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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
OperationsManagement
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
OperationsManagement
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
20
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
Kaplan and Norton’s Generic Strategy Map (Continued)
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
Kaplan and Norton’s Generic Strategy Map (Continued)
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
42
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
52
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
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
Expected Time = Opt. Time + 4(Most Likely Time) + Pess. Time
6Expected Time =
Opt. Time + 4(Most Likely Time) + Pess. Time
6
78
Example 2. Expected Time Calculations
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
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
Expected Time = Opt. Time + 4(Most Likely Time) + Pess. Time
6Expected Time =
Opt. Time + 4(Most Likely Time) + Pess. Time
6
79
Example 2. Expected Time Calculations
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
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
Expected Time = Opt. Time + 4(Most Likely Time) + Pess. Time
6Expected Time =
Opt. Time + 4(Most Likely Time) + Pess. 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
89
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
108
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
End of Technical Note 4
132
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
Importance to
Cust.
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Doesn’t leak in rain
No road noise
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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
157
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
Station 1 Station 2 Station 3
A (4.2-2=2.2)
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
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)
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
Station 1 Station 2 Station 3
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
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
Station 1 Station 2 Station 3
182
A
C
B
D E F
GH
2
3.25
1
1.2 .5
11.4
1
C (4.2-3.25)=.95
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
A (4.2-2=2.2)B (2.2-1=1.2)G (1.2-1= .2)
Idle= .2
Station 1 Station 2 Station 3
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
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
A (4.2-2=2.2)B (2.2-1=1.2)G (1.2-1= .2)
Idle= .2
Station 1 Station 2 Station 3
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
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
A (4.2-2=2.2)B (2.2-1=1.2)G (1.2-1= .2)
Idle= .2
Station 1 Station 2 Station 3
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
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
A (4.2-2=2.2)B (2.2-1=1.2)G (1.2-1= .2)
Idle= .2
Station 1 Station 2 Station 3
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
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
A (4.2-2=2.2)B (2.2-1=1.2)G (1.2-1= .2)
Idle= .2
Station 1 Station 2 Station 3
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
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
A (4.2-2=2.2)B (2.2-1=1.2)G (1.2-1= .2)
Idle= .2
Station 1 Station 2 Station 3
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
End of Technical Note 5
194
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
212
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
End of Technical Note 6
243
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
270
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
End of Technical Note 7
303
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
Operations Consulting Tool Kit: Category 2
Data Gathering
Plant tours/audits
Work sampling
Flow charts
Organizational charts
315
Operations Consulting Tool Kit: Category 3
Data Analysis and Solution Development
Problem analysis (SPC tools)
Bottleneck analysis
Computer simulation
Statistical tools
316
Operations Consulting Tool Kit: Category 4
Cost Impact and Payoff Analysis
Decision trees
Balanced scorecard
Stakeholder analysis
317
Operations Consulting Tool Kit: Category 5
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
324
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.
valueinventory aggregate Average
sold goods ofCost turnoverInventory
valueinventory aggregate Average
sold goods ofCost turnoverInventory
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
340
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
350
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
Year: 1 2 3 4Small (000s) 150 170 200 240Family (000s) 115 140 170 200
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
©The McGraw-Hill Companies, Inc., 2004
367
368
Year: 1 2 3 4Small (000s) 150 170 200 240Family (000s) 115 140 170 200
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
©The McGraw-Hill Companies, Inc., 2004
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
378
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
End of Technical Note 10
392
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
container theof Size
stockSafety timelead during demand Expected
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
container theof Size
stockSafety timelead during demand Expected
C
SDL
k
3or ,75.24
5(2)(1.1))(1
container theof Size
stockSafety timelead during demand Expected
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
425
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?
Assume you only have 3 weeks and 6 weeks of actual demand data for the respective forecasts
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:
©The McGraw-Hill Companies, Inc., 2004
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?
Assume you only have 3 weeks and 5 weeks of actual demand data for the respective forecasts
Question: What is the 3 week moving average forecast for this data?
Assume you only have 3 weeks and 5 weeks of actual demand data for the respective forecasts
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
465
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
476
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
Exhibit 13.1Exhibit 13.1
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
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
Below are the complete calculations for the remaining months in the six month planning horizon
Below are the complete calculations for the remaining months in the six month planning horizon
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
Below are the complete calculations for the remaining months in the six month planning horizon with the other costs included
Below are the complete calculations for the remaining months in the six month planning horizon with the other costs included
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
494
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
529
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:
©The McGraw-Hill Companies, Inc., 2001
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
Exhibit 15.5Exhibit 15.5
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
©The McGraw-Hill Companies, Inc., 2004
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
LT=1 Scheduled receipts Proj. avail. balance 25 25 25 25 25 25 25 25
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
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
LT=1 Scheduled receipts Proj. avail. balance 25 25 25 25 25 25 25 25
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
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)
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
LT=1 Scheduled receipts Proj. avail. balance 25 25 25 25 25 25 25 25
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 3 C’s for each A
It takes 3 C’s for each A
551
A(2) B(1)
C(2)
X
C(3)
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
LT=1 Scheduled receipts Proj. avail. balance 25 25 25 25 25 25 25 25
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 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)
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
LT=1 Scheduled receipts Proj. avail. balance 25 25 25 25 25 25 25 25
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 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
557
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
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
Jobs (in order Processing Due Dateof arrival) Time (days) (days hence)
A 4 5B 7 10C 3 6D 1 4
Answer: Shortest Operating Time Schedule
Jobs (in order Processing Due Date Flow Timeof arrival) Time (days) (days hence) (days)
D 1 4 1C 3 6 4A 4 5 8B 7 10 15
Suppose you have the four jobs to the right arrive for processing on one machine
Suppose you have the four jobs to the right arrive for processing on one machine
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
Do all the jobs get done on time?Do all the jobs get done on time?
569
Example of Job Sequencing: Earliest Due Date First
Jobs (in order Processing Due Dateof arrival) Time (days) (days hence)
A 4 5B 7 10C 3 6D 1 4
Answer: Earliest Due Date First
Jobs (in order Processing Due Date Flow Timeof arrival) Time (days) (days hence) (days)
D 1 4 1A 4 5 5C 3 6 8B 7 10 15
Suppose you have the four jobs to the right arrive for processing on one machine
Suppose you have the four jobs to the right arrive for processing on one machine
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
Do all the jobs get done on time?Do all the jobs get done on time?
570
Example of Job Sequencing: Critical Ratio Method
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
Suppose you have the four jobs to the right arrive for processing on one machine
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)
Do all the jobs get done on time?Do all the jobs get done on time?
571
Example of Job Sequencing:Last-Come First-Served
Jobs (in order Processing Due Dateof arrival) Time (days) (days hence)
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
Suppose you have the four jobs to the right arrive for processing on one machine
Suppose you have the four jobs to the right arrive for processing on one machine
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
582
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
End of Technical Note 16
603
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
Capacity Example Situation 2
Y X Market
Case B
X YBottleneck Nonbottleneck
Demand/month 200 units 200 unitsProcess time/unit 1 hour 45 minsAvail. time/month 200 hours 200 hours
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
Capacity Example Situation 3
X Y
Assembly
Market
Case C
X YBottleneck Nonbottleneck
Demand/month 200 units 200 unitsProcess time/unit 1 hour 45 minsAvail. time/month 200 hours 200 hours
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?
Yes, 25% in YYes, 25% in Y
619
Capacity Example Situation 4
X Y
Market Market
Case D
X YBottleneck Nonbottleneck
Demand/month 200 units 200 unitsProcess time/unit 1 hour 45 minsAvail. time/month 200 hours 200 hours
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?
Yes, 25% in YYes, 25% in Y
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
Exhibit 17.9Exhibit 17.9
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
634
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
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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)
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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
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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
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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…
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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:
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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:
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…
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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:
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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:
We can now state the constraints and non-negativity requirements as:
We can now state the constraints and non-negativity requirements as:
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.
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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
651
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
667
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)
671
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:
672
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)
673
Software Systems
• Computer-aided-design (CAD)– Computer-aided engineering (CAE)– Computer-aided process planning (CAPP)
• Automated manufacturing planning and control systems (MP & CS)
674
Computer Integrated Manufacturing (CIM)
• Product and process design
• Planning and control
• The manufacturing process
675
Cost Reduction Benefits from Adopting New Technologies
• Labor costs
• Material costs
• Inventory costs
• Transportation or distribution costs
• Quality costs
• Other costs
676
Other Benefits….
• Increased product variety
• Improved product features and quality
• Shorter cycle times
677
Risks
• Technological risks
• Organizational risks
• Environmental risks
• Market risks
678
End of Supplement C
679