4/9/2020
IENG 486
CH 7: 9, 10, 17, 25 D. H. JENSEN
PROBLEM 7.9 CONSIDER THE TWO PROCESSES SHOWN HERE (THE SAMPLE SIZE n = 5).
SPECIFICATIONS ARE AT 100  10. CALCULATE C
P
, C
PK
, AND C
PM
AND INTERPRET
THESE RATIOS. WHICH PROCESS WOULD YOU PREFER TO USE?
PROCESS A
‗
X
A
= 100
S
A
= 3
_
PROCESS B
‗
X
B
= 105
S
B
= 1
CONVERTING FROM s TO σ, AND COMPUTING C
P
FOR BOTH PROCESSES: SOLN.

C
C
A
P
P
A


 s c
4
USL

6
( 100



3
.
9400
LSL
10 )
6 ( 3

.
 3 .
19
( 100
19 )
,


10 )
B

 c s
4
20
19 .

14
.
9400

1
1 .
04
 1 .
06
C
P
B 
( 100  10 )  ( 100
6 ( 1 .
06 )
 10 )

20
6 .
36
 3 .
14
JUDGING FROM THE PROCESS CAPABILITY RATIO C
P
ALONE: PROCESS B HAS
THE HIGHER C
P
RATIO, AND THEREFORE THE BETTER POTENTIAL CAPABILITY.
COMPUTING C
PK
FOR BOTH PROCESSES:
C
PK
C
PK
A
C
PK
B



MIN
110
3 (
110
3 (
[ C

PL
3 .
19 )
1 .
 105
06
,
)
C
PU
]
100



9 .
MIN
10
57


1

3 .
5
18
 1 .

.
04
57
 LSL
3 
,
USL
3 
  

ASSESSING: PROCESS A HAS A C
P
RATIO THAT IS EQUAL TO ITS’ C
PK
RATIO,
INDICATING THAT THE PROCESS IS CENTERED AND MAY PRODUCE FEWER
DEFECTS.
COMPUTING C
PM
FOR BOTH PROCESSES:
C
PM

6
USL
 2 

( 
LSL
 T ) 2
C
C
PM
PM
A
B


6
6
( 100
3
( 100

.
19

1 .
06 2
10
2
10


)
)
(
(


100
(
( 100

100
105 


10
100 ) 2
10
100 )
)
2
)


20
19 .
14
20
30 .
67


1 .
04
0 .
652
1
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4/9/2020
IENG 486
CH 7: 9, 10, 17, 25 D. H. JENSEN
PROBLEM 7.9 (CONT) ASSESSING: PROCESS A HAS A C
P
RATIO THAT IS EQUAL TO ITS’ C
PK
RATIO AND C
PM
RATIO, INDICATING THAT THE PROCESS IS CENTERED.
PROCESS B HAS A C
PM
RATIO THAT INDICATES THAT IT IS CENTERED TO THE
RIGHT OF THE SPECIFICATION TARGET. NOTE THAT FOR A CENTERED PROCESS
WITH TWO-SIDED C
P
= 1.10, THERE WOULD BE APPROXIMATELY 967 PARTS PER
MILLION DEFECTIVE (SEE TEXTBOOK TABLE 7-2), AND FOR A ONE SIDED C
P
=
1.60, THERE WOULD BE APPROXIMATELY 1 PART PER MILLION DEFECTIVE.
SINCE PROCESS A WOULD GENERATE A HIGHER NUMBER OF DEFECTIVE UNITS,
PROCESS B WOULD BE PREFERRED EVEN THOUGH IT IS NOT CENTERED WITHIN
THE SPECIFICATION TOLERANCE.
PROBLEM 7.10 SUPPOSE THAT 20 OF THE PARTS MANUFACTURED BY THE PROCESS IN
EXERCISE 7-9 WERE ASSEMBLED SO THAT THEIR DIMENSIONS WERE
ADDITIVE; THAT IS, x = x
1
+ x
2
+ . . . + x
20
SPECIFICATIONS ON x ARE 2000  200. WOULD YOU PREFER TO PRODUCE THE
PARTS USING PROCESS A OR PROCESS B? WHY? DO THE CAPABILITY RATIOS
COMPUTED IN EXERCISE 7-7 PROVIDE ANY GUIDANCE FOR PROCESS
SELECTION?
SOLN. ASSUMING THAT WE CAN CHANGE THE CENTERING OF THE PROCESS, I WOULD
PREFER TO CONTINUE TO USE PROCESS B. WHILE PROCESS B WAS NOT
CENTERED, THE PROCESS IS TIGHT ENOUGH TO BE THE BEST CHOICE (FEWEST
% DEFECTIVE) AS ORIGINALLY ESTIMATED IN PROBLEM 7.9. SINCE THE
VARIANCES IN AN ADDITIVE ASSEMBLY ALSO ADD, THE PROCESS WILL
CONTINUE TO BE THE TIGHTER OF THE TWO.
2
6
NOTE: ALL OF THE GUIDANCE FOR THIS INFORMATION MAY BE DERIVED FROM THE
CAPABILITY RATIOS IN THIS EXERCISE. IT IS ALSO POSSIBLE TO COME TO
THE SAME CONCLUSIONS BY COMPUTING THE PROCESS FALLOUT (NON-
CONFORMING PROPORTION UNDER THE DISTRIBUTION CURVES), BUT IT
WOULD TAKE MORE EFFORT.

4/9/2020
IENG 486
CH 7: 9, 10, 17, 25 D. H. JENSEN
PROBLEM 7.17 THE FAILURE TIME IN HOURS OF 10 LSI MEMORY DEVICES IS SHOWN HERE.
PLOT THE DATA ON NORMAL PROBABILITY PAPER, AND IF APPROPRIATE,
ESTIMATE THE PROCESS CAPABILITY. IS IT SAFE TO ESTIMATE THE
PROPORTION OF CIRCUITS THAT FAIL BELOW 1200 HOURS?
1210 1275 1400 1695 1900
2105 2230 2250 2500 2625
SOLN. COMPUTING THE CHART VALUES FOR A NORMAL PROBABILITY PLOT:
4
5
6 j
1
2
3
7
8
9
10
(j - 0.5) / n
0.05
0.15
0.25
0.35
0.45
0.55
0.65
0.75
0.85
0.95
Zj
-1.6449
-1.0364
-0.6745
-0.3853
-0.1257
0.1257
0.3853
0.6745
1.0364
1.6449
Sorted Data
1210
1275
1400
1695
1900
2105
2230
2250
2500
2625
PLOTTING THE DATA:
2.00
1.60
1.20
0.80
0.40
0.00
-0.40
-0.80
-1.20
-1.60
-2.00
SINCE THE POINTS AT EACH END OF THE PLOT DO NOT PASS THE FAT PENCIL
TEST, THE DATA ARE NON-NORMAL. NO FURTHER ANALYSIS IS SUPPORTED.
3
6

4/9/2020
IENG 486
CH 7: 9, 10, 17, 25 D. H. JENSEN
PROBLEM 7.25 TEN PARTS ARE MEASURED THREE TIMES BY THE SAME OPERATOR IN A GAGE
CAPABILITY STUDY. THE DATA ARE SHOWN HERE (SEE TABLE, BELOW).
A.)
SOLN.
DESCRIBE THE MEASUREMENT ERROR THAT RESULTS FROM THE USE OF THIS GAGE.
EACH OF THE SETS OF MEASUREMENTS IN THE DATA TABLE HAS SUMMARY
STATISTICS (MEAN AND RANGE) COMPUTED, AND THE GRAND MEAN AND
AVERAGE RANGE ARE ALSO COMPUTED:
R = 2.30
‗
X = 98.2
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6
PART
NUMBER
1
2
3
4
1
100
MEASUREMENT
2
101
95
101
96
93
103
95
3
100
97
100
97
‗
X
100.3
95.0
101.3
96.0
_
R
1
4
3
2
8
9
10
5
6
7
98
99
95
100
100
100
98
98
97
99
100
98
96
98
98
98
97
99
97.3
98.3
96.7
99.0
99.0
99.0
2
3
2
2
1
3
NOTING THAT n = 3, THE CENTER AND UPPER AND LOWER CONTROL LIMITS FOR
THE X-BAR CHART ARE COMPUTED, AND THE DATA ARE PLOTTED:
‗ _
UCL = X + A
2
R = 98.2 + (1.023)(2.30) = 100.6
‗
CL = X = 98.2
‗ _
LCL = X - A
2
R = 98.2 - (1.023)(2.30) = 95.8
102.00
100.00
98.00
96.00
94.00
92.00
90.00
1
2 3 4 5 6 7 8 9 10
THE LACK OF OUT-OF-CONTROL POINTS (ONLY TWO!) INDICATES THAT THE
SYSTEM IS NOT VERY CAPABLE OF DETECTING DIFFERENCES IN THE PARTS.

4/9/2020
PROBLEM 7.25 (CONT.)
AGAIN USING n = 3, THE CENTER AND UPPER AND LOWER CONTROL LIMITS FOR
THE R – CHART ARE COMPUTED, AND THE DATA ARE PLOTTED:
_
UCL = D
4
R = (2.575)(2.30) = 5.92
_
CL = R = 2.30
_
LCL = D
3
R = (0)(2.30) = 0
7.00
6.00
5.00
4.00
3.00
2.00
1.00
0.00
1 2 3 4 5 6 7 8 9 10
THE LACK OF OUT-OF-CONTROL POINTS (NONE!) IN THIS CHART INDICATES
THAT THE OPERATOR IS CAPABLE IN USING THE GAGE APPROPRIATELY.
B.)
SOLN.
IENG 486
CH 7: 9, 10, 17, 25 D. H. JENSEN
ESTIMATE TOTAL VARIABLITY AND PRODUCT VARIABILITY.
FROM THE 30 INDIVIDUAL MEASUREMENTS IN THE DATA TABLE, THE
STANDARD DEVIATION IS COMPUTED, GIVING THE TOTAL VARIATION:

TOTAL
 2 .
17 , SO  2
TOTAL

 2 .
17  2  4 .
71
5
6
SINCE:

GAGE

R d
2



2 .
30
1 .
693


 1 .
36 , SO  2
GAGE

 1 .
36  2  1 .
85
C.)
SOLN.
GIVEN:
 2
TOTAL
  2
PRODUCT
  2
GAGE
, THEN 
PRODUCT
  2
TOTAL
  2
GAGE

WHAT PERCENTAGE OF TOTAL VARIABLITY IS DUE TO THE GAGE?
TOTAL VARIABILITY IS:


GAGE
TOTAL

1
2 .
.
36
17
 0 .
627 ,  OR  62 .
7 %
4 .
71  1 .
85  1 .
69

4/9/2020
IENG 486
CH 7: 9, 10, 17, 25 D. H. JENSEN
PROBLEM 7.25 (CONT.)
D.) IF SPECIFICATIONS ON THE PART ARE AT 100  15, FIND THE P/T RATIO FOR THIS
GAGE. COMMENT ON THE ADEQUACY OF THE GAGE.
SOLN. THE PRECISION TO TOLERANCE RATIO IS:
6 
USL
GAGE
 LSL

( 100 
6
15
(
)
1 .

36
(
)
100  15 )

8 .
16
30
 0 .
27
SINCE THIS RATIO IS MUCH GREATER THAN 0.10, THE GAGE IS INADEQUATE.
6
6