MIT 2.810 Fall 2015 Homework 9 Solutions

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MIT 2.810 Manufacturing Processes and Systems Homework 9 Solutions

Process Control

November 4, 2015 Problem 1. Control Charts The data shown in Table 1 are x̄ and R values for 24 samples of size n = 5 taken from a process producing bearings. The measurements are made on the inside diameter of the bearing, with only the last three decimals recorded (i.e. 34.5 should be 0.50345). (a) Set up x̄ and R charts on this process. Does the process seem to be in statistical control? (b) If specifications on the diameter are 0.503 ± 0.0010, find the percentage of non-conforming

bearings produced by this process. What assumption(s) do you have to make to determine this number?

Sample Number

x̄ R Sample Number

x̄ R

1 34.5 3 13 35.4 8 2 34.2 4 14 34.0 6 3 31.6 4 15 37.1 5 4 31.5 4 16 34.9 7 5 35.0 5 17 33.5 4 6 34.1 6 18 31.7 3 7 32.6 4 19 34.0 8 8 33.8 3 20 35.1 4 9 34.8 7 21 33.7 2 10 33.6 8 22 32.8 1 11 31.9 3 23 33.5 3 12 38.6 9 24 34.2 2

Table 1: Bearing diameter data Answer: The center line and limits for the x̄ and R charts are given below:

Chart Center Line Control Limits x̄ 𝑥 𝑥 ± 𝐴!𝑅 R 𝑅 UCL = D4𝑅, LCL = D3𝑅

MIT 2.810 Fall 2015 Homework 9 Solutions

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From the given data, we can calculate:

𝑥 =  𝑥!  + 𝑥!  +⋯+  𝑥!"  

24 =  34.5+ 34.2+⋯+ 34.2

24 ≅ 34.

𝑅 =  𝑅!  + 𝑅!  +⋯+  𝑅!"  

24 =  3+ 4+⋯+ 2

24 = 4.7. From a sample size of 5, the factors required to set the control limits are1: A2 = 0.577, D3 = 0 and D4 = 2.115. So the control limits are: x̄ chart: UCL = 𝑥 + 𝐴!𝑅 = 34+ 0.577 ∙ 4.7 = 36.719   LCL = 𝑥 − 𝐴!𝑅 = 34− 0.577 ∙ 4.7 = 31.288

1 Standard tables for estimating the necessary factors for plotting control charts can be found here: http://onlinelibrary.wiley.com/doi/10.1002/0471790281.app6/pdf

UCL

LCL

Center Line

27

29

31

33

35

37

39

41

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

MIT 2.810 Fall 2015 Homework 9 Solutions

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R chart: UCL = D4𝑅 = 2.115 ∙ 4.7 = 9.94 LCL = D3𝑅 = 0

From the x̄ chart, we can conclude that the process is out of control based on the measurements of the 12th and 15th samples. Out of 24 samples, we have 2 samples falling outside the 3σ limits or a probability of 8.3%. We know that for the normal distribution, the probability that a point will fall outside the 3σ limits is 0.27%. So this occurrence is very unusual and the process needs to be evaluated. Note that the R chart does not reveal any irregularities in the process. But the chart would still be useful if we were to apply the Western Electric rules (see Problem 3) here to give a warning signal of a process likely to go out of control.

UCL

LCL

Center Line

0

2

4

6

8

10

12

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

MIT 2.810 Fall 2015 Homework 9 Solutions

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Problem 2. Control Charts The fill volume of soft-drink beverage bottles is an important quality characteristic. The volume is measured (approximately) by placing a gauge over the crown and comparing the height of the liquid in the neck of the bottle against a coded scale. On this scale, a reading of zero corresponds to the correct fill height. Fifteen samples of size n = 10 have been analyzed, and the fill heights are shown in Table 2.

a. Set up x̄ and s control charts on this process. Does the process exhibit statistical control? If necessary, construct revised control limits.

b. Set up an R chart, and compare it with the s chart in part a. Sample Number

x1 x2 x3 x4 x5 x6 x7 x8 x9 x10

1 2.5 0.5 2.0 -1.0 1.0 -1.0 0.5 1.5 0.5 -1.5 2 0.0 0.0 0.5 1.0 1.5 1.0 -1.0 1.0 1.5 -1.0 3 1.5 1.0 1.0 -1.0 0.0 -1.5 -1.0 -1.0 1.0 -1.0 4 0.0 0.5 -2.0 0.0 -1.0 1.5 -1.5 0.0 -2.0 -1.5 5 0.0 0.0 0.0 -0.5 0.5 1.0 -0.5 -0.5 0.0 0.0 6 1.0 -0.5 0.0 0.0 0.0 0.5 -1.0 1.0 -2.0 1.0 7 1.0 -1.0 -1.0 -1.0 0.0 1.5 0.0 1.0 0.0 0.0 8 0.0 -1.5 -0.5 1.5 0.0 0.0 0.0 -1.0 0.5 -0.5 9 -2.0 -1.5 1.5 1.5 0.0 0.0 0.5 1.0 0.0 1.0 10 -0.5 3.5 0.0 -1.0 -1.5 -1.5 -1.0 -1.0 1.0 0.5 11 0.0 1.5 0.0 0.0 2.0 -1.5 0.5 -0.5 2.0 -1.0 12 0.0 -2.0 -0.5 0.0 -0.5 2.0 1.5 0.0 0.5 -1.0 13 -1.0 -0.5 -0.5 -1.0 0.0 0.5 0.5 -1.5 -1.0 -1.0 14 0.5 1.0 -1.0 -0.5 -2.0 -1.0 -1.5 0.0 1.5 1.5 15 1.0 0.0 1.5 1.5 1.0 -1.0 0.0 1.0 -2.0 -1.5

Table 2: Fill height data

MIT 2.810 Fall 2015 Homework 9 Solutions

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Answer: The center line and limits for the x̄ and S charts are given below:

Chart Center Line Control Limits x̄ 𝑥 𝑥 ± 𝐴!𝑠 S 𝑠 UCL = B4𝑠, LCL = B3𝑠

We have m = 15 samples each of size n = 10. The standard deviation for each sample is calculated using the formula:

𝑠 =  𝑥! − 𝑥 !!

!!!𝑛 − 1 .

And,

𝑠 =  𝑠!!

!!!

𝑚 . From the data, we get: 𝑥 =  −0.0033  and  𝑠 =  1.066. For n = 10, we have, A3 = 0.975, B3 = 0.284 and B4 = 1.716. So the control limits are: x̄ chart: UCL = 𝑥 + 𝐴!𝑠 = −0.0033+ 0.975 ∙ 1.066 = 1.036 LCL = 𝑥 − 𝐴!𝑠 = −0.0033− 0.975 ∙ 1.066.=  −1.042

-1.5

-1

-0.5

0

0.5

1

1.5

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Center Line UCL LCL

MIT 2.810 Fall 2015 Homework 9 Solutions

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S chart: UCL = B4𝑠 = 1.716 ∙ 1.066 = 1.829 LCL = B3𝑠 =  0.284 ∙ 1.066.=  0.302

Based on the control charts, the process is in control. We now plot the R chart to see if we get the same result. From the data, we get, R̄ = 3.2. For n = 10, we have, D4 = 1.777 and D3 = 0.223. The control limits are: UCL = D4𝑅 = 1.777 ∙ 3.2 = 5.686 LCL = D3𝑅 = 0.223 ∙ 3.2 = 0.713

The R chart also indicates that the process is in control.

0

0.4

0.8

1.2

1.6

2

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Center Line UCL LCL

0

1

2

3

4

5

6

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Center Line

MIT 2.810 Fall 2015 Homework 9 Solutions

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Problem 3. Out of Control Detection [Adapted from Douglas Montgomery’s ‘Introduction to Statistical Control’] The Western Electric Handbook (1956) suggests a set of decision rules for detecting nonrandom patterns on control charts. Specifically, it suggests concluding that the process is out of control if either:

1. One point plots outside the three-sigma control limits, or 2. Two out of three consecutive points plot beyond the two-sigma warning limits, or 3. Four out of five consecutive points plot at a distance of one-sigma or beyond from the

center line, or 4. Eight consecutive points plot on one side of the centerline.

Those rules apply to one side of the centerline at a time. Therefore, a point above the upper warning limit followed immediately by a point below the lower warning limit would not signal an out-of-control alarm. Calculate the probability of the first two patterns occurring assuming the data points are independent but identically distributed with a normal distribution. Hint: You can choose r points out of n consecutive points in nCr ways, where, nCr = !!

!! !!! !. Then

calculate the probability of a point falling outside the warning limit. We have r points falling beyond a warning limit and (n-r) points falling before it. Since the points are independent, the probability of a pattern of points is the product of their individual probabilities. Answer: From Prof. Hardt’s lecture notes, we know that the area under the normal distribution curve for various spreads, zσ, is approximately written as:

Within 1σ 0.68 Within 2σ 0.95 Within 3σ 0.997

Therefore, by symmetry, the area between 1σ and 2σ limits on either side is (0.95-0.68)/2 = 0.135. The area between 2σ and 3σ limits on either side is (0.997-0.95)/2 = 0.0235. 1. The probability that one point plots outside the 3σ limits is (1 – 0.997) = 0.003 or 0.3%. 2. The probability that a point will fall beyond the 2σ limit on one side of the center line is (1 -

0.95)/2 = 0.025. Therefore, the probability that two out of three consecutive points will fall outside the 2σ limit on either side is: 2 x 3C2 x 0.0252 x (1 – 0.025) = 0.0036 or 0.36%.

We can also determine the probabilities for the other two rules mentioned above:

MIT 2.810 Fall 2015 Homework 9 Solutions

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3. The probability that one point falls beyond the 1σ sigma limit on one side of the centerline is (1 – 0.68)/2 = 0.16. Therefore, the probability that four out of five consecutive points will fall outside the 1σ limit on either side is: 2 x 5C4 x 0.164 x (1 – 0.16) = 0.005 or 0.5%.

4. The area on either side of the centerline equals 0.5. The probability that eight consecutive points fall on one side of the center line is (0.5)8 = 0.0039 or. 0.39%.