10-asset liability management

8/13/2019 10-Asset Liability Management

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ASSET / LIABILITY

MANAGEMENT


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Asset / Liability Management

Also known as asset-liability management, gapmanagement

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 rates Banks 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 are 3 types of GAPs.

Dollar Gap, Funds Gap, Repricing Gap

Maturity Gap

Duration Gap


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GAP

GAPt= RSA tRSL t

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

060 6190 91120 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 = -105 + 1030

= - 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 day3 months

over 3 months6 monthsover 6 months1 year

over 1 year5 year

over 5 year


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GAP

Problems with GAP

1. Uses book-value approach: Focuses only on

income effect and not on capital gains effectfrom rate changes.

2. Aggregation: Ignores distribution of

assets/liabilities within buckets could still

have 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 r P P/ r< 0

100 + 0.10 100

1 + 0.10P = = =100

100 + 0.10 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 = + =100

110

1.10210

1.10

P = + =98.29

110

1.112

10

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

1.112

10

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 fall increases at a diminishing

rate as a function of matur i ty.

Maturity

P

1 2 3


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Maturity Gap

Now, these principles apply to

banks since they have portfoliosof interest-rate sensiti ve assets

and liabil i ties.


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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 assetj as 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

worthP= 99 / 1.11 = 89.19

Thus

Assets Liabilities

97.56 89.19

8.37

E = 97.5689.19E = A L =2.44(0.81)

E=1.63

E = 101.63 = 8.37


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Maturity Gap

Thus, equity must absorb interest-rate risk exposure.

Notice

MG = MAML= 31 = 2

By previous propositions

If MG > 0If r, then bank will LOSEIf r, then bank will GAIN

If MG < 0If r, then bank will GAINIf r, 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 r16% 12.07(4.66) =7.41

If r17% 15.47(5.38) =10.09 YES!

+ + 100[ 90] =1010

(1+x)210

1 + x110

(1+x)399

1 + x


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Maturity Gap

What if bank has matched with MG= 0, that is

invested in 1 yr bond, then

If r11% from 10%

A = 99.10100 =0.90

L = 89.1190 =0.89

If r12%

A = 98.21100 =1.79

L = 88.3990 =1.61

E =0.90 + 0.89 =0.01

E =1.79 + 1.61 =0.18


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Maturity Gap

Setting MG= 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 a Duration Gapmeasure 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

t PVtPVD11

Where N= last period of CF

CFt= cash flow at time t

PVt= CFt/ (1+R)tR= 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

tt

t

R

CP

1 )1(


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DURATION

N

tt

t

R

tC

dR

dP

R

P

11)1(

N

tt

t

R

C

R

D

R

P

1 )1(1

N

t

t

N

t

t PVtPVD11

Since andt

t

tR

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

PP =(4.99)(0.001/1.08) =0.000462 =0.0462%

If P=1000, price would fall to 999.538


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DURATION

Example (cont):

For semi-annual payments, the equation

must be modified:

R

RD

P

P

5.01

R

RD

P

P

1

Annual

Payment


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DURATION

Example:

2 yr 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:R

RD

P

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

RD

A

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

R

DL

L

L

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

i= MV of liability i / total MV ofassets


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Now, let

E = AL

= - (DAADLL) R / (1+R)

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

where DG = DADLL / 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 R 11%, what is effect on

net worth?


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Example (cont.) :

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

Thus, E : 10 million 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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Why? GAP in $ domain

DG in time domain

Want DG = 0 for fall protection, notice

DG = DADLL / A

= DADL(AK) / A where K = capital

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

DG = DADL+ DLk DG as k


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However, bank with more capital is better protected.

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

EE

= -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 MSo, K = 5, and k = 0.05

DG = DADL(1k) = 2.1

E = - 1.95

E : 5 3.05

E / E = - 2.151/0.050.01/1.10 = - 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

LoansFloat 400 0 LT CDs 150 2.5

LoansFixed 350 2 Equity 100

Total 1000 1000


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DURATION GAP

Example: (continue)

DA=0.15

0.5+0.1

3.5+0.4

0+0.35

2=1.125 year

DL= 0+ 0.4+ 2.5=0.572 year

DG= DADL = 1.1250.572 0.9 = 0.6102

400900

350900

150900

LA


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DURATION

Example: (continue)

If R= 0.08 0.08 0.09

E

A =DG

R

1 +R

E

A=0.6102 =0.00565

0.01

1 + 0.08

E =0.00565 1000 =5.65

E from 10094.35


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DURATION GAP

Although Duration Gap takes timing

of cash flows into account, there are

problems with its implementation and

use.


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DURATION GAP

Problems with DG

1. Not easy to manipulate DAand DL . (reason for

using artificial hedges such as swaps, options, orfutures)

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, DGoverpredicts Pdecrease


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DURATION GAP

If R< 0, DGunderpredicts Pincrease

P

P

RR

1

Actual Model

R


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DURATION GAP

Problems with DG (Continue)

Convexity =

It can be measured.

Convexity is good for banks. They do better

as a result.

measure of curvature

of duration curve


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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

RD

P

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 savingsDuration?

Must analyze runoff and turnover as well as rateelasticity.


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TYPES OF RISK FOR BANKS

1. Market risk

Equity price

Interest rate

2. Liquidity risk

3. Credit risk (default)

Use of credit derivatives

4. Operation risk Technology

Processing

Legal

Regulatory


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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

10-asset liability management

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