Question

Question 25 (1 point) A crew of mechanics at the Hamilton Highway Department garage repair vehicles that break down at an average of = 9.9 vehicles per day (approximately Poisson in nature). The mechanic crew can service an average of u = 11.9 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 9.0 hours per day. Note: 1- Only round your final answer. Round your final answer to the nearest Integer. Your Answer: Answer

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/ W = Average customer time in the queue = L / 2 Po = Probability O customers in the system = 1-2/ IP = Probability n customers in the system = (a/u po lp = Average number of busy servers (utilization rate) or | Average number customers being served = 1/u

Hence,

$L = \frac{\lambda}{\mu-\lambda} = \frac{9.9}{11.9-9.9} = 4.95\ Customer/ Hour$

Hence,

$L_{q} = L - \frac{\lambda}{\mu} = 4.95 - \frac{9.9}{11.9} = 4.1181\ Customer/ Hour \approx 4\ Customer/hour$

Hence,

There will be 4 customers waiting for service per hour.