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.