Analyzing a waiting line in which arrivals are affected by the number of customers waiting for service Consider a tool crib with two attendants (c = 2). Time studies indicate customers arrive in a Poisson fashion at a rate of 12 per hour, so long as no more than one customer is waiting to be served

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Analyzing a waiting line in which arrivals are affected by the number of customers waiting for service

Consider a tool crib with two attendants [latex](c=2)[/latex]. Time studies indicate customers arrive in a Poisson fashion at a rate of 12 per hour, so long as no more than one customer is waiting to be served. If two customers are waiting, the rate at which people enter the tool crib is reduced to eight per hour. If three customers are waiting, customers enter at a rate of four per hour. No additional customers enter if four customers are waiting. Hence, the maximum number of customers in the system [latex](N)[/latex] equals [latex]6 .[/latex] The time required to fill a customer's order is exponentially distributed with a mean of 10 minutes.

Accepted Answer

From Table [latex]10.34[/latex], it is seen that [latex]L=2.7211[/latex] and [latex]L_{q}=1.0923[/latex].

Because the service rate for an individual server, [latex]\mu[/latex], is unchanged for [latex]0

[latex]\begin{aligned}U &=(2.7211-1.0923) / 2 \&=0.8144, ext { or } 81.44 \%\end{aligned}[/latex]

Table 10.34 Computation of [latex] P_{n},L,[/latex] and [latex]L_{q}[/latex] for Example 10.63
n	[latex]\lambda_{n}[/latex]	[latex]\mu_{n}[/latex]	[latex]P_{n}[/latex]	[latex]P_{n}[/latex]	[latex]nP_{n}[/latex]	[latex](n-c) P_{n}[/latex]
0	12	0	[latex]1 P_{0}[/latex]	0.0928	---	---
1	12	6	[latex]2 P_{0}[/latex]	0.1856	0.1856	---
2	12	12	[latex]2 P_{0}[/latex]	0.1856	0.3712	---
3	12	12	[latex]2 P_{0}[/latex]	0.1856	0.5568	0.1856
4	8	12	[latex]2 P_{0}[/latex]	0.1856	0.7424	0.3712
5	4	12	[latex]\frac{4}{3} P_{0}[/latex]	0.1237	0.6185	0.3711
6	0	12	[latex]\frac{4}{9} P_{0}[/latex]	0.0411	0.2466	0.1644
			[latex]\frac{97}{9} P_{0}[/latex]=1	1.0000	L=2.7211	[latex]L_{q}[/latex]=1.0923

[latex]\begin{aligned}\hat{\lambda} &=\left(L-L_{q} ight) \mu \&=(2.7211-1.0923)(6) \&=9.7728 ext { customers per hour }\end{aligned}[/latex]

and

[latex]\begin{aligned}W &=L / \hat{\lambda} \&=2.7211 / 9.7728 \&=0.2784 ext { hour per customer } \&=16.706 ext { minutes per customer } \W_{q} &=\frac{L_{q}}{\hat{\lambda}} \&=\frac{1.0923}{9.7728} \&=0.1118 ext { hour per customer } \&=6.706 ext { minutes per customer }\end{aligned}[/latex]

Question 10.63: Analyzing a waiting line in which arrivals are affected by t...

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