THE APPLICATION OF QUEUEING THEORY:Example 1—How Many Repairers?

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EXAMPLES

Example 1—How Many Repairers?

SIMULATION, INC., a small company that makes gidgets for analog computers, has 10 gidget-making machines. However, because these machines break down and require repair frequently, the company has only enough operators to operate eight machines at a time, so two machines are available on a standby basis for use while other machines are down. Thus, eight machines are always operating whenever no more than two ma- chines are waiting to be repaired, but the number of operating machines is reduced by 1 for each additional machine waiting to be repaired.

The time until any given operating machine breaks down has an exponential distribu- tion, with a mean of 20 days. (A machine that is idle on a standby basis cannot break down.) The time required to repair a machine also has an exponential distribution, with a mean of 2 days. Until now the company has had just one repairer to repair these machines, which has frequently resulted in reduced productivity because fewer than eight machines are op- erating. Therefore, the company is considering hiring a second repairer, so that two ma- chines can be repaired simultaneously.

Thus, the queueing system to be studied has the repairers as its servers and the ma- chines requiring repair as its customers, where the problem is to choose between having one or two servers. (Notice the analogy between this problem and the County Hospital emergency room problem described in Sec. 17.1.) With one slight exception, this system fits the finite calling population variation of the M/M/s model presented in Sec. 17.6, where N = 10 machines, A = 1 customer per day (for each operating machine), and

IL = 1 customer per day. The exception is that the A

and A1 parameters of the birth-and death process are changed from A0 = 10A and A1 = 9A to A0 = 8A and A1 = 8A. (All the other parameters are the same as those given in Sec. 17.6.) Therefore, the Cn factors for calculating the Pn probabilities change accordingly (see Sec. 17.5).

Each repairer costs the company approximately $280 per day. However, the estimated lost profit from having fewer than eight machines operating to produce gidgets is $400 per day for each machine down. (The company can sell the full output from eight operating machines, but not much more.) The analysis of this problem will be pursued in Secs. 26.3 and 26.4.

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