Question 3 Multi-Server Queue A supermarket manager is trying to decide how many

Question

Question 3 Multi-Server Queue A supermarket manager is trying to decide how many cashers to employ for the peak time. The service times for check outs are exponentially distributed with a mean service time of 4 minutes. Customer arrivals to cashiers follow a Poisson arrival process with ar average 54 customers per hour (a) What is the minimum number of cashers that would be needed to have the utilization less than one? (b) What is the minimum number of cashers that would be needed to assure that the average waiting time is less than 2 minutes?

Explanation / Answer

Answer to question a :

Let the minimum number of cashiers = n1

Arrival rate ( @54 customers per hour ) = a = 54/60 customer per minute

Mean service time by 1 cashier ( @ 4 minutes per customer ) =   1/ 4 = 0.25 customers per minute

Hence cumulative mean service time by n1 cashiers = S = 0.25.n1 customers per minute

Therefore Utilization rate

= a/ S

= 54/( 60 x 0.25.n1)

= 3.6/n1

If utilization < 1

Therefore , 3.6/n1 < 1

              Or, n1 > 3.6

Hence n1 must be 4( the next higher whole number )

MINIMUM NUMBER OF CASHIERS THAT WOULD BE NEEDED WILL BE 4

Answer to question b:

Let the minimum number of cashiers that wold be needed = n1

Therefore , average service Time using n1 cashiers = S= 0.25.n1

Thus , average waiting time = a / S x ( S – a )

                                                    = (54/60)/ S x ( S – 54/60) minute

Since, Average waiting time < 2 minutes

Therefore,

( 54/60) / S x ( S – 54/60)    < 2

Or, S x ( S – 54/60) > 27/60

Or, S^2 – 54/60.S > 27/60

Or, S^2 - 0.9.S > 0.45

Or . S^2 – 0.9..S + ( 0.45) ^2 > 0.45 +0.2025

Or. ( S – 0.45)^2 > 0.6525

Or, ( S – 0.45)^2 > 0.6525

Or, S – 0.45 > 0.8077

Or, S > 1.2577

Since S = 0.25.n1

Therefore ,

0.25.n1 > 1.2577

Or, n1 > 5.0308

Therefore n1 = 6 ( rounding to next higher whole number )

MINIMUM 6 CASHIERS WOULD BE NEEDED

MINIMUM NUMBER OF CASHIERS THAT WOULD BE NEEDED WILL BE 4

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