# 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:

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

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

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

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

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

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

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