Question

A crew of mechanics at the Hamilton Highway Department garage repair vehicles that break down at an average of X = 9.6 vehicles per day (approximately Poisson in nature). The mechanic crew can service an average of IA = 11.6 vehicles per day with a repair time distribution that approximates an exponential distribution (the entire crew works on one vehicle at a time). How many vehicles are likely to be waiting for service at any one time?

Assume the centre operates 10.0 hours per day.

Answer 1

SOLUTION :

M/M/1 PERFORMANCE MEASURES . For the M/M1 system the performance measures are given by these simple formulas: L = Average # of customers in the system = 1/(1-2) Lo Average # of customers in the queue = L-1/ W = Average customer time in the system = L/R W = Average customer time in the queue = L / 2 Po = Probability O customers in the system = 1-2/ Pn - Probability n customers in the system = (a/"Po p = Average number of busy servers (utilization rate) or Average number customers being served = 1/1

given

λ = 9.6

µ = 11.6

L =  λ / (  µ - λ)

= 9.6 / ( 11.6 -9.6 )

= 9.6 / 2

= 4.8 customer / hour

L = L - ( λ / µ)

= 4.8 - ( 9.6 / 11.6 )

= 4.8 - 0.82

= 3.98 approximately = 4 customer / hour

therefore there will be 4 customers waiting for service per hour

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