Determining the size of a pool of machine operators

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A machine shop has 20 automatic machines that have to be monitored by a machine operator. If a product defect occurs, it continues occurring until an operator makes the necessary adjustments to the machine. Hence, the time required to respond to a call for service and make the required adjustments is critical and costly. It is estimated that it costs [latex]\$ 500 /[/latex] hour when a machine is producing defective product (waiting time) and it costs [latex]\$ 200[/latex] /hour due to lost production during the time the machine is being serviced by an operator (excluding the cost of the operator). The cost of each operator is [latex]\$ 60 /[/latex] hour. The average amount of time a machine runs without requiring the attention of the operator is 45 minutes; the average time required for an operator to make the necessary adjustments is 5 minutes. Both are exponentially distributed times.

Hence, [latex]C_{1}=\$ 500 /[/latex] hour, [latex]C_{2}=\$ 200 /[/latex] hour, [latex]C_{3}=C_{4}=\$ 60 /[/latex] hour, [latex]\lambda=1.333 /[/latex] hour, [latex]\mu=[/latex] [latex]12 /[/latex] hour, and [latex]X=1.333 /(1.333+12)=0.10 .[/latex] From Table [latex]10.37[/latex], with [latex]K=20[/latex] machines, for [latex]c=1, \ldots, 5, F=0.500,0.878,0.975,0.995[/latex], and [latex]0.999[/latex], respectively. Table [latex]10.41[/latex] contains the calculations for variable cost as a function of the number of servers. As seen, the optimum number of servers is 4 , with a variable cost of [latex]\$ 688[/latex].

Table 10.37 Results Obtained by Peck and Hazelwood for (M\|M\|c) : (GD\|K\|K)
K	C	D	F	K	C	D	F	K	C	D	F
		X=0.05		12	1	0.879	0.764	10	1	0.987	0.497
4	1	0.149	0.992		2	0.361	0.970		2	0.692	0.854
5	1	0.198	0.989		3	0.098	0.996		3	0.300	0.968
6	1	0.247	0.985		4	0.019	0.999		4	0.992	0.994
	2	0.023	0.999	14	1	0.946	0.690		5	0.020	0.999
7	1	0.296	0.981		2	0.469	0.954	12	1	0.998	0.416
	2	0.034	0.999		3	0.151	0.992		2	0.841	0.778
8	1	0.343	0.977		4	0.036	0.999		3	0.459	0.940
	2	0.046	0.999	16	1	0.980	0.618		4	0.180	0.986
9	1	0.391	0.972		2	0.576	0.935		5	0.054	0.997
	2	0.061	0.998		3	0.214	0.988	14	2	0.934	0.697
10	1	0.437	0.967		4	0.060	0.998		3	0.619	0.902
	2	0.076	0.998	18	1	0.994	0.554		4	0.295	0.973
12	1	0.528	0.954		2	0.680	0.909		5	0.109	0.993
	2	0.111	0.996		3	0.285	0.983		6	0.032	0.999
14	1	0.615	0.939		4	0.092	0.997	16	2	0.978	0.621
	2	0.151	0.995		5	0.024	0.999		3	0.760	0.854
	3	0.026	0.999	20	1	0.999	0.500		4	0.426	0.954
16	1	0.697	0.919		2	0.773	0.878		5	0.187	0.987
	2	0.195	0.993		3	0.363	0.975		6	0.066	0.997
	3	0.039	0.999		4	0.131	0.995		7	0.019	0.999
18	1	0.772	0.895		5	0.038	0.999	18	2	0.994	0.555
	2	0.243	0.991	25	2	0.934	0.776		3	0.868	0.797
	3	0.054	0.999		3	0.572	0.947		4	0.563	0.928
20	1	0.837	0.866		4	0.258	0.987		5	0.284	0.977
	2	0.293	0.988		5	0.096	0.997		6	0.118	0.993
	3	0.073	0.988		6	0.030	0.999		7	0.040	0.998
25	1	0.950	0.771	30	2	0.991	0.664	20	2	0.999	0.500
	2	0.429	0.978		3	0.771	0.899		3	0.938	0.736
	3	0.132	0.997		4	0.421	0.973		4	0.693	0.895
	4	0.032	0.999		5	0.187	0.993		5	0.397	0.963
30	1	0.992	0.663		6	0.071	0.998		6	0.187	0.988
	2	0.571	0.963		X = 0.20				7	0.074	0.997
	3	0.208	0.994	4	1	0.549	0.862		8	0.025	0.999
	4	0.060	0.999		2	0.108	0.988	25	3	0.996	0.599
	X = 0.10				3	0.008	0.999		4	0.920	0.783
4	1	0.294	0.965	5	1	0.689	0.801		5	0.693	0.905
	2	0.028	0.999		2	0.194	0.976		6	0.424	0.963
5	1	0.386	0.950		3	0.028	0.998		7	0.221	0.987
	2	0.054	0.997	6	1	0.801	0.736		8	0.100	0.995
6	1	0.475	0.932		2	0.291	0.961		9	0.039	0.999
	2	0.086	0.995		3	0.060	0.995	30	4	0.991	0.665
7	1	0.559	0.912	7	1	0.883	0.669		5	0.905	0.814
	2	0.123	0.992		2	0.395	0.941		6	0.693	0.913
	3	0.016	0.999		3	0.105	0.991		7	0.446	0.963
8	1	0.638	0.889		4	0.017	0.999		8	0.249	0.985
	2	0.165	0.989	8	1	0.937	0.606		9	0.123	0.995
	3	0.027	0.999		2	0.499	0.916		10	0.054	0.998
9	1	0.711	0.862		3	0.162	0.985		11	0.021	0.999
	2	0.210	0.985		4	0.035	0.998
	3	0.040	0.998	9	1	0.970	0.548
10	1	0.776	0.832		2	0.599	0.887
	2	0.258	0.981		3	0.227	0.978
	3	0.056	0.998		4	0.060	0.996
Source: Peck and Hazelwood [59].

Table 10.41 Determining the Optimum Number of Machine Operators for a Group of 20 Identical Machines
C	F	VC(c)
1	0.500	$ 5,260.00
2	0.878	$ 1,691.20
3	0.975	$ 820.00
4	0.995	$ 688.00
5	0.999	$ 709.60

[latex]C_{1}[/latex] can increase to [latex]\$ 770[/latex] /hour without increasing the optimum number of servers from 4 to [latex]5 ; C_{3}[/latex] can be as much as [latex]\$ 192[/latex] /hour without decreasing the number of servers from 4 to 3. Due to the limited number of values of [latex]X[/latex] in Table [latex]10.37[/latex], we are limited in the sensitivity analysis we can perform on [latex]\lambda[/latex]; however, for [latex]\lambda=0.632 /[/latex] hour [latex](X=0.05), c^{*}=2[/latex], and for [latex]\lambda=[/latex] [latex]3 /[/latex] hour [latex](X=0.2), c^{*}=7[/latex].

In the problems at the end of the chapter, we explore additional situations in which economic models of queuing systems have been developed. For additional information regarding cost models, see [latex][36],[37],[8],[39][/latex], and [latex][67][/latex], and the recent research literature.

In conclusion, having considered a number of waiting line models to use in analyzing facilities planning requirements for buffer spaces, waiting lines, and inprocess storage, we would be negligent if we did not admit that many situations do not fit the assumptions underlying the models we presented. In such a case, what should you do? Three alternatives come to mind: utilize a more advanced waiting line model, one that more closely fits your situation; utilize one of the simple models we presented, recognizing that the solutions obtained are, at best, approximations; or utilize a simulation model rather than a waiting line model. The last approach is considered in the next section.

Question 10.75: Determining the size of a pool of machine operators

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