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					       South Dakota

School of Mines & Technology

      Introduction to
  Probability & Statistics

   Industrial Engineering
       South Dakota

School of Mines & Technology

   Applications in Quality

   Industrial Engineering
       South Dakota

School of Mines & Technology

    Acceptance Sampling

   Industrial Engineering
Quality Methods
                                           Quality Engineering Methods

                         100
Percent of Application




                         75
                                       Acceptance
                                        Sampling                         Process Control
                         50



                         25

                                                                 Design of Experiments
                          0
                               0   1        2         3          4         5       6       7
                                                          Time
              Quality Design & Process Variation
                      Lower Spec         Upper Spec
                        Limit             Limit
Acceptance
Sampling


                 60    80          100        120     140
Statistical
Process
Control

                 60                                    140
Experimental
Design

                 60                                   140
    Acceptance Sampling
Suppose we buy parts in lot sizes of 5,000 and
we wish to inspect to ensure that the incoming
quality is sufficient. Obviously, we would not
like to do 100% inspection.

Idea: Select a portion, n, of the lot at random
and determine the number of defects in the
sample. If the number of defects, d, surpasses
some critical value, c, assume the entire lot is
defective and ship it back.
      Example
Let   N = lot size = 5,000
      n = sample size = 50
      c = critical value = 1
      d = number of defectives found in sample
           size n

If    d  1 , accept the entire lot
      d  2 , reject the entire lot
    Theory
We assume that each part in the lot is either
defective or “successful”. The probability of a
defective part is constant with at some
proportionate rate p. In our example we assume
that an acceptable p is given by

          c 1
      pa      .02  2%
          n 50

We call this the Acceptable Quality Level or AQL
    Theory (cont.)
Now, if p is constant from trial to trial in a
sample of n, then the probability of accepting a
lot given that the lot has acceptable quality level
is given by

      Pa = P{d  c}
              c
      Pa     n  p d (1 p ) n  d
                            
           d  0 d 
   Theory (cont.)
Then,
           1
   Pa     50 .02 d (.98) 50 d
        d  0 d 

          50!  .02 0 (.98) 50  50!  .02 1 (.98) 49
                                   
          50!0!                49 !1!

        = .7358
 Risks
                   Decision
              Ho          HA
               no
        Ho    error           
Truth




                           no
        HA               error




  P{reject lot | lot is good}

  P{accept lot | lot is bad}
    Risks (cont.)
    P{reject lot | lot is good}
    =1 - P(accept lot | lot is good}
    = 1 - Pa
    = 1 - .7358
    = .2642

     P{accept lot | lot is bad}

Problem: What constitutes a bad lot?
    Computing 
Let’s suppose that we consider a lot with 4%
(p=.04) to be a bad lot. This is Lot Tolerance
Percent Defective, LTPD. Then

     P{accept lot | lot is bad}

      =P{d < c | pb }
          1
              50 .04 d (.96) 50 d
        
        d  0 d 
    Computing 
Then

     50!  .04 0 (.96) 50  50!  .04 1 (.96) 49
                             
     50!0!                49 !1!

  = .4005

Suppose we wish to evaluate  for a number of
different LTPD values (pb).
                Operating Characteristic

                                 Operating Characteristic Curve

                1.0
                0.9
                0.8
                0.7
P{accept lot}




                0.6
                0.5
                0.4
                0.3
                0.2
                0.1
                0.0
                   0.00   0.02           0.04           0.06      0.08   0.10
                                                 Pb
   Acceptance Sampling
 Uses
   when   testing is destructive
   cost of 100% inspection is high
   inspection error rate is high
   cost of passing limited number of defective
    products is low
   the process is stable
    Acceptance Sampling
 Limitations
   does  not estimate lot quality
   does not provide for quality control
   risks associated with accepting bad lots and
    rejecting good lots
 Today’s   Trends
   high quality
   small lots


       FRO’s find Acceptance Sampling
              of little use
       South Dakota
School of Mines & Technology

     Introductory SPC
    The “Magnificent 7”

   Industrial Engineering
              Quality Design & Process Variation
                      Lower Spec         Upper Spec
                        Limit             Limit
Acceptance
Sampling


                 60    80          100        120     140
Statistical
Process
Control

                 60                                    140
Experimental
Design

                 60                                   140
   Statistical Process Control
                    u1 u2 u 3 u4




     Input                             Output



                     x1 x2 x3   x4



ui = random noise, unassignable causes of variation

xi = assignable causes of variation
  Statistical Process Control
                u1 u2 u3 u4




    Input                            Output



                 x1 x2 x3   x4


Objective: remove all assignable causes of
variation

      smallest variation in output
Deming Wheel
       P



  A            D



       C
 The Magnificent “7”
Problem Solving     Useful Tools
  Step

Understand Mess     Flow Charts
Find Facts          Check Sheets
Identify Problems   Pareto Diagrams
                    Histograms
Generate Ideas      Cause-and-Effect
Develop Solutions   Scatter Diagrams
Implementation      Control Charts
Problem
Identification/Analysis
Problem Identification       Problem Analysis

      Flow Chart                     Histogram
                    Pareto Chart Scatter Diagram
 Check Sheet
                   Cause & Effect     Process Capability
Nominal Group
 Technique          Stratification
                                       Control Chart

  Brainstorming     Run Chart
                                     Force Field
                                     Analysis
      Flow Charting

Mix Dry       Mix Wet                Mix Wet    Fold 10
Ingredients   Ingredients            & Dry      Minutes



              Extrude                Fold 2     Add Chips
Cut Dough
              Dough                  Minutes



                            OK
Bake 10          Qual.               Pack and
Minutes                              Ship

                        Poor
                                 Feed to Hogs
Check Sheet; Cookies

 Type           Check   Total

Burnt                    11

Crumbly                   5
Too Few Chips            14

Poor Taste                6

Other                     3
            Histograms
                         Defects by Category


            15




            10
Frequency




             5




             0
                 Burnt   Crumbly     Few Chips   Poor Taste   Other


                                   Type Defect
    Pareto Chart
                               Pareto Analysis

             1.0

             0.8
Percentage




             0.6

             0.4

             0.2

             0.0
                   Too Few   Burnt   Poor Taste   Crumbly   Other
                    Chips
                                      Category
   Histogram; Class Problem
 Now that we’ve decided that too few chips is
 our quality problem. We need to determine
 how many chips are in each cookie. Count
 your chips.
    Histogram; Class Problem
                     Chips per Cookie

            8
Frequency




            6
            4
            2
            0
                10   12    14     16    18   20
                           No. Chips
 Cause & Effect
              Mix Chips
Automate       in Dry

                Fold Longer
                                        Too Few
                                         Chips
      Place Chips on      Drop Liquid
      by hand              Chips
  Cause & Effect
People        Methods       Handling




                                       Effect




     Design
                    Tools
Idea; Cookies
    As a first step, let’s try
       Increasing
      Folding Time
Scatter Diagram
                        Chips vs Fold Time
          25



          20
# Chips




          15



          10



          5
               5   10          15            20   25


                             Fold Time
New Idea; Cookies

   Try Mixing Chips in
   With Dry Ingredients
      Flow Charting

Mix Dry                            Mix Wet       Mix Wet
              Add Chips
Ingredients                        Ingredients   & Dry



              Extrude              Fold 2        Fold 10
Cut Dough                                        Minutes
              Dough                Minutes



                          OK
Bake 10          Qual.             Pack and
Minutes                            Ship

                        Poor
                               Feed to Hogs
     Evaluate; Control Chart
 Avg Chips

18




15                         Time




12
       South Dakota

School of Mines & Technology

       Control Charts

   Industrial Engineering
  Control Charts


                         
We assume that the underlying distribution is normal
with some mean  and some constant but unknown
standard deviation .
Let          n xi
         x
           i 1 n
Distribution of x
Recall that x is a function of random variables,
so it also is a random variable with its own
distribution. By the central limit theorem, we
know that

         x  N (  , x )

where,

               x
      x 
                  n
Control Charts


                                 x


                



                                 x


      x           x
Control Charts

                                UCL
     x


       


     x

                                LCL


   UCL & LCL Set at 3 x
   Problem: How do we estimate  &  ?
Control Charts
     m
     x     i
x   i 1
                
       m
     m
     R
R   i 1        f ( )  
       m
Control Charts
     m
     x     i
x   i 1
                
       m
     m
     R
R   i 1        f ( )  
       m

  UCLx  x  A2 R              UCLR  D4 R
  LCLx  x  A2 R              LCLR  D3 R
   Example
 Suppose specialized o-rings are to be
 manufactured at .5 inches. Too big and they
 won’t provide the necessary seal. Too little
 and they won’t fit on the shaft. Twenty
 samples of 2 rings each are taken. Results
 follow.
Control Charts
Part   Measurements                    R Chart      X Chart
No.      1      2        x       R     UCL     R    UCL       LCL      Xbar
  1    0.502 0.504    0.503    0.002   0.0077 0.002 0.5052    0.4964    0.503
  2    0.495 0.497    0.496    0.002   0.0077 0.002 0.5052    0.4964    0.496
  3    0.492 0.496    0.494    0.004   0.0077 0.004 0.5052    0.4964    0.494
  4    0.501 0.498    0.500    0.003   0.0077 0.003 0.5052    0.4964    0.500
  5    0.507 0.508    0.508    0.001   0.0077 0.001 0.5052    0.4964    0.508
  6    0.504 0.504    0.504    0.000   0.0077 0.000 0.5052    0.4964    0.504
  7    0.497 0.496    0.497    0.001   0.0077 0.001 0.5052    0.4964    0.497
  8    0.493 0.496    0.495    0.003   0.0077 0.003 0.5052    0.4964    0.495
  9    0.502 0.501    0.502    0.001   0.0077 0.001 0.5052    0.4964    0.502
 10    0.498 0.500    0.499    0.002   0.0077 0.002 0.5052    0.4964    0.499
 11    0.505 0.507    0.506    0.002   0.0077 0.002 0.5052    0.4964    0.506
 12    0.502 0.499    0.501    0.003   0.0077 0.003 0.5052    0.4964    0.501
 13    0.495 0.497    0.496    0.002   0.0077 0.002 0.5052    0.4964    0.496
 14    0.499 0.496    0.498    0.003   0.0077 0.003 0.5052    0.4964    0.498
 15    0.503 0.507    0.505    0.004   0.0077 0.004 0.5052    0.4964    0.505
 16    0.507 0.509    0.508    0.002   0.0077 0.002 0.5052    0.4964    0.508
 17    0.503 0.501    0.502    0.002   0.0077 0.002 0.5052    0.4964    0.502
 18    0.497 0.493    0.495    0.004   0.0077 0.004 0.5052    0.4964    0.495
 19    0.504 0.508    0.506    0.004   0.0077 0.004 0.5052    0.4964    0.506
 20    0.505 0.503    0.504    0.002   0.0077 0.002 0.5052    0.4964    0.504
              Avg =   0.501    0.002
                         x       R
             Std. =   0.0047
     Control Charts
                                         X-Bar Chart

     0.510


     0.505
 x




     0.500


     0.495


     0.490
             1   2   3   4   5   6   7    8   9   10 11 12 13 14 15 16 17 18 19 20




Xbar charts can identify special causes of
variation, but they are only useful if the process
is stable (common cause variation).
Control Limits for Range
UCL = D4R = 3.268*.002 = .0065
LCL = D3 R = 0

                                             R Chart

         0.010

         0.008
 Range




         0.006

         0.004

         0.002

         0.000
                 1   2   3   4   5   6   7   8   9   10 11 12 13 14 15 16 17 18 19 20
                                                 Observation
    The P-Chart
        UCL  p  3 p
        LCL  p  3 p
where
             m
              di               p (1  p )
        p   i 1        p 
              nm                    n
   Example; P-Chart
 Operators of a sorting machine must read the
 zip code on a letter and diver the letter to the
 proper carrier route. Over one month’s time,
 25 samples of 100 letters were chosen, and the
 number of errors was recorded. Error counts
 for each of the 25 days follows.
Example
Day   Error   %Defect   Day   Error   %Defect
  1     3      0.03      14     2      0.02
  2     1      0.01      15     3      0.03
  3     0      0.00      16     3      0.03
  4     0      0.00      17     2      0.02
  5     2      0.02      18     0      0.00
  6     3      0.03      19     1      0.01
  7     5      0.05      20     5      0.05
  8     2      0.02      21     4      0.04
  9     4      0.04      22     3      0.03
 10     3      0.03      23     2      0.02
 11     0      0.00      24     4      0.04
 12     1      0.01      25     3      0.03
 13     4      0.04
Example

p .03.01.00. . . .03
            25
  Example

 p .03.01.00. . . .03
             25
              p (1  p )
UCL p  p  3
                  n
                 .024(1  .024)
       .024  3
                      100
       0.070
  Example

 p .03.01.00. . . .03
             25
              p (1  p )
UCL p  p  3
                  n
                 .024(1  .024)
       .024  3
                      100
       0.070
                  p (1  p )
LCL p  p  3                  .022  0.0
                      n
  Example; P-Chart
                     P-Chart; Mail Sort

           0.08
           0.06
% Errors




           0.04
           0.02
           0.00
           -0.02 0        10              20   30
                                Day
       South Dakota

School of Mines & Technology

     Process Capability

   Industrial Engineering
Process Capability
                        Precise but not Accurate

            7
            6
            5
Frequency




            4
            3
            2
            1
            0
                1   2      3     4    5      6     7   8   9
                                     Value
Process Capability
                        Accurate but not Precise

            7
            6
            5
Frequency




            4
            3
            2
            1
            0
                1   2      3     4    5      6     7   8   9
                                     Value
Process Capability
                        Precise and Accurate

            7
            6
            5
Frequency




            4
            3
            2
            1
            0
                1   2     3    4    5      6   7   8   9
                                   Value
Process Capability
    LSL           USL


                        x




                        x




          USL  LSL
     Cp 
            6 x
Process Capability
        LSL             USL


                                   x




                                   x



              USL   LSL   
 C pk    min        ,        
              3 x     3 x 
   Cookie Example
 Suppose we have the cookie example and it is
 desired to have 15 + 3 chips.
     Obs.    Chips     Obs.    Chips
       1      12         11        17
       2      18         12        16
       3      19         13        19
       4      16         14        13
       5      13        151        15
       6      11         16        14
       7      16         17        17
       8      14         18        11
       9      10         19        16
      10      12         20        15
                     Average      14.1
                     Std. Dev.     3.0
Cookie Example

          18  14.1 14.1 12 
Cp  min 
                   ,         
                              
          3(3)        3(3) 

    min(0.433, 0.233)


    0.233
       South Dakota

School of Mines & Technology

     Funnel Experiment

   Industrial Engineering
   Funnel Experiment
 TheFunnel Experiment is a famous
 experiment first developed by Dr. Deming. It
 was designed to show primarily two things:
  1. All processes have statistical variation
  2. Improving processes to reduce variation is
     infinitely better than trying to control process
     variation
Funnel Experiment
             Funnel Experiment
Funnel Experiment




                    Rule 1: Do nothing
Funnel Experiment




                    Rule 2: Move the funnel in an equal
                            but opposite direction
                    Rule 3: Move the funnel in an equal
Funnel Experiment




                            but opposite direction from (0,0)
                    Rule 4: Move the funnel to the last
Funnel Experiment




                            position where the last hit.
  The Funnel Experiment
Rule 1: (LTDTA) Set the funnel over the target
at (0,0) and leave the funnel fixed through all 50 drops.

                         Funnel Experiment; Rule 1

                                     4
                                     3
                                     2
                                     1
     Yk




                                     0
          -4   -3   -2        -1    -1 0      1      2   3   4
                                    -2
                                    -3
                                    -4
                                         Xk
  The Funnel Experiment
Rule 2: Move the funnel in an equal but opposite
direction from where the last marbel hit. The funnel
should be moved the distance (-xk,-yk) from its last
resting point.
                        Funnel Experiment; Rule 2

                                    4
                                    3
                                    2
                                    1
    Yk




                                    0
         -4   -3   -2        -1    -1 0      1      2   3   4
                                   -2
                                   -3
                                   -4
                                     Xk
    The Funnel Experiment
Rule 2: Move the funnel in an equal but
opposite direction from where the last marbel
hit. The funnel should be moved the distance
(-xk,-yk) from its last resting point.

     Machine/Process adjustment rules using control
     chart limits.

     Automatic compensating machinery
  The Funnel Experiment
Rule 3: Move the funnel in an equal but opposite
direction from target (0,0) to where the last marble hit.
Rule 3 states that the funnel should be moved a distance
(-xk,-yk) from the target (0,0).

                         Funnel Experiment; Rule 3

                                     4
                                     3
                                     2
                                     1
     Yk




                                     0
          -4   -3   -2        -1    -1 0      1      2   3   4
                                    -2
                                    -3
                                    -4
                                      Xk
    The Funnel Experiment
Rule 3: Move the funnel in an equal but
opposite direction from target (0,0) to where the
last marble hit. Rule 3 states that the funnel
should be moved a distance (-xk,-yk) from the
target (0,0).

     Management by Objectives

     Committees
  The Funnel Experiment
Rule 4: Move the funnel to the last position where the
last marble hit. Rule 4 states that the funnel should be
moved to the resting point, (xk,yk).

                        Funnel Experiment; Rule 4

                                    4
                                    3
                                    2
                                    1
    Yk




                                    0
         -4   -3   -2        -1    -1 0      1      2   3   4
                                   -2
                                   -3
                                   -4
                                     Xk
    The Funnel Experiment
Rule 4: Move the funnel to the last position
where the last marble hit. Rule 4 states that the
funnel should be moved to the resting point,
(xk,yk).

      Operator matches color from batch to batch
      without reference to original swatch.

      Operator attempts to make every piece like the
      last one.

      On the job training.
     Rule 2: Deadband Controlling


                                       }
                                        Xn
 o  e
             X1                              (o  X n )
     o
              X2          X3
  o  e


1. Deviation |Xn - o| is random; do nothing
2. Deviation |Xn - o| is shift; adjust by (o - Xn)
Infinitely Large Deadband
                    Xn
     X1

      X2       X3



Xn ~ N(,)
   Infinitesimally Small Deadband
                                     Xn
           X1                        }    (  X n )
   
            X2           X3




 n   n 1    X n
     n 1    (  n 1  Zn )
Infinitesimally Small Deadband
                          Xn
        X1                }    (  X n )

         X2          X3




    Xn ~ N(, 22)
 Process Variation
                 Process Variance vs Deadband

            2
Variance




           1.5



            1
                 0     1     2       3     4    5
                                 K
       South Dakota

School of Mines & Technology

       Loss Functions

   Industrial Engineering
        Cost of Quality
        Traditional View
Costs




                           Total Cost

         Cost of
         Control

                                        Cost of Failure




                                        Quality Level
        Cost of Quality
        Contemporary View
Costs




                              Total Cost

        Cost of
        Control


            Cost of Failure




                                           Quality Level
Traditional Loss Function
    LSL       USL


                     x

          T




    LSL   T   USL
Taguchi Loss Function


                        x

          T
Taguchi Loss Function


                        x

          T



                    x

          T
       Example (Sony, 1979)
Comparing cost of two Sony television plants in
 Japan and San Diego. All units in San Diego
 fell within specifications. Japanese plant had
 units outside of specifications.
   Loss per unit (Japan)     = $0.44
   Loss per unit (San Diego) = $1.33
   How can this be?

Sullivan, “Reducing Variability: A New Approach to Quality,” Quality
   Progress, 17, no.7, 15-21, 1984.
Example
 LSL                            USL



                                      x

                 T
   U.S. Plant (2 = 8.33)
   Japanese Plant (2 = 2.78)
 Taguchi Loss Function
            L(x)




k(x - T)2



                               x
                           T
        L(x) = k(x - T)2
    Estimating Loss Function
Suppose we desire to make pistons with
  diameter D = 10 cm. Too big and they create
  too much friction. Too little and the engine
  will have lower gas milage. Suppose
  tolerances are set at D = 10 + .05 cm. Studies
  show that if D > 10.05, the engine will likely
  fail during the warranty period. Average cost
  of a warranty repair is $400.
Estimating Loss Function
          L(x)




  400



                           10.05
                     10
 400 = k(10.05 - 10.00)2
        = k(.0025)
Estimating Loss Function
          L(x)




  400



                           10.05
                     10
 400 = k(10.05 - 10.00)2
        = k(.0025)

  k = 160,000
Expected Loss

Expected Loss: Piston Diameter
                  Probability    Probability
Diameter          Process A      Process B
  9.925             0.000          0.025
  9.950             0.200          0.075
  9.975             0.200          0.200
 10.000             0.200          0.400
 10.025             0.200          0.200
 10.050             0.200          0.075
 10.075             0.000          0.025
Expected Loss

Expected Loss: Piston Diameter
                  Probability    Probability
Diameter   Loss   Process A      Process B
  9.925    900      0.000          0.025
  9.950    400      0.200          0.075
  9.975    100      0.200          0.200
 10.000     0       0.200          0.400
 10.025    100      0.200          0.200
 10.050    400      0.200          0.075
 10.075    900      0.000          0.025
Expected Loss

Expected Loss: Piston Diameter
                  Probability Weighted Probability
Diameter   Loss   Process A    Loss    Process B
  9.925    900      0.000       0.0      0.025
  9.950    400      0.200      80.0      0.075
  9.975    100      0.200      20.0      0.200
 10.000     0       0.200       0.0      0.400
 10.025    100      0.200      20.0      0.200
 10.050    400      0.200      80.0      0.075
 10.075    900      0.000       0.0      0.025
Expected Loss

Expected Loss: Piston Diameter
                  Probability Weighted Probability Weighted
Diameter   Loss   Process A    Loss    Process B    Loss
  9.925    900      0.000       0.0      0.025      22.5
  9.950    400      0.200      80.0      0.075      30.0
  9.975    100      0.200      20.0      0.200      20.0
 10.000     0       0.200       0.0      0.400       0.0
 10.025    100      0.200      20.0      0.200      20.0
 10.050    400      0.200      80.0      0.075      30.0
 10.075    900      0.000       0.0      0.025      22.5
Expected Loss

Expected Loss: Piston Diameter
                   Probability Weighted Probability Weighted
Diameter   Loss    Process A    Loss    Process B    Loss
  9.925    900       0.000       0.0      0.025      22.5
  9.950    400       0.200      80.0      0.075      30.0
  9.975    100       0.200      20.0      0.200      20.0
 10.000     0        0.200       0.0      0.400       0.0
 10.025    100       0.200      20.0      0.200      20.0
 10.050    400       0.200      80.0      0.075      30.0
 10.075    900       0.000       0.0      0.025      22.5
                  Exp. Loss =   200.0                145.0
  Aside

  L( x)  k ( x  T )   2


If we replace T by , L(x) looks similar to
the variance function. Indeed, we can
show that

  L( x)  k (  ( x  T ) )
                2             2
Example (Sony)
 LSL                              USL



                                        x

                  T
  U.S. Plant (2 = 8.33)
  Japanese Plant (2 = 2.78)


E[LUS(x)] = 0.16 * 8.33 = $1.33

 E[LJ(x)] = 0.16 * 2.78 = $0.44
      Tolerance (Pistons)
Recall,            L(x)




           400



                                    10.05
                              10
          400 = k(10.05 - 10.00)2
                 = k(.0025)

           k = 160,000
         Tolerance
                                   L(x)

Suppose repair for an engine
which will fail during
warranty can be made for     400
only $200                    200

                                                     10.05
                                          LSL   10 USL
         Tolerance
                                   L(x)

Suppose repair for an engine
which will fail during
warranty can be made for     400
only $200                    200

                                                     10.05
200 = 160,000(tolerance)2                 LSL   10 USL
         Tolerance
                                   L(x)

Suppose repair for an engine
which will fail during
warranty can be made for     400
only $200                    200

                                                     10.05
200 = 160,000(tolerance)2                 LSL   10 USL




tolerance = (200/160,000)1/2

          = .0354

				
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