Analyzing a waiting line at a workstation

To illustrate the approach suggested for analyzing a waiting line problem, suppose a Workstation receives parts automatically from a conveyor. An accumulation line has been provided at the workstation and has a storage capacity for five parts $(N=6)$ . Parts arrive randomly at the switching junction for the workstation; if the accumulation line is full, parts are diverted to another workstation.

Question

Analyzing a waiting line at a workstation

To illustrate the approach suggested for analyzing a waiting line problem, suppose a Workstation receives parts automatically from a conveyor. An accumulation line has been provided at the workstation and has a storage capacity for five parts [latex](N=6)[/latex]. Parts arrive randomly at the switching junction for the workstation; if the accumulation line is full, parts are diverted to another workstation.

Accepted Answer

Parts arrive at a Poisson rate of 1 per minute; service time at the workstation is exponentially distributed with a mean of 45 seconds. Hence,

[latex]\lambda_{n}= \begin{cases}1 & n=0,1, \ldots, 5 \ 0 & n=6\end{cases}[/latex]

(Note: the system has a capacity of [latex]6 ;[/latex] the waiting line has a capacity of [latex]5 .)[/latex]

[latex]\mu_{n}= \begin{cases}0 & n=0 \ \frac{4}{3} & n=1, \ldots, 6\end{cases}[/latex]

From Equation [latex]10.141[/latex]

[latex]P_{n}=\frac{\lambda_{0} \lambda_{1} \ldots \lambda_{n-1}}{\mu_{1} \mu_{2} \ldots \mu_{n}} P_{0}[/latex] (10.141)

[latex]P_{1}=\frac{\lambda_{0}}{\mu_{1}} P_{0}=\frac{1}{4 / 3} P_{0}=\frac{3}{4} P_{0}[/latex]

Likewise,

[latex]P_{2}=\frac{\lambda_{0} \lambda_{1}}{\mu_{1} \mu_{2}} P_{0}=\frac{(1)(1)}{\left(\frac{4}{3} ight)\left(\frac{4}{3} ight)} P_{0}=\left(\frac{3}{4} ight)^{2} P_{0}[/latex]

Table 10.32 Computation of [latex] P_{n}[/latex] for Example 10.61
n	[latex]\lambda_{n}[/latex]	[latex]\mu_{n}[/latex]	[latex]P_{n}[/latex]	[latex]P_{n}[/latex]
0	1	0	[latex]1 P_{0}[/latex]	[latex]\frac{4,096}{14,197}[/latex]=0.2885
1	1	[latex]\frac{4}{3}[/latex]	[latex]\frac{\lambda_{0}}{\mu_{1}} P_{0}=\frac{3}{4} P_{0} [/latex]	[latex]\frac{3,072}{14,197}[/latex]=0.2164
2	1	[latex]\frac{4}{3}[/latex]	[latex]\frac{\lambda_{0}}{\mu_{1}} P_{1}=\left(\frac{3}{4} ight)^{2} P_{0}[/latex]	[latex]\frac{2,304}{14,197}[/latex]=0.1623
3	1	[latex]\frac{4}{3}[/latex]	[latex]\frac{\lambda_{2}}{\mu_{3}} P_{2}=\left(\frac{3}{4} ight)^{3} P_{0}[/latex]	[latex]\frac{1,728}{14,197}[/latex]=0.1217
4	1	[latex]\frac{4}{3}[/latex]	[latex]\frac{\lambda_{3}}{\mu_{4}} P_{3}=\left(\frac{3}{4} ight)^{4} P_{0}[/latex]	[latex]\frac{1,296}{14,197}[/latex]=0.0913
5	1	[latex]\frac{4}{3}[/latex]	[latex]\frac{\lambda_{4}}{\mu_{5}} P_{4}=\left(\frac{3}{4} ight)^{5} P_{0}[/latex]	[latex]\frac{972}{14,197}[/latex]=0.0685
6	0	[latex]\frac{4}{3}[/latex]	[latex]\frac{\lambda_{5}}{\mu_{6}} P_{5}=\left(\frac{3}{4} ight)^{6} P_{0}[/latex]	[latex]\frac{729}{14,197}[/latex]=0.0513
			[latex]\frac{14,197}{4,096} P_{0} [/latex]=1	1.000

Table [latex]10.32[/latex] summarizes the calculations involved in solving for [latex]P_{n}[/latex]. Notice, in terms of [latex]P_{0}[/latex], the sum of the probabilities equals [latex](14,197 / 4,096) P_{0}[/latex]; thus, [latex]P_{0}[/latex] equals [latex]4,096 / 14,197[/latex]. Given the value of [latex]P_{0}[/latex], the values of [latex]P_{n}[/latex] are computed for [latex]n=1, \ldots, 6[/latex].

From Table [latex]10.32[/latex], it is seen that [latex]P_{0}=0.2885[/latex] and [latex]P_{6}=0.0513[/latex]; hence, the workstation will be idle [latex]28.85 \%[/latex] of the time, and the accumulation line will be full [latex]5.13 \%[/latex] of the time. The average number of parts in the accumulation (waiting) line [latex]L_{q}[/latex] is given by

[latex]L_{q}=0\left(P_{0}+P_{1} ight)+1 P_{2}+2 P_{3}+3 P_{4}+4 P_{5}+5 P_{6}[/latex]

or

[latex]\begin{aligned}L_{q} &=1(0.1623)+2(0.1217)+3(0.0913)+4(0.0685)+5(0.0513) \&=1.2101 ext { parts in the accumulation line }\end{aligned}[/latex]

The average number of parts in the system, [latex]L[/latex], is given by

[latex]\begin{aligned}L &=\sum_{n=0}^{N} n P_{n} \&=0(0.2885)+1(0.2164)+2(0.1623)+3(0.1217)+4(0.0913)+5(0.0685)+6(0.0513) \&=1.9216 ext { parts in the system }\end{aligned}[/latex]

The rate at which parts actually enter the accumulation line is not 1 per minute because the accumulation line is sometimes full and parts are diverted elsewhere. The effective arrival rate [latex]\hat{\lambda}[/latex] is defined as the rate at which customers enter the system and is given by Equation [latex]10.146[/latex] or

[latex]\hat{\lambda}=\left(L-L_{q} ight) \mu[/latex] (10.146)

[latex]\begin{aligned}\hat{\lambda} &=(1.9216-1.2101)(4 / 3) \&=0.9487 ext { parts per minute }\end{aligned}[/latex]

Alternately, in this case

[latex]\begin{aligned}\hat{\lambda} &=\left(1-P_{N} ight) \lambda \&=0.9487 ext { part per minute }\end{aligned}[/latex] (10.151)

The percentage of arriving parts that do not enter the accumulation line is [latex]P_{N}=0.0513[/latex]. Suppose the production manager desires that no more than [latex]2 \%[/latex] of the arriving parts be diverted: how long should the accumulation line be? We will provide an answer to this question subsequently.

Question 10.61: Analyzing a waiting line at a workstation To illustrate the ...

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