Students arrive one at a time, completely at random, to an advice clinic at a ra
ID: 434849 • Letter: S
Question
Students arrive one at a time, completely at random, to an advice clinic at a rate of 10 per hour. Students take on average 5 minutes of advice but there is wide variation in the time they need; this variation may be well modeled by the exponential distribution.
a. Assume there is only one advisor serving in the clinic. Find the following expected measures of performance for this system: the expected time in the clinic, the expected time in the queue for advice, the expected number of students in the clinic, and the expected number of students waiting for advice.
b. Again assuming one advisor, what is the probability that there are more than ten students in the clinic at a random point in time? What is the probability that the time spent in the clinic exceeds 30 minutes?
c. Now suppose that a second advisor is hired. Repeat your answer for a. What are the advisors’ utilizations? Would you recommend this second advisor is hired?
d. What are some ways that the advice clinic could improve students’ experiences in the advice clinic without hiring more staff?
Explanation / Answer
Ans
In system = In clinic
In queue = waiting for advice
A.
C.
For MM2 system now we get:
Yes for better service and less wait time the second one should be hired
D. Other way of improving is scheduling the students, so th variability will reduce and so the nature will be easy to handle without variability. So if every 5 minute is uniform distribution its handled better.
So
Current system Kendall's notation M/M/1 Comment Arrival rate A 10 per hour Service rate S 12 60/5 U Utilization ratio U=A/S 0.83333 <1 Length of queue diminishing Ls Expected number of people in system Ls A/(S-A) 5.00 Ws Avg waiting time in the system Ws 1/(S-A) 0.50000 hours 30.0000 mins Wq Avg waiting time in the Line Wq A/S(S-A) 0.41667 hous 25.0000 mins Lq Avg no of customer waiting in line Lq A^2/S(S-A) 4.1667 Po Prob of 0 units in system Po 1-(A/S) 0.16667 16.67% Pn Prob of k units in system (A/S)^(K+1) 0.13459 Ans B K = 10