A small post office has two open windows. Customers arrive according to a Poisson distribution at the rate of 1 every 3 minutes. However, only 80% of them seek service at the windows. The service time per customer is exponential, with a mean of 5 minutes. All arriving customers form one line and access available windows on a FIFO basis.

(a) What is the probability that an arriving customer will wait in line?
(b) What is the probability that both windows are idle?
(c) What is the average length of the waiting line?
(d) Would it be possible to offer reasonable service with only one window? Explain.

A) probability that an arriving customer will wait in line is 67%

B)the probability that both windows are idle is 0.33

C) the average length of the waiting line is 1.33

D)it would not be possible to offer a reasonable service with only one window

Step-by-step explanation:

arrival rate: δ = 20 x 0.80 = 16 customers per hour

service rate: μ = 2 × (60/5) = 24 customers/hour

Utilization factor is given as;

Φ = δ/μ

So, Φ = 16/24 ≈ 0.67

A) the probability that an arriving customer will wait in line is;

16/24 x 100% ≈ 67%

B) probability that both windows are idle is;

P(x=0) = 1 - 0.67 = 0.33

C) The average number of customers in the post office will be;

L_s = Φ/(1 - Φ)

L_s = 0.67/(1 - 0.67)

L_s = 0.67/0.33

L_s ≈ 2 customers

Thus, the average length of the waiting line is;

L_w = L_s - Φ

L_w = 2 - 0.67

L_w = 1.33

D) this part demands that we find the utilization factor with only one window.

Thus;

arrival rate: δ = 20 x 0.80 = 16 customers per hour

And

service rate: μ = 1 × (60/5) = 12 customers/hour

Thus, Utilization factor = 16/12 = 1.33

Thus, it would not be possible to offer a reasonable service with only one window

answer: 0.26

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