SIX SIGMA MEASUREMENTS

METRICS ASSOCIATED WITH SIX SIGMA

DON’T HAVE TO USE ALL THESE METRIC

KNOWLEDGE IS NECESSARY TO INSURE GOOD COMMUNICATION

DON’T HAVE TO USE ALL THESE METRIC

KNOWLEDGE IS NECESSARY TO INSURE GOOD COMMUNICATION

SIX SIGMA RELATIONSHIPS

SIX SIGMA RELATIONSHIPS

DEFININITIONS

Number Of Operation Steps = m Defects = DUnit = UOpportunities For A Defect = OYield = Y

Nomenclature

Total Opportunities (TOP) = U X O

Defects Per Unit (DPU) = D/U

Defects Per Unit Opportunity (DPO) = DPU/O = D/(U X O)

Defects Per Million Opportunities (DPMO) = DPO X 106

Basic Relationships

DISTRIBUTION OF DEFECTS IN MANUFACTURED PRODUCT

Non-Randomly Occurring Defect

Randomly Occurring Defect

Example

Result

Conclusion

Wrong part in manualinsertion parts bin

Every board contains same wrong part insame location.

Defects are easier todetect/diagnose & are less likely to be sent tonext operation

Mixed parts in manualinsertion parts bin.

Probability of a boardcontaining wrong part isequal to proportion of wrong parts in bin

Defects are harder todetect/diagnose and are more likely to be sent to next operation

PROBABILITY OF DEFECTS IN MANUFACTURED PRODUCT

GIVEN: (1) Average DPU For This Product Is One (DPU=1)(2) Defects are randomly distributed

What Is The Probability of Zero Defects In This Unit?

GENERAL CHARACTERISTICS OF DPU

Directly Proportional To:• Parts Count • Lines of code• Die Area

Product Built In Plant Having Best Process Controls Will Have The Lowest Defects Per Unit Level

DEFECTS PER UNIT

Defect Prevents Product From Fulfilling Physical And Functional Requirements Of The Customer

A Unit• A Measure Of Volume Of Output• Observable & Countable• Is An Individual Measurement - Not An Average

Number of Defects Found at Any Review Point

Number of Units Processed Through That Review PointDefects Per Unit =

A Count of All Defects - Not a Measure of the Consequences of The Defect

PREDICTING THE QUALITY OF PRODUCTS

FTY = e-DPU DPU = -ln FTY

Estimation of units containing “q’ defects (where “q” is not limited to zero)

If we use the Poisson distribution

Becomes

(DPU)q e -DPU

q!

x e-

x!P{x} =

YIELD RELATIONSHIPS

Throughput Yield : YTP = e-DPU

Defects Per Unit: DPU = -ln(Y)

Rolled Throughput Yield: YRT = YTPi

Total Defects Per Unit: TDPU = -ln(YRT)

Normalized Yield: Ynorm =

Defects per Normalized Unit: DPUnorm = -ln(Ynorm)

i=1

mYRT

m

YIELDProbability with zero defects

Y = P(x = 0) =e-x

x! e-= = e-D/U = e-DPU

Specification Limit

Probability of Defect= 1 - e-DPU

Yield = e-DPU

THE HIDDEN FACTORYTHE HIDDEN FACTORY

THE HIDDEN FACTORY

Customer Quality

Operation

NotOK

Scrap

Verify

Rework

Operation

NotOK

Scrap

Verify

Rework

Ytp

Throughput Yield

Ytp

Throughput Yield

Non-Value Added(The Hidden Factory)

Value Added(The Visible Factory)

ProducersQuality

Supplier Quality

Yrt = Ytpi

m

i=10

IMPLICATIONS OF THE HIDDEN FACTORY

Y e eTP

DPU 3679

1 0 . = 1.6321

.

1.63 Equivalent Units Must Be Produced To Get Out 1 Good Unit

Every Occurrence Of A Defect Within The Manufacturing Process Requires Time To Verify, Analyze, Repair And Re-verify

Average Cycle Time Per Unit Is Directly Proportional To The Total Number Of Defects Per Unit

Cycle Time = Work In ProcessThroughput

#1

#6

#2

#7

#5

#9

#4

#8

#3

#10Final Insp..

Process Yield is 93.17% - Right??????

A FACTORY WITH 10 PROCESSES

NOT QUITE!!!

1200 Good & 88 BadItems [93.2%Found OK]

COMPARISON OF YIELDS

Y = SU

= 12001288

= .931677

S = Number of Units That PassU = Number of Units Tested

Throughput Yield Analysis Tells UsYTPI = .47774

.4774 .932Why Not???

ROLLED THROUGHPUT YIELD

Operation Defects UnitsDPU(D/U)

Throughput Yield(YTPI = e-D/U = e-DPU)

12345678910

57518726

2882703588

523851334

1202252243943894234

1200

0.009560.088130.053890.059900.023810.115230.086960.078300.149570.07333

0.990490.915640.947530.941860.976470.891160.916720.924690.861080.92929

Sum of

Operation Steps = 479 6676 0.73868 0.47774 (YRT)

47.9 667.6 0.07387 0.73868 =TDPU TDPU = -ln(YRT)

Avg.. ofOperation Steps

=

HOW MANY UNITS PRODUCED?

UNITS PRODUCED = 1 + (1 - e-DPU)

= 1 + (1 - e-73868)= 1 + (1 - .477744)= 1 + .5222558= 1.52

To Achieve 100 Conforming Units, 152 ( 1.52 X 100) Would Need To Be Produced

INSPECTIONS AND

TESTS

INSPECTIONS AND

TESTS

Inspection E = 0.8 Escaping

DPU Level

Observed DPU Level

Submitted DPU Level

QUANTIFICATION OF DEFECTS

Given: Observed Defects = DPU of 0.25Then: Submitted Defects = (0.25/0.8) = DPU of 0.31

Escaping Defects = (0.31-0.25) = DPU of 0.06

A TYPICAL PROCESS

Operation1

INSP TESTPrior Operation

do

d1d2 di dt

dn

Operation2

do = Defects Escaping Prior Processd1 = Defects Added in Operation 1d2 = Defects Added in Operation 2di = Defects Found by Inspection & Correcteddt = Defects Found by Test and Correcteddn = Defects Escaping to Next Operation

Op.1 INSP TESTOp

2

DPU = 0.349 DPU = 0.907 DPU = 0.420 DPU = 0.075DPU = 0.07

FromPriorProcess

DPU = 0.279 DPU = 0.558 DPU = 0.487 DPU = 0.345

PD = Part Defective = 800 PPMSC = Solder Defective = 300 PPMPA = Part Assembly Error = 1000 PPM

EFF: PA = .8 PD = .0 SC = .8

EFF: PA = .9 PD = .95 SC = .05

ToNextProcess

DEFECT-BASED PLANNINGFOR A TYPICAL PROCESS

Operation 1 Operation 2 Inspection Test

# Parts

PD

PA

SC

TDU

FTY

Submitted Created Submitted Created Submitted Observed Submitted Observed

100

330

0.080

0.100

0.099

0.279

210

600

0.05

0.02

0.00

0.07

0.13

0.120

0.099

0.349

0.168

0.210

0.180

0.558

0.298

0.330

0.279

0.907

0.000

0.264

0.223

0.487

61.44%

0.298

0.066

0.056

0.420

0.283

0.059

0.003

0.345

70.82%

Escaping

0.015

0.007

0.053

0.075

DEFECT-BASED PLANNINGFOR A TYPICAL PROCESS

Total Observed Defects Per Unit = 0.832

Rolled Yield = 43.52%

REVIEW THE THREE PHASES OF PRODUCT LIFE

Early Life Useful Life Wear Out

PROB. OF FAILURE

EARLY LIFE FAILURE RATE & LATENT DEFECTS

•The Degree Of Abnormality

•The Magnitude Of Applied Stress

•The Duration Of Applied Stress

PROBABLE TIME TO FAILURE & DEGREE OF ABNORMALITY

Gross

ModerateSlightD

egre

e of

Ab

nor

mal

ity

Probable Time to Cause Failure

MAJOR CAUSE OF LATENT DEFECTS

The Design And Execution Of Processes Is The Major Cause of Latent Defects

Cpk - Major Cause of the Degree of Abnormality

The Average Number of Latent Defects Per Unit - Determines the Shape of Early Life Failure Rate Curve

WHAT CAN BE DONE TO MINIMIZE LATENT DEFECTS?

•Design & Document Detailed Product Flow

•Use Simplest Operations Possible

•Use Operations Of Known Capability

•Provide Documented Information - Each Operation

•Perform Each Operation Identically

MINIMIZATION OFOTHER FAILURE TYPES?

• Use The Fewest Number Of Parts

• Use Parts Of Known Capability

• Use Lowest Possible Stress Levels

• Avoid Marginal Overstress

• Provide Maximum Possible Operating Margins And Mechanical Tolerances

TOTAL DEFECTS PER UNIT (TDU)

TDU = DPU For All Key Characteristics

+ DPU For All Component Part Defects

+ DPU For All Process Defects

WHERE:Key Characteristics Are Measured By Cp & Cpk

Component Part Defects Are Defects Found In Material From Suppliers.

Process Defects - Created As Product Goes Through Manufacturing Process

DPU AS A DEFECT BUDGETING TOOL

Defect Budgeting Is A Method To Ensure Robust Product Design

A Robust Design Will Be Insensitive To Long-term Process & Material Variations & Product Misuse

Goal Of Defect Budgeting For A New Product/process Is A Lower TDU

OVERVIEW OF THE DEFECT BUDGETING PROCESSStructured Process Flow Diagram

- All Inspection /Test Points- Flow Of Non-conforming Product

Determine Capability/Error Rate Of Existing Process

Establish Maximum DPU Level For Delivered Product

Work Backwards & Establish Maximum DPUs

Revise Process Flow - Add Inspection/test Points- Redesign Operations

BENCHMARKING AGAINST BEST IN CLASS

Determine Factors Critical To Long-term Success

Compare Performance With Toughest Competitors Or Others

Use Info to Develop Strategies And Functional Standards

The Goal Of Benchmarking Is:- Exceed The Competition- Make Your Business The Very Best

EQUIVALENT COMPARISONSIN BENCHMARKING

DPU Must Be Normalized - Provide An Equivalent Comparison

Opportunities For Error - Accounts For Varying Complexity

Examples Of Opportunities For Error• Part Count• Process Step Count• Total Part Count And Process Step Count• Lines Of Solder Per Square Inch Of Printed Circuit Board

Defects Per Million Opportunities (DPMO) Provides An Equivalent Comparison

DEFECTS PER MILLION OPPORTUNITIES (DPMO)

DPMO = DPU x 1,000,000

Parts CountDPMO =

DPU x 1,000,000Average Opportunities For Error in One Unit

BENCHMARKING THE PROCESS1. Define The Purpose Of The Project

2. Define A Unit

3. Determine The Number Of Opportunities For Error In A Unit

- For Manufactured Product- Other Than Manufactured Product

4. Measure The Total Defects Per Unit (DPU)

5. Calculate Defects Per Million Opportunities (DPMO)

6. Estimate Specification Width

0.001

0.01

0.1

1

10

100

1K

10K

100K

2 3 4 5 6 7.67 1.0 1.33 1.67 2.0

SigmaCpk

DP

MO

Specification Width

BENCHMARKING AGAINST BEST IN CLASS

1.5 Sigma Shift

Centered

PROCESS CYCLE TIMEPROCESS CYCLE TIME

PROCESS CYCLE TIME

Time To Do An Entire Process

Could Be Time From Material ArrivingTo Final Product In Customer’s Hands

Real Process Time Includes Waiting,Storage & In-between Operations Times.

Theoretical Process Time - No Waiting

Compare Real And Theoretical Process Time

Reducing Real Process Cycle Time Can Reduce The Number Of Defective Units And Improve Process Performance

THEORETICAL PROCESS CYCLE TIME

Theoretical Process Cycle Time

Real Daily Operating Time

Number of Units Required Daily

=

TEST/INSPECTION WORK LOAD NONREPARIABLE PRODUCT

Failure Of Test Or Inspection - Item Must Be Scrapped. To Yield One Acceptable Unit, We Must Test/inspect A Larger Number Of Units

1YIELD

Starts =

= e DPU

The Number of Parts Required to Start (T)

= e -DPU

1

Test/Inspection Cycle Time = T x Time Per Test= (e DPU) x time Per Test

TEST WORK LOAD REPARIABLE PRODUCT

One Test Required To Catch Each Defect

Another Test Required To Pass Repaired Unit

T = Number Of Tests Per Accepted UnitT = (1+DPU)

Test Cycle Time = T x Time Per Test= (1 + DPU) x Time Per Test

EXAMPLE OF TEST WORK LOADFOR REPAIRABLE PRODUCT

NumberOf Tests

Number Of Units

NumberOf Defects

Total NumberOf Tests

1

2

3

4

187

35

9

3

234

0

35

18

9

62

187

70

27

12

296Totals

FTY = 187/234 = 80%TDU = 62/234 = 0.265

FTY = e-0.265 = 77%TDU = -ln(0.80) = 0.233

Observed Results: Estimated From Observations

Results Close Enough - Defects Randomly Distributed & Poisson Ok

INSPECTION WORK LOAD REPARIABLE PRODUCT

• Inspection Is Capable Of Finding All Defects In An Assembly On One Pass

• Each Assembly Must Be Inspected The First Time

• All Assemblies Containing One Or More Defects Must Be Inspected A Second Time

I = Number Of Inspections Per Accepted Unit

I = 1 + (1 - Yield) = 1 + ( 1- e -DPU) = 2 - e -DPU

Inspection cycle time = I x Time Per Inspection= (2 - e -DPU) x Time Per Inspection

ANALYZE WORK LOADEvery Unit Must Be Worked On

Work Load Is Directly Proportional To Defects Per Unit

Failures Analyzed One At A Time:Work Load = A = Analysis Per Unit = DPU

Analysis Cycle Time = A x Time Per Analysis

= DPU X Time Per Analysis

Failures Analyzed All At One Time:Work Load = A = 1 - e -DPU

Analysis cycle Time = A x Time Per Analysis

= 1 - e -DPU x time Per Analysis

REPAIR WORK LOAD

For Failures Repaired One At a Time:Work Load = R = Repair Per Unit = DPURepair Cycle time = R x Time Per Repair = DPU x Time Per Repair

For Failure Repaired All At One Time:Work Load = R = 1 - e -DPU

Repair Cycle time = R x Time Per Repair = 1 - e -DPU x Time Per Repair

CYCLE TIME

Test

Analysis

Repair

Test Cycle Time = (1+ DPU ) x Time Per Test

Analysis Cycle Time= DPU x Time Per Analysis

or = 1- e-DPU x Time Per Analysis

Repair Time= DPU x Time Per Repairor= 1- e-DPU x Time Per Repair

StoresandBank Time Total Mfg..

Cycle Time

Queue, Run, and Changeover Time

Output(capacity)

Input

CYCLE TIME / INVENTORY RELATIONSHIPS

INVENTORY AND THROUGHPUT RATE

How Are Inventory and Cycle Time Related?

Cycle Time = Inventory

Throughput Rate

Number of Days = Total # of Pieces in Stock

Pieces Processed Per day

UNIVERSAL CYCLE TIME/INVENTORY RELATIONSHIP

Work -In-processInventory

= Throughput Rate x Cycle Time

Number Of Work Orders

Number Of Pieces

Dollars

Orders Per Day

Pieces Per Day

Dollars Per Day

# Of Days

# Of Days

# Of Days

Inventory Throughput Rate Cycle Time

Examples

CYCLE TIME CONTROLS WIP INVENTORY

The Longer A Unit Of Product Remains In A Production Area, The Higher The Work In Process (WIP) Inventory Is In That Area

Time Unit Is Expected To Remain In Area = Average Total Time

RawMaterial

StoresFabrication

FabricationStores

Sub-Assembly

Sub-Assembly

Stores

FinalAssembly

FinishedGoods

Inventory

Work-In-ProcessManufacturing Cycle Time

InventoryCarryingCost

TotalValue of Product

RawMaterials Acquisition

Finished GoodsShipment

TRUE COST OF INVENTORY$

CYCLE TIME AND INVENTORY TURNS

An Inventory Turn Is An Index Or Indicator Of HowOften A Company’s Entire Inventory Stock Is Replaced

Cost Of Goods Sold

Average Annual InventoryTurns =

1.25

1.44

5.00

10.00

20.00

40.00

288

250

72

36

18

9

Turns Turn Days

Note:Dell Has been Able toAchieve 1440 Turns

INVENTORY COST AT MOTOROLA

Inventory Carrying Cost Equals At Least 25% of the Value of Average Annual Inventory Levels

Year Sales Average AnnualInventory Levels

InventoryCarrying Costs

1987198919911995

$6,700,000,000$9,620,000,000$11,340,000,000$27,037,000,000

$909,000,000$1,158,000,000$1,242,000,000$3,528,000,000

$227,000,000$289,000,000$310,500,000$882,000,000

In 1991, This cost can becompared to taking $310 million and burning it , since it adds no value to the company

THE BOTTOM LINE IMPACTIn 1987, Motorola Discovered That

Poor Quality Accounted For Approximately

25% Of Their Annual Inventory Carrying Costs.

This Expense Added No Value, It Was Just Like Taking

$250,000,000 and pouring it down a drain

….Annually!

OVERALL MANUFACTURING COST AND DPU

Defects Cost Money as Result Of:

Diagnostic Time

Repair Time / Re-Inspection

Extra Labor

Extra Materials

Extra Capacity

Extra Support

Extra Management

Extra Inventory

Cost of Failure (Internal) =

DPU x Volume x Average Cost

Defect

REDUCING CYCLE TIME & INVENTORY SUMMARY

Cycle Time & Inventory Are Key Competitive Factors

Cycle Time Is The Inverse Of Inventory Turns (Shorter Cycle Times = Lower Inventories)

Defects Have A Major Controlling Effect

Defect Must Be Analyzed And Repaired

Two Important Work Load Measures - Test & Inspection, And Analyze & Repair

Capacity Of Test / Inspection Equipment Is Inversely Proportional To Work Load