For a body shop in Starkville, the arrivals of cars require service follow a Poi

Question

For a body shop in Starkville, the arrivals of cars require service follow a Poisson Process with 5 cars per day (eight hours a day). There are two repairmen in the body and the repair time for a car follows an exponential distribution with a mean of 2 hours. The body shop has a lot of space and all customers are willing to wait.

How many cars on average are waiting for repair in long run?

How many cars on average are in the body shop in long run?

What is the average time for which a car stays in the reworking area on average? (We do not count the evening and night. In other words, you consider eight hours a day)

Explanation / Answer

Xa Mean Inter arrival time 1.6 hrs Xs Mean service time 2 hrs Sa Std deviation of Inter arrival time 1.6 hrs Ss Std deviation of service time 2 hrs S No of servers 2 Ca Coefficient of variation of inter arrival time 1 (Xa/Sa) Cs Coefficient of variation of service time 1 (Xs/Ss) l Customer arrival rate 0.625 (1/Xa) u Customer service rate 0.5 (1/Xs) p Server utilization 0.625 (l/(s*u) a) Lq Average no in the queue 0.84 (p^(sqrt(2*S+1)/(1-p)*(Ca^2+Cs^2)/2 b) Ls Average no in the system 2.09 Lq+l/u Average time waiting in the queue 1.35 Lq/l c) Average time waiting in the system 3.35 Ls/l

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