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.

Question

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 [latex]2.4[/latex] minutes.

Accepted Answer

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 [latex]b=10, \lambda=2, \mu=25[/latex], and [latex]c=1[/latex]. Hence, [latex] ho=0.80[/latex], and

[latex]L=\frac{ ho(1+b)}{2(1- ho)}=\frac{0.8(11)}{2(0.2)}=22[/latex] unit loads

[latex]\begin{aligned}&L_{q}=L-\frac{b \lambda}{\mu}=22-\frac{10(2)}{25}=21.2 ext { unit loads } \&W=\frac{1+b}{2 \mu(1- ho)}=\frac{11}{2(25)(0.2)}=1.1 ext { hours per unit load } \&W_{q}=W-\frac{1}{\mu}=1.1-0.04=1.06 ext { hours per unit load }\end{aligned}[/latex]

Question 10.66: Analyzing congestion at a receiving dock Truckloads of mater...

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