A professor holds online office hours all day Saturday and Sunday the weekend prior to final exams. He is available from 8AM to 8PM both days to answer student questions. Students who arrive while the professor is busy with another student simply wait until their turn comes up. Students are processed in the order they arrive, and all students tolerate whatever wait is necessary to get their questions answered. The professor notes from past experience that students arrive randomly with questions - the average time between arrivals is 36 minutes and the coefficient of variation of interarrival times is 1. Similarly, the time required to answer student questions is randomly distributed with an average of 24 minutes and a coefficient of variation of 1.
a. On average how long does a student have to wait to get in to see the professor?
Coeff of variation CVa = Std Dev of Interarrival time / Mean of interarrival time
As CVa = 1, It is exponential distribution.
Coeff of variation CVp = Std Dev of Processing time / Mean of Processing time = 1
Utilization U = Activity time/ Interarrival time = 24/36 = 0.67
So 1-U = 1-0.67 = 0.33
So Avge waiting time = Activity time *(U/(1-U) * (CVa^2 + CVp^2)/2
= 24*(0.67/0.33)*(1^2+1^2)/2
= 48.73 Mins
Average waiting time (minutes) 48.73 Mins
b. Suppose the professor would prefer the average waiting time to be no more than 40 minutes. By how much would the average interarrival time have to grow in order to meet this standard?
Increase in Average Interarrival Time (minutes) __________
Avge waiting time = Activity time *(U/(1-U) * (CVa^2 + CVp^2)/2
We also know that CVa & CVp are not affected by arrival or processing time.
So (CVa^2 + CVp^2)/2 = (1^1+1^1)/2 = 1 will remain same
So Avge waiting time = Activity time *(U/(1-U)*1
Or (U/(1-U) = Avge waiting time / Activity time = 40/24 = 1.67
Now Utilization U = Activity time/ Interarrival time = a/p
So 1-U = 1 - a/p = (p-a)/p
So U/(1-U) = (a/p) / [(p-a)/p] = a / (p-a)
So p/(p-a) = 1.67
Solving for p, we get p = 1.67*(p-a)
Or 0.67p = 1.67*a = 1.67*24
So p = 1.67*24/0.67 = 38.4 mins
So If interarrivakl time is 38.4 mins, Avge waiting time will be no more than 40 mins.
So Increase in Interarrivale time = 38.4-36 = 3.4 mins
c. Assume again an average interarrival time of 36 minutes and suppose the professor is considering reducing student waiting time by answering questions faster. How much faster would the professor have to answer questions in order to reduce the average waiting time to 40 minutes?
Decrease in the Average Processing Time (minutes) __________
Avge waiting time = Activity time *(U/(1-U) * (CVa^2 + CVp^2)/2
We also know that CVa & CVp are not affected by arrival or processing time.
So (CVa^2 + CVp^2)/2 = (1^1+1^1)/2 = 1 will remain same
So Avge waiting time = Activity time *(U/(1-U)*1= 40 mins .....(i)
Now Utilization U = Activity time/ Interarrival time = a/p = a/36
So 1-U = 1 - a/p = (p-a)/p
So U/(1-U) = (a/p) / [(p-a)/p] = a / (p-a) = a/(36-a)
So (i) become a*U/(1-U) = 40
Or a*a/(36-a) = 40
Or a^2 = 40*(36-a) = 1440 - 40*a
Or a^2 + 40*a - 1440 = 0
Solving this, we get [-40 +Sqrt(40^2 - 4*1*(-1440))]/(2*a) = 22.90 Mins
Or [-40 - Sqrt(40^2 - 4*1*(-1440))]/(2*a) = -62.90 Mins
As Activity time can't be negative, Activity time a = 22.90 Mins
Current Activity time is 24 mins
So Decreas in processing time = 24-22.90 = 1.10 mins
d. Again assume an average interarrival time of 36 minutes and an average processing time of 24 minutes. What would be the average waiting time if the professor could clone himself (thereby creating a system with two servers and a single queue)?
For no of servers = m, U = (1/Interarrival time) / (m/activity time) = p/(a*m)
So 1-U = 1 - p/(a*m) = (a*m - p)/(a*m)
So U/(1-U) = [p/(a*m)] / [(a*m - p)/(a*m)] = p/(a*m - p)
Avge waiting time = (Activity time/m) *(U/(1-U) * (CVa^2 + CVp^2)/2
= (24/2)*(24/(2*36-24)*1
= 6 mins
Average Waiting Time (minutes) 6