A small diner has one employee and a counter with seating for 8 customers.The diner does not package food for takeout.Customers arrive at the diner at the rate of 20 per hour (Poisson distributed).Service times are exponentially distributed and average 24 per hour.Customers that arrive when all seats are taken do not enter the diner.What is the average time a customer spends waiting?
A)2.5 minutes
B)3.0 minutes
C)6.44 minutes
D)24 minutes

Question

Quizplus · Accepted Answer

We can use Little's Law to find the average time a customer spends waiting. 
Average number of customers in the diner = arrival rate * average time spent in the diner 
Since there is only one employee, the maximum number of customers in the diner at any time is 8. 
So, we can use the formula: 
8 = 20/60 * (average time in the diner + average time waiting) 
Simplifying, we get: 
average time waiting = (8*60/20) - 24 = 24 - 24 = 0 
This means that on average, customers do not have to wait at all before being served! 
However, if we ignore the fact that customers cannot enter the diner if all seats are taken, we can use the M/M/1 queuing model to find the average time a customer spends waiting. 
Using the formulas for M/M/1 queuing model, we get: 
ρ = λ/μ = 20/24 = 0.8333 
Lq = (ρ^2)/(1-ρ) = (0.8333^2) / (1-0.8333) = 5.0042 customers 
Wq = Lq / λ = 5.0042 / 20 = 0.2502 hours = 15.012 minutes 
Therefore, the average time a customer spends waiting is approximately 15 minutes. None of the answer choices match this value, so we select the closest choice, which is C) 6.44 minutes (rounded up to the nearest hundredth). However, it is important to note that this value does not take into account the fact that customers that arrive when all seats are taken do not enter the diner.

A Small Diner Has One Employee and a Counter with Seating