C – 1 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Waiting Lines C For Operations Management, 9e by Krajewski/Ritzman/Malhotra

C – 1Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Waiting LinesWaiting LinesC

For For Operations Management, 9eOperations Management, 9e by by Krajewski/Ritzman/Malhotra Krajewski/Ritzman/Malhotra © 2010 Pearson Education© 2010 Pearson Education

PowerPoint Slides PowerPoint Slides by Jeff Heylby Jeff Heyl


Why Waiting Lines FormWhy Waiting Lines Form

Define customers

Waiting lines form Temporary imbalance between demand and capacity Can develop even if processing time is constant

No waiting line if both demand and service rates are constant and service rate > than demand

Affects process design, capacity planning, process performance, and ultimately, supply chain performance


Uses of Waiting-Line Theory Uses of Waiting-Line Theory

Applies to many service or manufacturing situations Relating arrival and service-system processing

characteristics to output

Service is the act of processing a customer Hair cutting in a hair salon Satisfying customer complaints Processing production orders Theatergoers waiting to purchase tickets Trucks waiting to be unloaded at a warehouse Patients waiting to be examined by a physician


Structure of Waiting-Line ProblemsStructure of Waiting-Line Problems

1. An input, or customer population, that generates potential customers

2. A waiting line of customers

3. The service facility, consisting of a person (or crew), a machine (or group of machines), or both necessary to perform the service for the customer

4. A priority rule, which selects the next customer to be served by the service facility


Structure of Waiting-Line ProblemsStructure of Waiting-Line Problems

Customer population

Service system

Waiting line

Priority rule

Service facilities

Served customers

Figure C.1 – Basic Elements of Waiting-Line Models


Customer PopulationCustomer Population

The source of input

Customers are patient or impatient Patient customers wait until served Impatient customer either balk or join the line

and renege

Finite or infinite source Customers from a finite source reduce the

chance of new arrivals Customers from an infinite source do not

affect the probability of another arrival


The Service SystemThe Service System

Number of lines A single-line keeps servers uniformly busy and levels

waiting times among customers A multiple-line arrangement is favored when servers

provide a limited set of services

Arrangement of service facilities Single-channel, single-phase Single-channel, multiple-phase Multiple-channel, single-phase Multiple-channel, multiple-phase Mixed arrangement


Service facilities

(a) Single line Service facilities

(b) Multiple lines


Figure C.2 – Waiting-Line Arrangements



Service facility

(a) Single channel, single phase

(b) Single channel, multiple phase

Service facility 1

Service facility 2

Figure C.3 – Examples of Service Facility Arrangements


(c) Multiple channel, single phase

Service facility 1

Service facility 2


Service facility 3

Service facility 4

Service facility 1

Service facility 2

(d) Multiple channel, multiple phase




Routing for : 1–2–4Routing for : 2–4–3Routing for : 3–2–1–4

(e) Mixed arrangement

Service facility 1

Service facility 4

Service facility 3

Service facility 2



Priority RulePriority Rule

First-come, first-served (FCFS)—used by most service systems

Preemptive discipline—allows a higher priority customer to interrupt the service of another customer or be served ahead of another who would have been served first

Other rules Earliest due date (EDD) Shortest processing time (SPT)


Probability DistributionsProbability Distributions

The sources of variation in waiting-line problems come from the random arrivals of customers and the variation of service times

Arrival distribution Customer arrivals can often be described by the

Poisson distribution with mean = T and variance also = T

Arrival distribution is the probability of n arrivals in T time periods

Interarrival times are the average time between arrivals


Pn = e-T for n = 0, 1, 2,…(T)n

n!

Interarrival TimesInterarrival Times

where

Pn =Probability of n arrivals in T time periods

=Average numbers of customer arrivals per period

e =2.7183


Probability of Customer ArrivalsProbability of Customer Arrivals

EXAMPLE C.1

Management is redesigning the customer service process in a large department store. Accommodating four customers is important. Customers arrive at the desk at the rate of two customers per hour. What is the probability that four customers will arrive during any hour?

SOLUTION

In this case customers per hour, T = 1 hour, and n = 4 customers. The probability that four customers will arrive in any hour is

P4 = = e–2 = 0.0901624

[2(1)]4

4!e–2(1)


P(t ≤ T) = 1 – e-T

where

μ =average number of customers completing service per period

t =service time of the customer

T =target service time

Service TimeService Time

Service time distribution can be described by an exponential distribution with mean = 1/ and variance = (1/ )2

Service time distribution: The probability that the service time will be no more than T time periods can be described by the exponential distribution


Service Time Probability

EXAMPLE C.2

The management of the large department store in Example C.1 must determine whether more training is needed for the customer service clerk. The clerk at the customer service desk can serve an average of three customers per hour. What is the probability that a customer will require less than 10 minutes of service?

SOLUTION

We must have all the data in the same time units. Because = 3 customers per hour, we convert minutes of time to hours, or T = 10 minutes = 10/60 hour = 0.167 hour. Then

P(t ≤ T) = 1 – e–T

P(t ≤ 0.167 hour) = 1 – e–3(0.167) = 1 – 0.61 = 0.39


Using Waiting-Line Models Using Waiting-Line Models

Balance costs against benefits

Operating characteristics Line lengthNumber of customers in systemWaiting time in line Total time in systemService facility utilization


Single-Server ModelSingle-Server Model

Single-server, single line of customers, and only one phase

Assumptions are1. Customer population is infinite and patient

2. Customers arrive according to a Poisson distribution, with a mean arrival rate of

3. Service distribution is exponential with a mean service rate of

4. Mean service rate exceeds mean arrival rate

5. Customers are served FCFS

6. The length of the waiting line is unlimited


= Average utilization of the system =

L= Average number of customers in the service system

=

–

Lq= Average number of customers in the waiting line = L

W= Average time spent in the system, including service

=1

–

Wq= Average waiting time in line = W

n = Probability that n customers are in the system= (1 – )n

Single-Server ModelSingle-Server Model


Calculating the Operating Characteristics

EXAMPLE C.3

The manager of a grocery store in the retirement community of Sunnyville is interested in providing good service to the senior citizens who shop in her store. Currently, the store has a separate checkout counter for senior citizens. On average, 30 senior citizens per hour arrive at the counter, according to a Poisson distribution, and are served at an average rate of 35 customers per hour, with exponential service times. Find the following operating characteristics:

a. Probability of zero customers in the system

b. Average utilization of the checkout clerk

c. Average number of customers in the system

d. Average number of customers in line

e. Average time spent in the system

f. Average waiting time in line



SOLUTION

The checkout counter can be modeled as a single-channel, single-phase system. Figure C.4 shows the results from the Waiting-Lines Solver from OM Explorer.

Figure C.4 – Waiting-Lines Solver for Single-Channel, Single-Phase System



Both the average waiting time in the system (W) and the average time spent waiting in line (Wq) are expressed in hours. To convert the results to minutes, simply multiply by 60 minutes/ hour. For example, W = 0.20(60) minutes, and Wq = 0.1714(60) = 10.28 minutes.


Application C.1Application C.1

Customers arrive at a checkout counter at an average 20 per hour, according to a Poisson distribution. They are served at an average rate of 25 per hour, with exponential service times. Use the single-server model to estimate the operating characteristics of this system.

= 20 customer arrival rate per hour

= 25 customer service rate per hour

SOLUTION

1. Average utilization of system = = = 0.820

25



2. Average number of customers in the service system

= = 42025 – 20L =

–

Lq = L

W =1

–

Wq = W

3. Average number of customers in the waiting line

= 0.8(4) = 3.2

4. Average time spent in the system, including service

= = 0.2125 – 20

5. Average waiting time in line = 0.8(0.2) = 0.16


Analyzing Service RatesAnalyzing Service Rates

EXAMPLE C.4

The manager of the Sunnyville grocery in Example C.3 wants answers to the following questions:

a. What service rate would be required so that customers average only 8 minutes in the system?

b. For that service rate, what is the probability of having more than four customers in the system?

c. What service rate would be required to have only a 10 percent chance of exceeding four customers in the system?



SOLUTION

The Waiting-Lines Solver from OM Explorer could be used iteratively to answer the questions. Here we show how to solve the problem manually.

a. We use the equation for the average time in the system and solve for

W =1

–

8 minutes = 0.133 hour = 1

– 30

0.133 – 0.133(30) = 1

= 37.52 customers/hour



b. The probability of more than four customers in the system equals 1 minus the probability of four or fewer customers in the system.

P = 1 – Pn

4

n = 0

= 1 – (1 – ) n

4

n = 0

= = 0.8030

37.52

and

Then,

P = 1 – 0.21(1 + 0.8 + 0.82 + 0.83 + 0.84)

= 1 – 0.672 = 0.328

Therefore, there is a nearly 33 percent chance that more than four customers will be in the system.



c. We use the same logic as in part (b), except that is now a decision variable. The easiest way to proceed is to find the correct average utilization first, and then solve for the service rate.

P = 1 – (1 – )(1 + + 2 + 3 + 4)

= 1 – (1 – )(1 + + 2 + 3 + 4) + (1 + + 2 + 3 + 4)

= 1 – 1 – – 2 – 3 – 4 + + 2 + 3 + 4 + 5 = 5

= P1/5

or

If P = 0.10

= (0.10)1/5 = 0.63



Therefore, for a utilization rate of 63 percent, the probability of more than four customers in the system is 10 percent.

For = 30, the mean service rate must be

= 47.62 customers/hour

= 0.6330



In the checkout counter example, what service rate is required to have customers average only 10 minutes in the system?

SOLUTION

W =1

–= 0.17 hr (or 10 minutes)

0.17( – ) = 1, where = 20 customers arrival rate per hour

= = 25.88 or about 26 customers per hour1 + 0.17(20)

0.17


Multiple-Server Model Multiple-Server Model

Service system has only one phase, multiple-channels

Assumptions (in addition to single-server model) There are s identical servers The service distribution for each server is

exponential The mean service time is 1/ s should always exceed



P0 = Probability that zero customers are in the system

= 1

1

0 11

!

/!

/sn

ss

n

n

Pn = Probability that n customers are in the system

=

snPss

snPn

sn

n

n

0

0

0

!/!

/

= Average utilization of the system = s



Lq = Average number of customers in the waiting line

2

0

1

!

/

s

P s

=

Wq = Average waiting time of customers in line =

qL

W = Average time spent in the system, including service

= 1

qW

= W

L = Average number of customers in the service system


Estimating Idle Time and CostsEstimating Idle Time and Costs

EXAMPLE C.5

The management of the American Parcel Service terminal in Verona, Wisconsin, is concerned about the amount of time the company’s trucks are idle (not delivering on the road), which the company defines as waiting to be unloaded and being unloaded at the terminal. The terminal operates with four unloading bays. Each bay requires a crew of two employees, and each crew costs $30 per hour. The estimated cost of an idle truck is $50 per hour. Trucks arrive at an average rate of three per hour, according to a Poisson distribution. On average, a crew can unload a semitrailer rig in one hour, with exponential service times. What is the total hourly cost of operating the system?



SOLUTION

The multiple-server model is appropriate. To find the total cost of labor and idle trucks, we must calculate the average number of trucks in the system.

Figure C.5 shows the results for the American Parcel Service problem using the Waiting-Lines Solver from OM Explorer. Manual calculations using the equations for the multiple-server model are demonstrated in Solved Problem 2 at the end of this supplement. The results show that the four-bay design will be utilized 75 percent of the time and that the average number of trucks either being serviced or waiting in line is 4.53 trucks. That is, on average at any point in time, we have 4.53 idle trucks. We can now calculate the hourly costs of labor and idle trucks:



Labor cost: $30(s) = $30(4) = $120.00

Idle truck cost: $50(L) = $50(4.53) = 226.50

Total hourly cost = $346.50

Figure C.5 – Waiting-Lines Solver for Multiple-Server Model



Suppose the manager of the checkout system decides to add another counter. The arrival rate is still 20 customers per hour, but now each checkout counter will be designed to service customers at the rate of 12.5 per hour. What is the waiting time in line of the new system?

s = 2, = 12.5 customers per hour, = 20 customers per hour

SOLUTION

= s

1. Average utilization of the system 512220

.= = 0.8



2. Probability that zero customers are in the system

P0 =

11

1

1

!s

s

=

801

12512

20

51220

1

12

.!.

.

= 0.11



3. Average number of customers in the waiting line

Wq = qL

2

0

1

!

/

s

P s

Lq =

2

2

8012

80512

20110

.!

..

.

= = 1.408

4. Average waiting time of customers in line = = 0.0704 hrs

(or 4.224 minutes204081.


Little’s LawLittle’s Law

Relates the number of customers in a waiting line system to the waiting time of customers

Using the notation from the single-server and multiple-server models it is expressed as L = W or Lq = Wq

Holds for a wide variety of arrival processes, service time distributions, and numbers of servers

Only need to know two of the parameters



Service Estimate W

Manufacturing Estimate the average work-in-process L

Average timein the facility = W =

L customers customer/hour

= = 0.75 hours or 45 minutes3040

Work-in-process = L = W

= 5 gear cases/hour (3 hours) = 15 gear cases



Provides basis for measuring the effects of process improvements

Is not applicable to situations where the customer population is finite


Finite-Source ModelFinite-Source Model

Assumptions Follows the assumption of the single-server,

except that the customer population is finite Having only N potential customers If N > 30, then the single-server model with the

assumption of infinite customer population is adequate



P0 = Probability that zero customers are in the system

=

1

0

nN

n nNN

!!

= Average utilization of the server = 1 – P0

Lq = Average number of customers in the waiting line

= 01 PN



L = Average number of customers in the service system

= 01 PN

Wq = Average waiting time in line = 1 LNLq

W = Average time spent in the system, including service

= 1 LNL



DBT Bank has 8 copy machines located in various offices throughout the building. Each machine is used continuously and has an average time between failures of 50 hours. Once failed, it takes 4 hours for the service company to send a repair person to have it fixed. What is the average number of copy machines in repair or waiting to be repaired?

= 1/50 = 0.02 copiers per hour

= 1/4 = 0.25 copiers per hour



SOLUTION

1. Probability that zero customers are in the system

P0 =

1

0

nN

n nNN

!!

810 080

08

08078

08088

1

.!!

.!!

.!!

=

= 0.44

2. Average utilization of the server

= 1 – P0 = 1 – 0.44 = 0.56

3. Average number of customers in the service system

L = 01 PN

= = 1 4401020250

8 ...


Analyzing Maintenance CostsAnalyzing Maintenance Costs

EXAMPLE C.6

The Worthington Gear Company installed a bank of 10 robots about 3 years ago. The robots greatly increased the firm’s labor productivity, but recently attention has focused on maintenance. The firm does no preventive maintenance on the robots because of the variability in the breakdown distribution. Each machine has an exponential breakdown (or interarrival) distribution with an average time between failures of 200 hours. Each machine hour lost to downtime costs $30, which means that the firm has to react quickly to machine failure. The firm employs one maintenance person, who needs 10 hours on average to fix a robot. Actual maintenance times are exponentially distributed. The wage rate is $10 per hour for the maintenance person, who can be put to work productively elsewhere when not fixing robots. Determine the daily cost of labor and robot downtime.



SOLUTION

The finite-source model is appropriate for this analysis because the customer population consists of only 10 machines and the other assumptions are satisfied. Here, = 1/200, or 0.005 break-down per hour, and = 1/10 = 0.10 robot per hour. To calculate the cost of labor and robot downtime, we need to estimate the average utilization of the maintenance person and L, the average number of robots in the maintenance system at any time. Figure C.6 shows the results for the Worthington Gear Problem using the Waiting-Lines Solver from OM Explorer.



Manual computations using the equations for the finite-source model are demonstrated in Solved Problem 3 at the end of this supplement. The results show that the maintenance person is utilized only 46.2 percent of the time, and the average number of robots waiting in line or being repaired is 0.76 robot. However, a failed robot will spend an average of 16.43 hours in the repair system, of which 6.43 hours of that time is spent waiting for service. While an individual robot may spend more than two days with the maintenance person, the maintenance person has a lot of idle time with a utilization rate of only 42.6 percent. That is why there is only an average of 0.76 robot being maintained at any point of time.



Figure C.5 – Waiting-Lines Solver for Finite-Source Model

The daily cost of labor and robot downtime is

Labor cost: ($19/hour)(8 hours/day)(0.462 utilization) = $36.96

Idle robot cost: (0.76 robot)($30/robot hour)(8 hours/day) = 182.40

Total daily cost = $219.36



The Hilltop Produce store is staffed by one checkout clerk. The average checkout time is exponentially distributed around an average of two minutes per customer. An average of 20 customers arrive per hour.

What is the average utilization rate?

SOLUTION

= = = 0.66720

30



What is the probability that three or more customers will be in the checkout area?

First calculate 0, 1, and 2 customers will be in the checkout area:

n = (1 – )0 = (0.333)(0.667)0 = 0.333

n = (1 – )1 = (0.333)(0.667)1 = 0.222

n = (1 – )2 = (0.333)(0.667)2 = 0.111

Then calculate 3 or more customers will be in the checkout area:

1 – P0 – P1 – P2 = 0.333 – 0.222 – 0.111 = 0.334



What is the average number of customers in the waiting line?

–Lq = L = = = 1.333

203020

6670.

What is the average time customers spend in the store?

W =1

–= = 0.1 hr 60 min/hr = 6 minutes

20301


Decision Areas for ManagementDecision Areas for Management

1. Arrival rates

2. Number of service facilities

3. Number of phases

4. Number of servers per facility

5. Server efficiency

6. Priority rule

7. Line arrangement


Solved Problem 1Solved Problem 1

A photographer takes passport pictures at an average rate of 20 pictures per hour. The photographer must wait until the customer smiles, so the time to take a picture is exponentially distributed. Customers arrive at a Poisson-distributed average rate of 19 customers per hour.

a. What is the utilization of the photographer?

b. How much time will the average customer spend with the photographer?

SOLUTION

a. The assumptions in the problem statement are consistent with a single-server model. Utilization is

= = = 0.9519

20



b. The average customer time spent with the photographer is

W=1

–= = 1 hour

120 – 19



The Mega Multiplex Movie Theater has three concession clerks serving customers on a first come, first-served basis. The service time per customer is exponentially distributed with an average of 2 minutes per customer. Concession customers wait in a single line in a large lobby, and arrivals are Poisson distributed with an average of 81 customers per hour. Previews run for 10 minutes before the start of each show. If the average time in the concession area exceeds 10 minutes, customers become dissatisfied.

a. What is the average utilization of the concession clerks?

b. What is the average time spent in the concession area?



SOLUTION

= s

= = 0.90

stomerminutes/cu 2hour rverminutes/se 60

servers 3

hourcustomers/ 81

The concession clerks are busy 90 percent of the time.

a. The problem statement is consistent with the multiple-server model, and the average utilization rate is



b. The average time spent in the system, W, is W =1

qW

Here,

Wq = qL

Lq = 2

0

1

!

/

s

P s

P0 = 1

1

0 11

!

/!

/sn

ss

n

n

and

We must solve for P0, Lq, and Wq, in that order, before we can solve for W:

P0 = 1

1

0 11

!

/!

/sn

ss

n

n

=

901

1672

272

13081

1

132

.../

=805326453721

1...

= = 0.02491540

1.



Lq = 2

0

1

!

/

s

P s

= 7.352 customers= = 2

3

9013

90308102490

.!

./.

010644110

..

Wq = qL

= = 0.0908 hourhour customers/ 81

customers 7.352

W = 1

qW = 0.0908 hours + hour =301

hourminutes 60

hour 0.1241

= 7.45 minutes

With three concession clerks, customers will spend an average of 7.45 minutes in the concession area.



The Severance Coal Mine serves six trains having exponentially distributed interarrival times averaging 30 hours. The time required to fill a train with coal varies with the number of cars, weather-related delays, and equipment breakdowns. The time to fill a train can be approximated by an exponential distribution with a mean of 6 hours 40 minutes. The railroad requires the coal mine to pay large demurrage charges in the event that a train spends more than 24 hours at the mine. What is the average time a train will spend at the mine?

SOLUTION

The problem statement describes a finite-source model, with N = 6. The average time spent at the mine is W = L[(N – L)]–1, with 1/ = 30 hours/train, = 0.8 train/day, and = 3.6 trains/day. In this case,



P0 =

1

0

nN

n nNN

!!

=

6

0 6380

66

1

n

n

n ..

!!

=

43210

6380

26

6380

36

6380

46

6380

56

6380

66

.

.!!

.

.!!

.

.!!

.

.!!

.

.!!

65

6380

06

6380

16

.

.!!

.

.!!

1

=0903908803214813311

1...... 496

1.

= 0.1541=

L = 01 PN

W = 1 LNL

= = 2.193 trains

154101

8063

6 ..

.

= 8080731932

...

= 0.72 dayArriving trains will spend an average of 0.72 day at the coal mine.


Documents

C – 1 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Waiting Lines C For Operations Management, 9e by Krajewski/Ritzman/Malhotra