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.

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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.

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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.

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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:

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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

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(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

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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.

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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

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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

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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