Question

QUESTION 5 Past records indicate that each of the 5 laser computer printers at the U.S. Department of Energy needs repair after 20 hours of use. Breakdowns are Poisson distributed. The one technician on duty can service a printer in 2 hours average, following an exponential distribution. If po = .5640, find the following:

Answer 1

Given that,

Number of printers, M = 5

Arrival rate, $\lambda$ = 1/20 = 0.05 per hour

Service rate, $\mu$ = 1/2 = 0.5 per hour

P0 = 0.5640

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Average utilization, $\rho$ = 1 - P0 = 1-0.5640 = 0.4360

Average number waiting, Lq = N - ($\lambda$+$\mu$)$\rho$/$\lambda$ = 5 - (0.05+0.5)*0.436/0.05 = 0.204

Average number in the system, L = Lq+$\rho$ = 0.204+0.436 = 0.64

Average waiting time, Wq = Lq/((N-L)*$\lambda$) = 0.204/((5-0.64)*0.05) = 0.93 hour

Average number being serviced = $\rho$ = 0.436