A retail shop experiences stable demand all afternoon on Saturdays. During this

Question

A retail shop experiences stable demand all afternoon on Saturdays. During this period the shop has 3 open cash registers located in a unique check-out area, each requiring an average of 4 minutes to serve a customer. This activity time is subject to unpredictable variability however, and specifically has a coefficient of variation of 0.8. On average a customer arrives to the check-out area every two 2 minutes, but this inter-arrival time is also subject to unpredictable variability with a coefficient of variation equal to 1.

2. What would be the average customer waiting time if there were 3 separate queues for each one of the cash registers, each seeing a new customer arrival every 6 minutes on average, with a coefficient of variation equal to 1 (i.e. separate dedicated waiting lines)?

Explanation / Answer

Let the variables be assigned as follows:

n = number of customers in the system (both waiting and being served)

C = number of servers in the system = 3 = (both being idle and being busy serving)

Pn(t) = probability of having n customers in the system at time t

Pn = steady state probability of having n customers in the system

P0 = probability of having 0 customer in the system

Lq = average number of customers waiting in the queue

a = arrival rate of customers = one customer per 2 minutes = 60 minutes /2 = 30 customers per hour

Miu = Service rate of the server = 4 minutes per customer = 60 minutes /4 = 15= 15 customers per hour

Chi = Utilization factor of the server = a / Miu = 30 / 15 = 2

SD = Standard Deviation

COV = Co efficient of variation = SD / mean

COVr = Co efficient of Variation for a variable in random = variance / mean * mean

COVs = Co efficient of Variation for service time

COVa = Co efficient of Variation for inter arrival time of customers in to our Q system

vs = variance in the time taken to serve customers

qc = queue capacity

aws = average waiting time in the system (both being served and waiting to get served)

awq = average waiting time in the queue

n = a * aws

Lq = a * awq

aws = awq + 1 / Miu

The service time needed = time waiting in the Q + time needed to serve

2. awq = ?

Here there are 3 different queues – like one Q in front of every server –

a = once in 6 minutes = 60 minutes / 6 = 10 customers arrive per hour

a = arrival rate of customers = one customer per 6 minutes = 60 minutes /6 = 10 customers per hour

Miu = 15

Chi = service / arrival = Miu / a = 15 /10 = 1.5

COVa = 1

COVs = 0.8

aws = 6.56 ?

awq = aws * ( Chi / (1-Chi) ) * ( ( COVa * COVa) + ( COVs * COVs ) ) / 2

= aws * 1.5/(1-1.5) * ( 1 + COVs*COVs)/2

= aws * (-3 ) * (1 + (0.8*0.8) ) / 2 = aws * 2.46 (in terms of absolute values)

= 6.56 * 2.46 = 16.14 = 16 minutes

Hence awq = 16 minutes

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.