Chapter 6 – Part 4
Process Capability

Meaning of Process Capability
 The capability of a process is the ability of the process to meet the specifications.
 A process is capability of meeting the specification limits if at least 99.73
% of the product falls within the specification limits.
 This means that the fraction of product that falls outside the specification limits is no greater than
0.0027, or that no more that 3 out of 1,000 units is
“out of spec.”
 Our method of computing process capability assumes that the process is normally distributed.

Control Limits vs. Spec. Limits
 Control limits apply to sample means, not individual values.
 Mean diameter of sample of 5 parts, X bar
 Spec limits apply to individual values
 Diameter of an individual part, X

Control Limits vs. Spec. Limits
Sampling distribution,
X-bar
Process distribution,
X
LSL
Lower control limit
Mean=
Target
Upper control limit
USL

Requirements for Assessing
Process Capability
 To assess capability of a process, the process must be in statistical control.
 That is, all special causes of variation must be removed prior to assessing capability.
 Also, process performance characteristic
(e.g., diameter, bake time) must be normally distributed.

C p
Index
Cp

USL

6
 ˆ
LSL
USL = upper specification limit
LSL = Lower specification limit
 ˆ  estimated process standard deviation

C p
Index
6
 ˆ  spread of the process
USL - LSL
 width of the spec.
limits
• We want the spread (variability) of the process to be as ???
• If the spread of the process is very ????, the capability of the process will be very ????

C p
Index
LSL X
Width of spec limits = USL - LSL
USL
Spread of Process = USL - LSL
Process distribution,
X

Process is Barely Capable if Cp = 1
.9973
.00135
.00135
LSL USL
X
X
Spread of process matches the width of specs.
99.73% of output is within the spec. limits.

Process Barely Capable if Cp = 1
 spec. limits?
LSL =
USL =

Process Barely Capable if Cp = 1
Cp


X

USL

6
 ˆ
LSL
3



6
 ˆ
( X

3


)


X
6


6
 ˆ

3



6
 ˆ
X

1

3



Process is Capable if Cp > 1
>.9973
< .00135
USL
< .00135
X
LSL X
Spread of process is less than the width of specs.
More than 99.73% of output is within the spec. limits.

Process is Not Capable if Cp < 1
< .9973
> .00135
> .00135
LSL USL
X
X
Spread of process is greater than the width of specs.
Less than 99.73% of output is within the spec. limits.

LCL

X

3
 ˆ n
UCL

X

3
 ˆ n
LCL

X

A
2
R
UCL

X

A
2
R

3
 ˆ n

A
2
R
 ˆ 
A
2
R
3 n

Sugar Example Ch. 6 - 3
Day Hour X1 X2 X3 X
1 10 am 17 13 6 36/3 =12
1 pm 15 12 24 51/3 =17
4 pm 12 21 15 48/3 =16
2 10 am 13 12 17 42/3 =14
1 pm 18 21 15 54/3 =18
R
11
12
9
5
6
4 pm 10 18 17 45/3 =15
X = 92/6
= 15.33
8
R = 51/6
= 8.5

R

8 .
5 , n

3 , A
2

1 .
02
 ˆ 
A
2
R
3 n

( 1 .
02 )( 8 .
5 )
3

15 .
0169
3

5 .
0
3

Capability of Sugar Process
USL = 20 grams
LSL = 10 grams
 ˆ 
5 .
0
Cp

USL

6
 ˆ
LSL

20

10
6 ( 5 )

10
30

0 .
33

Capability of Sugar Process
 Since C p
<1, the process is not capability of meeting the spec limits.
 The fraction of defective drinks (drinks with either too much or not enough sugar) will exceed .0027.
 That is, more than 3 out of every 1000 drinks produced can be expected to be too sweet or not sweet enough.
 We now estimate the process fraction defective, p -bar.

Estimated Process Fraction Defective
 What is the estimated process fraction defective -the percentage of product out of spec?
p -bar = F1 + F2
F1
LSL USL
F2
Mean

Estimated Process Fraction Defective
 We can then use Cp to determine the p -bar because there is a simple relationship between Cp and z : z = 3 C p
(See last side for deviation of this result.)
•
Suppose, C p
= 0.627
z = 3(0.627) =1.88

Estimated Process Fraction Defective
 The z value tells us how many standard deviations the specification limits are away from the mean.
 A z value of 1.88 indicates that the USL is 1.88 standard deviations above the mean.
 The negative of z , -1.88, indicates that the LSL is
1.88 standard deviations below the mean.
 We let
Area( z ) be the area under the standard normal curve between 0 and z .

Process Fraction Defective
Area( z ) = Area(1.88) = 0.4699
LSL USL
F
2
0 z = 1.88
F
2
= % above USL = .5000 - 0.4699 = .0301

.
.
0.2
.09
1.8
z
0.0
0.1
z Table (Text, p. 652)
.00
.01
.02
.
.
.
.08
.09
.4699

Process Fallout pbar = 2[.5 – Area( z )] = F1 + F2
F1
LSL
0
0.4699
USL z = 1.88
F2 pbar = 2(.5 – .4699) = 2(.0301)=.0602

Process Fallout – Two Sided Spec.
Cp
0.25
z = 3 Cp Fallout =
2[.5 – Area(z)]
Defect Rate in
PPM (parts per million)
0.75
2[.5-.2734] = .4532 453,200 PPM
0.80
1.0
1.5
2.40
2[.5-.4918] = .0164
16,400 PPM
3
-4.5
From
Excel
2[.5-.4987] = .0026
2,600 PPM
7 PPM 2[Area(z )]=
2[.0000034]
=.0000068

Recommended Minimum Cp
Process Cp z = 3 Cp
Existing process
New process
1.25
3.75
Fallout
2[Area(-z)]=
2[.000088]
=.0001769
1.45
4.35
2[Area(-z)]=
2[.000006812]
=.0000136
PPM
176.9
13.6

Recommended Minimum Cp
Process Cp z = 3 Cp
Safety, existing process
Safety, new process
Fallout
1.45
4.35
2[Area(-z)]=
2[.000006812]
=.0000136
1.60
4.80
2[Area(-z)]=
2[.000000794]
=0.0000016
PPM
13.6
1.6

Soft Drink Example
Cp = 0.33
z = 3 Cp = 3(0.33) = 0.99
Area( z ) = Area(0.99) = 0.3389
p -bar = 2[.5 - Area(0.99)]
= 2[.5 - 0.3389]
= 0.3222

Capability Index Based on Target
• Limitation of C p is that it assumes that the process is mean is on target.
Process Mean = Target Value = (LSL + USL)/2

C
T
Capability Index
 With Cp, capability value is the same whether the process is centered on target or is way off.
 Cp is not affected by location of mean relative to target.
 We need capability index that accounts for location of the mean relative to the target as well as the variance.
 C
T is an index that accounts for the location of mean relative to target.

C
T
Capability Index
C
T

6
USL

LSL
 ˆ  
( X

Target)
2

C
T
Capability Index
If process is centered on target,
X

Target

( X

Target)
2 
0
C
T

??
If process is off target,
X

T arget
C
T

C p

Example of C
T
LSL = 10, USL = 20, estimated standard deviation =
5.0 and estimate process mean = 15.33. Compute
C
T
.
C
T

6
USL

LSL
 ˆ  
( X

Target)
2

6
20

10
5
 
( 15 .
33

15 )
2

.
3326

C
T
Capability Index
If process mean is adjusted to target,
C
T

6
USL

LSL
 ˆ  
( X

Target)
2

6 0 .
3

14

10

( 12

12 )
2

2 .
2

C p

C
T
Capability Index
 C p is the largest value that C
T can equal.
 Since C p
= 2.2 and C
T
= .44, the difference
D

C p

C
T

.
3333

.
3326

0 .
0007 is the maximum amount by which we can increase
C
T by adjusting the mean to the target value.

Derivation of z = 3Cp
C


 p
X

2 z
 ˆ z
6
 ˆ

USL z
 ˆ

6
 ˆ

(
6
 ˆ
LSL
X

3 z
 ˆ
) z

3 C p