Paul Fenster owns and manages a chili-dog and softdrink stand near the Kean U. c
ID: 361577 • Letter: P
Question
Paul Fenster owns and manages a chili-dog and softdrink stand near the Kean U. campus. While Paul can service 2727 customers per hour on the average (mu), he gets only 2121 customers per hour (lambda). Because Paul could wait on 2929% more customers than actually visit his stand, it doesn't make sense to him that he should have any waiting lines. Paul hires you to examine the situation and to determine some characteristics of his queue. After looking into the problem, you find it follows the six conditions for a single-channel waiting line. The average number of customers in the system = nothing customers (round your response to two decimal places).
Explanation / Answer
We assume that arrival rate follows Poisson distribution and service rate follows negative exponential distribution
It is thus a M/M/1 queuing system as per Kendal notation
Given average arrival rate = Lambda = 21 / hour
Average service rate = S = 27/ hour
Then
Average number of customers in the system
= Lambda^2/ S x ( S – lambda) + Lamda/ S
= ( 21x 21) / 27 x ( 27 – 21) + 21/27
= ( 21 x 21) / ( 27 x 6) + 21/27
73.5/27 + 21/27
= 94.5/27
= 3.5
AVERAGE NUMBER OF CUSTOMERS IN THE SYSTEM = 3.5
AVERAGE NUMBER OF CUSTOMERS IN THE SYSTEM = 3.5