Customers arrive at a fast food restaurant with one server according to a Poisson process at a mean rate of 30 per hour.The server has just resigned,and the two candidates for the replacement are X (fast but expensive)and Y (slow but inexpensive).Both candidates would have an exponential distribution for service times with X having a mean of 1.2 minutes and Y having a mean of 1.5 minutes.Restaurant revenue per month is given by $6,000/W where W is the expected waiting time (in minutes)of a customer in the system.Determine the upper bound on the difference in their monthly compensations that would justify hiring X rather than Y.

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The Answer is For both alternatives,we have an M/M/1 queueing system.Using one hour as the unit of time,we also have

\lambda

= 30 in both cases.With candidate X: µ = 50,so W = $For both alternatives,we have an M/M/1 queueing system.Using one hour as the unit of time,we also have \lambda = 30 in both cases.With candidate X: µ = 50,so W = = 3 minutes,Monthly revenue = = $2000.With candidate Y: µ = 40,so W = = 6 minutes,Monthly revenue = = $1000.Since the difference in monthly revenues is $1000,the upper bound on the difference in their monthly compensations that would justify hiring candidate X rather than candidate Y is $1000.$ = 3 minutes,Monthly revenue = $For both alternatives,we have an M/M/1 queueing system.Using one hour as the unit of time,we also have \lambda = 30 in both cases.With candidate X: µ = 50,so W = = 3 minutes,Monthly revenue = = $2000.With candidate Y: µ = 40,so W = = 6 minutes,Monthly revenue = = $1000.Since the difference in monthly revenues is $1000,the upper bound on the difference in their monthly compensations that would justify hiring candidate X rather than candidate Y is $1000.$ = $2000.With candidate Y: µ = 40,so W = $For both alternatives,we have an M/M/1 queueing system.Using one hour as the unit of time,we also have \lambda = 30 in both cases.With candidate X: µ = 50,so W = = 3 minutes,Monthly revenue = = $2000.With candidate Y: µ = 40,so W = = 6 minutes,Monthly revenue = = $1000.Since the difference in monthly revenues is $1000,the upper bound on the difference in their monthly compensations that would justify hiring candidate X rather than candidate Y is $1000.$ = 6 minutes,Monthly revenue = $For both alternatives,we have an M/M/1 queueing system.Using one hour as the unit of time,we also have \lambda = 30 in both cases.With candidate X: µ = 50,so W = = 3 minutes,Monthly revenue = = $2000.With candidate Y: µ = 40,so W = = 6 minutes,Monthly revenue = = $1000.Since the difference in monthly revenues is $1000,the upper bound on the difference in their monthly compensations that would justify hiring candidate X rather than candidate Y is $1000.$ = $1000.Since the difference in monthly revenues is $1000,the upper bound on the difference in their monthly compensations that would justify hiring candidate X rather than candidate Y is $1000.

Customers Arrive at a Fast Food Restaurant with One Server