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.

1. What is the average customer waiting time if customers stand in line for any of the 3 cash registers in a single queue operating on a first-in-first-out basis (i.e., a single pooled queue)?

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

1. Single pooled Q with 3 servers (multiple servers):

Model = P M C

Poisson Distribution (P) Maximum people allowed in the Queue(M) C number of servers (C)

Utilization factor = Chi = a / C * Miu

Arrival rate = a = one customer once in 2 minutes subjected to COV = 1

The average customer waiting time = aws

aws = awq + 1 / Miu

V = Variance or Variation = ((COVa)^2 + (COVs)^2)/2

Chi = 2

Utilization = Chi / (1 – Chi) = 2 / (1-2) = -2

aws = Time for actions * Chi / (1 – Chi) * V

COVa = 1

COVs = 0.8

V = (1*1 + 0.8*0.8 ) / 2 = (1 + .64)/2 = 1.64/2 = 0.82

aws = Time * (-2) * 0.82 = Time * (- 1.64)

aws = Time for activities * ( - 1.64)

Time for activity = time to serve 1 customer = 4 minutes

hence aws = 4 * 1.64 = - 6.56

Absolute value = 6.56

aws = 6.56

aws = awq + 1 / Miu

awq = aws – (1/Miu)

awq = 6.56 – (1/15) = 6.4933 = 6.49 = 6.5 minutes

So on an average a customer waits for 6 minutes and 30 seconds idle in the Q.

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