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

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

$$ Q = \frac = \frac = 2.5 \,\!$$

$$ w = \frac{\lambda } = \frac = 0.003571hours = 12.8571\sec \,\!$$

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

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

$$ t = \frac{1} = \frac{1} = 0.005\ hours = 18{\rm{seconds}} \,\!$$

/Solutions