Transcript PRODUCTIONS/OPERATIONS MANAGEMENT

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
Mean=
Target
LSL
Lower
control
limit
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.
Cp Index
USL  LSL
Cp 
6ˆ
USL = upper specification limit
LSL = Lower specification limit
ˆ  estimated process standard deviation
Cp 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 ????
Cp Index
Process
distribution,
X
LSL
X
Width of spec limits = USL - LSL
Spread of Process = USL - LSL
USL
Process is Barely Capable if Cp = 1
.9973
.00135
.00135
LSL
X
USL
X
Spread of process matches the width of specs.
99.73% of output is within the spec. limits.
Process Barely Capable if Cp = 1

If Cp=1, what does this imply regarding the
spec. limits?
LSL =
USL =
Process Barely Capable if Cp = 1
USL  LSL
Cp 
6ˆ


X  3  ( X  3 )

6ˆ


X  3  X  3

6ˆ

6

1
6ˆ
Process is Capable if Cp > 1
>.9973
< .00135
< .00135
LSL
X
USL
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
X
USL
X
Spread of process is greater than the width of specs.
Less than 99.73% of output is within the spec. limits.
Estimating the Standard Deviation
LCL  X  3
UCL  X  3
ˆ
n
ˆ
n
LCL  X  A2 R
UCL  X  A2 R
Estimating the Standard Deviation
3
ˆ
n
 A2 R
A2 R n
ˆ 
3
Sugar Example Ch. 6 - 3
Day Hour
1
10 am
2
X1
17
X2
13
X
X3
6 36/3 =12
R
11
1 pm
15
12
24 51/3 =17
12
4 pm
12
21
15 48/3 =16
9
10 am
13
12
17 42/3 =14
5
1 pm
18
21
15 54/3 =18
6
4 pm
10
18
17 45/3 =15
8
X = 92/6 R = 51/6
= 15.33
= 8.5
R  8.5, n  3,
A2  1.02
A2 R n
ˆ 
3
(1.02)(8.5) 3

3
15.0169

3
 5 .0
Capability of Sugar Process
USL = 20 grams
LSL = 10 grams
ˆ  5.0
USL  LSL
Cp 
6ˆ
20  10

6(5)
10

 0.33
30
Capability 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 Defective

What is the estimated process fraction defective -the percentage of product out of spec?
p-bar = F1 + F2
LSL
USL
F1
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 = 3Cp
(See last side for deviation of this result.)
• Suppose, Cp =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
F2
0
z =1.88
F2 = % above USL = .5000 - 0.4699 = .0301
z Table (Text, p. 652)
z
.00
.01
.02
.
.
.
.08
0.0
0.1
0.2
.
.
.09
1.8
.4699
.09
Process Fallout
p-bar = 2[.5 – Area(z)] = F1 + F2
0.4699
LSL
USL
F1
F2
0
z =1.88
p-bar = 2(.5 – .4699) = 2(.0301)=.0602
Process Fallout – Two Sided Spec.
Cp
z = 3Cp
Fallout =
2[.5 – Area(z)]
0.25
0.75
Defect Rate in
PPM (parts
per million)
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
-4.5
From
Excel
2[Area(-z)]=
2[.0000034]
=.0000068
7 PPM
Recommended Minimum Cp
Process
Cp
z = 3Cp
Existing
process
1.25
3.75
New
process
1.45
4.35
Fallout
2[Area(-z)]=
2[.000088]
=.0001769
2[Area(-z)]=
2[.000006812]
=.0000136
PPM
176.9
13.6
Recommended Minimum Cp
Process
Cp
z = 3Cp
Safety,
existing
process
1.45
4.35
Safety,
new
process
1.60
4.80
Fallout
2[Area(-z)]=
2[.000006812]
=.0000136
2[Area(-z)]=
2[.000000794]
=0.0000016
PPM
13.6
1.6
Soft Drink Example
Cp =0.33
z = 3Cp = 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 Cp is that it assumes that the
process is mean is on target.
Process Mean = Target Value = (LSL + USL)/2
CT 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.
 CT is an index that accounts for the location of
mean relative to target.

CT Capability Index
CT 
USL  LSL

2
ˆ
6   ( X  Target)
CT Capability Index
If process is centered on target,
X  Target  ( X  Target)
CT  ??
If process is off target,
X  Target
CT  C p
2
0
Example of CT
LSL = 10, USL = 20, estimated standard deviation =
5.0 and estimate process mean = 15.33. Compute
CT.
CT 

USL  LSL
6 ˆ   ( X  Target) 2
20  10
6 5   (15.33  15) 2
 .3326
CT Capability Index
If process mean is adjusted to target,
CT 

USL  LSL
6 ˆ   ( X  Target) 2
14  10
6 0.3   (12  12) 2
 2.2
 Cp
CT Capability Index

Cp is the largest value that CT can equal.
 Since Cp = 2.2 and CT = .44, the difference
D  C p  CT
 .3333  .3326
 0.0007
is the maximum amount by which we can increase
CT by adjusting the mean to the target value.
Conclusion?
Derivation of z = 3Cp
USL  LSL
Cp 
6ˆ
X  zˆ  ( X  zˆ )

6ˆ
2 zˆ

6ˆ
z

3
z  3C p