Analyzing congestion at a receiving dock
Truckloads of material arrive at a receiving dock according to a Poisson process at a rate of two per hour. The number of unit loads per truck equals 10 . The time required to remove an individual unit load from a truck is exponentially distributed with a mean of 2.4 minutes.
A single lift truck is used in the receiving area for transporting unit loads from the truck to a conveyor that delivers the unit load to storage.
In this situation b=10, \lambda=2, \mu=25, and c=1. Hence, \rho=0.80, and
L=\frac{\rho(1+b)}{2(1-\rho)}=\frac{0.8(11)}{2(0.2)}=22 unit loads
\begin{aligned}&L_{q}=L-\frac{b \lambda}{\mu}=22-\frac{10(2)}{25}=21.2 \text { unit loads } \\&W=\frac{1+b}{2 \mu(1-\rho)}=\frac{11}{2(25)(0.2)}=1.1 \text { hours per unit load } \\&W_{q}=W-\frac{1}{\mu}=1.1-0.04=1.06 \text { hours per unit load }\end{aligned}