Production & Operation Management

1. CHAPTER 13: MAINTENANCE MANAGEMENT – I
Responses to Questions
1. No; as a management discipline, maintenance management does not
differ from operations management. Maintenance work, by general
perception, is random. But, it has its own statistical characteristics,
predictabilities, planning decisions, quality issues, human relations
problems, long term policies and philosophical underpinnings like in Total
Productive Maintenance. It lacks popularity and needs some more work to
be done on it.
2. Engineering design may involve a lot of technical input, but it is there for a
purpose – to deliver quality output as required by the consumer. That
being so, it has a management dimension into it.
3. Statistics deals with a large population. When a machine is unique, it
becomes difficult to obtain and use large amount of data for statistical
purposes. However, it must be noted that nothing is really that unique and
hence some statistical analysis is possible even in such special cases.
4. Statistically 2 months’ life, in a situation where mean life is 24 months, can
be justified. But, looking at it from a customer’s point of view, we feel the
need for a ‘guaranteed life’ (refer to Weibull diagram) that is much greater
than 2 months. This guaranteed life is a function of (result of) the design
of the machine. There seems to be much scope for improving the design.
5. Choice of a replacement policy depends on the relative economics of the
various available replacement policies. It is possible that a ‘group
replacement’ would work out to be economical if the cost of replacing in a
‘group’ is small.

2. 2
Replacement of the preventive kind, which is what a group replacement is
all about, assumes implicitly that the failure follows a ‘wear-out’ mode.
6. Absolutely. In any management, a good (appropriate) organization is an
important factor. It is also true of maintenance management. Excellent
leadership, clear roles, well-defined responsibilities and authority
structures, fast lines of communication, highly motivating organizational
environment are essential in all functions including maintenance. As
technology advances, organizational aspects need increased attention.
Total Productive Maintenance (TPM) depends on company-wide
involvement which, in turn, depends on appropriate organization.
7. Yes. The assumption was that the failure of the maintenance spares
follows a particular statistical distribution – Normal or Poisson.
8. Work Sampling can give us valuable information on machine breakdowns
(fraction of total time lost, frequency, cost to the firm, etc) or machine
stoppages such as for cleaning. A work sampling study can also indicate
the frequency of a specific kind of delays/breakdowns. Such information is
useful in diagnosis of the faults / problems, in the analysis of costs and in
deciding upon the maintenance policies.
9. MTM as such may not find use in maintenance. But special data blocks
such as Universal Maintenance Standards (UMS) are quite useful. UMS is
an extension of the pre-determined motion time systems (PMTS).
10. A technician’s job can be enriched by giving him independent charge of
maintaining certain equipments. Thus, responsibility levels and growth
levels can be increased in the jobs.

3. 3
11. (a) Queuing theory has applications in facility and manpower allocations in
managing breakdown maintenance as given in the two examples below:
Given a pattern in which breakdown occur, how many maintenance
technicians or maintenance crews one should have? What are the delays
(waiting times) with 1 crew, 2 crews and more?
Given a pattern with which ships come for maintenance, how many
maintenance berths should there be? What are the cost and time
implications of 1 berth, 2 berths, etc?
(b) Simulation can help in complex queuing situations observed many a
time in maintenance.
12. The reliability of the three items in parallel (0.9, 0.5 and 0.7) is computed
as follows:
For the parallel system with reliabilities of 0.5 and 0.7, the reliability of the
system is:
P (t) = P1 (t) + P2 (t) – P (t) . P2 (t)
= 0.5 + 0.7 - (0.5) (0.7)
= 0.85
For the parallel system with reliabilities of 0.85 and 0.9, the system
reliability is:
P (t) = 0.85 + 0.9 - (0.85) (0.9)
= 0.985
For the series system with reliabilities of 0.8, 0.985 and 0.5, the system
reliability is:
P (t) = (0.8)(0.985)(0.5) = 0.394
The reliability of the total system is, therefore, 0.394.

4. 4
13. (i) λ = 2, µ =3, λ/µ = 2/3 = 0.667
With 1 crew (i.e. s = 1)
Mean waiting time Tw = λ / µ (µ –λ)
= 2/3(3 -2) = 0.667 hour
With 2 and 3 crews (i.e. s = 2 and 3)
We may use formulae given in Table 7.2. However readymade tables are
available. Referring to such a table, for values of λ / µ = 0.667 we have:
s = 2 Tw = 0.042
s = 3 Tw = 0.005
(ii) Economics of the three alternatives
Mean time in the system Ts = Tw + (1 / µ)
s = 1 Ts = (2/3) + (1/3) = 1 hour per day per arrival
s = 2 Ts = 0.042 + 0.333 = 0.375 hour per day per arrival
s = 3 Ts = 0.005 + 0.333 = 0.338 hour per day per arrival
Cost of unavailability (per day) = Ts x 100
Let us tabulate the results:

5. 5
No. of
crews
Time in system Cost of
unavailability
Cost of
labour
Total cost
s Ts Ts X 100 s x 4x 20
1 1 100 80 180
2 0.375 37.50 160 197.50
3 0.338 33.80 240 273.80
From the above table it is seen that having 1 crew is the most economical
policy.
14.The questioner is asking us to choose between the policies of:
(i) individual breakdown replacement and
(ii) individual preventive replacement.
(i) Individual breakdown replacement policy
Mean life of equipment = 1(0.05) + 2(0.15) + 3(0.30)
+ 4(0.30) + 5(0.20)
= 3.45 months
Cost per month = Rs 30 X 50 (no. of equipments) / 3.45
= Rs 434.78
(ii) Individual preventive replacement policy
Costs of this policy will have to be calculated for various replacement
periods of 1, 2, 3 and 4 months.
If preventive replacement period was 1 month, the costs are:
(a) for the possibility that the equipment
may fail before the replacement age: Rs. (30) (0.05) = 1.50

6. 6
(b) for the possibility the equipment
may not fail until its replacement age: Rs. (15) (0.95) = 14.25
-----------------------------------
Total cost of = 15.75
Unit replacement
------------------------------------
Cost per month for this policy = Total cost of replacement
Expected life of equipment
= (15.75)(50) = Rs. 787.50
1
If preventive replacement period was 2 months, the costs are:
(a) Rs. (30) (0.05 + 0.15) = 6.00
(b) Rs. (15) (0.80) = 12.00
---------------------------------
Total of unit
Replacement = 18.00
----------------------------------
Cost per month for this policy = Total cost of replacement
Expected life of the equipment
= (18)(50) = Rs.461.54
[1(0.05) + 2(0.95) ]
If the preventive replacement period was 3 months, we have:
a) Rs. 30 (0.05 + 0.15 + 0.30) = 15.00
b) Rs. 15 (0.5) = 7.50
--------------------------------------
Total cost of
unit replacement = 22.50
--------------------------------------
Expected life of the equipment = 1(0.05) + 2 (0.15) + 3 (0.80)
= 2.75 months

7. 7
Cost for this policy = Total cost of replacement
Expected life of the equipment
= (22.50)(50) = Rs. 409.09
2.75
If the preventive replacement period was 4 months, we have the costs:
(a) Rs. 30 (0.05 + 0.15 + 0.30 + 0.30) = 24.00
(b) Rs. 15 (0.20) = 3.00
---------------------------------------
Total cost of unit
replacement = 27.00
---------------------------------------
Expected life of the equipment = 1(0.05) + 2 (0.15) + 3 (0.30) + 4 (0.5)
= 3.25
Cost per month under this policy = Rs. (27) (50) = Rs. 415.38
3.25
We can summarize the costs per month under these different policies, as
follows:
Individual breakdown replacement: Rs. 434.78
Individual preventive replacement :
Period Cost per month
1 787.50
2 461.54
3 409.09
4 415.38
From the above, an Individual preventive replacement every 3 months is
the most economical option.

8. 8
CHAPTER 13: MAINTENANCE MANAGEMENT – I
Objective Questions
1. An electrical fuse that blows out is likely to represent a failure mode that
is:
a. Hyper-exponential
√b. Negative exponential
c. Standard Normal
d. None of the above
2. A bearing on a car wheel, when it breaks down, is likely to represent a
failure mode that is:
a. Hyper-exponential
b. Negative exponential
√c. Standard Normal
d. Poisson
3. Maintenance prevention is mainly concerned with:
√a. design
b. preventive replacement
c. condition monitoring
d. replacement policy
4. UMS in maintenance pertains to:
a. replacement policy
b. failure statistics
c. reliability engineering
√d. work measurement
5. Preventive replacement is generally a good policy for a case of:
a. Negative exponential failure mode with high MTTF
b. Negative exponential failure mode with low MTTF
c. Normal failure mode with low MTTF and high cost of replacement
√d. None of the above
6. If the ‘age specific failure rate’ of an item is constant, it indicates:
a. constant wear out problems
b. problems in the design of the item
c. problems in the maintainability
√d. external problems

9. 9
7. Choice of a maintenance policy depends upon:
a. failure mode
b. cost comparisons
√c. a & b
d. none of the above
8. Reliability is high when:
√a. probability of survival is high
b. age specific failure rate is constant
c. failure mode is negative exponential
d. failure mode is hyper-exponential
9. Condition monitoring maybe used when:
a. preventive replacement is very expensive
b. an equipment can be run until it fails
c. a & b
√d. none of the above
10. In Weibull’s probability distribution, a shape factor value of 1 indicates
√a. external cause of failure
b. wear-out as the cause of failure
c. infant mortality phenomenon
d. a problem with the design
11. Contingent maintenance is encountered in:
a. condition monitoring
√ b. continuous process industries
c. high MTTF situations
d. Job-shop production systems
12. The objective of Total Productive Maintenance (TPM) is to:
a. minimize equipment downtime.
b. enhance total factor productivities of manufacturing
c. introduce autonomous maintenance
√d. minimize life cycle costs of the equipment
13. Equipment failures occurring immediately after start-up are usually
characterized by:
a. Normal probability distribution
b. Poisson probability distribution
√c. Hyper-exponential distribution
d. Negative exponential distribution

10. 10
14. If a system has two components in parallel, each having a reliability of
0.90, the reliability of the system is:
√a. 0.99
b. 0.91
c. 0.81
d. 0.45
15. Previous experience shows that the mean time to failure of a wireless
signaling set is 2000 hours. If the set shows a mode of failure where the
age-specific failure rate is constant, what is the likelihood of running the
set for 2000 hours without failure?
a. 1.00
b. 0.63
c. 0.50
√d. 0.37
16. Components of A and B, both exhibiting a negative exponential failure
mode each with a mean failure rate of 0.01 per hour, are connected in
series. If the reliability of this two-component system is to be 0.990 over
one hour, then the reliability of the components must be redesigned to:
a. 0.9990
√b. 0.9950
c. 0.9801
d. 0.9750
17. Mean time to failure is one of the measures of reliability.
√a. True b. False