Cairo 01 Six Sigma Measure
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Cairo 01 Six Sigma Measure

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Cairo 01 Six Sigma Measure

Cairo 01 Six Sigma Measure

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Cairo 01 Six Sigma Measure Cairo 01 Six Sigma Measure Presentation Transcript

  • SIX SIGMA MEASUREMENTS
    • METRICS ASSOCIATED WITH SIX SIGMA
    DON’T HAVE TO USE ALL THESE METRIC KNOWLEDGE IS NECESSARY TO INSURE GOOD COMMUNICATION
  • SIX SIGMA RELATIONSHIPS
  • DEFININITIONS Number Of Operation Steps = m Defects = D Unit = U Opportunities For A Defect = O Yield = 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 10 6 Basic Relationships
  • DISTRIBUTION OF DEFECTS IN MANUFACTURED PRODUCT Non-Randomly Occurring Defect Randomly Occurring Defect Example Result Conclusion Wrong part in manual insertion parts bin Every board contains same wrong part in same location. Defects are easier to detect/diagnose & are less likely to be sent to next operation Mixed parts in manual insertion parts bin. Probability of a board containing wrong part is equal to proportion of wrong parts in bin Defects are harder to detect/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
    Defects Per Unit = A Count of All Defects - Not a Measure of the Consequences of The Defect Number of Defects Found at Any Review Point Number of Units Processed Through That Review Point
  • 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 : Y TP = e -DPU Defects Per Unit: DPU = -ln(Y) Rolled Throughput Yield: Y RT =  Y TP i Total Defects Per Unit: TDPU = -ln(Y RT ) Normalized Yield: Y norm =  Defects per Normalized Unit: DPU norm = -ln(Y norm ) i =1 m Y RT m
  • YIELD Probability with zero defects Specification Limit Probability of Defect = 1 - e -DPU Yield = e -DPU Y = P(x = 0) = e-   x x! e-  = = e -D/U = e -DPU
  • THE HIDDEN FACTORY
  • THE HIDDEN FACTORY Customer Quality Operation Not OK Scrap Verify Rework Operation Not OK Scrap Verify Rework Y tp Throughput Yield Y tp Throughput Yield Non-Value Added (The Hidden Factory) Value Added (The Visible Factory) Producers Quality Supplier Quality Y rt =  Y tpi m i=10
  • IMPLICATIONS OF THE HIDDEN FACTORY Y e e TP 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 Process Throughput
  • #1 #6 #2 #7 #5 #9 #4 #8 #3 #10 Final Insp.. Process Yield is 93.17% - Right?????? A FACTORY WITH 10 PROCESSES NOT QUITE!!! 1200 Good & 88 Bad Items [93.2% Found OK]
  • COMPARISON OF YIELDS Y = S U = = .931677 S = Number of Units That Pass U = Number of Units Tested Throughput Yield Analysis Tells Us Y TPI = .47774 .4774  .932 Why Not??? 1200 1288
  • ROLLED THROUGHPUT YIELD Operation Defects Units DPU (D/U) Throughput Yield (Y TPI = e -D/U = e -DPU ) 1 2 3 4 5 6 7 8 9 10 5 75 18 72 6 28 82 70 35 88 523 851 334 1202 252 243 943 894 234 1200 0.00956 0.08813 0.05389 0.05990 0.02381 0.11523 0.08696 0.07830 0.14957 0.07333 0.99049 0.91564 0.94753 0.94186 0.97647 0.89116 0.91672 0.92469 0.86108 0.92929 Sum of Operation Steps = 479 6676 0.73868 0.47774 (Y RT ) 47.9 667.6 0.07387 0.73868 =TDPU TDPU = -ln(Y RT ) Avg.. of Operation 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
  • Inspection E = 0.8 Escaping DPU Level Observed DPU Level Submitted DPU Level QUANTIFICATION OF DEFECTS Given: Observed Defects = DPU of 0.25 Then: Submitted Defects = (0.25/0.8) = DPU of 0.31 Escaping Defects = (0.31-0.25) = DPU of 0.06
  • A TYPICAL PROCESS Operation 1 INSP TEST Prior Operation d o d 1 d 2 di dt dn Operation 2 d o = Defects Escaping Prior Process d 1 = Defects Added in Operation 1 d 2 = Defects Added in Operation 2 d i = Defects Found by Inspection & Corrected d t = Defects Found by Test and Corrected d n = Defects Escaping to Next Operation
  • Op. 1 INSP TEST Op 2 DPU = 0.349 DPU = 0.907 DPU = 0.420 DPU = 0.075 DPU = 0.07 From Prior Process DPU = 0.279 DPU = 0.558 DPU = 0.487 DPU = 0.345 PD = Part Defective = 800 PPM SC = Solder Defective = 300 PPM PA = Part Assembly Error = 1000 PPM EFF: PA = .8 PD = .0 SC = .8 EFF: PA = .9 PD = .95 SC = .05 To Next Process DEFECT-BASED PLANNING FOR A TYPICAL PROCESS
  • DEFECT-BASED PLANNING FOR A TYPICAL PROCESS Total Observed Defects Per Unit = 0.832 Rolled Yield = 43.52% 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
  • 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 Moderate Slight Degree of Abnormality 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 OF OTHER 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 C p & C pk 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 PROCESS Structured 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 COMPARISONS IN 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 Count DPMO = DPU x 1,000,000 Average Opportunities For Error in One Unit
  • BENCHMARKING THE PROCESS 1. 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
  • BENCHMARKING AGAINST BEST IN CLASS 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 Sigma Cpk DPMO Specification Width  1.5 Sigma Shift Centered
  • PROCESS CYCLE TIME
  • PROCESS CYCLE TIME Time To Do An Entire Process Could Be Time From Material Arriving To 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 1 YIELD 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 Unit T = (1+DPU) Test Cycle Time = T x Time Per Test = (1 + DPU) x Time Per Test
  • EXAMPLE OF TEST WORK LOAD FOR REPAIRABLE PRODUCT 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 Number Of Tests Number Of Units Number Of Defects Total Number Of Tests 1 2 3 4 187 35 9 3 234 0 35 18 9 62 187 70 27 12 296 Totals
  • 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 LOAD Every 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 = DPU Repair 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 Repair or = 1- e -DPU x Time Per Repair
  • Stores and Bank 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 = Number of Days = Inventory Throughput Rate Total # of Pieces in Stock Pieces Processed Per day
  • UNIVERSAL CYCLE TIME/ INVENTORY RELATIONSHIP Work -In-process Inventory = 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
  • Raw Material Stores Fabrication Fabrication Stores Sub- Assembly Sub- Assembly Stores Final Assembly Finished Goods Inventory Work-In-Process Manufacturing Cycle Time Inventory Carrying Cost Total Value of Product Raw Materials Acquisition Finished Goods Shipment TRUE COST OF INVENTORY $
  • CYCLE TIME AND INVENTORY TURNS An Inventory Turn Is An Index Or Indicator Of How Often A Company’s Entire Inventory Stock Is Replaced Cost Of Goods Sold Average Annual Inventory Turns = 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 to Achieve 1440 Turns
  • INVENTORY COST AT MOTOROLA Inventory Carrying Cost Equals At Least 25% of the Value of Average Annual Inventory Levels In 1991, This cost can be compared to taking $310 million and burning it , since it adds no value to the company Year Sales Average Annual Inventory Levels Inventory Carrying Costs 1987 1989 1991 1995 $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
  • THE BOTTOM LINE IMPACT In 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