Question

An average of 10 cars/hour arrive at a car repair station with two servers. Assume that the avera...

An average of 10 cars/hour arrive at a car repair station with two servers. Assume that the average service for each customer is 4 minutes and both interarrival and service times are exponentially distributed. If this car repair station has a capacity of 4 cars a. Write the steady-state equations and solve them. Compare the results with those calculated in question 1 and draw a conclusion. b. What is the probability that the car repair station is idle? c. What is the average number of cars waiting in the line? d. What is the average amount of time a customer spends in the car repair station? e. On average, how many customers will be served by the repair station in an hour?

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Answer #1
  1. Average number of cars arrive in the station (Demand or\lambda ) = 10 cars / hr
  2. Average service time for each customer = 4 / hr
  3. Hence actual service rate = 60/4 = 15 cars in 1 hr
  4. Stated capacity of the car station (\mu) = 4 cars
  • According to steady state equation = p (n) = ( \lambda0\lambda1.....\lambda n-1 / \mu0\mu1.....\mun-1 ) * p0
  • where n = 1,2,....infinity
  • where \lambda is the arrival rate
  • \mu is the service rate
  • P0 is the probability in 0 state

At the Zero th state, p1 = (\lambda / \mu) p0 ; p1 /p0 = \lambda / \mu

Considering the actual service rate = 60 mins /4 mins of service = the station can handle 15 cars in 1 hr

  • Average utilization = 10/15 = 66.67%  
  • Probability of car station being idle = (1 - 66.67%) *66.67%= 0.2244 = 22.44%  
  • Average amount of time spent in the car repair = 1 / (15-10) = 12 mins
  • Average customers spent in the repair station = 10 /( 15-10) = 2 customers
  • Average customer waiting time = 66.67 % * 12 mins = 7.9 mins
  • Average cars in waiting line = 2* 66.67% = 1.33 cars
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