Question

Atlantic Video, a small video rental store in Philadelphia, is open 24 hours a day, and—due to its proximity to a major business school—experiences customers arriving around the clock. A recent analysis done by the store manager indicates that there are 20 customers arriving every hour, with a standard deviation of interarrival times of 2 minute(s). This arrival pattern is consistent and is independent of the time of day. The checkout is currently operated by one employee, who needs on average 2.70 minute(s) to check out a customer. The standard deviation of this checkout time is 3 minute(s), primarily as a result of customers taking home different numbers of videos.

(a) If you assume that every customer rents at least one video (i.e., has to go to the checkout), what is the average time a customer has to wait in line before getting served by the checkout employee, not including the actual checkout time (within one minute)?  (Round your answer to two decimal places.)

The average waiting time is ___ minutes?

(b) If there are no customers requiring checkout, the employee is sorting returned videos, of which there are always plenty waiting to be sorted. How many videos can the employee sort over an eight hour shift (assume no breaks) if it takes exactly 1.7 minutes to sort a single video?  (Round your answer to nearest whole number.)

Number of videos the employee can sort is ___ sorts?

(c) What is the average number of customers at the checkout desk (both waiting and being served)? (Round your answer to two decimal places.)

The average number of customer is ___?

(d) Now assume for this question only that 10 percent of the customers do not rent a video at all and therefore do not have to go through checkout. What is the average time a customer has to wait in line before getting served by the checkout employee, not including the actual checkout time? Assume that the coefficient of variation for the arrival process remains the same as before. (Round your answer to two decimal places.)

The average waiting time is ___ minutes?

Answer 1

Steps:

1. Determine arrival patterns - arrival time, inter-arrival time (a), standard deviation (SDa), and Coefficient of variation of arrival (CVa)

2. Determine service patterns – service time (p), standard deviation (SDp), and Coefficient of variation of service time (CVp)

3. Identify number of servers in system

4. Determine utilization rate (u) of servers

5. Determine the expected waiting time in queue (Tq) by formula mentioned below.

The arrival rate = 20 per hour

Inter-arrival time = 60 min per hour/20 per hour = 3 mins per customer

	Inter arrival rate	Service rate
Mean time	a = 3 min per customer	p = 2.7 min per customer
Standard Deviation	S.D. = 2 min	SDp = 3
Coefficient of variance	CVa = SDa/a CVa = 2/3 CVa = 0.667	CVp = SDp/p CVp = 3/2.7 CVp = 1.111
Number of servers/window	m = 1
Utilization (U)	u = (flow rate) / (Capacity) = (1/a) / (m/p) = p/(a x m) u = 2.7/(3 x 1) u = 0.9

The formula for determining average time the customer waits in queue is given as follow:

Tq = 2.7 x 9 x 0.839

Tq = 20.4 minutes

average time a customer has to wait in line = 20.4 minutes

b.

no. of CD that employee can short in 8 hours = time available in 8hrs / time require to short one CD

= (60 mins x 8) / (1.7) = 282.36 CDs

employee can short 282.36 CDsin 8 hours =

c.

Average number of customer waiting system: Ia = (1/a)Tq + p

Ia = (1/3) x 20.4 + 2.7 = 9.5 customers

d.

The arrival rate at checkout point has to be reduced by 10%,

New arrival rate = 20 customers/hr – (0.1)*(20 customers per hour) = 20 – 2 = 18 customers per hour

Inter-arrival time = 60 min per hour/18 per hour = 3.33 mins per customer

	Inter arrival rate	Service rate
Mean time	a = 3.3333 min per customer	p = 2.7 min per customer
Standard Deviation	S.D. = 2 min	SDp = 3
Coefficient of variance	CVa = SDa/a CVa = 2/3.3333 CVa = 0.600	CVp = SDp/p CVp = 3/2.7 CVp = 1.111
Number of servers/window	m = 1
Utilization (U)	u = (flow rate) / (Capacity) = (1/a) / (m/p) = p/(a x m) u = 2.7/(3.33 x 1) u = 0.810

The formula for determining average time the customer waits in queue is given as follow:

Tq = 2.7 x 4.2631 x 0.7972

Tq = 9.1772 minutes

average time a customer has to wait in line = 9.1772 minutes