In this model, the arrival rate of the customers is A., but maximum of s customers can be served simultaneously

If µ be the average number of services per unit time per server then we have

λn = λ, n = 0, 1, 2,

µ_{n} = nµ, 0 < n <,s

sµ, n ≥ s

When n < s, there is no queue

In this model the condition of existence of steady state solution is ρ**/**s < 1.

The **steady-state probabilities** are obtained as

The other measures are obtained as follows

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

Expected number of customers in the service = ρ

Expected time for which a server is busy = ρ/s

Expected time for which a server is idle = 1 –ρ/s

P (all servers are busy) = P_{s} + P_{s + 1} + ………

P (an arrival has to wait) = P (all servers are busy).

**Example 6.** A post office has two counters, which handles the business of money orders,

registration letters etc. It has been found that the service time distributions for both the counters are exponential with mean service time of 4 minutes per customer. The customers are found to come in each counter in a Poisson fashion with mean arrival rate of 11 per hour. Calculate

(a) Probability of having to wait for service of a customer.

(b) Average waiting time in the queue.

(c) Expected number of idle counters.

Solution. Here λ= 11 customers/hour.

µ=60/4=15 customers/hour.

s = 2.

(a) P (an arrival has to wait) = P (all servers are busy)

= (11/15)^{2}*1/2! (1-11/30)^{-1}*(0.463) =0.197

(b) S_{q }= ρ^{s+1}/(s-1)!*(s-ρ)^{ 2} *p_{0}

_{ } = (11/15)^{3}/ (2-11/15)^{2} *(0.463) =0.114

Average waiting time in queue (W_{q}) = L_{ q} /λ= 0.114/11 = 0.01 hour = 0.62 min.

(c) Expected number of idle counters

= 2*P_{0} + 1*P_{1 }= 1.266.