MIT 2.810 Manufacturing Processes and Systems Homework …web.mit.edu/2.810/www/files/homeworks/2016_home...2.1 Vertical Mill Face end(s) ½” end mill 2.2 Vertical Mill Hog-out ½”

MIT2.810Fall2016 Homework6Solutions

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MIT2.810ManufacturingProcessesandSystemsHomework6Solutions

ManufacturingSystems,TransferLines&TPS

November7,2016Problem1.ProcessPlanConsiderthedrawingofapartcalledarockerarmshownbelow.Thetoleranceonthethree1-inch diameter holes is ±0.001 in. For all other dimensions the tolerance is ±0.007 in. Thematerialis1018steelwithadensityof0.3lb/in3(8.9g/cm3).a) Assumingyoustartwithbarstockwiththenominaldimensionsofthecross-sectionforthis

part(youmayadjustthisslightly),pleasewritedownaprocessplan(i.e.,themachine,tool,and operation for each step, aswell as any re-fixturingwhen needed) tomake this partusingabandsawandasingleCNCmillingmachine.

b) Whatdoyouthinkwillbethelongeststepinyourprocessplan,andwhy?c) Nowconsiderhighervolumeproductionofthispartandsuggestalternativestoincreasethe

productionrateanddecreasethecycletime.

Figure 4. Rocker arm part isometric drawing.


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Answer:(a) Machines:horizontalsaw,verticalCNCmillingmachine.

Stock: 2”x2.75”CrossSectionsteelstock(oversizedby¼”toalloweasierfixturinginthemill).

Processstep

Machine Operation Tool

1 Horizontalsaw Cutto4.6” 2 VerticalMill Fixtureasshown(setup#1) 2.1 VerticalMill Faceend(s) ½”endmill2.2 VerticalMill Hog-out ½”endmill2.3 VerticalMill Profile ½”endmill2.4 VerticalMill Drill Centerdrill2.5 VerticalMill Drill Pilot½”2.6 VerticalMill Drill Pilot63/64”2.7 VerticalMill Drill Ream3. VerticalMill Re-fixtureonside(setup#2) 3.1 VerticalMill Profile ½”endmill3.2 VerticalMill Drill Centerdrill3.3 VerticalMill Drill Pilot½”3.4 VerticalMill Drill Pilot63/64”3.5 VerticalMill Drill Ream4. VerticalMill Re-fixtureonotherside 4.1 VerticalMill Profile ½”endmill4.2 VerticalMill Drill Centerdrill4.3 VerticalMill Drill Pilot½”4.4 VerticalMill Drill Pilot63/64”4.5 VerticalMill Drill Ream5.0 VerticalMill Re-fixture(setup#3) 5.1 VerticalMill Facebottom 1/2”endmill

Setup#1:Firstsetupforhoggingout,profilinganddrilling1”holeinbase.


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Setup#2:Re-fixturetoprofileanddrill1”hole,flipanddootherside.

Setup#3.Refixturetoremoveexcessmaterialonbottom.


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(b) Alternativesforhigher-volumeproduction:

- Thebulkof thetimetoproducetherockerarm is in the“hogout”step(2.2).Youcould consider starting with a C-shaped extrusion instead, which would all buteliminatetheroughmachiningpartofthisstep.Theextrusionperlbwouldbemorecostlythantheblock,butfewerlbswouldbeneeded.

- Considerusingformdrills;thiseliminatesallthetoolchangestomaketheholes.- Consider rotary axes (which can be retrofit to machine tools) to automatically

reorientthepartandeliminaterefixturing.- Loosenuptolerancesandsurfacefinishtoreducethefinishingsteps.- Consideramachinetoolwithafastertoolchanger.- Designacellwithsimplemachines.


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Problem2.Consideratransferlineasshowninfigure1below.

Figure1:Transferlinelayout

Themachineparametersareasgiveninthetablebelow.

M1 M2 M3 M4 M5 M6MTTF(hours) 30 40 20 20 10 50MTTR(hours) 1 1 1 1 1 1Processingtime(hours) 1 1 1 1 1 1

Thebufferhasinfinitestoragecapacity.Assumeoperation-dependentfailures.

(a) LineevaluationEstimatetheproductioncapacityofthissystem.Assumethattheprofitperpartis$10and that you sell every part youmake.Maintenance costs (not included in the profitcalculation)are$1perhour.Calculatethenetprofitforthissystem.Answer:The first three machines and the last three machines behave like zero-buffer three-machine lines.WeapplyBuzzacott’s formulatocalculatetheproductionrate foreachthree-machineline:

𝑃 =1𝜏 ×

11+ !""#

!""#!!

𝑃! =11×

1

1+ !!"

+ !!"

+ !!"

=1

1.1083 = 0.902

𝑃! =11×

1

1+ !!"

+ !!"

+ !!"

=1

1.1083 = 0.855

Wecannow treat these linesas twomachineswithan infinitebufferbetween them.Then,theproductionrateistheproductionrateofthebottleneck=P2=0.855.Profit=0.855(parts/hr)*10($/part)-1($/hour)=7.55($/hour).

(b) Lineimprovement:PartIIdentifythebottleneckmachineforthisline.Supposethatyouhaveabudgetof$40,000to improvethe line.Each$10,000 increasestheMTTFofamachineby10hours.Howwillyouallocatethisbudgetsothatyoucanachieveaproductionrateof0.91?


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Answer:Weimprovethesystembyidentifyingthebottleneckandimprovingitby10unitsuntilitnolongeristhebottleneck.Oneiterationcanbeasfollows:

1. Increase MTTF of machine M5 from 10 to 20. It is now tied with M4 as thebottleneck.Theproductionrateofthesecondthree-machine line(M4-M5-M6)is0.8928.

2. IncreaseMTTF ofmachineM4 from 20 to 30. NowM5 is the bottleneck. Theproductionrateofthesecondthree-machineline(M4-M5-M6)is0.9063.

3. IncreaseMTTFofmachineM5from20to30.Theproductionrateofthesecondthree-machineline(M4-M5-M6)is0.9202.

Now,thisthree-machineline iscapableofmeetingtherequireddemandof0.91partsperhour.However,theM1-M2-M3systemhasaproductionrateof0.9022,whichislessthanthedemand.Wecontinueiteratingasfollows:

4. ImprovetheMTTFofmachineM3from20to30.It isnowtiedwithM1asthebottleneck. Theproduction rateof the first three-machine line (M1-M2-M3) is0.916.

Thus, this system now meets the required demand rate. The cost of theseimprovementsis:$10,000*4=$40,000.Thus,wearewithinourbudget.

(c) Lineimprovement:PartIIEach0.005unitsdecrementinthefailurerateofamachineincreasesthemaintenancecostsby3%.Recallthatthefailurerate,p=1/MTTF.Forexample,ifyoureducethep-value for machineM2 from (1/40 = 0.025) by 0.005 to 0.02, the maintenance costswouldbecome1.03 ($/hour). If you could improveonlyonemachine,whichmachinewould you apply this improvement to? Set up the equation for the profit of such asystem,intermsofthenumberofdecrements,n.Whataretheupperandlowerlimitson the value of n? Set up an optimization problem to find the value of n whichmaximizesprofit.Extracredit:Plottheprofitfunction,andfind(manuallyorusingsoftware)theoptimumvalue of n, the resulting failure probability rate for the chosen machine, and themaximumprofit.


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Answer:M4-M5-M6 is thebottleneck three-machine line.Wecouldpickanymachine toapplythe improvement to since the denominator simply consists of a sumof p-values.WefocusourattentiononmachineM5.Ithasafailurerate,p5of0.1.Letnbethenumberof0.005decrementswemaketofailureratep5.Wecanwritetheproductionrateandprofitofthesystemasafunctionofnasfollows:

P(𝑛) =1𝜏 ×

1

1+ !.!"!

+ !.!!!×!.!!"!

+ !.!"!

=1

1.17− 0.005𝑛

Thentheprofitperhour(notincludingmaintenancecosts)isgivenby:

Pr(𝑛) =10

1.17− 0.005𝑛Notethatp5is0.1.Thus,ncanbeatmost20sothatp5isnon-negative,andncannotbenegative.Thus,nliesbetween0and20.Themaintenancecostincreasesby3%foreveryndecrementinp5.Thatis,themaintenancecostcanbewrittenas1.03n.Then,thenetprofitfunctionbecomes:

𝑁𝑒𝑡 Pr (𝑛) =10

1.17− 0.005𝑛 − 1.03!

ExtraCredit:Theprofitfunctionisplottedinfigure2below.Solvingthisproblemgivesn=10.16forwhich,p5becomes0.04917.Theproductionratebecomes0.8935andthenetprofit is$7.58/hour.

Thefunctionisconcaveintheregionofinterestfornin[0,20].Thus,wecanhavetwovaluesofnwhichgive thesamenetprofit.Ahighervalueofnmeansthat the failurerateisless.Inthatcase,wewouldexpectlessinventorytoaccumulateinthebuffer.


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Figure2:Thenetprofitasafunctionofn

(d) InventorycostsAs themarketmatures, we are looking to cut our production costs. Inventory costs,which were so far not considered in our profit calculations, are now consideredimportant.Inventorycostperpartperhouris$0.01.Youhaveestimatedtheproductionratesandaverageinventoryinthesystemforvarioussizesofbuffers.Thesearegiveninthetablebelow.

Buffersize Productionrate Averageinventory1 0.7866 0.6115 0.8006 3.07710 0.8127 6.21820 0.8274 12.7450 0.8449 34.345100 0.8522 76200 0.8545 170400 0.8547 370

1. Consideringtheinventorycosts,findthebuffersize(outoftheoptionsinthetable)

whichmaximizesyourprofit.2. What answer would you get for the optimum buffer size using Gershwin’s

Approximation?


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Answer:

1. Buffersize(A)

Prodrate(B)

Avg.inventory(C)

Profitnotconsideringinventorycosts(D=B*10-$1/hr)

Inventorycostsbasedon$0.01/part/hr(E=0.01*C)

NetProfit($/hr)(F=D–E)

1 0.7866 0.611 6.866 0.00611 6.85989

5 0.8006 3.077 7.006 0.03077 6.97523

10 0.8127 6.218 7.127 0.06218 7.0648220 0.8274 12.74 7.274 0.1274 7.1466

50 0.8449 34.345 7.449 0.34346 7.10554100 0.8522 76 7.522 0.76 6.762

200 0.8545 170 7.545 1.7 5.845400 0.8547 370 7.547 3.7 3.847

Basedontheinventorycosts,wefindthatabufferofsize20givesusthemaximumnetprofit.

Figure3:Inventorycostsandnetprofit

Note:Here, we have the first three-machine line producing parts at a faster rate than thesecond three-machine line. Recall from the “Time, Rate, and Variability” lecture that for anM/M/1queuewithλ>µ,the inventory inthesystemwithan infinitebuffer tendsto infinityovertime.Inthispartoftheproblem,weareconsideringfinitebuffers.Here,wewitnessthephenomenoncalledblocking.Whenthebuffergetsfull,theupstreamthree-machinelinegets“blocked”andcanonlymakepartsattherateatwhichthedownstreamthree-machinelinecanprocesspartsfromthebuffer.Thus,weexpecttheupstreamthree-machinelinetobeidlequite

0

1

2

3

4

5

6

7

8

400 200 100 50 20 10 5 1BufferSize

Profitnotincl.inventorycosts

InventoryCosts

NetProfit


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often. Because of the assumption of operation-dependent failures, the upstream three-machinelinefailslessfrequentlythanitwouldwithoutblocking.

2. Let us now use Gershwin’s Approximation to find the optimum buffer size. As per theapproximation,

𝑁∗ = 2 𝑡𝑜 6 ∗𝑀𝑇𝑇𝑅 ∗ 𝜇Here,MTTR1=1andMTTR2=1.Thus,themeanMTTRis1.Also,µforeachthree-machinelineis1.Thus,weget,N*tobebetween2to6.


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Problem3.ProductionFlowIssues(a) Estimate theproduction rate, inventory, and time in the system for the systemshown in

figure2,madeupofeight identicalprocessstepseachwhich iscapableofproducing100partsadaywhenoperating.Thetwobuffersareof infinitecapacity.TheMTTF is10daysandtheMTTRis1dayforeachmachine.

Figure2:Transferlinelayoutwithinfinitebuffers

(b) Pleaseestimatethenumberofpartsinbufferb1atanytimet.Stateassumptions.(c) Pleaseestimatethenumberofpartsinbufferb2atanytimet.Stateassumptions.

Answer:


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Problem4.ToyotaProductionSystem(TPS)CellConsider the TPS manufacturing cell shown below. There are 4 machines and 5 walkingsegmentseach5seconds.Themanualtime/machinetimesareindicatednexttoeachmachineinseconds.

Figure3:TPScellshowingwalkingtimesandmanual/machinetimes(inseconds)

(a) Pleaseestimateλ,LandWforthesystem.Stateanyassumptions.

(b) Inordertoobtainthisperformance,machine1(alathe)isrunningatthemaximummaterialremovalrate.Atthisrate,thecuttinginsertwearsoutevery11minutesofcuttingtimeandrequire 1 minute to replace. Due to preventative maintenance, the machines neverunexpectedly break down. Due to themuch longer tool replacement rates for the othermachines,theyallarereplacedduringscheduledmaintenance.Withthisnewinformation,re-estimatetheproductionrateofthesystem.Stateanyassumptions.


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(c) Yousuggestrunningthelatheataslowerspeed.Fromexperimentalresults,youfindthata

different setof cuttingparameters increases thecutting time from55 to65seconds,buttheinsertscannowbereplacedduringthescheduledmaintenance.Whatistheproductionrateofthesystematthesenewconditions?Stateanyassumptions.

The walking time + manual time don’t change from the calculation in part (a), so theaverage production rate is still 1 part/85 seconds = 0.71 parts/min. The cutting time formachine #1 increases from 55 seconds to 65 seconds, still remaining below the averagesystemproductionrateof85secondsperpart.Sincetheinsertdoesn’tneedtobereplacedoutsidescheduledmaintenance (whenallmachinesaredown),wedon’tneedtoaccountforthatreplacementtime.Thenthesystemproductionrateremainsat0.71parts/min.


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Problem5.OutputofPhotovoltaic(PV)SystemEstimatethepoweroutputforaPVsystemthatis20%efficient(sunlighttoelectricity).Duringsunny periods, the solar radiation is 1000 W/m2, for light clouds—500 W/m2, and heavyclouds—100W/m2.Meteorologicaldatashowsthatoverthelongtermthetimeratioforthethreeconditionsduringdaytimehoursis1hour/30minutes/10minutes,respectively.Whatistheaverageelectricpoweroutputofthesystemduringthedaytime?Whatwouldtheresultbeifweincludeddaytime(12hrs)andnighttime(12hrs)?Stateanyassumptions.Answer:Outputof20%efficientPVsystemduring:Sunnyperiods(60min):1000W/m2*0.20=200W/m2Lightclouds(30min):500W/m2*0.20=100W/m2Darkclouds(10min):100W/m2*0.20=20W/m2Weighteddaytimeaverage(100min):(60/100)*200W/m2+(30/100)*100W/m2+(10/100)*20W/m2=120+30+2=152W/m2.Thedayandnightaverage(assuming12hoursdurationforeach)wouldbe:152/2=76W/m2.

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MIT 2.810 Manufacturing Processes and Systems Homework …web.mit.edu/2.810/www/files/homeworks/2016_home...2.1 Vertical Mill Face end(s) ½” end mill 2.2 Vertical Mill Hog-out ½”