Fundamentals of Transportation/Queueing/Solution1

TProblem

Problem:

Application of Single-Channel Undersaturated Infinite Queue Theory to Tollbooth Operation. Poisson Arrival, Negative Exponential Service Time

• Arrival Rate = 500 vph,
• Service Rate = 700 vph

Determine

• Percent of Time operator will be free
• Average queue size in system
• Average wait time for vehicles that wait

Note: For operator to be free, vehicles must be 0

Example

Solution:

$\rho =\frac{\lambda }{\mu }=\left(\frac{500}{700}\right)=0.714$

$P\left(n\right)={\rho }^n\left(1-\rho \right)={\left(0.714\right)}^0\left(1-0.714\right)=28.6\mathrm{\%}$

$Q=\frac{\rho }{1-\rho }=\frac{0.714}{0.286}=2.5$

$w=\frac{\lambda }{\mu \left(\mu -\lambda \right)}=\frac{500}{700\left(700-500\right)}=0.003571hours=12.8571\sec$

$Q=\lambda w=500\ast 0.00357=1.785$

$ServiceTime=\frac{1}{\mu }=\frac{1}{700}=0.001429hours=5.142\sec$

$t=\frac{1}{\left(\mu -\lambda \right)}=\frac{1}{\left(700-500\right)}=0.005hours=18\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{s}$