将单通道非饱和无限队列理论应用于收费站运营。泊松到达，负指数服务时间
确定
注意：为了让操作员空闲，车辆必须为 0
ρ = λ μ = ( 500 700 ) = 0.714 {\displaystyle \rho ={\frac {\lambda }{\mu }}=\left({\frac {500}{700}}\right)=0.714\,\!}
P ( n ) = ρ n ( 1 − ρ ) = ( 0.714 ) 0 ( 1 − 0.714 ) = 28.6 % {\displaystyle P\left(n\right)=\rho ^{n}\left({1-\rho }\right)=\left({0.714}\right)^{0}\left({1-0.714}\right)=28.6\%\,\!}
Q = ρ 1 − ρ = 0.714 0.286 = 2.5 {\displaystyle Q={\frac {\rho }{1-\rho }}={\frac {0.714}{0.286}}=2.5\,\!}
w = λ μ ( μ − λ ) = 500 700 ( 700 − 500 ) = 0.003571 h o u r s = 12.8571 sec {\displaystyle w={\frac {\lambda }{\mu \left({\mu -\lambda }\right)}}={\frac {500}{700\left({700-500}\right)}}=0.003571hours=12.8571\sec \,\!}
Q = λ w = 500 ∗ 0.00357 = 1.785 {\displaystyle Q=\lambda w=500*0.00357=1.785\,\!}
S e r v i c e T i m e = 1 μ = 1 700 = 0.001429 h o u r s = 5.142 sec {\displaystyle ServiceTime={\frac {1}{\mu }}={\frac {1}{700}}=0.001429\ hours=5.142\sec \,\!}
t = 1 ( μ − λ ) = 1 ( 700 − 500 ) = 0.005 h o u r s = 18 s e c o n d s {\displaystyle t={\frac {1}{\left({\mu -\lambda }\right)}}={\frac {1}{\left({700-500}\right)}}=0.005\ hours=18{\rm {seconds}}\,\!}
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