Analyzing a fleet of delivery trucks A distribution center delivers material to the retail stores in its service region. Twelve trucks are used for delivery. The time required to load a truck is exponentially distributed with a mean of 40 minutes; the time required to deliver a load and return is e

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Analyzing a fleet of delivery trucks

A distribution center delivers material to the retail stores in its service region. Twelve trucks are used for delivery. The time required to load a truck is exponentially distributed with a mean of 40 minutes; the time required to deliver a load and return is exponentially distributed with a mean of 6 hours. There are two crews available for loading trucks. The distribution center operates continuously, that is, 24 hours per day and 7 days per week.

Accepted Answer

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].

Based on the data for the distribution center [latex]K=12, \lambda=1 / 6, \mu=2[/latex], and [latex]c=2[/latex]; hence, [latex]X=0.1[/latex]. From Table [latex]10.37, F=0.970 .[/latex] Therefore,

[latex]\begin{aligned}L &=K[1-F(1-X)]=12[1-0.970(1-0.1)] \&=1.524 ext { trucks } \L_{q} &=K(1-F)=12(1-0.970) \&=0.360 ext { truck } \W &=\frac{1-F(1-X)}{\mu F X}=\frac{1-0.970(1-0.1)}{1.5(0.970)(0.1)} \&=0.873 ext { hour per truck } \W_{q} &=\frac{1-F}{\mu F X}=\frac{1-0.970}{1.5(0.970)(0.1)} \&=0.206 ext { hour per truck }\end{aligned}[/latex]

The results given above are based on the assumption that either of the two crews can load any truck. It has been suggested that trucks [latex]1, \ldots, 6[/latex] be serviced by crew 1 and trucks [latex]7, \ldots, 12[/latex] be serviced by crew [latex]2 .[/latex] In such a case, [latex]K=6, \lambda=1 / 6, \mu=2, c=1[/latex], and [latex]X=0.1[/latex] for the first group of trucks. From Table [latex]10.37, F=0.932 .[/latex] Hence,

[latex]\begin{aligned}L &=6[1-0.932(1-0.1)]=0.9672 ext { truck } \L_{q} &=6(1-0.932)=0.408 ext { truck } \W &=\frac{1-0.932(1-0.1)}{1.5(0.932)(0.1)}=1.153 ext { hours per truck } \W_{q} &=\frac{1-0.932}{1.5(0.932)(0.1)}=0.486 ext { hour per truck }\end{aligned}[/latex]

Since there are two crews, the average number of trucks in the total system is [latex]L=2(0.9672)=[/latex] [latex]1.9344[/latex] trucks; likewise, the average number of trucks waiting to be loaded is [latex]L_{q}=2(0.408)=[/latex] [latex]0.816[/latex] truck. Under the proposed plan, a complete delivery cycle by a truck will require 6 [latex]+1.153=7.153[/latex] hours; therefore, a truck can be expected to make [latex]7(24) / 7.153=23.49[/latex] deliveries per week. Under the present plan, a truck can be expected to make [latex]7(24) / 6.873=[/latex] [latex]24.44[/latex] deliveries per week.

Question 10.68: Analyzing a fleet of delivery trucks A distribution center d...

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