10-Asset Liability Management[1]

8/3/2019 10-Asset Liability Management[1]

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1

ASSET / LIABILITY

MANAGEMENT


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Asset / Liability ManagementAlso known as asset-liability management, gap

management

Activity usually run in a Treasury Department of abank

Managed weekly or biweekly by a committee

Activity began in late 1970s as a result of high and

volatile interest ratesBanks assume much interest rate risk since they

borrow in one set of markets and lend in another


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Asset / Liability Management

Measuring interest rate risk

Focus is on GAP , there are3

types of GAPs. Dollar Gap, Funds Gap, Repricing Gap

Maturity Gap

Duration Gap


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GAP

GAPt= RSAt RSLt

where t = particular time intervalRSAt = $ of assets which are reset during

interval t, Rate-Sensitive-Assets

RSLt = $ of liabilities which are reset during

interval t, Rate-Sensitive-Liabilities


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GAP

Example:

Bank with assets & liabilities of following

maturities

Days

0 60 61 90 91 120 121 - 180

Assets 10 0 40 20

Liabilities 20 5 30 50

GAP (A-L) -10 -5 10 -30


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GAP

Example (cont.)

Cumulative GAP = C GAP

= GAP over whole period

C GAP = -10 5 + 10 30

= - 35

Note: If + GAP, then lose if rates fall

If GAP, then lose if rates rise


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GAP

Federal Reserve has required banks to report

quarterly the repricing GAPs (schedule RC-J) as

follows:1 day

2 day 3 months

over3

months 6 monthsover 6 months 1 year

over 1 year 5 year

over5 year


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GAP

Problems with GAP

1. Uses book-value approach: Focuses only on

income effect and not on capital gains effect

from rate changes.

2. Aggregation: Ignores distribution of

assets/liabilities within buckets could stillhave mismatch

3. Runoffs ignored: Interest and principal paid

plus loan prepaid must be invested. This

feature is ignored.


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Maturity Gap

Background

Consider a 1year bond with coupon 10% and YTM 10%

If rates increase to 11%

Conclude: If ro Pq (P/ (r< 0

100 + 0.10 v 100

1 + 0.10P = = = 100

100 + 0.10 v 1001 + 0.11

P = = = 99.10

110

1.10

1101.11


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Maturity Gap

Consider a 2 year bond


Price fell more than 1 year bond!

P = + = 1001101.102101.10

P = + = 98.29

110

1.11210

1.11


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Maturity Gap

Conclusion:

The longer the maturity, thegreater the fall in price for a

given level increase in interest

rates.


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Maturity Gap

Consider a 3 year bond


P = + + = 100101.102

101.10

1101.103

P = + + = 97.56

10

1.11210

1.11

110

1.113


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Maturity Gap

Notice Decline:

Time P0 Pn P0Pn Pn1Pn

1 yr 100 99.10 0.90 0.90

2 yr 100 98.29 1.71 0.813 yr 100 97.56 2.44 0.73


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Maturity Gap

Conclude: The fallincreasesatadiminishing

rateasa functionof maturity.

Maturity

(P

1 2 3


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Maturity Gap

Now, these principlesapply to

bankssince they have portfoliosof interest-ratesensitiveassets

andliabilities.


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Maturity Gap

Let MA = WA1MA1 + WA2MA2 + + WAnMAn

Where MA = average maturity of banks assetsMAj = maturity of assetj

WAj = market value of assetjas a % of total

asset market value

And ML = WL1ML1 + WL2ML2 + + WLnMLn

Where ML = average maturity of banks liabilities


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Maturity Gap

Then MG = MAML

For a minimum of interest rate risk, want: MG = 0

Typically, MG> 0 i.e. MA >ML

Ex) Bank borrows at 1 yr deposit of $90 paying

10% and invests in $100 3 yr bond at 10% with

$10 of equity.A L

B 100 90 D

10 E


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Maturity Gap

Suppose rates rise to 11%, then 3 yr bond is

worth $97.56 (as before) and deposit is

worth

P= 99 / 1.11 = 89.19

Thus

Assets Liabilities

97.56 89.19

8.37

E = 97.56 89.19(E= (A(L = 2.44 (0.81)

(E= 1.63

E = 10 1.63 = 8.37


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Maturity Gap

Thus, equity must absorb interest-rate risk exposure.

Notice

MG = MAML = 3 1 = 2

By previous propositions

If MG> 0

If ro, then bank will LOSE

If rq, then bank will GAIN

If MG< 0If ro, then bank will GAIN

If rq, then bank will LOSE


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Maturity Gap

At what rate change will bank become insolvent?

(E= 10 or (A(L = 10

Want:

If rp 16% 12.07 (4.66) = 7.41

If rp 17% 15.47 (5.38) = 10.09 YES!

+ + 100 [ 90] = 1010

(1+x)210

1 +x

110

(1+x)399

1 +x


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Maturity Gap

What if bank has matched withMG= 0, that is

invested in 1 yr bond, then

If rp 11% from 10%

(A = 99.10 100 = 0.90

(L = 89.11 90 = 0.89

If rp 12%

(A = 98.21 100 = 1.79

(L = 88.39 90 = 1.61

(E= 0.90 + 0.89 = 0.01

(E= 1.79 + 1.61 = 0.18


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Maturity Gap

SettingMG= 0 does NOT insure one

completely from interest-rate risk but does

work quite well.

Reasons why some risk remains:1. Amounts not matched (as before)

2. Timing of cash flows not considered3. Rates may not move exactly together

Using aDuration Gap measure will resolve #2.


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DURATION

Duration of an asset or liability is the

weighted-average time until cash flows are

received or paid.

The weights are the PV of each cash flow as

a % of the PV of all cash flows.


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DURATION

!!

!N

t

t

N

t

tPVtPVD

11

Where N= last period of CF

CFt= cash flow at time t

PVt= CFt / (1+R)t

R = yield on asset or liability


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DURATION

Example:

Duration of 8% $1,000 6 year Euro-bond,Eurobonds pay interest annually, yield is

8%.


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DURATION

Example (cont.)

T CFt 1/(1+R)t PVt PVt t

1 80 0.9259 74.07 74.07

2 80 0.8573 68.59 137.18

3 80 0.7938 63.51 190.53

4 80 0.7350 58.80 235.205 80 0.6806 54.45 272.25

6 1080 0.6302 680.58 4083.48

D = 4993.71 / 1000 = 4.993 years


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DURATION

Features of Duration

1. Duration increases with maturity at a

decreasing rate.

0)/(

(

(((

M

MD0"(( MD

2. Duration increases as yield decreases.

0/ (( RD

3. The higher the coupon, the lower the duration.

0/ (( CD


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DURATION

Consider a bond with annual coupon

payments C

or

NR

FC

R

C

R

CP

)1(...

)1(1 2

!

!

!N

t

t

t

R

CP

1 )1(


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DURATION

!

!!(

(N

t

t

t

R

tC

dR

dP

R

P

1

1

)1(

!

!(

(N

t

t

t

R

C

R

D

R

P

1 )1(1

!!

!N

t

t

N

t

tPVtPVD

11

Since andt

t

t

R

CFPV

)1( !

HenceR

RD

P

P

(!

(

1


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DURATION

R

RD

P

P

(!

(

1% Price Change P

P(

R

R

(

10

Slope = D


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DURATION

Example:

Consider 6 year Eurobond from before.Recall D = 4.99

If yields rise 10 basis points

P

P( = (4.99)v(0.001/1.08) = 0.000462 = 0.0462%

IfP=1000, price would fall to 999.538


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DURATION

Example (cont):

For semi-annual payments, the equationmust be modified:

R

RD

P

P

5.01

(!

(

R

RD

P

P

(!

(

1

Annual

Payment


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DURATION

Example:

2yr treasury with coupon of 8%, pays semi-annually with price of $964.54, with face

value of $1000.

964.54 = + + +

40

(1+0.5R)240

(1+0.5R)

40

(1+0.5R)31040

(1+0.5R)4

R = 0.10 yrs89.154.964

37.1818!!

!

P

tPVD

t


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DURATION GAP

Now we can apply these ideas to a bank.

Recall: RR

DP

P

(

!

(

1

Now consider a bank and let:

A = value of assets

( A = change in value of assetsL = value of liabilities, excluding equity

( L = change in value of liabilities


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DURATION GAP (Cont.)

Then,

)1( R

R

DA

A

A

(

!

(

Where, DA

= weighted-average duration of the assets

=[1D1 + [2D2 + + [nDn[i = MV of asset i /total MV ofassets


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And for liabilities, we have the same:

)1( R

RD

L

L

L

(!

(

Where, DL = [1D1 + [2D2 + + [mDm

[i = MV of liability i /total MV ofassets


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Now, let

(E = (A (L

= - (DAA DLL) (R / (1+R)

So, (E / A = - DG (R / (1+R)

where DG = DA DLL / A

Duration Gap

(E = -DG A (R / (1+R)


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Thus, the change in the net worth of a bank

depends on:1. The duration gap of the bank (DG)

2. The size of the bank (A)

3. The size of the interest rate shock

((R / (1+R))


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Example:

Bank with DA =5

years, DL =3

years, R = 0.10,A = $100 million, L = $90 million,

E = $10 million. If Rp 11%, what is effect on

net worth?


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Example (cont.) :

(E = - (DA DLL/A) A (R / (1+R) = -$2.09 million

Thus, E : 10 million p 7.91 million

Notice: (A = -DA A (R / (1+R) = -$4.55 million

(L = - DL L (R / (1+R) = -$2.46 million


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Example (cont.) :

A L A L

100 90 95.45 87.54

10 7.91

Note: Both A and L fall with interest rate rise .

DG = 2.3 years


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Thus,

If DG > 0 and Ro bank lose

DG > 0 and Rq bank gain

If DG < 0 and Ro bank gain

DG < 0 and Rq bank lose

Note: this is opposite to GAP = RSA RSLif GAP > 0 and Ro bank gain

Why?


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Why? GAP p in $ domain

DG p in time domain

Want DG = 0 for fall protection, notice

DG = DA DL L / A

= DA DL (A K) / A where K = capital

= DA DL (1 k) where k = K/ADuration depends directly on capital ratio!

DG = DA DL + DLk DG o as ko


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However, bank with more capital is better protected.

To see this, (E = -DG A (R / (1+R)

(E

E= -DG

AE

(R

(1+R)

(E

E= -DG

1

k

(R

(1+R)

Thus, the larger the k, the smaller the % change in

equity will be.


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Ex) In previous example,

(E / E = - 2.09 M / 10 M = -20.9%

Ex) Suppose same example except L = 95 M

So, K = 5, and k = 0.05

DG = DA DL (1 k) = 2.1

(E = - 1.9

5

E : 5p 3.05

(E / E = - 2.15 1/0.05 0.01/1.10 = - 39.1%

or (E / E = -1.95/5 = -39%


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DURATION GAP

Example: BANK

ASSETS AMT D LIABILITIES AMT DST Securities 150 0.5 DD 400 0

LT Securities 100 3.5 ST CDs 350 0.4

Loans Float 400 0 LT CDs 150 2.5

Loans Fixed 350 2 Equity 100

Total 1000 1000


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DURATION GAP

Example: (continue)

DA=0.15v0.5+0.1v3.5+0.4v0+0.35v2=1.125 year

DL= v0+ v0.4+ v2.5=0.572 year

DG=DADL = 1.125 0.572v 0.9 = 0.6102

400900

350900

150900

LA


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DURATION

Example: (continue)

IfR = 0.08 0.08 p 0.09

(E

A = DG(R

1 +R

(E

A = 0.6102

= 0.005

65

0.01

1 + 0.08(E= 0.00565v 1000 = 5.65

Efrom 100 p 94.35


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DURATION GAP

Although Duration Gap takes timingof cash flows into account, there are

problems with its implementation and

use.


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DURATION GAP

Problems withDG

1. Not easy to manipulateDA andDL. (reason for

using artificial hedges such as swaps, options, or

futures)

2. Immunization is a DYNAMIC problem. (i.e.,

requires constant rebalancing)3. Large rate changes and convexity (model only

applies to small changes)


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DURATION GAP

P

P(

R

R

(

1

ModelActual

We are here


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DURATION GAP

P

P(

RR

(1

Actual Model

(R +

If(R > 0, DGoverpredictsPdecrease


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DURATION GAP

If(R < 0, DGunderpredictsPincrease

P

P(

RR

(1

Actual Model

(R


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DURATION GAP

Problems withDG(Continue)

Convexity =

It can be measured.

Convexity is good for banks. They do betteras a result.

measure of curvature

of duration curve


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DURATION GAP


4. Flat Term Structure. (Notice all rates R, implies

flat term structure. There are models whichmake different assumptions.)

5. Non-Traded Assets. (Small business loans and

consumer loans have no market value estimatesasR changes.)

6. Not consider Default Risk or Prepayment Risk.


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DURATION GAP


7. Duration of Equity. (Should equity be included?

POSSIBLY.)

To see this, using dividend growth model

d1 = div in year 1k= required return

g= growth rate in dividend

P0 =d1

(kg)


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DURATION GAP


Recall

R

R

DP

P

(

!

(

1

orP

R

dR

dP

P

R

R

P

R

R

P

PD

!

(

(!

(

(!

111

but2

1

)( gk

d

dk

dP

dR

dP

!!


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DURATION GAP


So

)()1(

)(

)1(

)( 12

1

2

1

gkd

k

gk

d

P

k

gk

d

D

!

!

gkkD

! 1

Example:

Stock with k=10%,g=5%

D = = 22 years1.100.05


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DURATION GAP


8. DD and Passbook savings Duration?

Must analyze runoff and turnover as well as rate

elasticity.


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How to manage Interest Rate Risk

1. Do nothing

2. Attempt to set GAPs to zero

3. Derivatives Forward contracts

Interest rate futures contracts (e.g. Eurodollar, TBill)

Option contracts (exchange-traded)

Exotic options (OTC)4. Interest rate swaps

Plain-vanilla

Exotic

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10-Asset Liability Management[1]

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