At the start of football season, the ticket office gets very busy the day before
ID: 379754 • Letter: A
Question
At the start of football season, the ticket office gets very busy the day before the first game. Customers arrive at the rate of three every 10 minutes and the average time to transact business is 3 minutes. a. what us the average number of people in line? b. what is the average time that a person will spend at the ticket office? c. what proportion of time is the server busy? d. what is the average number of people receiving and waiting to receive tickets? e. what is the average time that a person will spend waiting in line at the ticket office?
Explanation / Answer
This is a single queue problem M/M/1 type as per Kendall’s notation where both arrival rate and service rate have Poisson / exponential distribution.
Given
Arrival rate ( @ 10 minutes per arrival) = a = 6 per hour
Service rate ( @ 3 minutes per hour ) = s = 20 per hour
Average number of people in the line
= a^2 / s x ( s – a )
= 6 x 6 / 20 x 14
= 0.128
Average time a person will spend at the ticket office ( i.e. waiting time + ticketing time at counter)
= a/ S x ( s – a ) + 1/s
= 6/ 20 x ( 20 – 6) + 1/20
= 6 / ( 20 x 14) hour + 1/20
= 6 x 60 / ( 20 x 14) + 60/20 minutes
= 1.285 minutes + 3 minutes
= 4.285 minutes
Proportion of time server is busy = a/s = 6/20 = 0.30
Average number of people waiting and receiving ticket
= Average number of people in the line + a/s
= 0.128 + a/s
= 0.128 + 6/20
= 0.128 + 0.3
= 0.428
Average time a person will spend waiting in line at the ticket office
= a/s x ( s- a)
= 6/ 20 x 14 hour
= 6 x 60/ 20 x 14 minutes
= 1.285 minutes
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.