Answer:
a. Average number of copiers in line is 0.275
b. Average number of copiers still in operation is 4.23
c. Average number being serviced is 0.496
d. No, the office should not do it
Explanation:
N = number of copy machines = 5
U = average time between unit service requirements = 1 hour = 60 minutes
T = average service time = 7 minutes
M = number of servers = 1
Cost of copier downtime = $20
Cost of attendant = $15
(a) Service factor, X = T / (T+U) = 7 / (60+7) = 0.105
From the Finite Queuing Tables for a Population of N = 5,
For X = 0.105 and M=1, the efficiency factor, F = 0.945
So, The average number waiting in line, L = N × (1 - F) = 5 × (1 - 0.945) = 0.275
(b)
Average number in operation, J = N×F×(1 - X) = 5×0.945×(1 - 0.105) = 4.23
(c) Average number being serviced, H = F×N×X = 0.945×5×0.105 = 0.496
(d) For M=1
The average number of copier down = N - J = 5 - 4.23 = 0.77
So, cost of downtime per hour = $20×0.77 = $15.4
Also, the cost of the server per hour = $15×M = $15×1 = $15
So, total cost = 15.4 + 15 = $30.4 per hour (i)
For M=2
From the Finite Queuing Tables for a Population of N = 5, with X = 0.105 and M=2, the efficiency factor, F = 0.997
J = N×F×(1 - X) = 5×0.997×(1 - 0.105) = 4.46
The average number of copier down = N - J = 5 - 4.46 = 0.54
So, cost of downtime per hour = $20×0.54 = $10.8
Also, the cost of the server per hour = $15×M = $15×2 = $30
So, total cost = 10.8 + 30 = $40.8 per hour (ii)
Comparing (i) and (ii), we can say that having another attendant is not cost-effective.
