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12.3 Single-Channel Queuing Model with Poisson Arrivals and Exponential Service Times (M/M /1)

In this section, we present an analytical approach to determine important measures of performance in a typical service system. After these numeric measures have been computed, it will be possible to add in cost data and begin to make decisions that balance desirable service levels with waiting line service costs.

Assumptions of the Model

The single-channel, single-phase model considered here is one of the most widely used and simplest queuing models. It involves assuming that seven conditions exist:

Arrivals are served on a FIFO basis.
Every arrival waits to be served, regardless of the length of the line; that is, there is no balking or reneging.
Arrivals are independent of preceding arrivals, but the average number of arrivals (the arrival rate) does not change over time.
Arrivals are described by a Poisson probability distribution and come from an infinite or very large population.
Service times also vary from one customer to the next and are independent of one another, but their average rate is known.
Service times occur according to the negative exponential probability distribution.
The average service rate is greater than the average arrival rate.

When these seven conditions are met, we can develop a series of equations that define the queue’s operating characteristics. The mathematics used to derive each equation is rather complex and outside the scope of this book, so we will just present the resulting formulas here.

Queuing Equations

We let

\begin{array}{l} λ & = & mean number of arrivals per time period \\ μ & = & mean number of people or items served per time period \end{array}

$\begin{array}{l} λ & = & mean number of arrivals per time period \\ μ & = & mean number of people or items served per time period \end{array}$

When determining the arrival rate $(λ)$ $(λ)$ and the service rate $(μ),$ $(μ),$ the same time period must be used. For example, if $λ$ $λ$ is the average number of arrivals per hour, then $μ$ $μ$ must indicate the average number that could be served per hour.

The queuing equations follow.

The average number of customers or units in the system, L—that is, the number in line plus the number being served:

$L = \frac{λ}{μ - λ}$ $L = \frac{λ}{μ - λ}$ (12-1)
The average time a customer spends in the system, W—that is, the time spent in line plus the time spent being served:

$W = \frac{1}{μ - λ}$ $W = \frac{1}{μ - λ}$ (12-2)
The average number of customers in the queue, $L_{q} :$ $L_{q} :$

$L_{q} = \frac{λ^{2}}{μ (μ - λ)}$ $L_{q} = \frac{λ^{2}}{μ (μ - λ)}$ (12-3)
The average time a customer spends waiting in the queue, $W_{q} :$ $W_{q} :$

$W_{q} = \frac{λ}{μ (μ - λ)}$ $W_{q} = \frac{λ}{μ (μ - λ)}$ (12-4)
The utilization factor for the system, $ρ$ $ρ$ (the Greek lowercase letter rho)—that is, the probability that the service facility is being used:

$ρ = \frac{λ}{μ}$ $ρ = \frac{λ}{μ}$ (12-5)
The percent idle time, $P_{0}$ $P_{0}$ —that is, the probability that no one is in the system:

$P_{0} = 1 - \frac{λ}{μ}$ $P_{0} = 1 - \frac{λ}{μ}$ (12-6)
The probability that the number of customers in the system is greater than $k, P_{n > k} :$ $k, P_{n > k} :$

$P_{n > k} = {(\frac{λ}{μ})}^{k + 1}$ $P_{n > k} = {(\frac{λ}{μ})}^{k + 1}$ (12-7)

Arnold’s Muffler Shop Case

We now apply these formulas to the case of Arnold’s Muffler Shop in New Orleans. Arnold’s mechanic, Reid Blank, is able to install new mufflers at an average rate of three per hour, or about one every 20 minutes. Customers needing this service arrive at the shop on the average of two per hour. Larry Arnold, the shop owner, studied queuing models in an MBA program and feels that all seven of the conditions for a single-channel model are met. He proceeds to calculate the numerical values of the preceding operating characteristics:

\begin{array}{l} λ & = & 2 cars arriving per hour \\ μ & = & 3 cars serviced per hour \\ L & = & \frac{λ}{μ - λ} = \frac{2}{3 - 2} = \frac{2}{1} = 2 cars in the system on the average \\ W & = & \frac{1}{μ - λ} = \frac{1}{3 - 2} = 1 hour that an average car spends in the system \\ L_{q} & = & \frac{λ^{2}}{μ (μ - λ)} = \frac{2^{2}}{3 | (3 - 2)} = \frac{4}{3 (1)} = \frac{4}{3} = 1 .33 cars waiting in line on the average \\ W_{q} & = & \frac{λ}{μ (μ - λ)} = \frac{2}{3 (3 - 2)} = \frac{2}{3} hour = 40 minutes = average waiting time per car \end{array}

$\begin{array}{l} λ & = & 2 cars arriving per hour \\ μ & = & 3 cars serviced per hour \\ L & = & \frac{λ}{μ - λ} = \frac{2}{3 - 2} = \frac{2}{1} = 2 cars in the system on the average \\ W & = & \frac{1}{μ - λ} = \frac{1}{3 - 2} = 1 hour that an average car spends in the system \\ L_{q} & = & \frac{λ^{2}}{μ (μ - λ)} = \frac{2^{2}}{3 | (3 - 2)} = \frac{4}{3 (1)} = \frac{4}{3} = 1 .33 cars waiting in line on the average \\ W_{q} & = & \frac{λ}{μ (μ - λ)} = \frac{2}{3 (3 - 2)} = \frac{2}{3} hour = 40 minutes = average waiting time per car \end{array}$

Note that W and $W_{q}$ $W_{q}$ are in hours, since $λ$ $λ$ was defined as the number of arrivals per hour.

\begin{array}{l} ρ & = & \begin{array}{l} \frac{λ}{μ} = \frac{2}{3} = 0.67 = percentage of time mechanic is busy, or the probability \\ that the server is busy \end{array} \\ P_{0} & = & 1 - \frac{λ}{μ} = 1 - \frac{2}{3} = 0.33 = probability that there are 0 cars in the system \end{array}

$\begin{array}{l} ρ & = & \begin{array}{l} \frac{λ}{μ} = \frac{2}{3} = 0.67 = percentage of time mechanic is busy, or the probability \\ that the server is busy \end{array} \\ P_{0} & = & 1 - \frac{λ}{μ} = 1 - \frac{2}{3} = 0.33 = probability that there are 0 cars in the system \end{array}$

A table shows the “k” values and the related probabilities of more than “k” cars in the system. — Probability of More Than k Cars in the System

12.3-3 Full Alternative Text

A screenshot of a spreadsheet shows the solution for Arnold’s Muffler shop. — Program 12.1 Excel QM Solution for Arnold’s Muffler Shop

Figure 12.1 Full Alternative Text

Using Excel QM On The Arnold’s Muffler Shop Queue

To use Excel QM for this problem, from the Excel QM menu, select Waiting Lines - Single Channel (M/M/1). When the spreadsheet appears, enter the arrival rate (2) and service rate (3). All the operating characteristics will automatically be computed, as demonstrated in Program 12.1.

A screenshot of a spreadsheet shows the solution for Arnold’s Muffler multichannel example. — Program 12.2 Excel QM Solution for Arnold’s Muffler Multichannel Example

Figure 12.2 Full Alternative Text

Introducing Costs Into The Model

Now that the characteristics of the queuing system have been computed, Arnold decides to do an economic analysis of their impact. The waiting line model was valuable in predicting potential waiting times, queue lengths, idle times, and so on. But it did not identify optimal decisions or consider cost factors. As stated earlier, the solution to a queuing problem may require management to make a trade-off between the increased cost of providing better service and the decreased waiting cost derived from providing that service. These two costs are called the waiting cost and the service cost.

The total service cost is

\begin{array}{l} Total service cost & = & (Number of channels)(Cost per channel) \\ Total service cost & = & m C_{s} \end{array}

$\begin{array}{l} Total service cost & = & (Number of channels)(Cost per channel) \\ Total service cost & = & m C_{s} \end{array}$ (12-8)

where

\begin{array}{l} m & = & number of channels \\ C_{s} & = & service cost (labor cost) of each channel \end{array}

$\begin{array}{l} m & = & number of channels \\ C_{s} & = & service cost (labor cost) of each channel \end{array}$

The waiting cost when the waiting time cost is based on time in the system is

\begin{array}{l} Total waiting cost & = & (Total time spent waiting by all arrivals)(Cost of waiting) \\ = & (Number of arrivals)(Average wait per arrival) C_{w} \end{array}

$\begin{array}{l} Total waiting cost & = & (Total time spent waiting by all arrivals)(Cost of waiting) \\ = & (Number of arrivals)(Average wait per arrival) C_{w} \end{array}$

Total waiting cost = (λ W) C_{w}

$Total waiting cost = (λ W) C_{w}$ (12-9)

If the waiting time cost is based on time in the queue, this becomes

Total waiting cost = (λ W_{q}) C_{w}

$Total waiting cost = (λ W_{q}) C_{w}$ (12-10)

These costs are based on whatever time units (often hours) are used in determining $λ .$ $λ .$ Adding the total service cost to the total waiting cost, we have the total cost of the queuing system. When the waiting cost is based on the time in the system, this is

\begin{array}{l} Total cost & = & Total service cost + Total waiting cost \\ Total cost & = & m C_{s} + λ W C_{w} \end{array}

$\begin{array}{l} Total cost & = & Total service cost + Total waiting cost \\ Total cost & = & m C_{s} + λ W C_{w} \end{array}$ (12-11)

When the waiting cost is based on time in the queue, the total cost is

Total cost = m C_{s} + λ W_{q} C_{w}

$Total cost = m C_{s} + λ W_{q} C_{w}$ (12-12)

At times, we may wish to determine the daily cost, and then we simply find the total number of arrivals per day. Let us consider the situation for Arnold’s Muffler Shop.

Arnold estimates that the cost of customer waiting time, in terms of customer dissatisfaction and lost goodwill, is $50 per hour of time spent waiting in line. (After customers’ cars are actually being serviced on the rack, customers don’t seem to mind waiting.) Because on the average a car has a $^{2} /_{3}$ $^{2} /_{3}$ hour wait and there are approximately 16 cars serviced per day (2 per hour times 8 working hours per day), the total number of hours that customers spend waiting for mufflers to be installed each day is $^{2} /_{3} \times 16 =^{32} /_{3},$ $^{2} /_{3} \times 16 =^{32} /_{3},$ or 10 $^{2} /_{3}$ $^{2} /_{3}$ hours. Hence, in this case,

Total daily waiting cost = (8 hours per day) λ W_{q} C_{w} = (8) (2) (^{2} /_{3}) ($ 50) = $ 533.33

$Total daily waiting cost = (8 hours per day) λ W_{q} C_{w} = (8) (2) (^{2} /_{3}) ($ 50) = $ 533.33$

The only other cost that Larry Arnold can identify in this queuing situation is the pay rate of Reid Blank, the mechanic. Blank is paid $15 per hour:

Total daily service cost = (8 hours per day) m C_{s} = 8 (1) ($ 15) = $ 120

$Total daily service cost = (8 hours per day) m C_{s} = 8 (1) ($ 15) = $ 120$

The total daily cost of the system as it is currently configured is the total of the waiting cost and the service cost, which gives us

Total daily cost of the queuing system = $ 533.33 + $ 120 = $ 653.33

$Total daily cost of the queuing system = $ 533.33 + $ 120 = $ 653.33$

Now comes a decision. Arnold finds out through the muffler business grapevine that the Rusty Muffler, a cross-town competitor, employs a mechanic named Jimmy Smith who can efficiently install new mufflers at the rate of four per hour. Larry Arnold contacts Smith and inquires as to his interest in switching employers. Smith says that he would consider leaving the Rusty Muffler but only if he were paid a $20 per hour salary. Arnold, being a crafty businessman, decides that it may be worthwhile to fire Blank and replace him with the speedier but more expensive Smith.

He first recomputes all the operating characteristics using a new service rate of four mufflers per hour:

\begin{array}{l} λ & = & 2 cars arriving per hour \\ μ & = & 4 cars serviced per hour \\ L & = & \frac{λ}{μ - λ} = \frac{2}{4 - 2} = 1 car in the system on the average \\ W & = & \frac{1}{μ - λ} = \frac{1}{4 - 2} = \frac{1}{2} hour in the system on the average \\ L_{q} & = & \frac{λ^{2}}{μ (μ - λ)} = \frac{2^{2}}{4 | (4 - 2)} = \frac{4}{8} = \frac{1}{2} car waiting in line on the average \\ W_{q} & = & \begin{matrix} \frac{λ}{μ (μ - λ)} = \frac{2}{4 (4 - 2)} = \frac{2}{8} = \frac{1}{4} hour= & \begin{array}{l} 15 minutes average waiting time \\ per car in the queue \end{array} \end{matrix} \\ ρ & = & \frac{λ}{μ} = \frac{2}{4} = 0.5 = percentage of time mechanic is busy \\ P_{0} & = & 1 - \frac{λ}{μ} = 1 - 0.5 = 0.5 = probability that there are 0 cars in the system \end{array}

$\begin{array}{l} λ & = & 2 cars arriving per hour \\ μ & = & 4 cars serviced per hour \\ L & = & \frac{λ}{μ - λ} = \frac{2}{4 - 2} = 1 car in the system on the average \\ W & = & \frac{1}{μ - λ} = \frac{1}{4 - 2} = \frac{1}{2} hour in the system on the average \\ L_{q} & = & \frac{λ^{2}}{μ (μ - λ)} = \frac{2^{2}}{4 | (4 - 2)} = \frac{4}{8} = \frac{1}{2} car waiting in line on the average \\ W_{q} & = & \begin{matrix} \frac{λ}{μ (μ - λ)} = \frac{2}{4 (4 - 2)} = \frac{2}{8} = \frac{1}{4} hour= & \begin{array}{l} 15 minutes average waiting time \\ per car in the queue \end{array} \end{matrix} \\ ρ & = & \frac{λ}{μ} = \frac{2}{4} = 0.5 = percentage of time mechanic is busy \\ P_{0} & = & 1 - \frac{λ}{μ} = 1 - 0.5 = 0.5 = probability that there are 0 cars in the system \end{array}$

Probability of More Than k Cars in the System
k	$P_{n > k} = {(^{2} /_{4})}^{k + 1}$ $P_{n > k} = {(^{2} /_{4})}^{k + 1}$
0	0.500
1	0.250
2	0.125
3	0.062
4	0.031
5	0.016
6	0.008
7	0.004

It is quite evident that Smith’s speed will result in considerably shorter queues and waiting times. For example, a customer would now spend an average of $^{1} /_{2}$ $^{1} /_{2}$ hour in the system and $^{1} /_{4}$ $^{1} /_{4}$ hour waiting in the queue, as opposed to 1 hour in the system and $^{2} /_{3}$ $^{2} /_{3}$ hour in the queue with Blank as mechanic. The total daily waiting time cost with Smith as the mechanic will be

Total daily waiting cost = (8 hours per day) λ W_{q} C_{w} = (8) (2) (\frac{1}{4}) ($ 50) = $ 200 per day

$Total daily waiting cost = (8 hours per day) λ W_{q} C_{w} = (8) (2) (\frac{1}{4}) ($ 50) = $ 200 per day$

Notice that the total time spent waiting for the 16 customers per day is now

(16 cars per day) \times (^{1} /_{4} hour per car) = 4 hours

$(16 cars per day) \times (^{1} /_{4} hour per car) = 4 hours$

instead of 10.67 hours with Blank. Thus, the waiting is much less than half of what it was, even though the service rate only changed from three per hour to four per hour.

The service cost will go up due to the higher salary, but the overall cost will decrease, as we see here:

\begin{array}{l} Service cost of Smith & = & 8 hours/day \times $20/hour = $160 per day \\ Total expected cost & = & Waiting cost + Service cost = $200 + $160 \\ = & $360 per day \end{array}

$\begin{array}{l} Service cost of Smith & = & 8 hours/day \times $20/hour = $160 per day \\ Total expected cost & = & Waiting cost + Service cost = $200 + $160 \\ = & $360 per day \end{array}$

Because the total daily expected cost with Blank as mechanic was $653.33, Arnold may very well decide to hire Smith and reduce costs by $$ 653.33 - $ 360 = $ 293.33$ $$ 653.33 - $ 360 = $ 293.33$ per day.

Enhancing the Queuing Environment

Although reducing the waiting time is an important factor in reducing the waiting time cost of a queuing system, a manager might find other ways to reduce this cost. The total waiting time cost is based on the total amount of time spent waiting (based on W or $W_{q}$ $W_{q}$ ) and the cost of waiting $(C_{w}) .$ $(C_{w}) .$ Reducing either of these will reduce the overall cost of waiting. Enhancing the queuing environment by making the wait less unpleasant may reduce $C_{w}$ $C_{w}$ , as customers will not be as upset by having to wait. There are magazines in the waiting room of doctors’ offices for patients to read while waiting. There are tabloids on display by the checkout lines in grocery stores, and customers read the headlines to pass time while waiting. Music is often played while telephone callers are placed on hold. At major amusement parks, there are video screens and televisions in some of the queue lines to make the wait more interesting. For some of these, the waiting line is so entertaining that it is almost an attraction itself.

All of these things are designed to keep the customer busy and to enhance the conditions surrounding the waiting so that it appears that time is passing more quickly than it actually is. Consequently, the cost of waiting $(C_{w})$ $(C_{w})$ becomes lower, and the total cost of the queuing system is reduced. Sometimes, reducing the total cost in this way is easier than reducing the total cost

In Action Ambulances in Chile Evaluate and Improve Performance Metrics Through Queuing Theory

Researchers in Chile turned to queuing theory to evaluate and improve ambulance services. They investigated the key performance indicators that are of value to an ambulance company manager (e.g., number of ambulances deployed, ambulance base locations, operating costs) and the key performance indicators that are of value to a customer (e.g., waiting time until service, queue prioritization system).

The somewhat surprising result that the researchers brought to light was that seemingly simple metrics such as waiting time in queue, W_q, were of the most value for ambulance operations. Reducing wait times saves the ambulance company money in gasoline and transportation costs, and, more importantly, it can save lives for its customers. In other words, sometimes simpler is better.

Source: Based on Marcos Singer and Patricio Donoso, “Assessing an Ambulance Service with Queuing Theory,” Computers & Operations Research 35, 8 (2008): 2549–2560, © Trevor S. Hale.

by lowering W or $W_{q} .$ $W_{q} .$ In the case of Arnold’s Muffler Shop, Arnold might consider remodeling the waiting room and putting in a television so customers feel more comfortable while waiting for their cars to be serviced.

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Table of Contents for 12.3 Single-Channel Queuing Model with Poisson Arrivals and Exponential Service Times (M/M /1)

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Table of Contents for
12.3 Single-Channel Queuing Model with Poisson Arrivals and Exponential Service Times (M/M /1)