen a product or service as a
candidate for purchase. Order qualifiers get a company’s “foot in the door.” Order winners are
what make the sale. Obviously, answers will vary for the order winners from your last purchase.
6. Pick a company that you are familiar with and describe its operations strategy and how it relates
to winning customers. Describe specific activities used by the company that support the strategy.
Student answers will vary widely based on their experiences and views. It might be helpful for a
classroom exercise to assign certain companies to a number of students/teams and compare their
answers in class.
7. At times in the past, the dollar showed relative weakness with respect to foreign currencies, such
as the yen, mark, and pound. This stimulated exports. Why would long-term reliance on a lower
valued dollar be at best a short-term solution to the competitiveness problem?
This approach is dependent on economic policies of other nations. This is a fragile dependency. A
long-term approach is to increase manufacturing and service industry productivity in order to
regain competitive advantage. At a national level, solutions appear to lie in reversing attitudes.
At a firm level, competitive weapons are consistent quality, high performance, dependable
delivery, competitive pricing, and design flexibility.
8. Identify an operations and supply chain - related "disruption" that recently impacted a company.
What could the company have done to have minimized the impact of this type of disruption prior
to it occurring?
The March 2011 tsunami that struck Japan was geographically concentrated but had global
impact on multiple firms, many of which had no physical presence at all in the affected area.
Examples include firms that had sole source agreements with suppliers in the affected area. The
tsunami left these companies scrambling to find new suppliers to feed into their supply chains.
These firms could have reduced the impact of the tsunami by having a few high-quality,
dependable suppliers located in different geographical regions. There are many other examples
that could be taken from this one event. A simple Internet search will provide plenty of material
for discussion.
9. What do we mean when we say productivity is a “relative” measure?
For productivity to be meaningful, it must be compared with something else. The comparisons
can be either intra-company as in the case of year-to-year comparisons of the same measure, or
intercompany as in the case of benchmarking. Intercompany comparisons of single factor
productivity measures can be somewhat tenuous due to differences in accounting practices
(especially when comparing with foreign competitors) and the balance of labor to capital
resources. Total factor productivity measures are somewhat more robust for comparison
purposes.
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Objective Questions
1. Shell Oil Company’s motto “People, Planet and Profit” is a real-world implementation of what
OSCM concept?
Triple bottom line
2. A firm’s strategy should describe how it intends to create and sustain value for
________________________.
its current shareholders
3. What is the term used to describe individuals or organizations that are influenced by the actions
of the firm?
Stakeholders
4. How often should a company develop and refine the operations and supply chain strategy.
At least yearly
5. What is the term used to describe product attributes that attract certain customers and can be
used to form the competitive position of a firm?
Competitive dimensions
6. What are the two main competitive dimensions related to product delivery?
Delivery speed and delivery reliability
7. What are the two characteristics of a product or service that define quality?
Design quality and process quality
8. A diagram that shows how a company’s strategy is delivered by a set of supporting activities is
called a _____________________________.
activity-system map
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9. In implementing supply chain strategy a firm must minimize its total cost without compromising
the needs of its ____________________________.
Customers
10. What is defined as the likelihood of disruption that would impact the ability of a company to
continuously supply products or services?
Supply chain risk
11. Risks caused by natural or manmade disasters, and therefore impossible to reliably predict, are
called ______________________.
Disruption risks
12. Match the following common risks with the appropriate mitigation strategy.
E
Country risks
A:
Detailed tracking, alternate suppliers
D
Regulatory risk
B:
Carefully select and monitor suppliers
A
Logistics failure
C:
Contingency planning, insurance
C
Natural disaster
D:
Good legal advice, compliance
B
Major quality failure
E:
Currency hedging, local sourcing
13. The assessment of the probability of a negative event against the aggregate severity of the
related loss is called _____________________________.
Risk mapping
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14. As Operations Manager, you are concerned about being able to meet sales requirements in the
coming months. You have just been given the following production report.
JAN
FEB
MAR
APR
Units Produced
2175
1675
2675
2875
Hours per Machine
307
186
382
307
4
6
5
6
Number of Machines
Find the average of the monthly productivity figures (units per machine hour).
To answer this we need to realize that the measure of hours given is per machine, so we have to
multiply that by the number of machines in each period to get the total machine hours in each
period. Those figures are used in the calculations below.
Average productivity: (2175/1228 + 1675/1116 + 2675/1910 + 2875/1842)/4
Average productivity (1.77+1.50+1.40+1.56)/4= 1.56 units per machine hour
15. Sailmaster makes high-performance sails for competitive windsurfers. Below is information
about the inputs and outputs for one model, the Windy 2000.
Units sold
Sale price each
Total labor hours
Wage rate
Total materials
Total energy
1,224
$1,707
46,681
$12/hour
$60,500
$4,012
Calculate the productivity in sales revenue/labor expense.
We have to do some interim calculations here. Sales revenue is calculated by multiplying
units sold by the unit sales price. Labor expense is calculated by multiplying labor hours by
the wage rate.
(1224*1707) / (46681*12) = 3.73
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16. Live Trap Corporation received the data below for its rodent cage production unit. Find the total
productivity?
Output
Input
49,000 cages
Production time
644 labor hours
Sales price: $3.90 per unit
Wages
$7.90 per hour
Raw materials (total cost)
$31,500
Component parts (total cost)
$15,250
Total productivity could be expressed two ways here based on how you express output: in
units sold, or dollars of sales.
Units sold:
49,000 / ((644 * $7.90) + 31,500 + 15,250) = 0.95 units sold per dollar input
Dollars of sales:
(49000*3.9) / ((644 * $7.90) + 31,500 + 15,250) = 3.69 dollars in sales per dollar input
17. Two types of cars (Deluxe and Limited) were produced by a car manufacturer last year.
Quantities sold, price per unit, and labor hours follow. What is the labor productivity for each car?
Explain the problem(s) associated with the labor productivity.
Deluxe car
Limited car
Labor, Deluxe
Labor, Limited
QUANTITY
4,000 units sold
6,000 units sold
20,000 hours
30,000 hours
$/UNIT
$8,000/car
$9,500/car
$12/hour
$14/hour
Labor Productivity – units/hour
Model
Deluxe Car
Output
in Units
4,000
Input
in Labor Hours
20,000
Productivity
(Output/Input)
0.20 units/hour
Limited Car
6,000
30,000
0.20 units/hour
Output
in Dollars
4,000($8,000)=
$32,000,000
Input
in Dollars
20,000($12.00)=
$240,000
Productivity
(Output/Input)
133.33
6,000($9,500)=
30,000($14.00)=
135.71
Labor Productivity – dollars
Model
Deluxe Car
Limited Car
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$57,000,000
$420,000
The labor productivity measure is a conventional measure of productivity. However, as a partial
measure, it may not provide all of the necessary information that is needed. For example, increases in
productivity could result from decreases in quality, and/or increases in material cost.
18.
A U.S. manufacturing company operating a subsidiary in an LDC (less-developed country) shows
the following results:
Sales (units)
Labor (hours)
Raw materials (currency)
Capital equipment (hours)
U.S.
90,000
20,105
$19,550
63,000
LDC
20,010
15,120
FC 21,000
5,120
a. Calculate partial labor and capital productivity figures for the parent and subsidiary. Do
the results seem misleading?
Labor Productivity
Country
U.S.
Output
in Units
90,000
Input
in Hours
20,105
Productivity
(Output/Input)
4.48 units/hour
LDC
20,010
15,120
1.32 units/hour
U.S.
Output
in Units
90,000
Input
in Hours
63,000
Productivity
(Output/Input)
1.43 units/hour
LDC
20,010
5,120
3.91 units/hour
Capital Equipment Productivity
Country
Yes. You might expect the capital equipment productivity measure to be higher in the U.S.
than in a LDC. Also, the measures seem contradictory. Each plant appears to be far more
productive than the other on one measure, but much worse on the other.
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Chapter 02 - Strategy
b. Compute the multifactor productivity figures for labor and capital together. Do the
results make more sense?
Multifactor – Labor and Capital Equipment
Country
U.S.
Output
in Units
90,000
Input
in Hours
20,105 + 63,000=
83,105
Productivity
(Output/Input)
1.08 units/hour
LDC
20,010
15,120 + 5,120=
20,240
0.99 units/hour
Yes, labor and equipment can be substituted for each other. Therefore, this multifactor
measure is a better indicator of productivity in this instance.
c. Calculate raw material productivity figures (units/$ where $1 = FC 10). Explain why these
figures might be greater in the subsidiary.
Raw Material Productivity
Country
U.S.
Output
in Units
90,000
Input
in Dollars
$19,550
Productivity
(Output/Input)
4.60 units/$
LDC
20,010
FC 20,010/$10 =
$2,100
9.53 units/$
The raw material productivity measures might be greater in the LDC due to a reduced cost
paid for raw materials, which is typical of LDC’s, especially if there are local sources for the
raw materials.
19.
Various financial data for the past two years follow. Calculate the total productivity measure
and the partial measures for labor, capital, and raw materials for this company for both years.
What do these measures tell you about this company?
Output:
Input:
Sales
Labor
Raw materials
Energy
Capital
Other
Last Year
$201,005
30,010
34,775
4,870
49,700
1,870
This Year
$202,015
40,010
44,785
5,890
49,700
3,005
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Chapter 02 - Strategy
Total Productivity
Year
Output
in Dollars
Input
in Dollars
Productivity
(Output/Input)
Last Year
$201,005
$30,010 + 34,775 +
4,870 + 49,700 + 1,870
= $121,225
1.66
This Year
$202,015
$40,010 + 44,785 +
5,890 + 49,700 +3,005
= $143,390
1.41
Last Year
Output
in Dollars
$201,005
Input
in Dollars
$30,010
Productivity
(Output/Input)
6.70
This Year
$202,015
$40,010
5.05
Partial Measure – Labor
Year
Partial Measure – Raw Materials
Year
Output
in Dollars
Last Year
$201,005
This Year
$202,015
Input
in Dollars
$34,775
Productivity
(Output/Input)
$44,785
4.51
Input
in Dollars
$49,700
Productivity
(Output/Input)
$49,700
4.06
5.78
Partial Measure – Capital
Year
Output
in Dollars
Last Year
$201,005
This Year
$202,015
4.04
The overall productivity measure is declining, which indicates a possible problem. The partial
measures can be used to indicate cause of the declining productivity. In this case, it is a
combination of declines in both labor and raw material productivity, which were somewhat
offset by an increase in the capital productivity. Further investigation should be undertaken
to explain the drops in both labor and raw material productivity. An increase in the cost of
both of these measures, without an accompanying increase in the selling price might explain
these measures.
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Chapter 02 - Strategy
20. An electronics company makes communications devices for military contracts. The company
just completed two contracts. The navy contract was for 2,600 devices and took 25 workers four
weeks (40 hours per week) to complete. The army contract was for 5,567 devices that were
produced by 37 workers in four weeks. On which contract were the workers more productive?
Contract
Output
in Units
Input
in Hours
Productivity
(Output/Input)
Navy
2600
25(4)40 = 4000
0.65
Army
5567
37(4)40 = 5920
0.94
The workers were more productive on the Army contract.
21.
22.
A retail store had sales of $45,100 in April and $55,200 in May. The store employs eight fulltime workers who work a 40-hour week. In April the store also had seven part-time workers at 13
hours per week, and in May the store had eight part-timers at 17 hours per week (assume four
weeks in each month). Using sales dollars as the measure of output, what is the percentage
change in productivity from April to May?
Month
Output
in Dollars
April
$45,100
May
$55,200
Input
in Hours
(8(40)+7(13))*4 =
1644
Productivity
(Output/Input)
1824
30.26
Percentage Change
27.43
(30.26-27.43)/27.43 = 10.32%
A parcel delivery company delivered 103,700 packages last year, when its average
employment was 82 drivers. This year the firm handled 112, 780 deliveries with 93 drivers. What
was the percentage change in productivity over the two years?
Year
Output
in Packages
Input
in Drivers
Productivity
(Output/Input)
Last
103,700
82
1264.63
This
112,780
93
1212.69
Percentage Change
(1212.69 -1264.63)/1264.63 = - 4.11%
2-10
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Chapter 02 - Strategy
23.
A fast-food restaurant serves hamburgers, cheeseburgers, and chicken sandwiches. The
restaurant counts a cheeseburger as equivalent to 1.28 hamburgers and chicken sandwiches as
0.79 hamburger. Current employment is eight full-time employees who work a 40-hour week. If
the restaurant sold 750 hamburgers, 915 cheeseburgers, and 510 chicken sandwiches in one
week, what is its productivity? What would its productivity have been if it had sold the same
number of sandwiches (2,175), but the mix was 725 of each type?
Part
750 Hamburgers
915 Cheeseburgers (1.28)
510 Chicken Sandwiches (.79)
725 Hamburgers
725 Cheeseburgers (1.28)
725 Chicken Sandwiches (.79)
Output in
Hamburger
Equivalents
Input
in Hours
Productivity
(Output/Input)
2324.10
320
7.26
2225.75
320
6.96
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Chapter 3 - Operation Management
Operation Management (Zhejiang University of Finance & Economics)
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Chapter 3 – Design of Products and Services
CHAPTER 3
DESIGN OF PRODUCTS AND SERVICES
Discussion Questions
1. Describe the generic product development process described in this chapter. How does this process change
for technology push products?
Products that are developed using the “technology push” would be more narrowly focused in phase 0 and
phase 1 of Marketing. The focus would be narrower because you would only look at market segments that
could benefit from the application of your technology. The rest of the generic process may be somewhat less
complex as well since the technology of the product currently exists in your manufacturing facilities
2. How does the QFD approach help? What are some limitations of this approach?
QFD helps to get the voice of the customer into the design process using interfunctional teams. The
limitations of QFD relate to the culture of the organization. In the United States, we tend to be vertically
oriented and try to promote breakthrough. This can work against interfunctional teamwork, which is needed
for QFD success. If a breakthrough culture can be maintained with a continuous improvement mentality
through interfunctional teams, this would lead to tremendous improvements in productivity.
3. Discuss the product design philosophy behind industrial design and design for manufacture and assembly.
Which one do you think is more important in a customer-focused product development?
Industrial design is concerned with designing a product from the end-user’s point of view, such as aesthetics
and user-friendliness of the product. Design for manufacturability, on the other hand, makes the product
design less complicated and easier to manufacture. Very often it results in fewer parts, smaller size,
increased reliability, and lower cost.
Both philosophies are equally important for a customer-focused product development. In order to attract
customers, the product must be aesthetically pleasing and user-friendly (industrial design). However, to
sustain customer interests, it should also have a lower cost and higher reliability (design for
manufacturability).
4. Discuss design-based incrementalism, which is frequent product redesign throughout the product’s life.
What are the pros and cons of this idea?
Pro: enhanced function, higher quality, and lower cost through continuously advancing technology.
Con: time and money spent on frequent product and process redesigns, low priority given in servicing the
existing and older products. Consumer reaction to frequent changes may be negative.
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Chapter 3 – Design of Products and Services
5. Do the concepts of complexity and divergence apply to an online sales company such as Dell Computer?
Due to the size of Dell and the number of market segments they serve, these concepts certainly apply.
Consider as one example the technical support process for an existing customer. Service can vary from very
simple like an individual customer needing a particular driver for her new computer to very complex in
troubleshooting a network load problem in a server farm for an Internet service provider. In the first
example there is a straightforward solution to the problem in emailing the customer a copy of the driver. In
the latter, the service process may be quite divergent, with the process being adapted based on the symptoms
of the problem and the skill of the customer’s technical workforce.
For a pure sales company like Amazon, the complexity and divergence would be much less.
6. What factors must be traded off in the product development process before introducing a new product?
The factors that need careful attention for new products are product performance, development speed,
product cost, and development program expense. Smith and Reinertsen identify six pairs of trade-offs in
their book. These include all possible pairs among the four factors noted above.
7. Coca-Cola is a well-established consumer products company with a strong position in the global market.
The sales of their core soda products have remained relatively stable for decades, yet the company has
continued to grow and has remained extremely profitable. Discuss Coca-Cola’s history in light of the
statement that “generating a steady stream of new products to market is extremely important to
competitiveness.” Does Coca-Cola’s success disprove that statement? Is the company an exception to the
rule or an example of its application?
By growing into a global company, one could argue they have been introducing new products into markets
they never served before, even though the products are not new to the company. They have also grown by
acquisition of other drink companies and introduction of new drink products outside of their core soda
business. As the products are not subject to great innovation or technological advances, Coca-Cola has
needed to expand their product lines and global reach to continue to grow.
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Chapter 3 – Design of Products and Services
Objective Questions
1. Which phase of the generic development process involves construction and evaluation of multiple
preproduction versions of the product?
Testing and refinement
2. A process that emphasizes cross-functional integration and concurrent development of a product and its
associated processes is known as _________________________.
Concurrent engineering
3. Match the following product types to the appropriate product development description.
Technology-push
B
products
Entail unusually large uncertainties about the technology or
A
market. The development process takes steps to address those
:
uncertainties.
E Platform products
B A firm with a new proprietary technology seeks out a market
: where that technology can be applied.
D
Process-intensive
products
C Uses a repeated prototyping cycle. Results from one cycle are
: used to modify priorities in the ensuing cycle.
A High-risk products
The production process has an impact on the product properties.
D
Therefore, product design and process design cannot be
:
separated.
C Quick-build products
E:
Products are designed and built around a preexisting
technological subsystem.
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Chapter 3 – Design of Products and Services
4. Designing products for aesthetics and with the user in mind is generally called what?
Industrial design
5. The first step in developing a house of quality is to develop a list of ______________________.
customer requirements for the product
6. The purpose of value analysis/value engineering is to _________________________________.
simplify products and processes
7. What is it about service processes that makes their design and operation so different from manufacturing
processes?
Direct customer involvement in the process
8. What are the three general factors that determine the fit of a new or revised service process?
Service experience fit, operational fit, and financial impact.
9. Measures of product development success can be organized in what three categories?
Time to market, productivity, quality.
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Chapter 3 – Design of Products and Services
10. Tuff Wheels Kiddy Dozer
a. Base case
Project
Schedule
Year 1
Year 2
Year 3
Year 4
Kiddy Dozer
Q1 Q2 Q3 Q4 Q1 Q2 Q3 Q4 Q1 Q2 Q3 Q4 Q1 Q2 Q3 Q4
Development
333
33 33
3 3
Pilot Testing
10 10
0 0
Ramp-up
20 20
0 0
Marketing and
Support
38
Production
Volume
38
38
38
38
38
38
38
38
38
38
38
38
15
15
15
15
15
15
15
15
15
15
15
15
Unit
Production
Cost
100 100 100 100 100 100 100 100 100 100 100 100
Production
Costs
150 150 150 150 150 150 150 150 150 150 150 150
0
0
0
0
0
0
0
0
0
0
0
0
Sales Volume
15
Unit Price
PV Year 1 r =
15
15
15
15
15
15
15
15
15
15
170 170 170 170 170 170 170 170 170 170 170 170
255 255 255 255 255 255 255 255 255 255 255 255
0
0
0
0
0
0
0
0
0
0
0
0
Sales Revenue
Period Cash
Flow
15
- - 101 101 101 101 101 101 101 101 101 101 101 101
33 63 33
3
3
3
3
3
3
3
3
3
3
3
3
333
3 3 8
-
-
-
- 935 917 899 881 864 847 831 814 798 783 767 752
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Chapter 3 – Design of Products and Services
8
333
Project NPV
850
3
32 60 31
7 9 8
b. The results are shown below for both scenarios. If sales are only 50,000 then the project is still worthwhile
since the NPV decrease to $6,759,000. If Tuff Wheels has under estimated the sales and it ends up being
70,000 per year then NPV will increase from $8,503,000 base case to $10,247,000 with the higher sales rate.
Sales Revised to 50,000 per Year
Project
Schedule
Year 1
Year 2
Year 3
Year 4
Kiddy Dozer
Q1 Q2 Q3 Q4 Q1 Q2 Q3 Q4 Q1 Q2 Q3 Q4 Q1 Q2 Q3 Q4
Development
333 333 333
Pilot Testing
100 100
Ramp-up
200 200
Marketing and
Support
Production
Volume
38
38
38
38
38
38
38
38
38
38
38
38
38
13
13
13
13
13
13
13
13
13
13
13
13
Unit
Production
Cost
100 100 100 100 100 100 100 100 100 100 100 100
Production
Costs
125 125 125 125 125 125 125 125 125 125 125 125
0
0
0
0
0
0
0
0
0
0
0
0
Sales Volume
Unit Price
Sales Revenue
13
13
13
13
13
13
13
13
13
13
13
13
170 170 170 170 170 170 170 170 170 170 170 170
212 212 212 212 212 212 212 212 212 212 212 212
5
5
5
5
5
5
5
5
5
5
5
5
3-6
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Chapter 3 – Design of Products and Services
Period Cash
Flow
- - - 838 838 838 838 838 838 838 838 838 838 838 838
333 333 633 338
PV Year 1 r =
8
- - - 774 759 744 729 715 701 687 674 660 647 635 622
333 327 609 318
Project NPV
675
9
Sales Revised to 70,000 per Year
Project
Schedule
Year 1
Year 2
Year 3
Year 4
Kiddy Dozer
Q1 Q2 Q3 Q4 Q1 Q2 Q3 Q4 Q1 Q2 Q3 Q4 Q1 Q2 Q3 Q4
Development
333
33 33
3 3
Pilot Testing
10
100
0
Ramp-up
20
200
0
Marketing and
Support
Production
Volume
38
38
38
38
38
38
38
38
38
38
38
38
38
18
18
18
18
18
18
18
18
18
18
18
18
Unit
Production
Cost
100 100 100 100 100 100 100 100 100 100 100 100
Production
Costs
175 175 175 175 175 175 175 175 175 175 175 175
0
0
0
0
0
0
0
0
0
0
0
0
Sales Volume
18
18
18
18
18
18
18
18
18
18
18
18
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Chapter 3 – Design of Products and Services
Unit Price
170 170 170 170 170 170 170 170 170 170 170 170
297 297 297 297 297 297 297 297 297 297 297 297
5
5
5
5
5
5
5
5
5
5
5
5
Sales Revenue
Period Cash
Flow
- - 118 118 118 118 118 118 118 118 118 118 118 118
-333 33 63
338
8
8
8
8
8
8
8
8
8
8
8
8
3 3
PV Year 1 r =
8
- - 109 107 105 103 101
-333 32 60
994 974 955 936 918 900 882
318
7
6
4
4
4
7 9
Project NPV
1024
7
c. The impact of changing the interest rate is shown below. There is still a positive NPV but it shrinks the
interest rate increases. This would be expected since a higher the interest rate reduces the present value of future
cash flows.
Base
Case
8% $8,503,043
9% $8,283,241
10% $8,069,666
11% $7,862,116
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Chapter 3 – Design of Products and Services
11. Perot Corporation Patay2 Chip.
a. In the base case the Patay2 Chip Project has a very good NPV of $10,460,000, see below.
Project Schedule
Year 1
Year 2
Patay2 Chip
1st
half
2nd
half
1st
half
2nd
half
Development Cost
5,000
5,000 5,000
5,000
Pilot Testing Cost
2,500
2,500
Debug Cost
1,500
1,500
Ramp-up Cost
3,000
Advance Marketing Cost
5,000
Ongoing Marketing and
Support
Year 3
1st half
Year 4
2nd
half
1st
half
2nd
half
500
500
500
500
Production Volume
125
125
75
75
Unit Production Cost
655
655
545
545
Production Costs
81,875 81,875 40,875 40,875
Sales Volume
125
125
75
75
Unit Price
820
820
650
650
102,50 102,50
48,750 48,750
0
0
Sales Revenue
Period Cash Flow
-5,000 -5,000 -9,000 -17,000 20,125 20,125 7,375
7,375
PV Year 1 r = 12
-5,000 -4,762 -8,163 -14,685 16,557 15,768 5,503
5,241
Project NPV
10,460
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Chapter 3 – Design of Products and Services
b. Additional 10 million for higher price is clearly worthwhile as it raises the NPV from $10.46 million to
$16.654 million. See results below.
Project Schedule
Year 1
Year 2
Patay2 Chip
1st
half
2nd
half
1st
half
2nd
half
Development Cost
7,500
7,500 7,500
7,500
Pilot Testing Cost
2,500
2,500
Debug Cost
1,500
1,500
Ramp-up Cost
3,000
Advance Marketing Cost
5,000
Ongoing Marketing and
Support
Year 3
1st half
Year 4
2nd
half
1st
half
2nd
half
500
500
500
500
Production Volume
125
125
75
75
Unit Production Cost
655
655
545
545
Production Costs
81,875 81,875 40,875 40,875
Sales Volume
125
125
75
75
Unit Price
870
870
700
700
108,75 108,75
52,500 52,500
0
0
Sales Revenue
Period Cash Flow
-7,500 -7,500
-19,500 26,375 26,375 11,125 11,125
11,500
PV Year 1 r = 12
-7,500 -7,143
-16,845 21,699 20,666 8,302
10,431
Project NPV
16,654
7,906
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Chapter 3 – Design of Products and Services
c. Reduced sales estimates have a significant impact on the NPV. It reduces the NPV all the way down to
$10,000. The success of the Patay2 Chip is very dependent on the sales estimates. It would be wise for
Perot to make sure that there is sufficient demand for Patay2 Chips.
Project Schedule
Year 1
Year 2
Patay2 Chip
1st
half
2nd
half
1st
half
2nd
half
Development Cost
5,000
5,000 5,000
5,000
Pilot Testing Cost
2,500
2,500
Debug Cost
1,500
1,500
Ramp-up Cost
3,000
Advance Marketing Cost
5,000
Ongoing Marketing and
Support
Year 3
1st
half
Year 4
2nd
half
1st
half
2nd
half
500
500
500
500
Production Volume
100
100
50
50
Unit Production Cost
655
655
545
545
Production Costs
65,500 65,500 27,250 27,250
Sales Volume
100
100
50
50
Unit Price
820
820
650
650
Sales Revenue
82,000 82,000 32,500 32,500
Period Cash Flow
-5,000 -5,000 -9,000 -17,000 16,000 16,000 4,750
4,750
PV Year 1 r = 12
-5,000 -4,762 -8,163 -14,685 13,163 12,536 3,545
3,376
Project NPV
10
12. Answers will vary based upon the product selected and the student. Issues that should be considered in the
design and manufacture of a product include design process (traditional vs. concurrent engineering),
customer needs and expectations, legal considerations (EPA, OSHA, etc.), service life, reliability,
appearance, standardization, any industry standards that should be considered (e.g., television set and the
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Chapter 3 – Design of Products and Services
type of signal received from stations), method of shipment, material cost and availability, stage of the
product life cycle, design for manufacturability, design for assembly, packaging, environmental, unit cost,
pricing, availability of purchased material, availability of capacity, availability of subcontractors, setup cost,
manufacturing time, volume, and expected product life.
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Chapter 3 – Design of Products and Services
13. Answers may vary. Following is just one possible result.
Tournament
Activities (per
text)
Grounds
maintenance
Helpful service
attendants
Fair handicapping
system
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Chapter 3 – Design of Products and Services
CASE: IKEA: Design and Pricing
Questions:
1. What are IKEA”s competitive priorities?
Probably the key priority is maximizing value to the customer in each product offered through their stores. Low
cost is certainly a major priority as well.
2. Describe IKEA’s process for developing a new product.
This is described in the case: (1) Pick a price, (2) Choose a manufacturer, (3) Design the product, (5) Ship it.
The key here is to recognize that they pick a price point early in the product development process and then work
with a manufacturer to ensure they can meet this price point. This is a very innovative approach to product
design.
3. What are additional features of the IKEA concept (beyond their design process) that contribute to
creating exceptional value for the customer?
Customer can easily view the product in a setting similar to theirs (i.e. apartments) in the IKEA store. Also, the
product can be easily brought home by the customer eliminating delivery and setup charges. This creates more
value in the product by eliminating these costs.
4. What would be important criteria for selecting a site for an IKEA store?
Need to be located in a high density area where many people live in smaller apartments where space is a
premium. The IKEA products really appeal to young, affluent customers.
CASE: Dental Spa
3-14
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Chapter 3 – Design of Products and Services
Questions:
1. Which one of the three new service requirements would a dental spa least likely pass: service experience fit,
operational fit, or financial impact? Why?
The dental spa is least likely to have operational fit because new personnel must be hired and the layout of
current operations must be changed in order to offer this new service. Although service experience fit might
seem logical, distracting the patient from possible pain would improve the service experience of the core
service, dentistry. Often the services are given away. Dental Associates must “buy in” to the new services and
see the obvious patient comfort benefits to the dental business and experience.
2. What are some of the main areas of complexity and divergence in this kind of operation relative to the
standard dental clinic?
Complexity: By providing two services at once, the complexity is automatically increased. For example, if
cleaning teeth requires two steps and a hand massage requires one, then together the new service at the dental
spa would take three steps. Besides just being a combination of two services normally provided separately,
timing or coordination issues could add extra steps to this new process. However some steps, such as billing,
would be combined in comparison to two services and it might be simple addition (often these services are free
and do not affect billing).
Divergence: Any customer service that can possibly inflict pain, such as a visit to a dentist, requires a large
degree of judgment on the part of the service provider of how best to deal with a painful situation. For example,
some patients need empathy. Almost completely opposite, a visit to a spa can be like going to a psychiatrist as a
customer tells their problems as they are getting their manicure. This allows for divergence as a customer
service provider must decide how to react to hearing all of the issues in someone’s life. However, being at the
dentist would limit this verbal interaction much like a dentist who only asks questions when he has instruments
in the patients mouth preventing a real answer.
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answer
Financial Management (Lingnan University)
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Chapter 04 – Project Management
CHAPTER 4
PROJECT MANAGEMENT
Discussion Questions
1. What was the most complex project that you have been involved in? Give examples of the following
as they pertain to the project: the work breakdown structure, tasks, subtasks, and work package.
Were you on the critical path? Did it have a good project manager?
Obviously, the answer will vary. Remember that the project could be in a non-profit environment as well.
School plays (especially musicals) are a good example, because there are many major tasks that need to
be broken down and scheduled in parallel, but all must be completed by the time opening night comes.
This would include selecting the play and obtaining the rights, auditions, rehearsals of the actors,
rehearsals of the musicians, construction of the sets, setting the lighting, printing tickets and programs,
staffing the theater, advertising and fund raising.
2. What are some reasons project scheduling is not done well?
Several problems with project scheduling are discussed at the end of the chapter. The uncertainties
inherent in the activities comprising the network of any project make it necessary to update the schedule
on a regular basis. Maintaining accurate time and cost estimates is often difficult and frustrating.
Managing this evolving process requires a discipline that is not always available.
3. Which characteristics must a project have for critical path scheduling to be applicable? What types
of projects have been subjected to critical path analysis?
Project characteristics necessary for critical path scheduling to be applicable are:
a. Defined project beginning and ending
b. Well-defined jobs whose completion marks the end of the project.
c. The jobs of tasks are independent in that they may be started, stopped, and conducted
separately within a given sequence.
d. The jobs or tasks are ordered in that they must follow each other in a given sequence.
e. An activity once started is allowed to continue without interruption until it is completed.
A wide variety of projects have used critical path analysis. Some industries that more commonly use this
approach include aerospace, construction, and computer software.
4. What are the underlying assumptions of minimum-cost scheduling? Are they equally realistic?
The underlying assumptions of minimum cost scheduling are that it costs money to expedite a project
activity and it costs money to sustain or lengthen the completion time of the project.
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Chapter 04 – Project Management
While both assumptions are generally realistic, it often happens that there are little or no out-of-pocket
costs associated with sustaining a project. Personnel are often shifted between projects, and in the short
run there may be no incentive to compete a project in “normal time.”
5. “Project control should always focus on the critical path.” Comment.
In many project situations, it is not the activities on the critical path which cause problems, but rather
noncritical activities, which, for various reasons, become critical. In the context of PERT, it may turn out
that the activities on the critical path have small variances associated with them and can be treated as
near certain. At the same time, activities off the critical path may have extremely large variances and, in
fact, if not closely monitored, may delay the project. Thus, while project control must keep track of
critical path activities, it may be more useful to focus on those activities which are not on the critical path
but, for one reason or another, have a high degree of uncertainty associated with them.
Along these lines, some authors have suggested that the critical path approach should be replaced by a
critical activity approach in which simulation is used to estimate which activities are likely to become
sources of project delay. These activities, rather than critical path, would become the focus of
managerial control.
Additionally, the critical path focuses on the time or schedule aspects of the projects. Certain activities
could be "critical" because of cost or quality considerations.
6. Why would subcontractors for a government project want their activities on the critical path?
Under what conditions would they try to avoid being on the critical path?
A subcontractor might want his activities on the critical path in situations where cost incentives are
provided for early project completion. Since the critical path ultimately determines project length, it
stands to reason that activities on the path would be the ones that would draw additional funds to
expedite completion. A subcontractor might want his activities off the critical path because of some
error on his part or because he doesn’t want to be bothered by the close monitoring of progress which
often goes with critical path activities.
7. Discuss the graphic presentations in Exhibit 4.11. Are there any other graphic outputs you would
like to see if you were project manager?
The various graphs and charts presented are typical of the graphical techniques for presenting the
necessary data. Most are adaptable to computer programming. The major requirements in the graphics
package include planned activities related to time, a milestone chart to show major achievements, a
breakdown to show how funds were spent plus a plot of actual completion versus planned.
8. Why is it important to use expected value management (EVM) in the overall management of
projects? Compare to the use of baseline and current schedules only.
Using schedules only captures the time aspect of project management. Costs and revenues are also
critical to efficient project management and the overall success of any project. A project may be ahead
of schedule but at an unacceptable cost.
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Chapter 04 – Project Management
9. Consider the EVM charts in Exhibit 4.12. Are there any other measures you might want to use in the
management of a project? What are some controllable variables that may affect the costs being
tracked?
Students will likely have multiple answers to these questions. One might be quality defects and their
costs in rework or scrap. Utilization figures for workers and expensive resources might also be useful.
10. What do you think might be some barriers to the successful, effective use of the project
management software packages discussed in the chapter?
Students will have varying answers here, but we would expect them to include training and hiring
personnel with the right technical aptitude as well as the cost of buying and upgrading such systems.
Objective Questions
1. What are the three types of projects based on the amount of change involved?
Derivative, breakthrough, platform.
2. What are the four major categories of projects based on the type of change involved?
Product change, process change, research & development, and alliance & partnership.
3. Match the following characteristics with their relevant project team organizational structures.
B
The project is housed within a functional division of the firm.
A: Pure project
C
A project manager leads personnel from different functional areas.
B: Functional project
A
Personnel work on a dedicated project team.
C: Matrix project
C
A team member reports to two bosses.
A
Team pride, motivation, and commitment are high.
B
Team members can work on several projects.
C
Duplication of resources is minimized.
4-3
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Chapter 04 – Project Management
4. What is the term for a group of project activities that are assigned to a single organizational unit?
Work package
5.
The following activities are part of a project to be scheduled using CPM:
a. Draw the network.
6
13
13
C (7)
6
0
15
15
D (2)
13
13
18
F (3)
15
16
19
19
6
G (7)
A (6)
0
26
19
6
6
9
15
B (3)
12
26
19
E (4)
15
15
19
b. What is the critical path?
A-C-D-E-G, also shown in the network above as the bold path.
c. How many weeks will it take to complete the project?
6+7+2+4+7 = 26 weeks
d. How much slack does activity B have?
Activity B has 6 weeks of slack – the difference between its early and late start times.
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Chapter 04 – Project Management
6.
Schedule the following activities using CPM:
a. Draw the network path.
1
5
5
B (4)
1
7
7
9
D (2)
5
5
9
F (2)
7
8
11
G (2)
10
10
12
12
0
15
1
H (3)
A (1)
12
0
15
1
1
4
7
C (3)
4
12
E (5)
7
7
12
b. What is the critical path?
A-B-D-E-H, also shown in the network above as the bold path.
c. How many weeks will it take to complete the project?
15 weeks, 1+4+2+5+3
d.
Which activities have slack, and how much?
C, 3 weeks; F, 1 week; and G, 1 week.
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Chapter 04 – Project Management
7.
The R&D department is planning to bid on a large project for the development of a new
communications system for commercial planes. The accompanying table shows the activities, times,
and sequences required:
a. Draw the network diagram.
3
5
5
B (2)
0
3
0
E (6)
7
9
3
7
A (3)
9
15
7
C (4)
3
13
13
F (6)
3
7
3
7
7
15
15
G (2)
13
7
D (4)
3
11
13
18
I (3)
15
15
18
10
H (3)
7
12
15
Note that G has both D and F as immediate predecessors. However, D is redundant because F also has D
as an immediate predecessor.
b. What is the critical path?
A-C-F-G-I, and A-D-F-G-I at 18 weeks.
c. Suppose you want to shorten the completion time as much as possible, and you have the option
of shortening any or all of B, C, D, and G each one week. Which would you shorten?
B is not on a critical path and has slack of 4; therefore, do not shorten as it will not change the
project completion time. Shorten C, D, and G one week each. C and D are on parallel critical
paths, reducing them both will only reduce project completion time by 1 week.
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Chapter 04 – Project Management
d. What is the new critical path and earliest completion time?
A-C-F-G-I; and A-D-F-G-I remain the critical paths. Project completion time is reduced from 18 to
16 weeks.
8.
The following represents a project that should be scheduled using CPM:
a. Draw the network.
TIMES (DAYS)
ACTIVITY
A
B
C
D
E
F
G
H
IMMEDIATE
PREDECESSORS
—
—
A
A
B
C, D
D, E
F, G
1
0
0
b
5
3
3
4
11
5
6
5
ET
3
2
2
3
5
4
3.833
3.833
5
7
F (4)
5
1
12
5
4
H (3.83)
9
4
4
7
12
12
15
G (3.83)
E (5)
7
11
7
B (2)
4
15
7
D (3)
1
0.4444
0.1111
0.1111
0.1111
1.7778
0.1111
0.6944
0.2500
5
1
1
m
3
2
2
3
4
4
4
4
C (2)
1
A (3)
a
1
1
1
2
3
3
1
2
7
10
12
12
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Chapter 04 – Project Management
b. What is the critical path?
B-E-G-H
c. What is the expected project completion time?
2.00+5.00+3.833+3.833 = 14.67
d. What is the probability of completing this project within 16 days?
Variance of project completion time is found by adding the variances of activities on the critical
path. .1111 + 1.7778 + .6944 + .2500 = 2.833
Z
(16  14.67)
= .79
2.833
P(T<16) = P(Z<.79) = .7852 (From Excel’s NORMSDIST() function)
9.
There is an 82 percent chance the project below can be completed in X weeks or less. What is X?
Most
likely
Most
pessimistic
Expected
Time
Variance
Activity
Most
optimistic
A
B
C
D
E
2
3
1
6
4
5
3
3
8
7
11
3
5
10
10
5.5
3
3
8
7
2.25
0
.444
.444
1
First find the value of Z that results in a probability of .82. Using Excel’s NORMSINV(.82) = .915. Then
find the critical path (ABD) and the variance on the critical path: 2.25+ 0 + .444 = 2.694. Finally, use
equation 5.3 to solve for D.
Paths
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Chapter 04 – Project Management
ABD 16.5
.915 
D  16.5
2.694
D = 18
ACE 15.5
10.
The following table represents a plan for a project:
a. Construct the appropriate network diagram.
Job No.
1
2
3
4
5
6
7
8
9
10
11
a
3
2
2
1
1
2
3
1
2
3
1
m
5
2
5
4
3
4
6
4
6
4
7
b
10
14
8
10
5
6
12
7
13
5
13
ET
5.50
4.00
5.00
4.50
3.00
4.00
6.50
4.00
6.50
4.00
7.00
2
1.361
4.000
1.000
2.250
0.444
0.444
2.250
1.000
3.361
0.111
4.000
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5.5
9.5
9.5
2 (4 )
5 (3)
11.5
7.5
12.5
11.5
14.5
18.5
14.5
8 (4)
14.5
0
5.5
5.5
1 (5.5 )
0
5.5
10.5
10.5
3 (5)
9 (6 .5)
18.5
6 (4)
10.5
10.5
14.5
5.5
10
10
16.5
4 (4.5)
14.5
25
25
16.5
32
11 (7 )
25
7 (6..5)
14.5
25
18.5
14.5
5.5
10
18.5
32
20.5
10 (4)
21
21
25
b. Indicate the critical path.
1-3-6-8-9-11
c. What is the expected completion time for the project?
5.50+5.00+4.00+4.00+6.50+7.00 = 32
d. You can accomplish any one of the following at an additional cost of $1,500:
(1)Reduce job 7 by two days.
(2) Reduce job 3 by three days.
(3) Reduce job 8 by four days.
If you save $1,000 for each day that the earliest completion time is reduced, which action, if any,
would you choose?
(1) Job 7 is not on the critical path; therefore, reducing its time by two days will not reduce project
completion time.
(2) If you reduce job 3 by three days, then path 1-2-5-8-9-11 becomes critical and the project
length is 30 days. You have saved $2,000 but paid $1,500.
(3) Task 8 is on the critical path, so if you reduce job 8 by four days, then the project length
becomes 28 days. You have saved $4,000 but paid $1,500.
e. What is the possibility that the project will take more than 35 days to complete?
First, you need to compute the variance of the C.P.: 1.361 + 1.000 + 0.444 + 1.000 + 3.361 + 4.000 =
11.167. Then use equation 5.3 to find the correct Z and look its value up in Appendix E.
Z
35  32
 0.90
11.167
P  .82
4-10
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11. A construction project is broken down into the following 10 activities:
a. Draw the network diagram.
3
4
5
2 (1)
0
3
0
9
3
5
9
5
8
14
19
12
6 (3)
7
9
3
8
19
19
9 (7)
9
12
8
4 (5)
3
15
8 ( 5)
14
3 (2)
3
10
5 (5)
8
1 (3)
10
12
25
10 (6)
19
19
25
12
7 (4)
8
8
12
b. Find the critical path.
1-4-7-9-10, length = 25 weeks
c. If activities 1 and 10 cannot be shortened, but activities 2 and 9 can be shortened to a
minimum of one week each at a cost of $10,000 per week, which activities would you
shorten to cut the project by four weeks?
The one option would be to cut activity 7 by 3 weeks, and then reduce activity 4 by one week.
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Another option would be to reduce activities 4 and 9 by a total of 3 weeks, and then reduce
activity 7 by one week.
12. Here is a CPM network with activity times in weeks:
a. Determine the critical path.
7
12
14
B (5)
9
0
C (6)
14
7
14
11
A (7)
0
20
20
14
20
G (3)
11
7
7
D (6)
14
11
20
11
E(4)
7
26
26
19
F (8)
11
12
20
A-E-G-C-D
b. How many weeks will the project take to complete?
26 weeks
c. Suppose F could be shortened by two weeks and B by one week. How would this affect the
completion date?
No difference in completion date. Neither B nor F is on the critical path.
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13. Here is a network with the activity times shown in days:
a. Find the critical path.
6
18
14
23
F (5)
B (8)
6
0
14
6
20
14
A (6)
0
25
18
23
D (4)
6
14
6
11
G (3)
18
25
18
C (5)
9
28
28
25
E (7)
14
18
25
A-B-D-E-G
b. The following table shows the normal times and the crash times, along with the associated costs
for each activity.
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If the project is to be shortened by four days, show which activities, in order of reduction would be
shortened and the resulting cost.
Solution:
Activity
A
B
C
D
E
F
G
Normal
Time (NT)
6
8
5
4
7
5
3
Crash Time
(CT)
5
7
3
3
6
4
2
Normal Cost
(NC)
$7,300
5,300
9,000
3,300
2,200
4,100
5,200
Crash Cost
(CC)
$8,300
6,700
10,600
4,100
3,600
7,300
8,300
NT-CT
1
1
2
1
1
1
1
Cost/day to
expedite
$1,000
1,400
800
800
1,400
3,200
3,100
First, the lowest cost activities to crash are D at $800 per day. Critical path remains the same. Second,
lowest cost activity on the critical path is A. Crash activity A. Third, B and E are next lowest cost activities
on the critical paths. Both have a cost of $1,400 per day. Select B then E or reverse the order (E then B).
Summary of steps to reduce project by four days:
Step
Activity to crash
Cost to crash
Days saved
1
D
$800
1
2
A
1,000
1
3
B (or E)
1,400
1
4
E (or B)
1,400
1
Total
$4,600
14. The home office billing department of a chain department stores prepares monthly inventory
reports for use by the stores’ purchasing agents. Given the following information, use the critical
path method to determine:
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a. How long the total process will take.
0
10
20
b (10 )
10
0
50
d ( 30 )
20
50
80
80
0
a (0 )
0
100
100
g ( 20 )
0
80
0
20
20
20
20
c ( 20)
0
40
40
40
40
e (20 )
100
h (0)
100
100
100
80
f ( 40)
80
100 hours
b. Which jobs can be delayed without delaying the early start of any subsequent activity?
Activities b and d are not on the critical path. Their start can be delayed without delaying the
start of any subsequent activities. Activity b can be delayed by 10 hours and d can be delayed by
30 hours without affecting the project completion date.
15. For the network shown:
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Chapter 04 – Project Management
a. Determine the critical path and the early completion time in weeks for the project.
18
21
D (3 )
6
18
18
21
B (12 )
6
0
18
6
14
A (6 )
0
17
21
E (3 )
6
6
25
G (4)
18
21
14
17
21
25
14
C (8)
10
18
F (3)
18
21
A-B-D-G, 6+12+3+4 = 25 weeks
b. For the data shown, reduce the project completion time by three weeks. Assume a linear cost
per week shortened, and show, step by step, how you arrived at your schedule.
4-16
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Solution:
Activity
A
B
C
D
E
F
G
Normal
Time (NT)
6
12
8
3
3
3
4
Normal Cost
(NC)
$7,400
12,200
5,800
4,600
3,200
7,000
8,000
Crash Time
(CT)
5
7
6
1
2
2
3
Crash Cost
(CC)
$13,500
18,300
7,000
6,000
6,700
7,500
8,400
NT-CT
1
5
2
2
1
1
1
Cost/week
to expedite
$6,100
1,220
600
700
3,500
500
400
First, reduce G (lowest cost activity on the critical path) by one week. Second, crash activity D by one
week. Critical paths remain the same. Third, crash activity D by one more week at a cost of $700, which
is the least expensive.
Summary of activities crashed:
Step
Activity
Cost to crash
Weeks reduced
1
G
$400
1
2
D
700
1
3
D
700
1
Total cost
$1,800
16. The following CPM network has estimates of the normal time in weeks listed for the activities:
4-17
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Chapter 04 – Project Management
a. Identify the critical path.
5
15
8
21
F (6)
B (3)
8
0
11
15
5
11
A (5)
0
21
15
21
D (4)
5
11
5
G (5)
15
21
11
15
C (6)
5
26
26
18
E (3)
11
18
21
A-C-D-F-G
b. What is the length of time to complete the project?
5+6+4+6+5 = 26 weeks
c. Which activities have slack, and how much?
B, 3 weeks; E, 3 weeks.
d. Here is a table of normal and crash times and costs. Which activities would you shorten to cut
two weeks from the schedule in a rational fashion? What would be the incremental cost? Is the
critical path changed?
Solution:
Activity
Normal
Time (NT)
Crash Time
(CT)
Normal Cost
(NC)
Crash Cost
(CC)
NT-CT
Cost/week
to expedite
A
5
4
$7,600
$8,700
1
1,100
B
3
2
5,700
8,400
1
2,700
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Chapter 04 – Project Management
C
6
5
7,800
9,400
1
1,600
D
4
3
2,650
3,800
1
1,150
E
3
2
2,500
4,400
1
1,900
F
6
4
5,800
8,600
2
1,400
G
5
4
3,800
5,500
1
1,700
First, shorten activity A by one week at a cost of $1,100. This is the lowest cost/week activity on the
critical path. Second, shorten activity D by one week at a cost of $1,150. This is the next lowest
cost/week activity on the critical path. The total cost is $2,250 and the critical path remains
unchanged.
17.
Bragg’s Bakery is building a new automated bakery in downtown Sandusky. Here are the activities
that need to be completed to get the new bakery built and the equipment installed.
a. Draw the project diagram.
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Chapter 04 – Project Management
B ( 3)
D (10)
A (4)
F (2)
C (5)
E (4 )
b. and c. What is the normal project length? What is the project length if all activities are crashed to
their minimum?
Solution:
Path
ABDF
ACDF
ACEF
Normal Length
19
21
15
Crashed Length
12
13
09
d. Bragg’s loses $4,000 in profit per week for every week the bakery is not completed. How many weeks
will the project take if we are willing to pay crashing cost as long as it is less than $4,000?
We would only crash the project until 14 weeks since the cost of crashing F is $5600 which is greater than
the $4000 in additional profit.
LENGTH
ABDF - 19
18
17
16
15
15
15
14
13
ACDF - 21
20
19
18
17
16
15
14
13
ACEF - 15
15
15
15
15
14
13
12
11
D
D
D
D
C
C
A
F
Crash cost
2000
2000
2000
2000
3800
3800
3900
5600
Cumulative Cost
2000
4000
6000
8000
11800
15600
19500
25100
Activity Crashed
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Chapter 04 – Project Management
18. Assume the network and data that follow:
a. Construct the network diagram.
Activity
Normal
Time (NT)
Crash Time
(CT)
Normal Cost
(NC)
Crash Cost
(CC)
NT-CT
Cost/week
to expedite
A
4
3
$140
$180
1
40
B
5
3
70
100
2
15
C
4
1
90
120
3
10
D
3
2
90
150
1
60
E
5
3
40
50
2
5
F
8
5
100
190
3
30
G
9
8
60
120
1
60
4
9
9
E (5)
B (5)
5
0
10
4
10
4
A (4)
0
14
15
15
8
C (4)
4
11
4
7
G (9)
15
15
7
D (3)
4
24
24
15
F (8)
7
7
15
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Chapter 04 – Project Management
b. Indicate the critical path when normal activity times are used.
The critical path is A-D-F-G, with a length of 4+3+8+9 = 24
c. Compute the minimum total direct cost for each project duration based on the cost associated
with each activity. Consider durations of 19, 20, 21, 22, 23, and 24 weeks.
The normal time project cost is $590 at 24 weeks. Minimum cost crashing to 19 weeks is shown
below.
LENGTH
ABEG - 23
23
22
21
20
19
ACG - 17
17
17
17
16
15
ADFG - 24
23
22
21
20
19
Activity Crashed
F
E+F
E+F
A
G
Crash cost
30
35
35
40
60
Cum. Crash Cost
30
65
100
140
200
Total Direct Cost
620
655
690
730
790
d.
If the indirect costs for each project duration are $400 (24 weeks), $350 (23 weeks), $300 (22
weeks), $250 (21 weeks), $200 (20 weeks), and $150 (19 weeks), what is the total project cost
for each duration? Indicate the minimum total project cost duration.
Computations build off the earlier table and are shown below. The total project cost in normal
time is $590 + $400 = $990.
LENGTH
ABEG - 23
23
22
21
20
19
ACG - 17
17
17
17
16
15
ADFG - 24
23
22
21
20
19
Activity Crashed
F
E+F
E+F
A
G
Crash cost
30
35
35
40
60
Cum. Crash Cost
30
65
100
140
200
Total Direct Cost
620
655
690
730
790
Indirect Cost
350
300
250
200
150
Total Project Cost
970
955
940
930
940
We can achieve minimum cost by reducing the project to 20 weeks. The duration has the total cost of
$930.
4-22
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Chapter 04 – Project Management
19.
Your project to obtain charitable donations is now 50 days into a planned 60-day project.
The project is divided into 3 activities. The first activity is designed to solicit individual
donations. It is scheduled to run the first 44 days of the project and to bring in $26,300. Even
though we are 50 days into the project, we still see that we have only 88 percent of this activity
complete. The second activity relates to company donations and is scheduled to run for 50 days
starting on day 6 and extending through day 56. We estimate that even though we should have
(44/50) 88 percent of this activity complete, it is actually only 55 percent complete. This part of
the project was scheduled to bring in $151,300 in donations. The final activity is for matching
funds. This activity is scheduled to run the last 10 days of the project and has not started. It is
scheduled to bring in an additional $52,600. So far $175,600 has actually been brought in on
the project.
Calculate the schedule variance, schedule performance index, and cost (actually value in
this case) performance index. How is the project going? Hint: Note that this problem is
different since revenue rather than cost is the relevant measure. Use care in how the measures
are interpreted.
Solution:
Activity
1. Solicit
2. Donations
3. Matching
Funds
Total
Expected
Revenue
$26,300
$151,300
Planned
Duration
44
50
Planned
Start
Date
0
6
$52,600
10
50
Planned
Comp.
Date
44
56
Expected
%
Complete
100%
88%
Actual %
Complete
88%
55%
60
0%
0%
$230,200
Actual
Rev. to
Date
$175,600
BCWS
Activity 1
Activity 2
Activity 3
100% of $26,300
88% of $151,300
0 % of 52,600
=
=
=
26,300
133,144
0
$159,444
BCWP
Activity 1
Activity 2
Activity 3
88% of 26,300
55% of 151,300
0% of $52,600
= 23,144
= 83,215
=0
106,359
4-23
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Chapter 04 – Project Management
Scheduled Variance = 106,359 - 159,444 = -53,085
Scheduled Performance Index = 106,359/159,444 = .667
Cost Variance = 106,359 - 175,600 = -69,241
Cost Performance = 106,359/175,000 = .606
Because we are working with revenues instead of costs, we have to invert the evaluation rules
listed in the text. Our performance measures here are actually good. Although we are behind
schedule on completing tasks 1 and 2, we have brought in more money than expected for the
amount of work we have completed.
20.
A project to build a new bridge seems to be going very well since the project is well
ahead of schedule and costs seem to be running very low. A major milestone has been reached
where the first two activities have been totally completed and the third activity is 70 percent
complete. The planners were only expecting to be 57 percent through the third activity at this
time. The first activity involves prepping the site for the bridge. It was expected that this would
cost $1,426,000 and it was done for only $1,306,000. The second activity was the pouring of
concrete for the bridge. This was expected to cost $10,506,000 but was actually done for
$9,006,000. The third and final activity is the actual construction of the bridge superstructure.
This was expected to cost a total of $8,506,000. To date they have spent $5,006,000 on the
superstructure.
Calculate the schedule variance, schedule performance index, and cost index for the
project to date. How is the project going?
Solution:
Activity
1. Site preparation
2. Pour concrete
3. Construction
Total
Expected
Cost
$1,426,000
$10,506,000
$8,506,000
$20,438,000
Expected
%
Complete
100%
100%
57%
Actual %
Complete
100%
100%
70%
Actual Cost
to Date
$1,306,000
$9,006,000
$5,006,000
$15,318,000
BCWS
Activity 1
Activity 2
Activity 3
100% of $1,426,000
100% of $10,506,000
57 % of $8,506,000
=
=
=
$1,426,000
$10,506,000
$4,848,420
$16,780,420
4-24
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Chapter 04 – Project Management
BCWP
Activity 1
Activity 2
Activity 3
100% of $1,426,000
100% of $10,506,000
70 % of $8,506,000
=
=
=
$1,426,000
$10,506,000
$5,954,200
$17,886,200
Scheduled Variance = 17,886,200-16, 780,420 = 1,105,780
Scheduled Performance Index = 17,886,200/16,780,420 = 1.066
Cost Performance = 17,886,200/15,318,000 = 1.168
Ahead of schedule and under budget.
21. What feature in project management information systems can be used to resolve overallocation
of project resources?
Leveling
22. What was the first major project management information system that is now commonly used
for managing very large projects?
Primavera Project Planner
23. What type of chart compares the current project schedule with the original baseline schedule so
that deviations from the original plan can be easily noticed?
Tracking Gantt chart
Analytics Exercise: Product Design Project
1, 2.
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Chapter 04 – Project Management
0
2
P1 (2)
0
2
2
6
P2 (4)
16
22
S2 (6)
19
2
6
6
11
22
6
25
37
V2 (2)
25
35
21
22
35
37
D1 (1)
11
16
16
21
23
24
21
22
S1 (5)
P4 (5)
11
35
V1 (10)
P3 (5)
11
32
16
16
23
27
D4 (4)
D3 (1)
21
27
23
24
21
23
24
31
I1 (3)
28
31
23
27
31
27
23
35
31
I3 (4)
I2 (4)
23
35
I4 (4)
28
D2 (2)
21
30
27
27
31
The project will take 37 weeks to complete.
The critical path is P1-P2-P3-P4-S1-D2-I2-I3-I4-V2.
3.
Slack for each activity is listed in the following table.
Major Subprojects/Activities
Project Specifications (P)
Market research
Overall product specification
Hardware
Software
Supplier specifications (S)
Hardware
Software
Product design (D)
Battery
Display
Camera
Outer cover
Product integration (I)
Hardware
User interface
Software coding
Prototype testing
Subcontracting (V)
Supplier selection
Contract negotiation
Activity
ID
Dependencies
P1
P2
P3
P4
-P1
P2
P3
S1
S2
P4
P4
D1
D2
D3
D4
S1
S1
S1
D1, D2, D3
I1
I2
I3
I4
D4
D2
I2
I1, I3
V1
V2
S1, S2
I4, V1
Duration
(Weeks)
16
2
4
5
5
6
5
6
6
1
2
1
4
12
3
4
4
4
12
10
2
ES, LS
Slack
0, 0
2, 2
6, 6
11, 11
-----
16, 16
16, 19
-3
21, 23
21, 21
21, 23
23, 24
2
-2
1
27, 28
23, 23
27, 27
31, 31
1
----
22, 25
35, 35
3
--
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Chapter 04 – Project Management
4.
Assume that the activity lengths remain the same, but the precedence relationships within subprojects no
longer apply as all activities in a subproject will be worked on in parallel. Therefore, the length of each
subproject will now be equal to the length of the longest activity in the subproject. The subprojects are
outlined in dashed lines in the network drawing above.
Subproject P will take 5 weeks to complete; subproject S, 6 weeks; D, 4 weeks; I, 4 weeks; and V, 10
weeks. Since all of these subprojects will completed in series now, the length of the project is the
sum of the subproject times: 5+6+4+4+10 = 29 weeks, a decrease of 8 weeks.
5.
The revised network diagram is shown below. Adding P5 extends the time of subproject P to 12 weeks while
subprojects D and I remain at 4 weeks each. By eliminating subprojects S and V, the project length is now
down to 20 weeks – 17 weeks less than the original schedule.
0
2
P1 (2)
10
12
0
12
4
P2 (4)
13
D1 (1)
16
8
12
0
5
0
12
15
16
12
13
0
D4 (4)
12
15
16
5
12
14
12
20
I4 (4)
12
14
16
17
20
16
20
16
16
16
20
20
I3 (4)
I2 (4)
D2 (2)
12 weeks
6.
16
I1 (3)
16
12
P4 (5)
7
12
19
P5 (12)
D3 (1)
P3 (5)
7
0
20
16
20
16
4 weeks
4 weeks
Having the team focus on a single subproject at a time will allow more collaboration on each subproject
as opposed to the team being split across several subprojects. This might result in higher quality of work
as the entire project team is focused on a single subproject at a time and you will have more input and a
wider variety of experiences working on each subproject than you would otherwise.
By combining subprojects S and V with P, Nokia can perhaps take advantage of supplier expertise in
designing the new phone. Nokia would better understand the technological capabilities of their
suppliers and include them in the phone design from the beginning.
One concern might be the feasibility of eliminating the original precedence relationships when
changing to the new project structures in parts 4 and 5. Assuming there were good reasons for those
relationships originally, eliminating them might cause problems in the project if Nokia does not fully
address those reasons in the new project structure. For example, if they select suppliers and
negotiate contracts before the product specifications are complete, and they do not include their
suppliers in the product specification process, they might end up with a supplier that cannot supply
the needed materials nor do so at the proper level of quality.
Assuming the technical/managerial precedence issues are properly addressed, the new project
structures make sense. In addition to reducing project time, there are other possible benefits to be
gained from the increased collaboration the new structures would bring.
4-27
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CHAPTER 5
STRATEGIC CAPACITY MANAGEMENT
Discussion Questions
1. What capacity problems are encountered when a new drug is introduced to the market?
The primary concerns come from uncertain demand for the drug and the high capital investment
typically needed for modern drug production. Being a new drug, there are no historical sales data on
which to base forecasts of future demand. If forecasts are too high, significant capital resources will
be underutilized. If forecasts are too low, there may be insufficient capital resources to meet the
actual demand, resulting in lost sales when the price for the new drug is typically highest.
2. List some practical limits to economies of scale; that is, when should a plant stop growing?
The obvious answer is that a plant should stop growing when its long-run average cost curve hits the
inflection point and starts increasing. Factors leading to this situation include difficulties coordinating
and managing a facility of that size, demand variations that can lead to regular periods of low capacity
utilization, and capacity imbalance within the facility.
3. What are some capacity balance problems faced by the following organizations or facilities?
a. An airline terminal
Congested flight arrival/departure scheduling typically leads to problems throughout the system,
including waiting areas, distances from boarding gates, ground crew requirements, runways, baggage
handling, etc.
b. A university computing lab
The number of computer workstations, the size of each workstation (room for student papers, etc.),
the mix of different computer types (Mac or PC), the number of printers, the capacity of the network
access, study space for students waiting. These problems are exacerbated by surges in demand
during certain points in the semester (e.g. finals week).
c.
A clothing manufacturer
Many manufacturers now use highly decentralized shops to make clothes. This means that capacity
of multiple sites must be accounted for in planning production.
4. At first glance, the concepts of the focused factory and capacity flexibility may seem to contradict
each other. Do they really?
This is not necessarily true. This will depend on the available technology of the facility and on the
type of industry it competes in. An FMS plant may, for example, use flexible processes to enlarge the
variety of products produced and delivered in a very short time. Therefore, it can choose to compete
on fast delivery of customized products rather than on cost. The PWP concept can capitalize on the
overall facility economies of scale while maintaining focus within each individual PWP.
5. Management may choose to build up capacity in anticipation of demand or in response to developing
demand. Cite the advantages and disadvantages of both approaches.
The strategy of building up capacity ahead of demand is a risk-taking stance. Investment is based on
projections. This investment involves costs for new facilities, equipment, human resources, and
overhead. If the demand materializes, the investment is worthwhile since the firm may capture a
large amount of market share. If it does not materialize, the firm must redirect the invested
resources. This strategy is most appropriate in high growth areas.
If the demand materializes, but the capacity planning strategy is risk averse, i.e., building capacity
only as demand develops, then most likely market share will be lost. The growth in demand will
encourage new entrants, resulting in more competition. The risk averse strategy may be most
appropriate for small firms who cannot afford to invest in unproven prospects. To prevent potential
loss of market share, firms may choose to incrementally increase capacity to match the increase in
demand.
6. What is capacity balance? Why is it hard to achieve? What methods are used to deal with capacity
imbalances?
In a perfectly balanced plant, the output of each stage provides the exact input requirement for the
subsequent stage. This continues throughout the entire operation. This condition is difficult to
achieve because the best operating levels for each stage generally differ. Variability in product
demand and the processes may also lead to imbalance, in the short run.
There are various ways of dealing with capacity imbalances. One is to add capacity to those stages
that are the bottlenecks. This can be achieved by temporary measures such as overtime, leasing
equipment, or subcontracting. Another approach is to use buffer inventories so that interdependence
between two departments can be loosened. A third approach involves duplicating the facilities of one
department upon which another is dependent.
7. What are some reasons for a plant to maintain a capacity cushion? How about a negative capacity
cushion?
A plant may choose to maintain a capacity cushion for a number of reasons. If the demand is highly
unstable, maintaining cushion capacity will ensure capacity availability at all times. Also, capacity
cushions can be useful if high service quality levels are established. Some organizations choose to
use capacity cushions as a competitive weapon to create barriers to entry for competitors.
Negative capacity cushions may be maintained when demand is expected to decrease rapidly and
capacity investment is high enough to discourage short run capacity acquisitions. It may also make
sense where capital investment needed to achieve a capacity cushion is extremely expensive, and
capacity can be easily increased in the short run by methods such as overtime or subcontracting.
8. Will the use of decision tree analysis guarantee the best decision for a firm? Why or why not? If not,
why bother using it?
No they cannot, due to the effect of future chance events. First, the probabilities are not known with
certainty, but are just estimates. However, even if the probabilities used are accurate, we are still just
computing expected values. For any one decision, there is no guarantee it will be the best possible
decision. Then why use it? For any one decision you are going with the “best odds” so to speak. For
a series of decisions over time, the best long-term results will come from decision tree analysis with
accurate probabilities.
9. Consider the example in Exhibit 5.5. Can you think of anything else you might do with that example
that would be helpful to the ultimate decision maker?
As with the probabilities in the prior question, the rate of return used in NPV analysis is only an
estimate. The analyst could repeat the decision tree analysis with multiple rates of return, performing
sort of a sensitivity analysis on the decision model with respect to the rate of return. If the same
solution results from all of the analyses, the decision maker can feel more confident in choosing the
recommend approach.
10. What are some major capacity considerations in a hospital? How do they differ from those of a
factory?
Some capacity considerations are size and composition of nursing staff (RNs vs. LPNs), balance
between operating room and intensive care units, emergency rooms, etc., and, of course, how many
beds are to be available. One of the differences in capacity considerations between a hospital and a
factory is that a hospital can add capacity rather quickly in the short run, through “simply” adding
more staff and more beds. A factory is usually technologically limited, and, therefore, must plan well
in advance to add major chunks of capacity. On the other hand, though, the general uncertainty
which surrounds the demand for hospital services on any given day is much greater than would be
faced by a factory. Additionally, factory management generally has the ability to backlog demand in
such a way as to achieve more efficient levels of capacity utilization than does a hospital. Sick and
injured patients cannot be put on a shelf and made to wait during periods of peak demand.
11. Refer to Exhibit 5.6. Why is it that the “critical zone” begins at a utilization rate of about 70 percent in
a typical service operation? Draw upon your own experiences as either a customer or a server in
common service establishments.
Uncertainty in the arrival and service rates is the key problem here. The utilization rate of 70 percent
is based on the average arrival rate and service rate. As most of us have observed, both can vary
widely throughout the day or even from one customer to the next. Sometimes things just “get busy”
as more customers than average arrive during a short time window. Also, a “problem customer” or
two can greatly extend the time to service them and consume valuable resources. Even though the
average utilization rate may be 70 percent, these issues can make the short term utilization rate
exceed 100 percent occasionally.
Objective Questions
1. A manufacturing shop is designed to operate most efficiently at an output of 550 units per day. In the
past month the plant produced 490 units. What was their capacity utilization rate last month?
𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑢𝑡𝑖𝑙𝑖𝑧𝑎𝑡𝑖𝑜𝑛 𝑟𝑎𝑡𝑒 =
490
550
= 89.1%
2. A company has a factory that is designed so that it is most efficient (average unit cost is minimized)
when producing 15,000 units of output each month. However, it has an absolute maximum output
capability of 17,250 units per month, and can produce as little as 7000 units per month without
corporate headquarters shifting production to another plant. If the factory produces 10,925 units in
October, what is the capacity utilization rate in October for this factory?
𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑢𝑡𝑖𝑙𝑖𝑧𝑎𝑡𝑖𝑜𝑛 𝑟𝑎𝑡𝑒 =
10,925
15,000
= 72.83%
3. Hoosier Manufacturing operates a production shop that is designed to have the lowest unit production
cost at an output rate of 100 units per hour. In the month of July, the company operated the
production line for a total of 175 hours and produced 16,900 units of output. What was its capacity
utilization rate for the month?
16,900
𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑢𝑡𝑖𝑙𝑖𝑧𝑎𝑡𝑖𝑜𝑛 𝑟𝑎𝑡𝑒 = (175)∗(100) = 96.57%
4. AlwaysRain Irrigation, Inc., would like to determine capacity requirements for the next four years.
Currently two production lines are in place for making bronze and plastic sprinklers. Three types of
sprinklers are available in both bronze and plastic: 90-degree nozzle sprinklers, 180-degree nozzle
sprinklers, and 360-degree nozzle sprinklers. Management has forecast demand for the next four
years as follows:
Both production lines can produce all the different types of nozzles. The bronze machines needed
for the bronze sprinklers require two operators and can produce up to 12,000 sprinklers. The
plastic injection molding machine needed for the plastic sprinklers requires four operators and
can produce up to 200,000 sprinklers. Three bronze machines and only one injection molding
machine are available. What are the capacity requirements for the next four years? (Assume that
there is no learning.)
Solution:
Plastic
Year 1
Year 2
Year 3
Year 4
97
115
136
141
48.5%
57.5%
68.0%
70.5%
.485
.575
.680
.705
1.94
2.30
2.72
2.82
Bronze
Year 1
Year 2
Year 3
Year 4
Demand for bronze sprinklers
21
24
29
34
Percentage of capacity used
58.3%
66.7%
80.6%
94.4%
Machine requirements
1.75
2.00
2.42
2.83
Labor requirements
3.50
4.00
4.83
5.66
Demand for plastic sprinklers
Percentage of capacity used
Machine requirements
Labor requirements
There is sufficient capacity to meet expected demand over the 4-year planning horizon. The only concern
might be year 4 on the bronze line. Capacity is approaching 100% in that year, and forecast error might
lead to an over-capacity situation. It is probably not a large concern at this point in time, but management
should pay special attention to that point in time as forecasts are updated in the future.
5.
Requirements for plastic remain unchanged.
Bronze
Year 1
Year 2
Year 3
Year 4
Demand for bronze sprinklers
32
36
41
52
Percentage of capacity used
88.9%
100.0%
113.9%
144.4%
Machine requirements
2.67
3
3.42
4.33
Labor requirements
5.33
6
6.83
8.67
It is obvious that not enough capacity is available after year two to meet the increased demand.
AlwaysRain will have to consider purchasing additional machines for the bronze operations.
6.
Bronze
Year 1
Year 2
Year 3
Year 4
Demand for bronze sprinklers
32
36
41
52
Percentage of capacity used
66.67%
75.00%
85.42%
108.33%
Machine requirements
2.67
3.00
3.42
4.33
Labor requirements
5.33
6.00
6.83
8.67
No. An additional machine will provide enough capacity cushion until the third year. AlwaysRain
must consider additional ways of meeting the fourth year demand. This can include purchasing or
leasing an additional machine, or outsourcing some of the demand.
7.
Labor requirements-bronze
Year 1
Year 2
Year 3
Year 4
5.33
6.00
6.83
8.67
Labor requirements-plastic
1.94
2.30
2.72
2.82
Total labor requirements
7.27
8.30
9.55
11.49
AlwaysRain will face a problem of not having enough trained personnel for running the equipment
after the third year. At that time, they will need to either hire new trained employees or initiate a
training program for existing employees from other workstations who can be utilized at the bronze or
plastic molding machines.
8.
High growth
P = .40
$12.0 million
$10.8 million
Build Small Factory
EV = $10.8 – 6.0 million
EV = $4.8 million
Low growth
P = .60
$10.0 million
Do Nothing, EV = $0
Build Large Factory
EV = $11.6 – 9.0 million
EV = $2.6 million
High growth
P = .40
$14.0 million
$11.6 million
Low growth
P = .60
$10.0 million
For the small facility,
NPV = .40 ($12 Million) + .60 ($10 Million) - $6 Million = $4.8 Million
Do nothing,
NPV = $0
For the large facility
NPV = .40($14 Million) + .60($10 Million) - $9 Million = $2.6 Million
Therefore, build the small facility.
9.
Do Nothing
EV = $0
Sell to dept. chain @ $4.0 million
P = .70
1 EV = $4.3 – 1.0 million = $3.3 million
P = .30
Sell to insurance co. @ $5.0 million
Rezoned
P = .60
Buy/Develop Property
EV = $2.94 – 2.0 million
EV = $940,000
3 EV = $2.94 million
High price
P = .60
1,500 apts. @ $3,000 = $4.5 million
2 EV = $3.9 - $1.0 million = $2.9 million
Low price
P = .40
1,500 apts. @ $2,000 = $3.0 million
Not Rezoned
P = .40
Build 600 homes @ $4,000 = $2.4 million
The “Do Nothing” option is included here for completeness.
Rezoned shopping center (includes $1.0 rezoning costs):
Point 1: Expected value = .70($4 Million) + .30($5 Million) - $1.0 million = $3.3 Million
Rezoned apartments:
Point 2: Expected value = .60($4.5 Million) + .40($3 Million) - $1.0 million = $2.9 Million
Since a shopping center has more value, prune the apartment choice. In other words, if rezoned, build a
shopping center with a revenue of $4.3 Million - $1 Million = $3.3 Million. (The purchase cost could be
included here if desired, but would need to be included in the calculations for all development options.
This solution shows it at the leftmost part of the tree.)
If not rezoned the revenue will be $2.4 million from building homes:
Point 3: Expected value of developing the land is .6*($3.3 million) + .4*($2.4 million) = $2.94 million.
Expected profit of buying and developing the land is $2.94 million - $2 million purchase cost = $940,000.
Since this is a positive expected value, prune the option of doing nothing.
10. A local restaurant is concerned about their ability to provide quality service as they continue to grow
and attract more customers. They have collected data from Friday and Saturday nights, their busiest
time of the week. During these time periods about 75 customers arrive per hour for service. Given
the number of tables and chairs, and the typical time it takes to serve a customer, they figure they can
serve on average about 100 customers per hour. During these nights, are they in the zone of service,
the critical zone, or the zone of non-service?
𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑢𝑡𝑖𝑙𝑖𝑧𝑎𝑡𝑖𝑜𝑛 𝑟𝑎𝑡𝑒 =
75
100
= 75%
According to the text, this restaurant is in the critical zone on these nights.
11. The restaurant in the prior problem anticipates that in one year, their demand will double as long as
they can provide good service to their customers. How much will they have to increase their service
capacity to stay out of the critical zone?
If demand doubles, they will be receiving about 150 customers per hour on average. Find the service
rate necessary to result in a utilization rate of 70%.
𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑢𝑡𝑖𝑙𝑖𝑧𝑎𝑡𝑖𝑜𝑛 𝑟𝑎𝑡𝑒 =
150
= 70% ⟹ 𝜇 = 214.3
𝜇
Therefore, the restaurant will have to increase capacity to at least 215 customers per hour to stay out of
the critical zone. That will be quite an expansion.
Case: Shouldice Hospital - A Cut Above
1. Mon. - Fri. Operations with 90 beds (30 patients per day)
Beds Required
Monday
Monday
Tuesday
Wednesday
30
30
30
30
30
30
30
30
30
30
30
Tuesday
Wednesday
Check-in Thursday
On
Friday
Thursday
Friday
Saturday
Sunday
30
Saturday
Sunday
30
30
30
Total
60
90
90
90
60
30
30
450
Utilization
66.7%
100.0%
100.0%
100.0%
66.7%
33.3%
33.3%
71.4%
Saturday
Sunday
2. Mon. - Sat. Operations with 90 beds (30 patients per day)
Beds Required
Monday
Monday
Tuesday
Wednesday
30
30
30
30
30
30
30
30
30
30
30
30
30
30
Tuesday
Wednesday
Check-in Thursday
On
Friday
Thursday
Friday
30
Saturday
Sunday
30
30
30
Total
60
90
90
90
90
60
60
540
Utilization
66.7%
100.0%
100.0%
100.0%
100.0%
66.7%
66.7%
85.7%
3. Mon. - Fri. Operations with 135 beds (minimum)
Beds Required
Monday
Monday
Tuesday
Wednesday
45
45
45
45
45
45
45
45
45
45
45
Tuesday
Wednesday
Check-in Thursday
On
Friday
Thursday
Friday
Saturday
Sunday
45
Saturday
Sunday
45
45
45
Total
90
135
135
135
90
45
45
675
Utilization
66.7%
100.0%
100.0%
100.0%
66.7%
33.3%
33.3%
71.4%
Can the capacity of the rest of Shouldice keep up?
One operating room can handle about 1 patient every hour. Since there are five operating rooms,
each must be able to handle 45/5 or 9 patients per day. This means they must be operated 9 hours a
day. In order to finish operating early enough for all patients to recover by the evening, Shouldice
would probably have to add operating room capacity although it might be easy to just start earlier in
the day. With 45 patients each day the total number of operations each week is 225. The 12
surgeons would need to do between 18 and 19 each week or between 3 and 4 a day. This should be
feasible and even if it were not Shouldice could hire some additional surgeons. These guys would be
making over $450,000/year (3 ops/day x 5 days/week x 50 weeks/yr x $600 = $450,000)!
4. Using the financial data given in the fourth discussion question it is easy to justify the expansion to
135 beds. The following is the analysis as presented in the spreadsheet. Based on average costs
and full capacity utilization, the hospital would pay back its investment in about 86 weeks, or 1.72
years.
Mon
Beds Required
Mon
Tues
Wed
45
45
45
45
45
45
45
45
45
45
45
45
135
100.0%
90
66.7%
45
33.3%
Tues
Wed
Check-in day
Thurs
Thurs
Fri
Sat
Sun
Fri
Sat
Total Beds
135
Sun
45
45
Total
Utilization
90
66.7%
135
135
100.0% 100.0%
Operating Rooms Operations
5
Oper/Room
Surgeons
45
9
45
45
33.3%
675
71.4%
12
Oper/Surg
3.75
Cost of expansion Beds
Cost/Bed
Total
45
$100,000
$4,500,000
Incremental
Revenue
Rev/Oper
$1,300
Surgeon
Incr Rev
$600
$700
Oper/Week
Rev/Week
Payback
75
$52,500
85.7
Additional
Weeks
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Operations Management (York University)
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Chapter 10 - Waiting Line Analysis and Simulation
CHAPTER 10
WAITING LINE ANALYSIS AND SIMULATION
Discussion Questions
1. Distinguish between a channel and a phase.
A channel is the initial service point of a queuing system. A phase refers to the number of
stages that the service points provide. It is possible to have single to multiple service
channels and single to multiple service phases.
2. In what way might the first-come, first-served rule be unfair to the customer waiting for
service in a bank or hospital?
In a bank, FCFS may be perceived to be unfair by customers who have large accounts, but
who must wait while the less "important” customers obtain service.
In a hospital, especially in an emergency room, FCFS is probably the exception rather than
the rule. FCFS would be unfair when a patient with a minor problem is treated before
another experiencing severe pain.
3. Define, in a practical sense, what is meant by an exponential service time.
An exponential service time means that most of the time, the service requirements are of
short time duration, but there are occasional long ones. Exponential distribution also means
that the probability that a service will be completed in the next instant of time is not
dependent on the time at which it entered the system. We can see that a barber, for
example, does not fit an exponential distribution in either case. The barber has an average
time for cutting hair, and a person who has been sitting in the chair getting a haircut for the
past 15 minutes has a higher probability of being completed in the next minute than a
person who just walked in and sat down.
4. Would you expect the exponential distribution to be a good approximation of service times
for
a. Buying an airline ticket at the airport?
Yes. Although certain customers will require special routings and payment methods,
most ticketing is straightforward and entails a short service time.
b. Riding a merry-go-round at a carnival?
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Chapter 10 - Waiting Line Analysis and Simulation
No. The Merry-go-round has a fixed cycle time and hence constant services rate.
c. Checking out of a hotel?
No. Probably a better approximation would be a normal distribution since fast and slow
checkouts are likely to be equally balanced.
d. Completing a midterm exam in your OM class?
No. From our experience, students require the entire class period (and then some) to
finish the typical mid-term. Thus, the service rate is close to constant.
5. Would you expect the Poisson distribution to be a good approximation of
a. Runners crossing the finish line in the Boston Marathon?
Yes. The arrival pattern typically shows a few runners arriving “early” and the majority
arriving in a bunch and the remainder spread out along the tail of the distribution.
b. Arrival times of the student in your OM class?
No. Arrivals are not random since there is a schedule to be met.
c. Arrival times of the bus to your stop at school?
No. Again, arrivals are not random since the bus follows a set schedule.
6. What it the major cost trade-off that must be made in managing waiting line situations?
The classic trade-off is between the cost of waiting for service versus the cost of providing
additional service capacity, e.g., the cost of idle WIP versus the cost of adding more workers
and machines to process the inventory.
7. Which assumptions are necessary to employ the formulas given for Model 1?
Poisson arrival rates, exponential service rates, which imply a purely random process, but
with a known mean (and hence known variance). Also assumed is that the process has
reached a point of stochastic equilibrium. In other words, steady state conditions prevail.
Infinite calling population and unlimited queue length.
8. Why is simulation often called a technique of last resort?
Simulation is called a technique of last resort because simulation models are time consuming
to build (flow charting, coding, etc.) and do not “guarantee” an optimal solution or indeed
any solution. Therefore, it makes sense to investigate other problem-solving methods such
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as linear programming or waiting line theory before embarking on simulation. There are
some easy to use “visual” simulation programs now available, and they are worth checking
into.
9. Must you use a computer to get good information from a simulation? Explain.
A computer is a must for any but the most simple simulation problems. Because simulation
is a sampling process, it stands to reason that a large number of observations is desirable,
and the computer is the only practical way of providing them. Of course, computerization is
no guarantee of “good” information. Simulating an invalid model on the computer will only
provide a larger volume of questionable data.
10. What methods are used to increment time in a simulation model? How do they work?
Time incrementing methods include fixed time increments and variable time increments.
With fixed time increments, uniform clock times are specified (minutes, hours, days, etc.) and
the simulation proceeds by fixed intervals from one time period to the next. At each point in
clock time, the system is scanned to determine if any events have occurred and time is
advanced; if none have, time is still advanced by one unit.
With variable time increments, the clock time is advanced by the amount required to initiate
the next event.
11. What are the pros and cons of starting a simulation with the system empty? With the
system in equilibrium?
The pros of starting a simulation with the system empty are that this enables evaluation of
the transient period in terms of time to reach steady-state and the activities which are
peculiar to the transient period.
One con is that it takes a longer period of time to perform the simulation. A second is that
the model will be biased by the set of initial values selected, since the time to achieve steadystate and the activities which take place during the transient period will be affected by the
initial values. Steps must be taken to remove these initial values if steady-state results are
needed.
The advantages of starting the system in equilibrium are that the run time may be greatly
reduced, and that the aforementioned bias may be eliminated.
The disadvantage of starting the simulation in equilibrium is, in essence, that it assumes that
the analyst has some idea of the range of output he is looking for. This, in a sense,
constitutes “beating” the model and may lead to incorrect conclusions from the simulation
run.
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12. Distinguish between known mathematical distributions and empirical distributions. What
information is needed to simulate using a known mathematical distribution?
A “known mathematical distribution” is one that can be generated mathematically and is
amenable to the laws of statistical probability. Examples of such distributions are the
normal, binomial, Poisson, Gamma, and hypergeometric.
An empirical distribution is one that is obtained from observing the probability of occurrence
of phenomena relating to a specific situation. While it may be possible to define the
moment generating function for such distributions, their applicability to other situations is
likely to be small.
The information required to simulate using a known distribution, of course, depends on the
known distribution selected. Generally speaking, however, at a minimum, the analyst must
be able to estimate the mean and standard deviation of the population to be sampled since
they are parameters of distributions. Alternatively, the analyst can use a discrete probability
table to represent the empirical distribution.
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Chapter 10 - Waiting Line Analysis and Simulation
Objective Questions
1. The exponential distribution is often used to model what in a queuing system?
The time between customer arrivals and/or service times.
2. If the average time between customer arrivals is 8 minutes, what is the hourly arrival rate?
7.5 customers per hour
3. How much time on average would a server need to spend on a customer to achieve a service
rate of 20 customers per hour?
Three minutes
4. What is the term used for the situation where a potential customer arrives at a service
operation and upon seeing a long line decides to leave?
Balking
5. What is the most commonly used priority rule for setting queue discipline, likely because it is
seen as most fair?
First-come, first-served (FCFS)
6. Use model 1.
λ=
μ=
4/hour
6/hour
λ
4
1−ρ=1− =1− =.3333
μ
6
a.
or 33.33%
2
Lq =
b.
2
λ
4
=
μ (μ−λ ) 6(6−4 )
W q=
=1.33,
Lq 1 .33
=
λ
4
= 1/3 hour or 20 minutes
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2
2
λ
4
Lq =
=
μ (μ−λ ) 6(6−4 )
= 1.33 students
c.
d. At least one other student waiting in line is the same as at least two in the system. This
probability is 1-(P0+P1).
n
( )( )
λ λ
Pn = 1−
μ μ
0
( )( )
P0 = 1−
4 4
6 6
= .3333
1
( )( )
P1= 1−
4 4
6 6
= .2222
Probability of at least one in line is 1-(.3333 + .2222) = .4444
7. Use model 2.
λ=
μ=
60/50 per minute
60/45 per minute
minutes.
Lq =
(60 /50)2
λ2
=
=
2 μ( μ−λ ) 2(60 /45)(60 /45−60/50)
4.05 cars
Ls =Lq + λ/ μ
= 4.05 + (60/50)/(60/45) = 4.95 cars
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Chapter 10 - Waiting Line Analysis and Simulation
W q=
W s=
Lq
λ
Ls
λ
=
4 .05
=3 . 375 minutes
(60/50)
=
4 . 95
(60/50)
= 4.125 minutes
8. Use Model 1.
λ=
μ=
100 per hour
Ls =
120 per hour
λ 100
=
μ−λ 120−100
a.
= 5 customers
W s=
Ls
λ
=
5
100
= .05 hours or 3 minutes
μ=
b. Now,
Ls =
180 per hour
λ 100
=
μ−λ 180−100
= 1.25 customers
W s=
Ls
λ
=
1 .25
100
= .0125 hours or .75 minutes or 45 seconds
λ=
c. Using model 3,
μ=
100 per hour
120 per hour
λ 100
ρ= = =. 8333
μ 120
S =2 , and
Lq
, from spreadsheet,
= .1756
λ
100
Ls =Lq + =. 1756+
μ
120
= 1.01 customers
10-7
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Chapter 10 - Waiting Line Analysis and Simulation
W s=
Ls 1 .01
=
λ 100
= .0101 hours or .605 minutes or 36.3 seconds
9. Use model 2.
λ=
μ=
10 per hour
12 per hour
2
Lq =
2
λ
10
=
2 μ( μ−λ ) 2(12 )(12−10)
a.
= 2.083 people
Ls =Lq + λ/ μ
= 2.083 + 10/12 = 2.917 people
b.
W q=
Lq 2 . 083
=
=.2083
λ 10
c.
hours.
W s=
Ls 2 . 917
=
λ 10
d.
= .2917 hours
2
Lq =
λ=
e. It will cause it to increase, at
12 per hour,
10. Use model 1
λ=
μ=
3 per minute
Ls =
a.
2
λ
12
=
→∞
2 μ( μ−λ ) 2(12 )(12−12)
4 per minute
λ
3
=
μ−λ 4−3
= 3 customers
10-8
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Chapter 10 - Waiting Line Analysis and Simulation
W s=
Ls 3
=
λ 3
b.
= 1 minute
λ 3
ρ= =
μ 4
c.
= .75 or 75%
d. Probability of 3 or more is equal to 1 – probability of 0, 1, 2
( )( )
3 3
P0 = 1−
4 4
( )( )
0
3 3
P1= 1−
4 4
( )( )
1
= .2500,
3 3
P2 = 1−
4 4
= .1875,
1406
Total of P0 + P1 + P2 = (.2500 + .1875 + .1406) = .5781
Therefore, the probability of three or more is 1 - .5781 = .4219
e. If a automatic vendor is installed, use model 2.
(a. revisited)
2
2
λ
3
Lq =
=
2 μ( μ−λ ) 2( 4 )(4−3 )
=1.125 customers
Ls =Lq + λ/ μ
= 1.125 + ¾ =1.875 customers
(b. revisited)
W q=
Lq 1 .125
=
=. 375 minutes
λ
3
W s=
Ls 1 .875
=
=.625 minutes
λ
3
10-9
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2
=.
lOMoARcPSD|12655597
Chapter 10 - Waiting Line Analysis and Simulation
By converting to constant service time, the number in line is reduce from 3 to 1.875
people (a reduction of 1.125, and time in system is reduced from 1 minute to .625
minutes (a reduction of .375 minutes or 22.5 seconds).
11. Use model 4.
N = 4, population of 4 engineers,
S = 1, one technical specialist,
T = 1, average time to help engineer,
U = 7, time between requests for help
X=
T
1
=
T +U 1+7
= .125, look up value of F in Exhibit 10.10
a.
F = .945, therefore, L = N(1-F) = 4(1-.945) = .22 engineers waiting
b.
Wq=
L( T +U ) .22( 1+ 7)
=
=¿ .466 hours or 28 minutes
N−L
4−.22
c. From Exhibit 10.10 at X=.125, and S=1, D = .362. In other words, 36.2% of the time an
engineer will have to wait for the specialist.
λ=
12. Use model 1.
Ls =
30 per hour
λ
20
=
μ−λ 30−20
a.
= 2 people in the system
W s=
b.
μ=
20 per hour
Ls 2
=
λ 20
= .10 hours or 6 minutes
c. Probability of 3 or more is equal to 1 – probability of 0, 1, 2
10-10
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Chapter 10 - Waiting Line Analysis and Simulation
n
( )( )
λ λ
Pn = 1−
μ μ
(
)( )
0
20 20
P0 = 1−
30 30
= .3333
( )( )
1
20 20
P1= 1−
30 30
= .2222
(
)( )
20 20
P2 = 1−
30 30
2
= .1481
Total of P0 + P1 + P2 = (.3333 + .2222 + .1481) = .7036
Therefore, the probability of three or more is 1 - .7036 = .2964
λ 20
ρ= =
μ 30
d.
= .67 or 67%
e. Use model 3.
λ 20
ρ= =
μ 30
=.6667
From Exhibit 10.9, Lq = .0093
Ls =Lq + λ/ μ
= .0093 + 20/30 = .676
W s=
Ls .676
=
λ 20
= .0338 hours or 2.03 minutes
13. Use model 4.
N = 4, population of 4 pieces of equipment,
10-11
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Chapter 10 - Waiting Line Analysis and Simulation
S = 1, one repairperson,
T = .0833, average time for repair,
U = .5, time between requests for repair (hours)
X=
T
. 0833
=
T +U . 0833+.5
= .1428, look up value of F in Exhibit 10.10
a.
F = .928, therefore, L = N(1-F) = 4(1-.928) = .288 machines waiting
b. J=NF(1-X) = 4(.928)(1-.1428) = 3.18 machines operating
c. H = FNX =.928(4).1428 = .53 machines being serviced
d. Ls =L + H = .288 + .53 = .818 machines in the system
Cost of downtime is .818 times $20 per hour
= $16.36 per hour
Cost of one serviceperson
= $ 6.00 per hour
Total cost per hour
= $22.36
With 2 repairpersons, S =2,
X=
T
. 0833
=
T +U . 0833+.5
= .1428, look up value of F in Exhibit 10.10
F = .995, therefore, L = N(1-F) = 4(1-.995) = .020 machines waiting
H = FNX =.995(4).1428 = .568 machines being serviced
Ls = L + H = .020 + .568 = .588
Cost of downtime is .588 times $20 per hour
= $11.76 per hour
10-12
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Chapter 10 - Waiting Line Analysis and Simulation
Cost of two serviceperson
= $12.00 per hour
Total cost per hour
= $23.76
No, added cost of $1.42 per hour would be added for second repairperson
14. Use model 1.
λ=
μ=
2 per hour
3 per hour
2
2
λ
2
Lq =
=
μ (μ−λ ) 3(3−2 )
a.
= 1.333 customers waiting
W q=
Lq 1 .333
=
λ
2
b.
= .667 hours or 40 minutes
Ls =
Ls 2
λ
2
=
=2 , W s = =
μ−λ 3−2
λ 2
c.
= 1 hour
λ 2
ρ= =
μ 3
d.
= .67 or 67% of the time
10-13
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Chapter 10 - Waiting Line Analysis and Simulation
15. Use model 1.
ρ=.55
λ=1.5/hour
1.5
.55=
μ
μ=2.727
λ
2.727
.85=¿ ¿
¿
¿ λ=2.318/ hour ¿¿
16. Use model 1.
λ=
μ=
6 per hour
Ls =
10 per hour
λ
6
=
μ−λ 10−6
a.
= 1.5 people
W s=
Ls 1 .5
=
λ
6
= .25 hours or 15 minutes
λ 6
ρ= =
μ 10
b.
c.
= .60 or 60%
Probability of more than 2 people is equal to 1 – probability of 0, 1, or 2
n
( )( )
λ λ
Pn = 1−
μ μ
( )( )
6 6
P0 = 1−
10 10
0
= .4000
10-14
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Chapter 10 - Waiting Line Analysis and Simulation
( )( )
1
6 6
P1= 1−
10 10
= .2400
(
)( )
6 6
P2 = 1−
10 10
2
= .1440
Total of P0 + P1 + P2 = (.4000 + .2400 + .1440) = .7840
Therefore, the probability of three or more is 1 - .7840 = .2160
d. Use model 3.
λ 6
ρ= =
μ 10
Lq
= .60, from Exhibit 10.9,
= .0593
Ls =Lq + λ/ μ=.0593+6/10=. 6593
W s=L s / λ
= .6593/6 = .1099 hours or 6.6 minutes
17. Use model 1.
λ=
μ=
25 per hour
30 per hour
λ 25
ρ= =
μ 30
a.
= .833 or 83.3%
Ls =
b.
λ
25
=
μ−λ 30−25
= 5.00 document in the system
10-15
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Chapter 10 - Waiting Line Analysis and Simulation
W s=
Ls
λ
=
5 . 00
25
c.
= .20 hours or 12 minutes
d.
Probability of 4 or more is equal to 1 – probability of 0, 1, 2, 3
n
( )( )
λ λ
Pn = 1−
μ μ
(
)( )
0
25 25
P0 = 1−
30 30
= .1667
(
P1= 1−
)( )
25 25
30 30
1
= .1389
(
)( )
2
25 25
P2 = 1−
30 30
= .1157
(
3
)( )
25 25
P3 = 1−
30 30
= .0965
Total of P0 + P1 + P2 + P3 = (.1667 + .1389 + .1157 + .0965) = .5178
Therefore, the probability of three or more is 1 - .5178 = .4822 or 48.22%
Lq =
λ2
302
=
→∞
μ (μ−λ ) 30(30−30)
e.
18. Use model 1.
λ=
μ=
4 per hour
6 per hour
10-16
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Chapter 10 - Waiting Line Analysis and Simulation
λ 4
ρ= =
μ 6
= .667 or 66.7%
a.
λ
4
=
μ−λ 6−2
Ls =
b.
= 2.00 students in the system
W s=
Ls
λ
=
2 . 00
4
c.
= .50 hours or 30 minutes
d.
Probability of 4 or more is equal to 1 – probability of 0, 1, 2, or 3
n
( )( )
λ λ
Pn = 1−
μ μ
0
( )( )
4 4
P0 = 1−
6 6
= .3333
( )( )
P1= 1−
4 4
6 6
1
= .2222
( )( )
2
4 4
P2 = 1−
6 6
= .1481
( )( )
4 4
P3 = 1−
6 6
3
= .0988
Total of P0 + P1 + P2 + P3 = (.3333 + .2222 + .1481 + .0988) = .8024
Therefore, the probability of four or more is 1 - .8024 = .1976 or 19.76%
Lq =
λ2
62
=
→∞
μ (μ−λ ) 6(6−6 )
e.
10-17
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Chapter 10 - Waiting Line Analysis and Simulation
19. Use model 1.
λ=
μ=
10 per hour
12 per hour
2
2
λ
10
Lq =
=
μ (μ−λ ) 12(12−10 )
a.
= 4.17 people
Ls =
λ 10
=
μ−λ 12−10
W s=
Ls
λ
=
5
10
= 5,
b.
= .5 hours or 30 minutes
λ 10
ρ= =
μ 12
c.
d.
= .833 or 83.3%
Probability of 3 or more is equal to 1 – probability of 0, 1, 2
n
( )( )
λ λ
Pn = 1−
μ μ
( )( )
0
10 10
P0 = 1−
12 12
= .1667
( )( )
P1= 1−
10 10
12 12
1
= .1389
( )( )
10 10
P2 = 1−
12 12
2
= .1157
Total of P0 + P1 + P2 + P3 = (.1667 + .1389 + .1157) = .4213
Therefore, the probability of three or more is 1 - .4213 = .5787 or 57.87%
10-18
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Chapter 10 - Waiting Line Analysis and Simulation
20. Use model 1. [Students may want to use model 4 because of the limited number of cars in
such an attraction, but there is insufficient information in the problem to solve it under
model 4. Kudos to students who note this.]
λ=
2 per hour
μ=
With one repair person:
Ls =
2
λ
2
=
→∞
μ−λ 2−2
μ=
With two repair people:
Ls =
3
2
λ
=
=2
μ−λ 3−2
cars
μ=
With three repair people:
Ls =
4
2
λ
=
=1
μ−λ 4−2
car
Service rate
per hour
ns
Cost of waiting
per hour1
Cost of service
per hour2
Total cost
per hour
2

$
$ 20
$ 
2
3
2
80
40
120
3
4
1
40
60
100
Number of repair
personnel
()
1
Note: 1 = cost of waiting is number in system times downtime cost of $40 per hour.
2 = cost of service is number of repair personnel times wage rate ($20 per hour).
We should use three repair persons.
10-19
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Chapter 10 - Waiting Line Analysis and Simulation
21. Use model 2.
λ=
μ=
750 per hour
900 per hour
W s=L s /λ
= 2.92/750 = .003889 hours or .2333 minutes or 14 seconds
a.
2
2
Lq =
λ
750
=
2 μ( μ−λ ) 2(900 )(900−750)
b.
= 2.083 cars
Ls =Lq + λ/ μ
= 2.083 + 750/900 = 2.92 cars
22. Use waiting line approximation because the service times do not follow an exponential
distribution. [Note: Student answers may vary slightly due to rounding differences.]
X̄ a =3
X̄ s=15
, Sa = 3,
C a=
, Ss = 7, S = 6
Sa
3
= =1
X̄ a 3
C s=
Ss 7
= =. 4667
X̄ s 15
,
λ=
,
1 1
= =. 3333
X̄ a 3
μ=
1 1
= =. 06667
X̄ s 15
,
ρ=
λ .3333
=
=. 8332
Sμ 6(. 06667)
,
a. On average how many customers would be waiting in line?
ρ√2( S +1 ) C a +C s . 8332√2( 6+1) 12 +. 46672
X
=
X
=1. 8443
Lq =
1−ρ
2
1−. 8332
2
2
2
b. On average how long would a customer spend in the bank?
Ls =Lq +S p =1. 848+(6 )(. 8332 )=6 . 8435
W s=
Ls 6 . 8468
=
=20 .5325
λ .3333
c. If a customer arrived, saw the line and decided not to get in line
that customer has ________________________________.
Balked
10-20
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Chapter 10 - Waiting Line Analysis and Simulation
d. A customer who enters the line but decides to leave the line before getting service is
said to have ________________________________.
Reneged
23. Use waiting line approximation because the service times do not follow an exponential
distribution.
X̄ a =4
X̄ s=7
, Sa = 4,
C a=
, Ss = 3, S = 2
Sa
4
= =1
X̄ a 4
C s=
Ss
X̄ s
3
= =. 4286
7
,
λ=
,
1 1
= =. 2500
X̄ a 4
μ=
1 1
= =. 1429
X̄ s 7
,
ρ=
λ .2500
=
=. 875
Sμ 2 (. 1429 )
,
[Note: Student answers may vary slightly due to rounding differences.]
10-21
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Chapter 10 - Waiting Line Analysis and Simulation
a. On average, how long will each line be at each of the cashier windows?
ρ√2( S +1 ) C a +C s . 8750√2( 2+1) 12 +. 42862
Lq =
X
=
X
=3 .5015
1−ρ
2
1−. 8750
2
2
2
b. On average how long will a customer spend in the bank (assume they enter, go directly to
one line and leave as soon as service is complete).
Ls =Lq +S p =3 .4138+(2 )(.8750 )=5 . 2509
W s=
Ls 5 . 1638
=
=21. 0036
λ .2500
You decide to consolidate all the cashiers so they can handle all types of customers without
increasing the service times.
c. What will happen to the amount of time each cashier spends idle? (increase, decrease, stay
the same, depends on _____ )
Stay the same
d. What will happen to the average amount of time a customer spends in the bank? (increase,
decrease, stay the same, depends on _____ )
Decrease
24. Use model 1.
Avg number in system (Ls) = 4
Avg time in system (Ws) = 1.176
1. 176=
4=
4
λ
λ=3. 4
3.4
μ=4 .25
( μ−3 . 4 )
10-22
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Chapter 10 - Waiting Line Analysis and Simulation
25.
Customer
number
Time between arrivals (minutes)
Probability
RN assignment
1
.08
00-07
2
.35
08-42
3
.34
43-76
4
.17
77-93
5
.06
94-99
Service Time (minutes)
Probability
RN assignment
1.0
.12
00-11
1.5
.21
12-32
2.0
.36
33-68
2.5
.19
69-87
3.0
.07
88-94
3.5
.05
95-99
RN
Interarrival
time
Arrival
time
Service
begins
RN
Service
time
Service
ends
Waiting
time
Idle
time
1
08
2
2
2.0
74
2.5
4.5
0.0
2.0
2
24
2
4
4.5
34
2.0
6.5
0.5
0.0
3
45
3
7
7.0
86
2.5
9.5
0.0
0.5
4
31
2
9
9.5
32
1.5
11.0
0.5
0.0
5
45
3
12
12.0
21
1.5
13.5
0.0
1.0
6
10
2
14
14.0
67
2.0
16.0
0.0
0.5
Average waiting time = 1.0/6 = 1/6 minute, and average teller idle time between customers =
4.0/6 = 4/6 minute or 40 seconds. Average teller idle time percent = 4.0/16 = .25 or 25%.
10-23
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Chapter 10 - Waiting Line Analysis and Simulation
26.
Machine
breakdown
number
RN
Interarrival
time
(hours)
Breakdown
time
(hours)
RN
Service
time
(hours)
Repairman 1
Repairman 2
begin
end
begin
1
30
1.0
1.0
81
3.0
1.0
4.0
2
02
0.5
1.5
91
4.0
3
51
1.0
2.5
08
0.5
4.0
4.5
2.0
4
28
0.5
3.0
44
1.0
4.5
5.5
2.5
5
86
3.0
6.0
84
3.0
6.0
9.0
3.0
3.0
1.5
Average down time is 14.5/5 = 2.9 hours
10-24
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end
Down
time
(hours)
5.5
4.0
lOMoARcPSD|12655597
Chapter 10 - Waiting Line Analysis and Simulation
Case: Community Hospital Evening Operations Room
1. Calculate the average customer arrival rate and service rate per hour.
The customer arrival rate lambda = 0.0212 patients per hour. This is calculated 62/(8 x
365) = 0.0212. The service rate mu = 0.7427 patients per hour. This calculated 60
min/hour divided by 80.79 minute/patient = 0.7427 patients per hour.
2. Calculate the probability of zero patients in the system (P0), probability of one patient (P1),
and the probability of two or more patients simultaneously arriving during the night shift.
P(0) = (1 – lambda/mu) = (1 - .0212/.7427) = 0.9714 the probability of no patients in the
system is over 97 percent.
P(1) = (1 – lambda/mu)(lambda/mu)1 = (1 - .0212/.7427)(.0212/.7427) = 0.0278 the
probability of exactly 1 patient in the system is 2.78 percent.
The probability that 2 or more patients are in the system is 1 – (P(1) + P(0)) = 1 – (.0278 +
0.9714) = 0.0008.
The probability of two or more patients occurring simultaneously on the night shift is
less than 0.1% (less than one chance in 1,000).
3. Using a criterion that if the probability is greater than 1 percent, a backup OR team should be
employed, make a recommendation to hospital administration.
A second OR is not needed at this hospital.
10-25
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Chapter 10 - Waiting Line Analysis and Simulation
ANALYTICS EXERCISE: Processing Customer Orders – Analyzing a Taco Bell Restaurant
1. Draw a diagram of the process using the format in Exhibit 9.5.
Customer arrives and
enters order queue
Greet
customer
Customer
places order
Customer enters pickup
window queue
Confirm order
and price
Customer departs
Service Champ
collects payment
Service Champ
serves food
Food Champ
prepares order
Service Champ
prepares drinks
2.
Consider a base case where a customer arrives every 40 seconds and the Customer
Service Champion can handle 120 customers per hour. There are two Food Champions each
capable of handling 100 orders per hour. How long should it take to be served by the restaurant
(from the time a customer enters the kiosk queue until her food is delivered)? Use queuing
models to estimate this.
 = 90, order = 120, prep = 100.
For the Customer Service Champion, the average service time is .0083 hours (40 seconds)
and the average time in line is .0250 hours (from queuing formulas) for a total of .0333 hrs.
(120 seconds). This is calculated using Model 1. For preparing the food for the two Food
Champions, the average service time is .01111 hours (40 seconds). Using model 3, and with
lambda/mu = .9, Lq from Exhibit 10.9 is 0.2285 customers. The time a customer spends in
the system for food to be prepared is .01254 hours or 45.1 seconds. The total time is 120 +
45.1 = 165.1 seconds or 2.75 minutes.
10-26
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Chapter 10 - Waiting Line Analysis and Simulation
3. On average, how busy are the Service Champion and the two Food Champions?
The Service Champion is busy  = 90/120 = 75% of the time and the Food Champions are
busy  = 90/200 = 45% of the time.
4. On average, how many cars do you expect to have in the drive-thru line? (Include those
waiting to place order and waiting for food.)
Lq due to the order taking = λ2/µ(µ - λ)
= 8100/3600 = 2.25 customers
Lq due to the food preparation = .2285 customers
Average total customers waiting in line = 2.4785 customers
Ls due to the order taking = 3 customers
Ls due to the food preparation = 1.1285 customers
Average number of customers in the system = 4.1285
5. If the restaurant runs a sale and the customer arrival rate increases by 20%, how would this
change the total time expected to serve a customer? How would this change the average
number of cars in the drive-thru line?
The current customer arrival rate is 90 per hour, this would go up to 90(1.2) = 108 per hour.
Ls due to order taking = (λ/µ – λ) = 108/(120-108) = 9.0 cars
10-27
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Chapter 10 - Waiting Line Analysis and Simulation
Ws = Ls/λ = 9.0/108 = .08333 hours = 5.0 minutes
Ls due to order preparation = Lq + λ/µ = .445 + 108/100 = 1.525 cars
(note λ/µ = 108/100 = 1.08, Lq = .445 from the spreadsheet)
Ws = Ls/λ = 1.525/108 = .01412 hours = .8472 minutes
Average total time in the system = 5.0 + .8742 = 5.8472 minutes
Average total number of cars in the drive thru = 9.0 + 1.525 = 10.525 cars
6. Currently, relatively few customers (less than ½ percent) order the Crunchwrap Supreme.
What would happen if we ran the sale and demand jumped on the Crunchwrap Supreme and
30% of our orders were for this item? Take a quantitative approach to answering this question.
Assume that the two processes remain independent.
Just to make this interesting, let’s assume that the customer arrival rate went to 108
customers per hour.
Assume the order taking rate stays the same at 120/hour.
For the order preparation, assume that 70% of the orders still take (60*60)/100 = 36
seconds on average to prepare, but 30% of them now take 72 seconds.
The average time to make an order would be .7(36) + .3(72) = 46.8 seconds
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Chapter 10 - Waiting Line Analysis and Simulation
So they could prepare (60*60)/46.8 = 76.92 orders per hour (each Food Champ).
Ls due to order taking = λ/(µ – λ) = 108/(120-108) = 9.0 cars
Ws = Ls/λ = 9.0/108 = .08333 hours = 5.0 minutes  this is the same as before.
Ls due to order preparation = Lq + λ/µ = 1.3449 + 108/76.92 = 2.749 cars
(note λ/µ = 108/76.92 = 1.4, Lq = 1.3449 from exhibit 10.9.)
Ws = Ls/λ = 2.749/108 = .02545 hours = 1.527 minutes
Average total number of cars in the drive thru = 11.749 cars
Average total time in the system = 6.527 minutes Not much of an impact.
7. For the type of analysis done in this case, what are the key assumptions? What would be the
impact on our analysis if these assumptions were not true?
A big assumption is that there is no interference between the two processes. This would
occur if one process was causing a delay in the other process for some reason. It does not
appear that this would be true in this process. We are also assuming the Food Champs are
working independent, not as a team. Other major assumptions are the distribution
associated with arrival and service rates, and that these rates are valid during the peak
noon period.
8. Could this type of analysis be used for other service type businesses? Give examples to
support your answer.
Yes, many examples can be given.
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