ABC Printing Store provides self-service machines so that students can make last
ID: 377563 • Letter: A
Question
ABC Printing Store provides self-service machines so that students can make last-minute copies of class presentations (either that, or printing 90 copies in the computer lab!). Arrival patterns follow a Poisson distribution. The mean rate is 15 arrivals per hour for this operation. With the current equipment, the average service time is 3 minutes. However ABC is thinking of buying new machines that are available that will reduce this to an average of 2 minutes.
(a) ABC currently leases one old copy machine for student self-service. What is the average number of students in system in the current system. If a new machine can be leased to replace the old machine, what will be the new average number of students in system?
(b) Instead of leasing the new machine, ABC could lease a second, old machine. What will be the average number of students in system in this case?
(c) The average value of time for the students requiring copying is $15 per hour. From the point of view of overall cost of leasing, the option of leasing two old machines is $3 per hour more expensive than the option of leasing one new machine. However, the ABC is mindful of customer’s waiting cost. It wants to choose the option with less total cost i.e (customer cost + service cost). Which is the better option, to lease a new machine or to lease one more old machine?
Explanation / Answer
(a) Arrival rate of students (A) = 15 students per hour
Average service time (t) = 3 minutes
Average service rate (S) = 60 / t = 60 / 3 = 20 per hour
As per poissons distribution,
Average Number of students in the system = (A / (S-A)) = (15 / (20-15))= 3 students
(b) If a simialr second machine is leased, then because two machines are available, service rate would be doubled. This means new service rate, S' = 20*2 = 40 per hour.
Now the number of students in the system = (A / (S' - A)) = (15 / (40 - 15)) = 0.6 students
(c) To solve this, we need to find the student waiting time in both cases.
Student waiting time = 1 / (S - A)
Case 1: Two old machines
As calculated earlier, in this case S= 40
So waiting time = 1 / (40 - 15) = 0.04 hours
So customer cost = waiting time*$15 = 0.04*15 = $0.6
As per the problem statement, If we consider service cost of leasing one new machine as X, then service cost of leasing two old machines will be X+3
So total cost = customer cost + service cost = 0.6 + X + 3 = $3.6 + X
Case 2: one new machine
Service time = 2 minutes
Therefore service rate S = 60 / 2 = 30 per hour
Student waiting time = 1 / (30 - 15) = 0.066 hours
Customer cost = 0.066 * 15 = $1
As stated earlier, service cost is taken as X
Total cost = $1 + X
Since X is fixed, we can see that Total cost of two old machines is higher that cost of one new machine.
So the option of one new machine is better!