operations management

2

Chapter 1

Overview: Introduction to the Field

3

• Operations Management

• Why Study Operations Management?

• Production System Defined

• Operations as a Service

• Plan of This Book

• Historical Development of OM

• Current Issues in OM

OBJECTIVES

4

What is Operations Management?Defined

Operations management (OM) is defined as the design, operation, and improvement of the systems that create and deliver the firm’s primary products and services

5

Why Study Operations Management?

Business Education

Systematic Approach to Org. Processes

Career Opportunities

Cross-Functional Applications

OperationsManagement

6

What is a Production System?Defined

A production system is defined as a user of resources to transform inputs into some desired outputs

7

Transformations

• Physical--manufacturing

• Locational--transportation

• Exchange--retailing

• Storage--warehousing

• Physiological--health care

• Informational--telecommunications

8

What is a Service and What is a Good?

• “If you drop it on your foot, it won’t hurt you.” (Good or service?)

• “Services never include goods and goods never include services.” (True or false?)

9

OM in the Organization Chart

OperationsOperations

Plant Manager

Plant Manager

OperationsManager

OperationsManager

DirectorDirector

Manufacturing, Production control, Quality assurance, Engineering,

Purchasing, Maintenance, etc

Manufacturing, Production control, Quality assurance, Engineering,

Purchasing, Maintenance, etc

Finance Marketing

10

Core services are basic things that customers want from products

they purchase

Core ServicesDefined

11

Core Services Performance Objectives


Flexibility

Quality

Speed

Price (or cost Reduction)

12

Value-added services differentiate the organization from competitors and build relationships that bind

customers to the firm in a positive way

Value-Added ServicesDefined

13

Value-Added Service Categories


Information

Problem Solving

Sales Support

Field Support

14

Plan of This Book

I. Operations Strategy and Managing Change

1. Introductionto the Field

2. Operations Strategy and Competitiveness

3. Project Management

III. SupplyChain Design

9. Supply Chain Strategy

10. Strategic Capacity

Management

11. Just-in-Time and Lean Systems

IV. Planning and Controlling the Supply Chain

12. Forecasting and Demand Management

13. Aggregate Sales and Operations Planning

14. Inventory Control

15. Materials Requirements Planning 8. Operations Consulting

and Reengineering

16. Operations Scheduling

II. Product Designand Process

Selection

4. ProcessAnalysis

5. ProductDesign and

Process Selection-Manufacturing

6. ProductDesign and

Process Selection-Services

7. QualityManagement

17. Synchronous Manufacturing and Theory of Constraints

15

Historical Development of OM

• JIT and TQC

• Manufacturing Strategy Paradigm

• Service Quality and Productivity

• Total Quality Management and Quality Certification

16

Historical Development of OM (cont’d)

• Business Process Reengineering

• Supply Chain Management

• Electronic Commerce

17

Current Issues in OM

• Effectively consolidating the operations resulting from mergers

• Developing flexible supply chains to enable mass customization of products and services

• Managing global supplier, production and distribution networks

• Increased “commoditization” of suppliers

18

Current Issues in OM (cont’d)

• Achieving the “Service Factory”

• Enhancing value added services

• Making efficient use of Internet technology

• Achieving good service from service firms

19

End of Chapter 1

21

Chapter 2

Operations Strategy and Competitiveness

22

• Operations Strategy

• Competitive Dimensions

• Order Qualifiers and Winners

• Strategy Design Process

• A Framework for Manufacturing Strategy

• Service Strategy Capacity Capabilities

• Productivity Measures

OBJECTIVES

23

Operations Strategy

ExampleStrategy Process

Customer Needs

Corporate Strategy

Operations Strategy

Decisions on Processes and Infrastructure

More Product

Increase Org. Size

Increase Production Capacity

Build New Factory

24

Competitive Dimensions

• Cost

• Product Quality and Reliability

• Delivery Speed

• Delivery Reliability

• Coping with Changes in Demand

• Flexibility and New Product Introduction Speed

• Other Product-Specific Criteria

25

Dealing with Trade-offs

Cost

Quality

DeliveryFlexibility

For example, if we improve customer service problem solving by cross-training personnel to deal with a wider-range of problems, they may become less efficient at dealing with commonly occurring problems.

For example, if we improve customer service problem solving by cross-training personnel to deal with a wider-range of problems, they may become less efficient at dealing with commonly occurring problems.

For example, if we reduce costs by reducing product quality inspections, we might reduce product quality.

For example, if we reduce costs by reducing product quality inspections, we might reduce product quality.

26

Order Qualifiers and WinnersDefined•Order qualifiers are the basic criteria that permit the firms products to be considered as candidates for purchase by customers

•Order winners are the criteria that differentiates the products and services of one firm from another

27

Service Breakthroughs

• A brand name car can be an “order qualifier”

Repair services can be “order winners”

Examples: Warranty, Roadside Assistance, Leases, etc

28

Strategy Design ProcessStrategy Map

Financial Perspective

Customer Perspective

Internal Perspective

Learning and Growth Perspective

Improve Shareholder Value

Customer Value Proposition

Build-Increase-Achieve

A Motivated and Prepared Workforce

What it is about!

29

Kaplan and Norton’s Generic Strategy Map

In the Kaplan and Norton’s Generic Strategy

Map, under the Financial Perspective, the

Productivity Strategy is generally made up

from two components:

1. Improve cost structure: Lower direct and indirect costs

2. Increase asset utilization: Reduce working and fixed capital

30

Kaplan and Norton’s Generic Strategy Map (Continued)

In the Kaplan and Norton’s Generic Strategy

Map, under the Financial Perspective, the

Revenue Growth Strategy is generally made

up from two components:1. Build the franchise: Develop new

sources of revenue2. Increase customer value: Work with

existing customers to expand relationships with company

31


In the Kaplan and Norton’s Generic Strategy Map,

under the Customer Perspective, there are three

ways suggested as means of differentiating a

company from others in a marketplace:

1. Product leadership2. Customer intimacy3. Operational excellence

32


In the Kaplan and Norton’s Generic Strategy Map,

under the Learning and Growth Perspective, there

are three principle categories of intangible assets

needed for learning:

1. Strategic competencies2. Strategic technologies3. Climate for action

33

Operations Strategy FrameworkCustomer Needs

New product : Old product

Competitivedimensions & requirements

Quality, Dependability, Speed, Flexibility, and Price

Operations & Supplier capabilities

R&D Technology Systems People Distribution

Support Platforms

Financial management Human resource management Information management

Enterprise capabilities

Operations and Supplier Capabilities

R&D Technology Systems People Distribution

34

Steps in Developing a Manufacturing Strategy

• 1. Segment the market according to the product group

• 2. Identify product requirements, demand patterns, and profit margins of each group

• 3. Determine order qualifiers and winners for each group

• 4. Convert order winners into specific performance requirements

35

Service Strategy Capacity Capabilities

• Process-based – Capacities that transforms material or information

and provide advantages on dimensions of cost and quality

• Systems-based – Capacities that are broad-based involving the

entire operating system and provide advantages of short lead times and customize on demand

• Organization-based– Capacities that are difficult to replicate and

provide abilities to master new technologies

36

What is Productivity?Defined

Productivity is a common measure on how well resources are being used. In the broadest sense, it can be defined as the following ratio:

Outputs Inputs

37

Total Measure Productivity

Total Measure Productivity = Outputs Inputs

or

= Goods and services produced

All resources used

38

Partial Measure Productivity

• Partial measures of productivity =

• Output or Output or Output or Output Labor Capital Materials Energy

39

Multifactor Measure Productivity

• Multifactor measures of productivity =

• Output . Labor + Capital + Energy

or

• Output . Labor + Capital + Materials

40

Example of Productivity Measurement

• You have just determined that your service employees have used a total of 2400 hours of labor this week to process 560 insurance forms. Last week the same crew used only 2000 hours of labor to process 480 forms.

• Which productivity measure should be used?• Answer: Could be classified as a Total Measure or

Partial Measure.• Is productivity increasing or decreasing?• Answer: Last week’s productivity = 480/2000 = 0.24,

and this week’s productivity is = 560/2400 = 0.23. So, productivity is decreasing slightly.

41

End of Chapter 2

43

Technical Note 2

Learning Curves

44

• Underlying Principles of Learning Curves

• Learning Curve Example• Types of Learning• From Learning Curves to Performance

Improvement

OBJECTIVES

45

Underlying Principles of Learning Curves

1. Each time you perform a task it takes less time than the last time you performed the same task

2. The extent of task time decreases over time

3. The reduction in time will follow a predictable pattern

46

Example of a Learning Curve

Suppose you start a term paper typing business. You time yourself on the first paper, then the second, and so on.

Suppose you start a term paper typing business. You time yourself on the first paper, then the second, and so on.

Term paper

1

2

3

4

5

6

Time (in Minutes)

100

90

84.62

81.00

78.30

76.16

Note that only 90 of 100 minutes are used in the second repetition. This is an example of a 90% learning curve.

Note that only 90 of 100 minutes are used in the second repetition. This is an example of a 90% learning curve.

47

Plotting the Learning Curve

All learning curves have this downward sloping curve.

All learning curves have this downward sloping curve.

90 % Learning Curve

020406080

100120

0 1000 2000 3000 4000 5000

Unit

Pro

du

cti

on

T

ime

(Min

ute

s)

48

Types of Learning

• Individual Learning

Improvement when individuals gain a skill or efficiency by repetition of a job

• Organizational LearningImprovement from the groups of individuals from repetition and changes in administration, equipment, and product design

49

From Learning Curves to Performance Improvement (Part 1)

• Proper selection of workers

• Proper training

• Motivation

• Work specialization

• Do one or very few jobs at a time

50

From Learning Curves to Performance Improvement (Part 2)

• Use tools or equipment that assists or supports performance

• Provide quick and easy access for help

• Allow workers to help redesign their tasks

51

End of Technical Note 2

53

Chapter 3

Project Management

54

• Definition of Project Management

• Work Breakdown Structure

• Project Control Charts

• Structuring Projects

• Critical Path Scheduling

OBJECTIVES

55

Project Management Defined

• Project is a series of related jobs usually directed toward some major output and requiring a significant period of time to perform

• Project Management are the management activities of planning, directing, and controlling resources (people, equipment, material) to meet the technical, cost, and time constraints of a project

56

Project Control Charts: Gantt Chart

Activity 1Activity 2Activity 3Activity 4Activity 5Activity 6

Time

Vertical Axis: Always Activities or Jobs

Vertical Axis: Always Activities or Jobs

Horizontal Axis: Always TimeHorizontal Axis: Always Time

Horizontal bars used to denote length of time for each activity or job.

Horizontal bars used to denote length of time for each activity or job.

57

Structuring Projects Pure Project: Advantages

Pure ProjectDefinedA pure project is where a self-contained team works full-time on the project

• The project manager has full authority over the project

• Team members report to one boss• Shortened communication lines• Team pride, motivation, and

commitment are high

58

Structuring Projects Pure Project: Disadvantages

• Duplication of resources• Organizational goals and policies are ignored• Lack of technology transfer• Team members have no functional area

"home"

59

Functional ProjectDefined

President

Research andDevelopment

Engineering Manufacturing

ProjectA

ProjectB

ProjectC

ProjectD

ProjectE

ProjectF

ProjectG

ProjectH

ProjectI

A functional project is housed within a functional division

Example, Project “B” is in the functional area of Research and Development.

Example, Project “B” is in the functional area of Research and Development.

60

Structuring Projects Functional Project: Advantages

• A team member can work on several projects

• Technical expertise is maintained within the functional area

• The functional area is a “home” after the project is completed

• Critical mass of specialized knowledge

61

Structuring Projects Functional Project: Disadvantages

• Aspects of the project that are not directly related to the functional area get short-changed

• Motivation of team members is often weak

• Needs of the client are secondary and are responded to slowly

62

Structuring Projects: Matrix Project Organization Structure

President

Research andDevelopment

Engineering Manufacturing Marketing

ManagerProject A

ManagerProject B

ManagerProject C

63

Structuring Projects Matrix: Advantages

• Enhanced communications between functional areas

• Pinpointed responsibility

• Duplication of resources is minimized

• Functional “home” for team members

• Policies of the parent organization are followed

64

Structuring Projects Matrix: Disadvantages

• Too many bosses

• Depends on project manager’s negotiating skills

• Potential for sub-optimization

65

Work Breakdown StructureDefined

Program

Project 1 Project 2

Task 1.1

Subtask 1.1.1

Work Package 1.1.1.1

Level

1

2

3

4

Task 1.2

Subtask 1.1.2

Work Package 1.1.1.2

A work breakdown structure defines the hierarchy of project tasks, subtasks, and work packages

66

Network-Planning Models• A project is made up of a sequence of

activities that form a network representing a project

• The path taking longest time through this network of activities is called the “critical path”

• The critical path provides a wide range of scheduling information useful in managing a project

• Critical Path Method (CPM) helps to identify the critical path(s) in the project networks

67

Prerequisites for Critical Path Methodology

A project must have:

well-defined jobs or tasks whose completion marks the end of the project;

independent jobs or tasks;

and tasks that follow a given sequence.

68

Types of Critical Path Methods• CPM with a Single Time Estimate

– Used when activity times are known with certainty– Used to determine timing estimates for the project, each

activity in the project, and slack time for activities

• CPM with Three Activity Time Estimates– Used when activity times are uncertain – Used to obtain the same information as the Single Time

Estimate model and probability information

• Time-Cost Models– Used when cost trade-off information is a major

consideration in planning– Used to determine the least cost in reducing total project

time

69

Steps in the CPM with Single Time

Estimate

• 1. Activity Identification

• 2. Activity Sequencing and Network Construction

• 3. Determine the critical path– From the critical path all of the project and

activity timing information can be obtained

70

Example 1. CPM with Single Time Estimate

Consider the following consulting project:

Develop a critical path diagram and determine the duration of the critical path and slack times for all activities.

Activity Designation Immed. Pred. Time (Weeks)Assess customer's needs A None 2Write and submit proposal B A 1Obtain approval C B 1Develop service vision and goals D C 2Train employees E C 5Quality improvement pilot groups F D, E 5Write assessment report G F 1

71

Example 1. CPM with Single Time Estimate

Consider the following consulting project:Activity Designation Immed. Pred. Time (Weeks)Assess customer's needs A None 2Write and submit proposal B A 1Obtain approval C B 1Develop service vision and goals D C 2Train employees E C 5Quality improvement pilot groups F D, E 5Write assessment report G F 1

Develop a critical path diagram and determine the duration of the critical path and slack times for all activities.

72

Example 1: First draw the network

A(2) B(1) C(1)

D(2)

E(5)

F(5) G(1)

A None 2

B A 1

C B 1

D C 2

E C 5

F D,E 5

G F 1

Act. Imed. Pred. Time

73

Example 1: Determine early starts and early finish times

ES=9EF=14

ES=14EF=15

ES=0EF=2

ES=2EF=3

ES=3EF=4

ES=4EF=9

ES=4EF=6

A(2) B(1) C(1)

D(2)

E(5)

F(5) G(1)

Hint: Start with ES=0 and go forward in the network from A to G.

Hint: Start with ES=0 and go forward in the network from A to G.

74

Example 1: Determine late

starts and late finish times

ES=9EF=14

ES=14EF=15

ES=0EF=2

ES=2EF=3

ES=3EF=4

ES=4EF=9

ES=4EF=6

A(2) B(1) C(1)

D(2)

E(5)

F(5) G(1)

LS=14LF=15

LS=9LF=14

LS=4LF=9

LS=7LF=9

LS=3LF=4

LS=2LF=3

LS=0LF=2

Hint: Start with LF=15 or the total time of the project and go backward in the network from G to A.

Hint: Start with LF=15 or the total time of the project and go backward in the network from G to A.

75

Example 1: Critical Path & Slack

ES=9EF=14

ES=14EF=15

ES=0EF=2

ES=2EF=3

ES=3EF=4

ES=4EF=9

ES=4EF=6

A(2) B(1) C(1)

D(2)

E(5)

F(5) G(1)

LS=14LF=15

LS=9LF=14

LS=4LF=9

LS=7LF=9

LS=3LF=4

LS=2LF=3

LS=0LF=2

Duration = 15 weeks

Slack=(7-4)=(9-6)= 3 Wks

76

Example 2. CPM with Three Activity Time Estimates

TaskImmediate

Predecesors Optimistic Most Likely PessimisticA None 3 6 15B None 2 4 14C A 6 12 30D A 2 5 8E C 5 11 17F D 3 6 15G B 3 9 27H E,F 1 4 7I G,H 4 19 28

77

Example 2. Expected Time Calculations

ET(A)= 3+4(6)+15

6

ET(A)= 3+4(6)+15

6

ET(A)=42/6=7ET(A)=42/6=7Task

Immediate Predecesors

Expected Time

A None 7B None 5.333C A 14D A 5E C 11F D 7G B 11H E,F 4I G,H 18

TaskImmediate


Expected Time = Opt. Time + 4(Most Likely Time) + Pess. Time

6Expected Time =

Opt. Time + 4(Most Likely Time) + Pess. Time

6

78


TaskImmediate

PredecesorsExpected

TimeA None 7B None 5.333C A 14D A 5E C 11F D 7G B 11H E,F 4I G,H 18

ET(B)=32/6=5.333ET(B)=32/6=5.333

ET(B)= 2+4(4)+14

6

ET(B)= 2+4(4)+14

6

TaskImmediate



6Expected Time =


6

79


TaskImmediate

PredecesorsExpected

TimeA None 7B None 5.333C A 14D A 5E C 11F D 7G B 11H E,F 4I G,H 18

ET(C)= 6+4(12)+30

6

ET(C)= 6+4(12)+30

6

ET(C)=84/6=14ET(C)=84/6=14

TaskImmediate



6Expected Time =


6

80

Example 2. Network

A(7)

B(5.333)

C(14)

D(5)

E(11)

F(7)

H(4)

G(11)

I(18)

Duration = 54 Days

81

Example 2. Probability Exercise

What is the probability of finishing this project in less than 53 days?

What is the probability of finishing this project in less than 53 days?

p(t < D)

TE = 54

Z = D - TE

cp2

Z = D - TE

cp2

tD=53

82

Activity variance, = (Pessim. - Optim.

6)2 2Activity variance, = (

Pessim. - Optim.

6)2 2

Task Optimistic Most Likely Pessimistic VarianceA 3 6 15 4B 2 4 14C 6 12 30 16D 2 5 8E 5 11 17 4F 3 6 15G 3 9 27H 1 4 7 1I 4 19 28 16

(Sum the variance along the critical path.)

2 = 41 2 = 41

83

There is a 43.8% probability that this project will be completed in less than 53 weeks.

There is a 43.8% probability that this project will be completed in less than 53 weeks.

p(Z < -.156) = .438, or 43.8 % (NORMSDIST(-.156)p(Z < -.156) = .438, or 43.8 % (NORMSDIST(-.156)

Z = D - T

=53- 54

41= -.156E

cp2

Z = D - T

=53- 54

41= -.156E

cp2

TE = 54

p(t < D)

t

D=53

84

Example 2. Additional Probability Exercise

• What is the probability that the project duration will exceed 56 weeks?

• What is the probability that the project duration will exceed 56 weeks?

85

Example 2. Additional Exercise Solution

tTE = 54

p(t < D)

D=56

Z = D - T

=56 - 54

41= .312E

cp2

Z = D - T

=56 - 54

41= .312E

cp2

p(Z > .312) = .378, or 37.8 % (1-NORMSDIST(.312)) p(Z > .312) = .378, or 37.8 % (1-NORMSDIST(.312))

86

Time-Cost Models

• Basic Assumption: Relationship between activity completion time and project cost

• Time Cost Models: Determine the optimum point in time-cost tradeoffs– Activity direct costs– Project indirect costs– Activity completion times

87

CPM Assumptions/Limitations • Project activities can be identified as entities (There

is a clear beginning and ending point for each activity.)

• Project activity sequence relationships can be specified and networked

• Project control should focus on the critical path• The activity times follow the beta distribution, with

the variance of the project assumed to equal the sum of the variances along the critical path

• Project control should focus on the critical path

88

End of Chapter 3

90

Chapter 4

Process Analysis

91

• Process Analysis

• Process Flowcharting

• Types of Processes

• Process Performance Metrics

OBJECTIVES

92

Process Analysis Terms

• Process: Is any part of an organization that takes inputs and transforms them into outputs

• Cycle Time: Is the average successive time between completions of successive units

• Utilization: Is the ratio of the time that a resource is actually activated relative to the time that it is available for use

93

Process Flowcharting

Defined • Process flowcharting is the use of a diagram to

present the major elements of a process

• The basic elements can include tasks or operations, flows of materials or customers, decision points, and storage areas or queues

• It is an ideal methodology by which to begin analyzing a process

94

Flowchart

Symbols Tasks or operations Examples: Giving an

admission ticket to a customer, installing a engine in a car, etc.

Examples: Giving an admission ticket to a customer, installing a engine in a car, etc.

Decision Points Examples: How much change should be given to a customer, which wrench should be used, etc.

Examples: How much change should be given to a customer, which wrench should be used, etc.

Purpose and Examples

95

Examples: Sheds, lines of people waiting for a service, etc.

Examples: Sheds, lines of people waiting for a service, etc.

Examples: Customers moving to a seat, mechanic getting a tool, etc.

Examples: Customers moving to a seat, mechanic getting a tool, etc.

Storage areas or queues

Flows of materials or customers

Flowchart

Symbols Purpose and Examples

96

Example: Flowchart of Student Going to

School

Yes

No

Goof off

Go to school today?

Walk to class

Drive to school

97

Types of Processes

Single-stage Process

Stage 1

Stage 1 Stage 2 Stage 3

Multi-stage Process

98

Types of Processes (Continued)

Stage 1 Stage 2

Buffer

Multi-stage Process with Buffer

A buffer refers to a storage area between stages where the output of a stage is placed prior to being used in a downstream stage

99

Other Process Terminology• Blocking

– Occurs when the activities in a stage must stop because there is no place to deposit the item just completed

– If there is no room for an employee to place a unit of work down, the employee will hold on to it not able to continue working on the next unit

• Starving– Occurs when the activities in a stage must stop because

there is no work

– If an employee is waiting at a work station and no work is coming to the employee to process, the employee will remain idle until the next unit of work comes

100

Other Process Terminology (Continued)

• Bottleneck– Occurs when the limited capacity of a process

causes work to pile up or become unevenly distributed in the flow of a process

– If an employee works too slow in a multi-stage process, work will begin to pile up in front of that employee. In this is case the employee represents the limited capacity causing the bottleneck.

• Pacing– Refers to the fixed timing of the movement of items

through the process

101

Other Types of Processes

• Make-to-order– Only activated in response to an actual order

– Both work-in-process and finished goods inventory kept to a minimum

• Make-to-stock– Process activated to meet expected or forecast

demand

– Customer orders are served from target stocking level

102

Process Performance Metrics

• Operation time = Setup time + Run time

• Throughput time = Average time for a unit tomove through the system

• Velocity = Throughput time

Value-added time

103

Process Performance Metrics (Continued)

• Cycle time = Average time betweencompletion of units

• Throughput rate = 1 . Cycle time

• Efficiency = Actual output Standard Output

104

Process Performance Metrics (Continued)

• Productivity = Output

Input

• Utilization = Time Activated

Time Available

105

Cycle Time Example

Suppose you had to produce 600 units in 80 hours to meet the demand requirements of a product. What is the cycle time to meet this demand requirement?

Suppose you had to produce 600 units in 80 hours to meet the demand requirements of a product. What is the cycle time to meet this demand requirement?

Answer: There are 4,800 minutes (60 minutes/hour x 80 hours) in 80 hours. So the average time between completions would have to be: Cycle time = 4,800/600 units = 8 minutes.

Answer: There are 4,800 minutes (60 minutes/hour x 80 hours) in 80 hours. So the average time between completions would have to be: Cycle time = 4,800/600 units = 8 minutes.

106

Process Throughput Time Reduction

• Perform activities in parallel

• Change the sequence of activities

• Reduce interruptions

107

End of Chapter 4

109

Technical Note 4

Job Design and Work Measurement

110

• Job Design Defined

• Job Design Decisions

• Trends in Job Design

• Work Measurement

• Basic Compensation Systems

• Financial Incentive Plans

OBJECTIVES

111

What is Job Design?Defined• Job design is the function of specifying the

work activities of an individual or group in an organizational setting

• The objective of job design is to develop jobs that meet the requirements of the organization and its technology and that satisfy the jobholder’s personal and individual requirements

112

Job Design DecisionsHowWhyWhenWhereWhatWho

Mental andphysicalcharacteristicsof the work force

Tasks to beperformed

Geographiclocale of theorganization;location of work areas

Time of day;time of occurrence inthe work flow

Organizationalrationale forthe job; object-ives and mot-ivation of theworker

Method of performanceandmotivation

UltimateJob

Structure

113

Trends in Job Design Quality control as part of the worker's job

Cross-training workers to perform multi skilled jobs

Employee involvement and team approaches to designing and organizing work

"Informating" ordinary workers through e-mail and the Internet

114

Trends in Job Design (Continued)

Extensive use of temporary workers

Automation of heavy manual work

Organizational commitment to providing meaningful and rewarding jobs for all employees

115

Behavioral Considerations in Job Design

Ultimate Job

Structure

Degree of

Specialization

Job Enrichment

(vs. Enlargement)

Balancing the specialization in a job and its content through enrichment can give us….

Balancing the specialization in a job and its content through enrichment can give us….

116

Sociotechnical Systems

Task VarietySkill VarietyFeedback

Task IdentityTask Autonomy

Process Technology

Needs

Worker/Group Needs

Focuses on the interaction between technology and the work group by looking at….

Focuses on the interaction between technology and the work group by looking at….

117

Physical Considerations in Job Design• Work physiology sets work-rest cycles

according to the energy expended in various parts of the job. The harder the work, the more the need for rest periods.

• Ergonomics is a term used to describe the study of the physical arrangement of the work space together with tools used to perform a task. Fit the work to the body rather than forcing the body to conform to the work.

118

Work Methods

Workers Interacting with Other Workers

A Production Process

Worker at a Fixed Workplace

Worker Interacting with Equipment

Ultimate Job Design

Ultimate

Job Design

119

Work Measurement Defined

• Work measurement is a process of analyzing jobs for the purpose of setting time standards• Why use it?

– Schedule work and allocate capacity– Motivate and measure work performance– Evaluate performance– Provide benchmarks

120

Time Study Normal Time Formulas

• Normal time(NT)=Observed performance time per unit x (Performance rating)*

*The Performance Rating is usually expressed in decimal form in these formulas. So a person working 10% faster than normal would have a Performance Rating of 1.10 or 110% of normal time. Working 10% slower, 0.90 or 90% of normal.

• NT= Time worked _ x (Performance rating)* Number of units produced

• Normal time(NT)=Observed performance time per unit x (Performance rating)*

*The Performance Rating is usually expressed in decimal form in these formulas. So a person working 10% faster than normal would have a Performance Rating of 1.10 or 110% of normal time. Working 10% slower, 0.90 or 90% of normal.

• NT= Time worked _ x (Performance rating)* Number of units produced

121

Time Study Standard Time Formulas

• Standard time = Normal time + (Allowances x Normal times)

• Standard time = NT(1 + Allowances)

• Standard time = NT .

1 - Allowances

• Standard time = Normal time + (Allowances x Normal times)

• Standard time = NT(1 + Allowances)

• Standard time = NT .

1 - Allowances

122

Time Study Example Problem

• You want to determine the standard time for a job. The employee selected for the time study has produced 20 units of product in an 8 hour day. Your observations made the employee nervous and you estimate that the employee worked about 10 percent faster than what is a normal pace for the job. Allowances for the job represent 25 percent of the normal time.

• Question: What are the normal and standard times for this job?

• You want to determine the standard time for a job. The employee selected for the time study has produced 20 units of product in an 8 hour day. Your observations made the employee nervous and you estimate that the employee worked about 10 percent faster than what is a normal pace for the job. Allowances for the job represent 25 percent of the normal time.

• Question: What are the normal and standard times for this job?

123

Time Study Example Solution

Normal time = Time worked x (Performance rating) Number of units produced

= (480 minutes/20) x (1.10)

= 26.4 minutes

Standard time = NT . 1 – Allowances

= (26.4)/(1-0.25)

= 35.2 minutes

124

Work Sampling• Use inference to make statements about work

activity based on a sample of the activity• Ratio Delay

– Activity time percentage for workers or equipment

• Performance Measurement– Relates work time to output (performance index)

• Time Standards– Standard task times

125

Advantage of Work Sampling over Time Study

• Several work sampling studies may be conducted simultaneously by one observer

• The observer need not be a trained analyst unless the purpose of the study is to determine a time standard

• No timing devices are required

• Work of a long cycle time may be studied with fewer observer hours

126

Advantage of Work Sampling over Time Study (Continued)

• The duration of the study is longer, which minimizes effects of short-period variations

• The study may be temporarily delayed at any time with little effect

• Because work sampling needs only instantaneous observations (made over a longer period), the operator has less chance to influence the findings by changing work method

127

Basic Compensation Systems

• Hourly Pay

• Straight Salary

• Piece Rate

• Commissions

128

Financial Incentive Plans

• Individual and Small-Group Plans– Output measures– Quality measures– Pay for knowledge

• Organization-wide Plans– Profit-sharing– Gain-sharing

• Bonus based on controllable costs or units of output• Involve participative management

129

Scanlon PlanBasic Elements

Ratio =Total labor cost

Sales value of productionRatio =

Total labor cost

Sales value of production• The ratio

– Standard for judging business performance

• The bonus– Depends on reduction in costs below the preset

ratio

• The production committee

• The screening committee

130

Pay-for-Performance

• Paying employees based on their performance works--improvements in productivity and quality

• Pay-for-performance will become increasingly common components of performance management strategies and systems

131


133

Chapter 5

Product Design & Process Selection-Manufacturing

134

• Typical Phases of Product Design Development

• Designing for the Customer

• Design for Manufacturability • Types of Processes

• Process Flow Structures

• Process Flow Design

• Global Product Design and Manufacturing

OBJECTIVES

135

Typical Phases of Product Design Development

• Concept Development

• Product Planning

• Product/Process Engineering

• Pilot Production/Ramp-Up

136

Concurrent EngineeringDefined

• Concurrent engineering can be defined as the simultaneous development of project design functions, with open and interactive communication existing among all team members for the purposes of reducing time to market, decreasing cost, and improving quality and reliability

137

Concurrent Engineering(Continued)

• Teams provide the primary integration mechanism in CE programs

• There are three types of teams– Program Management Team– Technical Team– Design-Build Teams

• Time savings of CE programs are created by performing activities in parallel

138

Designing for the Customer

Quality FunctionDeployment

Value Analysis/Value Engineering

Ideal Customer Product

House of Quality

139

Designing for the Customer: Quality Function Deployment

• Interfunctional teams from marketing, design engineering, and manufacturing

• Voice of the customer

• House of Quality

140

Designing for the

Customer: The House of Quality

Customer Requirements

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140

Customer requirements information forms the basis for this matrix, used to translate them into operating or engineering goals.

Customer requirements information forms the basis for this matrix, used to translate them into operating or engineering goals.

141

Designing for the Customer: Value Analysis/Value Engineering (VA/VE)

• Achieve equivalent or better performance at a lower cost while maintaining all functional requirements defined by the customer– Does the item have any design features that are

not necessary?– Can two or more parts be combined into one?– How can we cut down the weight?– Are there nonstandard parts that can be

eliminated?

142

Design for Manufacturability

• Traditional Approach– “We design it, you build it” or “Over the

wall”

• Concurrent Engineering– “Let’s work together simultaneously”

143

Design for Manufacturing and Assembly• Greatest improvements related to DFMA arise

from simplification of the product by reducing the number of separate parts:

1. During the operation of the product, does the part move relative to all other parts already assembled?

2. Must the part be of a different material or be isolated from other parts already assembled?

3. Must the part be separate from all other parts to allow the disassembly of the product for adjustment or maintenance?

144

Types of Processes

• Conversion (ex. Iron to steel)

• Fabrication (ex. Cloth to clothes)

• Assembly (ex. Parts to components)

• Testing (ex. For quality of products)

145

Process Flow Structures

• Job shop (ex. Copy center making a single copy of a student term paper)

• Batch shop (ex. Copy center making 10,000 copies of an ad piece for a business)

• Assembly Line (ex. Automobile manufacturer)

• Continuous Flow (ex. Petroleum manufacturer)

146

IV.Continuous

Flow

III.Assembly

Line

II.Batch

I.Job

Shop

LowVolume,One of a

Kind

MultipleProducts,

LowVolume

FewMajor

Products,HigherVolume

HighVolume,

HighStandard-

ization

CommercialPrinter

French Restaurant

HeavyEquipment

AutomobileAssembly

Burger King

SugarRefinery

Flexibility (High)Unit Cost (High)

Flexibility (Low)Unit Cost (Low)

Exhibit 5.10Exhibit 5.10

These are the major stages of product and process life cycles

These are the major stages of product and process life cycles

147

Virtual FactoryDefined

A virtual factory can be defined as a manufacturing operation where activities are carried out not in one central plant, but in multiple locations by suppliers and partner firms as part of a strategic alliance

148

Break-Even Analysis

• A standard approach to choosing among alternative processes or equipment

• Model seeks to determine the point in units produced (and sold) where we will start making profit on the process or equipment

• Model seeks to determine the point in units produced (and sold) where total revenue and total cost are equal

149

Break-Even Analysis (Continued)

This formula can be used to find any of its components algebraically if the other parameters are known

Break-even Demand=Break-even Demand=

Purchase cost of process or equipment Price per unit - Cost per unit or Total fixed costs of process or equipment Unit price to customer - Variable costs per unit

Purchase cost of process or equipment Price per unit - Cost per unit or Total fixed costs of process or equipment Unit price to customer - Variable costs per unit

150

Break-Even Analysis (Continued)• Example: Suppose you want to purchase a new computer

that will cost $5,000. It will be used to process written orders from customers who will pay $25 each for the service. The cost of labor, electricity and the form used to place the order is $5 per customer. How many customers will we need to serve to permit the total revenue to break-even with our costs?

• Break-even Demand: = Total fixed costs of process or equip.

Unit price to customer – Variable costs =5,000/(25-5) =250 customers

• Example: Suppose you want to purchase a new computer that will cost $5,000. It will be used to process written orders from customers who will pay $25 each for the service. The cost of labor, electricity and the form used to place the order is $5 per customer. How many customers will we need to serve to permit the total revenue to break-even with our costs?

• Break-even Demand: = Total fixed costs of process or equip.

Unit price to customer – Variable costs =5,000/(25-5) =250 customers

151

Process Flow DesignDefined

• A process flow design can be defined as a mapping of the specific processes that raw materials, parts, and subassemblies follow as they move through a plant

• The most common tools to conduct a process flow design include assembly drawings, assembly charts, and operation and route sheets

152

Example: Assembly Chart (Gozinto)

A-2SA-2

4

5

6

7

Lockring

Spacer, detent spring

Rivets (2)

Spring-detent

A-5Component/Assy Operation

Inspection

From Exhibit 5.14From Exhibit 5.14

153

Example: Process Flow Chart

Material Received from Supplier

Inspect Material for Defects Defects

found?

Return to Supplier for Credit

Yes

No, Continue…

154

Global Product Design and Manufacturing Strategies

• Joint Ventures

• Global Product Design Strategy

155

Measuring Product Development Performance

Measures

•Freq. Of new products introduced•Time to market introduction•Number stated and number completed•Actual versus plan•Percentage of sales from new products

•Freq. Of new products introduced•Time to market introduction•Number stated and number completed•Actual versus plan•Percentage of sales from new products

Time-to-marketTime-to-market

ProductivityProductivity

QualityQuality

•Engineering hours per project•Cost of materials and tooling per project•Actual versus plan

•Engineering hours per project•Cost of materials and tooling per project•Actual versus plan

•Conformance-reliability in use•Design-performance and customer satisfaction•Yield-factory and field

•Conformance-reliability in use•Design-performance and customer satisfaction•Yield-factory and field

Performance Dimension

156

End of Chapter 5

158

Technical Note 5

Facility Layout

159

• Facility Layout and Basic Formats

• Process Layout

• Layout Planning

• Assembly Line balancing

• Service Layout

OBJECTIVES

160

Facility Layout Defined

Facility layout can be defined as the process by which the placement of departments, workgroups within departments, workstations, machines, and stock-holding points within a facility are determined

This process requires the following inputs:

– Specification of objectives of the system in terms of output and flexibility

– Estimation of product or service demand on the system

– Processing requirements in terms of number of operations and amount of flow between departments and work centers

– Space requirements for the elements in the layout

– Space availability within the facility itself

161

Basic Production Layout Formats

• Process Layout (also called job-shop or functional layout)

• Product Layout (also called flow-shop layout)

• Group Technology (Cellular) Layout

• Fixed-Position Layout

162

Process Layout: Interdepartmental Flow

• Given– The flow (number of moves) to and from all

departments– The cost of moving from one department to

another– The existing or planned physical layout of the

plant• Determine

– The “best” locations for each department, where best means maximizing flow, which minimizing costs

163

Process Layout: CRAFT Approach• It is a heuristic program; it uses a simple rule of

thumb in making evaluations: – "Compare two departments at a time and

exchange them if it reduces the total cost of the layout."

• It does not guarantee an optimal solution

• CRAFT assumes the existence of variable path material handling equipment such as forklift trucks

164

Process Layout: Systematic Layout Planning

• Numerical flow of items between departments – Can be impractical to obtain– Does not account for the qualitative factors that may

be crucial to the placement decision

• Systematic Layout Planning– Accounts for the importance of having each

department located next to every other department– Is also guided by trial and error

• Switching departments then checking the results of the “closeness” score

165

Example of Systematic Layout Planning: Reasons for Closeness

Code

1

2

3

4

5

6

Reason

Type of customer

Ease of supervision

Common personnel

Contact necessary

Share same price

Psychology

166

Example of Systematic Layout Planning:Importance of Closeness

Value

A

E

I

O

U

X

ClosenessLinecode

Numericalweights

Absolutely necessary

Especially important

Important

Ordinary closeness OK

Unimportant

Undesirable

16

8

4

2

0

80

167

Example of Systematic Layout Planning: Relating Reasons and Importance

From

1. Credit department

2. Toy department

3. Wine department

4. Camera department

5. Candy department

6

I

--

U

4

A

--

U

--

U

1

I

1,6

A

--

U

1

X

1

X

To2 3 4 5

Area(sq. ft.)

100

400

300

100

100

Closeness rating

Reason for rating

Note here that the (1) Credit Dept. and (2) Toy Dept. are given a high rating of 6.

Note here that the (1) Credit Dept. and (2) Toy Dept. are given a high rating of 6.Letter

Number

Note here that the (2) Toy Dept. and the (5) Candy Dept. are given a high rating of 6.

Note here that the (2) Toy Dept. and the (5) Candy Dept. are given a high rating of 6.

168

Example of Systematic Layout Planning:Initial Relationship Diagram

1

2

4

3

5

U U

E

A

I

The number of lines here represent paths required to be taken in transactions between the departments. The more lines, the more the interaction between departments.

The number of lines here represent paths required to be taken in transactions between the departments. The more lines, the more the interaction between departments.

Note here again, Depts. (1) and (2) are linked together, and Depts. (2) and (5) are linked together by multiple lines or required transactions.

Note here again, Depts. (1) and (2) are linked together, and Depts. (2) and (5) are linked together by multiple lines or required transactions.

169

Example of Systematic Layout Planning:Initial and Final Layouts

1

2 4

3

5

Initial Layout

Ignoring space andbuilding constraints

2

5 1 43

50 ft

20 ft

Final Layout

Adjusted by squarefootage and buildingsize

Note in the Final Layout that Depts. (1) and (5) are not both placed directly next to Dept. (2).

Note in the Final Layout that Depts. (1) and (5) are not both placed directly next to Dept. (2).

170

Station 1

Minutes per Unit 6

Station 2

7

Station 3

3

Assembly Lines Balancing Concepts

Question: Suppose you load work into the three work stations below such that each will take the corresponding number of minutes as shown. What is the cycle time of this line?

Question: Suppose you load work into the three work stations below such that each will take the corresponding number of minutes as shown. What is the cycle time of this line?

Answer: The cycle time of the line is always determined by the work station taking the longest time. In this problem, the cycle time of the line is 7 minutes. There is also going to be idle time at the other two work stations.

Answer: The cycle time of the line is always determined by the work station taking the longest time. In this problem, the cycle time of the line is 7 minutes. There is also going to be idle time at the other two work stations.

171

Example of Line Balancing

• You’ve just been assigned the job a setting up an electric fan assembly line with the following tasks:

Task Time (Mins) Description PredecessorsA 2 Assemble frame NoneB 1 Mount switch AC 3.25 Assemble motor housing NoneD 1.2 Mount motor housing in frame A, CE 0.5 Attach blade DF 1 Assemble and attach safety grill EG 1 Attach cord BH 1.4 Test F, G

172

Example of Line Balancing: Structuring the Precedence Diagram

Task PredecessorsA None

A

B A

B

C None

C

D A, C

D

Task PredecessorsE D

E

F E

F

G B

G

H E, G

H

173

Example of Line Balancing: Precedence Diagram

A

C

B

D E F

GH

2

3.25

1

1.2 .5

11.4

1

Question: Which process step defines the maximum rate of production?

Question: Which process step defines the maximum rate of production?

Answer: Task C is the cycle time of the line and therefore, the maximum rate of production.

Answer: Task C is the cycle time of the line and therefore, the maximum rate of production.

175

Example of Line Balancing: Determine Cycle Time

Required Cycle Time, C = Production time per period

Required output per periodRequired Cycle Time, C =

Production time per period

Required output per period

C = 420 mins / day

100 units / day= 4.2 mins / unitC =

420 mins / day

100 units / day= 4.2 mins / unit

Question: Suppose we want to assemble 100 fans per day. What would our cycle time have to be?

Question: Suppose we want to assemble 100 fans per day. What would our cycle time have to be?

Answer: Answer:

176

Example of Line Balancing: Determine Theoretical Minimum Number of

Workstations

Question: What is the theoretical minimum number of workstations for this problem?

Question: What is the theoretical minimum number of workstations for this problem?

Answer: Answer: Theoretical Min. Number of Workstations, N

N = Sum of task times (T)

Cycle time (C)

t

t

Theoretical Min. Number of Workstations, N

N = Sum of task times (T)

Cycle time (C)

t

t

N = 11.35 mins / unit

4.2 mins / unit= 2.702, or 3t

N = 11.35 mins / unit

4.2 mins / unit= 2.702, or 3t

177

Example of Line Balancing: Rules To Follow for Loading Workstations

• Assign tasks to station 1, then 2, etc. in sequence. Keep assigning to a workstation ensuring that precedence is maintained and total work is less than or equal to the cycle time. Use the following rules to select tasks for assignment.

• Primary: Assign tasks in order of the largest number of following tasks

• Secondary (tie-breaking): Assign tasks in order of the longest operating time

178

A

C

B

D E F

GH

2

3.25

1

1.2 .5

11.4

1

Station 1 Station 2 Station 3

Task Followers Time (Mins)A 6 2C 4 3.25D 3 1.2B 2 1E 2 0.5F 1 1G 1 1H 0 1.4

179

A

C

B

D E F

GH

2

3.25

1

1.2 .5

11.4

1


A (4.2-2=2.2)


180

A

C

B

D E F

GH

2

3.25

1

1.2 .5

11.4

1

A (4.2-2=2.2)B (2.2-1=1.2)



181

A

C

B

D E F

GH

2

3.25

1

1.2 .5

11.4

1

A (4.2-2=2.2)B (2.2-1=1.2)G (1.2-1= .2)

Idle= .2



182

A

C

B

D E F

GH

2

3.25

1

1.2 .5

11.4

1

C (4.2-3.25)=.95


A (4.2-2=2.2)B (2.2-1=1.2)G (1.2-1= .2)

Idle= .2


183

C (4.2-3.25)=.95

Idle = .95

A

C

B

D E F

GH

2

3.25

1

1.2 .5

11.4

1


A (4.2-2=2.2)B (2.2-1=1.2)G (1.2-1= .2)

Idle= .2


184

C (4.2-3.25)=.95

Idle = .95

A

C

B

D E F

GH

2

3.25

1

1.2 .5

11.4

1

D (4.2-1.2)=3


A (4.2-2=2.2)B (2.2-1=1.2)G (1.2-1= .2)

Idle= .2


185

A

C

B

D E F

GH

2

3.25

1

1.2 .5

11.4

1

C (4.2-3.25)=.95

Idle = .95

D (4.2-1.2)=3E (3-.5)=2.5


A (4.2-2=2.2)B (2.2-1=1.2)G (1.2-1= .2)

Idle= .2


186

A

C

B

D E F

GH

2

3.25

1

1.2 .5

11.4

1

C (4.2-3.25)=.95

Idle = .95

D (4.2-1.2)=3E (3-.5)=2.5F (2.5-1)=1.5


A (4.2-2=2.2)B (2.2-1=1.2)G (1.2-1= .2)

Idle= .2


187

Which station is the bottleneck? What is the effective cycle time?

A

C

B

D E F

GH

2

3.25

1

1.2 .5

11.4

1

C (4.2-3.25)=.95

Idle = .95

D (4.2-1.2)=3E (3-.5)=2.5F (2.5-1)=1.5H (1.5-1.4)=.1Idle = .1


A (4.2-2=2.2)B (2.2-1=1.2)G (1.2-1= .2)

Idle= .2


188

Example of Line Balancing: Determine the Efficiency of the Assembly Line

Efficiency =Sum of task times (T)

Actual number of workstations (Na) x Cycle time (C)Efficiency =

Sum of task times (T)

Actual number of workstations (Na) x Cycle time (C)

Efficiency =11.35 mins / unit

(3)(4.2mins / unit)=.901Efficiency =

11.35 mins / unit

(3)(4.2mins / unit)=.901

189

Group Technology:Benefits

1. Better human relations

2. Improved operator expertise

3. Less in-process inventory and material handling

4. Faster production setup

190

Group Technology:Transition from Process Layout

1. Grouping parts into families that follow a common sequence of steps

2. Identifying dominant flow patterns of parts families as a basis for location or relocation of processes

3. Physically grouping machines and processes into cells

191

Fixed Position Layout

Question: What are our primary considerations for a fixed position layout?

Question: What are our primary considerations for a fixed position layout?

Answer: Arranging materials and equipment concentrically around the production point in their order of use.

Answer: Arranging materials and equipment concentrically around the production point in their order of use.

192

Retail Service Layout

• Goal--maximize net profit per square foot of floor space

• Servicescapes– Ambient Conditions– Spatial Layout and Functionality– Signs, Symbols, and Artifacts

193


195

Chapter 6

Product Design and Process Selection – Services

196

• Service Generalizations

• Service Strategy: Focus & Advantage

• Service-System Design Matrix

• Service Blueprinting

• Service Fail-safing

• Characteristics of a Well-Designed Service Delivery System

OBJECTIVES

197

Service Generalizations

1. Everyone is an expert on services

2. Services are idiosyncratic

3. Quality of work is not quality of service

4. Most services contain a mix of tangible and intangible attributes

198

Service Generalizations (Continued)

5. High-contact services are experienced, whereas goods are consumed

6. Effective management of services requires an understanding of marketing and personnel, as well as operations

7. Services often take the form of cycles of encounters involving face-to-face, phone, Internet, electromechanical, and/or mail interactions

199

Service BusinessesDefined

• Facilities-based services: Where the customer must go to the service facility

• Field-based services: Where the production and consumption of the service takes place in the customer’s environment

A service business is the management of organizations whose primary business requires interaction with the customer to produce the service

200

Internal ServicesDefined

Internal Supplier

Internal Supplier

InternalCustomer

ExternalCustomer

Internal services is the management of services required to support the activities of the larger organization. Services including data processing, accounting, etc

201

The Service TriangleExhibit 6.1Exhibit 6.1

TheCustomer

The ServiceStrategy

ThePeople

TheSystems

A philosophical view that suggests the organization exists to serve the customer, and the systems and the employees exist to facilitate the process of service.

A philosophical view that suggests the organization exists to serve the customer, and the systems and the employees exist to facilitate the process of service.

202

Applying Behavioral Science to Service Encounters

1. The front-end and back-end of the encounter are not created equal

2. Segment the pleasure, combine the pain

3. Let the customer control the process

4. Pay attention to norms and rituals

5. People are easier to blame than systems

6. Let the punishment fit the crime in service recovery

203

Service Strategy: Focus and AdvantagePerformance Priorities

• Treatment of the customer

• Speed and convenience of service delivery

• Price

• Variety

• Quality of the tangible goods

• Unique skills that constitute the service offering

204

Service-System Design MatrixExhibit 6.7Exhibit 6.7

Mail contact

Face-to-faceloose specs

Face-to-facetight specs

PhoneContact

Face-to-facetotal

customization

Buffered core (none)

Permeable system (some)

Reactivesystem (much)

High

LowHigh

Low

Degree of customer/server contact

Internet & on-site

technology

SalesOpportunity

ProductionEfficiency

205

Example of Service Blueprinting

Brushshoes

Applypolish

Failpoint

BuffCollect

payment

Cleanshoes Materials

(e.g., polish, cloth)

Select andpurchasesupplies

Standardexecution time

2 minutes

Total acceptableexecution time

5 minutes

30secs

30secs

45secs

15secs

Wrongcolor wax

Seen bycustomer 45

secs

Line ofvisibility

Not seen bycustomer butnecessary toperformance

206

Service Fail-safingPoka-Yokes (A Proactive Approach)

• Keeping a mistake from becoming a service defect

• How can we fail-safe the three Ts?

Task

TangiblesTreatment

207

Have we compromised

one of the 3 Ts?

1. Task

2. Treatment

3. Tangible

1. Task

2. Treatment

3. Tangible

208

Three Contrasting Service Designs

• The production line approach (ex. McDonald’s)

• The self-service approach (ex. automatic teller machines)

• The personal attention approach (ex. Ritz-Carlton Hotel Company)

209

Characteristics of a Well-Designed Service System

1. Each element of the service system is consistent with the operating focus of the firm

2. It is user-friendly

3. It is robust

4. It is structured so that consistent performance by its people and systems is easily maintained

210

Characteristics of a Well-Designed Service System (Continued)

5. It provides effective links between the back office and the front office so that nothing falls between the cracks

6. It manages the evidence of service quality in such a way that customers see the value of the service provided

7. It is cost-effective

211

End of Chapter 6

213

Technical Note 6

Waiting Line Management

214

• Waiting Line Characteristics

• Suggestions for Managing Queues

• Examples (Models 1, 2, 3, and 4)

OBJECTIVES

215

Components of the Queuing System

CustomerArrivals

Servers

Waiting Line

Servicing System

Exit

Queue or

216

Customer Service Population Sources

Population Source

Finite Infinite

Example: Number of machines needing repair when a company only has three machines.

Example: Number of machines needing repair when a company only has three machines.

Example: The number of people who could wait in a line for gasoline.

Example: The number of people who could wait in a line for gasoline.

217

Service Pattern

ServicePattern

Constant Variable

Example: Items coming down an automated assembly line.

Example: Items coming down an automated assembly line.

Example: People spending time shopping.

Example: People spending time shopping.

218

The Queuing System

Queue Discipline

Length

Number of Lines &Line Structures

Service Time Distribution

Queuing System

219

Examples of Line Structures

Single Channel

Multichannel

SinglePhase Multiphase

One-personbarber shop

Car wash

Hospitaladmissions

Bank tellers’windows

220

Degree of Patience

No Way!

BALK

No Way!

RENEG

221

Suggestions for Managing Queues

1. Determine an acceptable waiting time for your customers

2. Try to divert your customer’s attention when waiting

3. Inform your customers of what to expect

4. Keep employees not serving the customers out of sight

5. Segment customers

222

Suggestions for Managing Queues (Continued)

6. Train your servers to be friendly

7. Encourage customers to come during the slack periods

8. Take a long-term perspective toward getting rid of the queues

223

Waiting Line Models

Model LayoutSourcePopulation Service Pattern

1 Single channel Infinite Exponential

2 Single channel Infinite Constant

3 Multichannel Infinite Exponential

4 Single or Multi Finite Exponential

These four models share the following characteristics: Single phase Poisson arrival FCFS Unlimited queue length

224

Notation: Infinite Queuing: Models 1-3

linein tingnumber wai Average

server single afor

rate sevice torate arrival totalof Ratio = =

arrivalsbetween timeAverage

timeservice Average

rate Service =

rate Arrival =

1

1

Lg

linein tingnumber wai Average

server single afor

rate sevice torate arrival totalof Ratio = =

arrivalsbetween timeAverage

timeservice Average

rate Service =

rate Arrival =

1

1

Lg

225Infinite Queuing Models 1-3 (Continued)

linein waitingofy Probabilit

systemin units exactly ofy Probabilit

channels service identical ofNumber =

system in the units ofNumber

served) be to time(including

systemin time totalAverage

linein waiting timeAverage = g

served) being those(including

systemin number Average = s

Pw

nPn

S

n

Ws

W

L

linein waitingofy Probabilit

systemin units exactly ofy Probabilit

channels service identical ofNumber =

system in the units ofNumber

served) be to time(including

systemin time totalAverage

linein waiting timeAverage = g

served) being those(including

systemin number Average = s

Pw

nPn

S

n

Ws

W

L

226

Example: Model 1Assume a drive-up window at a fast food restaurant.Customers arrive at the rate of 25 per hour.The employee can serve one customer every two minutes.Assume Poisson arrival and exponential service rates.

Determine:A) What is the average utilization of the employee?B) What is the average number of customers in line?C) What is the average number of customers in the system?D) What is the average waiting time in line?E) What is the average waiting time in the system?F) What is the probability that exactly two cars will be in the system?

Determine:A) What is the average utilization of the employee?B) What is the average number of customers in line?C) What is the average number of customers in the system?D) What is the average waiting time in line?E) What is the average waiting time in the system?F) What is the probability that exactly two cars will be in the system?

227

= 25 cust / hr

= 1 customer

2 mins (1hr / 60 mins) = 30 cust / hr

= = 25 cust / hr

30 cust / hr = .8333

= 25 cust / hr

= 1 customer

2 mins (1hr / 60 mins) = 30 cust / hr

= = 25 cust / hr

30 cust / hr = .8333

Example: Model 1

A) What is the average utilization of the employee?

228

Example: Model 1

B) What is the average number of customers in line?

4.167 = 25)-30(30

(25) =

) - ( =

22

Lg 4.167 = 25)-30(30

(25) =

) - ( =

22

Lg

C) What is the average number of customers in the system?

5 = 25)-(30

25 =

- =

Ls 5 = 25)-(30

25 =

- =

Ls

229

Example: Model 1

D) What is the average waiting time in line?

mins 10 = hrs .1667 =

=

LgWg mins 10 = hrs .1667 =

=

LgWg

E) What is the average waiting time in the system?

mins 12 = hrs .2 = =Ls

Ws mins 12 = hrs .2 = =Ls

Ws

230

Example: Model 1

F) What is the probability that exactly two cars will be in the system (one being served and the other waiting in line)?

p = (1-n

n

)( )p = (1-n

n

)( )

p = (1- = 2

225

30

25

30)( ) .1157p = (1- =

2

225

30

25

30)( ) .1157

231

Example: Model 2

An automated pizza vending machine heats and dispenses a slice of pizza in 4 minutes.

Customers arrive at a rate of one every 6 minutes with the arrival rate exhibiting a Poisson distribution.

Determine:

A) The average number of customers in line.B) The average total waiting time in the system.

Determine:

A) The average number of customers in line.B) The average total waiting time in the system.

232

Example: Model 2

A) The average number of customers in line.

.6667 = 10)-(2)(15)(15

(10) =

) - (2 =

22

Lg .6667 = 10)-(2)(15)(15

(10) =

) - (2 =

22

Lg

B) The average total waiting time in the system.

mins 4 = hrs .06667 = 10)-51)(15(2

10 =

) - (2 =

Wg mins 4 = hrs .06667 = 10)-51)(15(2

10 =

) - (2 =

Wg

mins 8 = hrs .1333 = 15/hr

1 + hrs .06667 =

1 + =

WgWs mins 8 = hrs .1333 = 15/hr

1 + hrs .06667 =

1 + =

WgWs

233

Example: Model 3Recall the Model 1 example:Drive-up window at a fast food restaurant.Customers arrive at the rate of 25 per hour.The employee can serve one customer every two minutes.Assume Poisson arrival and exponential service rates.

If an identical window (and an identically trained server) were added, what would the effects be on the average number of cars in the system and the total time customers wait before being served?

If an identical window (and an identically trained server) were added, what would the effects be on the average number of cars in the system and the total time customers wait before being served?

234

Example: Model 3Average number of cars in the system

ion)interpolatlinear -using-TN6.10(Exhibit

1760= .Lgion)interpolatlinear -using-TN6.10(Exhibit

1760= .Lg

1.009 = 30

25 + .176 = + =

LgLs 1.009 = 30

25 + .176 = + =

LgLs

Total time customers wait before being served

)( = mincustomers/ 25

customers .176 = = Wait! No

LgWg mins .007

)( =

mincustomers/ 25

customers .176 = = Wait! No

LgWg mins .007

235

Notation: Finite Queuing: Model 4

channels service ofNumber

linein units ofnumber Average

)( system

queuingin thoseless source Population =

served being units ofnumber Average

linein wait tohaving

ofeffect theof measure a factor, Efficiency

linein must wait arrivalan y that Probabilit =

S

L

n-N

J

H

F

D

channels service ofNumber

linein units ofnumber Average

)( system

queuingin thoseless source Population =

served being units ofnumber Average

linein wait tohaving

ofeffect theof measure a factor, Efficiency

linein must wait arrivalan y that Probabilit =

S

L

n-N

J

H

F

D

236

Finite Queuing: Model 4 (Continued)

required timeservice of proportionor factor, Service

linein time waitingAverage

tsrequiremen servicecustomer between timeAverage

service theperform to timeAverage =

system queuingin units exactly ofy Probabilit

source populationin units ofNumber

served) being one the(including

system queuingin units ofnumber Average =

X

W

U

T

nPn

N

n

required timeservice of proportionor factor, Service

linein time waitingAverage

tsrequiremen servicecustomer between timeAverage

service theperform to timeAverage =

system queuingin units exactly ofy Probabilit

source populationin units ofNumber

served) being one the(including

system queuingin units ofnumber Average =

X

W

U

T

nPn

N

n

237

Example: Model 4

The copy center of an electronics firm has four copymachines that are all serviced by a single technician.

Every two hours, on average, the machines require adjustment. The technician spends an average of 10minutes per machine when adjustment is required.

Assuming Poisson arrivals and exponential service, how many machines are “down” (on average)?

The copy center of an electronics firm has four copymachines that are all serviced by a single technician.

Every two hours, on average, the machines require adjustment. The technician spends an average of 10minutes per machine when adjustment is required.

Assuming Poisson arrivals and exponential service, how many machines are “down” (on average)?

238

Example: Model 4N, the number of machines in the population = 4M, the number of repair people = 1T, the time required to service a machine = 10 minutesU, the average time between service = 2 hours

X =T

T + U

10 min

10 min + 120 min= .077X =

T

T + U

10 min

10 min + 120 min= .077

From Table TN6.11, F = .980 (Interpolation)From Table TN6.11, F = .980 (Interpolation)

L, the number of machines waiting to be serviced = N(1-F) = 4(1-.980) = .08 machines

L, the number of machines waiting to be serviced = N(1-F) = 4(1-.980) = .08 machines

H, the number of machines being serviced = FNX = .980(4)(.077) = .302 machines

H, the number of machines being serviced = FNX = .980(4)(.077) = .302 machines

Number of machines down = L + H = .382 machinesNumber of machines down = L + H = .382 machines

239

Queuing Approximation• This approximation is quick way to analyze a queuing situation. Now, both

interarrival time and service time distributions are allowed to be general.• In general, average performance measures (waiting time in queue, number

in queue, etc) can be very well approximated by mean and variance of the distribution (distribution shape not very important).

• This is very good news for managers: all you need is mean and standard deviation, to compute average waiting time

222

Define:

Standard deviation of Xcoefficient of variation for r.v. X =

Mean of XVariance

squared coefficient of variation (scv) = mean

x

x x

C

C C

240

Queue Approximation

2( 1) 2 2

1 2

Sa s

q

C CL

s qL L S

Compute S

2 2,a sC CInputs: S, , ,

(Alternatively: S, , , variances of interarrival and service time distributions)

as before, , and q sq s

L LW W

241

Approximation Example• Consider a manufacturing process (for example making plastic parts) consisting of a single stage with

five machines. Processing times have a mean of 5.4 days and standard deviation of 4 days. The firm operates make-to-order. Management has collected date on customer orders, and verified that the time between orders has a mean of 1.2 days and variance of 0.72 days. What is the average time that an order waits before being worked on?

Using our “Waiting Line Approximation” spreadsheet we get:Lq = 3.154 Expected number of orders waiting to be completed.

Wq = 3.78 Expected number of days order waits.

Ρ = 0.9 Expected machine utilization.

242


244

Chapter 7

Quality Management

245

• Total Quality Management Defined

• Quality Specifications and Costs

• Six Sigma Quality and Tools

• External Benchmarking

• ISO 9000

• Service Quality Measurement

OBJECTIVES

246

Total Quality Management (TQM)

Defined

• Total quality management is defined as managing the entire organization so that it excels on all dimensions of products and services that are important to the customer

247

Quality Specifications

• Design quality: Inherent value of the product in the marketplace

– Dimensions include: Performance, Features, Reliability, Durability, Serviceability, Response, Aesthetics, and Reputation.

• Conformance quality: Degree to which the product or service design specifications are met

248

Costs of Quality

External Failure Costs

Appraisal Costs

Prevention Costs

Internal FailureCosts

Costs ofQuality

249

Six Sigma Quality

• A philosophy and set of methods companies use to eliminate defects in their products and processes

• Seeks to reduce variation in the processes that lead to product defects

• The name, “six sigma” refers to the variation that exists within plus or minus three standard deviations of the process outputs

3

250

Six Sigma Quality (Continued)• Six Sigma allows managers to readily

describe process performance using a common metric: Defects Per Million Opportunities (DPMO)

1,000,000 x

units of No. x

unit per error for

iesopportunit ofNumber

defects ofNumber

DPMO 1,000,000 x

units of No. x

unit per error for

iesopportunit ofNumber

defects ofNumber

DPMO

251

Six Sigma Quality (Continued)

Example of Defects Per Million Opportunities (DPMO) calculation. Suppose we observe 200 letters delivered incorrectly to the wrong addresses in a small city during a single day when a total of 200,000 letters were delivered. What is the DPMO in this situation?

Example of Defects Per Million Opportunities (DPMO) calculation. Suppose we observe 200 letters delivered incorrectly to the wrong addresses in a small city during a single day when a total of 200,000 letters were delivered. What is the DPMO in this situation?

000,1 1,000,000 x

200,000 x 1

200DPMO

000,1 1,000,000 x

200,000 x 1

200DPMO

So, for every one million letters delivered this city’s postal managers can expect to have 1,000 letters incorrectly sent to the wrong address.

So, for every one million letters delivered this city’s postal managers can expect to have 1,000 letters incorrectly sent to the wrong address.

Cost of Quality: What might that DPMO mean in terms of over-time employment to correct the errors?

Cost of Quality: What might that DPMO mean in terms of over-time employment to correct the errors?

252

Six Sigma Quality: DMAIC Cycle

• Define, Measure, Analyze, Improve, and Control (DMAIC)

• Developed by General Electric as a means of focusing effort on quality using a methodological approach

• Overall focus of the methodology is to understand and achieve what the customer wants

• DMAIC consists of five steps….

253

Six Sigma Quality: DMAIC Cycle

(Continued)

1. Define (D)

2. Measure (M)

3. Analyze (A)

4. Improve (I)

5. Control (C)

Customers and their priorities

Process and its performance

Causes of defects

Remove causes of defects

Maintain quality

254

Analytical Tools for Six Sigma and

Continuous Improvement: Flow Chart No, Continue…

Material Received

from Supplier

Inspect Material for

DefectsDefects found?

Return to Supplier for Credit

Yes

Can be used to find quality problems

Can be used to find quality problems

255

Analytical Tools for Six Sigma and Continuous Improvement: Run Chart

Can be used to identify when equipment or processes are not behaving according to specifications

Can be used to identify when equipment or processes are not behaving according to specifications

0.440.460.48

0.50.520.540.560.58

1 2 3 4 5 6 7 8 9 10 11 12Time (Hours)

Dia

me

ter

256

Analytical Tools for Six Sigma and Continuous Improvement: Pareto

Analysis

Can be used to find when 80% of the problems may be attributed to 20% of thecauses

Can be used to find when 80% of the problems may be attributed to 20% of thecauses

Assy.Instruct.

Fre

quen

cy

Design Purch. Training Other

80%

257

Analytical Tools for Six Sigma and Continuous Improvement: Checksheet

Billing Errors

Wrong Account

Wrong Amount

A/R Errors

Wrong Account

Wrong Amount

Monday

Can be used to keep track of defects or used to make sure people collect data in a correct manner

Can be used to keep track of defects or used to make sure people collect data in a correct manner

258

Analytical Tools for Six Sigma and Continuous Improvement: Histogram

Nu

mb

er

of

Lo

ts

Data Ranges

Defectsin lot

0 1 2 3 4

Can be used to identify the frequency of quality defect occurrence and display quality performance

Can be used to identify the frequency of quality defect occurrence and display quality performance

259

Analytical Tools for Six Sigma and Continuous Improvement: Cause &

Effect Diagram

Effect

ManMachine

MaterialMethod

Environment

Possible causes:Possible causes: The results or effect

The results or effect

Can be used to systematically track backwards to find a possible cause of a quality problem (or effect)

Can be used to systematically track backwards to find a possible cause of a quality problem (or effect)

260

Analytical Tools for Six Sigma and Continuous Improvement: Control Charts

Can be used to monitor ongoing production process quality and quality conformance to stated standards of quality

Can be used to monitor ongoing production process quality and quality conformance to stated standards of quality

970

980

990

1000

1010

1020

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

LCL

UCL

261

Other Six Sigma Tools• Opportunity Flow Diagram used to graphically

show those activities that add value from those that are performed (and maybe could be reduced or removed) that do not add value to the finished product

• Failure Mode and Effect Analysis (DMEA) is a structured approach to identify, estimate, prioritize, and evaluate risk of possible failures at each stage in the process

• Design of Experiments (DOE) a statistical test to determine cause-and-effect relationships between process variables and output

262

Six Sigma Roles and Responsibilities

1. Executive leaders must champion the process of improvement

2. Corporation-wide training in Six Sigma concepts and tools

3. Setting stretch objectives for improvement

4. Continuous reinforcement and rewards

263

The Shingo System: Fail-Safe Design

• Shingo’s argument:– SQC methods do not prevent defects– Defects arise when people make errors– Defects can be prevented by providing workers with

feedback on errors

• Poka-Yoke includes:– Checklists– Special tooling that prevents workers from making

errors

264

ISO 9000

• Series of standards agreed upon by the International Organization for Standardization (ISO)

• Adopted in 1987

• More than 100 countries

• A prerequisite for global competition?

• ISO 9000 directs you to "document what you do and then do as you documented"

265

Three Forms of ISO Certification

1. First party: A firm audits itself against ISO 9000 standards

2. Second party: A customer audits its supplier

3. Third party: A "qualified" national or international standards or certifying agency serves as auditor

266

External Benchmarking Steps

1. Identify those processes needing improvement

2. Identify a firm that is the world leader in performing the process

3. Contact the managers of that company and make a personal visit to interview managers and workers

4. Analyze data

267

Service Quality Measurement:Servqual

• A perceived service quality questionnaire survey methodology

• Examines “Dimensions of Service Quality” including: Reliability, Responsiveness, Assurance, Empathy, and Tangibles (e.g., appearance of physical facilities, equipment, etc.)

268

Service Quality Measurement: Servqual (Continued)

• New version of this methodology is called “e-Service Quality” dealing service on the Internet

• Dimensions of Service Quality on the e-Service methodology include: Reliability, Responsiveness, Access, Flexibility, Ease of Navigation, Efficiency, Assurance/Trust, Security/Privacy, Price Knowledge, Site Aesthetics, and Customization/Personalization

269

End of Chapter 7

271

Technical Note 7

Process Capability and Statistical Quality Control

272

• Process Variation

• Process Capability

• Process Control Procedures– Variable data– Attribute data

• Acceptance Sampling– Operating Characteristic Curve

OBJECTIVES

273

Basic Forms of Variation

Assignable variation is caused by factors that can be clearly identified and possibly managed

Common variation is inherent in the production process

Example: A poorly trained employee that creates variation in finished product output.

Example: A poorly trained employee that creates variation in finished product output.

Example: A molding process that always leaves “burrs” or flaws on a molded item.

Example: A molding process that always leaves “burrs” or flaws on a molded item.

274

Taguchi’s View of Variation

IncrementalCost of Variability

High

Zero

LowerSpec

TargetSpec

UpperSpec

Traditional View

IncrementalCost of Variability

High

Zero

LowerSpec

TargetSpec

UpperSpec

Taguchi’s View

Exhibits TN7.1 & TN7.2

Exhibits TN7.1 & TN7.2

Traditional view is that quality within the LS and US is good and that the cost of quality outside this range is constant, where Taguchi views costs as increasing as variability increases, so seek to achieve zero defects and that will truly minimize quality costs.

Traditional view is that quality within the LS and US is good and that the cost of quality outside this range is constant, where Taguchi views costs as increasing as variability increases, so seek to achieve zero defects and that will truly minimize quality costs.

275

Process Capability

• Process limits

• Tolerance limits

• How do the limits relate to one another?

276

Process Capability Index, Cpk

3

X-UTLor

3

LTLXmin=C pk

Shifts in Process Mean

Capability Index shows how well parts being produced fit into design limit specifications.

Capability Index shows how well parts being produced fit into design limit specifications.

As a production process produces items small shifts in equipment or systems can cause differences in production performance from differing samples.

As a production process produces items small shifts in equipment or systems can cause differences in production performance from differing samples.

277

Types of Statistical Sampling

• Attribute (Go or no-go information)– Defectives refers to the acceptability of product

across a range of characteristics.– Defects refers to the number of defects per unit

which may be higher than the number of defectives.

– p-chart application

• Variable (Continuous)– Usually measured by the mean and the standard

deviation.– X-bar and R chart applications

278

UCL

LCL

Samples over time

1 2 3 4 5 6

UCL

LCL

Samples over time

1 2 3 4 5 6

UCL

LCL

Samples over time

1 2 3 4 5 6

Normal BehaviorNormal Behavior

Possible problem, investigatePossible problem, investigate

Possible problem, investigatePossible problem, investigate

Statistical Process Control (SPC) Charts

279

Control Limits are based on the Normal Curve

x

0 1 2 3-3 -2 -1z

Standard deviation units or “z” units.

Standard deviation units or “z” units.

280

Control Limits

We establish the Upper Control Limits (UCL) and the Lower Control Limits (LCL) with plus or minus 3 standard deviations from some x-bar or mean value. Based on this we can expect 99.7% of our sample observations to fall within these limits.

xLCL UCL

99.7%

281

Example of Constructing a p-Chart: Required Data

1 100 42 100 23 100 54 100 35 100 66 100 47 100 38 100 79 100 1

10 100 211 100 312 100 213 100 214 100 815 100 3

Sample

No.

No. of

Samples

Number of defects found in each sample

282

Statistical Process Control Formulas:Attribute Measurements (p-Chart)

p =Total Number of Defectives

Total Number of Observationsp =

Total Number of Defectives

Total Number of Observations

ns

)p-(1 p = p n

s)p-(1 p

= p

p

p

z - p = LCL

z + p = UCL

s

s

p

p

z - p = LCL

z + p = UCL

s

s

Given:

Compute control limits:

283

1. Calculate the sample proportions, p (these are what can be plotted on the p-chart) for each sample

1. Calculate the sample proportions, p (these are what can be plotted on the p-chart) for each sample

Sample n Defectives p1 100 4 0.042 100 2 0.023 100 5 0.054 100 3 0.035 100 6 0.066 100 4 0.047 100 3 0.038 100 7 0.079 100 1 0.01

10 100 2 0.0211 100 3 0.0312 100 2 0.0213 100 2 0.0214 100 8 0.0815 100 3 0.03

Example of Constructing a p-chart: Step 1

284

2. Calculate the average of the sample proportions2. Calculate the average of the sample proportions

0.036=1500

55 = p 0.036=1500

55 = p

3. Calculate the standard deviation of the sample proportion 3. Calculate the standard deviation of the sample proportion

.0188= 100

.036)-.036(1=

)p-(1 p = p n

s .0188= 100

.036)-.036(1=

)p-(1 p = p n

s

Example of Constructing a p-chart: Steps 2&3

285

4. Calculate the control limits4. Calculate the control limits

3(.0188) .0363(.0188) .036

UCL = 0.0924LCL = -0.0204 (or 0)UCL = 0.0924LCL = -0.0204 (or 0)

p

p

z - p = LCL

z + p = UCL

s

s

p

p

z - p = LCL

z + p = UCL

s

s

Example of Constructing a p-chart: Step 4

286

Example of Constructing a p-Chart: Step 5

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Observation

p

UCL

LCL

5. Plot the individual sample proportions, the average of the proportions, and the control limits

5. Plot the individual sample proportions, the average of the proportions, and the control limits

287

Example of x-bar and R Charts: Required Data

Sample Obs 1 Obs 2 Obs 3 Obs 4 Obs 51 10.68 10.689 10.776 10.798 10.7142 10.79 10.86 10.601 10.746 10.7793 10.78 10.667 10.838 10.785 10.7234 10.59 10.727 10.812 10.775 10.735 10.69 10.708 10.79 10.758 10.6716 10.75 10.714 10.738 10.719 10.6067 10.79 10.713 10.689 10.877 10.6038 10.74 10.779 10.11 10.737 10.759 10.77 10.773 10.641 10.644 10.72510 10.72 10.671 10.708 10.85 10.71211 10.79 10.821 10.764 10.658 10.70812 10.62 10.802 10.818 10.872 10.72713 10.66 10.822 10.893 10.544 10.7514 10.81 10.749 10.859 10.801 10.70115 10.66 10.681 10.644 10.747 10.728

288

Example of x-bar and R charts: Step 1. Calculate sample means, sample ranges,

mean of means, and mean of ranges.Sample Obs 1 Obs 2 Obs 3 Obs 4 Obs 5 Avg Range

1 10.68 10.689 10.776 10.798 10.714 10.732 0.1162 10.79 10.86 10.601 10.746 10.779 10.755 0.2593 10.78 10.667 10.838 10.785 10.723 10.759 0.1714 10.59 10.727 10.812 10.775 10.73 10.727 0.2215 10.69 10.708 10.79 10.758 10.671 10.724 0.1196 10.75 10.714 10.738 10.719 10.606 10.705 0.1437 10.79 10.713 10.689 10.877 10.603 10.735 0.2748 10.74 10.779 10.11 10.737 10.75 10.624 0.6699 10.77 10.773 10.641 10.644 10.725 10.710 0.13210 10.72 10.671 10.708 10.85 10.712 10.732 0.17911 10.79 10.821 10.764 10.658 10.708 10.748 0.16312 10.62 10.802 10.818 10.872 10.727 10.768 0.25013 10.66 10.822 10.893 10.544 10.75 10.733 0.34914 10.81 10.749 10.859 10.801 10.701 10.783 0.15815 10.66 10.681 10.644 10.747 10.728 10.692 0.103

Averages 10.728 0.220400

289

Example of x-bar and R charts: Step 2. Determine Control Limit Formulas and

Necessary Tabled Values

x Chart Control Limits

UCL = x + A R

LCL = x - A R

2

2

x Chart Control Limits

UCL = x + A R

LCL = x - A R

2

2

R Chart Control Limits

UCL = D R

LCL = D R

4

3

R Chart Control Limits

UCL = D R

LCL = D R

4

3

From Exhibit TN7.7From Exhibit TN7.7

n A2 D3 D42 1.88 0 3.273 1.02 0 2.574 0.73 0 2.285 0.58 0 2.116 0.48 0 2.007 0.42 0.08 1.928 0.37 0.14 1.869 0.34 0.18 1.82

10 0.31 0.22 1.7811 0.29 0.26 1.74

290

Example of x-bar and R charts: Steps 3&4. Calculate x-bar Chart and Plot Values

10.601

10.856

=).58(0.2204-10.728RA - x = LCL

=).58(0.2204-10.728RA + x = UCL

2

2

10.601

10.856

=).58(0.2204-10.728RA - x = LCL

=).58(0.2204-10.728RA + x = UCL

2

2

10.550

10.600

10.650

10.700

10.750

10.800

10.850

10.900

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Sample

Mea

ns

UCL

LCL

291

Example of x-bar and R charts: Steps 5&6. Calculate R-chart and Plot Values

0

0.46504

)2204.0)(0(R D= LCL

)2204.0)(11.2(R D= UCL

3

4

0

0.46504

)2204.0)(0(R D= LCL

)2204.0)(11.2(R D= UCL

3

4

0.000

0.100

0.200

0.300

0.400

0.500

0.600

0.700

0.800

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Sample

RUCL

LCL

292

Basic Forms of Statistical Sampling for Quality Control

• Acceptance Sampling is sampling to accept or reject the immediate lot of product at hand

• Statistical Process Control is sampling to determine if the process is within acceptable limits

293

Acceptance Sampling• Purposes

– Determine quality level– Ensure quality is within predetermined level

• Advantages– Economy– Less handling damage– Fewer inspectors– Upgrading of the inspection job– Applicability to destructive testing– Entire lot rejection (motivation for improvement)

294

Acceptance Sampling (Continued)

• Disadvantages– Risks of accepting “bad” lots and rejecting

“good” lots– Added planning and documentation– Sample provides less information than 100-

percent inspection

295

Acceptance Sampling: Single Sampling Plan

A simple goal

Determine (1) how many units, n, to sample from a lot, and (2) the maximum number of defective items, c, that can be found in the sample before the lot is rejected

296

Risk• Acceptable Quality Level (AQL)

– Max. acceptable percentage of defectives defined by producer

• The(Producer’s risk)– The probability of rejecting a good lot

• Lot Tolerance Percent Defective (LTPD)– Percentage of defectives that defines consumer’s

rejection point

• The (Consumer’s risk)– The probability of accepting a bad lot

297

Operating Characteristic Curve

n = 99c = 4

AQL LTPD

00.10.20.30.40.50.60.70.80.9

1

1 2 3 4 5 6 7 8 9 10 11 12

Percent defective

Pro

bab

ilit

y of

acc

epta

nce

=.10(consumer’s risk)

= .05 (producer’s risk)

The OCC brings the concepts of producer’s risk, consumer’s risk, sample size, and maximum defects allowed together

The OCC brings the concepts of producer’s risk, consumer’s risk, sample size, and maximum defects allowed together

The shape or slope of the curve is dependent on a particular combination of the four parameters

The shape or slope of the curve is dependent on a particular combination of the four parameters

298

Example: Acceptance Sampling Problem

Zypercom, a manufacturer of video interfaces, purchases printed wiring boards from an outside vender, Procard. Procard has set an acceptable quality level of 1% and accepts a 5% risk of rejecting lots at or below this level. Zypercom considers lots with 3% defectives to be unacceptable and will assume a 10% risk of accepting a defective lot.

Develop a sampling plan for Zypercom and determine a rule to be followed by the receiving inspection personnel.

Zypercom, a manufacturer of video interfaces, purchases printed wiring boards from an outside vender, Procard. Procard has set an acceptable quality level of 1% and accepts a 5% risk of rejecting lots at or below this level. Zypercom considers lots with 3% defectives to be unacceptable and will assume a 10% risk of accepting a defective lot.

Develop a sampling plan for Zypercom and determine a rule to be followed by the receiving inspection personnel.

299

Example: Step 1. What is given and what

is not?

In this problem, AQL is given to be 0.01 and LTDP is given to be 0.03. We are also given an alpha of 0.05 and a beta of 0.10.

In this problem, AQL is given to be 0.01 and LTDP is given to be 0.03. We are also given an alpha of 0.05 and a beta of 0.10.

What you need to determine is your sampling plan is “c” and “n.”

What you need to determine is your sampling plan is “c” and “n.”

300

Example: Step 2. Determine “c”

First divide LTPD by AQL.First divide LTPD by AQL.LTPD

AQL =

.03

.01 = 3

LTPD

AQL =

.03

.01 = 3

Then find the value for “c” by selecting the value in the TN7.10 “n(AQL)”column that is equal to or just greater than the ratio above.

Then find the value for “c” by selecting the value in the TN7.10 “n(AQL)”column that is equal to or just greater than the ratio above.

Exhibit TN 7.10Exhibit TN 7.10

c LTPD/AQL n AQL c LTPD/AQL n AQL0 44.890 0.052 5 3.549 2.6131 10.946 0.355 6 3.206 3.2862 6.509 0.818 7 2.957 3.9813 4.890 1.366 8 2.768 4.6954 4.057 1.970 9 2.618 5.426

So, c = 6.So, c = 6.

301

Example: Step 3. Determine Sample Size

c = 6, from Tablen (AQL) = 3.286, from TableAQL = .01, given in problem

c = 6, from Tablen (AQL) = 3.286, from TableAQL = .01, given in problem

Sampling Plan:Take a random sample of 329 units from a lot. Reject the lot if more than 6 units are defective.

Sampling Plan:Take a random sample of 329 units from a lot. Reject the lot if more than 6 units are defective.

Now given the information below, compute the sample size in units to generate your sampling plan

Now given the information below, compute the sample size in units to generate your sampling plan

n(AQL/AQL) = 3.286/.01 = 328.6, or 329 (always round up)n(AQL/AQL) = 3.286/.01 = 328.6, or 329 (always round up)

302


304

Chapter 8

Operations Consulting and Reengineering

305

• Operations Consulting Defined

• Operations Consulting and the 5 P’s

• Hierarchy Within a Consulting Organization

• Stages of Operations Consulting

• Operations Consulting Tool Kit

• Reengineering

OBJECTIVES

306

Operations ConsultingDefined

• Operations consulting involves assisting clients in developing operations strategies (i.e., product leadership, operational excellence, customer intimacy, etc.) and in improving production (and service delivery) processes.

307

Operations Consulting & the 5 Ps• Plants

– Adding and locating new plants– Expanding, contracting, or refocusing facilities

• Parts– Make or buy decisions– Vendor selection decisions

• Processes– Technology evaluation– Process improvement and reengineering

308

Operations Consulting & the 5 Ps (Continued)

• People– Quality improvement– Setting/revising work standards– Learning curve analysis

• Planning and Control Systems– Supply chain management– MRP– Shop floor control– Warehousing and distribution

309

Hierarchy within Consulting Firms

PartnersFinders

Who find new businessWho find new business

ManagersMindersWho manage the

business

Who manage the business

Consultants

GrindersWho actually do the work

Who actually do the work

A way of looking at the typical consulting firm’s organization

A way of looking at the typical consulting firm’s organization

310

Economics of Consulting Firms

• David H. Maister’s article on consulting draws an analogy between the consulting firm and a job shop operation. Three types of jobs:

• 1. Brain Surgery: Requiring innovation and creativity

• 2. Gray Hair: Requiring a great deal of experience (little innovation)

• 3. Procedures: Requiring activities similar to other existing projects (little innovation or experience)

311

When are Operations Consultants Needed

• When faced with major investment decision(s)

• When management believes it is not getting the maximum effectiveness from the organization’s productive capability

312

Stages in Operations Consulting Process

1. Sales and proposal development2. Analyze problem 3. Design, develop and test alternative

solutions4. Develop systematic performance measures5. Present final report6. Implement changes7. Assure client satisfaction8. Assemble learnings from the study

313

Operations Consulting Tool Kit: Category 1

Problem Definition

Issue trees

Customer surveys

Gap analysis

Employee surveys

Five forces model

In this scheme we have five categories of activities, starting with Problem Definition, that consultants perform and the supporting tools used to aid the consultant in performing that category

In this scheme we have five categories of activities, starting with Problem Definition, that consultants perform and the supporting tools used to aid the consultant in performing that category

314


Data Gathering

Plant tours/audits

Work sampling

Flow charts

Organizational charts

315


Data Analysis and Solution Development

Problem analysis (SPC tools)

Bottleneck analysis

Computer simulation

Statistical tools

316


Cost Impact and Payoff Analysis

Decision trees

Balanced scorecard

Stakeholder analysis

317


Implementation

Responsibility charts

Project management techniques

318

ReengineeringDefined

• Reengineering is defined as the fundamental rethinking and radical redesign of business processes to achieve dramatic improvements in critical, contemporary measures of performance such as cost, quality, service, and speed. As a engineering discipline, reengineering can be applied to any process in manufacturing and service businesses, education, and the government.

• Business process reengineering (BPR) is focused on reengineering business processes.

319

Key Words in the Reengineering Definition

• Fundamental– Why do we do what we do– Ignore what is and concentrate on what

should be

• Radical– Business reinvention vs. business

improvement

320

Key Words in the Reengineering Definition (Continued)

• Dramatic– Reengineering should be brought in “when a need

exits for heavy blasting”• Companies in deep trouble• Companies that see trouble coming• Companies that are in peak condition

• Business Process – a collection of activities that takes one or more

kinds of inputs and creates an output that is of value to a customer

321

Principles of Reengineering

• Organize around outcomes, not tasks

• Have those who use the output of the process perform the process

• Merge information-processing work into the real work that produces the information

• Treat geographically dispersed resources as though they were centralized

322

Principles of Reengineering (Continued)

• Link parallel activities instead of integrating their results

• Put the decision point where the work is performed, and build control into the process

• Capture information once and at the source

323

End of Chapter 8

325

Chapter 9

Supply-Chain Strategy

326

• Supply-Chain Management

• Measuring Supply-Chain Performance

• Bullwhip Effect

• Outsourcing

• Value Density

• Mass Customization

OBJECTIVES

327

• Supply-chain is a term that describes how organizations (suppliers, manufacturers, distributors, and customers) are linked together

What is a Supply-Chain?Defined

Suppliers

Inputs

Suppliers

Service support operations

Transformation

Manufacturing

Local service providers

Localization

Distribution

Customers

Output

Customers

Services

Supply networks

Manufacturing

328

What is Supply-Chain Management?Defined

• Supply-chain management is a total system approach to managing the entire flow of information, materials, and services from raw-material suppliers through factories and warehouses to the end customer

329

Formulas for Measuring Supply-Chain Performance

• One of the most commonly used measures in all of operations management is “Inventory Turnover”

• In situations where distribution inventory is dominant, “Weeks of Supply” is preferred and measures how many weeks’ worth of inventory is in the system at a particular time

valueinventory aggregate Average

sold goods ofCost turnoverInventory valueinventory aggregate Average

sold goods ofCost turnoverInventory

weeks52 sold goods ofCost

valueinventory aggregate Averagesupply of Weeks

weeks52

sold goods ofCost

valueinventory aggregate Averagesupply of Weeks

330

Example of Measuring Supply-Chain Performance

Suppose a company’s new annual report claims their costs of goods sold for the year is $160 million and their total average inventory (production materials + work-in-process) is worth $35 million. This company normally has an inventory turn ratio of 10. What is this year’s Inventory Turnover ratio? What does it mean?

Suppose a company’s new annual report claims their costs of goods sold for the year is $160 million and their total average inventory (production materials + work-in-process) is worth $35 million. This company normally has an inventory turn ratio of 10. What is this year’s Inventory Turnover ratio? What does it mean?

331

Example of Measuring Supply-Chain Performance (Continued)

= $160/$35 = 4.57

Since the company’s normal inventory turnover ration is 10, a drop to 4.57 means that the inventory is not turning over as quickly as it had in the past. Without knowing the industry average of turns for this company it is not possible to comment on how they are competitively doing in the industry, but they now have more inventory relative to their cost of goods sold than before.

= $160/$35 = 4.57

Since the company’s normal inventory turnover ration is 10, a drop to 4.57 means that the inventory is not turning over as quickly as it had in the past. Without knowing the industry average of turns for this company it is not possible to comment on how they are competitively doing in the industry, but they now have more inventory relative to their cost of goods sold than before.





332

Bullwhip Effect O

rder

Q

uan t

ity

Time

Retailer’s Orders

Ord

er

Qua

n tit

y

Time

Wholesaler’s Orders

Ord

er

Qua

n tit

y

Time

Manufacturer’s Orders

The magnification of variability in orders in the supply-chain

The magnification of variability in orders in the supply-chain

A lot of retailers each with little variability in their orders….

A lot of retailers each with little variability in their orders….

…can lead to greater variability for a fewer number of wholesalers, and…

…can lead to greater variability for a fewer number of wholesalers, and…

…can lead to even greater variability for a single manufacturer.

…can lead to even greater variability for a single manufacturer.

333

Hau Lee’s Concepts of Supply Chain Management

• Hau Lee’s approach to supply chain (SC) is one of aligning SC’s with the uncertainties revolving around the supply process side of the SC

• A stable supply process has mature technologies and an evolving supply process has rapidly changing technologies

• Types of SC’s– Efficient SC’s– Risk-Hedging SC’s– Responsive SC’s– Agile SC’s

334

Hau Lee’s SC Uncertainty Framework

Demand Uncertainty

Low (Functional products)

High (Innovative products)

Efficient SC

Ex.: Grocery

Responsive SC

Ex.: Computers

Risk-Hedging SC

Ex.: Hydro-electric power

Agile SC

Ex.: Telecom

Low(Stable Process)

High(Evolving Process)

Supply

Uncertainty

335

What is Outsourcing?Defined

Outsourcing is defined as the act of moving a firm’s internal activities and decision responsibility to outside providers

336

Reasons to Outsource

• Organizationally-driven

• Improvement-driven

• Financially-driven

• Revenue-driven

• Cost-driven

• Employee-driven

337

Value DensityDefined

• Value density is defined as the value of an item per pound of weight

• It is used as an important measure when deciding where items should be stocked geographically and how they should be shipped

338

Mass CustomizationDefined

• Mass customization is a term used to describe the ability of a company to deliver highly customized products and services to different customers

• The key to mass customization is effectively postponing the tasks of differentiating a product for a specific customer until the latest possible point in the supply-chain network

339

End of Chapter 9

341

Managerial Briefing 9

E-Ops

342

• Electronic Commerce and E-Ops Defined

• The Context of E-Ops

• Business Web Models

• E-Ops Applications

OBJECTIVES

343

Electronic Commerce and E-Ops Defined

• Electronic commerce (EC) is defined as “the use of computer applications communicated over networks to allow buyers and sellers to complete a transaction or part of a transaction”

• E-Ops is a term that refers to the application of the Internet and its attendant technologies to the field of operations management

344

The Context of E-Ops Business Model

“How we make our money?”

Operations

“How do we manage production of the product or service?”

Information System Architecture

“The set of tools used to support processes.”

Exhibit MB9.1Exhibit MB9.1

345

Business Web Models

B-Web

Model

Example

Marketplace

Ebay

Aggregator

E-Trade

Alliance

AOL

Value Chain

Dell Computers

Distributive Network

UPS

346

E-Ops Applications: A Make-to-Order Fulfillment Process

CustomersProduct Company

Factory

Suppliers

Step II: Build Plan

Develop Products

Orders sent

System provides information

Step I: Retailer

Factory updates customer

Step III: Logistics

Order fulfillment flows

Customer/Product info. flows

Exhibit MB9.3Exhibit MB9.3

347

Other E-Ops Applications

• Order Fulfillment in Aggregator Businesses

• Project Management

• Product and Process Design

• Purchasing

• Manufacturing Processes

348

Other E-Ops Applications (Continued)

• Inventory Management

• Services

• Quality Management

• Forecasting

• Operations Scheduling

• Reengineering and Consulting

349

End of Managerial Briefing 9

351

Chapter 10

Strategic Capacity Planning

352

• Strategic Capacity Planning Defined• Capacity Utilization & Best Operating

Level• Economies & Diseconomies of Scale• The Experience Curve• Capacity Focus, Flexibility & Planning• Determining Capacity Requirements• Decision Trees• Capacity Utilization & Service Quality

OBJECTIVES

353

Strategic Capacity PlanningDefined

• Capacity can be defined as the ability to hold, receive, store, or accommodate

• Strategic capacity planning is an approach for determining the overall capacity level of capital intensive resources, including facilities, equipment, and overall labor force size

354

Capacity Utilization

• Where• Capacity used

– rate of output actually achieved

• Best operating level– capacity for which the process was designed

level operating Best

usedCapacity rate nutilizatioCapacity

355

Best Operating Level

Example: Engineers design engines and assembly lines to operate at an ideal or “best operating level” to maximize output and minimize ware

Example: Engineers design engines and assembly lines to operate at an ideal or “best operating level” to maximize output and minimize ware

Underutilization

Best OperatingLevel

Averageunit costof output

Volume

Overutilization

356

Example of Capacity Utilization

• During one week of production, a plant produced 83 units of a product. Its historic highest or best utilization recorded was 120 units per week. What is this plant’s capacity utilization rate?

• During one week of production, a plant produced 83 units of a product. Its historic highest or best utilization recorded was 120 units per week. What is this plant’s capacity utilization rate?

Answer: Capacity utilization rate = Capacity used .

Best operating level = 83/120 =0.69 or 69%

Answer: Capacity utilization rate = Capacity used .

Best operating level = 83/120 =0.69 or 69%

357

Economies & Diseconomies of Scale

100-unitplant

200-unitplant 300-unit

plant

400-unitplant

Volume

Averageunit costof output

Economies of Scale and the Experience Curve workingEconomies of Scale and the Experience Curve working

Diseconomies of Scale start workingDiseconomies of Scale start working

358

The Experience Curve

As plants produce more products, they gain experience in the best production methods and reduce their costs per unit

As plants produce more products, they gain experience in the best production methods and reduce their costs per unit

Total accumulated production of units

Cost orpriceper unit

Yesterday

Today

Tomorrow

359

Capacity Focus

• The concept of the focused factory holds that production facilities work best when they focus on a fairly limited set of production objectives

• Plants Within Plants (PWP) (from Skinner)– Extend focus concept to operating level

360

Capacity Flexibility

• Flexible plants

• Flexible processes

• Flexible workers

361

Capacity Planning: Balance

Stage 1 Stage 2 Stage 3Unitsper

month6,000 7,000 5,000

Unbalanced stages of productionUnbalanced stages of production

Stage 1 Stage 2 Stage 3Unitsper

month6,000 6,000 6,000

Balanced stages of productionBalanced stages of production

Maintaining System Balance: Output of one stage is the exact input requirements for the next stage

362

Capacity Planning

• Frequency of Capacity Additions

• External Sources of Capacity

363

Determining Capacity Requirements

• 1. Forecast sales within each individual product line

• 2. Calculate equipment and labor requirements to meet the forecasts

• 3. Project equipment and labor availability over the planning horizon

364

Example of Capacity RequirementsA manufacturer produces two lines of mustard, FancyFine and Generic line. Each is sold in small and family-size plastic bottles.

The following table shows forecast demand for the next four years.

Year: 1 2 3 4FancyFine

Small (000s) 50 60 80 100Family (000s) 35 50 70 90Generic

Small (000s) 100 110 120 140Family (000s) 80 90 100 110

365

Example of Capacity Requirements (Continued): Product from a Capacity

Viewpoint

• Question: Are we really producing two different types of mustards from the standpoint of capacity requirements?

• Answer: No, it’s the same product just packaged differently.

• Question: Are we really producing two different types of mustards from the standpoint of capacity requirements?

• Answer: No, it’s the same product just packaged differently.

366

Example of Capacity Requirements (Continued) : Equipment and Labor

Requirements

Year: 1 2 3 4Small (000s) 150 170 200 240Family (000s) 115 140 170 200

•Three 100,000 units-per-year machines are available for small-bottle production. Two operators required per machine.

•Two 120,000 units-per-year machines are available for family-sized-bottle production. Three operators required per machine.

367


Small Mach. Cap. 300,000 Labor 6Family-size Mach. Cap. 240,000 Labor 6

Small

Percent capacity used 50.00%Machine requirement 1.50Labor requirement 3.00Family-size

Percent capacity used 47.92%Machine requirement 0.96Labor requirement 2.88

Question: What are the Year 1 values for capacity, machine, and labor?

Question: What are the Year 1 values for capacity, machine, and labor?

150,000/300,000=50%

At 2 operators for 100,000, it takes 3 operators for 150,000

At 1 machine for 100,000, it takes 1.5 machines for 150,000


367

368


Small Mach. Cap. 300,000 Labor 6Family-size Mach. Cap. 240,000 Labor 6

Small

Percent capacity used 50.00%Machine requirement 1.50Labor requirement 3.00Family-size

Percent capacity used 47.92%Machine requirement 0.96Labor requirement 2.88

Question: What are the values for columns 2, 3 and 4 in the table below?Question: What are the values for columns 2, 3 and 4 in the table below?

56.67%1.703.40

58.33%1.173.50

66.67%2.004.00

70.83%1.424.25

80.00%2.404.80

83.33%1.675.00

368


369

Example of a Decision Tree Problem

A glass factory specializing in crystal is experiencing a substantial backlog, and the firm's management is considering three courses of action:

A) Arrange for subcontractingB) Construct new facilitiesC) Do nothing (no change)

The correct choice depends largely upon demand, which may be low, medium, or high. By consensus, management estimates the respective demand probabilities as 0.1, 0.5, and 0.4.

A glass factory specializing in crystal is experiencing a substantial backlog, and the firm's management is considering three courses of action:

A) Arrange for subcontractingB) Construct new facilitiesC) Do nothing (no change)

The correct choice depends largely upon demand, which may be low, medium, or high. By consensus, management estimates the respective demand probabilities as 0.1, 0.5, and 0.4.

370

Example of a Decision Tree Problem (Continued): The Payoff Table

0.1 0.5 0.4Low Medium High

A 10 50 90B -120 25 200C 20 40 60

The management also estimates the profits when choosing from the three alternatives (A, B, and C) under the differing probable levels of demand. These profits, in thousands of dollars are presented in the table below:

The management also estimates the profits when choosing from the three alternatives (A, B, and C) under the differing probable levels of demand. These profits, in thousands of dollars are presented in the table below:

371

Example of a Decision Tree Problem (Continued): Step 1. We start by drawing

the three decisions

A

B

C

372

Example of Decision Tree Problem (Continued): Step 2. Add our possible

states of nature, probabilities, and payoffs

A

B

C

High demand (0.4)

Medium demand (0.5)

Low demand (0.1)

$90k$50k

$10k

High demand (0.4)

Medium demand (0.5)

Low demand (0.1)

$200k$25k

-$120k

High demand (0.4)

Medium demand (0.5)

Low demand (0.1)

$60k$40k

$20k

373

Example of Decision Tree Problem (Continued): Step 3. Determine the

expected value of each decision

High demand (0.4)High demand (0.4)

Medium demand (0.5)Medium demand (0.5)

Low demand (0.1)Low demand (0.1)

AA

$90k$90k

$50k$50k

$10k$10k

EVA=0.4(90)+0.5(50)+0.1(10)=$62kEVA=0.4(90)+0.5(50)+0.1(10)=$62k

$62k$62k

374

Example of Decision Tree Problem (Continued): Step 4. Make decision

High demand (0.4)

Medium demand (0.5)

Low demand (0.1)

High demand (0.4)

Medium demand (0.5)

Low demand (0.1)

A

B

CHigh demand (0.4)

Medium demand (0.5)

Low demand (0.1)

$90k$50k

$10k

$200k$25k

-$120k

$60k$40k

$20k

$62k

$80.5k

$46k

Alternative B generates the greatest expected profit, so our choice is B or to construct a new facility

Alternative B generates the greatest expected profit, so our choice is B or to construct a new facility

375

Planning Service Capacity vs. Manufacturing Capacity

• Time: Goods can not be stored for later use and capacity must be available to provide a service when it is needed

• Location: Service goods must be at the customer demand point and capacity must be located near the customer

• Volatility of Demand: Much greater than in manufacturing

376

Capacity Utilization & Service Quality

• Best operating point is near 70% of capacity

• From 70% to 100% of service capacity, what do you think happens to service quality?

377

End of Chapter 10

379

Technical Note 10

Facility Location

380

• Issues in Facility Location

• Various Plant Location Methods

OBJECTIVES

381

Competitive Imperatives Impacting Location

• The need to produce close to the customer due to time-based competition, trade agreements, and shipping costs

• The need to locate near the appropriate labor pool to take advantage of low wage costs and/or high technical skills

382

Issues in Facility Location

• Proximity to Customers• Business Climate• Total Costs• Infrastructure• Quality of Labor • Suppliers• Other Facilities

383

Issues in Facility Location

• Free Trade Zones • Political Risk• Government Barriers• Trading Blocs• Environmental Regulation• Host Community• Competitive Advantage

384

Plant Location Methodology: Factor Rating Method Example

Fuels in region 0 to 330Power availability and reliability 0 to 200Labor climate 0 to 100Living conditions 0 to 100Transportation 0 to 50Water supply 0 to 10Climate 0 to 50Supplies 0 to 60Tax policies and laws 0 to 20

Two refineries sites (A and B) are assigned the following range of point values and respective points, where the more points the better for the site location.

1231505424454855

Major factors for site location Pt. Range

156100639650545020

SitesA B

Total pts. 418 544

Best Site is B

Best Site is B

385

Plant Location Methodology: Transportation Method of Linear

Programming

• Transportation method of linear programming seeks to minimize costs of shipping n units to m destinations or its seeks to maximize profit of shipping n units to m destinations

386

Plant Location Methodology: Centroid Method

• The centroid method is used for locating single facilities that considers existing facilities, the distances between them, and the volumes of goods to be shipped between them

• This methodology involves formulas used to compute the coordinates of the two-dimensional point that meets the distance and volume criteria stated above

387

Plant Location Methodology: Centroid Method Formulas

C = d V

V x

ix i

i

C = d V

V x

ix i

i

Where:Cx = X coordinate of centroidCy = X coordinate of centroiddix = X coordinate of the ith locationdiy = Y coordinate of the ith locationVi = volume of goods moved to or from ith location

C = d V

Vy

iy i

i

C = d V

Vy

iy i

i

388

Plant Location Methodology: Example of Centroid Method

Question: What is the best location for a new Z-Mobile warehouse/temporary storage facility considering only distances and quantities sold per month?

Question: What is the best location for a new Z-Mobile warehouse/temporary storage facility considering only distances and quantities sold per month?

• Centroid method example– Several automobile showrooms are located according to the

following grid which represents coordinate locations for each showroom

S howroom No o f Z-Mo b ile s s o ld p e r mo nth

A 1250

D 1900

Q 2300X

Y

A(100,200)

D(250,580)

Q(790,900)

(0,0)

389Plant Location Methodology: Example of Centroid Method (Continued): Determining

Existing Facility CoordinatesTo begin, you must identify the existing facilities on a two-dimensional plane or grid and determine their coordinates.

To begin, you must identify the existing facilities on a two-dimensional plane or grid and determine their coordinates.

X

Y

A(100,200)

D(250,580)

Q(790,900)

(0,0)

You must also have the volume information on the business activity at the existing facilities.

You must also have the volume information on the business activity at the existing facilities.

S ho wro o m No o f Z-Mo b ile s s o ld p e r mo nth

A 1250

D 1900

Q 2300

390Plant Location Methodology: Example of Centroid Method (Continued): Determining the

Coordinates of the New Facility

C = 100(1250) + 250(1900) + 790(2300)

1250 + 1900 + 2300 =

2,417,000

5,450 = x 443.49C =

100(1250) + 250(1900) + 790(2300)

1250 + 1900 + 2300 =

2,417,000

5,450 = x 443.49

C = 200(1250) + 580(1900) + 900(2300)

1250 + 1900 + 2300 =

3,422,000

5,450 = y 627.89C =

200(1250) + 580(1900) + 900(2300)

1250 + 1900 + 2300 =

3,422,000

5,450 = y 627.89

S ho wro o m No o f Z-Mo b ile s s o ld p e r mo nth

A 1250

D 1900

Q 2300X

Y

A(100,200)

D(250,580)

Q(790,900)

(0,0)

You then compute the new coordinates using the formulas:You then compute the new coordinates using the formulas:

ZZ

New location of facility Z about (443,627)

New location of facility Z about (443,627)

You then take the coordinates and place them on the map:You then take the coordinates and place them on the map:

391


393

Chapter 11

Just-in-Time and Lean Systems

394

• JIT Defined

• The Toyota Production System

• JIT Implementation Requirements

• JIT in Services

OBJECTIVES

395

Just-In-Time (JIT)Defined

• JIT can be defined as an integrated set of activities designed to achieve high-volume production using minimal inventories (raw materials, work in process, and finished goods)

• JIT also involves the elimination of waste in production effort

• JIT also involves the timing of production resources (i.e., parts arrive at the next workstation “just in time”)

396

JIT and Lean Management• JIT can be divided into two terms: “Big JIT” and

“Little JIT”• Big JIT (also called Lean Management) is a

philosophy of operations management that seeks to eliminate waste in all aspects of a firm’s production activities: human relations, vendor relations, technology, and the management of materials and inventory

• Little JIT focuses more narrowly on scheduling goods inventory and providing service resources where and when needed

397

JIT Demand-Pull Logic

Customers

Sub

Sub

Fab

Fab

Fab

Fab

Vendor

Vendor

Vendor

Vendor

Final Assembly

Here the customer starts the process, pulling an inventory item from Final Assembly…

Here the customer starts the process, pulling an inventory item from Final Assembly…

Then sub-assembly work is pulled forward by that demand…

Then sub-assembly work is pulled forward by that demand…

The process continues throughout the entire production process and supply chain

The process continues throughout the entire production process and supply chain

398

The Toyota Production System

• Based on two philosophies:

• 1. Elimination of waste

• 2. Respect for people

399

Waste in Operations

1. Waste from overproduction

2. Waste of waiting time

3. Transportation waste

4. Inventory waste

5. Processing waste

6. Waste of motion

7. Waste from product defects

400Minimizing Waste:

Focused Factory Networks

CoordinationSystem Integration

These are small specialized plants that limit the range of products produced (sometimes only one type of product for an entire facility)

These are small specialized plants that limit the range of products produced (sometimes only one type of product for an entire facility)

Some plants in Japan have as few as 30 and as many as 1000 employees

Some plants in Japan have as few as 30 and as many as 1000 employees

401

Minimizing Waste: Group Technology (Part 1)

• Using Departmental Specialization for plant layout can cause a lot of unnecessary

material movement • Using Departmental Specialization for plant layout can cause a lot of unnecessary

material movement

Saw Saw

Lathe PressPress

Grinder

LatheLathe

Saw

Press

Heat Treat

Grinder

Note how the flow lines are going back and forthNote how the flow lines are going back and forth

402

Minimizing Waste: Group Technology (Part 2)

• Revising by using Group Technology Cells can reduce movement and improve product flow

• Revising by using Group Technology Cells can reduce movement and improve product flow

Press

Lathe

Grinder

Grinder

A

2

BSaw

Heat Treat

LatheSaw Lathe

PressLathe

1

403

Minimizing Waste: Uniform Plant Loading (heijunka)

Not uniform Jan. Units Feb. Units Mar. Units Total

1,200 3,500 4,300 9,000

Uniform Jan. Units Feb. Units Mar. Units Total

3,000 3,000 3,000 9,000

Suppose we operate a production plant that produces a single product. The schedule of production for this product could be accomplished using either of the two plant loading schedules below.

Suppose we operate a production plant that produces a single product. The schedule of production for this product could be accomplished using either of the two plant loading schedules below.

How does the uniform loading help save labor costs?How does the uniform loading help save labor costs?

or

404

Minimizing Waste: Just-In-Time Production

• Management philosophy• “Pull” system though the plant

WHAT IT IS

• Employee participation• Industrial engineering/basics• Continuing improvement• Total quality control• Small lot sizes

WHAT IT REQUIRES

• Attacks waste• Exposes problems and bottlenecks• Achieves streamlined production

WHAT IT DOES

• Stable environment

WHAT IT ASSUMES

405

Minimizing Waste: Inventory Hides Problems

Work in

process

queues

(banks)

Change

orders

Engineering design

redundancies

Vendor

delinquencies

Scrap

Design

backlogs

Machine

downtime

Decision

backlogsInspection

backlogs

Paperwork

backlog

Example: By identifying defective items from a vendor early in the production process the downstream work is saved

Example: By identifying defective work by employees upstream, the downstream work is saved

406

Minimizing Waste: Kanban Production Control Systems

Storage Part A

Storage Part AMachine

Center Assembly Line

Material Flow

Card (signal) Flow

Withdrawal kanban

Once the Production kanban is received, the Machine Center produces a unit to replace the one taken by the Assembly Line people in the first place

This puts the system back were it was before the item was pulled

The process begins by the Assembly Line people pulling Part A from Storage

Production kanban

407

Determining the Number of Kanbans Needed

• Setting up a kanban system requires determining the number of kanbans cards (or containers) needed

• Each container represents the minimum production lot size

• An accurate estimate of the lead time required to produce a container is key to determining how many kanbans are required

408

The Number of Kanban Card Sets

C

SDL

k

)(1

container theof Size

stockSafety timelead during demand Expected

C

SDL

k

)(1



k = Number of kanban card sets (a set is a card)D = Average number of units demanded over some time periodL = lead time to replenish an order (same units of time as demand)S = Safety stock expressed as a percentage of demand during leadtimeC = Container size

409

Example of Kanban Card Determination: Problem Data

• A switch assembly is assembled in batches of 4 units from an “upstream” assembly area and delivered in a special container to a “downstream” control-panel assembly operation

• The control-panel assembly area requires 5 switch assemblies per hour

• The switch assembly area can produce a container of switch assemblies in 2 hours

• Safety stock has been set at 10% of needed inventory

410

Example of Kanban Card Determination: Calculations

3or ,75.24

5(2)(1.1))(1



C

SDL

k

3or ,75.24

5(2)(1.1))(1



C

SDL

k

Always round up!

411

Respect for People

• Level payrolls

• Cooperative employee unions

• Subcontractor networks

• Bottom-round management style

• Quality circles (Small Group Involvement Activities or SGIA’s)

412

Toyota Production System’s Four Rules1. All work shall be highly specified as to content,

sequence, timing, and outcome

2. Every customer-supplier connection must be direct, and there must be an unambiguous yes-or-no way to send requests and receive responses

3. The pathway for every product and service must be simple and direct

4. Any improvement must be made in accordance with the scientific method, under the guidance of a teacher, at the lowest possible level in the organization

413

JIT Implementation Requirements: Design Flow Process

• Link operations

• Balance workstation capacities

• Redesign layout for flow

• Emphasize preventive maintenance

• Reduce lot sizes

• Reduce setup/changeover time

414

JIT Implementation Requirements: Total Quality Control

• Worker responsibility

• Measure SQC

• Enforce compliance

• Fail-safe methods

• Automatic inspection

415

JIT Implementation Requirements: Stabilize Schedule

• Level schedule

• Underutilize capacity

• Establish freeze windows

416

JIT Implementation Requirements: Kanban-Pull

• Demand pull

• Backflush

• Reduce lot sizes

417

JIT Implementation Requirements: Work with Vendors

• Reduce lead times

• Frequent deliveries

• Project usage requirements

• Quality expectations

418

JIT Implementation Requirements: Reduce Inventory More

• Look for other areas

• Stores

• Transit

• Carousels

• Conveyors

419

JIT Implementation Requirements: Improve Product Design

• Standard product configuration

• Standardize and reduce number of parts

• Process design with product design

• Quality expectations

420

JIT Implementation Requirements: Concurrently Solve Problems

• Root cause • Solve permanently

• Team approach

• Line and specialist responsibility

• Continual education

421

JIT Implementation Requirements: Measure Performance

• Emphasize improvement

• Track trends

422

JIT in Services (Examples)

• Organize Problem-Solving Groups

• Upgrade Housekeeping

• Upgrade Quality

• Clarify Process Flows

• Revise Equipment and Process Technologies

423

JIT in Services (Examples)

• Level the Facility Load

• Eliminate Unnecessary Activities

• Reorganize Physical Configuration

• Introduce Demand-Pull Scheduling

• Develop Supplier Networks

424

End of Chapter 11

426

Chapter 12

Forecasting

427

• Demand Management• Qualitative Forecasting Methods• Simple & Weighted Moving Average

Forecasts• Exponential Smoothing• Simple Linear Regression• Web-Based Forecasting

OBJECTIVES

428

Demand Management

A

B(4) C(2)

D(2) E(1) D(3) F(2)

Dependent Demand:Raw Materials, Component parts,Sub-assemblies, etc.

Independent Demand:Finished Goods

429

Independent Demand: What a firm can do to manage it?

• Can take an active role to influence demand

• Can take a passive role and simply respond to demand

430

Types of Forecasts

• Qualitative (Judgmental)

• Quantitative– Time Series Analysis– Causal Relationships– Simulation

431

Components of Demand

• Average demand for a period of time

• Trend

• Seasonal element

• Cyclical elements

• Random variation

• Autocorrelation

432

Finding Components of Demand

1 2 3 4

x

x xx

xx

x xx

xx x x x

xxxxxx x x

xx

x x xx

xx

xx

x

xx

xx

xx

xx

xx

xx

x

x

Year

Sal

es

Seasonal variationSeasonal variation

Linear

Trend

Linear

Trend

433

Qualitative Methods

Grass Roots

Market Research

Panel Consensus

Executive Judgment

Historical analogy

Delphi Method

Qualitative

Methods

434

Delphi Methodl. Choose the experts to participate representing a

variety of knowledgeable people in different areas2. Through a questionnaire (or E-mail), obtain

forecasts (and any premises or qualifications for the forecasts) from all participants

3. Summarize the results and redistribute them to the participants along with appropriate new questions

4. Summarize again, refining forecasts and conditions, and again develop new questions

5. Repeat Step 4 as necessary and distribute the final results to all participants

435

Time Series Analysis

• Time series forecasting models try to predict the future based on past data

• You can pick models based on:

1. Time horizon to forecast

2. Data availability

3. Accuracy required

4. Size of forecasting budget

5. Availability of qualified personnel

436

Simple Moving Average Formula

F = A + A + A +...+A

ntt-1 t-2 t-3 t-nF =

A + A + A +...+A

ntt-1 t-2 t-3 t-n

• The simple moving average model assumes an average is a good estimator of future behavior

• The formula for the simple moving average is:

Ft = Forecast for the coming period N = Number of periods to be averagedA t-1 = Actual occurrence in the past period for up to “n” periods

437

Simple Moving Average Problem (1)

Week Demand1 6502 6783 7204 7855 8596 9207 8508 7589 892

10 92011 78912 844

F = A + A + A +...+A

ntt-1 t-2 t-3 t-nF =

A + A + A +...+A

ntt-1 t-2 t-3 t-n

Question: What are the 3-week and 6-week moving average forecasts for demand?

Assume you only have 3 weeks and 6 weeks of actual demand data for the respective forecasts

Question: What are the 3-week and 6-week moving average forecasts for demand?


438

Week Demand 3-Week 6-Week1 6502 6783 7204 785 682.675 859 727.676 920 788.007 850 854.67 768.678 758 876.33 802.009 892 842.67 815.33

10 920 833.33 844.0011 789 856.67 866.5012 844 867.00 854.83

F4=(650+678+720)/3

=682.67F7=(650+678+720 +785+859+920)/6

=768.67

Calculating the moving averages gives us:


438

439

500

600

700

800

900

1000

1 2 3 4 5 6 7 8 9 10 11 12

Week

Dem

and

Demand

3-Week

6-Week

Plotting the moving averages and comparing them shows how the lines smooth out to reveal the overall upward trend in this example

Plotting the moving averages and comparing them shows how the lines smooth out to reveal the overall upward trend in this example

Note how the 3-Week is smoother than the Demand, and 6-Week is even smoother

Note how the 3-Week is smoother than the Demand, and 6-Week is even smoother

440

Simple Moving Average Problem (2) Data

Week Demand1 8202 7753 6804 6555 6206 6007 575

Question: What is the 3 week moving average forecast for this data?


Question: What is the 3 week moving average forecast for this data?


441

Simple Moving Average Problem (2) Solution

Week Demand 3-Week 5-Week1 8202 7753 6804 655 758.335 620 703.336 600 651.67 710.007 575 625.00 666.00

F4=(820+775+680)/3

=758.33F6=(820+775+680 +655+620)/5 =710.00

442

Weighted Moving Average Formula

F = w A + w A + w A +...+w At 1 t-1 2 t-2 3 t-3 n t-nF = w A + w A + w A +...+w At 1 t-1 2 t-2 3 t-3 n t-n

w = 1ii=1

n

w = 1ii=1

n

While the moving average formula implies an equal weight being placed on each value that is being averaged, the weighted moving average permits an unequal weighting on prior time periods

While the moving average formula implies an equal weight being placed on each value that is being averaged, the weighted moving average permits an unequal weighting on prior time periods

wt = weight given to time period “t” occurrence (weights must add to one)

wt = weight given to time period “t” occurrence (weights must add to one)

The formula for the moving average is:The formula for the moving average is:

443

Weighted Moving Average Problem (1) Data

Weights: t-1 .5t-2 .3t-3 .2

Week Demand1 6502 6783 7204

Question: Given the weekly demand and weights, what is the forecast for the 4th period or Week 4?

Question: Given the weekly demand and weights, what is the forecast for the 4th period or Week 4?

Note that the weights place more emphasis on the most recent data, that is time period “t-1”

Note that the weights place more emphasis on the most recent data, that is time period “t-1”

444

Weighted Moving Average Problem (1) Solution

Week Demand Forecast1 6502 6783 7204 693.4

F4 = 0.5(720)+0.3(678)+0.2(650)=693.4

445

Weighted Moving Average Problem (2)

Data

Weights: t-1 .7t-2 .2t-3 .1

Week Demand1 8202 7753 6804 655

Question: Given the weekly demand information and weights, what is the weighted moving average forecast of the 5th period or week?

Question: Given the weekly demand information and weights, what is the weighted moving average forecast of the 5th period or week?

446

Weighted Moving Average Problem (2) Solution

Week Demand Forecast1 8202 7753 6804 6555 672

F5 = (0.1)(755)+(0.2)(680)+(0.7)(655)= 672

447

Exponential Smoothing Model

• Premise: The most recent observations might have the highest predictive value

• Therefore, we should give more weight to the more recent time periods when forecasting

Ft = Ft-1 + (At-1 - Ft-1)Ft = Ft-1 + (At-1 - Ft-1)

constant smoothing Alpha

period epast t tim in the occurance ActualA

period past time 1in alueForecast vF

period t timecoming for the lueForcast vaF

:Where

1-t

1-t

t

448

Exponential Smoothing Problem (1) Data

Week Demand1 8202 7753 6804 6555 7506 8027 7988 6899 775

10

Question: Given the weekly demand data, what are the exponential smoothing forecasts for periods 2-10 using =0.10 and =0.60?

Assume F1=D1

Question: Given the weekly demand data, what are the exponential smoothing forecasts for periods 2-10 using =0.10 and =0.60?

Assume F1=D1

449

Week Demand 0.1 0.61 820 820.00 820.002 775 820.00 820.003 680 815.50 820.004 655 801.95 817.305 750 787.26 808.096 802 783.53 795.597 798 785.38 788.358 689 786.64 786.579 775 776.88 786.61

10 776.69 780.77

Answer: The respective alphas columns denote the forecast values. Note that you can only forecast one time period into the future.

Answer: The respective alphas columns denote the forecast values. Note that you can only forecast one time period into the future.

450

Exponential Smoothing Problem (1) Plotting

500

600

700

800

900

1 2 3 4 5 6 7 8 9 10

Week

Dem

and

Demand

0.1

0.6

Note how that the smaller alpha results in a smoother line in this example

Note how that the smaller alpha results in a smoother line in this example

451

Exponential Smoothing Problem (2) Data

Question: What are the exponential smoothing forecasts for periods 2-5 using a =0.5?

Assume F1=D1

Question: What are the exponential smoothing forecasts for periods 2-5 using a =0.5?

Assume F1=D1

Week Demand1 8202 7753 6804 6555

452

Exponential Smoothing Problem (2) Solution

Week Demand 0.51 820 820.002 775 820.003 680 797.504 655 738.755 696.88

F1=820+(0.5)(820-820)=820 F3=820+(0.5)(775-820)=797.75

453

The MAD Statistic to Determine Forecasting Error

MAD = A - F

n

t tt=1

n

MAD =

A - F

n

t tt=1

n

1 MAD 0.8 standard deviation

1 standard deviation 1.25 MAD

• The ideal MAD is zero which would mean there is no forecasting error

• The larger the MAD, the less the accurate the resulting model

454

MAD Problem Data

Month Sales Forecast1 220 n/a2 250 2553 210 2054 300 3205 325 315

Question: What is the MAD value given the forecast values in the table below?

Question: What is the MAD value given the forecast values in the table below?

455

MAD Problem Solution

MAD = A - F

n=

40

4= 10

t tt=1

n

MAD =

A - F

n=

40

4= 10

t tt=1

n

Month Sales Forecast Abs Error1 220 n/a2 250 255 53 210 205 54 300 320 205 325 315 10

40

Note that by itself, the MAD only lets us know the mean error in a set of forecasts

Note that by itself, the MAD only lets us know the mean error in a set of forecasts

456

Tracking Signal Formula

• The Tracking Signal or TS is a measure that indicates whether the forecast average is keeping pace with any genuine upward or downward changes in demand.

• Depending on the number of MAD’s selected, the TS can be used like a quality control chart indicating when the model is generating too much error in its forecasts.

• The TS formula is:

TS =RSFE

MAD=

Running sum of forecast errors

Mean absolute deviationTS =

RSFE

MAD=

Running sum of forecast errors

Mean absolute deviation

457

Simple Linear Regression Model

Yt = a + bx0 1 2 3 4 5 x (Time)

YThe simple linear regression model seeks to fit a line through various data over time

The simple linear regression model seeks to fit a line through various data over time

Is the linear regression modelIs the linear regression model

a

Yt is the regressed forecast value or dependent variable in the model, a is the intercept value of the the regression line, and b is similar to the slope of the regression line. However, since it is calculated with the variability of the data in mind, its formulation is not as straight forward as our usual notion of slope.

458

Simple Linear Regression Formulas for Calculating “a” and “b”

a = y - bx

b =xy - n(y)(x)

x - n(x2 2

)

a = y - bx

b =xy - n(y)(x)

x - n(x2 2

)

459

Simple Linear Regression Problem Data

Week Sales1 1502 1573 1624 1665 177

Question: Given the data below, what is the simple linear regression model that can be used to predict sales in future weeks?

Question: Given the data below, what is the simple linear regression model that can be used to predict sales in future weeks?

460

Week Week*Week Sales Week*Sales1 1 150 1502 4 157 3143 9 162 4864 16 166 6645 25 177 8853 55 162.4 2499

Average Sum Average Sum

b =xy - n(y)(x)

x - n(x=

2499 - 5(162.4)(3)=

a = y - bx = 162.4 - (6.3)(3) =

2 2

) ( )55 5 9

63

106.3

143.5

b =xy - n(y)(x)

x - n(x=

2499 - 5(162.4)(3)=

a = y - bx = 162.4 - (6.3)(3) =

2 2

) ( )55 5 9

63

106.3

143.5

Answer: First, using the linear regression formulas, we can compute “a” and “b”

Answer: First, using the linear regression formulas, we can compute “a” and “b”

460

461

Yt = 143.5 + 6.3x

180

Period

135140145150155

160165170175

1 2 3 4 5

Sal

es

Sales

Forecast

The resulting regression model is:

Now if we plot the regression generated forecasts against the actual sales we obtain the following chart:

461

462

Web-Based Forecasting: CPFR Defined

• Collaborative Planning, Forecasting, and Replenishment (CPFR) a Web-based tool used to coordinate demand forecasting, production and purchase planning, and inventory replenishment between supply chain trading partners.

• Used to integrate the multi-tier or n-Tier supply chain, including manufacturers, distributors and retailers.

• CPFR’s objective is to exchange selected internal information to provide for a reliable, longer term future views of demand in the supply chain.

• CPFR uses a cyclic and iterative approach to derive consensus forecasts.

463

Web-Based Forecasting: Steps in CPFR

• 1. Creation of a front-end partnership agreement

• 2. Joint business planning

• 3. Development of demand forecasts

• 4. Sharing forecasts

• 5. Inventory replenishment

464

End of Chapter 12

466

Managerial Briefing 12

Enterprise Resource Planning Systems

467

• Enterprise Resource Planning Defined

• R/3 System Components

• Reasons for Implementing R/3

OBJECTIVES

468

Enterprise Resource Planning (ERP) Systems Defined

• Enterprise Resource Planning Systems is a computer system that integrates application programs in accounting, sales, manufacturing, and other functions in the firm

• This integration is accomplished through a database shared by all the application programs

469

R/3 System Functional Components

R/3 SystemFunctional

ComponentsSales & Distribution Human Resources

Manufacturing & Logistics

Financial Accounting

470

Financial Accounting

• Financials

• Controlling

• Asset management

471

Human Resources

• Payroll

• Benefits administration

• Applicant data administration

• Personnel development planning

• Workforce planning

• Schedule & shift planning

• Time management

• Travel expense accounting

472

Manufacturing & Logistics

• Materials management

• Plant maintenance

• Quality management

• Production planning & control

• Project management system

473

Sales and Distribution

• Prospect & customer management

• Sales order management

• Configuration management

• Distribution

• Export controls

• Shipping and transportation management

• Billing, invoicing, and rebate processing

474

Reasons for Implementing SAP R/3

• Desire to standardize and improve processes

• To improve the level of systems integration

• To improve information quality

475

End of Managerial Briefing 12

477

Chapter 13

Aggregate Planning

478

• Sales and Operations Planning

• The Aggregate Operations Plan

• Examples: Chase and Level strategies

OBJECTIVES

479

Master scheduling

Material requirements planning

Order schedulingWeekly workforce andcustomer scheduling

Daily workforce and customer scheduling

Process planning

Strategic capacity planning

Sales and operations (aggregate) planning

Longrange

Intermediaterange

Shortrange

ManufacturingServices


Sales plan Aggregate operations plan

Forecasting & demand management

480

Sales and Operations Planning Activities

• Long-range planning– Greater than one year planning horizon– Usually performed in annual increments

• Medium-range planning– Six to eighteen months – Usually with monthly or quarterly increments

• Short-range planning– One day to less than six months– Usually with weekly increments

481

The Aggregate Operations Plan

• Main purpose: Specify the optimal combination of– production rate (units completed per unit of time)– workforce level (number of workers)– inventory on hand (inventory carried from

previous period)• Product group or broad category (Aggregation)• This planning is done over an intermediate-range

planning period of 6 to18 months

482

Balancing Aggregate Demandand Aggregate Production Capacity

0

2000

4000

6000

8000

10000

Jan Feb Mar Apr May Jun

45005500

7000

10000

8000

6000

0

2000

4000

6000

8000

10000

Jan Feb Mar Apr May Jun

4500 4000

90008000

4000

6000

Suppose the figure to the right represents forecast demand in units

Suppose the figure to the right represents forecast demand in units

Now suppose this lower figure represents the aggregate capacity of the company to meet demand

Now suppose this lower figure represents the aggregate capacity of the company to meet demand

What we want to do is balance out the production rate, workforce levels, and inventory to make these figures match up

What we want to do is balance out the production rate, workforce levels, and inventory to make these figures match up

483

Required Inputs to the Production Planning System

Planning for

production

External capacity

Competitors’behavior

Raw material availability

Market demand

Economic conditions

Currentphysical capacity

Current workforce

Inventory levels

Activities required for production

External to firm

Internal to firm

484

Key Strategies for Meeting Demand

• Chase

• Level

• Some combination of the two

485

Aggregate Planning Examples: Unit Demand and Cost Data

Materials $5/unitHolding costs $1/unit per mo.Marginal cost of stockout $1.25/unit per mo.Hiring and training cost $200/workerLayoff costs $250/workerLabor hours required .15 hrs/unitStraight time labor cost $8/hourBeginning inventory 250 unitsProductive hours/worker/day 7.25Paid straight hrs/day 8

Suppose we have the following unit demand and cost information:

Suppose we have the following unit demand and cost information:

Demand/mo Jan Feb Mar Apr May Jun

4500 5500 7000 10000 8000 6000

486

Jan Feb Mar Apr May JunDays/mo 22 19 21 21 22 20Hrs/worker/mo 159.5 137.75 152.25 152.25 159.5 145Units/worker 1063.33 918.33 1015 1015 1063.33 966.67$/worker $1,408 1,216 1,344 1,344 1,408 1,280

Productive hours/worker/day 7.25Paid straight hrs/day 8

Demand/mo Jan Feb Mar Apr MayJun

4500 5500 7000 10000 80006000

Given the demand and cost information below, whatare the aggregate hours/worker/month, units/worker, and dollars/worker?

Given the demand and cost information below, whatare the aggregate hours/worker/month, units/worker, and dollars/worker?

7.25x22

7.25x0.15=48.33 & 84.33x22=1063.33

22x8hrsx$8=$1408

Cut-and-Try Example: Determining Straight Labor Costs and Output

487

Chase Strategy(Hiring & Firing to meet demand)

JanDays/mo 22Hrs/worker/mo 159.5Units/worker 1,063.33$/worker $1,408

JanDemand 4,500Beg. inv. 250Net req. 4,250Req. workers 3.997HiredFired 3Workforce 4Ending inventory 0

Lets assume our current workforce is 7 workers.

Lets assume our current workforce is 7 workers.

First, calculate net requirements for production, or 4500-250=4250 units

Then, calculate number of workers needed to produce the net requirements, or 4250/1063.33=3.997 or 4 workers

Finally, determine the number of workers to hire/fire. In this case we only need 4 workers, we have 7, so 3 can be fired.

488

Jan Feb Mar Apr May JunDays/mo 22 19 21 21 22 20Hrs/worker/mo 159.5 137.75 152.25 152.25 159.5 145Units/worker 1,063 918 1,015 1,015 1,063 967$/worker $1,408 1,216 1,344 1,344 1,408 1,280

Jan Feb Mar Apr May JunDemand 4,500 5,500 7,000 10,000 8,000 6,000Beg. inv. 250Net req. 4,250 5,500 7,000 10,000 8,000 6,000Req. workers 3.997 5.989 6.897 9.852 7.524 6.207Hired 2 1 3Fired 3 2 1Workforce 4 6 7 10 8 7Ending inventory 0 0 0 0 0 0

Below are the complete calculations for the remaining months in the six month planning horizon


489

Jan Feb Mar Apr May JunDemand 4,500 5,500 7,000 10,000 8,000 6,000Beg. inv. 250Net req. 4,250 5,500 7,000 10,000 8,000 6,000Req. workers 3.997 5.989 6.897 9.852 7.524 6.207Hired 2 1 3Fired 3 2 1Workforce 4 6 7 10 8 7Ending inventory 0 0 0 0 0 0

Jan Feb Mar Apr May Jun CostsMaterial $21,250.00 $27,500.00 $35,000.00 $50,000.00 $40,000.00 $30,000.00 203,750.00Labor 5,627.59 7,282.76 9,268.97 13,241.38 10,593.10 7,944.83 53,958.62Hiring cost 400.00 200.00 600.00 1,200.00Firing cost 750.00 500.00 250.00 1,500.00

$260,408.62

Below are the complete calculations for the remaining months in the six month planning horizon with the other costs included

490

Level Workforce Strategy (Surplus and Shortage Allowed)

JanDemand 4,500Beg. inv. 250Net req. 4,250Workers 6Production 6,380Ending inventory 2,130Surplus 2,130Shortage

Lets take the same problem as before but this time use the Level Workforce strategy

Lets take the same problem as before but this time use the Level Workforce strategy

This time we will seek to use a workforce level of 6 workers

This time we will seek to use a workforce level of 6 workers

491

Jan Feb Mar Apr May JunDemand 4,500 5,500 7,000 10,000 8,000 6,000Beg. inv. 250 2,130 2,140 1,230 -2,680 -1,300Net req. 4,250 3,370 4,860 8,770 10,680 7,300Workers 6 6 6 6 6 6Production 6,380 5,510 6,090 6,090 6,380 5,800Ending inventory 2,130 2,140 1,230 -2,680 -1,300 -1,500Surplus 2,130 2,140 1,230Shortage 2,680 1,300 1,500

Note, if we recalculate this sheet with 7 workers we would have a surplus

Note, if we recalculate this sheet with 7 workers we would have a surplus



492

Jan Feb Mar Apr May Jun4,500 5,500 7,000 10,000 8,000 6,000

250 2,130 10 -910 -3,910 -1,6204,250 3,370 4,860 8,770 10,680 7,300

6 6 6 6 6 66,380 5,510 6,090 6,090 6,380 5,8002,130 2,140 1,230 -2,680 -1,300 -1,5002,130 2,140 1,230

2,680 1,300 1,500

Jan Feb Mar Apr May Jun$8,448 $7,296 $8,064 $8,064 $8,448 $7,680 $48,000.0031,900 27,550 30,450 30,450 31,900 29,000 181,250.002,130 2,140 1,230 5,500.00

3,350 1,625 1,875 6,850.00

$241,600.00



Note, total costs under this strategy are less than Chase at $260.408.62

Note, total costs under this strategy are less than Chase at $260.408.62

LaborMaterialStorageStockout

493

End of Chapter 13

495

Chapter 14

Inventory Control

496

• Inventory System Defined• Inventory Costs• Independent vs. Dependent Demand• Single-Period Inventory Model • Multi-Period Inventory Models: Basic Fixed-Order

Quantity Models• Multi-Period Inventory Models: Basic Fixed-Time

Period Model• Miscellaneous Systems and Issues

OBJECTIVES

497

Inventory SystemDefined• Inventory is the stock of any item or resource

used in an organization and can include: raw materials, finished products, component parts, supplies, and work-in-process

• An inventory system is the set of policies and controls that monitor levels of inventory and determines what levels should be maintained, when stock should be replenished, and how large orders should be

498

Purposes of Inventory

1. To maintain independence of operations

2. To meet variation in product demand

3. To allow flexibility in production scheduling

4. To provide a safeguard for variation in raw material delivery time

5. To take advantage of economic purchase-order size

499

Inventory Costs

• Holding (or carrying) costs– Costs for storage, handling, insurance, etc

• Setup (or production change) costs– Costs for arranging specific equipment setups, etc

• Ordering costs– Costs of someone placing an order, etc

• Shortage costs– Costs of canceling an order, etc

500

E(1)

Independent vs. Dependent Demand

Independent Demand (Demand for the final end-product or demand not related to other items)

Dependent Demand

(Derived demand items for

component parts,

subassemblies, raw materials,

etc)

Finishedproduct

Component parts

501

Inventory Systems• Single-Period Inventory Model

– One time purchasing decision (Example: vendor selling t-shirts at a football game)

– Seeks to balance the costs of inventory overstock and under stock

• Multi-Period Inventory Models– Fixed-Order Quantity Models

• Event triggered (Example: running out of stock)

– Fixed-Time Period Models • Time triggered (Example: Monthly sales call by

sales representative)

502

Single-Period Inventory Model

uo

u

CC

CP

uo

u

CC

CP

sold be unit will y that theProbabilit

estimatedunder demand ofunit per Cost C

estimatedover demand ofunit per Cost C

:Where

u

o

P

This model states that we should continue to increase the size of the inventory so long as the probability of selling the last unit added is equal to or greater than the ratio of: Cu/Co+Cu

This model states that we should continue to increase the size of the inventory so long as the probability of selling the last unit added is equal to or greater than the ratio of: Cu/Co+Cu

503

Single Period Model Example• Our college basketball team is playing in a

tournament game this weekend. Based on our past experience we sell on average 2,400 shirts with a standard deviation of 350. We make $10 on every shirt we sell at the game, but lose $5 on every shirt not sold. How many shirts should we make for the game?

Cu = $10 and Co = $5; P ≤ $10 / ($10 + $5) = .667

Z.667 = .432 (use NORMSDIST(.667) or Appendix E) therefore we need 2,400 + .432(350) = 2,551 shirts

504

Multi-Period Models:Fixed-Order Quantity Model Model

Assumptions (Part 1)

• Demand for the product is constant and uniform throughout the period

• Lead time (time from ordering to receipt) is constant

• Price per unit of product is constant

505

Multi-Period Models:Fixed-Order Quantity Model Model

Assumptions (Part 2)

• Inventory holding cost is based on average inventory

• Ordering or setup costs are constant

• All demands for the product will be satisfied (No back orders are allowed)

506

Basic Fixed-Order Quantity Model and Reorder Point Behavior

R = Reorder pointQ = Economic order quantityL = Lead time

L L

Q QQ

R

Time

Numberof unitson hand

1. You receive an order quantity Q.

2. Your start using them up over time. 3. When you reach down to

a level of inventory of R, you place your next Q sized order.

4. The cycle then repeats.

507

Cost Minimization Goal

Ordering Costs

HoldingCosts

Order Quantity (Q)

COST

Annual Cost ofItems (DC)

Total Cost

QOPT

By adding the item, holding, and ordering costs together, we determine the total cost curve, which in turn is used to find the Qopt inventory order point that minimizes total costs

By adding the item, holding, and ordering costs together, we determine the total cost curve, which in turn is used to find the Qopt inventory order point that minimizes total costs

508

Basic Fixed-Order Quantity (EOQ) Model Formula

H 2

Q + S

Q

D + DC = TC H

2

Q + S

Q

D + DC = TC

Total Annual =Cost

AnnualPurchase

Cost

AnnualOrdering

Cost

AnnualHolding

Cost+ +

TC=Total annual costD =DemandC =Cost per unitQ =Order quantityS =Cost of placing an order or setup costR =Reorder pointL =Lead timeH=Annual holding and storage cost per unit of inventory

TC=Total annual costD =DemandC =Cost per unitQ =Order quantityS =Cost of placing an order or setup costR =Reorder pointL =Lead timeH=Annual holding and storage cost per unit of inventory

509

Deriving the EOQ

Using calculus, we take the first derivative of the total cost function with respect to Q, and set the derivative (slope) equal to zero, solving for the optimized (cost minimized) value of Qopt

Using calculus, we take the first derivative of the total cost function with respect to Q, and set the derivative (slope) equal to zero, solving for the optimized (cost minimized) value of Qopt

Q = 2DS

H =

2(Annual D em and)(Order or Setup Cost)

Annual Holding CostOPTQ =

2DS

H =

2(Annual D em and)(Order or Setup Cost)

Annual Holding CostOPT

Reorder point, R = d L_

Reorder point, R = d L_

d = average daily demand (constant)

L = Lead time (constant)

_

We also need a reorder point to tell us when to place an order

We also need a reorder point to tell us when to place an order

510

EOQ Example (1) Problem Data

Annual Demand = 1,000 unitsDays per year considered in average

daily demand = 365Cost to place an order = $10Holding cost per unit per year = $2.50Lead time = 7 daysCost per unit = $15

Given the information below, what are the EOQ and reorder point?

Given the information below, what are the EOQ and reorder point?

511

EOQ Example (1) Solution

Q = 2DS

H =

2(1,000 )(10)

2.50 = 89.443 units or OPT 90 unitsQ =

2DS

H =

2(1,000 )(10)

2.50 = 89.443 units or OPT 90 units

d = 1,000 units / year

365 days / year = 2.74 units / dayd =

1,000 units / year

365 days / year = 2.74 units / day

Reorder point, R = d L = 2.74units / day (7days) = 19.18 or _

20 units Reorder point, R = d L = 2.74units / day (7days) = 19.18 or _

20 units

In summary, you place an optimal order of 90 units. In the course of using the units to meet demand, when you only have 20 units left, place the next order of 90 units.

In summary, you place an optimal order of 90 units. In the course of using the units to meet demand, when you only have 20 units left, place the next order of 90 units.

512

EOQ Example (2) Problem Data

Annual Demand = 10,000 unitsDays per year considered in average daily demand = 365Cost to place an order = $10Holding cost per unit per year = 10% of cost per unitLead time = 10 daysCost per unit = $15

Determine the economic order quantity and the reorder point given the following…

Determine the economic order quantity and the reorder point given the following…

513

EOQ Example (2) Solution

Q =2DS

H=

2(10,000 )(10)

1.50= 365.148 units, or OPT 366 unitsQ =

2DS

H=

2(10,000 )(10)

1.50= 365.148 units, or OPT 366 units

d =10,000 units / year

365 days / year= 27.397 units / dayd =

10,000 units / year

365 days / year= 27.397 units / day

R = d L = 27.397 units / day (10 days) = 273.97 or _

274 unitsR = d L = 27.397 units / day (10 days) = 273.97 or _

274 units

Place an order for 366 units. When in the course of using the inventory you are left with only 274 units, place the next order of 366 units.

Place an order for 366 units. When in the course of using the inventory you are left with only 274 units, place the next order of 366 units.

514Fixed-Time Period Model with Safety Stock Formula

order)on items (includes levelinventory current = I

timelead and review over the demand ofdeviation standard =

yprobabilit service specified afor deviations standard ofnumber the= z

demanddaily averageforecast = d

daysin timelead = L

reviewsbetween days ofnumber the= T

ordered be toquantitiy = q

:Where

I - Z+ L)+(Td = q

L+T

L+T

order)on items (includes levelinventory current = I

timelead and review over the demand ofdeviation standard =

yprobabilit service specified afor deviations standard ofnumber the= z

demanddaily averageforecast = d

daysin timelead = L

reviewsbetween days ofnumber the= T

ordered be toquantitiy = q

:Where

I - Z+ L)+(Td = q

L+T

L+T

q = Average demand + Safety stock – Inventory currently on handq = Average demand + Safety stock – Inventory currently on hand

515

Multi-Period Models: Fixed-Time Period Model:

Determining the Value of T+L

T+L di 1

T+L

d

T+L d2

=

Since each day is independent and is constant,

= (T + L)

i

2

T+L di 1

T+L

d

T+L d2

=

Since each day is independent and is constant,

= (T + L)

i

2

• The standard deviation of a sequence of random events equals the square root of the sum of the variances

516

Example of the Fixed-Time Period Model

Average daily demand for a product is 20 units. The review period is 30 days, and lead time is 10 days. Management has set a policy of satisfying 96 percent of demand from items in stock. At the beginning of the review period there are 200 units in inventory. The daily demand standard deviation is 4 units.

Given the information below, how many units should be ordered?

Given the information below, how many units should be ordered?

517

Example of the Fixed-Time Period Model: Solution (Part 1)

T+ L d2 2 = (T + L) = 30 + 10 4 = 25.298 T+ L d

2 2 = (T + L) = 30 + 10 4 = 25.298

The value for “z” is found by using the Excel NORMSINV function, or as we will do here, using Appendix D. By adding 0.5 to all the values in Appendix D and finding the value in the table that comes closest to the service probability, the “z” value can be read by adding the column heading label to the row label.

The value for “z” is found by using the Excel NORMSINV function, or as we will do here, using Appendix D. By adding 0.5 to all the values in Appendix D and finding the value in the table that comes closest to the service probability, the “z” value can be read by adding the column heading label to the row label.

So, by adding 0.5 to the value from Appendix D of 0.4599, we have a probability of 0.9599, which is given by a z = 1.75

So, by adding 0.5 to the value from Appendix D of 0.4599, we have a probability of 0.9599, which is given by a z = 1.75

518

Example of the Fixed-Time Period Model: Solution (Part 2)

or 644.272, = 200 - 44.272 800 = q

200- 298)(1.75)(25. + 10)+20(30 = q

I - Z+ L)+(Td = q L+T

units 645

or 644.272, = 200 - 44.272 800 = q

200- 298)(1.75)(25. + 10)+20(30 = q

I - Z+ L)+(Td = q L+T

units 645

So, to satisfy 96 percent of the demand, you should place an order of 645 units at this review period

So, to satisfy 96 percent of the demand, you should place an order of 645 units at this review period

519

Price-Break Model Formula

Cost Holding Annual

Cost) Setupor der Demand)(Or 2(Annual =

iC

2DS = QOPT

Based on the same assumptions as the EOQ model, the price-break model has a similar Qopt formula:

i = percentage of unit cost attributed to carrying inventoryC = cost per unit

Since “C” changes for each price-break, the formula above will have to be used with each price-break cost value

520

Price-Break Example Problem Data (Part 1)

A company has a chance to reduce their inventory ordering costs by placing larger quantity orders using the price-break order quantity schedule below. What should their optimal order quantity be if this company purchases this single inventory item with an e-mail ordering cost of $4, a carrying cost rate of 2% of the inventory cost of the item, and an annual demand of 10,000 units?

A company has a chance to reduce their inventory ordering costs by placing larger quantity orders using the price-break order quantity schedule below. What should their optimal order quantity be if this company purchases this single inventory item with an e-mail ordering cost of $4, a carrying cost rate of 2% of the inventory cost of the item, and an annual demand of 10,000 units?

Order Quantity(units) Price/unit($)0 to 2,499 $1.202,500 to 3,999 1.004,000 or more .98

521

Price-Break Example Solution (Part 2)

units 1,826 = 0.02(1.20)

4)2(10,000)( =

iC

2DS = QOPT

Annual Demand (D)= 10,000 unitsCost to place an order (S)= $4

First, plug data into formula for each price-break value of “C”

units 2,000 = 0.02(1.00)

4)2(10,000)( =

iC

2DS = QOPT

units 2,020 = 0.02(0.98)

4)2(10,000)( =

iC

2DS = QOPT

Carrying cost % of total cost (i)= 2%Cost per unit (C) = $1.20, $1.00, $0.98

Interval from 0 to 2499, the Qopt value is feasible

Interval from 2500-3999, the Qopt value is not feasible

Interval from 4000 & more, the Qopt value is not feasible

Next, determine if the computed Qopt values are feasible or not

522

Price-Break Example Solution (Part 3)Since the feasible solution occurred in the first price-break, it means that all the other true Qopt values occur at the beginnings of each price-break interval. Why?

Since the feasible solution occurred in the first price-break, it means that all the other true Qopt values occur at the beginnings of each price-break interval. Why?

0 1826 2500 4000 Order Quantity

Total annual costs

So the candidates for the price-breaks are 1826, 2500, and 4000 units

So the candidates for the price-breaks are 1826, 2500, and 4000 units

Because the total annual cost function is a “u” shaped function

Because the total annual cost function is a “u” shaped function

523

Price-Break Example Solution (Part 4)

iC 2

Q + S

Q

D + DC = TC iC

2

Q + S

Q

D + DC = TC

Next, we plug the true Qopt values into the total cost annual cost function to determine the total cost under each price-break

Next, we plug the true Qopt values into the total cost annual cost function to determine the total cost under each price-break

TC(0-2499)=(10000*1.20)+(10000/1826)*4+(1826/2)(0.02*1.20) = $12,043.82TC(2500-3999)= $10,041TC(4000&more)= $9,949.20

TC(0-2499)=(10000*1.20)+(10000/1826)*4+(1826/2)(0.02*1.20) = $12,043.82TC(2500-3999)= $10,041TC(4000&more)= $9,949.20

Finally, we select the least costly Qopt, which is this problem occurs in the 4000 & more interval. In summary, our optimal order quantity is 4000 units

Finally, we select the least costly Qopt, which is this problem occurs in the 4000 & more interval. In summary, our optimal order quantity is 4000 units

524

Miscellaneous Systems:Optional Replenishment System

Maximum Inventory Level, M

MActual Inventory Level, I

q = M - I

I

Q = minimum acceptable order quantity

If q > Q, order q, otherwise do not order any.

525

Miscellaneous Systems:Bin Systems

Two-Bin System

Full Empty

Order One Bin ofInventory

One-Bin System

Periodic Check

Order Enough toRefill Bin

526

ABC Classification System

• Items kept in inventory are not of equal importance in terms of:

– dollars invested

– profit potential

– sales or usage volume

– stock-out penalties

0

30

60

30

60

AB

C

% of $ Value

% of Use

So, identify inventory items based on percentage of total dollar value, where “A” items are roughly top 15 %, “B” items as next 35 %, and the lower 65% are the “C” items

527

Inventory Accuracy and Cycle CountingDefined

• Inventory accuracy refers to how well the inventory records agree with physical count

• Cycle Counting is a physical inventory-taking technique in which inventory is counted on a frequent basis rather than once or twice a year

528

End of Chapter 14

530

Chapter 15

Materials Requirements Planning

531

• Material Requirements Planning (MRP)

• MRP Logic and Product Structure Trees

• Time Fences

• MRP Example

• MRP II and Lot Sizing

OBJECTIVES

532

Material Requirements PlanningDefined• Materials requirements planning (MRP) is a

means for determining the number of parts, components, and materials needed to produce a product

• MRP provides time scheduling information specifying when each of the materials, parts, and components should be ordered or produced

• Dependent demand drives MRP• MRP is a software system

533

Example of MRP Logic and Product Structure Tree

B(4)

E(1)D(2)

C(2)

F(2)D(3)

A

Product Structure Tree for Assembly A Lead TimesA 1 dayB 2 daysC 1 dayD 3 daysE 4 daysF 1 day

Total Unit DemandDay 10 50 ADay 8 20 B (Spares)Day 6 15 D (Spares)

Given the product structure tree for “A” and the lead time and demand information below, provide a materials requirements plan that defines the number of units of each component and when they will be needed

Given the product structure tree for “A” and the lead time and demand information below, provide a materials requirements plan that defines the number of units of each component and when they will be needed

534

LT = 1 day

Day: 1 2 3 4 5 6 7 8 9 10A Required 50

Order Placement 50

First, the number of units of “A” are scheduled backwards to allow for their lead time. So, in the materials requirement plan below, we have to place an order for 50 units of “A” on the 9th day to receive them on day 10.

First, the number of units of “A” are scheduled backwards to allow for their lead time. So, in the materials requirement plan below, we have to place an order for 50 units of “A” on the 9th day to receive them on day 10.

535

Next, we need to start scheduling the components that make up “A”. In the case of component “B” we need 4 B’s for each A. Since we need 50 A’s, that means 200 B’s. And again, we back the schedule up for the necessary 2 days of lead time.

Next, we need to start scheduling the components that make up “A”. In the case of component “B” we need 4 B’s for each A. Since we need 50 A’s, that means 200 B’s. And again, we back the schedule up for the necessary 2 days of lead time.

Day: 1 2 3 4 5 6 7 8 9 10A Required 50

Order Placement 50B Required 20 200

Order Placement 20 200

B(4)

E(1)D(2)

C(2)

F(2)D(3)

A

SparesLT = 2

4x50=200

536

Day: 1 2 3 4 5 6 7 8 9 10A Required 50

LT=1 Order Placement 50B Required 20 200

LT=2 Order Placement 20 200C Required 100

LT=1 Order Placement 100D Required 55 400 300

LT=3 Order Placement 55 400 300E Required 20 200

LT=4 Order Placement 20 200F Required 200

LT=1 Order Placement 200

B(4)

E(1)D(2)

C(2)

F(2)D(3)

A

40 + 15 spares

Part D: Day 6

Finally, repeating the process for all components, we have the final materials requirements plan:

Finally, repeating the process for all components, we have the final materials requirements plan:


536

537

Master Production Schedule (MPS)

• Time-phased plan specifying how many and when the firm plans to build each end item

Aggregate Plan(Product Groups)

Aggregate Plan(Product Groups)

MPS(Specific End Items)

538

Types of Time Fences

• Frozen– No schedule changes allowed within this window

• Moderately Firm– Specific changes allowed within product groups as

long as parts are available• Flexible

– Significant variation allowed as long as overall capacity requirements remain at the same levels

539

Example of Time Fences

8 15 26

Weeks

FrozenModerately

Firm Flexible

Firm Customer Orders

Forecast and availablecapacity

Capacity


540

Material Requirements Planning System

• Based on a master production schedule, a material requirements planning system:– Creates schedules identifying the specific parts

and materials required to produce end items

– Determines exact unit numbers needed

– Determines the dates when orders for those materials should be released, based on lead times

541

From Exhibit 15.6From Exhibit 15.6

541


Firm orders from knowncustomers

Forecastsof demand

from randomcustomers

Aggregateproduct

plan

Bill ofmaterial

file

Engineeringdesign

changes

Inventoryrecord file

Inventorytransactions

Master productionSchedule (MPS)

Primary reportsSecondary reports

Planned order schedule for inventory and production control

Exception reportsPlanning reportsReports for performance control

Materialplanning(MRP

computer program)

542

Bill of Materials (BOM) FileA Complete Product Description

• Materials

• Parts

• Components

• Production sequence

• Modular BOM – Subassemblies

• Super BOM– Fractional options

543

Inventory Records File

• Each inventory item carried as a separate file– Status according to “time buckets”

• Pegging– Identify each parent item that created demand

544

Primary MRP Reports• Planned orders to be released at a future time• Order release notices to execute the planned

orders• Changes in due dates of open orders due to

rescheduling • Cancellations or suspensions of open orders due

to cancellation or suspension of orders on the master production schedule

• Inventory status data

545

Secondary MRP Reports

• Planning reports, for example, forecasting inventory requirements over a period of time

• Performance reports used to determine agreement between actual and programmed usage and costs

• Exception reports used to point out serious discrepancies, such as late or overdue orders

546

Additional MRP Scheduling Terminology

• Gross Requirements

• Scheduled receipts

• Projected available balance

• Net requirements

• Planned order receipt

• Planned order release

547

MRP Example

A(2) B(1)

D(5)C(2)

X

C(3)

Item On-Hand Lead Time (Weeks)X 50 2A 75 3B 25 1C 10 2D 20 2

Requirements include 95 units (80 firm orders and 15 forecast) of X in week 10

Requirements include 95 units (80 firm orders and 15 forecast) of X in week 10

548

A(2)

X

Day: 1 2 3 4 5 6 7 8 9 10X Gross requirements 95

LT=2 Scheduled receipts Proj. avail. balance 50 50 50 50 50 50 50 50 50 50

On- Net requirements 45hand Planned order receipt 4550 Planner order release 45A Gross requirements 90

LT=3 Scheduled receipts Proj. avail. balance 75 75 75 75 75 75 75 75

On- Net requirements 15 hand Planned order receipt 15 75 Planner order release 15 B Gross requirements 45


On- Net requirements 20 hand Planned order receipt 20 25 Planner order release 20 C Gross requirements 45 40

LT=2 Scheduled receipts Proj. avail. balance 10 10 10 10 10

On- Net requirements 35 40 hand Planned order receipt 35 40 10 Planner order release 35 40 D Gross requirements 100

LT=2 Scheduled receipts Proj. avail. balance 20 20 20 20 20 20 20

On- Net requirements 80 hand Planned order receipt 80 20 Planner order release 80

It takes 2 A’s for each X

It takes 2 A’s for each X

549












B(1)A(2)

X

It takes 1 B for each X

It takes 1 B for each X

550

A(2) B(1)

X

C(3)












It takes 3 C’s for each A

It takes 3 C’s for each A

551

A(2) B(1)

C(2)

X

C(3)












It takes 2 C’s for each B

It takes 2 C’s for each B

552

A(2) B(1)

D(5)C(2)

X

C(3)












It takes 5 D’s for each B

It takes 5 D’s for each B

553

Closed Loop MRP

Production PlanningMaster Production SchedulingMaterial Requirements PlanningCapacity Requirements Planning

Realistic?No

Feedback

Execute:Capacity PlansMaterial Plans

Yes

Feedback

554

Manufacturing Resource Planning (MRP II)

• Goal: Plan and monitor all resources of a manufacturing firm (closed loop):– manufacturing– marketing– finance– engineering

• Simulate the manufacturing system

555

Lot Sizing in MRP Programs

• Lot-for-lot (L4L)

• Economic order quantity (EOQ)

• Least total cost (LTC)

• Least unit cost (LUC)

• Which one to use? – The one that is least costly!

556

End of Chapter 15

558

Chapter 16

Operations Scheduling

559

• Work Center Defined• Typical Scheduling and Control Functions• Job-shop Scheduling • Examples of Scheduling Rules• Shop-floor Control• Principles of Work Center Scheduling • Issues in Scheduling Service Personnel

OBJECTIVES

560

Work CenterDefined• A work center is an area in a business in

which productive resources are organized and work is completed

• Can be a single machine, a group of machines, or an area where a particular type of work is done

561

Capacity and Scheduling

• Infinite loading (Example: MRP)

• Finite loading

• Forward scheduling

• Backward scheduling (Example: MRP)

562

Types of Manufacturing Scheduling Processes and Scheduling Approaches

Continuous process

Type of Process Typical Scheduling Approach

High-volume manufacturing

Med-volume manufacturing

Low-volume manufacturing

Finite forward of process, machine limited

Finite forward of line, machined limited

Infinite forward of process, labor and machined limited

Infinite forward of jobs, labor and some machine limited

563

Typical Scheduling and Control Functions

• Allocating orders, equipment, and personnel

• Determining the sequence of order performance

• Initiating performance of the scheduled work

• Shop-floor control

564

Work-Center Scheduling Objectives

• Meet due dates

• Minimize lead time

• Minimize setup time or cost

• Minimize work-in-process inventory

• Maximize machine utilization

565

Priority Rules for Job Sequencing

1. First-come, first-served (FCFS)

2. Shortest operating time (SOT)

3. Earliest due date first (DDate)

4. Slack time remaining (STR) first

5. Slack time remaining per operation (STR/OP)

566

Priority Rules for Job Sequencing (Continued)

6. Critical ratio (CR)

7. Last come, first served (LCFS)

8. Random order or whim

remaining days of Number

date) Current-date (DueCR

567

Example of Job Sequencing: First-Come First-Served

Jobs (in order Processing Due Date Flow Timeof arrival) Time (days) (days hence) (days)

A 4 5 4B 7 10 11C 3 6 14D 1 4 15

Answer: FCFS Schedule

Jobs (in order Processing Due Dateof arrival) Time (days) (days hence)

A 4 5B 7 10C 3 6D 1 4

Suppose you have the four jobs to the right arrive for processing on one machine


What is the FCFS schedule?What is the FCFS schedule?

No, Jobs B, C, and D are going to be late

No, Jobs B, C, and D are going to be late

Do all the jobs get done on time?Do all the jobs get done on time?

568

Example of Job Sequencing: Shortest Operating Time


A 4 5B 7 10C 3 6D 1 4

Answer: Shortest Operating Time Schedule


D 1 4 1C 3 6 4A 4 5 8B 7 10 15



What is the SOT schedule?What is the SOT schedule?

No, Jobs A and B are going to be late

No, Jobs A and B are going to be late


569

Example of Job Sequencing: Earliest Due Date First


A 4 5B 7 10C 3 6D 1 4

Answer: Earliest Due Date First


D 1 4 1A 4 5 5C 3 6 8B 7 10 15



What is the earliest due date first schedule?

What is the earliest due date first schedule?

No, Jobs C and B are going to be late

No, Jobs C and B are going to be late


570

Example of Job Sequencing: Critical Ratio Method


A 4 5B 7 10C 3 6D 1 4



What is the CR schedule?What is the CR schedule?

No, but since there is three-way tie, only the first job or two will be on time

No, but since there is three-way tie, only the first job or two will be on time

In order to do this schedule the CR’s have be calculated for each job. If we let today be Day 1 and allow a total of 15 days to do the work. The resulting CR’s and order schedule are:CR(A)=(5-4)/15=0.06 (Do this job last)CR(B)=(10-7)/15=0.20 (Do this job first, tied with C and D)CR(C)=(6-3)/15=0.20 (Do this job first, tied with B and D)CR(D)=(4-1)/15=0.20 (Do this job first, tied with B and C)


571

Example of Job Sequencing:Last-Come First-Served


A 4 5B 7 10C 3 6D 1 4

Answer: Last-Come First-Served ScheduleJobs (in order Processing Due Date Flow Time

of arrival) Time (days) (days hence) (days)D 1 4 1C 3 6 4B 7 10 11A 4 5 15

No, Jobs B and A are going to be late

No, Jobs B and A are going to be late



What is the LCFS schedule?What is the LCFS schedule?Do all the jobs get done on time?Do all the jobs get done on time?

572

Example of Job Sequencing: Johnson’s Rule (Part 1)

Suppose you have the following five jobs with time requirements in two stages of production. What is the job sequence using Johnson’s Rule?

Suppose you have the following five jobs with time requirements in two stages of production. What is the job sequence using Johnson’s Rule?

Time in HoursJobs Stage 1 Stage 2 A 1.50 1.25 B 2.00 3.00 C 2.50 2.00 D 1.00 2.00

573

Example of Job Sequencing: Johnson’s Rule (Part 2)

First, select the job with the smallest time in either stage.

That is Job D with the smallest time in the first stage. Place that job as early as possible in the unfilled job sequence below.

Drop D out, select the next smallest time (Job A), and place it 4th in the job sequence.

Drop A out, select the next smallest time. There is a tie in two stages for two different jobs. In this case, place the job with the smallest time in the first stage as early as possible in the unfilled job sequence.

Then place the job with the smallest time in the second stage as late as possible in the unfilled sequence.

Job Sequence 1 2 3 4

Job Assigned D A B C

Time in HoursJobs Stage 1 Stage 2 A 1.50 1.25 B 2.00 3.00 C 2.50 2.00 D 1.00 2.00

574

Shop-Floor Control:Major Functions

1. Assigning priority of each shop order

2. Maintaining work-in-process quantity information

3. Conveying shop-order status information to the office

575

Shop-Floor Control:Major Functions (Continued)

4. Providing actual output data for capacity control purposes

5. Providing quantity by location by shop order for WIP inventory and accounting purposes

6. Providing measurement of efficiency, utilization, and productivity of manpower and machines

576

Input/Output Control

Input Output

• Planned input should never exceed planned output

• Focuses attention on bottleneck work centers

WorkCenter

577

Principles of Work Center Scheduling

1. There is a direct equivalence between work flow and cash flow

2. The effectiveness of any job shop should be measured by speed of flow through the shop

3. Schedule jobs as a string, with process steps back-to-back

4. A job once started should not be interrupted

578

Principles of Job Shop Scheduling (Continued)

5. Speed of flow is most efficiently achieved by focusing on bottleneck work centers and jobs

6. Reschedule every day

7. Obtain feedback each day on jobs that are not completed at each work center

8. Match work center input information to what the worker can actually do

579

Principles of Job Shop Scheduling (Continued)

9. When seeking improvement in output, look for incompatibility between engineering design and process execution

10. Certainty of standards, routings, and so forth is not possible in a job shop, but always work towards achieving it

580

Personnel Scheduling in Services

• Scheduling consecutive days off

• Scheduling daily work times

• Scheduling hourly work times

581

End of Chapter 16

583

Technical Note 16

Simulation

584

• Definition of Simulation

• Simulation Methodology

• Proposing a New Experiment

• Considerations When Using Computer Models

• Types of Simulations

• Desirable Features of Simulation Software

• Advantages & Disadvantages of Simulation

OBJECTIVES

585

SimulationDefined

• A simulation is a computer-based model used to run experiments on a real system– Typically done on a computer– Determines reactions to different operating rules

or change in structure– Can be used in conjunction with traditional

statistical and management science techniques

586

Major Phases in a Simulation Study

Start

Define Problem

Construct Simulation Model

Specify values of variables and parameters

Run the simulation

Evaluate results

Validation

Propose new experiment

Stop From Exhibit TN16.1From Exhibit TN16.1

Lets look at each of these steps in turn…

Lets look at each of these steps in turn…

587

Simulation Methodology:Problem Definition

• Specifying the objectives

• Identifying the relevant controllable and uncontrollable variables of the system to be studied

588

Constructing a Simulation Model

• Specification of Variables and Parameters

• Specification of Decision Rules

• Specification of Probability Distributions

• Specification of Time-Incrementing Procedure

589

Data Collection & Random No. Interval Example Suppose you timed 20 athletes running the

100-yard dash and tallied the information into the four time intervals below

Seconds 0-5.996-6.997-7.998 or more

Tallies Frequency41042

You then count the tallies and make a frequency distribution

%20502010

Then convert the frequencies into percentages

Finally, use the percentages to develop the random number intervalsFinally, use the percentages to develop the random number intervals

RN Intervals00-1920-6970-8990-99

RN Intervals00-1920-6970-8990-99

Accum. %207090100

You then can add the frequencies into a cumulative distribution

590

Specify Values of Variables and Parameters

• Determination of starting conditions

• Determination of run length

591

Run the Simulation

• By computer

• Manually

592

Evaluate Results

• Conclusions depend on– the degree to which the model reflects the real

system– design of the simulation (in a statistical sense)

• The only true test of a simulation is how well the real system performs after the results of the study have been implemented

593

Validation

• Refers to testing the computer program to ensure that the simulation is correct

• To insure that the model results are representative of the real world system they seek to model

594

Proposing a New Experiment• Consider changing many of the factors:

– parameters– variables– decision rules– starting conditions– run length

• If the initial rules led to poor results or if these runs yielded new insights into the problem, then a new decision rule may be worth trying

595

Considerations When Using Computer Models

• Computer language selection

• Flowcharting

• Coding

• Data generation

• Output reports

• Validation

596

Types of Simulation Models

• Continuous– Based on mathematical equations– Used for simulating continuous values for all

points in time– Example: The amount of time a person spends in

a queue• Discrete

– Used for simulating specific values or specific points

– Example: Number of people in a queue

597

Desirable Features of Simulation Software• Be capable of being used interactively as well as allowing

complete runs

• Be user-friendly and easy to understand

• Allow modules to be built and then connected

• Allow users to write and incorporate their own routines

• Have building blocks that contain built-in commands

• Have macro capability, such as the ability to develop machining cells

598

Desirable Features of Simulation Software

• Have material-flow capability

• Output standard statistics such as cycle times, utilization, and wait times

• Allow a variety of data analysis alternatives for both input and output data

• Have animation capabilities to display graphically the product flow through the system

• Permit interactive debugging

599

Advantages of Simulation

• Often leads to a better understanding of the real system

• Years of experience in the real system can be compressed into seconds or minutes

• Simulation does not disrupt ongoing activities of the real system

• Simulation is far more general than mathematical models

• Simulation can be used as a game for training experience

600

Advantages of Simulation (Continued)

• Simulation provides a more realistic replication of a system than mathematical analysis

• Simulation can be used to analyze transient conditions, whereas mathematical techniques usually cannot

• Many standard packaged models, covering a wide range of topics, are available commercially

• Simulation answers what-if questions

601

Disadvantages of Simulation• There is no guarantee that the model will, in fact,

provide good answers• There is no way to prove reliability• Building a simulation model can take a great deal of

time• Simulation may be less accurate than mathematical

analysis because it is randomly based• A significant amount of computer time may be

needed to run complex models• The technique of simulation still lacks a standardized

approach

602


604

Chapter 17

Synchronous Manufacturing and the Theory of Constraints

605

• Goldratt’s Rules

• Goldratt’s Goal of the Firm

• Performance Measurement

• Capacity and Flow issues

• Synchronous Manufacturing

OBJECTIVES

606

Goldratt’s Rules of Production Scheduling

• Do not balance capacity balance the flow

• The level utilization of a nonbottleneck resource is not determined by its own potential but by some other constraint in the system

• Utilization and activation of a resource are not the same

• An hour lost at a bottleneck is an hour lost for the entire system

• An hour saved at a nonbottleneck is a mirage

607

Goldratt’s Rules of Production Scheduling (Continued)

• Bottlenecks govern both throughput and inventory in the system

• Transfer batch may not and many times should not be equal to the process batch

• A process batch should be variable both along its route and in time

• Priorities can be set only by examining the system’s constraints and lead time is a derivative of the schedule

608

Goldratt’s Theory of Constraints (TOC)

• Identify the system constraints

• Decide how to exploit the system constraints

• Subordinate everything else to that decision

• Elevate the system constraints

• If, in the previous steps, the constraints have been broken, go back to Step 1, but do not let inertia become the system constraint

609

Goldratt’s Goal of the Firm

The goal of a firm is to make money

610

Performance Measurement:Financial

• Net profit– an absolute measurement in dollars

• Return on investment– a relative measure based on investment

• Cash flow– a survival measurement

611

Performance Measurement:Operational

• 1. Throughput– the rate at which money is generated by the

system through sales

• 2. Inventory– all the money that the system has invested in

purchasing things it intends to sell

• 3. Operating expenses– all the money that the system spends to turn

inventory into throughput

612

Productivity

• Does not guarantee profitability– Has throughput increased?– Has inventory decreased?– Have operational expenses decreased?

613

Unbalanced Capacity

• In earlier chapters, we discussed balancing assembly lines– The goal was a constant cycle time across all

stations

• Synchronous manufacturing views constant workstation capacity as a bad decision

614

The Statistics of Dependent Events

• Rather than balancing capacities, the flow of product through the system should be balanced

Process Time (B)Process Time (A)

106 8 10 12 14

Process Time (B) Process Time (A)

10 6 8 10 12 14

(Constant)

(Constant)(Variable)

(Variable)

When one process takes longer than the average, the time can not be made up

615

Capacity Related Terminology• Capacity is the available time for production

• Bottleneck is what happens if capacity is less than demand placed on resource

• Nonbottleneck is what happens when capacity is greater than demand placed on resource

• Capacity-constrained resource (CCR) is a resource where the capacity is close to demand placed on the resource

616

Capacity Example Situation 1

X Y Market

Case A

X YBottleneck Nonbottleneck

Demand/month 200 units 200 unitsProcess time/unit 1 hour 45 minsAvail. time/month 200 hours 200 hours

There is some idle production in this set up. How much?There is some idle production in this set up. How much?

25% in Y25% in Y

617


Y X Market

Case B



Is there is going to be a build up of unnecessary production in Y?

Is there is going to be a build up of unnecessary production in Y?

Yes, 25% in YYes, 25% in Y

618


X Y

Assembly

Market

Case C



Is there going to be a build up in unnecessary production in Y?

Is there going to be a build up in unnecessary production in Y?


619


X Y

Market Market

Case D



If we run both X and Y for the same time, will we produce any unneeded production?

If we run both X and Y for the same time, will we produce any unneeded production?


620

Time Components of Production Cycle

• Setup time is the time that a part spends waiting for a resource to be set up to work on this same part

• Process time is the time that the part is being processed

• Queue time is the time that a part waits for a resource while the resource is busy with something else

621

Time Components of Production Cycle (Continued)

• Wait time is the time that a part waits not for a resource but for another part so that they can be assembled together

• Idle time is the unused time that represents the cycle time less the sum of the setup time, processing time, queue time, and wait time

622

Saving Time

Bottleneck Nonbottleneck

What are the consequences of saving time at each process?

What are the consequences of saving time at each process?

Rule: Bottlenecks govern both throughput and inventory in the system. Rule: An hour lost at a bottleneck is an hour lost for the entire system. Rule: An hour saved at a nonbottleneck is a mirage.

Rule: Bottlenecks govern both throughput and inventory in the system. Rule: An hour lost at a bottleneck is an hour lost for the entire system. Rule: An hour saved at a nonbottleneck is a mirage.

623

Drum, Buffer, Rope

A B C D E F

Bottleneck (Drum)

Inventorybuffer

(time buffer)Communication

(rope)

Market


624

Quality Implications

• More tolerant than JIT systems– Excess capacity throughout system

• Except for the bottleneck– Quality control needed before bottleneck

625

Batch Sizes

• What is the batch size?

• One?

• Infinity?

626

Bottlenecks and CCRs:Flow-Control Situations

• A bottleneck – (1) with no setup required when changing from

one product to another– (2) with setup times required to change from one

product to another

• A capacity constrained resource (CCR)– (3) with no setup required to change from one

product to another– (4) with setup time required when changing from

one product to another

627

Inventory Cost Measurement:Dollar Days

• Dollar Days is a measurement of the value of inventory and the time it stays within an area

department a withindays of Number

inventory of Value Days Dollar

628

Benefits from Dollar Day Measurement

• Marketing– Discourages holding large amounts of finished

goods inventory

• Purchasing– Discourages placing large purchase orders that on

the surface appear to take advantage of quantity discounts

• Manufacturing– Discourage large work in process and producing

earlier than needed

629

Comparing Synchronous Manufacturing to MRP

• MRP uses backward scheduling

• Synchronous manufacturing uses forward scheduling

630

Comparing Synchronous Manufacturing to JIT

• JIT is limited to repetitive manufacturing

• JIT requires a stable production level

• JIT does not allow very much flexibility in the products produced

631

Comparing Synchronous Manufacturing to JIT (Continued)

• JIT still requires work in process when used with kanban so that there is “something to pull”

• Vendors need to be located nearby because the system depends on smaller, more frequent deliveries

632

Relationship with Other Functional Areas

• Accounting’s influence

• Marketing and production

633

End of Chapter 17

635

Supplement A

Linear Programming Using Excel Solver

636

• Linear Programming Basics

• A Maximization Problem

• A Minimization Problem

OBJECTIVES

637

Linear Programming Essential Conditions

• Is used in problems where we have limited resources or constrained resources

• The model must have an explicit objective (function)

– Generally maximizing profit or minimizing costs subject to resource-based, or other, constraints

638

Linear Programming Essential Conditions (Continued)

• Linearity is a requirement of the model in both objective function and constraints

• Homogeneity of products produced (i.e., products must the identical) and all hours of labor used are assumed equally productive

• Divisibility assumes products and resources divisible (i.e., permit fractional values if need be)

639

Objective Function

Maximize (or Minimize) Z = C1X1 + C2X2 + ... + CnXnMaximize (or Minimize) Z = C1X1 + C2X2 + ... + CnXn

• Cj is a constant that describes the rate of contribution to costs or profit of (Xj) units being produced

• Z is the total cost or profit from the given number of units being produced

640

Constraints

A11X1 + A12X2 + ... + A1nXnB1

A21X1 + A22X2 + ... + A2nXn B2

:

:

AM1X1 + AM2X2 + ... + AMnXn=BM

A11X1 + A12X2 + ... + A1nXnB1

A21X1 + A22X2 + ... + A2nXn B2

:

:

AM1X1 + AM2X2 + ... + AMnXn=BM• Aij are resource requirements for each of the related (Xj) decision variables

• Bi are the available resource requirements• Note that the direction of the inequalities can

be all or a combination of , , or = linear mathematical expressions

641

Non-Negativity Requirement

X1,X2, …, Xn 0X1,X2, …, Xn 0

• All linear programming model formulations require their decision variables to be non-negative

• While these non-negativity requirements take the form of a constraint, they are considered a mathematical requirement to complete the formulation of an LP model

642

An Example of a Maximization Problem LawnGrow Manufacturing Company must determine the unit mix of its commercial riding mower products to be produced next year. The company produces two product lines, the Max and the Multimax. The average profit is $400 for each Max and $800 for each Multimax. Fabrication hours and assembly hours are limited resources. There is a maximum of 5,000 hours of fabrication capacity available per month (each Max requires 3 hours and each Multimax requires 5 hours). There is a maximum of 3,000 hours of assembly capacity available per month (each Max requires 1 hour and each Multimax requires 4 hours). Question: How many units of each riding mower should be produced each month in order to maximize profit?

Now let’s formula this problem as an LP model…Now let’s formula this problem as an LP model…

643

The Objective Function

Maximize Z=400X + 800 X

Where

Z =the monthly profit from Max and Multimax

X =the number of Max produced each month

X =the number of Multimax produced each month

1 2

1

2

Maximize Z=400X + 800 X

Where

Z =the monthly profit from Max and Multimax

X =the number of Max produced each month

X =the number of Multimax produced each month

1 2

1

2

If we define the Max and Multimax products as the two decision variables X1 and X2, and since we want to maximize profit, we can state the objective function as follows:

644

Constraints

Max (X1) Multimax (X2)Required Time/Unit Required Time/Unit Available Time/Month

3 5 5,000 Fab1 4 3,000 Assy

)negativity-(Non 0 X,X

(Assy.) 3,000 4X + X

(Fab.) 5,000 5X + 3X

21

21

21

Given the resource information below from the problem:Given the resource information below from the problem:

We can now state the constraints and non-negativity requirements as:


Note that the inequalities are less-than-or-equal since the time resources represent the total available resources for production

Note that the inequalities are less-than-or-equal since the time resources represent the total available resources for production

645

Solution

Produce 715 Max and 571 Multimax per monthfor a profit of $742,800

Produce 715 Max and 571 Multimax per monthfor a profit of $742,800

646

An Example of a Minimization Problem

HiTech Metal Company is developing a plan for buying scrap metal for its operations. HiTech receives scrap metal from two sources, Hasbeen Industries and Gentro Scrap in daily shipments using large trucks. Each truckload of scrap from Hasbeen yields 1.5 tons of zinc and 1 ton of lead at a cost of $15,000. Each truckload of scrap from Gentro yields 1 ton of zinc and 3 tons of lead at a cost of $18,000. HiTech requires at least 6 tons of zinc and at least 10 tons of lead per day. Question: How many truckloads of scrap should be purchased per day from each source in order to minimize scrap metal costs to HiTech?

Now let’s formula this problem as an LP model…Now let’s formula this problem as an LP model…

647

The Objective Function

Minimize Z = 15,000 X1 + 18,000 X2

WhereZ = daily scrap costX1 = truckloads from HasbeenX2 = truckloads from Gentro

Minimize Z = 15,000 X1 + 18,000 X2

WhereZ = daily scrap costX1 = truckloads from HasbeenX2 = truckloads from Gentro

HasbeenGentro

If we define the Hasbeen truckloads and the Gentro truckloads as the two decision variables X1 and X2, and since we want to minimize cost, we can state the objective function as follows:

If we define the Hasbeen truckloads and the Gentro truckloads as the two decision variables X1 and X2, and since we want to minimize cost, we can state the objective function as follows:

648

Constraints

1.5X1 + X2 > 6(Zinc/tons)

X1 + 3X2 > 10(Lead/tons)

X1, X2 > 0(Non-negativity)

1.5X1 + X2 > 6(Zinc/tons)

X1 + 3X2 > 10(Lead/tons)

X1, X2 > 0(Non-negativity)

Hasbeen (X1) Gentro (X2)Tons Tons Min Tons1.5 1 6 Zinc1 3 10 Lead

Given the demand information below from the problem:Given the demand information below from the problem:



Note that the inequalities are greater-than-or-equal since the demand information represents the minimum necessary for production.

Note that the inequalities are greater-than-or-equal since the demand information represents the minimum necessary for production.

649

Solution

Order 2.29 truckloads from Hasbeen and 2.57 truckloads from Gentro for daily delivery. The daily cost will be $80,610.

Order 2.29 truckloads from Hasbeen and 2.57 truckloads from Gentro for daily delivery. The daily cost will be $80,610.

Note: Do you see why in this solution that “integer” linear programming methodologies can have useful applications in industry?

Note: Do you see why in this solution that “integer” linear programming methodologies can have useful applications in industry?

650

End of Supplement A

652

Supplement B

Financial Analysis

653

• Cost Definitions• Expected Value• Depreciation • Activity-Based Costing• Investment Categories• Cost of Capital• Interest Rate Effects• Methods of Ranking Investments

OBJECTIVES

654

Cost Definitions

• Fixed costs are any expenses that remains constant regardless of the level of output

• Variable costs are expenses that fluctuate directly with changes in the level of output

• Sunk costs are past expenses or investments that have no salvage value and therefore should not be taken into account in considering investment alternatives

655

Cost Definitions (Continued)

• Opportunity cost is the benefit forgone, or advantage lost, that results from choosing one action over the best alternative course of action

• Avoidable costs include any expense that is not incurred if an investment is made but must be incurred if the investment is not made

656

Expected Value

• This analysis is used to include risk factors (probabilities) with payoff values for decision making

• Basic premise:

occuring outcome ofy Probabilit

x outcome Expected value Expected occuring outcome ofy Probabilit

x outcome Expected value Expected

657

Expected Value Problem Suppose you have to choose between one of three

processes (A, B, or C) with the following monthly profit and respective probabilities of those profits being realized. Compute expected values and choose a process.

occuring outcome ofy Probabilit

x outcome Expected value Expected

Process Payoffs Probabilities Pay x Prob. EV

A $6,000 90% 6,000x0.90 = $5,400

B $8,000 75% 8,000x0.75 = $6,000

C $9,000 65% 9,000x0.65 = $5,850

SelectProcess

B

658

Economic Life and Obsolescence

• Economic life of a machine is the period time over which it provides the best method for performing its task

• Obsolescence occurs when a machine is worn out

659

Depreciation

• Depreciation is a method for allocating costs of capital investment, including buildings, machinery, etc

• Depreciation procedures may not reflect an asset’s true value because obsolescence may at any time cause a large difference between the true value and book value

660

Depreciation Methods

• Straight-Line Method

• Sum-of-the-Years’-Digits (SYD) Method

• Declining-Balance Method

• Double-Declining-Balance Method

• Depreciation-by-Use Method

661

Traditional and Activity-Based Costing

Traditional Costing

End product cost

Total overhead

Labor-hourallocation

Activity-Based Costing

End product cost

Cost pools

Cost-driverallocation

Total overhead

Pooled based on activities

662

Choosing Among Investment Proposals:Investment Decision Categories

• Purchase of new equipment and/or facilities• Replacement of existing equipment or

facilities• Make-or-buy decisions• Lease-or-buy decisions• Temporary shutdowns or plant-abandonment

decisions• Addition or elimination of a product or

product line

663

Cost of Capital

• The cost of capital is calculated from a weighted average of debt and equity security costs

• Short-term debt

• Long-term debt

664

Interest Rate Effects

• Compound value of a single amount

• Compound value of an annuity

• Present value of a future single payment

• Present value of an annuity

• Discounted cash flow

665

Methods of Ranking Investments

• Net present value

• Payback period

• Internal rate of return

• Ranking investments with uneven lives

666

End of Supplement B

668

Supplement C

Operations Technology

669

• Hardware Systems

• Software Systems

• Formula for Evaluating Robots

• Computer Integrated Manufacturing

• Technologies in Services

• Benefits

• Risks

OBJECTIVES

670

Hardware Systems

• Numerically controlled (NC) machines

• Machining centers

• Industrial robots

• Automated material handling (AMH) systems

– Automated Storage and Retrieval Systems (AS/AR)

– Automate Guided Vehicle (AGV)

• Flexible manufacturing systems (FMS)

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Formula for Evaluating a Robot Investment

WhereP = Payback period in yearsI = Total capital investment required in robot and accessoriesL = Annual labor costs replaced by the robot (wage and

benefit costs per worker times the number of shifts per day)E = Annual maintenance cost for the robotZ = Annual depreciationq = Fractional speedup (or slowdown) factor (in decimals). Example:

If robot produces 150 % of what the normal worker iscapable of doing, the fractional speedup factor is 1.5.

Z)q(LE-LP

IZ)q(LE-L

P

I

The payback formula for an investment in robots is:

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Example of Evaluating a Robot Investment

Suppose a company wants to buy a robot. The bank wants to know what the payback period is before they will lend them the $120,000 the robot will cost. You have determined that the robot will replace one worker per shift, for a one shift operation. The annual savings per worker is $35,000. The annual maintenance cost for the robot is estimated at $5,000, with an annual depreciation of $12,000. The estimated productivity of the robot over the typical worker is 110%. What is the payback period of this robot?

P = I = 120,000 =1.47years L–E+q(L + Z) 35,000–5,000+1.1(35,000+12,000)

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Software Systems

• Computer-aided-design (CAD)– Computer-aided engineering (CAE)– Computer-aided process planning (CAPP)

• Automated manufacturing planning and control systems (MP & CS)

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Computer Integrated Manufacturing (CIM)

• Product and process design

• Planning and control

• The manufacturing process

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Cost Reduction Benefits from Adopting New Technologies

• Labor costs

• Material costs

• Inventory costs

• Transportation or distribution costs

• Quality costs

• Other costs

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Other Benefits….

• Increased product variety

• Improved product features and quality

• Shorter cycle times

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Risks

• Technological risks

• Organizational risks

• Environmental risks

• Market risks

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End of Supplement C

operations management

Documents

study operations management

operations scheduling17

service factory

project management

service plan

good service

quality assurance

product design