Solutions Manual for Operations
Management Sustainability and
Supply Chain Management 5th
Canadian Edition By Jay Heizer,
Render, Munson, Griffin
All Chapters 1-17, 100% Original
Verified, A+ Grade.
All Chapters Arranged Reverse: 17-1
This is the Only Original and
Complete Solutions Manual for 5th
Canadian Edition, All other Files in
The Market are Fake/Old/Wrong
Edition.
, 17
C H A P T E R
Maintenance and Reliability
DISCUSSION QUESTIONS Effectiveness of preventive maintenance as measured by:
1. The objective of maintenance and reliability is to maintain the Emergency maintenance hours
1−
capability of the system while controlling costs. Preventative maintenance hours
2. Candidates for preventive maintenance can be identified by 7. Machine design can ameliorate the maintenance problem by,
looking at the distributions for MTBF (mean time between fail- among other actions, stressing component reliability, simplicity of
ures). If the distributions have a small standard deviation, they are design and the use of common or standard components, simplicity
usually a candidate for preventive maintenance. of operation, and provision of appropriate product explanations and
3. Infant mortality refers to the high rate of failures that exists for user instructions.
many products when they are relatively new. 8. Information technology can play a number of roles in the
4. Simulation is an appropriate technique with which to investi- maintenance function, among them:
gate maintenance problems because failures tend to occur ran- Files of parts and vendors
domly, and the probability of occurrence is often described by a Management of data regarding failures
probability distribution that is difficult to employ in a closed-form
Active monitoring of system states
mathematical solution.
Problem diagnosis and tracking
5. Training of operators to perform maintenance may improve
morale and commitment of the individual to the job or organization. Via simulation—pretesting and evaluation of maintenance pol-
On the other hand, all operators are not capable of performing the icy
necessary maintenance functions or they may perform them less ef- Enabling more precise control to reduce the likelihood of
ficiently than a specialist. In addition, it is not always cost effective failure
to purchase the necessary special equipment for the operator’s use. Enabling improved system design
6. Some ways in which the manager can evaluate the effective-
9. The best response would probably be to enumerate the actual
ness of the maintenance function include:
costs, both tangible and intangible, for each practice.
Maintenance productivity as measured by: Costs of waiting until it breaks to fix it might include:
Units of production
Unnecessary damage to the machine
Maintenance hours
Significant down time on the production line
or
Maintenance hours Random interruption of the production schedule
Replacement cost of investment Ruined raw materials
or Poor quality of products produced in a time period prior to
Actual maintenance hours to do job breakdown
Standard maintenance hours to do job Frustration of employees
Machine utilization as measured by: Costs to repair the machine
( A − B) − (C + D) Costs of preventive maintenance would include primarily the
( A − B) cost to replace the machine component. Downtime could be sched-
where: uled so as to reduce its cost; and the frustration of employees, etc.,
A = total available operating hours would certainly be less than incurred when the breakdown occurs.
B = scheduled downtime 10. Only when preventive maintenance occurs prior to all outliers
C = scheduled mechanical downtime of the failure distribution will preventive maintenance preclude all
D = nonscheduled mechanical downtime failures. Even though most breakdowns of a component may
occur after time t, some of them may occur earlier. The earlier
breakdowns may not be eliminated by the preventive maintenance
policy. A distribution of natural causes exists.
250 Copyright © 2026 Pearson Canada Inc.
, CHAPTER 17 M A I N T E N A N C E AND RELIABILITY 251
ETHICAL DILEMMA END-OF-CHAPTER PROBLEMS
A number of ethical questions could be discussed in the context of
17.1 Using Figure 17.2: n = 50. Average reliability of compo-
the Walkerton situation. Was there adequate oversight? Why were
nents = 0.99. Average reliability of system = 0.9950 = 0.62. Actual
the staff members not adequately trained or qualified for the job
(calculated) reliability of the system = 0.605.
they were doing? Was it acceptable to release the drug for use while
it was still undergoing testing? Were any of the decisions made due 17.2 From Figure 17.2, about 13% overall reliability (or 0.995400)
to public pressure, political reasons, or cost-cutting? 17.3 E(breakdowns/year) = 0.1(0) + 0.1(1) + 0.25(2) + 0.2(3)
+ 0.25(4) + 0.1(5) + 0(6)
ACTIVE MODEL EXERCISES (AVAILABLE ON = 0.1 + 0.5 + 0.6 + 1.0 + 0.5
MYLAB OPERATIONS MANAGEMENT) = 2.7 breakdowns
ACTIVE MODEL 17.1: Series Reliability 17.4 E(daily breakdowns) = 0.1(0) + 0.2(1) + 0.4(2) + 0.2(3)
1. Would it be better to increase the worst clerk’s reliability from + 0.1(4)
0.8 to 0.81 or the best clerk’s reliability from 0.99 to 1? = 0 + 0.2 + 0.8 + 0.6 + 0.4 = 2.0
The worst clerk’s reliability from 0.8 to 0.81 Expected cost = 2($50) = $100 daily
2. Is it possible to achieve 90% reliability by focusing on only
one of the three clerks? 17.5 Let R equal the reliability of the components. Then R1 × R2
× R3 = Rs, the reliability of the overall system. Therefore, Rs = 0.98
No—the best we can do is 89.1% reliability even with R2 to
and each R ≅ 0.9933. Therefore, a reliability of approximately
100%.
99.33% is required of each component.
ACTIVE MODEL 17.2: Redundancy
1. If one additional clerk were available, which would be the best 17.6 (a) Percent of failures [FR(%)]
place to add this clerk as back-up? 5
%)
F (= = 0.05 = 5.0%
At R2, yielding a system reliability of 97.23%. 100
2. What is the minimum number of total clerks that need to be (b) Number of failures per unit hour [(FR(N)]:
added as back-up in order to achieve a system reliability of 99%? Number of failures
FR ( N ) =
3 more clerks—one more at each process. Total time − Nonoperating time
ACTIVE MODEL 17.3: Parallel Systems where
Total time = (5,000 hrs) × (100 units)
1. What is the reliability if only the top two parallel series are = 500,000 unit-hours
used?
Nonoperating time = (2,500 hrs) × (5 units) = 12,500
With the reliability of the bottom path set to 0.0, reliability is
.9975. 5 5
=FR( N ) =
2. What is the reliability if only the bottom two parallel series are 500,000 − 12,500 487,500
used? = 0.00001026 failure unit-hour
With the reliability of the top path set to 0.0, reliability is (c) Number of failures per unit year:
.9975. Failure/unit-year = FR(N) × 24 hr/day × 365 days/yr
3. What is the reliability if only the top and bottom components = 0.00001026 × 24 × 365 = 0.08985
are used? (d) Failures from 1100 installed units:
With the middle pair of components set to 0.0, reliability is Failures/year = 1100 units × 0.08985 failures/unit-year
.9975. = 98.83
4. What is the reliability if components 2 and 3 have reliability of 4
17.7 (a) FR (%= ) = 40%
only .95? 10
With the reliability of the middle pair set to .95 (all at .95), (b) FR ( N ) = 4 (10 × 60,000) − {(50,000 × 1) + (35,000 × 1)
reliability is .99976. + (15,000 × 2)}
5. What would the reliability of component 1 need to be to in- 4 [ 600,000 − 115,000] =
4 485,000
crease the overall reliability to .9999?
0.000008247 failures per unit hour
Component 1 would need to be set to .958 for the system to
have a reliability of .9999. (c) MTBF = 1 0.000008247 = 121, 256 hours
6. Suppose that components 2 and 3 both must have the same re-
liability. What does that need to be in order to have an overall reli-
ability of .9999?
To have a system reliability of .9999, components 2 and 3
must be set .979.
Copyright © 2026 Pearson Canada Inc.
, 252 CHAPTER 17 M A I N T E N A N C E AND RELIABILITY
17.8 The overall system has a reliability of 0.9941, or approxi- 17.13 The survival of bypass patients can be modelled as:
mately 99.4%.
Alternatively, 0.95 + (1 − 0.95) × 0.882 = 0.95 + 0.0441 = 0.9941
17.9 The overall system has a reliability of 0.9498, or approxi- These statistics indicate that a patient having bypass surgery
mately 95%. would have a probability of 0.649 of living more than one year after
the surgery.
17.14 Rsystem = 0.90 × [0.85 + 0.85(0.15)] × 0.90
= 0.9 × (0.9775) × 0.9 = 0.7918
17.15 The figure suggests that there are likely to be at least three
separate modes of failure; one or more causes of infant mortality,
and two modes of failure which occur at later times.
17.16
Reliability Probability Probability Probability
of a of first + of backup × of needing
=
system with component component backup
a backup working working component
(a) With a backup reliability of 0.90:
R ={0.90 + [0.90(1 − 0.90)]} × {0.92 + [0.90(1 − 0.92)]}
+ {0.94 + [0.90(1 − 0.94)]} + {0.96 + [0.90(1 − 0.96)]}
= {0.90 + 0.90 × 0.10} × {0.92 + 0.90 × 0.08}
× {0.94 + 0.90 × 0.06} × {0.96 + 0.90 × 0.04}
= {0.900 + 0.090} × {0.920 + 0.072}
× {0.940 + 0.054} ×{0.960 + 0.036}
= 0.990 × 0.992 × 0.994 × 0.996
17.10 The reliability of the system is given by: = 0.972
R = 1 − [(1 − 0.90) × (1 − 0.95) × (1 − 0.85)] = 0.99925, (b) With a backup reliability of 0.93:
or R ={0.90 + [0.93(1 − 0.90)]} × {0.92 + [0.93(1 − 0.92)]}
[0.9 + (0.1 × 0.95) + (0.1 × 0.05 × 0.85)] = 0.99925 + {0.94 + [0.93(1 − 0.94)]} + {0.96 + [0.93(1 − 0.96)]}
17.11 0.99 × 0.98 × 0.90 = 0.8732 ≈ 0.873 = {0.90 + 0.93 × 0.10} × {0.92 + 0.93 × 0.08}
17.12 (a) 0.99 × 0.98 × 0.90 = 0.8732 (top series) × {0.94 + 0.93 × 0.06} × {0.96 + 0.93 × 0.04}
0.99 × 0.98 × 0.90 = 0.8732 (bottom series) = {0.900 + 0.0903} × {0.920 + 0.074}
1 − [(1 − 0.873) × (1 − 0.873)] = 1 − (0.127 × 0.127) × {0.940 + 0.056} × {0.960 + 0.037}
= 1 − 0.0161 = 0.984 = 0.993 × 0.994 × 0.996 × 0.997
Alternatively, (0.873) + [0.873(1 – 0.873)] = 0.873 + = 0.9801
0.873 (0.127) = 0.984
(b) Reliability increases by 11.1%, from 0.873 to 0.984.
Copyright © 2026 Pearson Canada Inc.
Management Sustainability and
Supply Chain Management 5th
Canadian Edition By Jay Heizer,
Render, Munson, Griffin
All Chapters 1-17, 100% Original
Verified, A+ Grade.
All Chapters Arranged Reverse: 17-1
This is the Only Original and
Complete Solutions Manual for 5th
Canadian Edition, All other Files in
The Market are Fake/Old/Wrong
Edition.
, 17
C H A P T E R
Maintenance and Reliability
DISCUSSION QUESTIONS Effectiveness of preventive maintenance as measured by:
1. The objective of maintenance and reliability is to maintain the Emergency maintenance hours
1−
capability of the system while controlling costs. Preventative maintenance hours
2. Candidates for preventive maintenance can be identified by 7. Machine design can ameliorate the maintenance problem by,
looking at the distributions for MTBF (mean time between fail- among other actions, stressing component reliability, simplicity of
ures). If the distributions have a small standard deviation, they are design and the use of common or standard components, simplicity
usually a candidate for preventive maintenance. of operation, and provision of appropriate product explanations and
3. Infant mortality refers to the high rate of failures that exists for user instructions.
many products when they are relatively new. 8. Information technology can play a number of roles in the
4. Simulation is an appropriate technique with which to investi- maintenance function, among them:
gate maintenance problems because failures tend to occur ran- Files of parts and vendors
domly, and the probability of occurrence is often described by a Management of data regarding failures
probability distribution that is difficult to employ in a closed-form
Active monitoring of system states
mathematical solution.
Problem diagnosis and tracking
5. Training of operators to perform maintenance may improve
morale and commitment of the individual to the job or organization. Via simulation—pretesting and evaluation of maintenance pol-
On the other hand, all operators are not capable of performing the icy
necessary maintenance functions or they may perform them less ef- Enabling more precise control to reduce the likelihood of
ficiently than a specialist. In addition, it is not always cost effective failure
to purchase the necessary special equipment for the operator’s use. Enabling improved system design
6. Some ways in which the manager can evaluate the effective-
9. The best response would probably be to enumerate the actual
ness of the maintenance function include:
costs, both tangible and intangible, for each practice.
Maintenance productivity as measured by: Costs of waiting until it breaks to fix it might include:
Units of production
Unnecessary damage to the machine
Maintenance hours
Significant down time on the production line
or
Maintenance hours Random interruption of the production schedule
Replacement cost of investment Ruined raw materials
or Poor quality of products produced in a time period prior to
Actual maintenance hours to do job breakdown
Standard maintenance hours to do job Frustration of employees
Machine utilization as measured by: Costs to repair the machine
( A − B) − (C + D) Costs of preventive maintenance would include primarily the
( A − B) cost to replace the machine component. Downtime could be sched-
where: uled so as to reduce its cost; and the frustration of employees, etc.,
A = total available operating hours would certainly be less than incurred when the breakdown occurs.
B = scheduled downtime 10. Only when preventive maintenance occurs prior to all outliers
C = scheduled mechanical downtime of the failure distribution will preventive maintenance preclude all
D = nonscheduled mechanical downtime failures. Even though most breakdowns of a component may
occur after time t, some of them may occur earlier. The earlier
breakdowns may not be eliminated by the preventive maintenance
policy. A distribution of natural causes exists.
250 Copyright © 2026 Pearson Canada Inc.
, CHAPTER 17 M A I N T E N A N C E AND RELIABILITY 251
ETHICAL DILEMMA END-OF-CHAPTER PROBLEMS
A number of ethical questions could be discussed in the context of
17.1 Using Figure 17.2: n = 50. Average reliability of compo-
the Walkerton situation. Was there adequate oversight? Why were
nents = 0.99. Average reliability of system = 0.9950 = 0.62. Actual
the staff members not adequately trained or qualified for the job
(calculated) reliability of the system = 0.605.
they were doing? Was it acceptable to release the drug for use while
it was still undergoing testing? Were any of the decisions made due 17.2 From Figure 17.2, about 13% overall reliability (or 0.995400)
to public pressure, political reasons, or cost-cutting? 17.3 E(breakdowns/year) = 0.1(0) + 0.1(1) + 0.25(2) + 0.2(3)
+ 0.25(4) + 0.1(5) + 0(6)
ACTIVE MODEL EXERCISES (AVAILABLE ON = 0.1 + 0.5 + 0.6 + 1.0 + 0.5
MYLAB OPERATIONS MANAGEMENT) = 2.7 breakdowns
ACTIVE MODEL 17.1: Series Reliability 17.4 E(daily breakdowns) = 0.1(0) + 0.2(1) + 0.4(2) + 0.2(3)
1. Would it be better to increase the worst clerk’s reliability from + 0.1(4)
0.8 to 0.81 or the best clerk’s reliability from 0.99 to 1? = 0 + 0.2 + 0.8 + 0.6 + 0.4 = 2.0
The worst clerk’s reliability from 0.8 to 0.81 Expected cost = 2($50) = $100 daily
2. Is it possible to achieve 90% reliability by focusing on only
one of the three clerks? 17.5 Let R equal the reliability of the components. Then R1 × R2
× R3 = Rs, the reliability of the overall system. Therefore, Rs = 0.98
No—the best we can do is 89.1% reliability even with R2 to
and each R ≅ 0.9933. Therefore, a reliability of approximately
100%.
99.33% is required of each component.
ACTIVE MODEL 17.2: Redundancy
1. If one additional clerk were available, which would be the best 17.6 (a) Percent of failures [FR(%)]
place to add this clerk as back-up? 5
%)
F (= = 0.05 = 5.0%
At R2, yielding a system reliability of 97.23%. 100
2. What is the minimum number of total clerks that need to be (b) Number of failures per unit hour [(FR(N)]:
added as back-up in order to achieve a system reliability of 99%? Number of failures
FR ( N ) =
3 more clerks—one more at each process. Total time − Nonoperating time
ACTIVE MODEL 17.3: Parallel Systems where
Total time = (5,000 hrs) × (100 units)
1. What is the reliability if only the top two parallel series are = 500,000 unit-hours
used?
Nonoperating time = (2,500 hrs) × (5 units) = 12,500
With the reliability of the bottom path set to 0.0, reliability is
.9975. 5 5
=FR( N ) =
2. What is the reliability if only the bottom two parallel series are 500,000 − 12,500 487,500
used? = 0.00001026 failure unit-hour
With the reliability of the top path set to 0.0, reliability is (c) Number of failures per unit year:
.9975. Failure/unit-year = FR(N) × 24 hr/day × 365 days/yr
3. What is the reliability if only the top and bottom components = 0.00001026 × 24 × 365 = 0.08985
are used? (d) Failures from 1100 installed units:
With the middle pair of components set to 0.0, reliability is Failures/year = 1100 units × 0.08985 failures/unit-year
.9975. = 98.83
4. What is the reliability if components 2 and 3 have reliability of 4
17.7 (a) FR (%= ) = 40%
only .95? 10
With the reliability of the middle pair set to .95 (all at .95), (b) FR ( N ) = 4 (10 × 60,000) − {(50,000 × 1) + (35,000 × 1)
reliability is .99976. + (15,000 × 2)}
5. What would the reliability of component 1 need to be to in- 4 [ 600,000 − 115,000] =
4 485,000
crease the overall reliability to .9999?
0.000008247 failures per unit hour
Component 1 would need to be set to .958 for the system to
have a reliability of .9999. (c) MTBF = 1 0.000008247 = 121, 256 hours
6. Suppose that components 2 and 3 both must have the same re-
liability. What does that need to be in order to have an overall reli-
ability of .9999?
To have a system reliability of .9999, components 2 and 3
must be set .979.
Copyright © 2026 Pearson Canada Inc.
, 252 CHAPTER 17 M A I N T E N A N C E AND RELIABILITY
17.8 The overall system has a reliability of 0.9941, or approxi- 17.13 The survival of bypass patients can be modelled as:
mately 99.4%.
Alternatively, 0.95 + (1 − 0.95) × 0.882 = 0.95 + 0.0441 = 0.9941
17.9 The overall system has a reliability of 0.9498, or approxi- These statistics indicate that a patient having bypass surgery
mately 95%. would have a probability of 0.649 of living more than one year after
the surgery.
17.14 Rsystem = 0.90 × [0.85 + 0.85(0.15)] × 0.90
= 0.9 × (0.9775) × 0.9 = 0.7918
17.15 The figure suggests that there are likely to be at least three
separate modes of failure; one or more causes of infant mortality,
and two modes of failure which occur at later times.
17.16
Reliability Probability Probability Probability
of a of first + of backup × of needing
=
system with component component backup
a backup working working component
(a) With a backup reliability of 0.90:
R ={0.90 + [0.90(1 − 0.90)]} × {0.92 + [0.90(1 − 0.92)]}
+ {0.94 + [0.90(1 − 0.94)]} + {0.96 + [0.90(1 − 0.96)]}
= {0.90 + 0.90 × 0.10} × {0.92 + 0.90 × 0.08}
× {0.94 + 0.90 × 0.06} × {0.96 + 0.90 × 0.04}
= {0.900 + 0.090} × {0.920 + 0.072}
× {0.940 + 0.054} ×{0.960 + 0.036}
= 0.990 × 0.992 × 0.994 × 0.996
17.10 The reliability of the system is given by: = 0.972
R = 1 − [(1 − 0.90) × (1 − 0.95) × (1 − 0.85)] = 0.99925, (b) With a backup reliability of 0.93:
or R ={0.90 + [0.93(1 − 0.90)]} × {0.92 + [0.93(1 − 0.92)]}
[0.9 + (0.1 × 0.95) + (0.1 × 0.05 × 0.85)] = 0.99925 + {0.94 + [0.93(1 − 0.94)]} + {0.96 + [0.93(1 − 0.96)]}
17.11 0.99 × 0.98 × 0.90 = 0.8732 ≈ 0.873 = {0.90 + 0.93 × 0.10} × {0.92 + 0.93 × 0.08}
17.12 (a) 0.99 × 0.98 × 0.90 = 0.8732 (top series) × {0.94 + 0.93 × 0.06} × {0.96 + 0.93 × 0.04}
0.99 × 0.98 × 0.90 = 0.8732 (bottom series) = {0.900 + 0.0903} × {0.920 + 0.074}
1 − [(1 − 0.873) × (1 − 0.873)] = 1 − (0.127 × 0.127) × {0.940 + 0.056} × {0.960 + 0.037}
= 1 − 0.0161 = 0.984 = 0.993 × 0.994 × 0.996 × 0.997
Alternatively, (0.873) + [0.873(1 – 0.873)] = 0.873 + = 0.9801
0.873 (0.127) = 0.984
(b) Reliability increases by 11.1%, from 0.873 to 0.984.
Copyright © 2026 Pearson Canada Inc.
