IMSE561/EMGT525 - TOTAL QUALITY MANAGEMENT AND SIX SIGMA

HOMEWORK-6 Name: Shyam V.H Karumanchi UM ID 0833-3824 Question 1: Describe the design procedure for robust design REFERENCES: 1. Chapter 20 Quality engineering lecture notes Products and services should be designed to be inherently defect free and of high quality.
Robustness

The product or process performs its intended function well within user profiles and insensitive to the variation including: Variation in production Differences in materials Differences in users or operators aging of the product or process, and the product or process accomplishes this without major cost impact. General procedure for robust design 1. Problem Formulation: This step consists of identifying the main function, developing the P-diagram, defining the ideal function and S/N ratio, and planning the experiments. The experiments involve changing the control, noise and signal factors systematically using orthogonal arrays. 2. Data Collection/Simulation: The experiments may be conducted in hardware or through simulation. It is not necessary to have a full-scale model of the product for the purpose of experimentation. It is sufficient and more desirable to have an essential model of the product that adequately captures the design concept. Thus, the experiments can be done more economically. 3. Factor Effects Analysis: The effects of the control factors are calculated in this step and the results are analyzed to select optimum setting of the control factors. 4. Prediction/Confirmation: In order to validate the optimum conditions we predict the performance of the product design under baseline and optimum settings of the control factors. Then we perform confirmation

experiments under these conditions and compare the results with the predictions. If the results of confirmation experiments agree with the predictions, then we implement the results. Otherwise, the above steps must be iterated.

(Figure from text)

Question 2: exercise 3 chapter 20 Solution: given specification of electric device 24+/- 0.4(delta) Amp and average repair cost is 32$ (say A) Given 13 samples with readings and their basic statistic data is given below
Variable Amp N 13 N* 0 Mean 23.946 SE Mean 0.0584 StDev 0.211 Minimum 23.600 Q1 23.800 Median 23.900 Q3 24.150

K= A/∆2 = 32/0.4*0.4 = 200 Average loss L` = K( s2+(y- - t)2) = 200(0.2112+(23.946-24)2= 9.4874

Question 3: exe 4 CH20 Solution: given steel shaft with out of round condition with 8 sample readings
Variable readings N 8 N* 0 Mean 0.03625 SE Mean 0.00324 StDev 0.00916

a. At 0.03 average loss is 15$ loss function K= A/∆2 = 15/0.03*0.03 = 16666.67 b. Loss at 0.05 = 16666.67*0.052 = 41.667 c. Average loss L’ = k(y2+σ2) = 16666.67(0.03625*0.03625+ 0.00916*0.00916) = 23.29 Question 4: exe 7 Ch20 Solution: Given 3 two-level factor and 3 suspected two factor interactions …need to determine DOF and OA DOF = 3(2-1) + 3(2-1) +1 = 7 OA (ortho array) is given by OA8 with 7 factors TC 1 2 3 4 5 6 7 8 A 1 1 1 1 2 2 2 2 B 1 1 2 2 1 1 2 2 C 1 1 2 2 2 2 1 1 D 1 2 1 2 1 2 1 2 E 1 2 1 2 2 1 2 1 F 1 2 2 1 1 2 2 1 G 1 2 2 1 2 1 1 2

Question 5: ex11 Ch20 Solution: 3 factors with 3 interactions 1 (1)

6

3

3(4)

5

2(2)

Question 6: ex13 Ch20 Solution: New process has been developed and the average temperature is 21C and Sample SD is 0.8C where n= 5 a. S/N ratio = 10log10(y2/s2 – 1/n) = 28.38 b. Improvement = 7.38 c. Changing the units to Kelvin and we get S/N ratio for this problem as 10log10(2942/273.82- 1/5) = -0.21 For example problem we get 10log10(2942/2752 – 1/4) = -0.491 Conclude that changing units will not get the same results

Question 7: ex16 Ch20 Solution: Using the larger the better experimental design TC 1 2 3 4 5 6 7 8 A1 1 1 1 1 2 2 2 2 B2 1 1 2 2 1 1 2 2 C3 1 1 2 2 2 2 1 1 D4 1 2 1 2 1 2 1 2 E5 1 2 1 2 2 1 2 1 F6 1 2 2 1 1 2 2 1 G7 1 2 2 1 2 1 1 2 R1 19 20 24 22 26 25 25 25 R2 25 24 22 25 20 26 20 21 MEAN 22.0 22.0 23.0 23.5 23.0 25.5 22.5 23.0

Using MINITAB to get the response table as shown below Response Table for Means Level 1 2 Delta Rank A 1 22.63 23.50 0.88 2.5 B 2 23.13 23.00 0.13 7 C 3 22.38 23.75 1.38 1 D 4 22.63 23.50 0.88 2.5 E 5 23.38 22.75 0.63 4.5 F 6 22.88 23.25 0.38 6 G 7 23.38 22.75 0.63 4.5

Highest delta is 1.38 and half the difference is 0.69 • • • • All delta values which are equal or greater than 0.69 will be considered to be stronger effect Stronger the better performance...the highlighted levels in response table are strong factors Overall average from the means is 23.063 Confirmation run = C2+D2+A2-(n-1)*overall average = 24.624 Response Table for Signal to Noise Ratios Larger is better Level 1 2 Delta Rank A 1 26.98 27.29 0.31 3 B 2 27.12 27.15 0.03 7 C 3 26.84 27.43 0.59 1 D 4 26.93 27.34 0.42 2 E 5 27.27 27.00 0.27 4 F 6 27.03 27.24 0.21 6 G 7 27.25 27.02 0.22 5

Question 8: ex 17 Ch 20 Solution: Using minitab , from given table we obtain following mean and S/N ratio NOMINAL THE BEST tc A B C N1 N2 S N RATIO MEAN1

1 2 3 4

1 1 2 2

1 2 1 2

1 2 2 1

1.75 1.34 2.67 2.23

1.84 2.13 2.43 2.73

29.0067 9.8437 23.5369 16.9199

1.795 1.735 2.550 2.480

Response Table for Signal to Noise Ratios Nominal is best (10*Log10(Ybar**2/s**2)) Level 1 2 Delta Rank A 19.43 20.23 0.80 3 B 26.27 13.38 12.89 1 C 22.96 16.69 6.27 2

Response table for means obtained as below

Level 1 2 Delta Rank

A 1.765 2.515 0.750 1

B 2.172 2.107 0.065 2

C 2.138 2.143 0.005 3

PLOT FOR SN ratio
27.5
Variable B_1 * wa C_1 * wb A_1 * wc

25.0

22.5 Y-Data

20.0

17.5

15.0

1

B

2

3 X-Data

C

4

A5

Plot for means

2.6 2.5 2.4 2.3 Y-Data 2.2 2.1 2.0 1.9 1.8 1.7 1 2 3 X-Data 4 5

Variable A_2 * wa B_2 * wb C_2 * wc

Highlighted delta in each response table shows the strong effect

Question 9: ex20 Ch 20 Solution: given the results of a smaller the better experimental design tc 1 2 3 4 5 6 7 8 b 1 1 1 1 2 2 2 2 a 1 1 2 2 1 1 2 2 ab 1 1 2 2 2 2 1 1 c 1 2 1 2 1 2 1 2 ux 1 2 1 2 2 1 2 1 ac 1 2 2 1 1 2 2 1 d 1 2 2 1 2 1 1 2 s/n 32.1 33.6 32.8 31.7 31.2 33.7 32.3 33.6 MEAN1 32.1 33.6 32.8 31.7 31.2 33.7 32.3 33.6

Response Table for Means Level 1 2 Delta Rank b 32.55 32.70 0.15 6 a 32.65 32.60 0.05 7 ab 32.90 32.35 0.55 4 c 32.10 33.15 1.05 1 ux 33.05 32.20 0.85 3 ac 32.15 33.10 0.95 2 d 32.45 32.80 0.35 5

• • •

Larger difference is C 1.05 Stronger effects are C, UX, AC and AB For interactions between A-C: a1c1 = 32.15 A2c1 = 32.5 A1c2 = 33.65 A2c2 = 32.65 Average of level 1 & level 2 A-B : a1b1 = 32.85 A1b2 = 32.45 A2b1 = 32.25 A2b2 = 33.95

33.8 33.6 33.4 33.2 Y-Data 33.0 32.8 32.6 32.4 32.2 32.0 1.0 1.2 1.4 w 1.6 1.8 2.0

Variable a1 a2

Red C1 at a1 and a2 Black C2

34.0

Red b1 at a1 and a2
33.5

Variable a-1 a-2

Y-Data

33.0

Similarly black is b2 at a1 and a2

32.5

32.0 1.0 1.2 1.4 w 1.6 1.8 2.0

A1B2, A1C2 gives better results.

Question 10: Make the following decision using AHP method: o Objective: Select graduate study programs between UMD or UM-Ann Arbor o Attributes/Criteria: Cost, Convenience, Timing o Alternatives: UMD, UM-AA Solution: numerical rating 1: equally preferred 3: moderately preferred 5: strongly preferred 7: Very strongly preferred 9: Extremely preferred verbal assignment

Step1 pair wise comparison COST UMD UM-AA UMD 1 1/7 1.14 UM-AA 7 1 8 Timing UMD UM-AA Convenience UMD UMD UM-AA 1 1/3 1.33 UM-AA 3 1 4 1.2 6 UMD 1 1/5 UM-AA 5 1

Step 2 Matrix normalization and priority vector Cost UMD UM-AA UMD .88 .12 `UMAA .875 .125 AVG .877 .122 Timing UMD UMAA UMD .83 .16 UM AA ,83 .16 AVG .83 .16

Convenience UMD UMD UMAA .75 .24

UMAA .75 .25

AVG .75 .245

Pair wise comparison of three criteria Attribute Cost Timing Convenience Cost 1 1/7 1/5 1.34 Timing 7 1 1/3 8.33 Convenience 5 3 1 9

Weighing function Attribute Cost Timing Convenience Cost .75 .1 .15 Timing .84 .12 .04 Convenience .55 .33 .11 Weighted function .71 .18 .1

Overall UMD priority .71*.877+.18*.83+.1*.75 = .8381 + .1*.245 = .14 UMD preferred over UM-AA considering given attributes

UM-AA priority .71*.122 + .18*.16