Students arrive at the Administrative Services Office according to a Poisson dis
ID: 449092 • Letter: S
Question
Students arrive at the Administrative Services Office according to a Poisson distribution at a rate of 12 students per hour. Judy, the only clerk who works at the Office, works 6 hours per day. Students' requests take on average 3.75 minutes each to be processed. There have been complaints that the average wait time (in line) has been too long in the Office. Management is considering to hire a new temporary clerk (Mary) to work in parallel with Judy, working at the same rate as Judy's. Give management an estimate of the reduction in waiting time (in line) that would result from the addition of one clerk (Mary) to the office.
Explanation / Answer
As per MM1 system, arrivals follow Poisson and service follows exponential = 12 per hour µ = 16 per hour (3.75 minutes per student) Average time a student is in the queue = Wq = pWs p = / µ = 12/16 = 0.75 Ws = 1 /µ- = 1/16-12 = 0.25 hours = 0.75 * 0.25 hours = 0.1875 hours or 11.25 minutes When Mary gets added to system to perform same task as Judy, the reduction in waiting time can be calculated as follows - The system become Multiple server model Average Utilization of the system p = / Sµ = 12/(2*16) = 0.375 Probability that there are 0 students in the system (P0 = 1 / [1 + / µ + {( / µ)^s / s!} * (1 / 1-p)] = 1 / [ 1 + 12/16 + (12/16^2 / 2!) * (1 / 1 - 0.375) ] = 1 / [ 1 + 0.75 + (0.28125 * 1.6)] = 1 / (1.75 + 0.45) = 1 / 2.2 = 0.4545 Average number of students in waiting line = Lq = P0*(/µ)^s*p / s! (1 -p)^2 = (0.4545 * 0.75^2 * 0.375) / (2! * 0.625^2) = 0.095881 / 0.78125 = 0.1227 customers Average waiting time of students in line = Lq / = 0.1227/12 = 0.010227 hours or 0.6136 minutes Reduction is waiting time = 11.25 minutes - 0.6136 minutes = 10.64 minutes