Chapter 6 – Part 4 Chapter 6 – Part 4

Process Capability

Meaning of Process CapabilityMeaning 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. LimitsControl 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. LimitsControl Limits vs. Spec. LimitsSamplingdistribution, X-bar

Processdistribution, X

Mean=Target

Lowercontrol

limit

Uppercontrol

limit

LSL USL

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.

Requirements for Assessing Requirements for Assessing Process CapabilityProcess Capability

CCpp Index Index

6LSLUSLCp

deviation standard process estimatedˆ

USL = upper specification limit

LSL = Lower specification limit

CCpp Index Index

limits spec. theofwidth

process theof spreadˆ6

USL - LSL

• 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 ????

CCpp Index IndexProcessdistribution, X

LSL USLX

Width of spec limits = USL - LSL

Spread of Process = USL - LSL

Process is Barely Capable if Cp = 1Process is Barely Capable if Cp = 1

X

.00135 .00135

LSL USL

.9973

X

Spread of process matches the width of specs. 99.73% of output is within the spec. limits.

Process Barely Capable if Cp = 1Process Barely Capable if Cp = 1 If , what does this imply regarding the

spec. limits? Cp=1

LSL =

USL =

1ˆ6

6

ˆ633

ˆ6)3(3

ˆ6

XX

XX

LSLUSLCp

Process Barely Capable if Cp = 1Process Barely Capable if Cp = 1

Process is Capable if Cp > 1Process is Capable if Cp > 1

X

< .00135 < .00135

LSL USL

>.9973

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 < 1Process is Not Capable if Cp < 1

X

> .00135 > .00135

LSL USL

< .9973

X

Spread of process is greater than the width of specs. Less than 99.73% of output is within the spec. limits.

nXUCL

nXLCL

ˆ3

ˆ3

RAXUCL

RAXLCL

2

2

Estimating the Standard DeviationEstimating the Standard Deviation

RAn 2ˆ

3

3ˆ 2 nRA

Estimating the Standard DeviationEstimating the Standard Deviation

Day Hour X1 X2 X3 R1 10 am 17 13 6 36/3 =12 11

1 pm 15 12 24 51/3 =17 124 pm 12 21 15 48/3 =16 9

2 10 am 13 12 17 42/3 =14 51 pm 18 21 15 54/3 =18 64 pm 10 18 17 45/3 =15 8

= 92/6= 15.33

= 51/6= 8.5

Sugar Example Ch. 6 - 3Sugar Example Ch. 6 - 3

X R

X

0.530169.15

33)5.8)(02.1(

3ˆ 2

nRA

02.1,3,5.8 2 AnR

Capability of Sugar ProcessCapability of Sugar Process

33.03010

)5(61020

ˆ6

LSLUSLCp

USL = 20 grams

LSL = 10 grams

0.5ˆ

Capability of Sugar ProcessCapability of Sugar Process Since Cp <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 DefectiveEstimated Process Fraction Defective What is the estimated process fraction defective --

the percentage of product out of spec?

p-bar = F1 + F2

Mean

LSL USLF2F1

Estimated Process Fraction DefectiveEstimated Process Fraction Defective We can then use Cp to determine the p-bar because

there is a simple relationship between Cp and z:

z = 3Cp

(See last side for deviation of this result.)

• Suppose, Cp =0.627

z = 3(0.627) =1.88

Estimated Process Fraction DefectiveEstimated 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 DefectiveProcess Fraction Defective

LSL USL

z =1.88

Area(z) = Area(1.88) = 0.4699

0

F2 = % above USL = .5000 - 0.4699 = .0301

F2

zz Table (Text, p. 652) Table (Text, p. 652)z .00 .01 .02 . . . .08 .09

0.0

0.1

0.2

.

.

.09

1.8 .4699

Process FalloutProcess Fallout

LSL USL

F2F1

0

p-bar =

p-bar = 2(.5 – .4699) = 2(.0301)=.0602

z =1.88

2[.5 – Area(z)] = F1 + F2

0.4699

Process Fallout – Two Sided Spec.Process Fallout – Two Sided Spec.

Cp z = 3Cp Fallout =2[.5 – Area(z)]

Defect Rate in PPM (parts per million)

0.25 0.75 2[.5-.2734] = .4532 453,200 PPM

0.80 2.40 2[.5-.4918] = .0164 16,400 PPM

1.0 3 2[.5-.4987] = .0026 2,600 PPM

1.5 -4.5From Excel

2[Area(-z)]=2[.0000034] =.0000068

7 PPM

Recommended Minimum Recommended Minimum CpCp

Process Cp z = 3Cp Fallout PPM

Existing process

1.25 3.75 2[Area(-z)]=2[.000088] =.0001769

176.9

New process

1.45 4.35 2[Area(-z)]=2[.000006812] =.0000136

13.6

Process Cp z = 3Cp Fallout PPM

Safety, existing process

1.45 4.35 2[Area(-z)]=2[.000006812] =.0000136

13.6

Safety, new process

1.60 4.80 2[Area(-z)]=2[.000000794] =0.0000016

1.6

Recommended Minimum Recommended Minimum CpCp

Soft Drink ExampleSoft Drink Example

Area(z) = Area(0.99) = 0.3389

Cp =0.33

z = 3Cp = 3(0.33) = 0.99

p-bar = 2[.5 - Area(0.99)]

= 2[.5 - 0.3389]

= 0.3222

Capability Index Based on TargetCapability Index Based on Target• Limitation of Cp is that it assumes that the

process is mean is on target.

Process Mean = Target Value = (LSL + USL)/2

CCT T Capability IndexCapability 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.

CT is an index that accounts for the location of mean relative to target.

2Target)(ˆ6

X

LSLUSLCT

CCT T Capability IndexCapability Index

??

0Target)(Target 2

TC

XX

If process is centered on target,

CCT T Capability IndexCapability Index

If process is off target,

pT CC

TX

arget

3326.

)1533.15(56

1020

Target)(ˆ6

2

2

X

LSLUSLCT

LSL = 10, USL = 20, estimated standard deviation =5.0 and estimate process mean = 15.33. ComputeCT.

Example of CExample of CTT

p

T

C

X

LSLUSLC

2.2

)1212(3.06

1014

Target)(ˆ6

2

2

If process mean is adjusted to target,

CCT T Capability IndexCapability Index

CCT T Capability IndexCapability Index Cp is the largest value that CT can equal. Since Cp = 2.2 and CT = .44, the difference

is the maximum amount by which we can increase CT by adjusting the mean to the target value.

0007.0

3326.3333.

Tp CCD

Conclusion?Conclusion?

p

p

Cz

z

z

zXzX

LSLUSLC

3

3

ˆ6ˆ2

ˆ6)ˆ(ˆ

ˆ6

Derivation of Derivation of z z = 3= 3CpCp