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.
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,
\lambda_{n}= \begin{cases}1 & n=0,1, \ldots, 5 \\ 0 & n=6\end{cases}(Note: the system has a capacity of 6 ; the waiting line has a capacity of 5 .)
\mu_{n}= \begin{cases}0 & n=0 \\ \frac{4}{3} & n=1, \ldots, 6\end{cases}From Equation 10.141
P_{n}=\frac{\lambda_{0} \lambda_{1} \ldots \lambda_{n-1}}{\mu_{1} \mu_{2} \ldots \mu_{n}} P_{0} (10.141)
P_{1}=\frac{\lambda_{0}}{\mu_{1}} P_{0}=\frac{1}{4 / 3} P_{0}=\frac{3}{4} P_{0}Likewise,
P_{2}=\frac{\lambda_{0} \lambda_{1}}{\mu_{1} \mu_{2}} P_{0}=\frac{(1)(1)}{\left(\frac{4}{3}\right)\left(\frac{4}{3}\right)} P_{0}=\left(\frac{3}{4}\right)^{2} P_{0}| Table 10.32 Computation of P_{n} for Example 10.61 | ||||
| n | \lambda_{n} | \mu_{n} | P_{n} | P_{n} |
| 0 | 1 | 0 | 1 P_{0} | \frac{4,096}{14,197}=0.2885 |
| 1 | 1 | \frac{4}{3} | \frac{\lambda_{0}}{\mu_{1}} P_{0}=\frac{3}{4} P_{0} | \frac{3,072}{14,197}=0.2164 |
| 2 | 1 | \frac{4}{3} | \frac{\lambda_{0}}{\mu_{1}} P_{1}=\left(\frac{3}{4}\right)^{2} P_{0} | \frac{2,304}{14,197}=0.1623 |
| 3 | 1 | \frac{4}{3} | \frac{\lambda_{2}}{\mu_{3}} P_{2}=\left(\frac{3}{4}\right)^{3} P_{0} | \frac{1,728}{14,197}=0.1217 |
| 4 | 1 | \frac{4}{3} | \frac{\lambda_{3}}{\mu_{4}} P_{3}=\left(\frac{3}{4}\right)^{4} P_{0} | \frac{1,296}{14,197}=0.0913 |
| 5 | 1 | \frac{4}{3} | \frac{\lambda_{4}}{\mu_{5}} P_{4}=\left(\frac{3}{4}\right)^{5} P_{0} | \frac{972}{14,197}=0.0685 |
| 6 | 0 | \frac{4}{3} | \frac{\lambda_{5}}{\mu_{6}} P_{5}=\left(\frac{3}{4}\right)^{6} P_{0}
|
\frac{729}{14,197}=0.0513
|
| \frac{14,197}{4,096} P_{0} =1 | 1.000 | |||
Table 10.32 summarizes the calculations involved in solving for P_{n}. Notice, in terms of P_{0}, the sum of the probabilities equals (14,197 / 4,096) P_{0}; thus, P_{0} equals 4,096 / 14,197. Given the value of P_{0}, the values of P_{n} are computed for n=1, \ldots, 6.
From Table 10.32, it is seen that P_{0}=0.2885 and P_{6}=0.0513; hence, the workstation will be idle 28.85 \% of the time, and the accumulation line will be full 5.13 \% of the time. The average number of parts in the accumulation (waiting) line L_{q} is given by
L_{q}=0\left(P_{0}+P_{1}\right)+1 P_{2}+2 P_{3}+3 P_{4}+4 P_{5}+5 P_{6}or
\begin{aligned}L_{q} &=1(0.1623)+2(0.1217)+3(0.0913)+4(0.0685)+5(0.0513) \\&=1.2101 \text { parts in the accumulation line }\end{aligned}The average number of parts in the system, L, is given by
\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 \text { parts in the system }\end{aligned}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 \hat{\lambda} is defined as the rate at which customers enter the system and is given by Equation 10.146 or
\hat{\lambda}=\left(L-L_{q}\right) \mu (10.146)
\begin{aligned}\hat{\lambda} &=(1.9216-1.2101)(4 / 3) \\&=0.9487 \text { parts per minute }\end{aligned}Alternately, in this case
\begin{aligned}\hat{\lambda} &=\left(1-P_{N}\right) \lambda \\&=0.9487 \text { part per minute }\end{aligned} (10.151)
The percentage of arriving parts that do not enter the accumulation line is P_{N}=0.0513. Suppose the production manager desires that no more than 2 \% of the arriving parts be diverted: how long should the accumulation line be? We will provide an answer to this question subsequently.