Question

There are 3 servers in the checkout area. The interarrival time of customers is 4 minutes. The processing time is 10 minutes. The coefficient of variation for the arrival process is 1 and the standard deviation of the processing times is 20 minutes. What is the average number of customers that are waiting?

Answer 1

Question: There are 3 servers in the checkout area. The interarrival time of customers is 4 minutes. The processing time is 10 minutes. The coefficient of variation for the arrival process is 1 and the standard deviation of the processing times is 20 minutes. What is the average number of customers that are waiting?

Answer: The average number of customers that are waiting = 9 (rounded)

Explanation:

Given that:

Arrival Time = a = 4 minutes

Service Time = p = 10 minutes

M = 3

CV_a = 1

CV_p = 20 / 10 = 2

Now,

Utilization with m server is given by:

u = p / a x m

Therefore:

u = 10 / 4 x 3

u = 0.83

Now,

Let us find average time customer waiting = T_q, is given by:

T_q = (p / m) x (u^√2(m+1)-1 / 1 - u) x ( (CV_a² + CV_p² ) / 2 )

Therefore, by substituting the values we get: T_q

T_q = (10/3) x (0.83^√2(3+1)-1 / 1-3) x (( 1 + 4 ) / 2)

T_q = (10/3) x (0.83^1.83 / 1 - 0.83) x 2.5

T_q = 3.33 x 4.18 x 2.5

T_q = 34.79

Now,

The average number of customers that are waiting I_q, is given by:

I_q = (1/a) x Tq = Tq / a

Therefore:

I_q = 34.79 / 4

I_q = 8.69 = 9 (rounded)