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.
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 \lambda=2 parts per minute and let each server operate at a rate of 3 / 4 parts per minute. Hence, N=7 (waiting space for 5 parts, plus 2 being serviced) and
\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}\right.Notice that if both servers are busy (n=2), then the rate at which departures (services) occur is twice the service rate for a single server.
From Table 10.33 it is seen that P_{0}=0.1613, P_{1}=0.2420, and P_{7}=0.0430.
| Table 10.33 Computation of P_{n} for Example 10.62 | ||||
| n | \lambda_{n} | \mu_{n} | P_{n} | P_{n} |
| 0 | 2 | 0 | 1 P_{0} | 0.1613 |
| 1 | 2 | \frac{4}{3} | \frac{\lambda_{0}}{\mu_{1}} P_{0}=\frac{6}{4} P_{0} | 0.2420 |
| 2 | 2 | \frac{8}{3} | \frac{\lambda_{1}}{\mu_{2}} P_{1}=\frac{6}{4}\left(\frac{6}{8}\right) P_{0} | 0.1815 |
| 3 | 2 | \frac{8}{3} | \frac{\lambda_{2}}{\mu_{3}} P_{2}=\frac{6}{4}\left(\frac{6}{8}\right)^{2} P_{0} | 0.1361 |
| 4 | 2 | \frac{8}{3} | \frac{\lambda_{3}}{\mu_{4}} P_{3}=\frac{6}{4}\left(\frac{6}{8}\right)^{3} P_{0} | 0.1021 |
| 5 | 2 | \frac{8}{3} | \frac{\lambda_{4}}{\mu_{5}} P_{4}=\frac{6}{4}\left(\frac{6}{8}\right)^{4} P_{0} | 0.0766 |
| 6 | 2 | \frac{8}{3} | \frac{\lambda_{5}}{\mu_{6}} P_{5}=\frac{6}{4}\left(\frac{6}{8}\right)^{5} P_{0} | 0.0574 |
| 7 | 0 | \frac{8}{3} | \frac{\lambda_{6}}{\mu_{7}} P_{6}=\frac{6}{4}\left(\frac{6}{8}\right)^{6} P_{0}
|
0.0430
|
| (3,250,112 / 524,288) P_{0}=1 | 10000 | |||
Hence, both workstations are idle 16.13 \% of the time, exactly one workstation is busy 24.20 \% of the time, and the accumulation line is full 4.30 \% of the time. The average number of parts in the accumulation line, L_{q}, and the average number of parts in the system, L, are found to be
\begin{aligned}L_{q} &=1 P_{3}+2 P_{4}+3 P_{5}+4 P_{6}+5 P_{7} \\&=1.0146 \text { parts in the accumulation }\end{aligned}and
\begin{aligned}L &=\sum_{n=0}^{7} n P_{n} \\&=2.4501 \text { parts in the system }\end{aligned}