Question

Students arrive at the Administrative Services Office at an average of one every 15 minutes, and...

Students arrive at the Administrative Services Office at an average of one every 15 minutes, and their requests take on average 10 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? (Round your answer to 1 decimal place.)

Percentage of time             %

b. How much time, on average, does a student spend waiting in line? (Do not round intermediate calculations. Round your answer to the nearest whole number.)

Average time             minutes

c. How long is the (waiting) line on average? (Round your answer to 2 decimal places.)

Average length of the waiting line             customers

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? (Do not round the intermediate calculations. Round your answer to 2 decimal places.)

Probability   %

This is a M/M/1 system problem:

Arrival rate (l) = 4 students per hour

Service rate (m) = 6 students per hour

No of servers (s) = 1

Average utilization (p) =l/m = 4/6 = 0.667

a) Idle time = 1- utilization time = 1-p = 0.333 or 33.33%

b) The average time in the line, in minutes (Wq) = p * 1/(m-l) = 0.667 * (1/2) = 0.3335 hr = 0.3335 * 60 =20.01 mins

c) The average number of students waiting in the line (Lq) = p*l / (m-l) = (0.667*4)/ (6-4) = 2.668/2 = 1.334 students = 1.33 students

d) The probability that there are 1 or more people the system = 1 – 0.333(1 + 0.667) = 1 – 0.333 (1.667) = 1- 0.555 = 0.4451 = 44.51%

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