ch14_summary.htm

Waiting Line Costs (Analytical models of waiting lines.)
Find a happy medium between many service facilities and a minimum number of facilities.
Tradeoff between the cost of providing good service and the cost of customer waiting time.
Total expected Cost = sum of expected service costs + expected waiting times
As service increases in speed, the cost of customers waiting in lines decreases

Evaluating a Service Facility
One means of evaluating a service facility is thus to look at a total expected cost (see the figure above). Total expected cost is the sum of expected service costs and the expected waiting costs.

Service costs are seen to increase as a firm attempts to raise its level of service. For example, if three teams of stevedores, instead of two, are employed to unload a cargo ship, service costs are increased by the additional price of wages. As service improves in speed, however, the cost of time spent waiting in lines decreases. This waiting cost may reflect lost productivity of workers while their tools or machines are awaiting repairs or may simply be an estimate of the costs of customers lost because of poor service and long queues.

Example) Three Rivers Shipping Company Three Rivers run a huge docking facility.
Approximately 5 ships arrive to unload their cargoes of steel and ore during every 12-hour work shift.
Each hour that a ship sits idle in line waiting to be unloaded costs the firm $1,000 per hour.
If one team of stevedores is on duty to handle the unloading work, each ship will wait an average of 7 hours to be unloaded.
If two teams are working, the average waiting time drops to 4 hours; for three teams, it's 3 hours; and for four teams of stevedores, only 2 hours.

Three Rivers' superintendent would like to determine the optimal number of teams of stevedores to have on duty each shift.
The objective is to minimize total expected costs. This analysis is summarized in the table below.
To minimize the sum of service costs and waiting costs, the firm makes the decision to employ two teams of stevedores each shift.

Characteristics of a Queuing System Parts
Part 1) the arrivals or inputs to the system (sometimes referred to as the calling population)
Part 2) the queue or the waiting line itself
Part 3) the service facility.

Part 1) Arrival Characteristics
(1) The arrivals or input source that generates arrivals or customers for the service system has three major characteristics:
(i) the size of the calling population, (ii) the pattern of arrivals at the queuing system, and (iii) the behavior of the arrivals.

(i) Size The calling population is considered unlimited when the number of customers or arrivals on hand at any given moment is just a small portion of potential arrivals.
Examples of unlimited populations include cars arriving at a highway tollbooth, shoppers arriving at a supermarket, etc.
When this is not the case, modeling becomes much more complex. (An example of a finite population is a shop with only eight machines that might break down and require service.)

(ii) Pattern Customers either arrive at a service facility according to some known schedule (for example, one patient every 15 minutes) or else they arrive randomly.
Arrivals are considered random when they are independent of one another and their occurrence cannot be predicted exactly.
Frequently in queuing problems, the number of arrivals per unit of time can be estimated by a probability distribution known as the Poisson distribution.
For any given arrival rate, such as two customers per hour, or four trucks per minute, a discrete Poisson distribution can be established by using the formula

	l = average arrival rate
P(X) = probability of X arrivals	X = number of arrivals per unit of time = average arrival rate.

These probabilities, as well as others for l = 2 and l = 4, are shown in the Figure below.
Notice that the chances that 9 or more customers will arrive in a particular time period are virtually nil when the average arrivals are 2 and 4.
Arrivals are not always Poisson (they may follow some other distribution)
They should be examined to make certain that they are well approximated by Poisson.
This usually involves observing arrivals, plotting the data, and applying statistical measures of goodness of fit, a topic discussed in more advanced texts (like Marketing Research Statistics, MTH 334).

Poisson Probability Distributions for Arrival Times (figure above)

(iii) Behavior (of the Arrivals) Patient customers are people or machines that wait in the queue until they are served and do not switch between lines.
Balking refers to customers who refuse to join the waiting line because it is too long to suit their needs or interests.
Reneging customers are those who enter the queue but then become impatient and leave without completing their transaction.

Part 2) Waiting Line Characteristics (i) The length of a line and (ii) queue discipline.
(i) The length of a line can be either limited or unlimited.
A queue is limited when it cannot, by law of physical restrictions, increase to an infinite length (10 tables and 50 diners an evening).
Analytic queuing models are treated in this chapter under an assumption of unlimited queue length.
A queue is unlimited when its size is unrestricted, as in the case of the tollbooth serving arriving automobiles.

(ii) Queue discipline. This refers to the rule by which customers in the line are to receive service.
Most systems use a queue discipline known as the first-in, first-out rule (FIFO).
(But) Patients who are critically injured will move ahead in treatment priority over patients with broken fingers or noses.
Shoppers with fewer than 10 items may be allowed to enter the express checkout queue but are then treated as first-come, first-served.
Computer programming runs may operate under priority scheduling (for paychecks)

Part 3) Service Facility Characteristics (i) The basic queuing system configuration and (ii) the service time distribution (pattern of service times).

(i) Basic Queuing System Configurations Service systems are usually classified in terms of
Number of channels, or number of servers:
A single-channel system, with one server, is typified by the drive-in bank that has only one open teller
A multi-channel system would exist if the bank had several tellers on duty and each customer waited in one common line for the first available teller

Number of phases, or number of service stops, that must be made.
A single-phase system is one in which the customer receives service from only one station and then exits the system (a fast-food restaurant in which the person who takes your order also brings you the food and takes your money is a single-phase system).
A multiphase system example is a restaurant that requires you to place your order at one station, pay at a second, and pick up food at a third.

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(ii) Service Time Distribution. Service patterns are like arrival patterns in that they can be either constant or random.
If service time is constant, it takes the same amount of time to take care of each customer.
This is the case in a machine-performed service operation such as an automatic car wash.
More often, service times are randomly distributed.
In many cases it can be assumed that random service times are described by the negative exponential probability distribution.
This is a mathematically convenient assumption if arrival rates are Poisson distributed.
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The figure below illustrates that if service times follow an exponential distribution, the probability of any very long service time is low.
For example, when an average service time is 20 minutes, seldom if ever will a customer require more than 90 minutes in the service facility.
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Many of the models' theoretical underpinnings are based on the assumption of Poisson arrivals and exponential services.
Before they are applied, however, one should check by observing, collecting, and plotting service time data.
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Two Examples of Exponential Distributions for Service Times (figure above)

Identifying Models Using Kendall Notation The basic three-symbol Kendall notation is in the form
arrival distribution / service time distribution / number of service channels open
where specific letters are used to represent probability distributions.
M = Poisson distribution for number of occurrences (or exponential times)
D = Constant (deterministic) rate
G = General distribution with mean and variance known
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A single channel model with Poisson arrivals and exponential service times would be represented by
M / M /1
When a second channel is added, we would have
M /M / 2
If there are m distinct service channels in the queuing system with Poisson arrivals and exponential service times, the Kendall notation would be
M / M / m
A three-channel system with Poisson arrivals and constant service time would be identified as
M / D / 3
A four-channel system with Poisson arrivals and service times that are normally distributed would be identified as
M / G / 4.
There is a more detailed notation with additional terms that indicate the maximum number in the system and the population size.
When these are omitted, it is assumed there is no limit to the queue length or the population size.
Most of the models we study here will have those 2 above properties.
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SINGLE-CHANNEL QUEUING MODEL WITH POISSON ARRIVALS AND EXPONENTIAL SERVICE TIMES (M / M / 1)
Performance in a typical service system.
After these numeric measures (below) 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.
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Assumptions of the Model
The single-channel, single-phase model considered here is one of the most widely used and simplest queuing models.
Assume that seven conditions exist:

2. Every arrival waits to be served regardless of the length of the line; that is, there is no balking or reneging.

3. Arrivals are independent of preceding arrivals, but the average number of arrivals (the arrival rate) does not change over time.

4. Arrivals are described by a Poisson probability distribution and come from an infinite or very large population.

5. Service times also vary from one customer to the next and are independent of one another, but their average rate is known.

6. Service times occur according to the negative exponential probability distribution.

7. The average service rate is greater than the average arrival rate.
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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 complex, so the resulting formulas only follow.
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Queuing Equations
(Note: All values can be computed on your TI-83, 86, and/or 89 by simply entering l and m.
Get the programs from me.)

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 3 per hour, or about 1 every 20 minutes. Customers needing this service arrive at the shop on the average of 2 per hour. Larry Arnold, 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.

(Note: All values can be computed on your TI-83, 86, and/or 89 by simply entering l and m.
Get the programs from me.)

Introducing Costs into the Model - Economic analysis of the impact the characteristics of the queuing system.
Identify optimal decisions or consider cost factors.
May require management to make a trade-off between the increased cost of providing better service and the decreased waiting costs derived from providing that service.

    total service cost = m C_s(14-9)
    where
    m = number of channels
    C_S = service cost (labor cost) of each channel

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

total waiting cost = (total time spent waiting by all arrivals) x (cost of waiting)
= ((number of arrivals) x (average wait per arrival)) x C_w

total waiting cost = (l W) C_w (14-10)
If the waiting time cost is based on time in the queue, this becomes

These costs are based on whatever time units (often hours) are used in determining l.
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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

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 $10 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 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 x 16 = 32/3, or 10²/₃hours. Hence, in this case,

total daily waiting cost = (8 hours per day) x (l W_q) C_w = (8)(2)(²/₃)($10) = $106.67

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 $7 per hour.

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

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 4 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 $9 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.
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He first re-computes all the operating characteristics using a new service rate of 4 mufflers per hour..

Probability of more than k cars in the system

k P_n>k = (l / m)^{k + 1}= (2 / 4)^{k + 1}

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 ½ hour in the system and ¼ hour waiting in the queue, as opposed to 1 hour in the system and 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) x (l W_q) C_w = (8)(2)(¼)($10) = $40.00 per day
Notice that the total time spent waiting for the 16 customers per day is now
    (16 cars per day) x (¼ 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 3 per hour to 4 per hour.
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The service cost will go up due to the higher salary, but the overall cost will decrease, as we see here:
    service cost of Smith = 8 hours/day x $9/hour = $72 per day
    total expected cost   = waiting cost + service cost = $40 + $72 = $112 per day
Because the total daily expected cost with Blank as mechanic was $162, Arnold may very well decide to hire Smith and reduce costs by $162 -$112 = $50 per day.

Enhancing the queuing environment by making the wait less unpleasant may reduce C_w as customers will not be as upset by having to wait. Sometimes, reducing the total cost in this way is easier than reducing the total cost by lowering W or W_q. In the case of Arnold's Muffler Shop, Arnold might consider putting a television in the waiting room and remodeling this room so customers feel more comfortable while waiting for their cars to be serviced.

MULTIPLE-CHANNEL QUEUING MODEL WITH POISSON ARRIVALS AND EXPONENTIAL SERVICE TIMES (M I M I m)

Multiple-channel queuing system, in which two or more servers or channels are available to handle arriving customers.

Assume that customers awaiting service form one single line and then proceed to the first available server.

An example of such a multichannel, single-phase waiting line is found in many banks today.

A common line is formed and the customer at the head of the line proceeds to the first free teller.

Assume that arrivals follow a Poisson probability distribution and service times are distributed exponentially.

Service is first come, first served, and all servers are assumed to perform at the same rate.

Equations for the Multichannel Queuing Model
(Note: All values can be computed on your TI-83, 86, and/or 89 by simply entering m, l and m.
Get the programs from me.)

These equations are used in exactly the same fashion and provide the same type of information as did the simpler model.

Earlier, Larry Arnold examined two options. He could retain his current mechanic, Reid Blank, at a total expected cost of $162 per day; or he could fire Blank and hire a slightly more expensive but faster worker named Jimmy Smith. With Smith on board, service system costs could be reduced to $112 per day.

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A third option is now explored. Arnold finds that at minimal after-tax cost he can open a second garage bay in which mufflers can be installed. Instead of firing his first mechanic, Blank, he would hire a second worker. The new mechanic would be expected to install mufflers at the same rate as Blank - about m = 3 per hour. Customers, who would still arrive at the rate of l = 2 per hour, would wait in a single line until one of the two mechanics is free. To find out how this option compares with the old single-channel waiting line system, Arnold computes several operating characteristics for the m = 2 channel system.

(Note: All values can be computed on your TI-83, 86, and/or 89 by simply entering m, l and m.
Get the programs from me.)

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These data are compared with earlier operating characteristics in the next table (below). The increased service from opening a second channel has a dramatic effect on almost all characteristics. In particular, time spent waiting in line drops from 40 minutes with one mechanic (Blank) or 15 minutes with Smith down to only 2½ minutes! Similarly, the average number of cars in the queue falls to 0.083 (about 1/12 of a car).But does this mean that a second bay should be opened?

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To complete his economic analysis, Arnold assumes that the second mechanic would be paid the same as the current one, Blank, namely, $7 per hour. The daily waiting cost now will be
total daily waiting cost = (8 hours per day) x (l W_q) C_w = (8)(2)(0.0415)($10) = $6.64
Notice that the total waiting time for the 16 cars per day is (16 cars/day) x (0.0415 hour/car) = 0.664 hours per day instead of the 10.67 hours with only one mechanic.

The service cost is doubled, as there are two mechanics, so this is
total daily service cost = (8 hours per day) m C_s= (8)2($7) = $112

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

As you recall, total cost with just Blank as mechanic was found to be $162 per day. Cost with just Smith was just $112. Although opening a second channel would be likely to have a positive effect on customer goodwill and hence lower the cost of waiting time, it means an increase in the cost of providing service. Look back at the first figure in this document and you will see that such trade-offs are the basis of queuing theory. Arnold's decision is to replace his present worker with the speedier Smith and not to open a second service bay.
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Operating Characteristic	One Mechanic Blank m = 3	Two Mechanics m = 3 for each	One Fast Mechanic Smith m = 4

Number or teams of Stevedores Working	1 2 3 4