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, who works eight hours per day. Assume Poisson arrivals and exponential service times for both options and the Administrative Service Office opens for 8 hrs a day. The manager of the Administrative Services Office estimate that the time a student spends waiting in line costs them (due to goodwill loss and so on) $10 per hour. To reduce the time a student spends waiting, they know that they need to improve Judy's processing time (See problem #3). They are currently considering the following two options: Option 1: Install a computer system, with which Judy expects to be able to complete a student request 40 percept faster (i.e., 6 minutes/request on average). The computer costs $99.50 to operate per day Option 2: Hire another temporary clerk, who will work at the same rate as Judy: The temporary clerk gets paid $75 per day What is the total additional cost of Option 2? Answer: M The queuing model: M Total additional cost = waiting cost + additional service cost Waiting cost = number in line * goodwill loss/hour* number of hours/day To get the number in line, let's get . pa 16 - 67 . SE But, we cannot find the exact match in the M/M/s table. Instead, nl - 0.0 67 - with p = 0.65 and nl - 0.0 76 with p -0.70 Using a linear approximation, we can assume that nl increases by 0.0 09 when pincreases 0.05 (from p - 0.65 to 0/70), therefore: nl increases by 0.4 8 when p increases 1. number in line (nl when p = 0. 67) = 0.0 67 + 0.4 8 0.017 = 0.0 38 goodwill loss/hour = $10/hour number of hours/day = hours/day Therefore, Waiting cost = = $ .70 / day Total additional cost = waiting cost + $75 / day = $ .70/day Therefore Option is better