An One-server Queueing System
Consider a queueing system with one server and N - 1 waiting space.
The queueing discipline is first-come-first-serve. W
hen there are N customers in the system, any further arrival of customer will be rejected by the system.
The queueing system is observed every minute at time pointst = 0, 1, 2, ...,
and we assume that the arrival process and the service process at the server are independent.
Moreover, in each time interval (i, i+1], i = 0, 1, ..., we have
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Prob (one arrival of customer) = a |
| and |
Prob ( no arrival of customer) = 1 - a |
and if the server is busy then
|
Prob (one departure of customer) = b |
| and |
Prob (no departure of customer) = 1 - b |
The system is said in state i if there are i customers in the system.
Therefore for 0 < i < N we have the followings:
From state i to state i + 1 in one minute, the probability is p = a(1 - b).
From state i to state i - 1 in one minute, the probability is q = b(1 - a).
Remain in state i in one minute, the probability is 1 - p - q = (1 - b)(1 - a) + ab.
When the system is in state 0 we have
From state 0 to state 1 in one minute, the probability is a.
Remain in state 0 in one minute, the probability is 1 - a.
When the system is in state N we have
From state N to state N - 1 in one minute, the probability is b.
Remain in state N in one minute, the probability is 1 - b.
This is again a random walk (see Figure 2) with non-absorbing boundary 0 and N. In
this case, the walker will keep on walking forever. In the long run, what is the probability
that you find i customers in the system (the walker is in position i)?
Figure. 2. The Queueing System.
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