供应链管理英文Chapter12Exerciseanswers

1

Chapter12:ManagingUncertaintyinaSupplyChain:SafetyInventory

ExerciseSolutions

1.

L

D=LD=(2)(300)=600

L=LD=2(200)=283

ss=FS-1(CSL)L=FS-1(0.95)283=465(where,FS-1(0.95)=NORMSINV(0.95))

ROP=DL+ss=600+465=1065

ExcelWorksheet12-1illustratesthesecomputations

2.

DL=(T+L)D=(2+3)(300)=1500

L=T+L=2+3(200)=447

D

ss=FS-1(CSL)L=FS-1(0.95)447=736(where,FS-1(0.95)=NORMSINV(0.95))

OUL=D(T+L)+ss=1500+736=2236

ExcelWorksheet12-2illustratesthesecomputations

3.

DL=LD=(2)(300)=600

L=LD=2(200)=283

ESC=(1–fr)Q=(1–0.99)500=5

Weusethefollowingexpressiontodeterminethesafetystock(ss):

ESC=ss[1FS(ss)]+LfS(ss)

LL

ESC=–ss[1–NORMDIST(ss/L,0,1,1)]+LNORMDIST(ss/L,0,1,0).

WeutilizetheGOALSEEKfunctioninEXCELindeterminingsafetystock(ss)byusingssasthechangingvaluethatresultsinanESCvalueof5.

ExcelWorksheet12-3illustratesthesecomputations.

2

GoalSeekset-up:

SETCELL:C29

TOVALUE:5

BYCHANGINGCELL:C27

Thisresultsinassvalueof477andthereorderpointof:

ROP=DL+ss=600+477=1,077

4.

L

D=LD=(2)(250)=500

L=LD=2(150)=212

ss=ROP–ss=600–500=100

CSL=F(DL+ss,DL,L)=F(600,500,212)=NORMDIST(600,500,212,1)=0.68

ESC=ss[1FS(ss)]+LfS(ss)

LL

ESC=–ss[1–NORMDIST(ss/L,0,1,1)]+LNORMDIST(ss/L,0,1,0)=43.86

Fillrate(fr)=1-(ESC/Q)=1-(43.86/1000)=0.96

ExcelWorksheet12-4illustratesthesecomputations

5.

L

D=LD=(2)(250)=500

LDL

=L2+D2s2=21502+25021.52=431

ss=FS-1(CSL)L=FS-1(0.95)431=709(where,FS-1(0.95)=NORMSINV(0.95))

ESC=ss[1FS(ss)]+LfS(ss)

LL

ESC=–ss[1–NORMDIST(ss/L,0,1,1)]+LNORMDIST(ss/L,0,1,0)=9

Fillrate(fr)=1-(ESC/Q)=1-(9/1000)=0.991

3

Standarddeviationofleadtime

1.50

1.00

0.50

0.00

Requiredsafetyinventory

709

539

405

349

ExcelWorksheet12-5illustratesthesecomputations

6.

FollowingaretheevaluationsfortheKhakipants:

DisaggregatedOption:

L

D=LD=(4)(800)=3200

QL=LQD=4(100)=200

Coefficientofvariation=Q/=100/800=0.13

ssperstore=FS-1(CSL)QL=FS-1(0.95)200=329(where,FS-1(0.95)=NORMSINV(0.95))Totalsafetyinventory=(329)(900)=296,074

Totalvalueofsafetyinventory=(296,074)(30)=$8,882,210

Totalannualsafetyinventoryholdingcost=(8,882,210)(0.25)=$2,220,552

Holdingcostperunitsold=2220552/(800)(900)=$3.08

AggregatedOption:

DC=kD=(900)(800)=720000

QC=kQ=900(100)=3000

D

DL=LDC=(4)(800)(900)=2,880,000

QL=LQ=4(3000)=6000

ss=FS-1(CSL)QL=FS-1(0.95)6000=9869(where,FS-1(0.95)=NORMSINV(0.95))

Totalsafetyinventory=9869

4

Totalvalueofsafetyinventory=(9869)(30)=$296,070

Totalannualsafetyinventoryholdingcost=(296070)(0.25)=$74,018

Holdingcostperunitsold=74018/(800)(900)=$0.1

Savingsintheholdingcostperunitsoldfromaggregation=$3.08-$0.1=$2.98

FollowingaretheevaluationsfortheCashmereSweaters:

DisaggregatedOption:

L

D=LD=(4)(50)=200

QL=LQD=4(50)=100

Coefficientofvariation=Q/=50/50=1

ssperstore=FS-1(CSL)QL=FS-1(0.95)100=164(where,FS-1(0.95)=NORMSINV(0.95))Totalsafetyinventory=(164)(900)=147,600

Totalvalueofsafetyinventory=(147,600)(100)=$14,760,000

Totalannualsafetyinventoryholdingcost=$14,760,000)(0.25)=$3,690,000

Holdingcostperunitsold=$3,690,0001/(50)(900)=$82

Note:theabovearealsoincorrectontheworksheet.

AggregatedOption:

DC=kD=(900)(50)=45000

QC=kQ=900(50)=1500

D

DL=LDC=(4)(50)(900)=180000

QL=LQ=4(1500)=3000

ss=FS-1(CSL)QL=FS-1(0.95)3000=4935(where,FS-1(0.95)=NORMSINV(0.95))

Totalsafetyinventory=4935

Totalvalueofsafetyinventory=(4935)(100)=$493,456

5

Totalannualsafetyinventoryholdingcost=(493456)(0.25)=$123,364

Holdingcostperunitsold=123364/(50)(900)=$2.74

Savingsintheholdingcostperunitsoldfromaggregation=$82.84-$2.74=$79.50Centralizationresultsinsavingsforbothproducts,butitisevidentthatsavingsinholdingcostperunitsoldfromaggregatingCashmereSweatersishigherthanKhakipants.So,CashmereSweatersarebetterforcentralization.

ExcelWorksheet12-6illustratesthesecomputations.

7.

DisaggregatedOption:

France:

L

D=LD=(8)(3000)=24000

L=LD=82000=5657

ssatFrance=FS-1(CSL)L=FS-1(0.95)5657=9305

(where,FS-1(0.95)=NORMSINV(0.95))

Thessattheotherfivecountriesisevaluatedinasimilarmanner,whichresultsinatotalssfor

Europeof48,384

AggregatedOption:

ii=1

DC=6D=3000+4000+2000+2500+1000+4000=16500

62=

i

i=1

20002+22002+14002+16002+8002+24002=4445.22

C=

D

DL=LDC=(8)(16500)=132,000

L=L=8(4445.22)=12573

ss=FS-1(CSL)L=FS-1(0.95)12573=20,681(where,FS-1(0.95)=NORMSINV(0.95))

6

Inventorysavingsfromaggregation=48,384–20,681=27,704

ExcelWorksheet12-7illustratesthesecomputations.

8.

(a)

DisaggregatedOption:

Fromthepreviousproblem,weknowthatthetotalssforEuropeis48,384

Holdingcost=(200)(0.25)(48384)=$2,419,200

AggregatedOption:

ss=20,681

Holdingcost=(200)(0.25)(20681)=$1,034,036

Savingsfromaggregation=$2,419,200-$1,034,036=$1,385,164

(b)Ifthe$5/unitadditionalcostofassemblyfromcentralizationthenthetotaladditionalcosts=(132000)(52)(5)=

Savings=$1,385,164-So,itisnoteconomicaltoaggregate

(c)Iftheleadtimechangesto4weeks,weevaluatethesafetystocksandassociatedcostsinasimilarmanner.

Theholdingcostfromthedisaggregatedoption=(200)(0.25)(34213)=$1,710,650

Theholdingcostfromtheaggregatedoption=(200)(0.25)(14623)=$731,150

Savingsfromaggregation=$1,710,650-$731,150=$979,500

ExcelWorksheet12-8illustratesthesecomputations.

9.

Sincethedemandatvariouslocationsisnotindependent,weutilizethefollowingexpressionsfortheaggregatedoption:

DC==1Di

var(DC)=σ+2pijij

i=1i>j

=var(DC)

7

Forp=0.2

=var(DC)=20002+...+24002+2(0.2){(2000)(2200)+...+(800)(2400)

L=L=20002+...+24002+2(0.2){(2000)(2200)+...+(800)(2400)}8

=17307

ss=FS-1(CSL)L=FS-1(0.95)17307=28467(where,FS-1(0.95)=NORMSINV(0.95))Inventorysavingsfromaggregation=48,384–28,467=19,918

Holdingcostsavingsfromaggregation=(200)(0.25)(19918)=$995,900

Thefollowingtableshowsthesavingsasthecorrelationcoefficientincreasesfrom0to1withincrementsof0.2

ReplenishmentLeadTime(Weeks)

4

$979,474$704,191$489,462$307,211$146,047

$0

6

$1,199,606$862,454$599,466$376,255$178,870

$0

8

$1,385,185$995,876$692,203$434,462$206,542

$0

10

$1,548,684$1,113,424 $773,907 $485,743$230,921

$0

12

$1,696,498$1,219,694 $847,772 $532,105$252,961

$0

Corr.Coeff.

995876.3

0

0.2

0.4

0.6

0.8

1

ExcelWorksheet12-9illustratesthesecomputations.

10.

UsingSeaTransportation:

Averagebatchsize=DT=(5000)(20)=100,000

L=LD=36(4000)=24,000

ss=FS-1(CSL)L=FS-1(0.99)24000=55832(where,FS-1(0.99)=NORMSINV(0.99))

Daysofsafetyinventory=55832/5000=11.17days

Averagecycleinventory=batchsize/2=100000/2=50,000

Daysofcycleinventory=50000/5000=10days

8

Totalinventorycost(cycle+safety)=(50000+55832)(100)(0.2)=$2,116,640

Transportationcostperyear=(0.5)(5000)(365)=$912,500

AnnualHoldingCost+TransportationCost=$2,116,640+$912,500=$3,029,140

In-TransitInventory=DL=(5000)(36)=180,000

CostofHoldingIn-TransitInventory=(180000)(100)(0.2)=$3,600,000

TotalCosts(includingin-transitinventory)=$3,029,140+$3,600,000=$6,629,140

UsingAirTransportation:

Averagebatchsize=DT=(5000)(1)=5,000

L=LD=4(4000)=8,000

ss=FS-1(CSL)L=FS-1(0.99)8000=18,611(where,FS-1(0.99)=NORMSINV(0.99))

Daysofsafetyinventory=18611/5000=3.72days

Averagecycleinventory=batchsize/2=5000/2=2,500

Daysofcycleinventory=2500/5000=0.5days

Totalinventorycost(cycle+safety)=(2500+18611)(100)(0.2)=$422,220

Transportationcostperyear=(1.5)(5000)(365)=$2,737,500

AnnualHoldingCost+TransportationCost=$422,220+$2,737,500=$3,159,720

In-TransitInventory=DL=(5000)(4)=20000

CostofHoldingIn-TransitInventory=(20000)(100)(0.2)=$400,000

TotalCosts(includingin-transitinventory)=$3,159,720+$400,000=$3,559,720

Basedontheresultsairtransportationwouldbetheoptimalchoice,butifMotoroladoesnothavetheownershipofin-transitinventorythenseatransportationistheoptimalchoice.

ExcelWorksheet12-10illustratesthesecomputations.

9

11.

UsingSeaTransportation:

Averagebatchsize=DT=(5000)(20)=100,000

L=L+T=36+20(4000)=29,933

D

ss=FS-1(CSL)L=FS-1(0.99)29933=69,635(where,FS-1(0.99)=NORMSINV(0.99))Daysofsafetyinventory=69635/5000=13.93days

OUL=D(T+L)+ss=5000(36+20)+69635=349,635

Averagecycleinventory=batchsize/2=100000/2=50,000

Daysofcycleinventory=50000/5000=10days

Totalinventorycost(cycle+safety)=(50000+69635)(100)(0.2)=$2,392,700

Transportationcostperyear=(0.5)(5000)(365)=$912,500

AnnualHoldingCost+TransportationCost=$2,392,700+$912,500=$3,305,200

In-TransitInventory=DL=(5000)(36)=180,000

CostofHoldingIn-TransitInventory=(180000)(100)(0.2)=$3,600,000

TotalCosts(includingin-transitinventory)=$3,029,140+$3,600,000=$6,905,200

UsingAirTransportation:

Averagebatchsize=DT=(5000)(1)=5,000

L=L+T=4+1(4000)=8,944

D

ss=FS-1(CSL)L=FS-1(0.99)8944=20,807(where,FS-1(0.99)=NORMSINV(0.99))

Daysofsafetyinventory=20807/5000=4.16days

OUL=D(T+L)+ss=5000(1+4)+20807=45,807

Averagecycleinventory=batchsize/2=5000/2=2,500

Daysofcycleinventory=2500/5000=0.5days

10

Totalinventorycost(cycle+safety)=(2500+20807)(100)(0.2)=$466,150

Transportationcostperyear=(1.5)(5000)(365)=$2,737,500

AnnualHoldingCost+TransportationCost=$466,150+$2,737,500=$3,203,650

In-TransitInventory=DL=(5000)(4)=20000

CostofHoldingIn-TransitInventory=(20000)(100)(0.2)=$400,000

TotalCosts(includingin-transitinventory)=$3,203,650+$400,000=$3,603,650

Basedontheresultsairtransportationwouldbetheoptimalchoice.EvenifMotoroladoesnothavetheownershipofin-transitinventory,airtransportationistheoptimalchoice.

ExcelWorksheet12-11illustratesthesecomputations.

12.

ss=ROP–DL=750–300(2)=750-600=150

L=LD=2(100)=141.42

CSL=F(DL+ss,DL,L)=F(750,600,141.42)=NORMDIST(ss/L,0,1,1)=85.56%

ESC=一ss[1一FS(ss)]+LfS(ss)

LL

ESC=–ss[1–NORMDIST(ss/L,0,1,1)]+LNORMDIST(ss/L,0,1,0)=10

Fillrate(fr)=1-(ESC/Q)=1-(10/1500)=0.993

IftheROPincreasedfrom750to800thefillratewillincreaseto0.996

ExcelWorksheet12-12illustratesthesecomputations.

13.

Fillrate(fr)=1-(ESC/1500)=0.999

So,ESC=1.5

ESC=一ss[1一FS(ss)]+LfS(ss)]

LL

ESC=–ss[1–NORMDIST(ss/L,0,1,1)]+LNORMDIST(ss/L,0,1,0)=1.5

11

WeusetheGOALSEEKfunctionindeterminingthesafetystock(ss)byusingssasthechangingvaluethatresultsinanESCvalueof1.5.

GoalSeekset-up:

SETCELL:A15

TOVALUE:1.5

BYCHANGINGCELL:D12

Thisresultsinanssvalueof271andareorderpointof=300(2)+271=871

ExcelWorksheet12-13illustratesthesecomputations.

14.

(a)

DisaggregatedOption:

=L+T=3+7(50)=158

L+TD

ssperstore=FS-1(CSL)L=FS-1(0.99)158=367.83

(where,FS-1(0.99)=NORMSINV(0.99))

Totalsafetyinventory=(367.83)(25)=9195.7

AggregatedOption:

DC=kD=(25)(300)=7500

var(DC)=σ+2pijij

k

i=1i>j

=var(DC)

C=k=25(50)=250(weareassumingthatp=0.Ifpisnot0thenthecovariance

D

termshavetobeincluded)

=

L+T

L+TC=3+7(250)=791

D

ss=FS-1(CSL)L=FS-1(0.99)791=1839.14(where,FS-1(0.99)=NORMSINV(0.99))

12

Unitssavingsfromaggregation=9195.7–1839.14=7356.56

Inventorysavings=(7356.56)(10)=$73,566

Annualholdingcostsavings=(73,566)(0.2)=$14,712

Increaseindeliverycost=(300)(25)(365)(0.02)=$54,750

Sincetheincreaseintransportationcostsoutweighsthesavingsreceivedfromaggregation,wedonotrecommendaggregationforthiscase.

(b)

Weutilizethesameapproachasin(a)bychangingthedailydemandmeanandstandarddeviationto5and4,respectively

Unitssavingsfromaggregation=735.66–147.13=588.52

Inventorysavings=(588.52)(10)=$5885.2

Annualholdingcostsavings=($5885.2)(0.2)=$1177

Increaseindeliverycost=(5)(25)(365)(0.02)=$913

Sincetheincreaseintransportationcostsdoesnotoutweighthesavingsreceivedfrom

aggregation,werecommendaggregationforthiscase.

(c)Yes.Thebenefitfromaggregationdecreasesaspincreases.Whenp=0.5,wedonotrecommendaggregationinbothcases.

ExcelWorksheet12-14illustratesthesecomputations.

15.

(a)

PopularVariantatLargeDealer:

Decentralized:

ss(ateachlargedealer)=FS-1(CSL)LD=FS-1(0.95)4(15)=49.35

ss(acrossalllargedealers)=(5)(49.35)=246.73

PopularVariantatSmallDealer:

Decentralized:

ss(ateachsmalldealer)=FS-1(CSL)LD=FS-1(0.95)4(5)=16.45

ss(acrossallsmalldealers)=(30)(16.45)=493.46

(b)

PopularVariantallInventoriesCentralized:

Lowvolumevariantwithoutcomponentcommonality:

13

Demandperperiod=demandatlargedealers+demandatsmalldealers

=(50)(5)+(10)(30)=550

Standarddeviationofdemandperperiod=5(15)2+30(5)2=43.30

ss(atregionalwarehouse)=FS-1(CSL)LD=FS-1(0.95)4(43.30)=142.45reductioninsafetyinventoryfromcompleteaggregation=246.73+493.46–142.45=597.74

holdingcostsavingsperyear=(597.74)(20000)(0.2)=$2,390,942.52

production+transportationcostincreaseperyear=(550)(100)(52)=$2,860,000

(c)

PopularVariantonlySmallDealerInventoriesCentralized:

Demandperperiod=demandatsmalldealers

=(10)(30)=300

Standarddeviationofdemandperperiod=30(5)2=27.39

ss(atregionalwarehouse)=FS-1(CSL)LD=FS-1(0.95)4(27.39)=90.09

reductioninsafetyinventoryfromsmalldealercentralization=493.46–90.09=403.36

holdingcostsavingsperyear=(403.36)(20000)(0.2)=$1,613,440

production+transportationcostincreaseperyear=(300)(100)(52)=$1,560,000

(d)Centralizinginventoriesfromsmalldealersanddecentralizingatlargedealersistheoptimalstrategy

(e)Similaranalysiscanbeperformedfortheuncommonvariant(SeeEXCELWorksheet12-15formoredetails)

(f)Forthepopularvariant,centralizeinventoriesfromsmalldealersanddecentralizeatlargedealers.Fortheuncommonvariant,centralizeallinvento