Assuming that the arrival rate is 9 cars per hour and the average washing time i
ID: 419560 • Letter: A
Question
Assuming that the arrival rate is 9 cars per hour and the average washing time is 6 minutes. Also, assume Poisson arrivals and exponential service. Answer the following questions assuming that the system is in in steady-state (or in other words, in equilibrium):
1. What is the average number of cars waiting in station?
2. If the car wash costs $15 and the station works from 8:00am to 8:00 pm how much money is collected per day on average? How much does the station lose?
3. On average how much it take for a customer until he leaves with his car washed?
4. Management decided to buy another machine if the old machine works more than 85% of the time. Will the management buy a new machine?
Explanation / Answer
L = average arrival rate = 9 cars per hour
M = average service rate = 1 in 6 minutes = 10 per hour
1.
Average number of cars waiting in the system (Ls) = L / (M - L) = 9 / (10 - 9) = 9
2.
Average number of cars waiting in the queue (Lq) = L2 / (M*(M - L)) = 81 / 10(10 - 9) = 8.1
Arrivals in the 12 hours = 9 x 12 = 108 cars. So, potential money that could have been collected = 108 x $15 = $1,620.
But on an average, there will be 8 cars in the queue and that should be at the end of service as well. So, money collected will be (108 - 8) x $15 = $1,500 and money lost will be $120.
3.
Total lead time for a customer (Ws) = Ls / L (from Little's Law) = 9 / 9 = 1 hours
4.
Utilization (u) = L / M = 9/10 = 90%. So, the management should buy a new machine.