IE2130 Quality Engineering I (Semester 2, AY AY2015-16)
Tutorial Tutorial Problems __________________________________________________________________________________________ _______________________________________________________ ___________________________________
Guidelines and e!"mmendati"ns e!"mmendati"ns
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Solve Solve the problems problems yourself yourself and try to to understan understand d the the underlyi underlying ng concepts concepts..
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Use Microso Microsoft ft Ece Ecell or any other other soft! soft!are are if if neede needed. d.
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There is is no need need to submi submitt the solution solutionss for grading grading ecept for #uestion #uestion $ of Top Topic ic ".
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(ou may include comments on %ssign %ssignmen mentt 1 deals deals !ith !ith #uesti #uestion on $ of To Topic " &in &in red font'. font '. (ou the contribution of each group member if you so !ish. )ther!ise all team members !ill receive the same mar*s.
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The probl problems ems !ill !ill be discussed discussed in tutor tutorial ial classes. classes. Stand Standard ard hard hard or soft copy copy solutio solutions ns !ill not be provided. ,or similar problems !ith solutions- refer to the relevant eamples in the tetboo* tetboo*.. Study Study these these eample eampless !ill !ill help help you you to better better underst understand and the topic topic covered.
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%r"du!t&$r"!ess !'ara!teriati"n
&omments/ 0ead hapter " of the tetboo*- especially some of the eamples. (ou should have no problem !ith these uestions. The main issue here is ho! they are lin*ed to uality control.'
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(ield strengths of a component are measured. The first yields &in *' are as follo!s/ 134- 56- 132- 13$- 1+6- 126- 124 and 1+3. ,ind the median and the range. alculate the sample average- sample variance and sample standard deviation by hand. Then use Microsoft Ecel &or any other soft!are' to generate these 7summary statistics7. 8hich of these five sample statistics does not have the same unit of measurement as all the other9 8hat can you infer from the results obtained9 Answers: 117, 60, 121.25, 511.93, 22.63.
2.
The output voltage of a po!er supply is normally distributed !ith mean + : and standard deviation 3.32 :. ;f the lo!er and upper specifications for voltage are $.5+ : and +.3+ : respectively- !hat is the probability that a po!er supply selected at random !ill conform to the specifications on voltage9 Answer: 98.8%.
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The life of a car battery is normally distributed !ith mean 533 days and standard deviation "+ days. 8hat fraction of the batteries !ould be epected to survive beyond 1333 days9 Answer: 0.21%.
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% light bulb has a normally distributed light output !ith mean +333 end foot
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Problem 2 to $ illustrate ho! the normal distribution can be used to model the uality characteristic of a process and to determine *ey uality parameters based on sample information. 8hat if the distributions of the uality characteristics are not normal9
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>ive belo! are the data in Slide 1$ of Topic 2/ Mon Tue 8ed Thu ,ri Sat Sun
226.4 "62.4 1++.$ 2+4.3 2?6.+ $36." "2?.4
"42.? 2$".2 1??.? "52.5 +31.2 "$1." """.?
"$?.1 ""2.2 22$.6 1$$.4 "5+.2 "34.6 161.?
2?1.1 "25.+ $2".6 2+?.6 "16.4 $3+.3 +35.+
$1?.6 "3$.2 "+2.1 +34.2 2+3.1 2"6.1 "3?.4
&a' Use Microsoft Ecel &or any other soft!are' to generate the summary statistics in Slide 15- plot the histogram and PP as sho!n in Slide $$- and obtain the results for the three eercises in Slide $4. &b' @o! do you see the process9 ;s it good enough9 8hat are the possible actions to improve it if this is needed9 &c' 0efer to Slide +" and try to appreciate the steps in A% Systematic Problem
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"ntr"l !'arts *"r +ariales
&omments/ Some of the uestions are ta*en from hapter 6 of the tetboo*. #uestion $ is based on the data given in the lecture notes < see slide 24 of Topic $.' 1.
The data as sho!n are X and R values for 2$ samples of siCe n D + ta*en from a process producing a certain component. The measurements are made on the inside diameter of the component- !ith only the last three decimals recorded- i.e. "$.+ should be 3.+3"$+. Sample no. 1 2 " $ + 6 ? 4 5 13 11 12
R " $ $ $ + 6 $ " ? 4 " 5
X
"$.+ "$.2 "1.6 "1.+ "+.3 "$.1 "2.6 "".4 "$.4 "".6 "1.5 "4.6
Sample no. 1" 1$ 1+ 16 1? 14 15 23 21 22 2" 2$
X
"+.$ "$.3 "?.1 "$.5 "".+ "1.? "$.3 "+.1 "".? "2.4 "".+ "$.2
R 4 6 + ? $ " 4 $ 2 1 " 2
&a' Set up X and R charts on this process. ;s the process in statistical control9 ;f necessary- revise the trial control limits. &b' ;f specifications on this diameter are 3.+3"3 3.3313- find the percentage of nonconforming components produced by the process. %ssume that diameter is normally distributed. Answers: (a) Not in contro, e!c"#e sa$e no. 12 an# 15. Reot an# &ina contro c'arts 36.25, 33.65, 31.06 9.52, 4.5, 0. (*) 0.05%.
2.
Samples of siCe n D 6 items each are ta*en from a process at regular intervals. % uality characteristic is measured- and X and R values are calculated for each sample. %fter +3 samples- !e have +3
+3
∑ X = 2333 and ∑ R = 233 i
i =1
i
i =1
%ssume the uality characteristic is normally distributed and the process is in control. &a' ompute control limits for the X and R control charts. &b' The specification limits are $1 +. %ssume that if an item eceeds the upper specification limit it can be re!or*ed and if it is belo! the lo!er specification limit it must be scrapped. 8hat percent scrap and re!or* is the process producing9
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&c' Ma*e suggestions as to ho! the process performance could be improved. Answers: (a) 41.932 an# 38.068, 8.016 an# 0 (*) 0.57% an# 72 $.
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%n X chart is used to control the mean of a normally distributed uality characteristic. ;t is *no!n that F D 6.3 and n D $. The centre line D 233- UG D 235- and GG D 151. ;f the process mean shifts to 144- find the probability that this shift is detected on the first subseuent sample. Answer: 84.1%.
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The data set refers to the ball shear strengths & X ' of a product at a bonding temperature of 2$3 / °
Mon +1 +1 +2 +2 Tue +1 +1 +2 +2 8ed +1 +1 +2 +2 Thu +1 +1 +2 +2 ,ri +1 +1 +2 +2 Sat +1 +1 +2 +2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
4?.+ 43.1 ??.6 ?+.4 44.? ?$.4 4".? ?5.4 44.4 ?6.5 42.6 4+.? 66.5 4".$ ?6.5 6$.1 43.1 ?"." ?+." 42.1 ?".? ?".+ ?".+ 4?.3
46.+ 64.3 4$.3 ?+.? +6.$ 62.5 ?1.5 66.2 6".1 ?".? 63.4 +1.6 65.5 +2.6 ?$.+ 6".$ 64.+ 6?.3 ??.+ ?1.4 ?$.2 6?.1 ?1.+ 6".1
66." 6?.+ 64.1 65.2 62.? 61.1 +4.5 +6." +4." 6+.+ 6+.+ ?1.$ +$.1 65.$ 6".4 66.+ 6?.1 4$.2 66.$ 65.2 ?6.$ 6$.3 ?$.6 65.2
?".1 4$.2 63.3 6?.4 6?.6 ?3.1 ??.6 +$.+ ?$.6 +5.$ ?4.$ +?." ?1.$ ?$.5 ?3.6 +5.3 6+." 62.+ ?5.3 ?+.6 6".6 +$.3 ?4.1 +?."
65.$ 4?.+ 64.1 6+.$ 6".4 +4.$ ?3.1 +2.3 ?+.6 ?3.1 6?.6 63.3 6$.3 ++.1 ?$.4 63.? 6+.+ 62.4 61.4 65.+ ?3.3 +5.2 65." 43.6
&a' Ta*ing all the 123 observations together- find X $in- X $a!- R- and the mean and standard deviation. Answers: 51.6, 88.8, 37.2, 69.72, 8.90.
&b' Plot the histogram for X . @o! does a change in the number of intervals affect the plot9 Plot the normal probability plot &PP'. &c' >iven that GSG D $3- compute the epected ,. +
Answer: 419 $.
&d' Plot the X and R charts. %pply the synthesiCing rules. ;s the process in control9 Answers: 81.42, 69.72, 58.02 42.91, 20.29, 0.
&e' Hased on the 2$ X Is in &d'- plot the histogram and the PP. ompare the distributions of X and X . ,ind the mean and standard deviation of X . Answer: entra -i$it 'eore$ aies 69.72, 4.13.
&f' ,rom the dataJresults in &a'- &d' and &e'- sho! empirically !hether the follo!ing relationships hold/ σ !
≈
σ !
J
n
σ !
≈
R J # 2
8hat is the practical significance of the relationships in SP9
&g' %ssume that the process mean shifts to ?4. ,or the X chart in &d'- !hat is the probability of not detecting the shift in the first sample plotted after the shift9 8hat is the %0G in this case9 Use only sensitiCing rule 1. Answers: o *e ro/i#e# a&ter s"*$ission o& Assin$ent 1.
&h' ,or the same set of data- construct the X and + charts. ompare the plotted charts !ith the X and R charts obtained in &d'. ;s there any difference in the conclusion reached for the process9 Answers: 81.54, 69.72, 57.91 17.30, 8.28, 0.
&i' 8ith n D 1 and 2$ samples- construct the X and R charts. Use the larst column of the data set &leave out the first four columns'. ;n general- !hat are the differences bet!een these charts and those in &d'9 Answers: 91.21, 66.72, 42.23, 30.09, 9.21, 0.
&K' Use the first three columns of the original data set- i.e. n D " !ith 2$ samples and leave out the last t!o columns- construct the X and R charts. ,or this set of control chartsans!er the same uestions in part &d' and part &g'. ;n general and from the SP vie!point- eplain the differences bet!een the t!o sets of control charts and their performances- i.e. one !ith n D + and the other !ith n D ". Answers: o *e ro/i#e# a&ter s"*$ission o& Assin$ent 1.
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%r"!ess !a$aility analysis
&omments/ Some of the uestions are ta*en from hapter 4 of the tetboo*.' 1.
Perform a process capability analysis using the revised in Topic ".
X and R
charts for #uestion 1
Answers: 1.73 , 1.10.
2.
% process is in statistical control !ith X() D 155 and R D ".+. The control chart uses a sample siCe of n D $. Specifications are at 233 4. The uality characteristic is normally distributed. &a' Estimate the and . &b' @o! much improvement could be made in process performance if the mean could be centered at the nominal value9 Answers: (a) 1.57 an# 1.37 (*) ro$ 19 $ to 2.6 $.
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% process is in statistical control !ith X() D ?+ and + D 2. The control chart uses a sample siCe of n D +. Specifications are at 43 4. The uality characteristic is normally distributed. &a' Estimate the and . &b' @o! much could process fallout be reduced by shifting the mean to the nominal dimension9 Answers: (a) 1.25 an# 0.47 (*) ro$ 7.9% to 172 $.
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onsider the follo!ing t!o processes !hich produce the same product. The sample siCe is n D + and the processes are in statistical control. Process %
Process H
entre line of the X chart
133
13+
entre line of the + chart
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Specifications are at 133 13. alculate the and and interpret the values. 8hich process !ould you prefer to use. The uality characteristic is normally distributed. Answers: rocess A, D D 1.3$+L rocess , D ".1""- D 1.+66.
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.easurement systems analysis
&omments/ Use the techniue described in the lecture notes to solve the problems.' 1.
>iven the follo!ing data for readings by " appraisals on 6 parts !ith 2 trials. Part number 1 2 " $ + 6
%ppraiser % Trial 1 Trial 2 13.6+ 13.63 11.33 11.33 13.4+ 13.43 13.4+ 13.5+ 13.++ 13.$+ 11.33 11.33
%ppraiser H Trial 1 Trial 2 13.++ 13.++ 11.3+ 13.5+ 13.43 13.?+ 13.43 13.?+ 13.$3 13.$3 11.33 11.3+
%ppraiser Trial 1 Trail 2 13.+3 13.++ 11.3+ 11.33 13.43 13.43 13.43 13.43 13.$+ 13.+3 11.33 11.3+
&a' Using >00 Method 1- calculate gauge repeatability and reproducibility. ;f the specifications are at 13 1.$3- calculate the PJT ratio &use the default 600 Method 2- estimate the 00 &= of total variation'. omment on the gauge capability. Answers: (a) 0.0369, 0.0320, 10.45%, 8.97% (*) 21.14%.
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;n a >00 study- 2 operators use the same gauge to measure 13 parts " times each/ Part number 1 2 " $ + 6 ? 4 5 13
)perator % Measurements 1 2 " +3 $5 +3 +2 +2 +1 +" +3 +3 $5 +1 +3 $4 $5 $4 +2 +3 +3 +1 +1 +1 +2 +3 $5 +3 +1 +3 $? $6 $5
)perator H Measurements 1 2 " +3 $4 +1 +1 +1 +1 +$ +2 +1 $4 +3 +1 $4 $5 $4 +2 +3 +3 +1 +3 +3 +" $4 +3 +1 $4 $5 $6 $? $4
&a' Using >00 Method 1- estimate gauge repeatability and reproducibility. ;f the specifications are at +3 13- calculate the PJT ratio &use the default 600 Method 2- estimate the 00 &= of total variation'. omment on the gauge capability. Answers: (a) 1.181, 0.151, 35.7%, 30.7% (*) 64.18%. 4
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"ntr"l !'arts *"r attriutes &0efer to hapter ? of the tetboo*.'
The number of nonconforming items in samples of siCe 1+3 are as sho!n. onstruct a chart. ;s the process in control9 ;f not- assume that assignable causes can be found for all points outside the control limits and calculate the revised control limits.
Sample no. 1 2 " $ + 6 ? 4 5 13
o. of nonconforming items 4 1 " 3 2 $ 3 1 13 6
Sample no. 11 12 1" 1$ 1+ 16 1? 14 15 23
o. of nonconforming items 6 3 $ 3 " 1 1+ 2 " 3
Answers: ina re/ise# c'art i/es - 0.0141, - 0.0430, -- 0. 2.
;n designing a chart- !ith centre line at 3.23 and "
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The number of nonconformities found on final inspection of a device is as sho!n. ;s the process is in statistical control9 8hat centre line and control limits !ould you recommend for controlling future production9
Sample no. 1 2 " $ + 6 ? 4 5
o. of nonconformities 3 1 1 3 2 1 1 " 2
Sample no. 13 11 12 1" 1$ 1+ 16 16 14
o. of nonconformities 1 3 " 2 + 1 2 1 1
Answers: - 1.5, - 5.17 an# -- 0. rocess in contro. $.
,ind the "
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A!!e$tan!e sam$ling &omments/ 0efer to hapter 1+ of the tetboo*.'
Suppose that a single
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Suppose that a supplier ships components in lots of siCe +333. % single
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Using %S;J%S# O1.$- determine the single sampling plans for the follo!ing information/ ;nspection level &a' &b' &c' &d'
;; ; ;;; ;;;
;nspection
%#G
Got siCe
Tightened ormal 0educed ormal
1.+= 6+ 3.$3= 2.+=
1-$33 11+ 163-333 2?
Answers: (a) n 125, c 3, r 4 (*) n 8, c 10, r 11 (c) n 500, c 5, r 8 (#) n 20, c 1, r 2.
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Eplain the meaning of the sampling plan in #uestion "&c' if for a sample ta*en &i' # D 6&ii' # D 4- and &iii' # D $. Answers: (i) -ot accete#, N; reinstate# (ii) -ot re
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% product is supplied in lots of siCe N D 13-333. The %#G has been specified at 3.13=. ,ind the normal- tightened and reduced single
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