Q1 (Total 20 points). A car wash line is serving customers (here you can think t
ID: 345486 • Letter: Q
Question
Q1 (Total 20 points). A car wash line is serving customers (here you can think that customers in the system are cars). Arrivals to the system follow a Poisson distribution at a rate of 5 per minute. In serving them, the customers take about 10 seconds (service time per customer), exponentially distributed. (4*5 = 20 points) a) How many total customers would you expect to see on average in the system b) On average, how long would it take to complete the car wash service (including c) What percentage of time is the car wash line being used? (either serving themselves or waiting in line)? the waiting time)? d) What is the probability that 2 or more customers are waiting in line for service?Explanation / Answer
Arrival rate (A) = 5 per minute or 5*60 = 300 per hour
Service rate (S) = 10 seconds or 60*60/10 = 360 per hour
Lq = A^2/(S*(S-A)) = 300^2/(360*(360-300)) = 4.17
a) Number of customers waiting = Lq + A/S = 4.17 + 300/360 = 5
b) Time = Lq/A + 1/S = 4.17/300 + 1/360 = 0.0167 hours or 0.0167*60 = 1 minute
c) % of time used = A/S = 300/360*100 = 83.33%
d) Probability of zero customer (Po) = 1-A/S = 1-300/360 = 0.167
Probability of one customer = Po*(A/S)^1 = 0.167*(300/360) = 0.139
Probability of two or more customers = 1- 0.167 - 0.139 = 0.694
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