2.810ManufacturingProcessesandSystems
2016PracticeQuiz2Solutions
Openbook,opennotes,calculatorsandcomputerswithinternetturnedoff.Presentyourworkclearlyandstateallasssumptions.
Problems:1. GasShortage2. ToyotaProductionCellDesign3. UnreliableMachines4. FusedDepostionModeling5. Literature
Problem1.GasShortage(a) Supposethatallcarownersfilluptheirtankswhentheyareexactlyhalffull.Atthepresent
time, an average of 6 customers per hour arrive at a single pump gas station. It takes anaverageof5minutestoserviceacar.Pleaseestimatethetotaltimeeachcarisinthestation,including thewaiting time,and thenumberof cars in thequeue.Youmayassumethat thetimedistributionsareexponential.We can use the M/M/1 queue model (since we are told that the time distributions areexponential)combinedwithLittle’sLaw(𝐿 = 𝜆𝑊)tofindtherequiredquantities.Averagearrivalrate:𝜆 = 6 !"#$
!!= 0.1 !"#$
!"#
Averageservicerate:𝜇 = !
!!"#$!"#
= 0.2 !"#$!"#
Then,fromM/M/1queue,theaveragenumberofcarsinthequeueis:𝐿 = !
!!!= !.!
!.!!!.!= 1 𝑐𝑎𝑟
FromLittle’sLaw,theaveragewaitingtimeis:𝑊 = !
!= !
!.!= 10 𝑚𝑖𝑛
(b) Suppose there is a gasoline shortage and panic buying takes place. To model this
phenomenon,supposethatallcarownersnowpurchasegaswhentheirtanksareexactly3/4full.Now,sinceeachcarownerisputtinglessgasintohertankduringeachvisittothestation,weassume that theaverage service timehasbeen reduced to4.5minutes.Howhaspanicbuyingaffectedthewaitatthegasstationandthesizeofthequeue?
Thedriversarenowfillingupwhentheirtanksare¾full,which istwiceasoftenasbefore.Therefore,thearrivalratedoubles:Averagearrivalrate:𝜆 = 12 !"#$
!!= 0.2 !"#$
!"#
Theservicerateincreasesslightly:Averageservicerate:𝜇 = !
!.!!"#$!"#
= 0.222 !"#$!"#
Asbefore,usingM/M/1queueandLittle’sLaw:
𝐿 =𝜆
𝜇 − 𝜆=
0.20.222 − 0.2
= 9 𝑐𝑎𝑟𝑠
𝑊 =𝐿𝜆=
90.2
= 45 𝑚𝑖𝑛Wecan see that theboth thenumberof cars in thequeueand thewaiting time increasedsignificantly.
Problem2.ToyotaProductionCellDesignsConsiderthecelldesignshowninthediagrambelowandanswerthefollowingquestions.Notethataboxrepresentsaprocessandholdsonepartatatime,andadarkenedcirclerepresentsa“decoupler”whichcanholdonepart.Thetimesgivenbyeachprocessareinminutesasmanualtime/machine time. Assume a 3 second walking time between machines (including rawmaterialsasa“machine”).Stateallassumptions.(a) Whatisthecurrentproductionrate,inventory,andtimeinthesystemforthesystemshown
belowwithoneoperator?
Indicatesautomaticpartejectionandconveyancetonext station
Thebluesquaresrepresentpartsinthesystem.Part18istherawmaterialonitswaytomachine1.3seconds=0.05min
Segment ManualTime(min)
WalkingTime(min)
ProcessTime*(min)
1(RawMat’ltoMachine1)
1 0.05 4
2 1 0.05 3.23 2 0.05 1.74 1 0.05 3.55 2 0.05 76 1 0.05 47 1 0.05 1.58 1 0.05 19 0.5 0.05 4.5
FinishedPartstoRawMatl’s
0.05
10.5min 0.5min
1/4 1/3.2 2/1.7 1/3.5
Rawmat’l
2/7Finishedparts
0.5/4.5 1/1 1/1.5 1/4
1
2
53
4 6
9
8
7
18
17
16 14
15 13
12
11
10
*Foranymachinei,theMachineTimeMTi=(ProcessTime)i+(ManualTime)iCycleTime(CT)=TotalManualTime+WalkingTime=11minCheckthatCT>MTiforallmachines.Iftrue,thenCycleTimedeterminestheproductionrate.Little’sLawisL=λW,whereL=unitsinsystem(inventory)λ=arrivalrate(whichforsteadystate=productionrate)W=timeinsystemProductionRate:𝜆 = ! !"#$
!!!"#= 0.091 !"#$%
!"#
Formostofthe11minutecycletime,thereare18partsinthesystem(oneateachmachine,oneateachdecouplerbetweenthemachines,andoneintheoperator’shand).Forthetimebetweenwhenoperatorremovesthefinishedpart(part17inthediagram)fromthelastmachineandputsitinthefinishedpartscollection(notpartofthesystem)tillwhenshearrivesattherawmaterialsstation,shedoesnothaveapartinherhand,sofor0.05minthetotalnumberofpartsinthesystemisonly17.Thisyieldsthefollowingequation:Inventory:𝐿 = 17 !.!"!"#
!!!"#+ 18 !".!"!"#
!!!"#= 17.995 ≈ 18 𝑝𝑎𝑟𝑡𝑠
ProductionRate:𝑊 = !
!= 17.995𝑝𝑎𝑟𝑡𝑠 !!"#$
!!!"#= 197.95 𝑚𝑖𝑛
(b) Modifythesystemsothat itproducesonepartevery6minutes.Markyourchanges
clearlyonthediagrambelow.Alsoseparatelylistwhatyouhavechangedtomeetthisgoal.Estimate:
1) Theproductionrate2) Inventoryinthesystem3) Timeinthesystemforyournewsystemafteryoumakeyourmodifications.
Timesareagaingiveninminutes.Walkingtimebetweenmachinesisasinpart(a).
ChangestoSystem:1.Addanadditionalmachine:Theprocesstimeforthe5thmachineisgreaterthanthedesiredcycletime(7min>6min).Toalleviatethisproblem,simplyaddasecond,identicalmachine.WorkerBwillalternatebetweenmachinesiandii,sotheapparentprocesstimeforthismachineis7/2=3.5min.TheMachineTimeisnow3.5+2=5.5min.2.Addanadditionalworker:Theoriginalcycletimewithoneworkeris11minutes.Toreducethistoabout6minutes,addasecondworker,anddividethecellintotwoparts.WorkerA: CycleTime=(1+1+2+1+0.5)min+(0.05*6)min=5.8minWorkerB: CycleTime=(1+2+1+1)min+(0.05*4)=5.2minSincethecycletimeforworkerBis5.2min,andthelargestmachinetimeis5.5min,thenewcycletime(thetimetomakeonepart)isthecycletimeforworker1,5.8minutes.UsingthesameLittle’sLawequationasinparta,butnowwith19totalpartsinsteadof18duetotheduplicatedmachine:ProductionRate:𝜆 = ! !"#$
!.!!"#= 1.72 !"#$%
!"#
Inventory: 𝐿 = 18 !.!"!"#
!.!!"#+ 19 !.!"!"#
!.!!"#= 18.99 𝑝𝑎𝑟𝑡𝑠
ProductionRate:𝑊 = !
!= 18.99𝑝𝑎𝑟𝑡𝑠 !!"#$
!.!!"#= 110.1 𝑚𝑖𝑛
1/3.21/4 2/1.7 1/3.5
2/7
1/40.5/4.5 1/1 1/1.5
FinishedParts
RawMaterial
1
2
3
4
5
6
7
8
9
2/7
10
i ii11
12
13
14
15
16
17
18
19JA
JB
Problem3.UnreliableMachinesYouaremakingpartsona transfer line consistingof4machines. Eachmachinehasadifferentmeantimetofailureandmeantimetorepair,listedinthetable(inminutes).Theoperationtimeoneverymachineis1minuteperpart.Thereisnobufferbetweenanyofthemachines.
M1 M2 M3 M4MTTF 500 100 200 250MTTR 20 10 10 20
(a) Assume thatmachine failuresoccurasa resultof toolbreakage,whichonlyhappenswhen
themachinesareoperating.Whatistheaverageproductionrateofthetransferline?
Thisquestionassumestheoperation-dependentfailuremodel,i.e.,themachinecanonlyfailwhenitisoperating.Thus,Buzacott’sformulafortheproductionrateofalinewithnobufferscanbeused.
𝑃 =1𝜏
1
1 + 𝑀𝑇𝑇𝑅𝑀𝑇𝑇𝐹
𝑃 = 1 !!! !.!"!!.!"!!.!"!!.!"
= 0.787 !"#$%!"#
M1 M2 M3 M4
(b) Using the same operation-dependent failure model as in part a, you can now place oneinfinite buffer betweenany twomachines.Wherewould youplace it in order to achieve aminimumproductionrateof0.85parts/minute?Assumetheoriginal (operation-dependent) failuremodelapplies.Youwouldgenerallywanttoplacebuffersaround thebottleneckmachine,which in this case ismachine2 (ithas thelowestproductionrateonitsown).Ifwetryplacinganinfinitebufferafterthismachine,weget a line of machines 1+2 with no buffer, followed by the line of machines 3+4. UsingBuzacott’sformulaforeachofthetwolines:
𝑃!,! =1
1 + 0.04 + 0.10=
11.14
= 0.8772𝑝𝑎𝑟𝑡𝑠𝑚𝑖𝑛
𝑃!,! =1
1 + 0.05 + 0.08=
11.13
= 0.8850𝑝𝑎𝑟𝑡𝑠𝑚𝑖𝑛
With the infinite buffer, the production rate of the entire line will be the bottleneck(minimumproductionrate)amongthesetwoparts:
𝑃 = min 𝑃!,!,𝑃!,! = min 0.877, 0.885 = 0.877𝑝𝑎𝑟𝑡𝑠𝑚𝑖𝑛
,whichexceedstherequirementof0.85parts/minute.
Note: Ifyouplacethebufferbetweenmachines1and2,yougetaproductionrateof0.813parts/min. If you place it between machines 3 and 4, you get a production rate of 0.84parts/min.
(c) Assume you are using machines with the sameMTTF and MTTR as above, but now theirfailuresarecausedonlybythefailureofthecontrolsystem(which isalwaysonwhetherornotthemachineisoperating).Whatistheaverageproductionrateofthetransferline?This problem asks to use a different failuremodel, i.e. time-dependent failure, rather thanoperation-dependentfailureasinpart(a),whichwaspresentedonslides27-29ofthe“TimeAnalysis for Manufacturing Systems” lecture. The equation for production rate with thismodelisslightlydifferent:
𝑃 =1𝜏𝑃!𝑃!𝑃!𝑃! =
1𝜏
1
1 +𝑀𝑇𝑇𝑅!𝑀𝑇𝑇𝐹!
1
1 +𝑀𝑇𝑇𝑅!𝑀𝑇𝑇𝐹!
1
1 +𝑀𝑇𝑇𝑅!𝑀𝑇𝑇𝐹!
1
1 +𝑀𝑇𝑇𝑅!𝑀𝑇𝑇𝐹!
𝑃 = 11
1.0411.1
11.05
11.08
= 0.9615 0.9091 0.9524 0.9259 = 0.771𝑝�𝑟𝑡𝑠𝑚𝑖𝑛
Problem4.MaximumExtrusionRateforFusedDepositionModelingThe figurebelow,basedonProfessor JohnHart’s lectureslides, shows thenozzle for theFusedDepositionModeling(FDM)AdditiveManufacturingprocessforthermoplastics.AnABSfilamentapproximately1mmindiameterispushedthroughtheheatingzone(ZoneI)whereitmelts,andisthenreduced inareaforprinting.Weareconcernedwiththemaximumrateatwhichplasticcanbepushed through this zoneandstillbeguaranteed that theoutput is fullymelted (at thepointwhereitentersZoneII).Referringtothefigure,assumethediameteroftheheatingzoneisdl=1.0mmanditslengthisll=1.25 cm. Its walls aremaintained at 260°C with an electric resistance heater. Pleasemake anestimate for themaximumrateatwhich theABS filament canbe fed into theheating zone (inmeters/second).Note: The chart from the textbook by Lienhard for conductive heat transfer in cylinders isincludedattheendoftheexamforreference.
Transienttemperaturedistributionsinalongcylinderofradiusroatthreepositions.r/ro=0isthecenterline.(Lienhard,Fig5.8)
ThisproblemcanbeapproachedbyconsideringtheABSfilamentasacylinderthat’sbeingheatedby the nozzle walls that are held at constant temperature. Initially, the filament is at roomtemperature (Ti = 25℃). Thewalls remain at Tw = 𝑇!= 260℃. In order to ensure that the ABSmeltsbefore itentersthetransitionzone,wecanlookatthetimeneededforthecenterofthefilamenttoreachmeltingtemperature.BecauseABS is an amorphous polymer, it does not have awell-definedmelting pointwhere itchangesphaseandthusnolatentheatoffusion;instead,itchangesfromsolidtorubberyafteritpasses the glass transition temperature and then to a fluidwith decreasing viscosity as it risesbeyondthemeltingpoint.FromJohnHart’sotherslidesonFusedDepositionModelingandfromBoothroyd’s Ch. 8, Table 8.5 on injection molding parameters, the recommended formingtemperatureforABSis intherangeof240-260℃.(ThissolutionassumesT=250℃;other𝜃’s intherange0-0.1areOK.)Thetransienttemperaturedistributionchartprovidedaboveprovidesagraphicalwaytoestimatethetimeneededtoreachagiventemperatureatagivenradialpositionwithinthecylinder.Weneedtocalculatethedimensionlesstemperature(θ)andBiotnumber(Bi),andusethetopchart(forr0/r=0,sinceweare lookingatthecenterofthecylinder)tofindtheFouriernumber(Fo).Notethatthechartsareprovidedforabodythat’sbeingcooled;inordertoadaptthemtoabodybeing heated, we can just change the signs in the temperature calculation so that θ remainsbetween0and1:
𝜃 =𝑇! − 𝑇𝑇! − 𝑇!
=260 − 𝑇260 − 25
=260 − 250
235= 0.043
For the inverseBiotnumber,Bi-1 = k/hr0, knowing that thermal conductivityk is very low (~0.2W/mKforpolymers),wecanassumethatBi-1~0.AttheintersectionofthisθandBi-1,theFouriernumber on the x-axis is approximately𝐹𝑜 = 0.45, which we can use to calculate time (usingtypicalthermaldiffusivity𝛼forpolymers):
𝑡 =𝑟!!𝐹𝑜𝛼
=0.05 𝑐𝑚 !(0.45)
10!! 𝑐𝑚!
𝑠
= 1.1 𝑠
Thusthefilamentneedstospendaminimumof1.1secondsinsidetheheatingelement.Thenthemaximumfeedratecanbecalculatedfrom(lengthofheatingzone/time):
𝑣 =𝑙𝑡=1.25 𝑐𝑚1.1𝑠
= 1.13𝑐𝑚𝑠= 0.01
𝑚𝑠
Alternatively:Thesimplestwaytosolvethisproblemisbyanalogywithourone-dimensionalanalysisforinjectionmolding.Thesameequationappliesforheatingandforcooling.Remember
wegotthat!!~ !
!!
!andthat𝑡 =!!
!
!wastheexactsolutionwhen𝜃 = 0.1 forconstantwall
temperatureandBi-1=0(InjectionMoldinglecture,Slide14).Wehaveaverysimilarsituationhere,exceptthatthegeometryisnowacylinder.
Problem5.Literature(a) AccordingtoMaccoby,whichjobsaremostprizedbyfactoryworkers?
FromMaccoby,page162:“Evenwhensupervisorshaveallowedteamworkandopportunitiesforproblemsolvingandparticipation(allofwhichworkersappreciate),assemblyworkremainsmonotonousandstressful…Thejobsmostprizedbyemployeesaretheonesthatallowthemostautonomy,rangingfromhigh-payingjobssuchasprogrammingrobotstolower-payingjanitorialwork.”
(b) PleaselistfourreasonsexplainingwhyFord’schangeoverfromtheModelTtotheModelAwassodifficult.Hounshell’sChapter7(pp.280-292)identifiesmultiplechallengesencounteredduringthechangeovertotheModelAin1927,whichrequireda6-monthcompleteshutdownoftheplant.Someexamples:(1) HenryFordwantedtostopusingstampingsforthecarbodyanduseforgingsinstead.
ForgingwasusedonlyminimallyonthepreviousModelT,somostforgingcapabilityhadbeeneliminatedfromtheRougeplantandtheywerepurchasedfromoutsidesuppliers.Thiscausedprocurementdelaysandalargecostincrease.
(2) DelaysindesignduetodisagreementsbetweenFordandhisengineers.(3) Ford’sdecisiontomakethegasolinetankintegralwiththecowloftheModelA,which
causedissuesinproductionwithseamwelding.(4) Forddecidedtoincreasetheprecisionofallmachiningworkinordertoensurethe
highestpossiblequalityfortheautomobile.Smallertolerancesonmostpartsrequiredmorefrequentuseofgauging,scaling,andbalancinginproductionandthusmoretime.
(c) Please list threepossiblesourcesofwaste inamanufacturingsystemanddescribehowtheToyotaProductionSystemaddressedeachone.
ReferringtotheMondenreading,severalsourcesofwastecanbepointedout,suchas:(1)excessiveworkforce,(2)excessiveinventory,(3)excessivecapitalinvestment.TheTPSaddressestheseasfollows:a. Excessiveworkforcecanbeeliminatedbyre-allocatingworkeroperationsandensuringa
flexiblemulti-functionalworkforce.b. ExcessiveinventoryiscontrollerbyimplementingJust-in-timeproduction,i.e.,production
regulatedbythevelocityofsales,whichismaintainedusingthepullsystem.c. Excessivecapitalinvestmentcanbereducedbyeliminatingexcessiveinventory,since
extrainventoryrequiresinvestmentinextrawarehousesandadditionaltransportequipment.
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