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Non-Poisson Queuing Models

This is called s-server model with finite system capacity.

Arrival rate   

The steady-state probability are given as

The other characteristics of this model are given below:

Expected number of busy servers = Expected number of customers in service

Proportion of busy time for a server =

Special Cases : (I) M/M/s : FIFO/s

In this model s = N, the steady state probabilities are given by

Pn = ρn / n!. P0  , 0<n<s

=0,                               n > s

(II) M/M/∞ : FIFO/∞ (Self-service queueing model)

 

In this model a customer joining the system becomes a server. So this is called self-service system. The steady state probabilities are given by

 

Pn = en= 0, 1, 2, . ... (Poisson distribution) n n !

 

λ= λ , Ls = ρ, Ws = Ls/λ = 1/µ

Lq = 0, wq = 0

 

Example 7. A barber shop has two barbers and four chairs for customers. Assume that customers arrive in a Poisson fashion at a rate of 4 per hour and that each barber services customers according to an exponential distribution with mean of 18 minutes. Further, if a customer arrives and there are no empty chairs in the shop he will leave. Calculate the following:

 

(a)    Probability that the shop is empty.

(b)   Effective arrival rate.

(c)    Expected number of busy servers.

(d)   Expected number of customers in queue.

 

Solution. Here,                        s = 2, N = 4, λ= 4/60 = 1/15 customers per minute

 

µ = 1/18 customers per minute.

Ρ = λ/ µ=18/ 15=1.2, ρ/s =0.6

=          [3.6112]

=         0.28

P (shop is empty) = P0 = 0.28

 

λ = λ(l - P4) = 1/15(1 - 0.07) = 0.062

 

Expected no. of busy servers = λ/µ = 1 8 x 0. 062 = 1.116

 

Lq = ∑ (n - 2). Pn        = l .P3 + 2.P4

=0.12 + 2.(0.07) = 0.26.