Download - Solutions Review Questions Session 3
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7/25/2019 Solutions Review Questions Session 3
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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.