# Fundamentals of Transportation/Queueing/Solution1

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

Solution:

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

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

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

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

$Q= \lambda w =500*0.00357=1.785\,\!$

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

$t = \frac{1}{{\left( {\mu - \lambda } \right)}} = \frac{1}{{\left( {700 - 500} \right)}} = 0.005\ hours = 18{\rm{seconds}} \,\!$