Useful Notes on the Assessment of Maintenance Capacity

It is necessary to compute the manpower requirement depending on the maintenance load. This has been illustrated below:

A company has given you the following information, based on which you have been asked to suggest them what should be the optimum number of workers. A machine breaks down at a mean rate of 0.4 per day. One worker can repair one machine per day, i.e., u = 1. The machine’s nature is such that repair time gets proportionately reduced with the increase of workers.

However, the number of workers cannot be increased beyond three, as logistic support is inadequate. If at any point of time, the company desires to increase manpower beyond three, it is required to develop a second line.

Maintenance workers are paid Rs 300 per day, whereas, the machine’s idle cost is Rs 650 per day. Consider the optimum number of workers assuming no other cost is involved.

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Solution:

Total cost of one repair, assuming one worker is deployed, would then be:

Rs [1(300) + 0.66(650)] = Rs (300 + 429) = Rs 729

Assuming two workers are deployed, the mean rate of failure would be 0.4/(2 – .4) = 0.25

The cost of repair would then be:

Rs [2(300) + 0.25 (650)] = Rs 762.5

Assuming three workers are deployed, mean rate of failure would be;

0.4/ (3 – 0.4) = .15

The cost of repair would then be:

Rs [3(300) + 0.15(650)] = Rs 997.5

From this problem, it can clearly be seen that the company, in reality, gets benefited only when it employs one worker, since the proportionate cost goes up with the increase in number of workers. Hence, under the prevailing circumstances, the company should only use one worker for machine maintenance jobs.