Question

Records indicated that each of the two printers at the U.S. Department of Commerce, in Washington, DC, needs repair after about 360 minutes of use. Breakdowns have been determined to be Poisson distributed. The one technician on duty can service a printer at an average of 3 hours, following an exponential distribution. If printer downtime costs $100 per hour and the technician is paid $30 per hour,

1. What is λ per hour?

b) What is μ per hour?

c) what is Po?

d) What is the total cost per hour?

Answer 1

Arrival time = 360 minutes

a) Arrival rate (lambda or A) = 60/Arrival time per hour = 60/360 per hour = 0.167 per hour

Service time = 3 hours

b) Service rate (Mu or S) = 1/Service time = 1/3 = 0.33 per hour

c) Po = 1-A/S = 1-0.167/0.333 = 0.50

d) Number of printer waiting for Service (Lq) = A^2/(S*(S-A)) = (0.167)^2/(0.333*(0.333-0.167)) 0.5

Number of printer in system (Ls) = Lq + A/S = 0.5 + 0.167/0.333 = 1.0

Printer downtime cost per hour = 100 $

Number of technician = 1

Technician salary per hour = 30 $

Total cost per hour = Ls*Printer downtime cost per hour + Number of technician*Technician salary per hour = 1*100 + 1*30 = 130 $