Sunday, January 29, 2017

MBA Operations Management - Test Bank - Chapter 17

Operations Management, 12e (Heizer/Render/Munson)
Chapter 17  Maintenance and Reliability 

Section 1   The Strategic Importance of Maintenance and Reliability

1) The Orlando Utilities Commission uses a computerized maintenance management program and devotes significant dollar and labor resources to power plant maintenance, because the costs of unexpected failure are incredibly high.
Answer:  TRUE
Diff: 1
Key Term:  Maintenance
AACSB:  Information technology

2) Maintenance includes all activities involved in keeping a system's equipment in working order.
Answer:  TRUE
Diff: 1
Key Term:  Maintenance

3) Reliability is the probability that a machine part or product will function properly for a specified time regardless of conditions.
Answer:  FALSE
Diff: 2
Key Term:  Reliability

4) The objective of maintenance and reliability is to maintain the capability of the system.
Answer:  TRUE
Diff: 2

5) Which of the following statements about maintenance at the Orlando Utilities Commission is FALSE?
  1. A) There are at least two types of preventive maintenance, including an annual maintenance and a less frequent overhaul schedule.
  2. B) Its preventive maintenance program has earned the company top rankings and its competitive advantage.
  3. C) Each power-generating unit is taken off-line every three years for a complete overhaul.
  4. D) Each of its power-generating units is taken off-line for maintenance every one to three weeks.
  5. E) Costs associated with breakdowns are several times higher than costs arising from preventive maintenance.
Answer:  D
Diff: 2

6) The objective of maintenance and reliability is to:
  1. A) ensure that breakdowns do not affect the quality of the products.
  2. B) ensure that no breakdowns will ever occur.
  3. C) ensure that preventive maintenance costs are kept as low as possible.
  4. D) maintain the capability of the system.
  5. E) ensure that maintenance employees are fully utilized.
Answer:  D
Diff: 2

7) What is the probability that a product will function properly for a specified time under stated conditions?
  1. A) functionality
  2. B) maintenance
  3. C) durability
  4. D) reliability
  5. E) fitness for use
Answer:  D
Diff: 2
Key Term:  Reliability

8) Which of the following is a reliability tactic?
  1. A) improving individual components
  2. B) increasing repair speed
  3. C) providing redundancy
  4. D) A and C
  5. E) A, B, and C
Answer:  D
Diff: 2
Key Term:  Reliability
Objective:  LO 17.1 Describe how to improve system reliability

9) ________ consists of all activities involved in keeping a system's equipment in working order.
Answer:  Maintenance
Diff: 1
Key Term:  Maintenance

10) ________ is the probability that a machine part or product will function properly for a specified time under stated conditions.
Answer:  Reliability
Diff: 2
Key Term:  Reliability

11) Describe how the Orlando Utilities Commission obtains competitive advantage through its maintenance practices.
Answer:  The company recognizes the importance of preventive maintenance to the business it is in; this is driven in part by the very large cost of forced power outages. Their maintenance plan involves annual checkups for each plant and complete overhauls every three years. The firm has well-developed and highly detailed preventive maintenance procedures. Maintenance work orders are computerized for efficiency.
Diff: 3
Key Term:  Preventive maintenance

12) Identify the two reliability tactics and the two maintenance tactics.
Answer:  The reliability tactics are: (1) improving individual components and (2) providing redundancy. The maintenance tactics are: (1) implementing or improving preventive maintenance and (2) increasing repair capabilities or speed.
Diff: 2

13) Define reliability.
Answer:  Reliability is the probability that a machine part or product will function properly for a specified time under stated conditions.
Diff: 3
Key Term:  Reliability

14) Define maintenance.
Answer:  Maintenance is all activities involved in keeping a system's equipment in working order.
Diff: 3
Key Term:  Maintenance

Section 2   Reliability

1) The product failure rate FR(%) is the percent of failures among the total number of products tested.
Answer:  TRUE
Diff: 2
Objective:  LO 17.3 Determine mean time between failures (MTBF)

2) The MTBF (mean time between failures) is calculated as the reciprocal of the number of failures during a period of time.
Answer:  TRUE
Diff: 2
Key Term:  Mean time between failures (MTBF)
Objective:  LO 17.3 Determine mean time between failures (MTBF)

3) If the mean time between failures has been calculated to be 2,000 hours, then FR(N) = 0.0005 failures/unit-hour.
Answer:  TRUE
Diff: 2
Key Term:  Mean time between failures (MTBF)
AACSB:  Analytical thinking
Objective:  LO 17.3 Determine mean time between failures (MTBF)

4) The reliability of a system in which each individual component must function in order for the entire system to function, and in which each component has its own unique reliability, independent of other components, is the product of the probabilities of each of those components.
Answer:  TRUE
Diff: 2
Key Term:  Reliability
Objective:  LO 17.2 Determine system reliability

5) Adding an additional part to a component or product ordinarily reduces reliability by introducing an additional source of failure.
Answer:  TRUE
Diff: 2
Key Term:  Reliability
AACSB:  Reflective thinking
Objective:  LO 17.2 Determine system reliability

6) A redundant part or component increases reliability because it is connected in parallel, not in series.
Answer:  TRUE
Diff: 2
Key Term:  Redundancy
Objective:  LO 17.1 Describe how to improve system reliability


7) A redundant part decreases reliability if the reliability of the redundant part is lower than that of the part it is backing up.
Answer:  FALSE
Diff: 2
Key Term:  Redundancy
AACSB:  Reflective thinking
Objective:  LO 17.2 Determine system reliability

8) What is the reliability of a four-component product, with components in series, and component reliabilities of .90, .95, .98, and .99?
  1. A) 0.8295
  2. B) 0.90
  3. C) 0.955
  4. D) 0.99
  5. E) 0.945
Answer:  A
Diff: 2
Key Term:  Reliability
AACSB:  Analytical thinking
Objective:  LO 17.2 Determine system reliability

9) A system is composed of three components A, B, and C. All three must function for the system to function. There are currently no backups in place. The system has a reliability of 0.966. If a backup is installed for component A, the new system reliability will be:
  1. A) unchanged.
  2. B) less than 0.966.
  3. C) greater than it would be if a backup were also installed for component B.
  4. D) greater than 0.966
  5. E) none of the above
Answer:  D
Diff: 2
Key Term:  Redundancy
AACSB:  Analytical thinking
Objective:  LO 17.2 Determine system reliability

10) A system has three components in series with reliabilities 0.9, 0.7, and 0.5. What is the system reliability?
  1. A) 0.315
  2. B) 0.500
  3. C) 0.700
  4. D) 0.900
  5. E) 2.100
Answer:  A
Diff: 2
Key Term:  Reliability
AACSB:  Analytical thinking
Objective:  LO 17.2 Determine system reliability

11) A system has three components in parallel with reliabilities 0.9, 0.7, and 0.5. What is the system reliability?
  1. A) 0.315
  2. B) 0.700
  3. C) 0.900
  4. D) 0.985
  5. E) 2.100
Answer:  D
Diff: 3
Key Term:  Redundancy
AACSB:  Analytical thinking
Objective:  LO 17.2 Determine system reliability

12) Components A, B, and C are connected in series. Component D is connected in parallel to component B. Which of the following statements is true?
  1. A) The system works only if A works, B or C works, and D works.
  2. B) Component B must work for the system to work.
  3. C) The system works when A works, C, works, and either B or D works.
  4. D) Components B and C are backups to A.
  5. E) The system works if D works, and any of A, B, or C works.
Answer:  C
Diff: 2
Key Term:  Reliability
AACSB:  Analytical thinking
Objective:  LO 17.2 Determine system reliability


13) A job consists of a series of three tasks. Task 1 is performed correctly 98% of the time, task 2 is performed correctly 99% of the time, and task 3 is performed correctly 97% of the time. What is the reliability of this job?
  1. A) 91.27%
  2. B) 94.11%
  3. C) 97.00%
  4. D) 98.00%
  5. E) 99.00%
Answer:  B
Diff: 2
Key Term:  Reliability
AACSB:  Analytical thinking
Objective:  LO 17.2 Determine system reliability
14) As the number of components in a system connected in a series decreases, all other things being equal, the reliability of the system usually:
  1. A) increases.
  2. B) stays the same.
  3. C) decreases.
  4. D) increases, then decreases.
  5. E) decreases, then increases.
Answer:  A
Diff: 2
Key Term:  Reliability
AACSB:  Analytical thinking
Objective:  LO 17.1 Describe how to improve system reliability

15) What is the reliability of the three components connected in series shown below?



  1. A) 0.799425
  2. B) 0.85000
  3. C) 0.91333
  4. D) 0.95000
  5. E) 2.79
Answer:  A
Diff: 2
Key Term:  Reliability
AACSB:  Analytical thinking
Objective:  LO 17.2 Determine system reliability


16) Consider a product with three components in series, with reliabilities of 0.90, 0.80, and 0.99 for components A, B, and C, respectively. Furthermore, component B uses a backup that also has a reliability of 0.80. What is the reliability of the system?
  1. A) 0.50000
  2. B) 0.71280
  3. C) 0.80000
  4. D) 0.85536
  5. E) 3.49000
Answer:  D
Diff: 2
Key Term:  Reliability
AACSB:  Analytical thinking
Objective:  LO 17.2 Determine system reliability
17) Ten high-technology batteries are tested for 200 hours each. One failed at 20 hours; all others completed the test. FR(%) is ________ and MTBF is ________.
  1. A) 10%; 1/1820
  2. B) 90%; 1/1820
  3. C) 10%; 1820 hours
  4. D) 10%; 1980 hours
  5. E) 10%; 2000 hours
Answer:  C
Diff: 2
Key Term:  Mean time between failures (MTBF)
AACSB:  Analytical thinking
Objective:  LO 17.3 Determine mean time between failures (MTBF)

18) Ten high-technology batteries are tested for 200 hours each. One failed at 50 hours; all others completed the test. FR(%) is ________ and FR(N) is ________.
  1. A) 10%; 1/1850
  2. B) 10%; 1/2000
  3. C) 25%; 1850 hours
  4. D) 90%; 1/2000
  5. E) 10%; 1/1950
Answer:  A
Diff: 2
AACSB:  Analytical thinking
Objective:  LO 17.3 Determine mean time between failures (MTBF)


19) Ten high-technology batteries are tested for 200 hours each. One failed at 20 hours; another failed at 140 hours; all others completed the test. FR(%) is ________ and MTBF is ________.
  1. A) 20%; 880 hours
  2. B) 10%; 1980 hours
  3. C) 20%; 1760 hours
  4. D) 20%; 1000 hours
  5. E) 80%; 920 hours
Answer:  A
Diff: 2
Key Term:  Mean time between failures (MTBF)
AACSB:  Analytical thinking
Objective:  LO 17.3 Determine mean time between failures (MTBF)

20) MTBF measures the average:
  1. A) calendar time between failures.
  2. B) operating time between failures.
  3. C) number of failures per unit time.
  4. D) number of operations between failures.
  5. E) downtime per breakdown.
Answer:  B
Diff: 2
Key Term:  Mean time between failures (MTBF)
AACSB:  Reflective thinking
Objective:  LO 17.3 Determine mean time between failures (MTBF)
21) A system is composed of four parts, J, K, L, and M. All four must function for the system to function. The four component reliabilities are .99, .98, .992, and .998. The designers are considering putting an 80% reliable backup at K. This backup will change the system reliability from ________ to ________.
  1. A) 0.9762; 0.9605
  2. B) 0.9605; 0.9762
  3. C) 0.9605; some smaller value
  4. D) 0.9605; 0.996
  5. E) 0.98; 0.99
Answer:  B
Diff: 2
Key Term:  Reliability
AACSB:  Analytical thinking
Objective:  LO 17.2 Determine system reliability


22) Suppose that a process is comprised of five components, each having a 75% success rate. By how much would reliability change if the process switched from having the components in series to having the components in parallel?
  1. A) no change
  2. B) 100% increase
  3. C) 76.2% increase
  4. D) 76.2 % decrease
  5. E) 50% increase
Answer:  C
Diff: 2
Key Term:  Reliability
AACSB:  Analytical thinking
Objective:  LO 17.2 Determine system reliability

23) A two-component process has an 81% success rate in series and a 99% success rate in parallel. If each component has the same reliability, what is the reliability of an individual component?
  1. A) 50%
  2. B) 90%
  3. C) 99%
  4. D) 18%
  5. E) 81%
Answer:  B
Diff: 2
Key Term:  Reliability
AACSB:  Analytical thinking
Objective:  LO 17.2 Determine system reliability
24) How is MTBF related to FR(N)?
  1. A) MTBF is measured in hours, while FR(N) is measured in years.
  2. B) MTBF is normally distributed, with FR(%) as its mean and FR(N) as its standard deviation.
  3. C) MTBF is the reciprocal of FR(N).
  4. D) Both MTBF and FR(N) increase when breakdown maintenance is replaced by preventive maintenance.
  5. E) MTBF and FR(N) are unrelated concepts.
Answer:  C
Diff: 2
Key Term:  Mean time between failures (MTBF)
Objective:  LO 17.3 Determine mean time between failures (MTBF)

25) ________ is the expected time between a repair and the next failure of a component, machine, process, or product.
Answer:  Mean time between failures or MTBF
Diff: 2
Key Term:  Mean time between failures (MTBF)
Objective:  LO 17.3 Determine mean time between failures (MTBF)

26) The inverse of the mean time between failures is the ________.
Answer:  number of failures during a period of time or FR(N)
Diff: 2
Key Term:  Mean time between failures (MTBF)
Objective:  LO 17.3 Determine mean time between failures (MTBF)

27) ________ is the use of a component in parallel to raise reliabilities.
Answer:  Redundancy
Diff: 2
Key Term:  Redundancy
Objective:  LO 17.1 Describe how to improve system reliability

28) What is the impact on system reliability of adding parts or components in parallel?
Answer:  This will increase the reliability of the system by introducing redundancy.
Diff: 1
Key Term:  Redundancy
Objective:  LO 17.1 Describe how to improve system reliability

29) Increasing the number of parts or components in a product tends to reduce its reliability. Why is this true only when adding components in series?
Answer:  Adding parts in series involves an additional multiplication by a value less than one, so that reliability must fall. Adding parts in parallel (the redundancy concept) increases reliability because only one part of the parallel system needs to function.
Diff: 2
Key Term:  Reliability
AACSB:  Reflective thinking
Objective:  LO 17.2 Determine system reliability
30) Explain carefully how redundancy improves product reliability.
Answer:  A redundant part or component is connected in parallel with the primary part or component. "In parallel" means that either the original part or its backup needs to work, not that both must work at the same time. Redundancy increases reliability by providing an additional path (through the redundant part) to provide system reliability.
Diff: 2
Key Term:  Redundancy
Objective:  LO 17.1 Describe how to improve system reliability


31) "High reliability can be achieved in a product without having high reliability in the component parts. In fact, any reliability target, no matter how high, can be achieved with only mediocre parts, so long as enough of them are present." Discuss; an example may help.
Answer:  This carries the redundancy concept one step beyond the textbook, to having multiple redundancies on the same component. If only one part in a parallel system needs to work to provide reliability of that part of the system, then several parts in parallel should offer very high reliability. Example: four parts in parallel, each with only 0.50 reliability, provide reliability of .50 + .25 + .125 +. 0625 = .9375. Additional parts in parallel continue to improve reliability.
Diff: 3
Key Term:  Redundancy
AACSB:  Reflective thinking
Objective:  LO 17.1 Describe how to improve system reliability

32) What is FR(N)? How is it calculated? How are FR(N) and MTBF related?
Answer:  FR(N) is the number of failures during a period of time. It is the ratio of number of failures to number of unit-hours of operation time. MTBF is the reciprocal of FR(N).
Diff: 2
Key Term:  Mean time between failures (MTBF)
Objective:  LO 17.3 Determine mean time between failures (MTBF)

33) Ten high-intensity bulbs are tested for 100 hours each. One failed at 40 hours; all others completed the test. Calculate FR(%) and FR(N).
Answer:  FR(%) = 1/10 or 10%; operating hours = 9 × 100 + 40 = 940; FR(N) = 1/940 or 0.00106
Diff: 2
AACSB:  Analytical thinking
Objective:  LO 17.3 Determine mean time between failures (MTBF)

34) Ten high-intensity bulbs are tested for 100 hours each. One failed at 10 hours; all others completed the test. Calculate FR(%), FR(N) and MTBF.
Answer:  FR(%) = 1/10 or 10%; operating hours = 9 × 100 + 10 = 910; FR(N) = 1/910 or .0011, and
MTBF = 910 hours
Diff: 2
Key Term:  Mean time between failures (MTBF)
AACSB:  Analytical thinking
Objective:  LO 17.3 Determine mean time between failures (MTBF)

35) Ten high-intensity bulbs are tested for 100 hours each. One failed at 40 hours; another failed at 70 hours; all others completed the test. Calculate FR(%), FR(N), and MTBF.
Answer:  FR(%) = 2/10 or 20%; FR(N) = 2/910 or .0022; MTBF = 455 hours
Diff: 2
Key Term:  Mean time between failures (MTBF)
AACSB:  Analytical thinking
Objective:  LO 17.3 Determine mean time between failures (MTBF)
36) A product is composed of a series connection of four components with the following reliabilities. What is the reliability of the system?

Component 1 2 3 4
Reliability .90 .95 .97 .88

Answer:  The reliability of the system is R = (0.90)(0.95)(0.97)(0.88) = 0.7298
Diff: 2
Key Term:  Reliability
AACSB:  Analytical thinking
Objective:  LO 17.2 Determine system reliability

37) 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

Answer:  (0.90)(0.95)(0.90)(0.99) = .7618
Diff: 2
Key Term:  Reliability
AACSB:  Analytical thinking
Objective:  LO 17.2 Determine system reliability


38) A system consists of four components in series. The reliability of each component is 0.96. What is the reliability of the system?
Answer:  The reliability of the system is Rs = (0.96)4 = 0.8493
Diff: 2
Key Term:  Reliability
AACSB:  Analytical thinking
Objective:  LO 17.2 Determine system reliability

39) A system has six components in series. Each component has a reliability of 0.99. What is the reliability of the system?
Answer:  (0.99)6 = 0.9415
Diff: 2
Key Term:  Reliability
AACSB:  Analytical thinking
Objective:  LO 17.2 Determine system reliability
40) 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"?


Answer:  Reliability = (.85)(.90)(.95)(.98)(.995)(.85)(.98)(.60) = 0.3542
Diff: 2
Key Term:  Reliability
AACSB:  Analytical thinking
Objective:  LO 17.2 Determine system reliability

41) 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.
Answer:  FR(%) = 3/10 or 30%. FR(N) = 3/[(72)(7) + 40 + 62 + 70] = 3/676 = .00444. MTBF is 1/FR(N) = 225.3 hours.
Diff: 2
Key Term:  Mean time between failures (MTBF)
AACSB:  Analytical thinking
Objective:  LO 17.3 Determine mean time between failures (MTBF)


42) 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.
Answer:  FR(%) = 4/5 or 80%. FR(N) = 4/(72 + 26 + 32 + 50 + 62) = 4/242 = .0165. The MTBF is 1/FR(N) = 60.5 months.
Diff: 2
Key Term:  Mean time between failures (MTBF)
AACSB:  Analytical thinking
Objective:  LO 17.3 Determine mean time between failures (MTBF)

43) 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 that accompany these services are 95%, 90%, 99%, 98%, and 99%, respectively, what is the "reliability" of the entire product from the student's perspective, i.e., the probability that all five will work according to plan?
Answer:  Reliability is (.95)(.90)(.99)(.98)(.99) = .8212
Diff: 2
Key Term:  Reliability
AACSB:  Analytical thinking
Objective:  LO 17.2 Determine system reliability
44) 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.
Answer:  (0.9998)(0.9992)(0.9999) = 0.9989
Diff: 2
Key Term:  Reliability
AACSB:  Analytical thinking
Objective:  LO 17.2 Determine system reliability

45) 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 at least 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.
Answer:  Since no backup is possible, individual components must be redesigned. The windings represent the weak link, and represent the obvious choice (windings have to be improved at least some or overall reliability will not be able to exceed 0.980). If only the windings are improved, their new reliability must be at least 0.980 / [(0.992)(0.999)] = 0.988893. (The last decimal was rounded up to ensure at least .980 for the system.)
Diff: 2
Key Term:  Reliability
AACSB:  Analytical thinking
Objective:  LO 17.2 Determine system reliability


46) A product has four components A, B, C, and D. The finished product must have a reliability of at least .95. The first three components come from a supplier, and they have reliabilities of .99, .98, and .995, respectively. The fourth component is being designed now. What must the reliability of component D be in order to meet the product reliability condition?
Answer:  System reliability must be at least .99 × .98 × .995 × D = .95. Component D's reliability must be at least 0.95 / [(0.99)(0.98)(0.995)] = 0.98411. (The last decimal was rounded up to ensure at least .95 for the system.)
Diff: 2
Key Term:  Reliability
AACSB:  Analytical thinking
Objective:  LO 17.2 Determine system reliability

47) A product has three components X, Y, and Z. Product X has reliability of 0.991, and 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.
Answer:  The product has reliability of (0.991)(0.993)(0.991) = 0.9752. No--the required component reliability is impossible because X and Y together already fall below the target: (0.991)(0.993) = 0.984 < 0.99.
Diff: 2
Key Term:  Reliability
AACSB:  Analytical thinking
Objective:  LO 17.2 Determine system reliability
48) 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? (Hint: for the inferior components, you first need to determine the number of them that would be needed to ensure overall reliability of .9925.)
Answer:  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.
Diff: 3
Key Term:  Reliability
AACSB:  Analytical thinking
Objective:  LO 17.2 Determine system reliability

49) 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?
Answer:  2: 0.91 3: 0.973 4: 0.9919; one, two, or three is not enough; four will easily achieve the .98 reliability.
Diff: 3
Key Term:  Reliability
AACSB:  Analytical thinking
Objective:  LO 17.2 Determine system reliability


50) Suppose that a three-stage process had reliability ratings of .7, .8, and .9 at each station and that a failure at any station represented a failure for the entire process. If each station is given a redundant check with .9 reliability, what is the increase in system reliability?
Answer:  Old = (.7)(.8)(.9) = .504
New = [1 - (.3)(.1)][1 - (.2)(.1)][1 - (.1)(.1)] = (.97)(.98)(.99) = .941
New - Old = .437 or 43.7% improvement
Diff: 3
Key Term:  Reliability
AACSB:  Analytical thinking
Objective:  LO 17.2 Determine system reliability
51) Suppose that a three-stage process in a nuclear reactor had reliability ratings of .98 at each station and that only one stage needed to be successful for the process to work. If the plant wants to establish 6-sigma standards (≤ 3.4 defects per million), what % reliable redundant controls need to be added to each station.
Answer:  For 6-sigma standards, the reliability must be at least 0.9999966.
This problem essentially has backups for backups.
Without any redundant controls, reliability would be 1 - (.02)3 = 0.999992, which is not high enough.
Each stage with its backup will have a reliability equal to [1 - (.02)(1 - X )], where X is the reliability of the redundant control.
Thus, 1 - {1 -[1 - (.02)(1 -X)]}3 ≥ 0.9999966, or
0.0000034 ≥ {1 -[1 - (.02)(1 -X)]}3 or
0.0155036946 ≥ {1 -[1 - (.02)(1 -X)]}, or
0.0155036946 ≥ (.02)(1 -X), or
0.751847298 ≥ 1 - X, or
X ≥ 0.2482
Thus, each redundant control must be at least 24.82% reliable.
Diff: 3
Key Term:  Reliability
AACSB:  Analytical thinking
Objective:  LO 17.2 Determine system reliability

Section 3   Maintenance

1) Preventive maintenance is reactive.
Answer:  FALSE
Diff: 2
Key Term:  Preventive maintenance
Objective:  LO 17.4 Distinguish between preventive and breakdown maintenance

2) Preventive maintenance is nothing more than keeping the equipment and machinery running.
Answer:  FALSE
Diff: 2
Key Term:  Preventive maintenance
Objective:  LO 17.4 Distinguish between preventive and breakdown maintenance


3) Preventive maintenance implies that we can determine when a system needs service or will need repair.
Answer:  TRUE
Diff: 2
Key Term:  Preventive maintenance
Objective:  LO 17.4 Distinguish between preventive and breakdown maintenance

4) Preventive maintenance implies that machine sensors to detect variations from normal are inadequate.
Answer:  FALSE
Diff: 2
Key Term:  Preventive maintenance
Objective:  LO 17.4 Distinguish between preventive and breakdown maintenance
5) Electronic sensors have proven so effective that preventive maintenance is no longer necessary.
Answer:  FALSE
Diff: 2
Key Term:  Preventive maintenance
Objective:  LO 17.4 Distinguish between preventive and breakdown maintenance

6) Infant mortality refers to the high failure rate often encountered in the very early stages of the lifetime of a product.
Answer:  TRUE
Diff: 2
Key Term:  Infant mortality
Objective:  LO 17.4 Distinguish between preventive and breakdown maintenance

7) The MTBF distributions of products, machines, or processes that have "settled in," or gone beyond the infant mortality phase, often follow the normal distribution.
Answer:  TRUE
Diff: 2
Key Term:  Mean time between failures (MTBF)
Objective:  LO 17.4 Distinguish between preventive and breakdown maintenance

8) Recording the maintenance history of processes, machines, or equipment is important for preventive maintenance, but largely irrelevant for breakdown maintenance.
Answer:  FALSE
Diff: 1
Key Term:  Maintenance
Objective:  LO 17.5 Describe how to improve maintenance

9) The "full cost view of maintenance" results in more firms choosing a policy of breakdown maintenance, when compared to the "traditional view of maintenance."
Answer:  FALSE
Diff: 1
Key Term:  Maintenance
Objective:  LO 17.6 Compare preventive and breakdown maintenance costs


10) Small standard deviations in the MTBF distribution of a machine tend to support a policy of breakdown maintenance for that machine.
Answer:  FALSE
Diff: 2
Key Term:  Maintenance
Objective:  LO 17.4 Distinguish between preventive and breakdown maintenance

11) When identifying the optimal maintenance policy, the cost of inventory maintained to compensate for the downtime is a cost often ignored.
Answer:  TRUE
Diff: 2
Key Term:  Maintenance
Objective:  LO 17.6 Compare preventive and breakdown maintenance costs
12) An optimal maintenance policy strikes a balance between the costs of breakdown and preventive maintenance so that the total cost of maintenance is at a minimum.
Answer:  TRUE
Diff: 2
Key Term:  Maintenance
Objective:  LO 17.6 Compare preventive and breakdown maintenance costs

13) While breakdowns occur randomly, their frequency is somewhat predictable through such tools as the product failure rate, MTBF, and the breakdown costs model.
Answer:  TRUE
Diff: 2
Key Term:  Maintenance
Objective:  LO 17.5 Describe how to improve maintenance

14) The normal distribution is an appropriate model of:
  1. A) the high initial failure rates of product, machine, or processes.
  2. B) system reliability where components are connected in series.
  3. C) system reliability where components are connected in parallel.
  4. D) the MTBF distribution of products, machines, or processes that have "settled in."
  5. E) the full cost view of maintenance.
Answer:  D
Diff: 2
Key Term:  Mean time between failures (MTBF)
Objective:  LO 17.4 Distinguish between preventive and breakdown maintenance

15) Infant mortality refers to which one of the following examples?
  1. A) high frequency on the left side of the MTBF distribution
  2. B) failure of items used in the nursery ward of a hospital
  3. C) failure of products with a very short life cycle
  4. D) market failure of brand new products
  5. E) high failure rate often encountered in the very early stages of the lifetime of a product
Answer:  E
Diff: 2
Key Term:  Infant mortality
Objective:  LO 17.4 Distinguish between preventive and breakdown maintenance


16) Which one of the following statements about maintenance is TRUE?
  1. A) The optimal degree of preventive maintenance is associated with zero breakdowns.
  2. B) Breakdown maintenance is proactive.
  3. C) Preventive maintenance is reactive.
  4. D) Preventive maintenance is for emergency repairs.
  5. E) Human resources are a major component of effective maintenance management.
Answer:  E
Diff: 2
Key Term:  Maintenance
Objective:  LO 17.4 Distinguish between preventive and breakdown maintenance
17) What is the process that involves repair on an emergency or priority basis?
  1. A) breakdown maintenance
  2. B) emergency maintenance
  3. C) failure maintenance
  4. D) preventive maintenance
  5. E) priority maintenance
Answer:  A
Diff: 2
Key Term:  Breakdown maintenance
Objective:  LO 17.4 Distinguish between preventive and breakdown maintenance

18) As a firm's maintenance commitment increases:
  1. A) the breakdown maintenance costs increase, and the preventive maintenance costs decrease.
  2. B) both the breakdown maintenance costs and the preventive maintenance costs decrease.
  3. C) the breakdown maintenance costs decrease, and the preventive maintenance costs increase.
  4. D) both the breakdown maintenance costs and the preventive maintenance costs increase.
  5. E) None of the above is true.
Answer:  C
Diff: 2
Key Term:  Maintenance
Objective:  LO 17.6 Compare preventive and breakdown maintenance costs

19) An oil change is to ________ maintenance as repair of a broken alternator is to ________ maintenance.
  1. A) preventive; breakdown
  2. B) breakdown; preventive
  3. C) preventive; preventive
  4. D) breakdown; breakdown
  5. E) preventive; infant mortality
Answer:  A
Diff: 2
Key Term:  Maintenance
Objective:  LO 17.4 Distinguish between preventive and breakdown maintenance



20) Suppose that the introduction of a preventive maintenance program is deemed cost effective. Which of the following is most likely true if the cost to actually perform preventive maintenance is very expensive?
  1. A) high standard deviation in MTBF
  2. B) low standard deviation in MTBF
  3. C) low cost to repair
  4. D) Repairs carry little nonmonetary consequences.
  5. E) The system never fails.
Answer:  B
Diff: 2
Key Term:  Preventive maintenance
Objective:  LO 17.5 Describe how to improve maintenance
21) Sophisticated sensors in a maintenance system can detect:
  1. A) the slightest unusual vibration.
  2. B) low standard deviation in MTBF.
  3. C) low cost to repair options.
  4. D) repairs that carry little nonmonetary consequences.
  5. E) full cost view of maintenance.
Answer:  A
Diff: 2
Key Term:  Preventive maintenance
Objective:  LO 17.5 Describe how to improve maintenance

22) A manager is comparing the total maintenance (preventive plus breakdown) cost curves for two scenarios. The graphs have maintenance commitment on the x-axis and costs on the y-axis. Scenario A includes the direct breakdown maintenance costs, while Scenario B includes the "full" (direct and indirect) breakdown maintenance costs. Which of the following should the manager notice when going from Scenario A to Scenario B?
  1. A) Total cost is decreased.
  2. B) Optimal point is moved to the left.
  3. C) Optimal point is moved to the right.
  4. D) Preventative maintenance cost slope is increased.
  5. E) None of the above
Answer:  C
Diff: 2
Key Term:  Maintenance
Objective:  LO 17.6 Compare preventive and breakdown maintenance costs


23) As far as maintenance costs are concerned:
  1. A) for low levels of maintenance commitment, breakdown maintenance costs exceed preventive maintenance costs.
  2. B) for low levels of maintenance commitment, preventive maintenance costs exceed breakdown maintenance costs.
  3. C) for high levels of maintenance commitment, breakdown maintenance costs exceed preventive maintenance costs.
  4. D) preventive maintenance is always more economical than breakdown maintenance.
  5. E) breakdown maintenance is always more economical than preventive maintenance when the "full cost of breakdowns" is taken into consideration.
Answer:  A
Diff: 2
Key Term:  Maintenance
Objective:  LO 17.6 Compare preventive and breakdown maintenance costs

24) Which one of the following is NOT necessary to identify the optimal maintenance policy?
  1. A) historical data on maintenance costs
  2. B) cost of performing the analysis
  3. C) breakdown probabilities
  4. D) breakdown occurrences
  5. E) repair times
Answer:  B
Diff: 2
Key Term:  Maintenance
Objective:  LO 17.5 Describe how to improve maintenance
25) DuLarge Marine manufactures diesel engines for shrimp trawlers and other small commercial boats. One of their CNC machines has caused several problems. Over the past 30 weeks, the machine has broken down as indicated below.

Number of breakdowns per week 0 1 2 3 4
Frequency (Number of weeks that breakdowns occurred) 8 3 5 9 5

What is the expected number of breakdowns per week?
  1. A) 1
  2. B) 2
  3. C) 6
  4. D) 10
  5. E) 30
Answer:  B
Diff: 2
Key Term:  Breakdown maintenance
AACSB:  Analytical thinking
Objective:  LO 17.6 Compare preventive and breakdown maintenance costs


26) DuLarge Marine manufactures diesel engines for shrimp trawlers and other small commercial boats. One of their CNC machines has caused several problems. Over the past 30 weeks, the machine has broken down as indicated below. Each time the machine breaks down, the firm loses an average of $3,000 in time and repair expenses.

Number of breakdowns per week 0 1 2 3 4
Frequency (Number of weeks that breakdowns occurred) 8 3 5 9 5

What is the expected breakdown cost per week?
  1. A) $1,000
  2. B) $2,000
  3. C) $6,000
  4. D) $10,000
  5. E) $60,000
Answer:  C
Diff: 2
Key Term:  Breakdown maintenance
AACSB:  Analytical thinking
Objective:  LO 17.6 Compare preventive and breakdown maintenance costs
27) DuLarge Marine manufactures diesel engines for shrimp trawlers and other small commercial boats. One of their CNC machines has caused several problems. Over the past 30 weeks, the machine has broken down as indicated below. Each time the machine breaks down, the firm loses an average of $3,000 in time and repair expenses. If preventive maintenance were implemented, it is estimated that an average of only one breakdown per week would occur. The cost of preventive maintenance is $1,000 per week.

Number of breakdowns per week 0 1 2 3 4
Frequency (Number of weeks that breakdowns occurred) 8 3 5 9 5

What is the weekly total maintenance cost of this program?
  1. A) $1,000
  2. B) $3,000
  3. C) $4,000
  4. D) $6,000
  5. E) $8,000
Answer:  C
Diff: 2
Key Term:  Breakdown maintenance
AACSB:  Analytical thinking
Objective:  LO 17.6 Compare preventive and breakdown maintenance costs


28) When depot service is compared to operator maintenance:
  1. A) operator maintenance is slower than depot service.
  2. B) competence is higher, but costs may also be higher, with depot service.
  3. C) depot service is often the weak link in the chain because of the lack of specific training.
  4. D) depot service occurs on-site, while operator maintenance occurs off-site.
  5. E) depot service is better for systems connected in parallel, while operator maintenance is preferred for systems connected in series.
Answer:  B
Diff: 2
Key Term:  Maintenance
Objective:  LO 17.7 Define autonomous maintenance

29) Which of the following increases as repairs move from the maintenance department to the depot service?
  1. A) cost
  2. B) replacement time
  3. C) competence
  4. D) operator ownership
  5. E) A, B, and C
Answer:  E
Diff: 2
Key Term:  Maintenance
Objective:  LO 17.5 Describe how to improve maintenance
30) Autonomous maintenance occurs when:
  1. A) employees are empowered to observe, check, adjust, clean, and notify.
  2. B) a remote computer system signals the need for breakdown maintenance.
  3. C) a remote computer system signals the need for preventative maintenance.
  4. D) employees perform their own breakdown maintenance.
  5. E) none of the above
Answer:  A
Diff: 2
Key Term:  Autonomous maintenance
Objective:  LO 17.7 Define autonomous maintenance

31) ________ is a plan that involves routine inspections, servicing, and keeping facilities in good repair to prevent failure.
Answer:  Preventive maintenance
Diff: 2
Key Term:  Preventive maintenance
Objective:  LO 17.4 Distinguish between preventive and breakdown maintenance

32) ________ is the failure rate early in the life of a product or process.
Answer:  Infant mortality
Diff: 2
Key Term:  Infant mortality
Objective:  LO 17.4 Distinguish between preventive and breakdown maintenance


33) The ________ takes into account such costs as deteriorated customer relations and lost sales.
Answer:  full cost view of maintenance
Diff: 2
Key Term:  Maintenance
Objective:  LO 17.6 Compare preventive and breakdown maintenance costs

34) Why is it that many cases of infant mortality of products are not due to product failure?
Answer:  In many cases the failure is not of the product, but of its improper use. Thus, there is a need for good after-sales service that might include installation and training, as well as good instructions for use.
Diff: 2
Key Term:  Infant mortality
Objective:  LO 17.4 Distinguish between preventive and breakdown maintenance

35) What is breakdown maintenance?
Answer:  Breakdown maintenance is the remedial maintenance that occurs when equipment fails and must be repaired on an emergency or priority basis.
Diff: 2
Key Term:  Breakdown maintenance
Objective:  LO 17.4 Distinguish between preventive and breakdown maintenance
36) Explain why a small standard deviation of the MTBF distribution makes a product, machine, or process a good candidate for preventive maintenance while a large standard deviation does not.
Answer:  A small standard deviation means that there is a relatively narrow range for breakdowns. With this narrow range, it is easier to predict when failure might occur and, therefore, take preventive action in advance of failure. A large standard deviation implies a wider range of values over which failure might occur, making failure less predictable and preventive maintenance less advantageous.
Diff: 3
Key Term:  Mean time between failures (MTBF)
Objective:  LO 17.5 Describe how to improve maintenance

37) Why is it that many cost curves associated with maintenance rarely consider the full cost of a breakdown?
Answer:  Many costs are ignored because they are not directly related to the immediate breakdown. For example, some of these costs include the cost of inventory, employee morale, and impact on delivery schedules and customer relations.
Diff: 2
Key Term:  Breakdown maintenance
Objective:  LO 17.6 Compare preventive and breakdown maintenance costs

38) Is there an optimal amount of preventive maintenance? What caution should be exercised before calculating this optimal amount?
Answer:  Too little preventive maintenance causes breakdown costs to rise sharply, adding more to cost than is saved by less preventive maintenance; too much preventive maintenance reduces breakdowns, but by an amount insufficient to offset the added cost of preventive maintenance. Operations managers should assure that all costs of breakdowns have been properly included in the calculations. There may be a history of not including indirect and subjective breakdown cost elements, which leads to performing too little preventive maintenance.
Diff: 3
Key Term:  Preventive maintenance
Objective:  LO 17.6 Compare preventive and breakdown maintenance costs

39) How do many electronic firms deal with infant mortality in their products?
Answer:  They "burn in" their products prior to shipment; they execute a variety of tests to detect "startup" problems prior to shipment.
Diff: 3
Key Term:  Infant mortality
Objective:  LO 17.4 Distinguish between preventive and breakdown maintenance
40) 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

Answer: 
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
Diff: 2
Key Term:  Maintenance
AACSB:  Analytical thinking
Objective:  LO 17.6 Compare preventive and breakdown maintenance costs

41) 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

Answer: 
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
Diff: 2
Key Term:  Maintenance
AACSB:  Analytical thinking
Objective:  LO 17.6 Compare preventive and breakdown maintenance costs

42) 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?
Answer: 
Number of breakdowns 0 1 2 3 4 Total
Monthly probability 0.45 0.10 0.20 0.20 0.05 1.00

Expected # of breakdowns per month = (0)(.45) + (1)(.10) + (2)(.20) + (3)(.20) + (4)(.05) = 1.30 Expected cost of breakdowns per month with breakdown maintenance = (1.30)($400) = $520
Preventive maintenance cost per month = (.5)($400) + $200 = $400
Preventive maintenance is more cost-effective.
Diff: 3
Key Term:  Maintenance
AACSB:  Analytical thinking
Objective:  LO 17.6 Compare preventive and breakdown maintenance costs

43) 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?
Answer: 
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.
Diff: 3
Key Term:  Maintenance
AACSB:  Analytical thinking
Objective:  LO 17.6 Compare preventive and breakdown maintenance costs

Section 4   Total Productive Maintenance

1) TPM (total productive maintenance) is an application of TQM (total quality management) principles to the area of maintenance.
Answer:  TRUE
Diff: 2
Key Term:  Total productive maintenance (TPM)

2) Simulation models and expert systems are useful tools for improving total productive maintenance.
Answer:  TRUE
Diff: 2
Key Term:  Total productive maintenance (TPM)
AACSB:  Information technology

3) Which of the following is true regarding total productive maintenance (TPM)?
  1. A) TPM is concerned with machine operation, not machine design.
  2. B) Field service and depot service perform virtually all maintenance and repair activities.
  3. C) Operators run their machines, but maintenance departments maintain them.
  4. D) TPM reduces variability through autonomous maintenance and excellent maintenance practices.
  5. E) TPM views maintenance and repair as tactical issues, not strategic ones.
Answer:  D
Diff: 2
Key Term:  Total productive maintenance (TPM)

4) Techniques for improving total productive maintenance can include which of the following?
  1. A) simulation
  2. B) expert systems
  3. C) sensors
  4. D) A and C
  5. E) A, B, and C
Answer:  E
Diff: 2
Key Term:  Total productive maintenance (TPM)
AACSB:  Information technology

5) ________ combines total quality management with a strategic view of maintenance from process equipment design to preventive maintenance.
Answer:  Total preventive maintenance or TPM
Diff: 2
Key Term:  Total productive maintenance (TPM)

6) What is the primary concept of total productive maintenance (TPM)? Identify the four elements that productive maintenance includes.
Answer:  TPM's primary concept is reducing variability through autonomous maintenance and excellent maintenance practices. Four other ingredients of TPM are:
  1. Designing machines that are reliable, easy to operate, and easy to maintain
  2. Emphasizing total cost of ownership when purchasing machines, so that service and maintenance are included in the cost
  3. Developing preventive maintenance plans that utilize the best practices of operators, maintenance departments, and depot service
  4. Training for autonomous maintenance so operators maintain their own machines and partner with maintenance personnel
Diff: 3
Key Term:  Total productive maintenance (TPM)

----------------------------------

OPERATIONS MANAGEMENT - 2017 - COLLECTION
FREE DOWNLOAD
EBOOKS
Operations Management, 2015, 12th Edition, William J. Stevenson - Free Download Link
Operations Management, Sustainability and Supply Chain Management, 11th Edition, 2014, Jay Heizer, Barry Render - Free Download Link
Operations Management: Sustainability and Supply Chain Management, 12th Edition, Jay Heizer, Barry Render, Chuck Munson, 2017
Principles of Operations Management: Sustainability and Supply Chain Management, 10th Edition, 2017
Operations Research: An Introduction, 10th Edition, Hamdy A. Taha, 2017
Introduction to Operations and Supply Chain Management, 4th Edition, Cecil B. Bozarth, Robert B. Handfield, 2016 
Operations Management: Processes and Supply Chains, 11th Edition, Lee J. Krajewski, Manoj K. Malhotra, Larry P. Ritzman, 2016
Free Online Course Materials
1. Operations Management Ebooks - Free Downloads
2. Slides - 11th Edition - Free Downloads
3. Slides - 12nd Edition - Free Downloads
4. Full List of Videos Case Studies - Link
5. All articles about Operations Management
http://top20mba.com/mba-cases/94-mba-operation-management
2017 Updates
6. QUIZ, Multiple choice questions and answers
7. CASE STUDY GUIDES
8. Video Case study guides

For Test Bankz, Quiz Answers and Case study Guides, email to: wincapsimtips@gmail.com
All Free downloads - LINK

Buy All Collections for USD 19.99 – Buy now LINK (Click here)
Any inquiry and support, please email: wincapsimtips@gmail.com
FREE DOWNLOADS:
1. Operations Management: Sustainability and SCM, 12th Edition, Jay Heizer, 2017
2. Operations Management, Sustainability and SCM, 11th Ed, 2014, Jay Heizer
3. Principles of Operations Management: Sustainability and SCM, 10th Ed, Jay Heizer, 2017
Free Download Link
4. Operations Management, 2017, 12th Edition, William J. Stevenson
Free Download Link
5. Operations Management: Processes and Supply Chains, 11th Edition, Lee J. Krajewski, 2016
6. Operations Research: An Introduction, 10th Edition, Hamdy A. Taha, 2017
7. Introduction to Operations and SCM, 4th Edition, Cecil B. Bozarth, Robert B. Handfield, 2016 
8. Hotel Operations Management, 3rd Edition, David K. Hayes, 2017Ebooks – Free Download Links
Videos – Youtube channel links

Good Luck and Success, Enjoy Your Study !
Posted by at

1 comment:

Subscribe to: Post Comments (Atom)