Mod 2 ProgMgmt COPQ Sep 03

8/12/2019 Mod 2 ProgMgmt COPQ Sep 03

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1© Visteon Corporation BB Mod #2 Program Management Rev 3.1 9/03.

Program ManagementBlack Belt Training

Module #2


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Project Management TheoryProject Value

Rolled Throughput YieldCost of Poor Quality

Instructor:Brian Edelman


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Philosophy

• Stop me immediately when a concept is unclear — odds are,

if you are confused about something, at least one other

person in the room is also.

• I will either:

– Try to rephrase/clarify the point

– Ask you to hold on for a few moments until we get to a

downstream slide

– Defer it to 1:1 discussion.

• Dual purposes of the material: – Knowledge to do projects.

– Knowledge to pass certification exam.

• Zip the Monkey.


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Project Management Theory

• Purpose

• Deliverables

– Problem Statement

• Objectives• Cost/Benefit Analysis

• Team Makeup

• Process Owner/Process Sponsor

– Project Closure


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Project Management Theory Agenda(cont.)

• Tools

– Timing

• Critical Path Method

• PERT

• Compression

– Crashing

– Fast Tracking


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Purpose—Why ProjectManagement?

• To establish the boundaries, or ‗scope‘, of the

project.

• To determine what the goals of the project are.

• To identify and obtain commitment for theresources necessary for project completion.

• To complete the project in the most efficient and

timely manner possible.


7/1627© Visteon Corporation BB Mod #2 Program Management Rev 3.1 9/03.

Deliverables

• Problem Statement — A single paragraph that

captures the ‗pain‘ the organization feels from the

problem and what the Six Sigma project will do to

improve it.• Objective — This lists the primary metric and the

degree to which the project will improve the

metric. As a baseline, improvement is typically

targeted at 70% improvement from current status.After the completion of the Define phase, this is

the ‗contract‘ for successful project completion.



Deliverables (continued)

• Benefits — This item is not the primary metric and

goal of the project , but instead is a listing of the

benefits that will accrue from achieving the goal.

– Financial Impacts — What is the bottom lineimprovement to the organization?

– Others — May include floor space, goodwill, TGW

reductions, or other ‗soft‘ savings.

– Memo: Hard savings are generally things that will ‗hitthe bottom line.‘ As a rule of thumb they can be

entered into the Cost Savings System (CSS). Soft

savings are those things that may enable savings later

but do not immediately affect the bottom line.




• Team — To successfully complete the Define

phase of a Six Sigma project requires that a

preliminary team be identified. This team must

include a representative from the local controller‘soffice, as well as all key stakeholders in the

process.




• Process Sponsor/Process Owner — The manager

who will ‗own‘ the process once the project is

complete. He/she is responsible for making sure

Visteon ‗sustains the gain‘ from the Six Sigma project once the project team completes its work.



Problem Statement

• A well-crafted problem statement is key to

successful project completion. It must align with

the project objectives and metrics.

• Example: – Problem Statement: Defects on Line 1 have averaged

121,000 ppm for the last four months. This does not

allow us to meet our corporate cost and quality

objectives – Objective: Reduce ppm on Line 1 from an average of

121,000 ppm to 50,000 ppm by July 1, 2002.


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Project Closure

• The project can be closed when it is verified that thegoals/objectives for the project have been achieved.

• The most recent closure procedures are available on theSix Sigma website. (Home page:

http://hub.visteon.com/6sigma/)• Project closure requires approval of the Black Belt, the

Project Champion, the responsible FinanceAnalyst/Controller, and the divisional DeploymentDirector.

• Certification requires two successfully closed projects,demonstrated competency in a statistical proficiencyexamination, and recommendation of the divisionalDeployment Director.

http://hub.visteon.com/6sigma/http://hub.visteon.com/6sigma/


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Project Management Tools

• ―Project Management‖ generally refers to all

aspects of the project: timing, cost, quality,

resource availability, etc.

• For purposes of this section, we will focus on project time management.


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Project Scheduling

• Once a project scope has been established, a

schedule of work can be developed.

• Typically, a series of tasks is identified, along with

the precedence relationships, constraints, and whois responsible

• This list of tasks needed to accomplish the project

is sometimes called the Work Breakdown

Structure (WBS).


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Project Scheduling (cont.)

• A ‗task‘ is any activity with an identifiable

beginning and end.

– Ex: Purchase prototype tool

– Ex: Run DOE• Precedence relationships define how tasks relate to

one another.

– Ex: Cannot PSW a part until a drawing for it is

released.


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Example WBS and Gantt ChartMonth 1 Month 2 Month 3 Month 4

w1 w2 w3 w4 w1 w2 w3 w4 w1 w2 w3 w4 w1 w2 w3

14-May 21-May 28-May 4-Jun 11-Jun 18-Jun 25-Jun 2-Jul 9-Jul 16-Jul 23-Jul 30-Jul 6-Aug 13-Aug 20-Aug 27

Project Selection Form (from Project Champion)

Define

Develop Project Statement

Identify Customer

Identify CTQ's

Develop Y/X Diagram

Map the Process

Develop Fishbone Diagram

Develop Cause & Effect Matrix

Perform or Review FMEA

Confirm Project Scope & Goal

Refine Problem Statement

Measure

Develop Data Collection Plan

Perform Measurement System Analysis

Operational Definitions

Gage R&R

Conduct Data Collection

Perform Graphical Analysis

Pareto Chart

Run ChartHistogram

Box Plot

Scatter Plot

Conduct Baseline Capability Analysis

Cp /CPk

Rolled Throughput Yield

DPMO

Sigma Level

Determine Cost of Poor Quality

Confirm Project Scope & Goal

Analyze

Confirm Data Type - Attribute or Variable

Attribute Data Analysis

1 Proportion

2 Proportion

Chi- Square

Variable Data Analysis

Analysis of Variation

F-Test

ANOVA

Analysis of Means

1- Sample Test

2- Sample Test

ANOVA for Means

Correlation Regression

Improve

Brainstorm Alternatives

Conduct Design Of Experiments

Create "should be" Process Map

Conduct FMEA

Perform Cost / Benefit Analysis

Conduct Pilot

Validate Improvement

Control

Mistake Proofing

Long Term Measurement System Analysis Plan

Develop Statistical Process Control Chart

Develop Preventative Maintenance Plan

Develop Reaction Plan

Updated Standard Operating ProceduresCongratulate the team


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• Precedence relationships are generally, but notalways, finish-to-start relationships:

– Finish-to-start: Cannot begin the next task until the

preceding one is finished.

– Start-to-start: Cannot start the next task until the preceding one starts.

– Finish-to-finish: Cannot finish the next task until the

preceding one finishesGantt Chart

from MS

Project


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• Constraints can tie activities to specific dates.

– Ex: Must do equipment installation during contractual

shutdown.

– Ex: Must file legal documents on/by specific dates.


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• Analysis Tools — the most common general means

of analyzing schedules include:

– Critical Path Method

– Program Evaluation and Review Technique (PERT) – Memo: As an aside, CPM uses node-based information

(duration), where PERT uses information on the links.

From a practical matter, this makes little difference, and

for the rest of this presentation, we will use node-based

information, similar to Microsoft Project conventions.


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Critical Path Method (cont.)

Tasks with zero float are on a critical path.

• Float greater than zero is an indication of how

much this task can slip from its earliest date before

it also becomes critical path.• Having multiple critical paths is very bad — any

slippage of any activity on any of the critical paths

will delay the project completion. Or, in other

words, Murphy has more places to strike.


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Critical Path Method

• This technique, by identifying the critical path(s),

emphasizes which tasks will affect overall

execution time of the project if they slip (or finish

early).• This technique also allows tracking of progress

versus a baseline to determine if the overall

project is on track.


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Critical Path Method Example

• A recruiting event is planned at a university.

• To be successful, it is necessary to do thefollowing: – Reserve the facility — 6 weeks

– Obtain brochures and other giveaways — 2 weeks

– Arrange personnel — 4 weeks

– Hold the event — 1 week

– Select candidates — 1 week

– Follow up with appropriate candidates — 1 week

Let‘s map out the network diagram.


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• The network diagram:

Reserve facility

(6 weeks)

Get brochures, etc.

(2 weeks)

Arrange Personnel

(4 weeks)

Start

Hold event

(1 week)

Select Candidates

(1 week)

Follow Up

(1 week)

Done


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CPM Example (continued)

Task Duration Early Start Early

Finish

Reserve

Facility

6

ObtainBrochures

2

Arrange

Personnel

4

Hold Event 1

Select

Candidates

1

Follow Up 1

Early start is

the latest of

the

predecessoractivities‘

early finishes

Early finish is

the early start

plus the

duration


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Finish

Reserve

Facility

6 0

Obtain

Brochures

2 0

Arrange

Personnel

4 0

Hold Event 1

Select

Candidates

1

Follow Up 1

These items

have no

predecessors,so they can

start

immediately


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Finish

Reserve

Facility

6 0 6

ObtainBrochures

2 0 2

Arrange

Personnel

4 0 4

Hold Event 1

Select

Candidates

1

Follow Up 1

+

++

=

==


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Finish

Reserve

Facility

6 0 6

ObtainBrochures

2 0 2

Arrange

Personnel

4 0 4

Hold Event 1 6

Select

Candidates

1

Follow Up 1

The latest‗early

finish‘ date

of a

predecessor

is the ‗early

start‘ date

of the

successor


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Finish

Reserve

Facility

6 0 6

ObtainBrochures

2 0 2

Arrange

Personnel

4 0 4

Hold Event 1 6 7

Select

Candidates

1

Follow Up 1

+ =

Repeat

through the

chart


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Finish

Reserve

Facility

6 0 6

ObtainBrochures

2 0 2

Arrange

Personnel

4 0 4

Hold Event 1 6 7

Select

Candidates

1 7 8

Follow Up 1 8 9

+ =

+ =


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Task Duration Early

Start

Early

Finish

Late

Finish

Late

Start

Float

Reserve

Facility

6 0 6

ObtainBrochures

2 0 2

Arrange

Personnel

4 0 4

Hold Event 1 6 7

Select

Candidates

1 7 8

Follow Up 1 8 9

Late Start

= Late

Finish -

Duration

Earliest

late start of

thesuccessor

tasks


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Task Duration Early

Start

Early

Finish

Late

Finish

Late

Start

Float

Reserve

Facility

6 0 6

ObtainBrochures

2 0 2

Arrange

Personnel

4 0 4

Hold Event 1 6 7

Select

Candidates

1 7 8

Follow Up 1 8 9 9

The late finishis the latest

time a task can

finish without

slipping the

project.


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Task Duration Early

Start

Early

Finish

Late

Finish

Late

Start

Float

Reserve

Facility

6 0 6

ObtainBrochures

2 0 2

Arrange

Personnel

4 0 4

Hold Event 1 6 7

Select

Candidates

1 7 8

Follow Up 1 8 9 9 8

9 - 1 = 8


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Task Duration Early

Start

Early

Finish

Late

Finish

Late

Start

Float

Reserve

Facility

6 0 6

ObtainBrochures

2 0 2

Arrange

Personnel

4 0 4

Hold Event 1 6 7

Select

Candidates

1 7 8 8

Follow Up 1 8 9 9 8


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Task Duration Early

Start

Early

Finish

Late

Finish

Late

Start

Float

Reserve

Facility

6 0 6

ObtainBrochures

2 0 2

Arrange

Personnel

4 0 4

Hold Event 1 6 7

Select

Candidates

1 7 8 8 7

Follow Up 1 8 9 9 8

8 - 1 = 7


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Task Duration Early

Start

Early

Finish

Late

Finish

Late

Start

Float

Reserve

Facility

6 0 6 6 0

ObtainBrochures

2 0 2 6 4

Arrange

Personnel

4 0 4 6 2

Hold Event 1 6 7 7 6

Select

Candidates

1 7 8 8 7

Follow Up 1 8 9 9 8


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Task Duration Early

Start

Early

Finish

Late

Finish

Late

Start

Float

Reserve

Facility

6 0 6 6 0

ObtainBrochures

2 0 2 6 4

Arrange

Personnel

4 0 4 6 2

Hold Event 1 6 7 7 6

Select

Candidates

1 7 8 8 7

Follow Up 1 8 9 9 8

Float (or Slack) = Late Finish –

Early Finish

= Late Start – Early Start


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Task Duration Early

Start

Early

Finish

Late

Finish

Late

Start

Float

Reserve

Facility

6 0 6 6 0 0

ObtainBrochures

2 0 2 6 4 4

Arrange

Personnel

4 0 4 6 2 2

Hold Event 1 6 7 7 6 0

Select

Candidates

1 7 8 8 7 0

Follow Up 1 8 9 9 8 0


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Task Duration Early

Start

Early

Finish

Late

Finish

Late

Start

Float

Reserve

Facility

6 0 6 6 0 0

ObtainBrochures

2 0 2 6 4 4

Arrange

Personnel

4 0 4 6 2 2

Hold Event 1 6 7 7 6 0Select

Candidates

1 7 8 8 7 0Follow Up 1 8 9 9 8 0

Tasks with zerofloat are on the

critical path. If they

slip, the project

completion slips.


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Reserve facility

(6 weeks)

Get brochures, etc.

(2 weeks)

Arrange Personnel

(4 weeks)

Start

Hold event

(1 week)

Select Candidates

(1 week)

Follow Up

(1 week)

Done


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PERT (continued)

• PERT assumes a probability distribution for

possible durations.

• As a practical matter, PERT uses optimistic,

pessimistic, and most-likely durations as inputs tothe calculation.

– Best way to get these estimates is simply to talk with

people who have completed similar projects.

PERT-type algorithms then use a weighted

average duration.

6

) _ *4( c pessimistilikelymost optimisticduration


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Examples

• Assume we‘re in charge of building the newVisteon headquarters.

– We have a series of high-level tasks and we‘ve

identified their precedence relationships.

– We have also determined what a best-case, realistic,

and worst-case duration for each step is.


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Examples (cont.)

Task ID Description Predecessors Best-case

duration

Likely

Duration

Worst-case

Duration

Weighted

Average

Duration

Std. Dev

(Worst –

Best) / 6

A Select Site none 0.2 years 0.75 years 2 years

B Design

building

A 1 year 1.5 years 2.5 years

C Obtain permits B 0.1 years 0.1 years 0.3 years

D Contract

builders

B 0.3 years 0.4 years 0.5 years

E Arrange

financing


F Build

headquarters

C, D, E 0.7 years 0.9years 1.0 years

G Pay contractors F 0.3 years 0.4 years 1.0 years


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Examples (cont.)


duration

Likely

Duration

Worst-case

Duration

Weighted

Average

Duration

Std. Dev

(Worst –

Best) / 6

A Select Site none 0.2 years 0.75 years 2 years 0.87 yr.

B Design

building

A 1 year 1.5 years 2.5 years

C Obtain permits B 0.1 years 0.1 years 0.3 years

D Contract

builders


E Arrange

financing


F Build

headquarters

C, D, E 0.7 years 0.9years 1.0 years

G Pay contractors F 0.3 years 0.4 years 1.0 years

87.06

275.0*42.0


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Examples (cont.)


duration

Likely

Duration

Worst-case

Duration

Weighted

Average

Duration

Std. Dev

(Worst –

Best) / 6


B Design

building

A 1 year 1.5 years 2.5 years 1.58 yr.

C Obtain permits B 0.1 years 0.1 years 0.3 years 0.13 yr.

D Contract

builders

B 0.3 years 0.4 years 0.5 years 0.4 yr.

E Arrange

financing


F Build

headquarters

C, D, E 0.7 years 0.9years 1.0 years 0.88 yr.

G Pay contractors F 0.3 years 0.4 years 1.0 years 0.48 yr.

Completing

the column.


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Exercise


duration

Likely

Duration

Worst-case

Duration

Weighted

Average

Duration

Std. Dev

(Worst –

Best) / 6


B Design

building

A 1 year 1.5 years 2.5 years 1.58 yr.

C Obtain permits B 0.1 years 0.1 years 0.3 years 0.13 yr.

D Contract

builders


E Arrange

financing


F Build

headquarters

C, D, E 0.7 years 0.9years 1.0 years 0.88 yr.

G Pay contractors F 0.3 years 0.4 years 1.0 years 0.48 yr.

Exercise: Draw the network diagram and find

the critical path.


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Exercise—Network Diagram

Select Site

ObtainPermits

ArrangeFinancing

PayContractors

DesignBuilding

ContractBuilders

BuildHQ


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Exercises—Critical Path

Task ID Description Predecessors Weighted

Average

Duration

Early Start Early

Finish

Late Finish Late Start Float

A Select Site none 0.87 yr. 0 0.87 .87 0 0

B Design

building

A 1.58 yr. 0.87 2.45 2.45 0.87 0

C Obtain

permits

B 0.13 yr. 2.45 2.58 2.85 2.72 0.27

D Contract

builders

B 0.4 yr. 2.45 2.85 2.85 2.45 0

E Arrange

financing

B 0.28 yr. 2.45 2.73 2.85 2.57 0.12

F Build

headquarters

C, D, E 0.88 yr. 2.85 3.73 3.73 2.85 0

G Pay

contractors

F 0.48 yr. 3.73 4.21 4.21 3.73 0


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Exercises—Critical Path

Task ID Description Predecessors Weighted

Average

Duration

Early Start Early

Finish

Late Finish Late Start Float

A Select Site none 0.87 yr. 0 0.87 .87 0 0

B Design

building

A 1.58 yr. 0.87 2.45 2.45 0.87 0

C Obtain

permits

B 0.13 yr. 2.45 2.58 2.85 2.72 0.27

D Contract

builders

B 0.4 yr. 2.45 2.85 2.85 2.45 0

E Arrange

financing

B 0.28 yr. 2.45 2.73 2.85 2.57 0.12

F Build

headquarters

C, D, E 0.88 yr. 2.85 3.73 3.73 2.85 0

G Pay

contractors

F 0.48 yr. 3.73 4.21 4.21 3.73 0


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Exercise—Critical Path

Select Site

ObtainPermits

ArrangeFinancing

PayContractors

DesignBuilding

ContractBuilders

BuildHQ


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Examples (cont.)

• From the exercise, the overall duration is the sum

of the expected durations on the critical path.

– Overall duration = 0.87 + 1.58 + 0.4 + 0.88 + 0.48 =

4.21 years.

• To determine the variability in the critical path

estimate, we assume that the durations of each task

are statistically independent of the durations of theothers.


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Examples (cont.)

• Because variances (2) of independent random

events are additive,

• Variance is the square of the standard deviation,

which, for each event, is calculated as:

path taskscriticalall

2

task

2

projectoverall

6

_ _ durationoptimisticdurationc Pessimisti


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Examples (cont.)


duration

Likely

Duration

Worst-case

Duration

Weighted

Average

Duration

Std. Dev

(Worst –

Best) / 6

A Select Site none 0.2 years 0.75 years 2 years 0.87 yr. 0.3 yr.

B Design

building

A 1 year 1.5 years 2.5 years 1.58 yr. 0.25 yr.

C Obtain permits B 0.1 years 0.1 years 0.3 years 0.13 yr. 0.03 yr

D Contract

builders

B 0.3 years 0.4 years 0.5 years 0.4 yr. 0.03 yr.

E Arrange

financing

B 0.1 years 0.3 years 0.4 years 0.28 yr. 0.05 yr.

F Build

headquarters

C, D, E 0.7 years 0.9years 1.0 years 0.88 yr. 0.05 yr.

G Pay contractors F 0.3 years 0.4 years 1.0 years 0.48 yr. 0.12 yr.

Filling inthe column


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Examples (cont.)

• Because variances (2) of independent randomevents are additive,

path taskscriticalall

2

task

2

projectoverall

41.0

17.0

12.005.003.025.03.0

projectoverall

2

projectoverall

222222

projectoverall

Note: Be careful to add the variances, not the standard deviations.


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Examples (cont.)

• With the expected value and standard deviation of

the critical path, one can use basic statistics to

determine the probability that a project will be

completed within a certain length of time. – We‘ll cover Continuous Distributions in Week Two.

– We‘ll practice this application in Week Four.


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Compression

• Compression involves finding ways to shorten theduration of a project without changing its scope.

– Crashing: Compressing timing by looking at cost and

schedule tradeoffs to determine how to reduce the most

time for the least money.

– Fast tracking: Doing activities in parallel that are

normally done sequentially. This reduces timing at a

significant risk of rework.


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Crashing

• By spending extra money to reduce timing oncritical path items, overall project completion

timing can be reduced.

For purposes of the exam, it will be necessary toknow the cost per unit time to crash each activity.

ID Predecessors Normal Time Crash Time Cost per

Week to

Crash

A none 4 weeks 1 week $8,000

B A 5 weeks 2 weeks $10,000

C A 3 weeks 2 weeks $1,000

D B 6 weeks 4 weeks $3,500

E C and D 8 weeks 3 weeks $25,000


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Crashing (cont.)

Questions asked typically include:Project cost and duration without crashing.

Project cost and duration at maximum crashing.

Project cost and duration for minimum cost project.

For non-crash situations, the critical path analysis

will give project duration, and the cost per unit

time of the project can be used to determine its

total cost.


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Example—no crashing

• Assume it costs $15,000 per week for a given project. Also, assume the information below.

Find the non-crashed duration and cost.

ID Predecessors NormalTime

CrashTime

Cost perWeek to

Crash







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Example—Network Diagram

A

B

C

D

E


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Example—no crashing

• Duration is 23 weeks without crashing.

• Cost is $15,000 per week x 23 weeks = $345,000


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Crashing

For a maximum crashing situation, it is necessary todetermine which critical path task has the least crash cost

per unit time.

This task is crashed to the maximum it can be until the critical path

changes.

If the critical path has not changed, the next least expensive crash

cost on the critical path is chosen and the above step is repeated.

If the critical path changes, the new critical paths are identified.

The least expensive crash costs for each critical path are identified.

These are reduced to the point the critical path changes. Note thatif one task is on both critical paths, it is not counted twice when

tallying crash costs.


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Crashing (cont.)

This continues until no further crashing is possible

for items on the critical path(s). The result is the

maximum crash condition.


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Example—Maximum crashing

• Find the maximum crash duration and cost for the prior example.


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ID Normal

Time

Early

Start

Early

Finish

Late

Finish

Late

Start

Float

A 4 weeks 0 4 4 0 0B 5 weeks 4 9 9 4 0

C 3 weeks 4 7 15 12 8

D 6 weeks 9 15 15 9 0

E 8 weeks 15 23 23 15 0

Identify Critical

Path


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Example—Max Crash.

A

B

C

D

E


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CrashTime

Cost perWeek to

Crash






The least cost critical path

item to crash is D


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CrashTime

Cost perWeek to

Crash






D can be crashed up to 2

weeks.


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ID Normal

Time

Early

Start

Early

Finish

Late

Finish

Late

Start

Float

A 4 weeks 0 4 4 0 0

B 5 weeks 4 9 9 4 0

C 3 weeks 4 7 15 12 8

D 6 weeks 9 15 15 9 0

E 8 weeks 15 23 23 15 0

There is eight weeks of float on all of the

non-critical path items, so we know the

critical path will not change if we shortedtask D by 2 weeks.

T k D i h d Th i i l h d ‘


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ID Predecessors Time CrashTime

Cost perWeek to

Crash




D B 4 weeks 4 weeks


Task D is crashed. The critical path doesn‘t

change. We find the next least expensive

critical path item to crash. This is task A.


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Cost perWeek to

Crash




D B 4 weeks 4 weeks


A can be crashed up to 3 weeks.


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A

B

C

D

E

A is on all the paths, so

crashing it will not change

the critical path.

T k A i h d 3 k W fi d th


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Cost perWeek to

Crash

A none 1 week 1 week



D B 4 weeks 4 weeks


Task A is crashed 3 weeks.. We find the

next least expensive critical path item to

crash. This is task B.


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Cost perWeek to

Crash




D B 4 weeks 4 weeks


Task B can be crashed up to 3 weeks.


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ID Normal

Time

Early

Start

Early

Finish

Late

Finish

Late

Start

Float

A 1 week 0 1 1 0 0

B 5 weeks 1 6 6 1 0

C 3 weeks 1 4 10 7 6

D 4 weeks 6 10 10 6 0

E 8 weeks 10 18 18 10 0

The only non-critical path task has 6 weeks of

float, so crashing task B will not change the

critical path.


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Cost perWeek to

Crash


B A 2 weeks 2 weeks


D B 4 weeks 4 weeks


Task B is fully crashed. The next least

expensive task on the critical path is Task E,

which can be crashed up to 5 weeks.


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A

B

C

D

E

E is on all the paths, so

crashing it will not change

the critical path.


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Cost perWeek to

Crash


B A 2 weeks 2 weeks


D B 4 weeks 4 weeks

E C and D 3 weeks 3 weeks

Task E is crashed as much as possible. Theonly task that is able to be crashed now is

task C, which is not on the critical path. We

are at the maximum crash condition.


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• Our project duration is now calculated at 10weeks.

• Our project crash costs are:Task Weeks Crashed Cost per time to

Crash

Total Cost

A 3 $8,000 $24,000

B 3 $10,000 $30,000

D 2 $3,500 $7,000

E 5 $25,000 $125,000

TOTAL: $186,000


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• The crashing costs were $186,000. This is addedto our ongoing project costs.

• Ongoing project costs = 10 weeks duration x

$15,000 per week = $150,000

• Our total project cost was $336,000.

• Note that this amount may be more or less than the

‗normal‘ project duration costs.


85/162


Crashing—Minimum cost

To determine the minimum cost project, an

iterative approach is used.

The premise of this technique is that ongoing costsof the project ($x/month) are more than some of

the costs to crash tasks ($y/month). By selectively

identifying the tasks that reduce total cost, the

project cost is optimized.


86/162


Crashing (continued.)

Procedure for the least cost project:1. Identify all of the critical paths.

2. Identify the tasks on the critical path(s) that have a lowercrash cost than the ongoing cost of the project.

3. Choose the least crash cost task(s) that reduce all of thecritical paths (and therefore the project). The total crashcost (per unit time) of these must be less than the ongoing

project cost. If not, no further cost reductions are possible.

4. Crash them until they can no longer be crashed or untilthe critical path changes.

5. Repeat steps 2 through 5.


87/162


Example—Minimum cost

• Using the prior example, find the minimum costduration. How much does this cost?


88/162




CrashTime

Cost perTime to

Crash






The least cost critical path

item to crash is D


89/162




CrashTime

Cost perTime to

Crash






It can be crashed 2 weeks, and the

critical path (based on the last

exercise) will not change.


90/162




CrashTime

Cost perTime to

Crash






With an ongoing project cost of $15,000 per week, any crash

in Task D (as long as Task D remains on the critical path)

will reduce total cost.


91/162




CrashTime

Cost perTime to

Crash




D B 4 weeks 4 weeks


D is crashed as much as possible. The

next least expensive critical path item

is Task A.


92/162




CrashTime

Cost perTime to

Crash




D B 4 weeks 4 weeks


From the prior exercise, we know the critical path

will not change if task A is crashed. As $8,000

crash cost is less than the $15,000 ongoing costs,

we crash task A as much as possible.


93/162


94/162




CrashTime

Cost perTime to

Crash




D B 4 weeks 4 weeks


Task B‘s crash costs are less than the ongoing

project costs, so it is crashed as much as possible.


95/162




CrashTime

Cost perTime to

Crash


B A 2 weeks 2 weeks


D B 4 weeks 4 weeks


The only remaining critical path item is task E,

which has a crash cost higher than the ongoing

cost of the project. Therefore, no further costreductions are possible on this project.


96/162



ID Normal

Time

Early

Start

Early

Finish

Late

Finish

Late

Start

Float

A 1 week 0 1 1 0 0

B 2 weeks 1 3 3 1 0

C 3 weeks 1 4 7 4 3

D 4 weeks 3 7 7 3 0

E 8 weeks 7 15 15 7 0


97/162


98/162



• The crashing costs were $61,000. This is added toour ongoing project costs.

• Ongoing project costs = 15 weeks duration x

$15,000 per week = $225,000

• Our total project cost was $286,000 (versus

$345,000 for the non-crashed project).

• Note that more complex examples will result in

multiple critical paths.

P j M


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Project Management

• That concludes the section on ProjectManagement.

• Any questions?


100/162

P j t V l


101/162


Project Value

• Evaluation of a project must take account of the―time value of money.‖

• The premise behind ‗time value of money‘ is that

$1 today is worth more than $1 tomorrow.

• Why?

– Inflation.

– Lost investment opportunity.

P j t V l ( t )


102/162


Project Value (cont.)

• To account for this, a discount factor is applied tomoney we will receive (or spend) in the future.

This discount factor is usually called the ‗rate of

return‘ and is analogous to the interest rate on a

loan or deposit.

• Example: If the interest rate r is 6% per annum

and we will receive $1000 in one year, then the

value of this in today‘s dollars is:

40.939$

06.1

1000

1

1000$ todayValue

r

P j t V l ( t )


103/162


Project Value (cont.)

• Terms – Rate of return is also called ‗cost of capital‘.

• An organization‘s established ‗hurdle rate‘ may also be used in

the same calculations.

• The terms TARR (Time Adjusted Rate of Return) and IRR(Internal Rate of Return) are also used.

– Value today is generally called the present value or net

present value of the decision.

– Payback period (or simply ‗payback‘) is the length of

time it will take the company to recoup its investment.

P j t V l D i i M ki


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Project Value—Decision Making

• Typical Decision making critieria1. If a project‘s TARR is greater than an established

‗hurdle rate‘, the project is approved.

2. If the project‘s net present value is greater than zero,

the project is approved.3. If the project‘s payback is less than some period of

time, the project is approved.

P j t V l ( t)


105/162


Project Value (cont)

• Many projects result in a stream of savings, ratherthan just a ‗one time‘ save.

– Example: Improving the process saves us $0.43 for

each fuel tank produced.

– Example: Negotiating a lower price for our computerservices saves us $500,000 per year.

• Each year‘s savings get discounted back into

today‘s dollars to determine the value of the

project.

321

Savings3Year

1

Savings2Year

1

Savings1YeartodayValue

r r r


106/162

P j t V l ( t)


107/162



• The NPV formula for an infinite stream of payments of amount ‗A‘ given an investment and

a rate of return ‗r‘ is:

• In our example, the net present value is:

r

A Investment- NPV

000,000,210.0000,300000,000,1 NPV

P j t V l ( t)


108/162



• For a finite stream of payments, the Excel function NPV() can be used.

• The syntax of this is NPV(rate, value, value,

value)

• Example: If you invest $10,000 today and receive

annual payments of $8,000 for three years, what is

the net present value of this transaction, assuming

a 6% cost of capital? present value = -10000 + NPV(0.06, 8000,8000,8000)

present value = $11,384.10

Project Val e (cont)


109/162



• Given a stream of payments, it is also possible tocalculate the internal rate of return.

• For a stream of n payments, one simply solves

for the roots of an nth order polynomial.

Or…

• One uses the IRR() function in Excel to do it for

you.

• The syntax for IRR is (value, value, …, guess). Note that the guess is optional.



110/162



• Example: – Assume investing $1,000,000 for a new machine lowers

piece prices sufficiently that the company saves$250,000 per year. Also assume that the product has a 5year life (and no salvage value for the machine). What

is the internal rate of return for this investment?

internal rate of return = IRR(-1000000,250000,250000,250000,250000,250000)

internal rate of return = 0.079 or 7.9%

– If this was greater than or equal to the organization‘shurdle rate, this project would be considered a goodinvestment.



111/162



• The third Excel function that is useful is the payment function, PMT(). The PMT function

takes as input the present value, the rate of

return, and the number of payments. It outputs

the periodic payment that results.

• Syntax: Payment PMT = (rate, nper, pv) where

rate is the rate of return, nper is the number of

periods, and pv is the present value.


112/162


113/162


114/162



The Classical Perspective of Yield


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0.95

0.97

0.99

1.01

1 4 7 10 13 16 19

Y = SU

Where:

Y = Yield

S = Number of Units that PassU = Number of Units Tested

Where:

Y = Yield

S = Number of Units that Pass

U = Number of Units Tested

Y.Y.

Time PeriodTime Period

The Classical Perspective of Yield

Traditionally, the

output of a process is

compared to a

pass/fail criteria to

determine yield.

Six Sigma Breakthrough Challenge


116/162


Test Room Fai lu re RateTest Room F ailur e Rate

Data: Jan 92 through Dec 92

Data Courtesy of Asea Brown Boveri , Ltd .

Each data point is a factory

Y.FT = 1- Failure Rate

Profit

Margin

Profit

Margin

0

.01

.02

.03

.04

.05

.06

.07

.08

.09

-.03 0 .025 .05 .075 .1 .125 .15 .175 .2 .225

Failure Rate

S c r a p R a t e

S…

0.01

.02

.03

.04

.05

.06

.07

.08

.09

-.03 0 .025 .05 .075 .1 .125 .15 .175 .2 .225

Failure Rate

S c r a p R a t e

S…

Test Room Fail ure RateTest Room F ailur e Rate

Scrap

Rate

Scrap

Rate

Data Courtesy of Asea Brown Boveri , Ltd .

Each data point is a factory

Y.FT = 1- Failure Rate


Intuition would tell us that yield should be correlated to profitmargin and scrap rate…but the data do not show this.

First Time Yield


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First Time Yield

• First time yield (also called first pass yield or firsttime through) is another measurable. First time

yield does not count rework.

Finished

Product

85

Units

A100

Units

20 Units Rework

15 Units

Scrap

“Hidden

Factory”

Non-Value

Added

Activities

20 Units

65 Units

First Time Yield is Process Yield Without Rework…

Rework is the „Hidden Factory‟

Traditional yield

would be85/100 or 85%.

First-time yield

is only 65%


118/162



119/162


RightFirstTime

Rev. 1.1 1/99

Receive parts from Supplier

45,000 PPM wasted

46,652 PPM wasted

Following Receiving

Inspection and Line Fall-out...

From Machining Operations

At Test Stands on

first attempt

113,617 parts per millionwasted opportunities

21,965 PPM wasted

95.5% Yield

97% Yield

94.4% Yield

YRT = .955*.97*.94.4 = 87.4%


Traditional Yield View Exercise


120/162

120© Visteon Corporation BB Mod #2 Program Management Rev 3.1 9/03.Rev. 1.1 1/99

What’s the Yield?

_________%

100 units

85 units

Traditional Yield View—Exercise

Simple First Time Yield = Traditional Yield


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Simple First Time Yield = Traditional Yield

• What‘s the Yield?

• 1 x .9 x .97 x .98

• 85 %

100 documents

85 units

Scrap 10

Scrap 3

Scrap 2

100

90

87

100/100=1

90/100=.9

87/90=.97

85/87=.98

First Time Yield (YFT) Gives Credit Only for Product not scrapped the First Time



122/162


Product A is Produced in 3 Consecutive (Independent) Steps

Step 3

YFT = 90%

Step 1 Step 2

Rolled Throughput Yield is the probability that anygiven part will go through the process with zerodefects (no rework)

YFT = 80% YFT = 70%

A

Rolled Throughput Yield = Product of the First Pass Yields

YRT = (.80)(.70)(.90) = .504 = 50.4%



123/162



124/162


Verify

Rework

Scrap

Product

N s

Classic Yield =s

NRolled Throughput Yield = P(Op1)*P(Op2)….

= e -dpu

Operation

The Hidden Factory and Rolled Yield


Rolled Throughput Yieldcan also be stated in terms

of average defects per unit

(dpu)



125/162


99% 99% 98%

Op 1 OutputOp 2x =

Without Inspection or Test Without Inspection or Test Without Inspection or Test

Process Centered Process Centered

ORORY

RT= e – DPU

There are two operations (OPs) for a unit, and each OP has a 0.01probability of a defect. Therefore:

DPU = 2 OPs per unit x .01 defect per OP= .02 defects per unit

e-.02

= .98019or 98%


Complexity


126/162


# of Parts (Steps) ± 3 ± 4 ± 5 ± 61 93.32% 99.379% 99.9767% 99.99966

7 61.63 95.733 99.839 99.9976

10 50.08 93.96 99.768 99.9966

20 25.08 88.29 99.536 99.993240 6.29 77.94 99.074 99.9864

60 1.58 68.81 98.614 99.979680 0.40 60.75 98.156 99.9728

100 0.10 53.64 97.70 99.966150 -- 39.38 96.61 99.949

200 -- 28.77 95.45 99.932

300 -- 15.43 93.26 99.898400 -- 8.28 91.11 99.864

500 -- 4.44 89.02 99.830600 -- 2.38 86.97 99.796

700 -- 1.28 84.97 99.762

800 -- 0.69 83.02 99.729900 -- 0.37 81.11 99.695

1000 -- 0.20 79.24 99.661

1200 -- 0.06 75.88 99.593

3000 -- -- 50.15 98.98517000 -- -- 0.02 94.384

38000 -- -- -- 87.880

70000 -- -- -- 78.820

150000 -- -- -- 60.000

(Distribution Shifted ± 1.5 )

Process

Capability

Complexity(part count or process steps)

Rolled

Yield

Complexity

4 capability in each of 100 steps produces a rolled throughput yield of

.99379 to the 100th power = 0.5254 or 52.54%

Complexity


127/162


The GOAL is to REDUCE the number of opportunities

andconcurrently INCREASE the capability of each opportunity that remains.

NO!

Is Complexity better?

Complexity

Formulas to Know


128/162


DPU (Defects per unit) = Defects / Unit

TOP (Total Opportunities) = Units * opportunities

DPO (Defects per Opportunity) = Defects / TOPProbability the Opportunity is defective = DPO

Pr(ND) (Probability the Opportunity is Not defective)

Pr(ND) = 1 - DPO

Rolled Yield ] (The likelihood that any give unit of product willcontain 0 defects)

YRT= Pr(ND) # of OpportunitiesYRT = Pr(ND) * Pr(ND) * Pr(ND) *......Pr(ND) n

] Recommend using when you know the yield for each process element or opportunity

Formulas to Know

Calculating Transactional Yield


129/162


Assumption1 unit = 1 invoiceThe invoice has one process step that is CTQ for the customer :1) Total invoice price = customer’s PO price

• One opportunity for a defect exists for each invoice.• There is one opportunity for a defect in the invoicing process

To evaluate actual invoicing performance:

Process # of # # Opp YieldImpacting Units of of for for

CTQ Produced Defects Successes Process Process

Ready for Pick Up 500 35 465 1 0.93

YRT= .93




130/162


Assumptions:

1 Unit = 1 Invoice

Invoicing to the customer has 3 process steps that are CTQ’s for the customer 1) Correct Quantity2) Correct Price3) Invoice Received in a Timely Manner

• 3 opportunities for a defect exist for each invoice.• There is one opportunity for a defect in each process

To evaluate actual invoicing performance:

YRT = .93 x .7 x .85 = .55

Process # of # # Opp YieldImpacting Units of of for for

CTQ Produced Defects Successes Process Process

Correct Quantity 500 465 1 0.93Correct Price 500 350 1 0.7Invoice Received On Time 500 425 1 0.85

260

35150

75


Hidden Factory


131/162


Hidden Factory

What is it?• Unnecessary activities—We will cover more on this in the Lean Techniques Module.• Rework loops.

Learning Exercise

ABC Electric has demanded that the quality and timeliness of the invoices received into the Accounts PayableDepartment be at a 6 level. The CTQ elements for ABC Electric are an invoice where quantity and price matchthe PO that was submitted. Also, the invoice must be received by ABC Electric within 3 days of shipping forreceiving purposes.

On the following page are the steps that your invoicing department has put in place to reach the level ofquality the customer is expecting.

Hidden Factory

Hidden Factory


132/162


If no defects were in the system, which activities could be eliminated?

How can defects on upstream steps affect downstream steps?

Operator 1enters thequantity

shipped into thedata entrysystem

Operator 2inputs the priceof the Productinto the system

The confirmationis check by

Operator 2 foraccuracy

Was the inputaccurate?

Yes

No

Invoice

Operator 3inspects invoice

for accuracy

Operator 3mails invoice

to thecustomer

Customerinspects invoice

to determineaccuracy of theinvoice and if it

was received ontime.

Is pricecorrect?

Yes

No

Does invoicemeet CTQ'S?

Pay invoiceand chargea $10 penalty /

invoice

Pay invoice

Yes

No

confirmation

Hidden Factory

Hidden Factory


133/162


If no defects were in the system, which activities could be eliminated?

How can defects on upstream steps affect downstream steps?

Operator 1enters thequantity

shipped into thedata entrysystem

Operator 2inputs the priceof the Productinto the system

The confirmationis check by

Operator 2 foraccuracy

Was the inputaccurate?

Yes

No

Invoice

Operator 3inspects invoice

for accuracy

Operator 3mails invoice

to thecustomer

Customerinspects invoice

to determineaccuracy of theinvoice and if it

was received ontime.

Is pricecorrect?

Yes

No

Does invoicemeet CTQ'S?

Pay invoiceand chargea $10 penalty /

invoice

Pay invoice

Yes

No

confirmation

Hidden Factory

Defects and the Hidden Factory


134/162


Six Sigma is about identi fying/quantifyingand eliminating the H idden Factory through

Defect Reduction and ul timately Design for 6 Sigma

90% Yield After

Inspection or Test

Each defect must be detected, repaired and placed back

in the process. This results in Wasted:

• Time

• Money

• Resources• Floor Space

Scrap

Rework

Hidden Factory

NOT

OK

OperationInputs Inspect First TimeYield

OK90%

Customer Quality

Defects and the Hidden Factory

• This classic approach to

calculating yield does not account

for the hidden factory.

Described below is any process that meets a customer need

Rolled Throughput Yield Versus FirstTime Yield


135/162


Time Yield

66% 90%

... why not?

Scrap

90%Customer Quality

Rework

Hidden FactoryNOTOK

Yield AfterInspection or Test

OperationInputs Inspect First Time Yield =

OK

ProcessA B C

90% Yield

90% Yield

81 % 73 %

Final TestD

Using “final test (or first time) yield” ignores the hidden factory. Final test

performance is a function of inspection & test not actual defect data.

Rolled-Throughput Yield Classical First-Time Yield

90% Yield

66 %

90% Yield

Rolled Yield

Six Sigma, Metrics andContinuous Improvement


136/162


Continuous Improvement

• Six Sigma is characterized by: – defining critical business metrics

– tracking them

– improving them using proactive process improvement

• Six Sigma‘s primary metric is Defects per Unit,

which is directly related to Rolled Throughput

Yield (Yrt)

• Yrt = e-dpu

• Cost of Poor Quality and Cycle Time

(Throughput) are two other key metrics

Cost of Poor Quality


137/162


Cos o oo Qua y

Cost of (Poor) Quality


138/162


( ) Q y

• The ‗hidden factory‘ in any process is one key partof the cost of poor quality (COPQ).

– Direct cost to rework product.

– Cost of scrap (including disposal costs).

– Foregone throughput/productivity.

• Hidden factory costs are insidious — they become

part of the day-to-day business and go unnoticed.

• Tracking COPQ is not easy with traditional costaccounting systems.


139/162

What is Cost of Poor Quality?


140/162


y

• In addition to the direct costs associated with finding

and fixing defects, ―Cost of Poor Quality‖ also

includes:

– The hidden cost of failing to meet customer expectations

the first time

– The hidden opportunity for increased efficiency – The hidden potential for higher profits

– The hidden loss in market share

– The hidden increase in production cycle time

– The hidden labor associated with ordering replacement

material

• In almost every company, the COPQ exceeds the

profit margin.



141/162


• Direct Costs of Failures – Internal Failure

• Costs associated with defects found BEFORE the customer receives a product or service

– External Failure

• Costs associated with defects found AFTER the customer receives a product or service

• Costs to Control or Avoid Failures – Appraisal Cost

• Costs associated with measuring, evaluating, or auditing products orservices to assure conformance to quality requirements

– Prevention Cost• Costs of activities specifically designed to prevent poor quality



142/162


Optimum Quality Costs

Total

Quality

CostsInternal &

External

Failure Costs

Minimal COPQ

Quality of Conformance, %0

(100%

Defective)

Appraisal &

Prevention

Costs

100

(Zero

Defects)

C o s t p e r g o o d u n i t o f p r o

d u c t

The modern COPQ model depicted

here shows failure costs going to zero

at 100% conformance (zero defects).

Accordingly, appraisal and prevention

costs reach a maximum at zero defects.

The total COPQ is obtained by adding

the two sets of costs at a given

conformance level. Adapted fromJuran‘s Quality Control Handbook

(p.4.19), by J.M. Juran and F.M. Gryna

(Eds.), 1998, New York: McGray-Hill


143/162

“Traditional” Cost of Poor Quality


144/162


• Customer Returns

• Testing Costs

• Inspection Costs

• Scrap

• Quality Rejects

4% of Visteon’s

$18.5B in sales is

$740M

COPQ Data


145/162


Source: Journal for Tooling & Production. December, 1994.

Manufacturing IndustryPoor Quality Cost

asPercent of Sales

Median Profitsas

Percent of Sales

4.3 7.1 2.34.4 7.0 1.2

5.2 7.1 2.25.2 7.3 2.95.9 6.9 1.65.9 7.2 3.05.9 7.2 2.94.3 7.0 2.16.3 8.2 4.17.0 8.2 3.66.1 7.2 3.36.1 8.1 2.76.9 7.8 1.5

Aluminum Extruded Products

Steel Wire & Related Products

Aluminum Die Castings

Bolts, Nuts, Rivets & WashersFoundries: AluminumForgings-NonferrousMetal Stampings

Automotive StampingsElectronic ComponentsIndustrial MachineryMetal-Cutting Machine ToolsMotor Vehicles, PartsPlastics ProductsScrew Machine Products 5.3 7.0 2.2

Low High

Waste, Spoilage, Rejects & Rework As Percent of Sales


146/162

COPQ Example: Operational


147/162


• Air conditioningcondensers go

through an end-of-

line ‗mass

spectrometer‘ leak

check. Those thatfail are ‗bubble

tested‘ in a dunk

tank and repaired.

Mass Spec

TestingLeak? Pack and ship

No

Dunk tank test

to find the leak

Yes

Find Leak?

3rd try?

No

No

Yes

Scrap partYes

Hidden

Factory

Repair

COPQ Example: Operational (cont’d)


148/162


• COPQ includes: – Equipment, floor space, and personnel to dunk tank test.

– Equipment, floor space, and personnel to repair parts.

– Costs to move/handle the material, including costs of parts

damaged during movement.

– Costs of extra dunnage to accommodate parts in the hiddenfactory.

– Costs of scrap condensers (net of reclaimed aluminum.)

– Lost production due to retest, including overtime costs to meet

customer requirements.

– Other inventory carrying costs on product being reworked.

COPQ Example: Transactional


149/162


• A CPARS prototype tooling order has an incorrect buyer code. It is rejected back to the requisitioner,

who must then spend time correcting and

resubmitting.

– Costs include the time of the requisitioner, the buyer,and anyone else who handled the order.

– Other costs may accrue due to this delay:

• premium tooling funds to meet deadlines

• delayed incorporation of material cost savings• overtime to test the prototypes and still meet timing

Visteon Opportunities

Vi t 4% f


150/162

Visteon 4% ofsales = $740 Million

Visteon 20% of sales= $3.7 Billion

VISTEON CONFIDENTIAL


151/162

Visteon COPQ Estimates


152/162


• Items Related to QR‘s $97.3M

• Launch Costs $35M

• Stop Ships/Field Actions $22.4M

• Financial Controls $72.3M

• Supplier spills/premium freight $40.5M

• Plant QA $22.7M

• Scrap and Excess Inventory $71.8M

• Manufacturing Costs $361M

• Equipment and Process Quality $95M

• Other Support Costs $192M

Total

$1.01B

Cost Element Estimated Annual Cost



153/162


• Our material and equipment suppliers also have COPQ

• This COPQ is included in the prices Visteon pays formaterial and equipment

– This COPQ is not shown in the $1B estimate

• The $AVE program is our primary initiative for reducingCOPQ with our material and equipment suppliers

• The $AVE program target is to reduce our costs by $1.2 B

over three years



154/162


• Another major category of COPQ is presently being paid by our OEM customers, and by consumers …the costs of replacing defective products in the field

• A major portion of this category of COPQ iswarranty

– Our major customer – Ford – presently pays more than$400 Million per year in warranty costs associated withVisteon products



155/162


• If we add these three categories of COPQ, we can seea PARTIAL estimate of Visteon COPQ:

– Initial internal estimates: $1 Billion

– $AVE 3-year target: $1.2 Billion – Annual warranty (partial): $0.4 Billion

• Partial est. of Visteon COPQ: $2.6 Billion (14% ofsales)

• So … how do we get the money?

How is Visteon Attacking COPQ?


156/162


• Visteon Breakaway Improvement

• Aggregated improvement measures (COPQ, Revenue)

• Improvement Initiatives

• Tools, Techniques, Processes

Breakaway Objectives for COPQ


157/162


Cost of Poor Quality ~ $1850 M

2 0 0 2 P r o f i t s

V o l u m e / E x d h a n g e / P r i c i n g / E c o n .

I n f o r m a t i o n T e c h n o l o g y

N e w B u s i n e s s P r o b a b i l i t y

M a t e r i a l E f f i c i e n c i e s

M a n u f a c t u r i n g E f f i c i e n c i e s

R e s t r u c t u r i n g S a v i n g

A d m

i n , S e l l i n g , M k t , & E n g .

E f f i c i e n c i e s

P r i c e a b l e D e s i g n

P r o f i t a b i l i t y G a p

P r o g r e s s T o D a t e

COPQ Improvement Initiatives


158/162


• $AVE Program

• Manufacturing Efficiencies

• Engineering/A&S Efficiencies

• TVA/TVM

Note: $AVE, Manufacturing Efficiencies, and Engineering/A&S Efficiencies

are internally driven initiatives. TVA/TVM is externally driven.

How Does a COPQ Project Work?


159/162


• Scope/define the project

• Trace the usage of resources – Identify quality-related activities – Determine costs associated with those activities

• Identify/prioritize specific improvementopportunities

• Initiate improvement projects

• Keep track of improvement projects – Results as expected? – Further opportunities?



160/162


Does this method look familiar?

• Scope/define the project … Define

• Trace the usage of resources … Measure +

Value Stream Map

• Identify/prioritize opportunities … Analyze

• Initiate improvement projects … Improve

• Keep track … Control



161/162


• What tools and techniques can we use to get the money?

– In house: Lean Six Sigma (DMAIC)

– Supply base: $AVE

Cost of Poor Quality


162/162

• Any questions?

Documents

Mod 2 ProgMgmt COPQ Sep 03