FinalExam-Solutions
Module1
SmoothChinDeviceCompany SC1.Whichstationisthebottleneckoftheprocess?
Theactivitycapacitiesare: A=1/44razor/sec B=1/40 C=1/55 D=1/60 E=1/50 F=1/40 SoD,theslowest,isthebottleneck. .Whatistheprocesscapacity(youcanigno pacity(youcanignoreanystart-uporempt reanystart-uporemptysystemeffects)? ysystemeffects)? SC2.Whatistheprocessca ThisisthecapacityofStationD:1/60*3600razors/hour. 60razor(s)/hour .Whatarethedirectlaborcostsassociated rcostsassociatedwithproducingoner withproducingonerazor? azor? SC3.Whatarethedirectlabo 1minute*6stations*$25/60perminute=$2.5perrazor SC4.Whatistheaveragelaborutilizationofthesixworkers?
289/(289+71)=0.802
IndustrialBakingProcess IB1.Inventory=4000units/hrx1/5hr=800units
Save-A-LotRetailers
SL1.UsingLittle’sLaw:Flowtime=In UsingLittle’sLaw:Flowtime=Inventory/FlowratewhereFlow ventory/FlowratewhereFlowrate=COGS(costofthe rate=COGS(costofthe
goodsthat“flows”peryear) Substituting,Flowtime=$5,743MM/$53,962MM/year=0.106years
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Indays,Flowtime=0.106years*365days/year=38.69days SL2.
First,wecalculateinventorycostsforSave-A-Lot.Aproductstays38.69dayson First,wecalculateinventorycostsforSave-A-Lot.Aprod uctstays38.69daysonaveragefor averagefor Save-A-Lot.Theinventorycostsfora$50householdcleanerare$50*0.2*38.69/365=$1.06 ForWally’sMart,theaveragenumberofdaysininventoryis ($40,694MM/$316,606MM)*365=46.91days.Theinventorycostsfora ($40,694MM/$316,606MM)*365=46.91da ys.Theinventorycostsfora$50householdcleaner $50householdcleaner are$50*0.2*46.91/365=$1.29 Theinventorycostfora$50householdcleaneris$0.23lowerforSave-A-Lot.
GreatValleyIncomeTaxAdvice
Resource
Capacity (min/mth)
Workload for groups (min/mth)
Total workload (min/mth)
Admin
9600
7.5(70)+2.5(120)+25(50)+15(100) 7.5(70)+2.5(120)+25 (50)+15(100)
3575
37%
Senior
9600
7.5(50)+2.5(150)+25(5)+15(30) 7.5(50)+2.5(150)+25 (5)+15(30)
1325
14%
Junior
9600
7.5(120)+25(300)+25(80)+15(200) 7.5(120)+25(300)+25 (80)+15(200)
6650
69%
Admin w/WP
9600
7.5(45)+2.5(80)+25(35)+15(70) 7.5(45)+2.5(80)+25( 35)+15(70)
2462.5
26%
Implied utilization
GV1.Junioraccountant GV2.Senioraccountant:14% GV3.Junioraccountant:69% GV4.Administrator:37%
Module2
WindTunnelTesting–CarbonBikeFrames CBF1.
Onaverage,2outof7customersperdayarerefits Onaverage,2outof7cu stomersperdayarerefits,whichleaves5newcus ,whichleaves5newcustomersperday. tomersperday.
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CBF2.
OEE=(100minvalue-addtime/customerx5newcustomers/dayx9days/cycle)/(60m OEE=(100minvalue-addtime/customerx5newcusto mers/dayx9days/cycle)/(60min/hrx in/hrx 24hr/dayx10days/cycle)=4500/14,400=0.31
TastyTim’s TT1.
30+20+240+180+10+15+40+5+5+5+10+20=580seconds=9minutesand40seconds TT2.
1/demand=3600secs/120units=30sec/unit TT3.
Laborcontent/takt=580/30=19.3->roundup=20workers TT4.
Takt=3600/40=90sec/unit Laborcontent/takt=580/90=6.4->roundup=7employees
FastBusInc. FB1.
#ofcustomersservedperyear=(28seatssoldperone-waytrip)x(14one-ways #ofcustomersservedperyear=(28seatssoldperone-w aytrip)x(14one-waystripsperday)x tripsperday)x (2buses)x(365days)=286,160customers FB2.
0.3(seeExcelbelow,column1) FB3.
0.66(seeExcelbelow,column2)
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Module3
PTests
PT1.
Capacity=300/(30+(300x0.2))=200samples/hr PT2.
Capacity=2.5samples/min=B/(30+(Bx0.2)) B=150samples
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PowerForAll(PFA) PFA1.
Totaldemand=12kg/hr.Totalsetuptime=1hr.Processingtimep=1/20hr/kg. Settingcapacity=demandandsolvingforBgivesB=30 Settingcapacity=demandandsolvingforBgivesB=30kg.Therefore,theamountofsoyb kg.Therefore,theamountofsoybased ased proteinthatshouldbeproduced=10/12*30=25kg. PFA2.
Totaldemand=15kg/hr.Totalsetuptime=1.5hr.Processingtimeremainsthesame. Settingcapacity=demandandsolvingforBgivesB=90 Settingcapacity=demandandsolvingforBgivesB=90.Therefore,theamountofsoyb .Therefore,theamountofsoybased ased proteinthatshouldbeproduced=10/15*90=60kg.
Module4
TomOpim
TO1.
Utilization=flowrate/capacity=(20calls/hr)/(30calls/hr)=2/3=0.6667 Idletime=1–utilization=1/3=20min/hr Inan8-hourshift,Tomwillhave8*20=160mintoread160pages TO2.
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Usewaittimeformulaforoneresource: Averagewaittime=2.5min TO3.
Linecharge=$5.00/hr=$0.083/min Averagecustomertakes2.5minwaittime+2minprocessingtime=4.5min Thereforetheaveragelinecostpercustomer=$0.083*4.5=$0.375 Flowrateperday=(20customers/hour)*8hour Flowrateperday=(20cu stomers/hour)*8hours=160customers/day s=160customers/day Linecostperday=$0.375*160=$60
PhillyBarberShops PBS1.
Usewaittimeformulaformultipleresources: Activitytime=30min m=4 Utilization=flowrate/capacity=(4customers/hr)/(8customers/hr)=0.5 CVa=1 CVp=1 Thereforeaveragewaittime=3.35min PBS2.
Usewaittimeformulaforoneresource: Activitytime=30min Utilization=flowrate/capacity=(1customer/hr)/(2customers/hr)=0.5 CVa=1 CVp=1 Thereforeaveragewaittime=30min PBS3.
Totaltime=20minwalking+3.35minwaittime+30minhaircut=53.35min PBS4.
Newflowrate=5customers/hr Thereforenewinter-arrivaltime=12min PBS5.
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Usewaittimeformulaformultipleresources: Activitytime=30min m=5 Utilization=flowrate/capacity=(5customers/hr)/(10customers/hr)=0.5 CVa=1 CVp=1 Thereforeaveragewaittime=2.17min.Thewaitingtimeislowercomparedtothepre-merger Thereforeaveragewaittime=2.17min.Thewaitingtimeislowercomp aredtothepre-merger scenariobecausetheonlyvariablethathas scenariobecausetheonlyvariablethathaschangedism(thenu changedism(thenumberofbarbers);thegr mberofbarbers);thegreatest eatest impactofthischangeinthewaittimeformulaisin impactofthischangeinthewaittimeformulaisinreducingthesizeofthefirstt reducingthesizeofthefirsttermofthe ermofthe formula(activitytime/m).
HospitalTraumaBays HTB1.
m=6 r=p/a=90min/30min=3 FromErlanglosstable,probabilitythatalltrauma FromErlanglosstable,pro babilitythatalltraumabaysarefullandpatien baysarefullandpatientissenttoanother tissenttoanother hospital=0.0522 Thus,5.22%ofpatientsendupsenttoanotherhospital HTB2.
Average#ofpatientstreatedintraumabaysperd Average#ofpatientstreat edintraumabaysperday=(1–0.0522)*(demandof2p ay=(1–0.0522)*(demandof2patients/hr) atients/hr) *(24hrs/day)=45.5patients HTB3.
Answer:b.Processingtimechanges,whichgivesusanewr=p/a=60min/30min=2 Answer:b.Processingtimechanges,whichgivesu sanewr=p/a=60min/30min=2.Thus, .Thus, theprobabilitythatalltraumabaysarefulldeclines theprobabilitythatalltraumabaysarefulldeclinesto0.0121,andmorepatien to0.0121,andmorepatientswillhave tswillhave immediateaccesstoatraumabay. HTB4.
ConsultingtheErlanglosstablerevealsthattheminimumnumberoftraumab ConsultingtheErlanglosstablerevealsthattheminimu mnumberoftraumabaysneededto aysneededto ensurealossprobabilityof<0.1is6.
ToDoList Answer:C(homework).Dotheshortesttaskfirst.Theseq eshortesttaskfirst.Thesequenceisthefollowing:Friend uenceisthefollowing:Friend TDL1.Answer:C(homework).Doth (finishesin10minfromnow),food(30minfromnow),mot (finishesin10minfromnow) ,food(30minfromnow),mother(60minfromnow),Fac her(60minfromnow),Facebook ebook
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(120minfromnow),homework(200minfromnow),nap,snowshoes.Since90m (120minfromnow),homework(200minfromnow),na p,snowshoes.Since90min<120min, in<120min, youwillbedoinghomeworkin190minutesfromthetimethatyoustart.
YourNurse YN1.Answer:C
Module5
50-StepAssemblyLine 50SAL1.Probabilityofdefect:1-(0.99^50)=0.39499 50SAL2.Probabilityofshippingadefectiveproduct=0.39499*0.1*0.1=0.0039499
ProcesswithScrap AssumethedemandisDunits.ThatimpliesthatDu units.ThatimpliesthatDunitshavetoflowthrou nitshavetoflowthroughsteps3and ghsteps3and PS1.AssumethedemandisD 4.However,becauseofthescraprate,5Dunitsh 4.However,becauseofthes craprate,5Dunitshavetoflowthroughreso avetoflowthroughresources1and2. urces1and2. PS2.Thecapacityif1/5,1/4,2/20,1/12unitsperminuterespectively.
Calculateimpliedutilization(IU)foreachresourceas(demandatthatresou Calculateimpliedutilization(IU)foreachresourceas(dema ndatthatresource)/(capacityat rce)/(capacityat thatresource): Resource1:IU=5D/(1/5)=25D Resource2:IU=5D/(1/4)=20D Resource3:IU=D/(2/20)=10D Resource4:IU=D/(1/12)=12D Thehighestimpliedutilizationisatresource1,whichmakes Thehighestimpliedutilizationisat resource1,whichmakesthisresourcethebott thisresourcethebottleneck. leneck.
LeanBurgers LB1.CapabilityScore=(95.5-94.5)/(6*0.25)=0.66667
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LB2.Whatistheprobabilitythatbeefpattyistooheavy?Thisis
1-Normdist(95.5,95,0.25,1)=0.02275 Whatistheprobabilitythatthatthebeefpattyistoolight?Thisis Normdist(95.5,95,0.25,1)=0.02275 Sothedefectprobabilityis:2*0.02275=0.0455 LB3.Togetthenewstandarddeviation,wesolve:(95.5-94.5)/(6*stdev)=1=>stdev=1/6=
0.16666
Toyota–Jidoka TJ1.Answer:G.Jidokareferstomakingp Answer:G.Jidokareferstomakingproductionproblemsvisib roductionproblemsvisibleandstoppingprod leandstoppingproduction uction
upondetectionofdefects
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