Analyzing an accumulation line serving two workstations

The previous example involved a single server $(c=1)$ . We consider now an example in which two servers are present $(c=2)$ . Suppose, in the previous situation, the accumulation line supplies parts for two workstations, as depicted in Figure 10.61.

Question

Analyzing an accumulation line serving two workstations

The previous example involved a single server [latex](c=1)[/latex]. We consider now an example in which two servers are present [latex](c=2)[/latex]. Suppose, in the previous situation, the accumulation line supplies parts for two workstations, as depicted in Figure 10.61.

Accepted Answer

Parts are removed from the accumulation line, worked on, and placed on the conveyor, which delivers the parts to the packaging department. In this case, we assume [latex]\lambda=2[/latex] parts per minute and let each server operate at a rate of [latex]3 / 4[/latex] parts per minute. Hence, [latex]N=7[/latex] (waiting space for 5 parts, plus 2 being serviced) and

[latex]\lambda_{n}=\left\{\begin{array}{ll}2 & n=0,1, \ldots, 6 \0 & n=7\end{array} \quad \mu_{n}= \begin{cases}0 & n=0 \ \frac{4}{3} & n=1 \ \frac{8}{3} & n=2,3, \ldots, 7\end{cases} ight.[/latex]

Notice that if both servers are busy [latex](n=2)[/latex], then the rate at which departures (services) occur is twice the service rate for a single server.

From Table [latex]10.33[/latex] it is seen that [latex]P_{0}=0.1613, P_{1}=0.2420[/latex], and [latex]P_{7}=0.0430[/latex].

Table 10.33 Computation of [latex] P_{n}[/latex] for Example 10.62
n	[latex]\lambda_{n}[/latex]	[latex]\mu_{n}[/latex]	[latex]P_{n}[/latex]	[latex]P_{n}[/latex]
0	2	0	[latex]1 P_{0}[/latex]	0.1613
1	2	[latex]\frac{4}{3}[/latex]	[latex]\frac{\lambda_{0}}{\mu_{1}} P_{0}=\frac{6}{4} P_{0} [/latex]	0.2420
2	2	[latex]\frac{8}{3}[/latex]	[latex]\frac{\lambda_{1}}{\mu_{2}} P_{1}=\frac{6}{4}\left(\frac{6}{8} ight) P_{0}[/latex]	0.1815
3	2	[latex]\frac{8}{3}[/latex]	[latex]\frac{\lambda_{2}}{\mu_{3}} P_{2}=\frac{6}{4}\left(\frac{6}{8} ight)^{2} P_{0}[/latex]	0.1361
4	2	[latex]\frac{8}{3}[/latex]	[latex]\frac{\lambda_{3}}{\mu_{4}} P_{3}=\frac{6}{4}\left(\frac{6}{8} ight)^{3} P_{0}[/latex]	0.1021
5	2	[latex]\frac{8}{3}[/latex]	[latex]\frac{\lambda_{4}}{\mu_{5}} P_{4}=\frac{6}{4}\left(\frac{6}{8} ight)^{4} P_{0}[/latex]	0.0766
6	2	[latex]\frac{8}{3}[/latex]	[latex]\frac{\lambda_{5}}{\mu_{6}} P_{5}=\frac{6}{4}\left(\frac{6}{8} ight)^{5} P_{0}[/latex]	0.0574
7	0	[latex]\frac{8}{3}[/latex]	[latex]\frac{\lambda_{6}}{\mu_{7}} P_{6}=\frac{6}{4}\left(\frac{6}{8} ight)^{6} P_{0}[/latex]	0.0430
			[latex](3,250,112 / 524,288) P_{0}=1[/latex]	10000

Hence, both workstations are idle [latex]16.13 \%[/latex] of the time, exactly one workstation is busy [latex]24.20 \%[/latex] of the time, and the accumulation line is full [latex]4.30 \%[/latex] of the time. The average number of parts in the accumulation line, [latex]L_{q}[/latex], and the average number of parts in the system, [latex]L[/latex], are found to be

[latex]\begin{aligned}L_{q} &=1 P_{3}+2 P_{4}+3 P_{5}+4 P_{6}+5 P_{7} \&=1.0146 ext { parts in the accumulation }\end{aligned}[/latex]

and

[latex]\begin{aligned}L &=\sum_{n=0}^{7} n P_{n} \&=2.4501 ext { parts in the system }\end{aligned}[/latex]

Question 10.62: Analyzing an accumulation line serving two workstations The ...

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