Waiting Lines 1

To accompany Quantitative Analysis for Management, 8e by Render/Stair/Hanna

14-1 © 2003 by Prentice Hall, Inc. Upper Saddle River, NJ 07458

Waiting Lines and Queuing Waiting Lines and Queuing Theory ModelsTheory Models



IntroductionIntroduction

• The study of waiting lines, called queuing theory, is one of the oldest and most widely used Operations Research techniques.

• Waiting lines are an everyday occurrence, affecting people shopping for groceries, buying gasoline, or making a bank deposit.

• Queues, another term for waiting lines, may also take the form of machines waiting to be repaired, trucks in line to be unloaded, or airplanes lined up on a runway waiting for permission to take off.

• The three basic components of a queuing process are arrivals, service facilities, and the actual waiting line.In this chapter we discuss how analytical models of waiting lines can help managers evaluate the cost and effectiveness of service systems.



Waiting Line CostsWaiting Line Costs

• Most waiting line problems are centered on the question of finding the ideal level of services that a firm should provide.

"one of the goals of queuing analysis is finding the best level of service for an organization".

• In most cases, this level of service is an option over which management has control. An extra teller, for example, can be borrowed from another chore or can be hired and trained quickly if demand warrants it. This may not always be the case, though. A plant may not be able to locate or hire skilled mechanics to repair sophisticated electronic machinery.



Waiting Line Costs\continuedWaiting Line Costs\continued

• When an organization does have control, its objective is usually to find a happy medium between two extremes:

1. On the one hand, a firm can retain a large staff and provide many service facilities. This may result in excellent customer service, with seldom more than one or two customers in a queue. Customers are kept happy with the quick response and appreciate the convenience. This, however, can become expensive.

2. The other extreme is to have the minimum possible number of checkout lines, gas pumps, or teller windows open. This keeps the service cost down but may result in customer dissatisfaction."As the average length of the queue increases and poor service results, customers and goodwill may be lost."



Waiting Line Costs\continuedWaiting Line Costs\continued

"Managers must deal with the trade-off between the cost of providing good service and the cost of customer waiting time. The latter may be hard to quantify."

• One means of evaluating a service facility is thus to look at a total expected cost, a concept illustrated in the figure (the next slide). Total expected cost is the sum of expected service costs plus expected waiting costs.

• Service costs are seen to increase as a firm attempts to raise its level of service. As service improves in speed, however, the cost of time spent waiting in lines decreases.

"The goal is to find the service level that minimizes total expected cost."



Queuing Costs and Queuing Costs and Service LevelsService Levels

Cos

t of

Ope

rati

ng

Ser

vice

Fac

ilit

y

Service Level

Total Expected

Cost

Cost of Providing Service

Cost of Waiting

Time

Optimal Service Level



Waiting Line Costs\continuedWaiting Line Costs\continuedThree Rivers Shipping Company ExampleThree Rivers Shipping Company Example

• Three Rivers run a huge docking facility located on the Ohio River near Pittsburgh. Approximately five 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 a great deal of money, about $1,000 per hour. From experience, management estimates that 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 four teams of stevedores, only 2 hours. But each additional team of stevedores is also an expensive proposition, due to union contacts.

• 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 next table:



Waiting Line Cost AnalysisWaiting Line Cost AnalysisThree Rivers ShippingThree Rivers Shipping

Number of Stevedore Teams1 2 3 4

5 5 5 5

7 4 3 2

35 20 15 10

$1,000 $1,000 $1,000 $1,000

35,000 29,000 $15,000 $10,000

$6,000 $12,000 18,000 $24,000

$41,000 $32,000 $33,000 $34,000

Average waiting time per shipTotal ship hours lost

Est. cost per hour of idle ship timeValue of ships' lost timeStevedore teams salaryTotal Expected Cost

Avg. number of ships arriving per shift



Characteristics Of A Queuing Characteristics Of A Queuing SystemSystem

• We will study three parts of a queuing system:

1. The arrivals or inputs to the system (sometimes referred to as the calling population),

2. The queue or the waiting line itself, and

3. The service facility



1. Arrival Characteristics1. Arrival Characteristics

• The input source that generates arrivals or customers for the service system has three major characteristics.

It is important to consider:• the size of the calling population, • the pattern of arrivals at the queuing system, and • the behavior of the arrivals.



1. Arrival Characteristics1. Arrival Characteristics\continued\continued

Size of the Calling Population• Population sizes are considered to be either unlimited

(essentially infinite) or limited (finite).• For practical purposes, examples of unlimited population

includes cars arriving at a highway tollbooth, shoppers arriving at a supermarket, or students arriving to register for classes at large university."Most queuing models assume such an infinite calling population".

• An example of a finite population is a shop with only eight machines that break down and require service.




Pattern of Arrivals at the System• Customers either arrive at service facility according to

some known schedule (for example, one patient every 15 minutes or one student for advising every half hour) 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.

"The Poisson probability distribution is used in many queuing models to represent arrival patterns."



Poisson Distribution for Poisson Distribution for Arrival TimesArrival Times

! P(X)

X

e x

For X=0,1,2,3,4,……..

P(X)= probability of X arrivals

X= number of arrivals per unit of time

= average arrival rate

e= 2.7183



Poisson Distribution for Arrival Poisson Distribution for Arrival TimesTimes

! P(X)

X

e x

.00

.05

.10

.15

.20

.25

.30

.35

0 1 2 3 4 5 6 7 8 9 10 11X

P(X

)P(X), = 2

.00

.05

.10

.15

.20

.25

.30

0 1 2 3 4 5 6 7 8 9 10 11X

P(X

)

P(X), = 4




Behavior of the Arrival• Most queuing models assume that an arriving customer is a

patient customer.• Patient customers are people or machines that wait in the

queue until they are served and do not switch between lines.• Unfortunately, life and operations research are complicated

by the fact that people have been known to balk or renege.• 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.



2. Waiting Line Characteristics2. Waiting Line Characteristics

We will study the:

• Length of the queue

• Service priority/Queue discipline



2. Waiting Line Characteristics2. Waiting Line Characteristics\\continuedcontinued

Length of the queue• The length of a line can be either limited or unlimited.

• A queue is limited when it cannot increase to an infinite length. This may be the case in a small restaurant that has only 10 tables and can serve no more than 50 diners an evening.

• Analytic queuing models are treated in this chapter under an assumption of unlimited queue length.



2. Waiting Line Characteristics2. Waiting Line Characteristics\\continuedcontinued

Service priority/Queue discipline• Queue discipline refers to the rule by which customers in

the line are to receive service.

• Customers may be served according to one of the next disciplines:

1. first-in, first-out rule (FIFO)

2. priority discipline

3. last-come first-served (LCFS)

4. service in random order (SIRO)



3. Service Facility Characteristics3. Service Facility Characteristics

• It is important to examine two basic properties:

1. the configuration of the service system

2. the pattern of service times



3. Service Facility 3. Service Facility CharacteristicsCharacteristics\continued\continued

Basic Queuing System Configuration• Service systems are usually classified in terms of

number of channels and number of phases in service.

1. Number of channels• A single-channel system, with one server, is typified by

the drive-in bank that has only one open teller.

• A multi-channel system, when the bank has several tellers on duty and each customer waits in one common line for the first available teller.




2. Number of phases in service• A single-phase system, in this case the customer receives

service from only one station and then exists the system.

• A multiphase system, for example, if the restaurant require you to place your order at one station, pay at a second, and pick up the food at a third service stop.



Basic Queuing System Basic Queuing System ConfigurationsConfigurations

Single Channel, Single Phase

Queue

Servicefacility

Facility 1

Facility2

Single Channel, Multi-Phase

Queue Service Facility



Basic Queuing System Basic Queuing System ConfigurationsConfigurations

Servicefacility 1

Servicefacility 2

Servicefacility 3

Multi-Channel, Single Phase

Queue

Multi-Channel, Multiphase Phase

Queue Type 1 Service Facility

Type 1 Service Facility






The Pattern of Service Times• 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.




• The next figure illustrates that if service times follow an exponential distribution, the probability of any very long service time is low.

Pro

babi

lity

(for

Int

erva

ls o

f 1

Min

ute)

30 60 90 120 150 180

Average Service Time of 1 Hour

Average Service Time of 20 Minutes

X

MinutePer ServedNumber Average μ

0 μ 0, for x μxμef(x)



Identifying Models Using Identifying Models Using Kendall NotationKendall Notation

• D.G. Kendall developed a notation that has been widely accepted for specifying the pattern of arrivals, the service time distribution, and the number of channels in a queuing model.

• 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. The following letters are commonly used in Kendall notation:

• M= Poisson distribution for number of occurrences (or exponential times)

• D= Constant (deterministic) rate• G= General distribution with mean and variance known



Identifying Models Using Identifying Models Using Kendall NotationKendall Notation\continued\continued

• Thus, 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

(An M/M/2 model has Poisson arrivals, exponential service times, and two channels)

• If there are m distinct service channels in the queuing system with Poisson arrivals and exponential service, the Kendall notation would be M/M/m.



Identifying Models Using Identifying Models Using Kendall NotationKendall Notation\continued\continued

• 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.



Single-Channel Queuing Model With Single-Channel Queuing Model With Poisson Arrivals And Exponential Poisson Arrivals And Exponential

Service Times (MService Times (M//MM//1)1)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:

1. Arrivals are served on a FIFO basis.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.




Service Times (MService Times (M//MM//1)1)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.

When these seven conditions are met, we can develop a series of equations that define the queue's operating characteristics.



Single-Channel Queuing Model Single-Channel Queuing Model With Poisson Arrivals And With Poisson Arrivals And

Exponential Service Times (MExponential Service Times (M//MM//1)1)Performance Measures of Queuing Systems:• Average time each customer spends in the queue• Average length of the queue• Average time each customer spends in the system• Average number of customers in the system• Probability that the service facility will be idle• Utilization factor for the system• Probability of a specific number of customers in

the system




Service Times (MService Times (M//MM//1)1)Queuing Equations

We let

• λ = means number of arrivals per time period (for example, per hour)

• µ = means number of people or items served per time period

• When determining the arrival rate (λ) and the service rate (µ), the same time period must be used.




Service Times (MService Times (M//MM//1)1)

Queuing Equations The queuing equations follow:

-

L system,in number Average

-

1 W system,in time waitingAverage

- L queue,in number Average

2

q

- Wqueue, in the time waitingAverage q




Service Times (MService Times (M//MM//1)1)Queuing Equations The queuing equations follow:

(The utilization factor for the system, , that is, the probability that the service facility is being used)

Factor,n Utilizatio

1P Idle,Percent 0

(The percent idle time, P0, that is, the probability that no one is in the system)

1

k

knP (The probability that the number of

customers in the system is greater than

K, Pn>k)



Example: Arnold's Muffler Example: Arnold's Muffler Shop CaseShop Case

• 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, the shop owner, feels that all seven of the conditions for a single-channel model are met. He proceeds to calculate the performance measures of the systems.

Solution:

λ = 2 cars arriving per hour

µ = 3 cars serviced per hour



Example: Arnold's Muffler Example: Arnold's Muffler Shop CaseShop Case//solutionsolution

223

2

- L

cars in the system on the average

123

1

-

1 W

hour that an average car spends in the system

33.13

4

)23(3

2

- L

22

q

cars waiting on line on the average

3

2

)23(3

2

- Wq

hour = 40 minutes = average waiting

time per car




33.03

211P0

= probability that there are 0 cars in the system

• Calculate the probability that more than three cars are in the system:

1k

knP

198.03

213

implies that there is a 19.8% chance that more than three cars are in the system.



Introducing Costs into the Introducing Costs into the ModelModel

• As stated earlier, the solution to queuing problem may require management to make trade-off between the increased cost of providing better service and the decreased waiting costs derived from providing that service. These two costs are called the waiting cost and the service cost.

• The total service cost is

total service cost = (number of channels) (cost per channel)

total service cost = mCs

where

m = number of channels

Cs = service cost (labor cost) of each channel



Introducing Costs into the Introducing Costs into the ModelModel\\continuedcontinued

• 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) (cost of waiting)

= (number of arrivals) (average wait per arrival) CwSo, total waiting cost = (W)Cw

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

total waiting cost = (Wq)Cw



Introducing Costs into the Introducing Costs into the ModelModel\\continuedcontinued

The total costs of the queuing system:

Total cost = total service cost + total waiting cost

Total cost = mCs + (W)Cw (waiting time is based on the time in the system)

Total cost = mCs + (Wq)Cw (waiting time is based on the time in the queue)




• 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.

• 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 *16=32/3 hours. Hence, in this case,

total daily waiting cost = (8 hours per day) WqCw = (8)(2)(2/3)($10) = $106.67



Example: Arnold's Muffler Example: Arnold's Muffler Shop CaseShop Case\\continuedcontinued

• 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

total daily service cost = (8 hours per day) mCs = 8(1)($7)=$56

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 = $106.67 + $56 = $162.67




• 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.

• He first recomputes all the operating characteristics using new service rate of 4 mufflers per hour.




cars in the system on the average

hour that an average car spends in the system

cars waiting on line on the average

hour = 15 minutes = average waiting time per car

124

2

- L

2

1

24

1

-

1 W

2

1

8

4

)24(4

2

- L

22

q

4

1

8

2

)24(4

2

- Wq




= probability that there are 0 cars in the system

5.05.011P0

• It is quite evident that Smith's speed will result in considerably shorter queues and waiting times. For example, a customer would now spend and average of 1/2 hour in the system and 1/4 hour waiting in the queue, as opposed to 1 hour in the system and 2/3 hour in the queue with Blank as mechanic.



Example: Arnold's Muffler Example: Arnold's Muffler Shop CaseShop Case\\solutionsolution

The total daily waiting time cost with Smith as the mechanic will betotal daily waiting cost = (8 hours per day) WqCw = (8)(2)(1/4)($10)=$40 per day

service cost of Smith = 8 hours/day * $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.



• In this case two or more servers or channels are available to handle arriving customers.

• 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.

• The multiple-channel system presented here again assumes that arrivals follow a Poisson probability distribution and that service times are distributed exponentially.

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

• Other assumptions listed earlier for the single-channel model apply as well.

Multiple-Channel Queuing Model Multiple-Channel Queuing Model With Poisson Arrivals And With Poisson Arrivals And

Exponential Service Times (MExponential Service Times (M//MM//m)m)




Exponential Service Times (MExponential Service Times (M//MM//m)m)Equations for the Multichannel Queuing Model

If we let

• m = number of channel open

• λ = average arrival rate

• µ = average service rate at each channel

The following formulas may be used in the waiting line analysis.

1. The probability that there are zero customers or units in the system:

mm

mn

PmMn

n

n

!1

!1

11

0

0




Exponential Service Times (MExponential Service Times (M//MM//m)m)2. The average number of customers or units in the system:

3. The average time a unit spends in the waiting line or being serviced (in the system):

4. The average number of customers or units in line waiting for service

02 P !1

LMM

M

W L

LqL




Exponential Service Times (MExponential Service Times (M//MM//m)m)5. The average time a customer or unit spends in the queue

waiting for service

5. Utilization factor

1

Wq qL

W

M



Arnold’s Muffler Shop Arnold’s Muffler Shop Example (multichannel)Example (multichannel)

Suppose that Arnold opened 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 µ = 3 per hour. Customers, who would still arrive at the rate of λ = 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



Arnold’s Muffler Shop Example Arnold’s Muffler Shop Example (multichannel)(multichannel)//solutionsolution

5.0

2)3(2)3(2

32

!21

32

!1

121

0

0

n

n

n

n

P

75.03

2(0.5)

2)3(2!13

2)3(2 L 2

2

minutes 22.52

75.0 W

083.03

275.0Lq hour 0.0415

2

083.0Wq



Arnold’s Muffler Shop Example Arnold’s Muffler Shop Example (multichannel)(multichannel)//solutionsolution

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) λWqCw

=(8)(2)(0.0415)($10)= $6.64

total daily service cost = (8hours)mCs= (8)(2)($7)=$112

Total daily cost of the queuing system = $6.64 + $112= $118.64



Effect of service level on Arnold’s Effect of service level on Arnold’s Operating CharacteristicsOperating Characteristics

Operating characteristics

Level of ServiceOne mechanic

(Blank)µ=3Two

mechanics

µ=3 for each

One mechanic

(Smith)µ=4

P0 0.33 0.50 0.50

L 2 cars 0.75 car 1 car

W 60 minutes 22.5 min. 30 min.

Lq 1.33 cars 0.083 car 0.50

Wq 40 minutes 2.5 min. 15 min.

Total costs $162 $118.64 $112



Constant Service Time Model Constant Service Time Model (M(M//DD//1)1)

• Some service systems have constant service times instead of exponentially distributed times.

• When customers or equipment are processed according to a fixed cycle, as in the case of an automatic car wash, constant service rates are appropriate.

• Because constant rates are certain, the values for Lq, Wq, L, and W are always less than they would be in the models we have just discussed, which have variable service times.



Constant Service Time Model Constant Service Time Model (M(M//DD//1)1)

Equations for the Constant Service Time Model

1 W

L

2W

2L

q

2

q

q

q

W

L



Constant service modelConstant service model(Grocia-Golding Recycling Example)(Grocia-Golding Recycling Example)

GGR collects and compacts aluminum cans and glass bottles in New York city. Their truck drivers, who arrive to unload these materials for recycling, currently wait an average of 15 minutes before emptying their loads. The cost while waiting in queue is $ 60 /hr.

A new automated compactor can be purchased that will process truck loads at a constant rate of 12 trucks / hr.

Trucks arrive according to a Poisson dist. at an average rate of 8 /hr. If the new compactor is put in use, its cost will be amortized at a rate of $3 /truck unloaded. Evaluate the costs versus benefits of the purchase.



Constant service modelConstant service model(Grocia-Golding Recycling (Grocia-Golding Recycling

Example)Example)//solutionsolutionCurrent waiting cost/trip=(1/4 hour waiting now)($60/hour cost)

= $15/trip

New system: λ = 8 trucks/hour arriving

µ = 12 trucks/hour served

Waiting cost/trip with new compactor= (1/12 hr wait) ($ 60/hr cost) = $ 5/trip

Savings with new equipment = $ 15 (current) - $ 5 (new system) = $ 10/trip

Cost of new equipment amortized =$ 3/trip

Net savings = $ 10 - $ 3 =$ 7/trip

hrWq121

)812)(12(28

)(2



Finite Population Model Finite Population Model (M(M//MM//1 with finite source)1 with finite source)

• When there is a limited population of potential customers for a service facility, we need to consider a different queuing model.

• The limited population model permits any number of servers to be considered.

• The reason this model differs from the three earlier queuing models is that there is now a dependent relationship between the length of the queue and the arrival rate.

• To illustrate the extreme situation, if your factory had five machines and all were broken and awaiting repair, the arrival rate would drop to zero.

• In general, as the waiting line becomes longer in the limited population model, the arrival rate of customers or machines drops lower.




• In this case, we describe a finite calling population model that has the following assumptions:

1. There is only one server

2. The population of units seeking service is finite

3. Arrivals follow a Poisson distribution, and service times are exponentially distributed.

4. Customers are served on a first-comes, first-served basis.




• Equations for the Finite Population Modelusing

λ = mean arrival rate, µ = mean service rate, N= size of the population

N

n

n

nNN

0

0

)!(!

1P

0

0q

1 L

1L

PL

PN

q




0n

q

!

! P N)n P(n,

1W

W

PnN

N

W

LN

L

n

q

q



Finite Population modelFinite Population model(Department of Commerce Example)(Department of Commerce Example)

• Past records indicate that each of the five high-speed “page” printers at the U.S. Department of commerce, in Washington, needs repair after about 20 hours of use. Breakdowns have been determined to be Poisson distributed. The one technician on duty can service a printer in average of 2 hours, following an exponential dist.

To compute the system’s operation characteristics we first note that the mean arrival rate is λ = 1/20=0.05 printer/hour. The mean service rate is µ = ½=0.50 printer/hour. Then:

564.0)

5.005.0

()!5(

!51

)!(!

1P 5

00

0

n

nN

n

n

nnNN




.2.0)564.01)(05.0

5.005.0(51L 0q prinPN

.64.0)564.01(2.01 L 0 prinPLq

hrLN

Lq 91.0)05.0)(64.05(

2.0Wq

hrWq 91.25.0

191.0

1W




• If printer downtime costs $ 120/hr and the technician is paid $ 25/hr:

Total hurly cost = (average number of printers down) (cost per downtime hour) + cost / technician hr

=(0.64)($ 120) + $25= $ 101.8

Waiting Lines 1

Documents

Transcript of Waiting Lines 1