Fundamentals of Transportation/Queueing/Solution1

From Wikibooks, open books for an open world
Jump to: navigation, search
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\% 
\,\!


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}}
\,\!