Analyzing congestion at an in-process inspection station
Parts are supplied by a conveyor to an inspection station from a numerically controlled milling machine. Because of the insignificant variation in machining time and conveyor speed, parts arrive at the inspection station in a deterministic fashion at a rate of six per minute.
Inspection is performed manually, and the inspection times are exponentially distributed with a mean of 7.5 seconds. In this situation, \lambda=6, \mu=8, c=1, and \rho=0.75. From Figure 10.67, \theta=0.55 . Therefore, if an estimate for \theta is desired such that \left|\theta_{k}-\theta_{k-1}\right|< 0.001, then, for the example with \rho=0.75, the following iterative process can be used:
\theta_{k}=e^{-\left(1-\theta_{k-1}\right) / 0.75}Calculations show that if \theta_{0}=0.55, then \theta_{1}=0.5488116, \theta_{2}=0.5479427, and \theta_{3}=0.5473083. Therefore, substituting 0.5473083 for \theta in Equations 10.176 through 10.179 yields
\begin{aligned} L =\frac{\theta}{1-\theta}\end{aligned} (10.176)
\begin{aligned} L_{q} =\frac{\theta^{2}}{1-\theta}\end{aligned} (10.177)
\begin{aligned} W =\frac{1}{\mu(1-\theta)}\end{aligned} (10.178)
\begin{aligned} W_{q} =\frac{\theta}{\mu(1-\theta)} \end{aligned} (10.179)
\begin{gathered}L=\frac{0.5473083}{0.4526917}=1.209 \text { parts } \end{gathered}\begin{gathered}L_{q}=\frac{0.5473083}{0.4526917}=0.662 \text { part }\end{gathered}
\begin{aligned} W =\frac{1}{8(0.4526917)}=0.276 \text { minute per part } \end{aligned}
\begin{aligned} W_{q} &=\frac{0.5473083}{8(0.4526917)}=0.151 \text { minutes per part } \end{aligned}
