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

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

P_{n} = ρ^{n} / n!. P_{0} , 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

P_{n} = e^{-ρ}.ρ^{n}= 0, 1, 2, . ... (Poisson distribution) n n !

λ= λ , L_{s }= ρ, W_{s} = L_{s}/λ = 1/µ

L_{q} = 0, w_{q} = 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.

= [3.6112]

= 0.28

P (shop is empty) = P_{0} = 0.28

λ = λ(l - P_{4}) = 1/15(1 - 0.07) = 0.062

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

Lq = ∑ (n - 2). P_{n} = l .P_{3} + 2.P_{4}

=0.12 + 2.(0.07) = 0.26.