Assignment 1_solutions

# Assignment 1_solutions - University of Toronto Department...

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Unformatted text preview: University of Toronto Department of Mechanical and Industrial Engineering MIE364S: Methods of Quality Control and Improvement Assignment 1 Solutions Due: Tuesday, Feb 23, 2010 (Lecture Time) Instructions: First, solve the problems without using Minitab, and then develop also Minitab solutions (where appropriate). Submit both. Marks 1. Samples of 4 items each are taken from a manufacturing process at regular intervals. 200 and the standard deviation 10. Suppose the process mean 10 12 14 a. Find both the 3 sigma control limits and the probability limits ( 0.002) to control the process mean. b. Find the average run length and the standard deviation of the run length for both charts (in a.), in 209. the following cases: i) process is in control, ii) process mean shifts to c. For both charts in a., what is the required sample size to detect this shift in the process mean on the first or second sample following the shift with the probability 0.9? For which of the two charts is the required sample size bigger and why? d. Assume that the following rule is applied to the control chart with 3 sigma limits constructed in a. The process is declared out of control when either a point plots outside the 3 sigma limits or two consecutive points plot between the two and three sigma control limits on the same side of 209, what is the probability that the shift the control chart. If the process mean shifts to will be detected on the first sample following the shift? 14 Question 1 10 a) 4 10 ̂ . 200 3.09 10 √4 10 √4 3-sigma control limits: 3 3 √4 √4 200 200 3 3 185 215 Probability limits ( . . 0.002): 200 200 3.09 3.09 10 √4 10 √4 √4 √4 184.55 215.45 1 12 b) 3-sigma control limits: 0.0027 1 370.37 · √1 369.87 209 0.9 10 3 0.9 2 3 0.9 2 1 8.69 1 0.8849 · 8.175 0.002): 200 1.2 0.8849 Probability limits ( 1 1 0.998 500 · 209 200 499.5 10 3.09 1 14 c) 0.9 3.09 10.15 9.64 0.9 2 1.29 0.90147 0.9 2 1 0.90147 · P detect on 1st or 2nd β 0.316 1 β β1 β 1 β1 β 1 β 0.9 3-sigma: 0.48 3 0.9√ n 14.95 n 15 Probability limits ( 0.002): 0.48 3.09 0.9√ n 15.73 n 16 Probability limits 0.002 are wider than 3-sigma limits and require greater n to detect a shift in the process. The reason why the probability limits are wider is the smaller value of . 2 14 d) Let the following 4 regions of the control chart be defined: A: out of three sigma control limits B: between two and three sigma control limits, below centre line C: within two sigma limits D: between two and three sigma control limits, above centre line The probabilities given that the process mean is α 0.0027 PA 0.0214 PB 0.9545 PC 0.0214 PD 200 : 209 are: The probabilities given that the process mean shifts to 1 0.11507 PA 2 3 3.8 4.8 0.00007 PB √ √ PC 2 2 0.2 3.8 0.57919 √ √ PD 3 2 1.2 0.2 0.30567 √ √ P detect out of control on the first sample after the shift P BP B P DP D PA 0.11507 0.0214 0.00007 0.0214 0.30567 0.121612836 2. A production process was designed for mass production of ball bearings 0.5 in. in diameter. A quality control engineer randomly sampled four ball bearings every hour for 15 hours. The following data resulted: Hour 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Diameter of ball bearing (in.) 0.481 0.511 0.463 0.495 0.507 0.486 0.491 0.491 0.495 0.483 0.48 0.508 0.511 0.453 0.539 0.494 0.479 0.51 0.554 0.499 0.488 0.479 0.497 0.52 0.511 0.508 0.501 0.51 0.507 0.522 0.521 0.515 0.494 0.521 0.503 0.497 0.522 0.482 0.524 0.477 0.509 0.501 0.53 0.52 0.518 0.523 0.47 0.486 0.49 0.473 0.48 0.5 0.464 0.498 0.473 0.521 0.464 0.476 0.513 0.491 3 Assume that the specification limits are: LSL=0.42 in. and USL=0.58 in. 12 a. Check, if the process is in statistical control. Use and R charts with probability limits, 0.002. If necessary, revise the trial control limits. Then, check normality by plotting the incontrol data on the attached normal paper. b. Check the claim that 1.3 at 0.05 significance level and estimate the fraction nonconforming below the LSL and above the USL. Find the 95% confidence interval for . Is and . the process centered at the target value? Compute 12 Question 2 12 a) ( , ) with probability limits ( First calculate for the R Chart: 0.048 0.041 0.029 1 15 4 . 0.002): 0.021 0.01 0.053 0.04087 4 0.04087 0.199446 2.059 0.04087 5.308804 2.059 0.028 0.015 0.027 0.086 0.027 0.057 0.075 0.047 0.049 0.00396 4 . 4 0.10538 4 0.01985 For X Chart: 0.48750 0.49600 0.51500 1 15 . . 0.49375 0.50750 0.49925 0.49150 0.51625 0.48575 0.49925 0.50375 0.48900 0.51050 0.50125 0.48600 0.49882 0.49882 0.49882 3.09 3.09 0.01985 √4 0.01985 √4 4 0.46815175 0.52948825 √4 √4 ̂ 0.49882 Process is in statistical control. R Chart of C1, ..., C4 0.10 UCL=0.10538 0.08 Sample Range 0.06 _ R=0.0420 0.04 0.02 0.00 1 2 3 4 5 6 7 8 9 Sample 10 11 12 13 14 15 LCL=0.00396 5 Xbar Chart of C1, ..., C4 0.53 0.52 0.51 Sample Mean 0.50 0.49 0.48 0.47 0.46 1 2 3 4 5 6 789 Sample 10 11 12 13 14 15 _ _ X=0.49882 UCL=0.52948825 LCL=0.46815175 Probability Plot: Probability Plot of Diameter Normal - 95% CI 99.9 99 95 90 80 70 60 50 40 30 20 10 5 1 0.1 Mean StDev N AD P-Value 0.4988 0.02032 60 0.249 0.738 Percent 0.42 0.44 0.46 0.48 0.50 CDF 0.52 0.54 0.56 0.58 Normality assumption is acceptable. 6 12 b) ∑ = 0.02032 0.58 0.42 6 0.02032 1.3123 6 Claim Cp > 1.3 : 1.3 : 1.3 . , 59 1 2 9 · 59 1 1.65 2 9 · 59 42.2939163 · . , 1.3 · 59 42.2939163 1.535431363 C> Therefore, fail to reject . , , do not accept supplier’s claim. 2 9 · 59 2 9 · 59 59 1 2 9 · 59 2 9 · 59 1.96 39.6536055 . , 59 1 1.96 82.12091777 95% CI for Cp · · 3 , 3 0.0203 0.0203344 6 7 0.16 6 0.0203344 min 1.2927,1.3320 0.498817 1.3114 0.58 2 1.2927 0.42 0.0004135 . , 1 . , 1.3123 · 39.6536055 59 1.3123 · 82.12091777 59 1.075841923 1.548224801 1 16 3. Assume that in addition to the data in Problem 2, the following measurements are available: Hour Diameter of ball bearing (in.) 16 0.461 0.458 0.472 17 0.501 0.495 0.511 0.505 0.515 18 0.485 0.492 0.502 19 0.512 0.497 0.499 20 0.482 0.49 0.495 Set up the and S charts, both with 3 sigma limits, using the complete set of data (i.e., the data collected in 20 hours). Estimate the process mean and standard deviation. Question 3 16 a) S-Chart with 3-sigma control limits: 1 ∑ 1 77 20 1 58 1 0.487500 0.0204206 4 6 0.496000 0.0176068 4 11 0.515000 0.0126754 4 16 0.463667 0.0073711 3 2 0.493750 0.0091424 4 7 0.507500 0.0045092 4 12 0.499250 0.0254738 4 17 0.505400 0.0079246 5 3 0.491500 0.0127671 4 8 0.516250 0.0068981 4 13 0.485750 0.0117863 4 18 0.493000 0.0085440 3 4 0.499250 0.0359850 4 9 0.503750 0.0120934 4 14 0.489000 0.0257294 4 19 0.502667 0.0081445 3 5 0.510500 0.0317122 4 10 0.501250 0.0252108 4 15 0.486000 0.0211187 4 20 0.489000 0.0065574 4 8 ∑ ∑ 58 4 58 4 58 1 0.018385 1 3 1.013333333 0.018385 1.013333333 ̂ ̂ 3· 3· 0.018143092 1 1 For 4 4, 0.9213 ̂ 58 ̂ 58 3, 0.8862 ̂ 58 ̂ 58 5, 0.9400 ̂ 58 ̂ 58 4 4 3· 3· 1 1 4 4 0 0.037880194 For 3 3 3 3· 3· 1 1 3 3 0 0.041295649 For 5 5 5 3· 3· 1 1 5 5 0 0.035624385 9 S Chart of Diam 0.04 UCL=0.04203 0.03 Sample StDev 0.02 _ S=0.01636 0.01 0.00 1 3 5 7 9 11 Sample 13 15 17 19 LCL=0 Tests performed with unequal sample sizes -Chart with 3-sigma control limits: ∑ 0.49741561 ∑ 3 3 For 4, 3 √4 3 √4 0.4702009722 0.524630249 For 3, 3 √3 3 √3 0.465990853 0.528840368 For 5, 3 √5 3 √5 0.473074098 0.521757123 1 0 Xbar Chart of Diam 0.53 0.52 0.51 Sample Mean 0.50 0.49 0.48 0.47 LCL=0.465990853 0.46 1 3 5 7 9 11 Sample 13 15 1 UCL=0.528840368 _ _ X=0.49742 17 19 Tests performed with unequal sample sizes Delete sample 16. n LCL 4 0.470201 3 0.465991 Original UCL LCL New UCL 0.52463 0.471569 0.525998 0.52884 0.467359 0.530209 5 0.473074 0.521757 0.474442 0.523125 1 1 Xbar Chart of Diam 0.54 0.53 0.52 Sample Mean 0.51 0.50 0.49 0.48 0.47 0.46 1 3 5 7 9 11 Sample 13 15 17 19 LCL=0.46735904 _ _ X=0.49878 UCL=0.530208555 Tests performed with unequal sample sizes ̂ 0.49878 0.018143092 10 4. To ensure chemical purity of a commercial organic chemical, measurements of the level of certain intermediate chemical material are taken every 4 hours. Data from 22 samples appear below: Sample Number Level Sample Number Level 1 15.3 12 15.9 2 15.7 13 14.7 3 14.4 14 15.2 4 14 15 14.6 5 15.2 16 13.7 6 15.8 17 12.9 7 16.7 18 13.2 8 16.6 19 14.1 9 15.9 20 14.2 10 17.4 21 13.8 11 19.7 22 14.8 Analyze the data using the control charts with probability limits ( 0.002 for both charts). Revise, if necessary. Find the in-control estimates of the process mean and standard deviation. 1 2 Question 4 10 For MR Chart, i=1 0.4 8 0.7 15 0.9 1 21 2 data points: 2 3 1.3 0.4 9 10 1.5 2.3 16 17 0.8 0.3 0.94762 4 1.2 11 3.8 18 0.9 5 0.6 12 1.2 19 0.1 6 0.9 13 0.5 20 0.4 7 0.1 14 0.6 21 1 0.94762 1.128 0.840088652 Probability limits ( 2 ·. 2 ·. For I Chart 1 22 3.09 3.09 0.002): 0.840088652 · 0.001772 0.840088652 · 4.653508 0.001488637 3.909359265 15.17272727 15.17272727 3.09 0.840088652 15.17272727+3.09 0.840088652 12.57685334 17.7686012 Moving Range Chart of Level 4 1 UB=3.9 3 Moving Range 2 1 __ MR=0.948 0 1 3 5 7 9 11 13 Observation 15 17 19 21 LB=0.002 1 3 I Chart of Level 20 19 18 Individual Value 17 16 15 14 13 12 1 3 5 7 9 11 13 Observation 15 17 19 21 +3.1SL=17.769 1 _ X=15.173 -3.1SL=12.577 I chart not in control => revisions required. After deleting sample 11, process is in control. 12.362, 17.553 ̂ 14.957 0.840088652 1 4 ...
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## This note was uploaded on 07/24/2010 for the course MIE MIE364 taught by Professor Makis during the Spring '10 term at University of Toronto- Toronto.

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