金融市场与机构(第六版)教师手册M24-MISH1438-06-IM-C

PartVII

TheManagementofFinancialInstitutions

Chapter24

RiskManagementinFinancialInstitutions

ManagingCreditRisk

ScreeningandMonitoring

Long-TermCustomerRelationships

LoanCommitments

Collateral

CompensatingBalances

CreditRationing

ManagingInterest-RateRisk

IncomeGapAnalysis

DurationGapAnalysis

ExampleofaNonbankingFinancialInstitution

SomeProblemswithIncomeandDurationGapAnalysis

ThePracticingManager:StrategiesforManagingInterest-RateRisk

Riskmanagementhasbecomeamajorconcernformanagersoffinancialinstitutionsinrecentyears.Thischapterprovidesanintroductiontothissubjectwhichisusefulforbusinessstudentswhowillnottakejobsinthefinancialinstitutionsindustry,butwhichwillalsoprovideasolidgroundingforstudentswhowillgoontopursuemoreadvancedcoursesinriskmanagementinfinancialinstitutions.

Thesectiononmanagingcreditriskisanexcellentapplicationofthebasicconceptsofadverseselectionandmoralhazardtoexplainmanagerialpracticesinthefinancialinstitutionsindustry.

Thesectiononmanaginginterestrateriskintroducesstudentstohowinterest-rateriskismeasuredandthenusesaPracticingManagerapplicationtooutlinestrategiesforhowtomanageinterest-raterisk.IfdurationGAPanalysisiscovered,thenitisnecessarythatthePracticingManagerapplicationinChapter3ondurationbecoveredearlierinthecourse.

Thischapterisreallyonebigapplicationofconceptsintroducedearlierinthecourse.Notonlydoesithelpsolidifystudents’understandingoftheseconcepts,butitshowshowtheseconceptsareusefulforsolvingmanagerialproblemsthatbusinessstudentsarelikelytoencounterintherealworld.

Chapter24RiskManagementinFinancialInstitutions137

1.Yes.Bywarningborrowersthattheywillnotbeabletogetfutureloansiftheyengageinriskyactivities,borrowerswillbelesslikelytoengageintheseactivitiesinordertohaveaccesstoloansinthefuture.

2.Securedloansareanimportantmethodoflendingforfinancialinstitutionsbecauseiftheborrowerdefaults,thefinancialinstitutioncantaketitletothecollateral,sellitoff,andusetheproceedstooffsetanylossesontheloan.Thusthefinancialinstitutioncanworrylessabouttheadverseselectionproblembecauseithassomeprotectioneveniftheborrowerwasabadcreditrisk.

3.Uncertain.Insomecases,raisingratesmaygeneratemoreincomeandthusincreaseprofits.However,thehigherinterestratesdoincreaseadverseselectioninwhichtherisk-proneborrowersaremorelikelytoseekouttheloans.Thustherateofdefaultmightgoupandbankprofitscouldsuffer.

4.Toreduceadverseselection,abankerneedstoscreenoutbadcreditrisksbylearningasmuchaspossibleaboutpotentialborrowers.Similarly,tominimizemoralhazard,thebankermustcontinuallymonitorborrowerstoseethattheyarecomplyingwithrestrictiveloancovenants.Henceitpaysforthebankertobynosy.

5.Compensatingbalancescanactascollateral.Theyalsohelpestablishlong-termcustomerrelationships,whichmakeiteasierforthebanktocollectinformationaboutprospectiveborrowers,thusreducingtheadverseselectionproblem.Compensatingbalanceshelpthebankmonitortheactivitiesofaborrowingfirm,sothatitcanpreventthefirmfromtakingontoomuchrisk,therebynotactingintheinterestofthebank.

6.False.Althoughdiversificationisadesirablestrategyforabank,itmaystillmakesenseforabanktospecializeincertaintypesoflending.Forexample,abankmayhavedevelopedexpertiseinscreeningandmonitoringaparticularkindofloan,therebyimprovingitsabilitytohandleproblemsofadverseselectionandmoralhazard.

1.Abankissuesa$100,000variable-rate,30-yearmortgagewithanominalannualrateof4.5%.Iftherequiredratedropsto4.0%afterthefirstsixmonths,whatistheimpactontheinterestincomeforthefirst12months?

Solution:At4.5%,therequiredpaymentiscalculatedas:

PV=100,000,I=4.5/12,N=360,FV=0

ComputePMT.PMT=506.685

Ifrateremainat4.5%,themortgagebalanceafter12monthsis:

PMT=506.685,N=348,I=4.5/12,FV=0

ComputePV.PV=98,386.71,or$1,613.29ofthepaymentswenttowardprincipal.

Thetotalpayments=506.68512=$6,080.22.

Interestincomefortheyearis$6080.22$1613.29=$4,466.93

Ifratesdropto4%forthelast6monthsoftheyear:

First,calculatetheinterestforthefirstsixmonths:

PMT=506.685,N=354,I=4.5/12,FV=0

ComputePV.PV=99,202.38,or$797.62ofthepaymentswenttowardprincipal.

138Mishkin/Eakins•FinancialMarketsandInstitutions,SixthEdition

Thetotalpayments=506.6856=$3,040.11.

Interestincomefortheyearis$3040.11一$797.62=$2,242.49ofinterestincome.Next,calculatetheinterestforthelastsixmonths:

At4.0%,therequiredpaymentiscalculatedas:

PV=99,202.38,I=4.0/12,N=354,FV=0

ComputePMT.PMT=477.772

PMT=477.772,N=348,I=4.0/12,FV=0

ComputePV.PV=98,312.41,or$889.97ofthepaymentswenttowardprincipal.Thetotalpayments=477.7726=$2,866.63.

Interestincomefortheyearis$2866.63一$889.97=$1,976.66ofinterestincome.Totalinterestincome=$2,242.49+$1,976.66=$4,219.15

Interestincomehasfallenby$247.78,or5.5%,belowwhatitwouldhavebeenhadtheinterestratenotfallen.

2.Abankissuesa$100,000fixed-rate,30-yearmortgagewithanominalannualrateof4.5%.Iftherequiredratedropsto4.0%immediatelyafterthemortgageisissued,whatistheimpactonthevalueofthemortgage?

Solution:At4.5%,therequiredpaymentiscalculatedas:

PV=100,000,I=4.5/12,N=360,FV=0

ComputePMT.PMT=506.685

Ina4.0%market,thevalueofthemortgageis:

PMT=506.685,I=4.0/12,N=360,FV=0

ComputePV.PV=106,131.

Thevalueofthemortgagehasincreasedby$6,131,or6.131%.

3.Calculatethedurationofa$100,000fixed-rate,30-yearmortgagewithanominalannualrateof7.0%.Whatistheexpectedpercentagechangeinvalueiftherequiredratedropsto6.5%immediatelyafterthemortgageisissued?

Solution:Thedurationcalculationshouldbecompletedusingaspreadsheet.Althoughthetechnique

isthesame,thereare360monthstohandle.Doingthis,thedurationcanbecalculatedas10.15years.

Pi

=一Duration=一10.15一0.005/1.07=4.74%P1+i

4.Thevalueofa$100,000fixed-rate,30-yearmortgagefallsto$89,537wheninterestratesmovefrom5%to6%.Whatistheapproximatedurationofthemortgage?

Solution:

Pi1+iP

=一Duration,orDuration=一

P1+iiP

Duration=一=10.98years0.01100,000

1.05一10,463

Chapter24RiskManagementinFinancialInstitutions139

5.Calculatethedurationofacommercialloan.Thefacevalueoftheloanis$2,000,000.Itrequiressimpleinterestyearly,withanAPRof8%.Theloanisdueinfouryears.Thecurrentmarketrateforsuchloansis8%.

Solution:Theannualinterestis0.082,000,000=160,000

01234

2,000,000

160,000

148,148

160,000

137,174

160,000

127,013

2,160,000

1,587,664

0.137174

0.074074

3.577097

3.175329

0.19052

Thisloanhasadurationof3.57years.

6.Abank’sbalancesheetcontainsinterest-sensitiveassetsof$280millionandinterest-sensitiveliabilitiesof$465million.Calculatetheincomegap.

Solution:GAP=$280$465=$185million

Anotherwaytostatethisisthebankhasaliability-sensitivegapof$185million.

7.Calculatetheincomegapforafinancialinstitutionwithrate-sensitiveassetsof$20millionandrate-sensitiveliabilitiesof$48million.Ifinterestratesrisefrom4%to4.8%,whatistheexpectedchangeinincome?

Solution:GAP=RSARSL

=$20million$48million

=$28million

I=APi

=$28million0.008

=$224,000million

8.Calculatetheincomegapgiventhefollowingitems:

•$8millioninreserves

•$25millioninvariable-ratemortgages

•$4millionincheckabledeposits

•$2millioninsavingsdeposits

•$6millionof2yearCD’s

Solution:GAP=RSARSL

RSA=$25million

RSL=$4million+$2million=$6million

GAP=$25million$6million

GAP=$19million

140Mishkin/Eakins•FinancialMarketsandInstitutions,SixthEdition

9.Thefollowingfinancialstatementisforthecurrentyear.Fromthepast,youknowthat10%offixed-ratemortgagesprepayeachyear.Youalsoestimatethat10%ofcheckabledepositsand20%ofsavingsaccountsareratesensitive.

SecondNationalBank

Assets

Liabilities

Reserves

$

1,500,000

CheckableDeposits

$15,000,000

Securities

MoneyMarketDeposits

$5,500,000

1Year

$

6,000,000

SavingsAccounts

$8,000,000

1to2Years

$

8,000,000

CDs

>2years

$

12,000,000

Variables-rate

$15,000,000

ResidentialMortgages

1Year

$22,000,000

Variables-rate

$

7,000,000

1to2Years

$5,000,000

Fixed-rate

$

13,000,000

>2years

$2,500,000

CommercialLoans

Fedfunds

$5,000,000

1Year

$

1,500,000

Borrowings

1to2Years

$

18,500,000

1Year

$12,000,000

>2years

$

30,000,000

1to2Years

$3,000,000

Buildings,etc.

$

2,500,000

>2years

BankCapital

$2,000,000$5,000,000

Total

$

100,000,000

Total

$100,000,000

WhatisthecurrentIncomeGAPforSecondNationalBank?Whatwillhappentothebank’scurrentnetinterestincomeifratesfallby75basispoints?

Solution:RSA=6M+7M+(0.1013M)+1.5M=$15.8million

RSL=(0.1015M)+5.5M+(0.208M)+15M+22M+5M+12M=$62.6millionGAP=$15.8million$62.6million

GAP=$46.8million

I=46.8million(0.0075)

=$+351,000

10.ChicagoAvenueBankhasthefollowingassets:

Asset

Value

Duration(inyears)

T-Bills

$100,000,000

0.55

ConsumerLoans

$40,000,000

2.35

CommercialLoans

$15,000,000

5.90

WhatisChicagoAvenueBank’sassetportfolioduration?

Solution:Totalassets=155M

Assetduration=0.55(100/155)+2.35(40/155)+5.90(15/155)=1.53years

Chapter24RiskManagementinFinancialInstitutions141

11.Abankaddedabondtoitsretainedportfolio.Thebondhasadurationof12.3yearsandcost$1,109.Justafterbuyingthebond,thebankdiscoveredthatmarketinterestratesareexpectedtorisefrom8%to8.75%.Whatistheexpectedchangeinthebond’svalue?

Solution:

编编

=-Duration根1+

编=-12.3根根$1,109=-$94.73

12.Calculatethechangeinthemarketvalueofassetsandliabilitieswhentheaveragedurationofassetsis3.60,theaveragedurationofliabilities0.88,andinterestratesincreasefrom5%to5.5%.

Solution:Assets:

%编A=-3.60根(0.005/1.05)=-1.71%

Liabilities:

%编L=-0.88根(0.005/1.05)=-0.42%

13.SpringerCountyBankassetstotaling$180millionwithadurationof5years,andliabilitiestotaling$160millionwithadurationof2years.Ifinterestratesdropfrom9%by75basispoints,whatisthechangeinbank’scapitalizationratio?

Solution:Priortotheratechange,thecapitalizationratio=20/180=11.11%

Thedurationgap=5-(160/180)根2=3.22

Changeinequity=-3.22根(-0.0075/1.09)根$180million=3.988millionChangeinassets=-5根(-0.0075/1.09)根$180million=6.192millionNewcapratio=(20+3.988)/(180+6.192)=12.88%

14.ThemanagerforTylerBankandTrusthasthefollowingassetstomanage:

Asset

Value

Duration(inyears)

Liability

Value

Duration(inyears)

Bonds

115,000,000

9.00

DemandDeposits

690,000,000

1.00

ConsumerLoans

Commercial

345,000,000

2.00

SavingAccounts

??

0.50

Loans

575,000,000

5.00

Ifthemanagerwantsadurationgapof3.00,whatlevelofSavingAccountsshouldthebankraise?Assumethatanydifferencebetweenassetsandliabilitiesisheldascash(duration=0).

Solution:

DURa=(9.00根115/1035)+(2.00根345/1035)+(5.00根575/1035)=4.44AssumethattheSavingAccountvalue=Y.

DURl=[1.00根690/(690+Y)]+[0.50根Y/(690+Y)]

()

DURgap=DURa-|(根DURl)|

3.00=4.44-[(690+Y)/1035]根[1.00根690/(690+Y)+0.50根Y/(690+Y)]

142Mishkin/Eakins•FinancialMarketsandInstitutions,SixthEdition

Y=1,600,000,000

Solution:

DURa=(9.00〉75/1650)+(2.00〉875/1650)+(5.00〉700/1650)=3.59

AssumethattheSavingAccountsvalue=Y.

DURl=[1.00〉300/(300+Y)]+[0.50〉Y/(300+Y)]

DURgap=DURa-(L/A〉DUR)l

3.00=3.59-[(300+Y)/1650]〉[1.00〉300/(300+Y)+0.50〉Y/(300+Y)]Y=1347

15.Thefollowingfinancialstatementisforthecurrentyear.

SecondNationalBank

AssetsDurationLiabilitiesDuration

Reserves

5,000,000

0.00

CheckableDeposits

15,000,000

2.00

Securities

MoneyMarketDeposits

5,000,000

0.10

<1Year

5,000,000

0.40

SavingsAccounts

15,000,000

1.00

1to2Years

5,000,000

1.60

CDs

>2years

10,000,000

7.00

Variables-rate

10,000,000

0.50

ResidentialMortgages

<1Year

15,000,000

0.20

Variables-rate

10,000,000

0.50

1to2Years

5,000,000

1.20

Fixed-rate

10,000,000

6.00

>2years

5,000,000

2.70

CommercialLoans

InterbankLoans

5,000,000

0.00

<1Year

15,000,000

0.70

Borrowings

1to2Years

10,000,000

1.40

<1Year

10,000,000

0.30

>2years

25,000,000

4.00

1to2Years

5,000,000

1.30

Buildings,etc.

5,000,000

0.00

>2years

5,000,000

3.10

BankCapital

5,000,000

Total

100,000,000

Total

$100,000,000

Calculatethedurationgapforthebank.

Solution:

DURa=(0.40〉0.05)+(1.60〉0.05)+(7.00〉0.10)+(0.50〉0.10)

+(6.00〉0.10)+(0.70〉0.15)+(1.40〉0.10)+(4.00〉0.25)

DUR=2.695

a

Bythesametechnique,usingtotalliabilitiesof95,000,000,

DUR=1.03

l

(L)

gapa(Al)

DUR=DUR-|〉DUR|=2.695-(95/100〉1.03)=1.7165

Chapter24RiskManagementinFinancialInstitutions143

16.Ratesensitiveassetsincreaseby$10million,sotheGAPrisesby$10millionto一$7.5million.Ifinterestratesfallby3percentagepoints,profitsrisenextyearbyI=GAPi=一$7.5(一0.03)=0.225million=$225,000.

144Mishkin/Eakins•FinancialMarketsandInstitutions,SixthEdition

17.Rate-sensitiveassetsincreaseby$4million,soGAPgoesfrom一$17.5millionto一$13.5million.(Ofthe$5millionoffixed-ratemortgages,$1millionisrate-sensitivebecause20%arerepaidwithinayear,soconvertingthe$5millionoffixed-ratemortgagesintovariable-ratemortgagesproducesanetincreaseofrate-sensitiveassetsof$4million.)BecauseGAPfallsinabsolutevalue,theeffectofchangesininterestratesonitsprofitsandhenceonitsinterest-rateriskissmaller.

18.Shereducesherestimateofrate-sensitiveassetsby$1million(10%of$10million)sotheGAPgoesfrom一$17.5millionto一$18.5million.Ifinterestratesfallby2percentagepoints,profitsnextyearrisebyI=GAPi=一$18.5(一0.02)=+$0.37million=$370,000.

19.Themanagerraisestheestimateofrate-sensitiveliabilitiesby$2.25millionsothatGAPgoesfrom一$17.5millionto一$19.75million.BecauseGAPrisesinabsolutevalue,theeffectofchangesininterestratesonitsprofitsandhenceitsinterest-rateriskislarger.Ifinterestratesriseby5percentagepoints,profitsnextyearchangebyI=GAPi=一$19.75million0.05=一$0.9875million.

20.Thepercentagechangeinnetworthasapercentageofassetsis%NW=一DURGAPi/(1+i)=一1.720.10/(1+0.10)=一0.156=一15.6%.With$100millionofassetsthismeansnetworthdeclinesby$15.6million,from$5millionto一$10.6million.Sincethebanknowhasnegativenetworth,itisinsolventandwillgooutofbusiness.

21.ThedurationgapisnowDURgap=DURA一(L/ADUR)=4L一(95/1002)=2.1years.Thechangeinnetworthasapercentageofassetsis%NW=一DURGAPi/(1+i)=一2.10.02/(1+0.10)=一0.038=一3.8%.With$100millionofassets,networthdeclinesby$3.8million,from$5millionto$1.2million.

22.Itshouldsolvethefollowingequation:0=DURA一[95/1002].Thisyield