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So the utilization of the server is: ρ=20/30=2/3. Using the metrics shown above, the results are as following: 1) Average number in line L<sub>Q</sub>= 0.6667; 2) Average number in system L =1.333; 3) Average time in line ω<sub>Q</sub> = 0.033 hour; 4) Average time in system ω = 0.067 hour.
===
For an equilibrium M/G/1 queue, the expected value of the time W spent by a customer in the queue are given by Pollaczek-Khintchine formula as below<ref name=":0">{{Cite book|title=Introduction to Queuing Theory|last=Cooper|first=Robert B.|publisher=Elsevier Science Publishing Co.|year=1981|isbn=0-444-00379-7|___location=|pages=189}}</ref>:
<math>E(W)=\frac{\rho\tau}{2(1-\rho)}(1+\frac{\sigma^2}{\tau^2})</math>
where τ is the mean service time; σ<
For M/M/1 queue, the service times are exponentially distributed, then σ<
<math>{W_M}=\frac{\rho\tau}{1-\rho}</math>
Using this, the corresponding equation for M/D/1 queue can be derived, assuming constant service times. Then the variance of service time becomes zero, i.e. σ<
<math>{W_D}=\frac{\rho\tau}{2(1-\rho)}</math>
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==Stationary distribution==
The number of jobs in the queue can be written as
:<math>\begin{align}\pi_0 &= 1-\lambda \\
\pi_1 &= (1-\lambda)(e^\lambda - 1)\\
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==Delay==
Define ''ρ'' = ''λ''/''μ'' as the utilization; then the mean delay in the system in
::<math>\frac{1}{2\mu}\cdot\frac{2-\rho}{1-\rho}.</math>
and in the queue:
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</math>
The mean waiting time W
<math>W_N=(N-1-\frac{\sum_{k=0}^{N-1}b_k-N}{\rho b_{N-1}})T</math>
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