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Hi, could you please answer this question by using excle with full details and s

ID: 380939 • Letter: H

Question

Hi, could you please answer this question by using excle with full details and showin formulas please. thank u

You're measuring the voltage drop across a resistor circuit and the engineers have indicated that the specification limits are 120V +/- 2.5 V. If the voltage exceeds those specifications, the circuit must be scrapped at a cost of $500.00 per circuit. You've collected the following samples.

a. compute the value of k in the Taguchi loss function

b. Based on the data provided, what is the expected loss from this process?

measurements 124.0 120.0 117.0 120.0 118.0 117.0 122.0 119.0 119.0 118.0 120.0 116.0 116.0 122.0 117.0 122.0 118.0 116.0 122.0 119.0 123.0 120.0 120.0 119.0 123.0 120.0 120.0 118.0 119.0 119.0 118.0 119.0 116.0 121.0 121.0 118.0 121.0 119.0 119.0 123.0 119.0 116.0 119.0 117.0 122.0 117.0 120.0 120.0 118.0 121.0 117.0 121.0 122.0 122.0 120.0 123.0 119.0 119.0 120.0 119.0 116.0 119.0 122.0 117.0 123.0

Explanation / Answer

Taguchi loss function

a. compute the value of k in the Taguchi loss function

K=80

b. Based on the data provided, what is the expected loss from this process?

=$23903.36

measurements, n=65

x

x-P.M

(x-P.M)^2

1

124

4.52

20.4304

2

117

-2.48

6.1504

3

120

0.52

0.2704

4

122

2.52

6.3504

5

123

3.52

12.3904

6

120

0.52

0.2704

7

118

-1.48

2.1904

8

118

-1.48

2.1904

9

119

-0.48

0.2304

10

117

-2.48

6.1504

11

117

-2.48

6.1504

12

123

3.52

12.3904

13

116

-3.48

12.1104

14

120

0.52

0.2704

15

122

2.52

6.3504

16

116

-3.48

12.1104

17

118

-1.48

2.1904

18

120

0.52

0.2704

19

120

0.52

0.2704

20

119

-0.48

0.2304

21

121

1.52

2.3104

22

116

-3.48

12.1104

23

120

0.52

0.2704

24

121

1.52

2.3104

25

119

-0.48

0.2304

26

119

-0.48

0.2304

27

117

-2.48

6.1504

28

119

-0.48

0.2304

29

116

-3.48

12.1104

30

116

-3.48

12.1104

31

120

0.52

0.2704

32

118

-1.48

2.1904

33

116

-3.48

12.1104

34

119

-0.48

0.2304

35

119

-0.48

0.2304

36

120

0.52

0.2704

37

122

2.52

6.3504

38

119

-0.48

0.2304

39

122

2.52

6.3504

40

120

0.52

0.2704

41

119

-0.48

0.2304

42

122

2.52

6.3504

43

122

2.52

6.3504

44

119

-0.48

0.2304

45

119

-0.48

0.2304

46

121

1.52

2.3104

47

119

-0.48

0.2304

48

117

-2.48

6.1504

49

118

-1.48

2.1904

50

122

2.52

6.3504

51

120

0.52

0.2704

52

117

-2.48

6.1504

53

118

-1.48

2.1904

54

118

-1.48

2.1904

55

117

-2.48

6.1504

56

119

-0.48

0.2304

57

123

3.52

12.3904

58

119

-0.48

0.2304

59

121

1.52

2.3104

60

123

3.52

12.3904

61

122

2.52

6.3504

62

121

1.52

2.3104

63

120

0.52

0.2704

64

119

-0.48

0.2304

65

123

3.52

12.3904

S(x-process mean)2

282.22

SUM

7766

S.D

PROCESS MEAN                             =sum/n=

119.48

2.08

               

PROCESS MEAN

=sum/n

=7766/65

=119.48

S.D

=Sqrt ( S (x-process mean)2 / n )

= Sqrt ( 282.22 / 65 )

=2.08

65 samples (constituting a single sample set or a process) are collected from the resistor circuit

T=120V

X=120+ 2.5 V =122.5 V


The average of the points or samples, PM= 119.48

Standard deviation =2.08

Find k

If the voltage exceeds those specifications in any of the sample, the circuit must be scrapped at a cost of $500.00 per circuit.

Loss at a point /loss if a sample exceeds the specification limit:

L(x) = k*(x-t) ^2

Where,

k = loss coefficient

x = measured value

t = target value

If the circuit voltage exceeds the specification limit, then it has to be scrapped for $500

$500.00= k*(122.5- 120)2

K=80

Average Loss of a sample set/ process:

L = k*(s2 + (pm - t)2)

Where,

s = standard deviation of sample

PM= process mean

L = 80 * (2.082 + (119.48- 120)2) = $367.74

So, the average loss per sample in this set is $8.73.For the loss of the total 30 parts produced

Expected loss from the process

Total Loss form the process

= Avg. Loss * number of samples

=$367.74 * 65

=$23903.36

measurements, n=65

x

x-P.M

(x-P.M)^2

1

124

4.52

20.4304

2

117

-2.48

6.1504

3

120

0.52

0.2704

4

122

2.52

6.3504

5

123

3.52

12.3904

6

120

0.52

0.2704

7

118

-1.48

2.1904

8

118

-1.48

2.1904

9

119

-0.48

0.2304

10

117

-2.48

6.1504

11

117

-2.48

6.1504

12

123

3.52

12.3904

13

116

-3.48

12.1104

14

120

0.52

0.2704

15

122

2.52

6.3504

16

116

-3.48

12.1104

17

118

-1.48

2.1904

18

120

0.52

0.2704

19

120

0.52

0.2704

20

119

-0.48

0.2304

21

121

1.52

2.3104

22

116

-3.48

12.1104

23

120

0.52

0.2704

24

121

1.52

2.3104

25

119

-0.48

0.2304

26

119

-0.48

0.2304

27

117

-2.48

6.1504

28

119

-0.48

0.2304

29

116

-3.48

12.1104

30

116

-3.48

12.1104

31

120

0.52

0.2704

32

118

-1.48

2.1904

33

116

-3.48

12.1104

34

119

-0.48

0.2304

35

119

-0.48

0.2304

36

120

0.52

0.2704

37

122

2.52

6.3504

38

119

-0.48

0.2304

39

122

2.52

6.3504

40

120

0.52

0.2704

41

119

-0.48

0.2304

42

122

2.52

6.3504

43

122

2.52

6.3504

44

119

-0.48

0.2304

45

119

-0.48

0.2304

46

121

1.52

2.3104

47

119

-0.48

0.2304

48

117

-2.48

6.1504

49

118

-1.48

2.1904

50

122

2.52

6.3504

51

120

0.52

0.2704

52

117

-2.48

6.1504

53

118

-1.48

2.1904

54

118

-1.48

2.1904

55

117

-2.48

6.1504

56

119

-0.48

0.2304

57

123

3.52

12.3904

58

119

-0.48

0.2304

59

121

1.52

2.3104

60

123

3.52

12.3904

61

122

2.52

6.3504

62

121

1.52

2.3104

63

120

0.52

0.2704

64

119

-0.48

0.2304

65

123

3.52

12.3904

S(x-process mean)2

282.22

SUM

7766

S.D

PROCESS MEAN                             =sum/n=

119.48

2.08

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