7/25/2019 Solutions Review Questions Session 3

1/2

Solutions

to

Review

Questions

Session

3

QuickLubeInc.

A)Utilization==3/4=0.75;Wq=15*(0.75/(10.75))**(1+1)=45minutes(or0.75hrs)

B)FromLittleslaw:WIP=TR*TT=3cars/hrx0.75hr=2.25cars

CD)

If

every

job

lasts

exactly

15

min,

the

process

has

no

variability,

therefore

after

purchasing

the

newmachineCVs=0.Boththeaveragewaitingtimeandtheaveragenumberofcarsinline

woulddecrease(youcanseethisbylookingatWq(CVa=0)andLittlesLaw(TTdecreasesso

doesWIP).

Mr.Ks

a)Youcanserve3customersperhour,and60/25customersarriveperhour,hencetheutilization

rateis(60/25)/3=80%.TheaveragewaitingtimeWq=60/3*0.8^(Sqrt(2*(3+1))1)/(10.8)*

(1.2^2+1.2^2)/2=95.8minutes.UsingLittleslaw,theaveragenumberofcustomerswaitingis:

WIP=TT*TR=95.8min*1/25customers/min=3.83.

b)Ifyouhireoneadditionalserver(m=4)thewaitingtimeis17.9minutes,ifyouhiretwoadditional

servers(m=5)thewaitingtimeis5.45minutes.Thus,weneedtwoadditionalservers,ifwewanta

waitingtime

of

less

than

10

minutes.

Localbranchofabank

GiventermsareWIP=7[customers]andThroughputRate=60/2=30[customers/hour].

UsingLittlesformula:Throughputtime=WIP/TR=7[customers]/30[customers/lunchhour]*

60[minutes/hour]=14[min].

Therefore,theestimateof6minutesappearstobeunrealistic.

Inventoryturns

Givenarethroughputtime=1/40[year]andthroughputrate=100,000[units/year].

WIP=throughputtime*throughputrate= (1/40)[years]*100,000[units/year]= 2,500[units]

Thus,theannualholdingcostsare15[$/unit]*2500[units]=$37,500.

Maketoorderenvironment

Nottrue.Reducingleadtimeisequivalenttoreducingwaitingtimeinaqueue.Thisobjectivecanbe

achievedalsobyreducingvariabilityinthesystemandbyincreasingthepaceofoperations(for

instancereplacingslowmachineswithfasterworkstations)toreduceutilization.

NewStationonJubileeLine

ConfigurationB.

Theproblemcanbeanalyzedbyconsideringthetubestationasacombinationoftwoqueuing

systems,wheretheescalatorsaretheservers.Inthefirstsystemincomingcustomersqueueto

accessthedownwardescalator,whereasinthesecondsystemoutgoingcustomersqueuetoreach

thehallupstairsandexitthestation.

Theobjectiveofthedesignershouldclearlybetomaintaintheaveragequeuelengthbelowan

acceptablelevel.Giventhattheplatformandthehallhavethesamesurfacetheycantolerate

approximatelythesamenumberofcustomerswaiting,hencetheacceptablequeuelengthisthe

sameinthetwocases.Mattshouldthereforeanalyzethetwosystems(upwardanddownward)

whenoperatedwithonlyoneescalator,andallocatetheadditionalescalatortotheonethatismore

likelytogeneratethelongestqueue.

Rememberthatqueuelengthisproportionaltoscale,utilizationandvariability.Clearlyinthiscase

scaleandutilizationarethesameforthetwosystems(sameescalatorspeedandsameamountof

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incomingandoutgoingcustomers).Conversely,thevariabilityofoutgoingcustomersislargerthan

thevariabilityofincomingcustomers.Toseethisconsiderthatwhereasincomingcustomersarrive

atthehallonebyone,followingapoissonprocess(coefficientofvariation=1),outgoingcustomers

arriveattheplatforminlargebatches,wheneveratrainstops.Thisisaburstyarrivalprocessand

hasacoefficientofvariationlargerthanone.

Therefore,everythingelsebeingequalthesysteminwhichcustomerqueuetoexitthestationis

likelyto

generate

the

longest

queue

and

should

be

given

the

additional

escalator.