In this model the system capacity is restricted to N. Therefore, (N + l )th customer will not join and the difference-differential equations of the previous model are valid if n < N. Then for n = N, we have

PN (t + h) = P_{N}(t) (1 - µh) + P_{N-1}(t)(λh)(1 - µh)

On simplification, the additional difference-differential equation is obtained as

P'_{N}(t) = - µ.P_{N}(t) + λP_{N-1}(t).

Under steady-state this equation reduces to

0 = - µPN + λP_{N-1}(t).

Hence we have three difference equations

λP_{N+1}= (λ+ µ)Pn - λP_{n-1}

µP1 = λP_{0}, n = 0

µPN = λP_{n-1 ,}n = N.

As before, the first two equations give

P_{n} = ρ^{n}.P_{0}

This equation satisfies the third difference equation for n = N.

To determine P_{0}, we use, ∑P_{n} = 1.

So in this model, p can be > 1 or < 1 .

Let λ = Effective arrival rate.

Here λ_{n} = 0 for n ≥ N.

The other measures are obtained as follows :

All will give two values i.e., one for p :t; 1 and the other for p = 1, e.g.,

**Example 4.** Assume that the trucks with goods are coming in a market yard at the rate of 30 trucks per day and suppose that the inter-arrival times follows an exponential distribution. The time to unload the trucks is assumed to be exponential with an average of 42 minutes. If the market yard can admit 1 0 trucks at a time, calculate P (the yard is empty) and find the average queue length.

If the unload time increases to 48 minutes, then again calculate the above two questions.

**Example 5 .** Cars arrive in a pollution testing Centre according to Poisson distribution at an Average rate of 15 cars per hour. The e testing Centre can accommodate at maximum 15 cars. The service time (i. e., testing time) per car is an exponential distribution with mean rate 10 per hour.

(a) Find the effective arrival rate at the pollution testing Centre.

(b) What is the probability that an arriving car has not to wait for testing.

(c) What is the probability that an arriving car will find a vaccant place in the testing Centre.

(d) What is the expected waiting time until a car is left from the testing centre.

** **

**Solution.**

λ= 15 cars/hour

µ= 1 0 cars/hour

ρ= λ/ µ = 1.5, N = 15

(a) λ = λ (1 - P_{N}) = 15( 1 - P_{15}) = 15( 1 - 0.333) = 10 cars/hour.

(b) P (arriving car has not to wait for testing)