MIT2.810Fall2016 Homework3Solutions

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MIT2.810ManufacturingProcessesandSystems

Homework3Solutions-ProcessControl-2016

Problem1.ControlChartsThedatashowninTable1arex̄andRvaluesfor24samplesofsizen=5takenfromaprocessproducingbearings.Themeasurementsaremadeonthe insidediameterofthebearing,withonlythelastthreedecimalsrecorded(i.e.34.5shouldbe0.50345).(a) Setupx̄andRchartsonthisprocess.Doestheprocessseemtobeinstatisticalcontrol?(b) Ifspecificationsonthediameterare0.503±0.0010,findthepercentageofnon-conforming

bearingsproducedbythisprocess.Whatassumption(s)doyouhavetomaketodeterminethisnumber?

SampleNumber

x̄ R SampleNumber

x̄ R

1 34.5 3 13 35.4 82 34.2 4 14 34.0 63 31.6 4 15 37.1 54 31.5 4 16 34.9 75 35.0 5 17 33.5 46 34.1 6 18 31.7 37 32.6 4 19 34.0 88 33.8 3 20 35.1 49 34.8 7 21 33.7 210 33.6 8 22 32.8 111 31.9 3 23 33.5 312 38.6 9 24 34.2 2

Table1:BearingdiameterdataAnswer:Thecenterlineandlimitsforthex̄andRchartsaregivenbelow:

Chart CenterLine ControlLimitsx̄ 𝑥 𝑥 ± 𝐴!𝑅R 𝑅 UCL=D4𝑅,LCL=D3𝑅

MIT2.810Fall2016 Homework3Solutions

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Fromthegivendata,wecancalculate:

𝑥 = 𝑥! + 𝑥! +⋯+ 𝑥!"

24 = 34.5+ 34.2+⋯+ 34.2

24 ≅ 34.

𝑅 = 𝑅! + 𝑅! +⋯+ 𝑅!"

24 = 3+ 4+⋯+ 2

24 = 4.7.Fromasamplesizeof5,thefactorsrequiredtosetthecontrollimitsare1:A2=0.577,D3=0andD4=2.115.Sothecontrollimitsare:x̄chart:UCL=𝑥 + 𝐴!𝑅 = 34+ 0.577 ∙ 4.7 = 36.719 LCL=𝑥 − 𝐴!𝑅 = 34− 0.577 ∙ 4.7 = 31.288

1 Standardtablesforestimatingthenecessaryfactorsforplottingcontrolchartscanbefoundhere: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

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Rchart:UCL=D4𝑅 = 2.115 ∙ 4.7 = 9.94LCL=D3𝑅 = 0

From the x̄ chart, we can conclude that the process is out of control based on themeasurements of the 12th and 15th samples. Out of 24 samples, we have 2 samples fallingoutside the3σ limitsoraprobabilityof8.3%.Weknow that for thenormaldistribution, theprobabilitythatapointwillfalloutsidethe3σlimitsis0.27%.Sothisoccurrenceisveryunusualandtheprocessneedstobeevaluated.NotethattheRchartdoesnotrevealanyirregularitiesintheprocess.ButthechartwouldstillbeusefulifweweretoapplytheWesternElectricrules(seeProblem3)heretogiveawarningsignalofaprocesslikelytogooutofcontrol.

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

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Problem2.ControlChartsThefillvolumeofsoft-drinkbeveragebottlesisanimportantqualitycharacteristic.Thevolumeismeasured (approximately)byplacingagaugeover thecrownandcomparing theheightofthe liquid in the neck of the bottle against a coded scale. On this scale, a reading of zerocorrespondstothecorrect fillheight.Fifteensamplesofsizen=10havebeenanalyzed,andthefillheightsareshowninTable2.

a. Setupx̄andscontrolchartsonthisprocess.Doestheprocessexhibitstatisticalcontrol?Ifnecessary,constructrevisedcontrollimits.

b. SetupanRchart,andcompareitwiththeschartinparta.SampleNumber

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 -1.52 0.0 0.0 0.5 1.0 1.5 1.0 -1.0 1.0 1.5 -1.03 1.5 1.0 1.0 -1.0 0.0 -1.5 -1.0 -1.0 1.0 -1.04 0.0 0.5 -2.0 0.0 -1.0 1.5 -1.5 0.0 -2.0 -1.55 0.0 0.0 0.0 -0.5 0.5 1.0 -0.5 -0.5 0.0 0.06 1.0 -0.5 0.0 0.0 0.0 0.5 -1.0 1.0 -2.0 1.07 1.0 -1.0 -1.0 -1.0 0.0 1.5 0.0 1.0 0.0 0.08 0.0 -1.5 -0.5 1.5 0.0 0.0 0.0 -1.0 0.5 -0.59 -2.0 -1.5 1.5 1.5 0.0 0.0 0.5 1.0 0.0 1.010 -0.5 3.5 0.0 -1.0 -1.5 -1.5 -1.0 -1.0 1.0 0.511 0.0 1.5 0.0 0.0 2.0 -1.5 0.5 -0.5 2.0 -1.012 0.0 -2.0 -0.5 0.0 -0.5 2.0 1.5 0.0 0.5 -1.013 -1.0 -0.5 -0.5 -1.0 0.0 0.5 0.5 -1.5 -1.0 -1.014 0.5 1.0 -1.0 -0.5 -2.0 -1.0 -1.5 0.0 1.5 1.515 1.0 0.0 1.5 1.5 1.0 -1.0 0.0 1.0 -2.0 -1.5

Table2:Fillheightdata

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Answer:Thecenterlineandlimitsforthex̄andSchartsaregivenbelow:

Chart CenterLine ControlLimitsx̄ 𝑥 𝑥 ± 𝐴!𝑠S 𝑠 UCL=B4𝑠,LCL=B3𝑠

We havem = 15 samples each of size n = 10. The standard deviation for each sample iscalculatedusingtheformula:

𝑠 = 𝑥! − 𝑥 !!

!!!𝑛 − 1 .

And,

𝑠 = 𝑠!!

!!!

𝑚 .Fromthedata,weget:𝑥 = −0.0033 and 𝑠 = 1.066.Forn=10,wehave,A3=0.975,B3=0.284andB4=1.716.Sothecontrollimitsare:x̄chart:UCL=𝑥 + 𝐴!𝑠 = −0.0033+ 0.975 ∙ 1.066 = 1.036LCL=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

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

Basedonthecontrolcharts,theprocessisincontrol.WenowplottheRcharttoseeifwegetthesameresult.Fromthedata,weget,R̄=3.2.Forn=10,wehave,D4=1.777andD3=0.223.Thecontrollimitsare:UCL=D4𝑅 = 1.777 ∙ 3.2 = 5.686LCL=D3𝑅 = 0.223 ∙ 3.2 = 0.713

TheRchartalsoindicatesthattheprocessisincontrol.

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

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Problem3.OutofControlDetection[AdaptedfromDouglasMontgomery’s‘IntroductiontoStatisticalControl’]TheWesternElectricHandbook(1956)suggestsasetofdecisionrulesfordetectingnonrandompatternsoncontrolcharts.Specifically,itsuggestsconcludingthattheprocessisoutofcontrolifeither:

1. Onepointplotsoutsidethethree-sigmacontrollimits,or2. Twooutofthreeconsecutivepointsplotbeyondthetwo-sigmawarninglimits,or3. Fouroutoffiveconsecutivepointsplotatadistanceofone-sigmaorbeyondfrom

thecenterline,or4. Eightconsecutivepointsplotononesideofthecenterline.

Thoserulesapplytoonesideofthecenterlineatatime.Therefore,apointabovetheupperwarninglimitfollowedimmediatelybyapointbelowthelowerwarninglimitwouldnotsignalanout-of-controlalarm.Calculate the probability of the first two patterns occurring assuming the data points areindependentbutidenticallydistributedwithanormaldistribution.

Hint:YoucanchooserpointsoutofnconsecutivepointsinnCrways,where,nCr=

!!!! !!! !

.Then

calculate theprobabilityof apoint fallingoutside thewarning limit.Wehave r points fallingbeyondawarninglimitand(n-r)pointsfallingbeforeit.Sincethepointsareindependent,theprobabilityofapatternofpointsistheproductoftheirindividualprobabilities.Answer:FromProf.Hardt’slecturenotes,weknowthattheareaunderthenormaldistributioncurveforvariousspreads,zσ,isapproximatelywrittenas:

Within1σ 0.68Within2σ 0.95Within3σ 0.997

Therefore, by symmetry, theareabetween1σand2σ limitsoneither side is (0.95-0.68)/2=0.135.Theareabetween2σand3σlimitsoneithersideis(0.997-0.95)/2=0.0235.1. Theprobabilitythatonepointplotsoutsidethe3σlimitsis(1–0.997)=0.003or0.3%.2. Theprobabilitythatapointwillfallbeyondthe2σlimitononesideofthecenterlineis(1-

0.95)/2=0.025.Therefore,theprobabilitythattwooutofthreeconsecutivepointswillfalloutsidethe2σlimitoneithersideis:2x3C2x0.025

2x(1–0.025)=0.0036or0.36%.

Wecanalsodeterminetheprobabilitiesfortheothertworulesmentionedabove:

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3. Theprobabilitythatonepointfallsbeyondthe1σsigmalimitononesideofthecenterlineis(1–0.68)/2=0.16.Therefore,theprobabilitythatfouroutoffiveconsecutivepointswillfalloutsidethe1σlimitoneithersideis:2x5C4x0.16

4x(1–0.16)=0.005or0.5%.4. Theareaoneithersideofthecenterlineequals0.5.Theprobabilitythateightconsecutive

pointsfallononesideofthecenterlineis(0.5)8=0.0039or.0.39%.