# TM 665 Project Planning _ Control.ppt

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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

Cut Dough
Dough                  Minutes

OK
Bake 10          Qual.               Pack and
Minutes                              Ship

Poor
Feed to Hogs

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
Histogram; Class Problem

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
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

Try Mixing Chips in
With Dry Ingredients
Flow Charting

Mix Dry                            Mix Wet       Mix Wet
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
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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

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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 
 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

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

 min(0.433, 0.233)

 0.233
South Dakota

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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.

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.

}
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)
Xn
X1

X2       X3

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

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

Xn ~ N(, 22)
Process Variation

2
Variance

1.5

1
0     1     2       3     4    5
K
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Loss Functions

Industrial Engineering
Cost of Quality
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
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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