IENG 4861

11/4/2018CH7: 9, 10, 17, 25D.H. JENSEN6

PROBLEM7.9CONSIDER THE TWO PROCESSES SHOWN HERE (THE SAMPLE SIZE n = 5). SPECIFICATIONS ARE AT 100  10. CALCULATE CP, CPK, AND CPM AND INTERPRET THESE RATIOS. WHICH PROCESS WOULD YOU PREFER TO USE?

PROCESS APROCESS B

‗‗

XA = 100XB = 105

SA = 3SB = 1

_

SOLN.CONVERTING FROM s TO σ, AND COMPUTING CP FOR BOTH PROCESSES:

JUDGING FROM THE PROCESS CAPABILITY RATIO CP ALONE: PROCESS B HAS THE HIGHER CP RATIO, AND THEREFORE THE BETTER POTENTIAL CAPABILITY.

COMPUTING CPK FOR BOTH PROCESSES:

ASSESSING: PROCESS A HAS A CP RATIO THAT IS EQUAL TO ITS’ CPK RATIO, INDICATING THAT THE PROCESS IS CENTERED AND MAY PRODUCE FEWER DEFECTS.

COMPUTING CPM FOR BOTH PROCESSES:

PROBLEM 7.9 (CONT) ASSESSING: PROCESS A HAS A CP RATIO THAT IS EQUAL TO ITS’ CPK RATIO AND CPM RATIO, INDICATING THAT THE PROCESS IS CENTERED. PROCESS B HAS A CPM RATIOTHAT INDICATES THAT IT IS CENTERED TO THE RIGHT OF THE SPECIFICATION TARGET. NOTE THAT FOR A CENTERED PROCESS WITH TWO-SIDED CP = 1.10, THERE WOULD BE APPROXIMATELY 967 PARTS PER MILLION DEFECTIVE (SEE TEXTBOOK TABLE 7-2), AND FOR A ONE SIDED CP = 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.

PROBLEM7.10SUPPOSE THAT 20 OF THE PARTS MANUFACTURED BY THE PROCESS IN EXERCISE 7-9 WERE ASSEMBLED SO THAT THEIR DIMENSIONS WERE ADDITIVE; THAT IS,

x = x1 + x2 + . . . + x20

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.

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.

PROBLEM7.17THE 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?

12101275140016951900

21052230225025002625

SOLN.COMPUTING THE CHART VALUES FOR A NORMAL PROBABILITY PLOT:

j / (j - 0.5) / n / Zj / Sorted Data
1 / 0.05 / -1.6449 / 1210
2 / 0.15 / -1.0364 / 1275
3 / 0.25 / -0.6745 / 1400
4 / 0.35 / -0.3853 / 1695
5 / 0.45 / -0.1257 / 1900
6 / 0.55 / 0.1257 / 2105
7 / 0.65 / 0.3853 / 2230
8 / 0.75 / 0.6745 / 2250
9 / 0.85 / 1.0364 / 2500
10 / 0.95 / 1.6449 / 2625

PLOTTING THE DATA:

SINCE THE POINTS AT EACH END OF THE PLOT DO NOTPASS THE FAT PENCIL TEST, THE DATA ARE NON-NORMAL. NO FURTHER ANALYSIS ISSUPPORTED.

PROBLEM 7.25TEN PARTS ARE MEASURED THREE TIMES BY THE SAME OPERATOR IN A GAGE CAPABILITY STUDY. THE DATA ARE SHOWN HERE (SEE TABLE, BELOW).

A.)DESCRIBE THE MEASUREMENT ERROR THAT RESULTS FROM THE USE OF THIS GAGE.

SOLN.EACH OF THE SETS OF MEASUREMENTS IN THE DATA TABLE HAS SUMMARY STATISTICS (MEAN AND RANGE) COMPUTED, AND THE GRAND MEAN AND AVERAGERANGE ARE ALSO COMPUTED:

‗

X = 98.2R = 2.30

PART NUMBER / MEASUREMENT / ‗
X / _
R
1 / 2 / 3
1 / 100 / 101 / 100 / 100.3 / 1
2 / 95 / 93 / 97 / 95.0 / 4
3 / 101 / 103 / 100 / 101.3 / 3
4 / 96 / 95 / 97 / 96.0 / 2
5 / 98 / 98 / 96 / 97.3 / 2
6 / 99 / 98 / 98 / 98.3 / 1
7 / 95 / 97 / 98 / 96.7 / 3
8 / 100 / 99 / 98 / 99.0 / 2
9 / 100 / 100 / 97 / 99.0 / 3
10 / 100 / 98 / 99 / 99.0 / 2

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 + A2R = 98.2 + (1.023)(2.30) = 100.6

‗

CL = X = 98.2

‗ _

LCL = X - A2R = 98.2 - (1.023)(2.30) = 95.8

THE LACK OF OUT-OF-CONTROL POINTS (ONLY TWO!) INDICATES THATTHE SYSTEM IS NOT VERY CAPABLE OF DETECTING DIFFERENCES IN THE PARTS.

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 = D4R = (2.575)(2.30) = 5.92

_

CL = R = 2.30

_

LCL = D3R = (0)(2.30) = 0

THE LACK OF OUT-OF-CONTROL POINTS (NONE!) IN THIS CHART INDICATES THAT THE OPERATOR IS CAPABLE IN USING THE GAGE APPROPRIATELY.

B.)ESTIMATE TOTAL VARIABLITY AND PRODUCT VARIABILITY.

SOLN.FROM THE 30 INDIVIDUAL MEASUREMENTS IN THE DATA TABLE, THE STANDARD DEVIATION IS COMPUTED, GIVING THE TOTAL VARIATION:

SINCE:

GIVEN:

C.)WHAT PERCENTAGE OF TOTAL VARIABLITY IS DUE TO THE GAGE?

SOLN.TOTAL VARIABILITY IS:

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:

SINCE THIS RATIO IS MUCH GREATER THAN 0.10, THE GAGE IS INADEQUATE.