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93.Given the following data, find the expected breakdown cost. The cost per breakdown is $200.

Number of breakdowns per week

0

 

1

 

2

 

 

3

4

Weekly frequency

5

 

12

 

10

 

 

18

5

 

 

 

 

 

 

 

 

 

 

 

Number of breakdowns per week

0

 

1

2

 

3

 

4

 

Total

Weekly Probability

.10

 

.24

.20

 

.36

.10

 

1.00

Expected number of breakdowns = (0 * .10) + (1 * .24) + (2 * .20) + (3 * .36) + (4 * .10) = 2.12 Expected cost of breakdowns = 2.12 * $200 = $424. (Maintenance, moderate) {AACSB: Analytic Skills}

94.Given the following data, find the expected breakdown cost. The cost per breakdown is $100.

Number of breakdowns

0

 

1

 

2

3

 

 

 

Monthly frequency

5

 

20

23

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Number of breakdowns

0

 

 

 

1

 

2

 

3

Total

Monthly probability

.10

 

 

.40

 

.46

 

.04

1.000

Expected number of breakdowns per month = (0 * .10) + (1 * .40) + (2 * .46) +

(3 * .04) = 1.44; Expected cost of breakdowns = 1.44 * $100 = $144 (Maintenance, moderate) {AACSB: Analytic Skills}

95. Great Southern Consultants Group's computer system has been down several times over the past few months, as shown below.

Number of breakdowns

0

1

2

3

4

Monthly frequency

9

2

4

4

1

Each time the system is down, the firm loses an average of $400 in time and service expenses. They are considering signing a contract for preventive maintenance. With preventive maintenance, the system would be down on average only 0.5 per month. The monthly cost of preventive maintenance would be $200 a month. Which is cheaper, breakdown or preventive maintenance?

Number of breakdowns

0

1

2

3

4

Total

Monthly probability

0.45

0.10

0.20

0.20

0.05

1.00

Expected number of breakdowns per month = (0 * .45) + (1 * .10) + (2 * .20) + (3 * .20) + (4 * .05) = 1.30; Expected cost of breakdowns per month = 1.30 * $400 = $520

Preventive maintenance cost per month = (.5 * $400) + $200 = $400; Preventive maintenance is more cost-effective. (Maintenance, moderate) {AACSB: Analytic Skills}

471

96.

 

A system has four components in a series. What is the reliability of the system?

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Component

 

1

 

2

 

3

 

 

 

 

4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Reliability

 

.90

 

.95

 

.90

 

 

 

.99

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0.90 * 0.95 * 0.90 * 0.99 = .7618 (Reliability, moderate) {AACSB: Analytic Skills}

 

97.

 

A system consists of four components in series. The reliability of each component is 0.96. What is

 

 

the reliability of the system?

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

The reliability of the system is R = (0.96)4 = 0.8493 (Reliability, moderate) {AACSB: Analytic

 

 

Skills}

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

98.

 

A system has six components in series. Each component has a reliability of 0.99. What is the

 

 

reliability of the system?

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(0.99)6 = 0.9415 (Reliability, moderate) {AACSB: Analytic Skills}

 

 

 

 

99.

 

Tiger Island Fabricators, which builds offshore oil platforms, has been experiencing problems with

 

 

its profiling machine, a computer-driven device that cuts the ends of pipe so that it can be welded

 

 

to another pipe, as shown in the data below.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Number of breakdowns

0

 

1

 

 

2

 

 

3

 

 

4

5

 

 

 

 

 

 

 

 

 

 

Breakdown frequency

 

 

2

 

2

 

 

2

 

 

6

 

 

7

1

 

 

 

 

 

 

 

 

 

 

Each time a machine breaks down, the company loses about $3,000. If the company implements

 

 

preventive maintenance, it will be able to reduce the number of breakdowns to one per month.

 

 

Preventive maintenance costs would be $500 a month. Is preventive maintenance a cost-effective

 

 

option?

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Number of breakdowns

0

 

1

 

 

2

 

 

3

 

 

4

5

 

Total

 

 

 

 

 

 

Breakdown frequency

.10

 

.10

 

.10

 

 

.30

 

.35

.05

 

1.00

 

 

 

 

 

 

 

Expected number of breakdowns = (0 * .10) + (1 * .10) + (2 * .10) + (3 * .30) + (4 * .35) +

 

 

(5 * .05) = 2.85; Expected cost of breakdowns per month = 2.85 * $3,000 = $8,550; Cost of

 

 

preventive maintenance = (1 * $3,000) + $500 = $3,500. It is about three times more expensive

 

 

to suffer breakdowns than to perform preventive maintenance. (Maintenance, moderate)

 

 

{AACSB: Analytic Skills}

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

100.

 

The diagram below identifies the elements of service as provided by a soft drink vending machine.

 

 

Each element has an estimate of its own reliability, independent of the others. What is the

 

 

 

reliability of the "system"?

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Took my

 

 

Made

 

 

Dispensed

 

 

 

Dispensed

 

 

 

Power

 

Out of

 

 

Couldn't

 

 

Wouldn't

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

make

 

 

take my

 

 

money

 

wrong

 

 

 

wrong

 

 

 

 

warm

 

 

 

failed

 

 

stock

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

change

 

 

dollar bill

 

 

.85

 

change

 

 

beverage

 

 

 

beverage

 

 

 

.995

 

 

.85

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

.90

 

 

 

.95

 

 

 

 

.98

 

 

 

 

 

 

 

 

 

.98

 

 

.60

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Reliability = .85 * .90 * .95 * .98 * .995 * .85 * .98 * .60=0.3542 (Reliability, moderate) {AACSB: Analytic Skills}

472

101.Century Digital Phone advertises phone battery life (on standby) of up to three days. The standard deviation is thought to be five hours. Tina Talbot, an employee at CDP, tested 10 of these batteries for 72 hours. One failed at 40 hours; one failed at 62 hours; one failed at 70 hours. All others

completed the test. Calculate FR(%), FR(N), and MTBF.

FR(%) = 3/10 or 30%. FR(N) = 3/[(72*10)-32-10-2] = 3/676 = .00444. MTBF is 1/FR(N) =

225.3hours. (Reliability, moderate) {AACSB: Analytic Skills}

102.The Everstart is a battery with an intended design life of 72 months. Stephanie Bradley recently put five of these batteries through accelerated testing (the company couldn't wait six years) to simulate failure patterns. The test results had one failure at 26 months, one failure at 32 months, one failure at 50 months, and one failure at 62 months. Calculate FR(%), FR(N), and MTBF.

FR(%) = 4/5 or 80%. FR(N) = 4/(72+26+32+50+62) =4/242 = .0165. The MTBF is 1/FR(N) =

60.5(Reliability, moderate) {AACSB: Analytic Skills}

103.The academic service commonly referred to as "registration" consists of several smaller components: advising, registration for courses, fee assessment, financial aid calculations, and fee payment. Each of these modules operates independently and has some probability of failure for each student. If the five probabilities which accompany these services are 95%, 90%, 99%, 98%, and 99%, what is the "reliability" of the entire product from the student's perspective—the probability that all five will work according to plan?

Reliability is .95 * .90 * .99 * .98 * .99 = .82123 (Reliability, moderate) {AACSB: Analytic Skills}

104.A simple electrical motor has three components: windings, armature, and housing. These three components have reliabilities of .9998, .9992, and .9999. There is no possibility of redundant parts. What is the reliability of the motor? Round your answer to four decimal places.

0.9998 * 0.9992 * 0.9999 = 0.9989 (Reliability, moderate) {AACSB: Analytic Skills}

105.A simple electrical motor has three components: windings, armature, and housing. These three components have reliabilities of .97, .992, and .999. There is no possibility of redundant parts. The motor must have an overall reliability of 0.980, according to the product line manager who will use the motor as an input. What would you do to redesign the motor to meet this specification? Discuss, including a recalculation to meet the standard.

Since no backup is possible, individual components must be redesigned. The windings represent the weak link, and are the obvious choice. If only the windings are improved, their new reliability must be at least 0.980 / (0.992* 0.999) = 0.98889 (Reliability, moderate) {AACSB: Analytic Skills}

106.A product has four components A, B, C, and D. The finished product must have a reliability of .95. The first three components come from a supplier, and have reliabilities of .99, .98, and .995. The fourth component is being designed now. What must the reliability of component D be in order to meet the product reliability condition?

System reliability must be at least .99 x .98 x .995 x D = .95. Component D's reliability must be at least 0.95 / ( 0.99 * 0.98 * 0.995) = 0.9841 (Reliability, moderate) {AACSB: Analytic Skills}

473

107.A product has three components X, Y, and Z. X has reliability of 0.991; Y has reliability of 0.993. If Z has reliability of 0.991, what is the reliability of the entire product? Can Z be redesigned to be reliable enough for the entire product to have reliability of 0.99? Explain.

The product has reliability 0.991 * 0.993 * 0.991 = 0.9752; No: the required component reliability is impossible. 0.99 / (0.991 * 0.993) = 1.006 (Reliability, moderate) {AACSB: Analytic Skills}

108.A component must have reliability of .9925. Two technologies are available for this component: one produces a component with .999 reliability at a cost of $2000. Another produces a component with .73 reliability at a cost of $450. Which is cheaper: one high quality component or a parallel set

of inferior components?

First determine how many of the 0.73 reliable components are needed: 2: 0.9271 3: 0.980317 4: 0.994686 Four are needed, which will cost $1800. This is cheaper than the $2000 single high-quality component. (Reliability, difficult) {AACSB: Analytic Skills}

109.General Grant must send orders to General Butler. Carrier pigeons are the medium of choice. A single pigeon has a .7 probability of arriving at the proper destination in a timely fashion. How many pigeons, each carrying an identical set of orders, must Grant send in order for him to have 98% confidence that the orders reached General Butler?

2: 0.91 3: 0.973 4: 0.9919 One, two, or three is not enough; four will easily achieve the .98 reliability. (Reliability, difficult) {AACSB: Analytic Skills}

110. Which product design below has the higher system reliability?

 

 

 

A

 

 

 

 

 

 

 

B

 

 

 

 

R1

 

 

R2

 

 

R3

 

R1

 

 

R2

 

 

R3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

.98

 

 

.99

 

 

.985

 

.98

 

 

.98

 

 

.98

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

.95

 

.95

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

A's reliability is .98 * .99 * .985 = .9556. B's reliability is .98 * (.98 + .02 * .95) * (.98 + .02 *

.95) = .9780. B has a higher reliability. (Reliability, moderate) {AACSB: Analytic Skills}

474

111. Which product design below has the higher system reliability?

 

 

 

A

 

 

 

 

 

 

 

B

 

 

 

 

R1

 

 

R2

 

 

R3

 

R1

 

 

R2

 

 

R3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

.99

 

 

.99

 

 

.99

 

.99

 

 

.95

 

 

.96

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

.95

 

.95

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

A's reliability is .99 * .99 * .99 = .9703. B's reliability is .99 * (.95 + .05 * .95) * (.96 + .04 *

.95) = .9856. B has a higher reliability. (Reliability, moderate) {AACSB: Analytic Skills}

112.Consider a product that is "settled in." Its MTBF distribution has been found to be normal with a mean of 10,000 hours and a standard deviation of 100 hours. What is the probability of a breakdown before 8,000 hours? Before 9,000 hours? Would you prefer a policy of preventive

maintenance, or a policy of breakdown maintenance, on this product? Explain your choice.

Both of the specified times are far below the mean: 8,000 hours is twenty standard deviations away, and 9,000 hours is ten standard deviations away. In both instances, the normal areas table computes the probability of failure to be approximately zero. Most students will opt for preventive maintenance; the standard deviation is quite small compared to the mean, so that preventive maintenance can pretty reliably be used. (Maintenance, moderate) {AACSB: Analytic Skills}

475

113.A product design team is preparing to build a new doohickey. A doohickey consists of one A module, one B module, and one C module. There are different versions of these modules available in the company's design library. For example, there are two choices for A: A1

is .99 reliable, while A2 is .975 reliable. The table below details the choices available, along with the cost of each choice.

Module variation

Reliability

Cost, each

A1

.99

$17

A2

.975

$10

B1

.995

$4

B2

.992

$3

C1

.98

$2

C2

.90

$0.50

C3

.60

$0.25

Help the design team by selecting the least costly version of a doohickey that has system reliability of at least .96. Draw a schematic of your finished design.

Students will need to solve this problem by trial and error. Here are some possible designs.

(I) A10.99 * B10.995 * C10.98 = 0.96535 meets the reliability specification but costs $23. (II) A20.975 * B20.992 * C10.98 = 0.947856 costs $15 but does not meet the specification.

There are a dozen design alternatives which have no redundant parts. There are many more choices with one or more redundancies built in. The reliability of A2 + B2 is 0.975 * 0.992 = 0.9672, which means that redundancies at C are an option.

(III) A20.975 * B20.992 * C2+C20.99 = 0.957528 costs $16 but does not meet the test. (IV) A20.975 * B20.992 * C1+C30.992 = 0.95946 costs $15.25 but does not meet the test.

(V) A20.975 * B20.992 * C2+C3+C3+C30.9936 = 0.96101 costs $14.25 and meets the test. It is possible that even cheaper solutions exist. The schematic for this design appears below.

A

 

 

B

 

 

 

C

A2

 

 

B2

 

 

 

C2

 

.975

 

 

.992

 

 

.90

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

C3

 

 

 

 

 

 

.60

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

C3

 

 

 

 

 

 

.60

 

C3

.60

(Reliability, difficult) {AACSB: Analytic Skills}

476

114.Consider a product that is "settled in." Its MTBF distribution has been found to be normal with a mean of 1,000 hours and a standard deviation of 250 hours. What is the probability of a breakdown before 750 hours? Before 500 hours? Would you prefer a policy of preventive maintenance, or a

policy of breakdown maintenance, on this product? Explain your choice.

Failure probabilities are not negligible. The value 750 hours is one standard deviation away from the mean, and the area under the normal curve left of that point is 0.15866. The value 500 hours is two standard deviations away, and the area representing failure is 0.02275. Most students will opt for breakdown maintenance. The standard deviation is relatively large compared to the mean; preventive maintenance would have to be performed relatively early and often. (Maintenance, moderate) {AACSB: Analytic Skills}

115.A high school senior is seeking admission into her state's flagship university. The admission requirements are as follows. The student must meet at least ONE of the criteria A, B, C. The student must meet criterion D. These criteria are detailed below. And the student must file a complete application, file a medical form, and pay an application fee by a stated date.

A:score a composite 28 on the ACT exam

B:have a high school GPA of 3.0 or greater

C:place in the top ten percent of the high school class

D:complete all courses in the state-mandated "core college preparatory" curriculum (CCPC); OR complete all but one course in CCPC with a 3.5 GPA on all CCPC courses taken.

Using the tools of reliability analysis with redundancy, translate these conditions into the appropriate reliability schematic.

The connection to reliability is clear: each element in the system has some chance of failure, and the probability of system success is driven by the probability of success of individual components. Students must treat the "or" conditions as redundancies, or parallel items; and they must treat "and" items as items in series. Criterion D is really two items in parallel, which might be labeled D1 and D2. The elements of application, medical, and payment are each separate events, and might be labeled E, F, and G. A possible solution appears below.

A

 

D1

 

E

 

F

 

G

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

B D2

C

(Reliability, moderate) {AACSB: Analytic Skills}

477

MODULE A: DECISION-MAKING TOOLS

TRUE/FALSE

1.Analytic decision making is based on logic and considers all available data and possible alternatives.

True (The decision process in operations, easy) {AACSB: Analytic Skills}

2.The last step in the analytic decision process clearly defines the problem and the factors that influence it.

False (The decision process in operations, easy) {AACSB: Analytic Skills}

3.A state of nature is an occurrence of a situation over which the decision maker has little or no control.

True (Fundamentals of decision making, easy)

4.In a decision tree, a square symbol represents a state of nature node.

False (Fundamentals of decision making, moderate)

5.If a decision maker can assign probabilities of occurrence to the states of nature, then the decision is being made under conditions of uncertainty.

False (Types of decision-making environments, moderate)

6.An example of a conditional value would be the payoff from selecting a particular alternative when a particular state of nature occurs.

True (Types of decision-making environments, moderate)

7.The maximax criterion of decision making requires that all decision alternatives have an equal probability of occurrence.

False (Types of decision-making environments, easy)

8.The maximin criterion is pessimistic, while the maximax criterion is optimistic.

True (Types of decision-making environments, easy)

9.If a decision maker knows for sure which state of nature will occur, he/she is making a decision under certainty.

True (Types of decision-making environments, moderate)

10.The expected value with perfect information assumes that all states of nature are equally likely.

False (Types of decision-making environments, easy)

11.An example of expected monetary value would be the payoff from selecting a particular alternative when a particular state of nature occurs.

False (Decision tables, moderate)

12.The expected monetary value of a decision alternative is the sum of all possible payoffs from the alternative, each weighted by the probability of that payoff occurring.

True (Types of decision-making environments, easy)

478

13.If a decision maker has to make a certain decision only once, expected monetary value is a good indication of the payoff associated with the decision.

False (Types of decision-making environments, moderate)

14.The expected value of perfect information is the same as the expected value with perfect information.

False (Decision tables, moderate)

15.Decision trees and decision tables can both solve problems requiring a single decision, but decision tables are the preferred method when a sequence of decisions is involved.

False (Decision trees, easy)

16.In a decision tree, the expected monetary values are computed by working from right to left.

True (Decision trees, moderate)

MULTIPLE CHOICE

17.Which of the following is not considered a step in the decision-making process?

a. Clearly identify the problem.

c.Develop objectives.

d.Evaluate alternatives.

b. Select the best alternative.

e. Minimize costs whenever possible.

e (The decision process in operations, moderate)

18.The first step, and a key element, in the decision-making process is to

a.consult a specialist

b.clearly define the problem

c.develop objectives

d.monitor the results

e.select the best alternative

b (The decision process in operations, easy)

19.In terms of decision theory, an occurrence or situation over which the decision maker has no control is called a(n)

a.decision under uncertainty

b.decision tree

c.state of nature

d.alternative

e.none of the above

c (Fundamentals of decision making, easy)

479

20.A tabular presentation that shows the outcome for each decision alternative under the various possible states of nature is called a(n)

a.isoquant table

b.payback period matrix

c.payoff table

d.feasible region

e.decision tree

c (Decision tables, easy)

21.The decision criterion that would be used by an optimistic decision maker solving a problem under conditions of uncertainty would be the

a.expected monetary value criterion

b.equally likely criterion

c.maximax criterion

d.maximin criterion

e.minimin criterion

c(Types of decision-making environments, moderate)

22.A decision maker who uses the maximin criterion when solving a problem under conditions of uncertainty is

a.an optimist

b.a pessimist

c.an economist

d.an optometrist

e.making a serious mistake; maximin is not appropriate for conditions of uncertainty

b (Types of decision-making environments, moderate)

23.Expected monetary value is most appropriate for problem solving that takes place

a.when conditions are average

b.when all states of nature are equally likely

c.when all alternatives are equally likely

d.under conditions of uncertainty

e.under conditions of risk

e(Types of decision-making environments, moderate)

24.The expected value with perfect information

a.is appropriate when solving problems under conditions of certainty

b.requires that each decision alternative have a known probability of occurrence

c.is an input into the calculation of the expected value of perfect information

d.is the average of the maximax and the maximin

e.none of the above

c (Types of decision-making environments, moderate)

480

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