10-Asset Liability Management
Transcript of 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
If rates increase to 11%
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
If rates increase to 11%
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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DURATION GAP (Cont.)
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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DURATION GAP (Cont.)
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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DURATION GAP (Cont.)
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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DURATION GAP (Cont.)
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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DURATION GAP (Cont.)
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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DURATION GAP (Cont.)
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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DURATION GAP (Cont.)
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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DURATION GAP (Cont.)
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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DURATION GAP (Cont.)
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
Problems with DG (Continue)
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
Problems with DG (Continue)
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
Problems with DG (Continue)
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
Problems with DG (Continue)
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