Consider a post office with two employees and a single waiting line. On average,

Question

Consider a post office with two employees and a single waiting line. On average, the inter-arrival time is 6 minutes. Standard deviation of inter-arrival time is 3.94 minutes. Average service time is 5 minutes. Standard deviation of service time is 2.83 minutes.

What is the average number of customers in the system?

What is the average number of customers waiting?

What is the average time a customer waits for service?

What is the average time a customer spends in the system?

What is the system utilization?

Explanation / Answer

L = the average number of customers in the system
W = the average time a customer spends in the system.

a = the average rate of customer arrivals =6
h = the average rate at which customers can be served =5
1/a = the average time between successive arrivals =1/6
1/h = the average service time per customer.=1/5

ratio of customer arrival rate to customer service rate, x = a/h

x=a/h =6/5 =1.2

L = x/(1 – x) = a/(h – a) = 1.2/(1-1.2) = 6

W = the average time a customer spends in the system.

W=L/a =6/6 =1 minute

S=system utilization

S=avg arrival rate/avg service rate = 6/5 =1.2

Z=average time a customer waits for service

Z=SL= 1.2*6 = 7

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