Students arrive at the Administrative Services Office at an average of one every 6 minutes, and...

Students arrive at the Administrative Services Office at an average of one every 6 minutes, and their requests take on average 5 minutes to be processed. The service counter is staffed by only one clerk, Judy Gumshoes, who works eight hours per day. Assume Poisson arrivals and exponential service times.

a. What percentage of time is Judy idle?

b. How much time, on average, does a student spend waiting in line?

c. How long is the (waiting) line on average?

d. What is the probability that an arriving student (just before entering the Administrative Services Office) will find at least one other student waiting in line?

Solutions

Expert Solution

The arrival rate is 1 every 6 minutes or 10 per hour. The Service time is 5 minutes or 60/5 = 12 per hour.

Hence = 10 and = 12

a. Percentage of time is Judy idle, P0

b. The time, on average, a student spend waiting in line, Wq

c. The length of the (waiting) line on average, Lq

d. What is the probability that an arriving student (just before entering the Administrative Services Office) will find at least one other student waiting in line?

The probability that there would be no students in the line is P0 and the probability that there is 1 student in the system is P1

We know that P0 is 0.1667.

P1 = (P0)(/) = 0.1667(10/12) = 0.1389

Hence the probability that there is at least 1 student waiting in the line is 1-(probability of no student+probability of 1 student in the system) = 1-(0.1667+0.1389) = 1-0.3056 = 0.6944

keosha answered 1 month ago

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