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Solved Problems

Solved Problem 12-1 The Maitland Furniture store gets an average of 50 customers per shift. The manager of Maitland wants to calculate whether she should hire 1, 2, 3, or 4 salespeople. She has determined that average waiting times will be 7 minutes with 1 salesperson, 4 minutes with 2 salespeople, 3 minutes with 3 salespeople, and 2 minutes with 4 salespeople. She has estimated the cost per minute that customers wait at $1. The cost per salesperson per shift (including benefits) is $70.

How many salespeople should be hired?

Solution

The manager’s calculations are as follows:

	NUMBER OF SALESPEOPLE
	1	2	3	4
(a) Average number of customers per shift	50	50	50	50
(b) Average waiting time per customer (minutes)	7	4	3	2
(c) Total waiting time per shift $(a \times b)$ $(a \times b)$ (minutes)	350	200	150	100
(d) Cost per minute of waiting time (estimated)	$1.00	$1.00	$1.00	$1.00
(e) Value of lost time $(c \times d)$ $(c \times d)$ per shift	$ 350	$ 200	$ 150	$ 100
(f) Salary cost per shift	$ 70	$ 140	$ 210	$ 280
(g) Total cost per shift	$ 420	$ 340	$ 360	$ 380

Because the minimum total cost per shift relates to 2 salespeople, the manager’s optimum strategy is to hire 2 salespeople.

Solved Problem 12-2 Marty Schatz owns and manages a chili dog and soft drink store near the campus. Although Marty can service 30 customers per hour on the average $(μ),$ $(μ),$ he gets only 20 customers per hour $(λ) .$ $(λ) .$ Because Marty could wait on 50% more customers than actually visit his store, it doesn’t make sense to him that he should have any waiting lines.

Marty hires you to examine the situation and to determine some characteristics of his queue. After looking into the problem, you find this to be an $M / M / 1$ $M / M / 1$ system. What are your findings?

Solution

\begin{array}{l} L & = & \frac{λ}{μ - λ} = \frac{20}{30 - 20} = 2 customers in the system on the average \\ W & = & \frac{1}{μ - λ} = \frac{1}{30 - 20} = 0.1 \begin{array}{l} hour (6 minutes) = average time the customer spends in \\ the total system \end{array} \\ L_{q} & = & \frac{λ^{2}}{μ (μ - λ)} = \frac{20^{2}}{30 (30 - 20)} = 1.33 customers waiting for service in line on the average \\ W_{q} & = & \frac{λ}{μ (μ - λ)} = \frac{20}{30 (30 - 20)} = \begin{array}{l} \frac{1}{15} hour = (4 minutes) = average waiting time of \\ customer in the queue awaiting service \end{array} \\ ρ & = & \frac{λ}{μ} = \frac{20}{30} = 0.67 = Percentage of the time that Marty is busy waiting on customers \\ P_{0} & = & 1 - \frac{λ}{μ} = 1 - ρ = 0.33 = \begin{array}{l} Probability that there are no customers in the system \\ (being waited on or waiting in the queue) at any given time \end{array} \end{array}

$\begin{array}{l} L & = & \frac{λ}{μ - λ} = \frac{20}{30 - 20} = 2 customers in the system on the average \\ W & = & \frac{1}{μ - λ} = \frac{1}{30 - 20} = 0.1 \begin{array}{l} hour (6 minutes) = average time the customer spends in \\ the total system \end{array} \\ L_{q} & = & \frac{λ^{2}}{μ (μ - λ)} = \frac{20^{2}}{30 (30 - 20)} = 1.33 customers waiting for service in line on the average \\ W_{q} & = & \frac{λ}{μ (μ - λ)} = \frac{20}{30 (30 - 20)} = \begin{array}{l} \frac{1}{15} hour = (4 minutes) = average waiting time of \\ customer in the queue awaiting service \end{array} \\ ρ & = & \frac{λ}{μ} = \frac{20}{30} = 0.67 = Percentage of the time that Marty is busy waiting on customers \\ P_{0} & = & 1 - \frac{λ}{μ} = 1 - ρ = 0.33 = \begin{array}{l} Probability that there are no customers in the system \\ (being waited on or waiting in the queue) at any given time \end{array} \end{array}$

Probability of k or More Customers Waiting in Line and/or Being Waited On
k	$p_{n > k} = {(\frac{λ}{μ})}^{k + 1}$ $p_{n > k} = {(\frac{λ}{μ})}^{k + 1}$
0	0.667
1	0.444
2	0.296
3	0.198

Solved Problem 12-3 Refer to Solved Problem 12-2. Marty agreed that these figures seemed to represent his approximate business situation. You are quite surprised at the length of the lines and elicit from him an estimated value of the customer’s waiting time (in the queue, not being waited on) at 10 cents per minute. During the 12 hours that he is open he gets $(12 \times 20) = 240$ $(12 \times 20) = 240$ customers. The average customer is in a queue 4 minutes, so the total customer waiting time is $(240 \times 4 minutes) = 960 minutes .$ $(240 \times 4 minutes) = 960 minutes .$ The value of 960 minutes is $($ 0.10) (960 minutes) = $ 96.$ $($ 0.10) (960 minutes) = $ 96.$ You tell Marty that not only is 10 cents per minute quite conservative but also he could probably save most of that $96 of customer ill will if he hired another salesclerk. After much haggling, Marty agrees to provide you with all the chili dogs you can eat during a weeklong period in exchange for your analysis of the results of having two clerks wait on the customers.

Assuming that Marty hires one additional salesclerk whose service rate equals Marty’s rate, complete the analysis.

Solution

With two people working, the system now has two channels, or $m = 2.$ $m = 2.$ The computations yield

$\begin{array}{l} P_{0} & = & \frac{1}{{\sum_{n = 0}^{n = m - 1} \frac{1}{n!} [\frac{20}{30}]}^{n}] + \frac{1}{2!} {[\frac{20}{30}]}^{2} [\frac{2 (30)}{2 (30) - 20}]} \\ = & \frac{1}{(1) {(2 / 3)}^{0} + (1) {(2 / 3)}^{1} + (1 / 2) (4 / 9) (6 / 4)} = 0.5 \\ = & probability of no customers in the system \\ L & = & [\frac{(20) (30) {(20 / 30)}^{2}}{(2 - 1)! {[(2) (30 - 20)]}^{2}}] 0.5 + \frac{20}{30} = 0.75 customer in the system on the average \\ W & = & \frac{L}{λ} = \frac{3 / 4}{20} = \frac{3}{80} hour = 2 .25 minutes = average time the customer spends in the total system \\ L_{q} & = & L - \frac{λ}{μ} = \frac{3}{4} - \frac{20}{30} = \frac{1}{12} = 0 .083 customer waiting for service in line on the average \\ W_{q} & = & \frac{L_{q}}{λ} = \frac{\frac{1}{2}}{20} = \frac{1}{240} hour= \frac{1}{4} minute= \begin{array}{l} Average waiting time of a customer in the queue \\ itself (not being serviced) \end{array} \\ ρ & = & \frac{λ}{m μ} = \frac{20}{2 (30)} = \frac{1}{3} = 0.33 = Utilization rate \end{array}$ $\begin{array}{l} P_{0} & = & \frac{1}{{\sum_{n = 0}^{n = m - 1} \frac{1}{n!} [\frac{20}{30}]}^{n}] + \frac{1}{2!} {[\frac{20}{30}]}^{2} [\frac{2 (30)}{2 (30) - 20}]} \\ = & \frac{1}{(1) {(2 / 3)}^{0} + (1) {(2 / 3)}^{1} + (1 / 2) (4 / 9) (6 / 4)} = 0.5 \\ = & probability of no customers in the system \\ L & = & [\frac{(20) (30) {(20 / 30)}^{2}}{(2 - 1)! {[(2) (30 - 20)]}^{2}}] 0.5 + \frac{20}{30} = 0.75 customer in the system on the average \\ W & = & \frac{L}{λ} = \frac{3 / 4}{20} = \frac{3}{80} hour = 2 .25 minutes = average time the customer spends in the total system \\ L_{q} & = & L - \frac{λ}{μ} = \frac{3}{4} - \frac{20}{30} = \frac{1}{12} = 0 .083 customer waiting for service in line on the average \\ W_{q} & = & \frac{L_{q}}{λ} = \frac{\frac{1}{2}}{20} = \frac{1}{240} hour= \frac{1}{4} minute= \begin{array}{l} Average waiting time of a customer in the queue \\ itself (not being serviced) \end{array} \\ ρ & = & \frac{λ}{m μ} = \frac{20}{2 (30)} = \frac{1}{3} = 0.33 = Utilization rate \end{array}$

You now have $(240 customers) \times (1 / 240 hour) = 1$ $(240 customers) \times (1 / 240 hour) = 1$ hour total customer waiting time per day.

$Total cost of 60 minutes of customer waiting time = (60 minutes) ($ 0.10 per minute) = $ 6$ $Total cost of 60 minutes of customer waiting time = (60 minutes) ($ 0.10 per minute) = $ 6$

Now you are ready to point out to Marty that the hiring of one additional clerk will save $96 $-$ $-$ $6 $=$ $=$ $90 of customer ill will per 12-hour shift. Marty responds that the hiring should also reduce the number of people who look at the line and leave as well as those who get tired of waiting in line and leave. You tell Marty that you are ready for two chili dogs, extra hot.
Solved Problem 12-4 Vacation Inns is a chain of hotels operating in the southwestern part of the United States. The company uses a toll-free telephone number to take reservations for any of its hotels. The average time to handle each call is 3 minutes, and an average of 12 calls are received per hour. The probability distribution that describes the arrivals is unknown. Over a period of time, it is determined that the average caller spends 6 minutes either on hold or receiving service. Find the average time in the queue, the average time in the system, the average number in the queue, and the average number in the system.

Solution

The probability distributions are unknown, but we are given the average time in the system (6 minutes). Thus, we can use Little’s Flow Equations:

$\begin{array}{l} W & = & 6 minutes = 6/60 hour = 0 .1 hour \\ λ & = & 12 per hour \\ μ & = & 60/3 = 20 per hour \\ Average time in queue & = & W_{q} = W - 1 / μ = 0.1 - 1 / 20 = 0.1 - 0.05 = 0.05 hour \\ Average number in system & = & L = λ W = 12 (0.1) = 1.2 callers \\ Average number in queue & = & L_{q} = λ W_{q} = 12 (0.05) = 0.6 caller \end{array}$ $\begin{array}{l} W & = & 6 minutes = 6/60 hour = 0 .1 hour \\ λ & = & 12 per hour \\ μ & = & 60/3 = 20 per hour \\ Average time in queue & = & W_{q} = W - 1 / μ = 0.1 - 1 / 20 = 0.1 - 0.05 = 0.05 hour \\ Average number in system & = & L = λ W = 12 (0.1) = 1.2 callers \\ Average number in queue & = & L_{q} = λ W_{q} = 12 (0.05) = 0.6 caller \end{array}$

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