Statistical Process Control
Operations Management
Dr. Ron Lembke
Designed Size
10
11
12
13
14
15
16
17
18
19
20
Natural Variation
14.5 14.6 14.7 14.8 14.9 15.0 15.1 15.2 15.3 15.4 15.5
Theoretical Basis of Control Charts
Properties of normal distribution
95.5% of allX
fall within ± 2
X 
Theoretical Basis of Control Charts
Properties of normal distribution
99.7% of allX
fall within ± 3
X 
16
14
12
10
8
6
4
2
0
Skewness


Lack of symmetry
Pearson’s coefficient of
skewness:
3( x  Median )
s
Positive Skew > 0
16
14
12
10
8
6
4
2
0
Skewness = 0
Negative Skew < 0
16
14
12
10
8
6
4
2
0
0.45
Kurtosis
0.4
0.35
0.3
0.25

Amount of peakedness
or flatness
4
(x  x)


ns 4
Kurtosis < 0
0.2
0.15
0.1
0.05
0
-6
-4
-2
0
2
Kurtosis = 0
Kurtosis > 0
4
6
Design Tolerances

Design tolerance:





Determined by users’ needs
UTL -- Upper Tolerance Limit
LTL -- Lower Tolerance Limit
Eg: specified size +/- 0.005 inches
No connection between tolerance and 

completely unrelated to natural variation.
Process Capability and 6
LTL
UTL
3


LTL
UTL
6
A “capable” process has UTL and LTL 3 or more
standard deviations away from the mean, or 3σ.
99.7% (or more) of product is acceptable to
customers
Process Capability
Capable
Not Capable
LTL
UTL
LTL
UTL
LTL
UTL
LTL
UTL
Process Capability



Specs: 1.5 +/- 0.01
Mean: 1.505 Std. Dev. = 0.002
Are we in trouble?
Process Capability

Specs: 1.5 +/- 0.01



LTL = 1.5 – 0.01 = 1.49
UTL = 1.5 + 0.01 = 1.51
Mean: 1.505 Std. Dev. = 0.002


LCL = 1.505 - 3*0.002 = 1.499
UCL = 1.505 + 0.006 = 1.511
Process
Specs
1.49
1.499
1.51 1.511
Capability Index



Capability Index (Cpk) will tell the position of
the control limits relative to the design
specifications.
Cpk>= 1.0, process is capable
Cpk< 1.0, process is not capable
Process Capability, Cpk

Tells how well parts
produced fit into specs
C pk
 X  LTL UTL  X 
 min 
or

3 
 3
Process
Specs
LTL
3
X
3
UTL
Process Capability

Tells how well parts produced fit into specs
C pk



 X  LTL UTL  X 
 min 
or

3 
 3
For our example:
1.505  1.49 1.51  1.505 
C pk  min 
or
0.006 
 0.006
Cpk= min[ 0.015/.006, 0.005/0.006]
Cpk= min[2.5,0.833] = 0.833 < 1 Process not capable
Process Capability: Re-centered


If process were properly centered
Specs: 1.5 +/- 0.01



LTL = 1.5 – 0.01 = 1.49
UTL = 1.5 + 0.01 = 1.51
Mean: 1.5 Std. Dev. = 0.002


LCL = 1.5 - 3*0.002 = 1.494
UCL = 1.5 + 0.006 = 1.506
Process
Specs
1.49
1.494
1.506 1.51
If re-centered, it would be Capable
C pk
C pk
1.5  1.49 1.51  1.5 
 min 
,

0.006 
 0.006
 0.01 0.01 
 min 
,
  1.67
 0.006 0.006 
Process
Specs
1.49
1.494
1.506 1.51
Packaged Goods



What are the Tolerance Levels?
What we have to do to measure capability?
What are the sources of variability?
Production Process
Make Candy
Make Candy
Make Candy
Mix
Package
Put in big bags
Wrong wt.
Wrong wt.
Make Candy
Mix %
Make Candy
Make Candy
Candy irregularity
Processes Involved

Candy Manufacturing:




Mixing:


Is proper color mix in each bag?
Individual packages:



Are M&Ms uniform size & weight?
Should be easier with plain than peanut
Percentage of broken items (probably from printing)
Are same # put in each package?
Is same weight put in each package?
Large bags:


Are same number of packages put in each bag?
Is same weight put in each bag?
Weighing Package and all candies




Before placing candy
on scale, press
“ON/TARE” button
Wait for 0.00 to appear
If it doesn’t say “g”,
press Cal/Mode button
a few times
Write weight down on
form
Candy colors
1.
2.
3.
4.
5.
6.
7.
Write Name on form
Write weight on form
Write Package # on form
Count # of each color and
write on form
Count total # of candies
and write on form
(Advanced only): Eat
candies
Turn in forms and
complete wrappers
The effects of rounding
25.00
0.80
24.00
g - rounded
oz - rounded
23.00
0.7 Ounces
0.70
Rounded Weight - Ounces
Rounded Weight - grams
22.00
21 grams
21.00
20 grams
20.00
0.6 Ounces
0.60
19 grams
19.00
18 grams
18.00
17.00
0.50
14.5
15.0
15.5
16.0
16.5
17.0
17.5
18.0
18.5
19.0
Original Weight in grams
19.5
20.0
20.5
21.0
21.5
22.0
22.5
Peanut Candy Weights
Avg. 2.18, stdv 0.242, c.v. = 0.111
Peanut Individuals
9
8
7
6
Count

5
4
3
2
1
0
1.5
1.6
1.7
1.8
1.9
2
2.1
2.2
2.3
Mass (g)
2.4
2.5
2.6
2.7
2.8
2.9
3
Plain Candy Weights
Avg 0.858, StDev 0.035, C.V. 0.0413
Individual Plain Candies
16
14
12
10
8
6
4
2
0
0.
78
0.
79
0.
8
0.
81
0.
82
0.
83
0.
84
0.
85
0.
86
0.
87
0.
88
0.
89
0.
9
0.
91
0.
92
0.
93
0.
94
0.
95
0.
96
0.
97
Count

Mass (g)
Peanut Color Mix






Brown 17.7%
Yellow 8.2%
Red
9.5%
Blue 15.4%
Orange 26.4%
Green 22.7%
website
20%
20%
20%
20%
10%
10%
Plain Color Mix






Brown
Yellow
Red
Blue
Orange
Green
Class
12.1%
14.7%
11.4%
19.5%
21.2%
21.2%
website
30%
20%
20%
10%
10%
10%
So who cares?




Dept. of Commerce
National Institutes of Standards & Technology
NIST Handbook 133
Fair Packaging and Labeling Act
Acceptable?
Package Weight




“Not Labeled for Individual Retail Sale”
If individual is 18g
MAV is 10% = 1.8g
Nothing can be below 18g – 1.8g = 16.2g
Goal of Control Charts



collect and present data visually
allow us to see when trend appears
see when “out of control” point occurs
Process Control Charts

X
Graph of sample data plotted over time
60
50
40
30
20
10
0
UCL
LCL
1 2 3 4 5 6 7 8 9 10 11 12
Time
Process
Average
± 3
Process Control Charts

Graph of sample data plotted over time
X 60
Assignable
Cause
Variation
Natural
Variation
50
40
30
20
10
0
UCL
LCL
1 2 3 4 5 6 7 8 9 10 11 12
Time
Definitions of Out of Control
1.
2.
3.
4.
No points outside control limits
Same number above & below center line
Points seem to fall randomly above and
below center line
Most are near the center line, only a few are
close to control limits
1.
2.
3.
8 Consecutive pts on one side of centerline
2 of 3 points in outer third
4 of 5 in outer two-thirds region
Attributes vs. Variables
Attributes:
 Good / bad, works / doesn’t
 count % bad (P chart)
 count # defects / item (C chart)
Variables:
 measure length, weight, temperature (x-bar
chart)
 measure variability in length (R chart)
Attribute Control Charts

Tell us whether points in tolerance or not




p chart: percentage with given characteristic
(usually whether defective or not)
np chart: number of units with characteristic
c chart: count # of occurrences in a fixed area of
opportunity (defects per car)
u chart: # of events in a changeable area of
opportunity (sq. yards of paper drawn from a
machine)
p Chart Control Limits
UCLp  p  z 
p  1 p
n
k
p
X
i
i1
k
n
i1
i
# Defective
Items in
Sample i
Sample i
Size
p Chart Control Limits
p  1  p 
UCLp  p  z 
n
k
k
n
# Samples
n
i1
k
i
p
X
i
i1
k
n
i1
i
z = 2 for
95.5% limits;
z = 3 for
99.7% limits
# Defective
Items in
Sample i
Sample i
Size
p Chart Control Limits
p  1  p 
UCLp  p  z 
n
p  1  p 
LCLp  p  z 
n
k
k
n
# Samples
n
i1
k
i
p
X
i
i1
k
n
i1
i
z = 2 for
95.5% limits;
z = 3 for
99.7% limits
# Defective
Items in
Sample i
Sample i
Size
p Chart Example
You’re manager of a 500room hotel. You want to
achieve the highest level
of service. For 7 days,
you collect data on the
readiness of 200 rooms. Is
the process in control (use
z = 3)?
© 1995 Corel Corp.
p Chart Hotel Data
Day
1
2
3
4
5
6
7
No.
Rooms
200
200
200
200
200
200
200
No. Not
Ready Proportion
16 16/200 = .080
7
.035
21
.105
17
.085
25
.125
19
.095
16
.080
p Chart Control Limits
k
n
n
i1
k
i
1400

 200
7
p Chart Control Limits
k
k
n
n
i1
k
i
1400

 200
7
p
X
i
i1
k
n
i1

16 + 7 +...+ 16
i
121

 0.0864
1400
p Chart Solution
k
k
n
n
i1
k
i
1400

 200
7
p
X
16 + 7 +...+ 16
i
i1
k
n
121

 0.0864
1400
i
i1
p z
p  1 p
0.0864  1 0.0864 
 0.0864  3
n
200

p Chart Solution
k
k
n
n
i1
k
i
1400

 200
7
p
X
16 + 7 +...+ 16
i
i1
k
n
121

 0.0864
1400
i
i1
p z
p  1 p
0.0864  1 0.0864 
 0.0864  3
n
200
 0.0864
 3* 0.01984  0.0864  0.01984
 0.1460, and 0.0268
p Chart
0.15
P
UCL
0.10
0.05
LCL
0.00
1
2
3
4
Day
5
6
7
R Chart

Type of variables control chart


Shows sample ranges over time



Interval or ratio scaled numerical data
Difference between smallest & largest values
in inspection sample
Monitors variability in process
Example: Weigh samples of coffee &
compute ranges of samples; Plot
Hotel Example
You’re manager of a 500room hotel. You want to
analyze the time it takes to
deliver luggage to the room.
For 7 days, you collect data
on 5 deliveries per day. Is
the process in control?
Hotel Data
Day
Delivery Time
1
7.30 4.20 6.10 3.45
2
4.60 8.70 7.60 4.43
3
5.98 2.92 6.20 4.20
4
7.20 5.10 5.19 6.80
5
4.00 4.50 5.50 1.89
6 10.10 8.10 6.50 5.06
7
6.77 5.08 5.90 6.90
5.55
7.62
5.10
4.21
4.46
6.94
9.30
R &X Chart Hotel Data
Day
Delivery Time
1 7.30 4.20 6.10 3.45 5.55
Sample
Mean Range
5.32
7.30 + 4.20 + 6.10 + 3.45 + 5.55
Sample Mean =
5
R &X Chart Hotel Data
Day
Delivery Time
1 7.30 4.20 6.10 3.45 5.55
Largest
Smallest
Sample
Mean Range
5.32
3.85
Sample Range = 7.30 - 3.45
R &X Chart Hotel Data
Day
1 7.30
2 4.60
3 5.98
4 7.20
5 4.00
6 10.10
7 6.77
Delivery Time
4.20 6.10 3.45
8.70 7.60 4.43
2.92 6.20 4.20
5.10 5.19 6.80
4.50 5.50 1.89
8.10 6.50 5.06
5.08 5.90 6.90
5.55
7.62
5.10
4.21
4.46
6.94
9.30
Sample
Mean Range
5.32
3.85
6.59
4.27
4.88
3.28
5.70
2.99
4.07
3.61
7.34
5.04
6.79
4.22
R Chart Control Limits
UCLR  D4  R
From Exhibit 6.13
LCLR  D3  R
k
R
 Ri
i 1
k
Sample Range
at Time i
# Samples
Control Chart Limits
n
A2
D3
D4
2
1.88
0
3.278
3
1.02
0
2.57
4
0.73
0
2.28
5
0.58
0
2.11
6
0.48
0
2.00
7
0.42
0.08
1.92
R Chart Control Limits
k
R
 Ri
i 1
k
3.85  4.27  4.22

 3.894
7
R Chart Solution
k
R
 Ri
i 1
k
3.85  4.27  4.22

 3.894
7
UCLR  D4  R  (2.11) (3.894)  8.232
LCLR  D3  R  (0)(3.894) 0
From 6.13
(n = 5)
R Chart Solution
R, Minutes
8
6
4
2
0
1
2
UCL
3
4
Day
5
6
7
X Chart Control Limits
UCLX  X  A2  R
Sample
Mean at
Time i
k
X
X
i 1
k
k
i
R
R
i 1
k
i
Sample
Range
at Time i
# Samples
X Chart Control Limits
From
Table 6-13
UCLX  X  A2  R
LCLX  X  A2  R
k
X 
 Xi
i 1
k
k
R
 Ri
i 1
k
X Chart Control Limits
From 6.13
UCLX  X  A2  R
Sample
Mean at
Time i
LCLX  X  A2  R
k
X 
 Xi
i 1
k
k
R
 Ri
i 1
k
Sample
Range
at Time i
# Samples
Exhibit 6.13 Limits
n
A2
D3
D4
2
1.88
0
3.278
3
1.02
0
2.57
4
0.73
0
2.28
5
0.58
0
2.11
6
0.48
0
2.00
7
0.42
0.08
1.92
R &X Chart Hotel Data
Day
1 7.30
2 4.60
3 5.98
4 7.20
5 4.00
6 10.10
7 6.77
Delivery Time
4.20 6.10 3.45
8.70 7.60 4.43
2.92 6.20 4.20
5.10 5.19 6.80
4.50 5.50 1.89
8.10 6.50 5.06
5.08 5.90 6.90
5.55
7.62
5.10
4.21
4.46
6.94
9.30
Sample
Mean Range
5.32
3.85
6.59
4.27
4.88
3.28
5.70
2.99
4.07
3.61
7.34
5.04
6.79
4.22
X Chart Control Limits
k
X 
Xi

i 1
k
5.32  6.59    6.79

 5.813
7
k
R
 Ri
i 1
k
3.85  4.27    4.22

 3.894
7
X Chart Control Limits
k
X 
 Xi
i 1
k
5.32  6.59    6.79

 5.813
7
k
R
 Ri
i 1
k
From 6.13
3.85  4.27    4.22

 3.894 (n = 5)
7
UCLX  X  A2  R  5.813  0.58 * 3.894  8.060
X Chart Solution
k
X 
 Xi
i 1
k
k
R
 Ri
i 1
k
5.32  6.59    6.79

 5.813
7
From 6.13
(n = 5)
3.85  4.27    4.22

 3.894
7
UCLX  X  A2  R  5.813  (0 .58) (3.894) = 8.060
LCLX  X  A2  R  5.813  (0 .58) (3.894) = 3.566
X Chart Solution*
X, Minutes
8
6
4
2
0
1
2
3
UCL
LCL
4
Day
5
6
7
Thinking Challenge
You’re manager of a 500room hotel. The hotel owner
tells you that it takes too
long to deliver luggage to the
room (even if the process
may be in control). What do
you do?
N
© 1995 Corel Corp.
Solution


Redesign the luggage delivery process
Use TQM tools



Cause & effect diagrams
Process flow charts
Pareto charts
Method
People
Too
Long
Material
Equipment