Mafinrisk – 2010
Market Risk
The duration gap model and
clumping
Session 2
Andrea Sironi
Agenda
 Market value versus historical cost
accounting
 The duration gap model
 The Clumping Model
2
The Duration Gap Model



“Market Value” model  target variable =
market value of shareholders’ equity
Focus on impact of interest rate changes
on the market value of assets and
liabilities
Gap = difference between the change in
the market value of assets and the market
value of liabilities
3
Market Value vs Historical Value
Dec. 31, 2008
€m
ASSETS
LIABILITIES
Fixed rate (5%) 10 Y Mortgages
100 Fixed Rate (3%) 2 y Notes
Shareholders’ Equity
Total
100 Total
€m
90
10
100
NII2009  II2009  IE2009  5% 100  3%  90  5  2.7  2.3
Dec. 31, 2009
ASSETS
Cash
Fixed rate (5%) 10 Y Mortgages
Total
€m
LIABILITIES
2.3 Fixed Rate (3%) 2 y Notes
100 Shareholders’ Equity
102.3 Total
€m
90
12.3
102.3
4
follows
Dec. 31, 2010
€m
ASSETS
Cash
Fixed rate (5%) 10 Y Mortgages
Total
LIABILITIES
4.6 Fixed Rate (3%) 2 y Notes
100 Shareholders’ Equity
104.6 Total
€m
90
14.6
104.6
On the 1/1/2009 the ECB increase the interest rates of 100 bp
Nothing changes in the FS of the bank
ROE 2009 
2.3
 23% 
10
ROE 2010 
2.3
 18.7%
12.3
5
follows
In 2011 the bank has to finance the 10Y Mortgages with a new fixed
rate note issued at the new market rate: 4%
NII2011  II2011  IE2011  5% 100  4%  90  5  3.6  1.4
ROE 2011 
1.4
 9.59%
14.6
The effect of the increase of the interest rates on the profitability of
the bank appears only two years after the variation itself.
6
follows
This problem can be solved using in the FS the market value of A/L
instead of the historical value
9
MVMortgage 5%   
t 1
5
100

93.2
t
9
1  6% 1  6%
92.7
MVNote3%  
 89.13
1  4%
Dec. 31, 2009
€m
ASSETS
LIBILITIES
Cash
Fixed rate (5%) 10 Y Mortgages
2.3 Fixed Rate (3%) 2 y Notes
93.2 Shareholders’ Equity
Total
95.5 Total
€m
89.13
6.37
95.5
P / L  MVA  MVL  MVSE
P / L2009  93.2  5 100  89.13 2.7  90  1.8 1.83  3.63
7
follows
Next Year (2010)
8
5
100
MVMortgage 5%   

93.79
t
8
1  6%
t 1 1  6%
Notes (maturity)  90
Dec. 31, 2010
ASSETS
€m
Cash
Fixed rate (5%) 10 Y Mortgages
4,6 Fixed Rate (3%) 2 y Notes
93,79 Shareholders’ Equity
90,00
8,39
Total
98,39 Total
98,39
LIBILITIES
€m
P / L2010  5  93,79  93.2  90  2.7  89.13  5.59  3.57  2.02
8
Agenda
 Market value versus historical cost
accounting
 The duration gap model
 The Clumping Model
9
The Duration gap
The same result could be obtained using the duration gap
MVA
DA

 i A
1  iA 
MVA
MVA  MVA 
DA
 i  MVA  MDA  i A
1  iA  A
MVL
DL

 iL
1  iL 
MVL
MVL   MVL 
DL
 i   MVL  MVL  iL
1  iL  L
MVE  MVA  MVL   MVA  MDA  iA    MVL  MDL  iL 
MVE  MVA  MDA  MVL  MDL  i
MVE
 MD A  LEV  MD L   i
MVA
MVE  MDA  LEV  MDL  MVA  i
10
The Duration gap
MVE  MDA  LEV  MDL  MVA  i
The change in the market value of Shareholders’ Equity is a function of
three variables:
1.
The difference between the modified duration of assets and the
modified duration of the liabilities corrected for the bank’s leverage
(“leverage adjusted duration gap”)  duration gap (DG)
2.
The size of the intermediation activity of the bank measured by the
market value of total assets
3.
The size of the interest rates change
MVE   DG  MVA  i
11
Immunization
•
If MVA = MVL  MVE is not sensitive to interest rates
changes if MDA = MDL.
•
If MVA > MVL  MVE is not sensitive to interest rates
changes if DG=0, i.e. MDA < MDL. In this case the
higher sensitivity of liabilities will compensate the initial
lower market value and the change in the absolute value
of assets and liabilities will be equal.

MVE
MD A  LEV  MD P   MVA

 i
MVE
MVE
12
The example again
Let’s go back to our bank
CFt
5
105
t
t
10
10
9



1 i
1.05
1.05
DMortgage 5%    t 
 t 
 10 
 8.108
MV
100
100
t 1
t 1
MDMortgage 5%  
DMortgage 5% 
1  i 
8.108

 7.722
1.05
CFt
92.7
2.7
t
2
t



1 i
1.03
1.03
DNote 3%    t 
 1
 2
 1.971
MV
90
90
t 1
MDNote3%  
DNote3% 
1  i 
1.971

 1.914
1.03
13
follows
DG  MDA  LEV  MDL   7.722 0.901.914  6
MVE   DG  MVA  i  6 100 1%  6
For an interest rates increase of 100 bp the market
value of shareholders’ equity would decrease by 6 m€
(60% of the original value)
14
Some remarks
The result (-6) is different form what we got before
(-3.63) for three main reasons:

-6 m€ is an instantaneous decrease estimated at
the time of the int. rates change (January 1st
2004);

In the – 3.63 m€ we also have 2.3 m€ of interest
margin

The duration is just a first order approximation
15
Duration gap: problems and
limits
Duration (and duration gap) changes every
1.
instant, when interest rate change, or simply
because of the passage of time

Duration (and duration gap) is based on a
linear approximation
2.

3.
Immunization policies based on duration gap
should be updated continuously
Impact not estimated precisely
The model assumes uniform interest rate
changes (i) of assets and liabilities interest
rates
16
Problem 1:
duration changes
 Every time market
interest rate change,
duration needs to be
computed again wuth
new weights (PV of
cash flows)
 Even if rates do not
change, duration
decreases: linearly
with “jumps” related
to coupon payments
Duration
t1
t2
t3
time
Coupon payments
17
Answer to problem 2:
convexity
Rather than proxying % change in value with the first derivative only
MVA dMVA MVA

 i
MVA
di
…we could add the second term in Taylor
(or McLaurin) including second derivative:
MVA dMVA MVA
d 2 MVA MVA (i) 2

 i 

2
MVA
di
di
2
See following slides
18
Answer to problem 2:
convexity
Second derivative of VMA to i
N
d 2 MVA d N
t 1
t  2






t

CF
1

i


t
(

t

1
)
CF
1

i



t
t
2
di
di t 1
t 1
1

1  i 2
N
CFt
(t  t )

t


1

i
t 1
2
Dividing both terms by MVA :
CFt 1  i 
d 2 MVA MVA
1 N 2

(t  t )
2 
2
di
MVA
1  i  t 1
t
Convexity, C
Modified convexity MC
19
Answer to problem 2:
duration gap and convexity gap
Substituting duration and convexity in the second order expansion
MVA
(i) 2
 MDA  i  MCA 
MVA
2
Multiplying both terms by MVA:
(i) 2
MVA   MDA  i  MVA  MCA 
 MVA
2
Same for liabilities:
(i) 2
MVL  MDL  i  MVL  MCL 
 MVL
2
The change in market value of the bank’s equity can now be better estimated:
(i) 2
MVE  MDA  L  MDL   i  MVA  MCA  L  CM L  
 MVA
2
duration gap
convexity gap
20
Duration gap and convexity gap:
our example
MC A  MCmortgage 
t
9
 8 2
1
5 1  5% 
105 1  5%  
 (t  t )
  69.79

 (81 9)
2 

100
100
1  5%  t 1

CM L  CM CD 
1
92.7 1  3%
2
1

1
 0.97
2
90
1  3%


(1%)2
VM B   DG 1% 100 69.79  0.9  0.97 
100  5.93
2
First order
proxy, -6.23
convexity gap,
equal to 61.6
Very close to
the true change
(-5.94)
21
Answer to problem 3:
beta-duration gap
Similar to standardized repricing gap. For each asset (liability) estimate:
iA   A  i
iP   P  i
Then substitute in the change of the value of the bank
MVB  MVA  MVL 
  MVA  MDA  i A    MVL  MDL  iL  
 MVA  MDA   A  MVL  L  MDL   L   i 
 MDA   A  L  MDL   L   MVA  i
beta-duration gap
The impact of an interest rate change depends on 4 factors:
 average MD of assets and liabilities
 average sensitivity of assets and liabilities interest rates to the base rate
(beta)
 financial leverage L
22
 size of the bank (MVA)
Residual Problems

Assumption of a uniform change of assets and
liabilities’ interest rates.

Assumption of a uniform change of interest
rates for different maturities.

The model does not consider the effect of a
variation of interest rates on the volume of
financial assets and liabilities
23
Questions & Exercises
1. Which of the following does not represent a limitation of
the repricing gap model which is overcome by the
duration gap model?
A) Not taking into account the impact of interest rates
changes on the market value of non sensitive assets
and liabilities
B) Delay in recognizing the impact of interest rates
changes on the economic results of the bank
C) Not taking into account the impact on profit and loss
that will emerge after the gapping period
D) Not taking into account the consequences of interest
rate changes on current account deposits
24
Questions & Exercises
2.
A bank’s assets have a market value of 100
million euro and a modified duration of 5.5
years. Its liabilities have a market value of 94
million euro and a modified duration of 2.3
years. Calculate the bank’s duration gap and
estimate which would be the impact of a 75
basis points interest rate increase on the
bank’s equity (market value).
25
Questions & Exercises
3. Which of the following statements is NOT correct?
A. The convexity gap makes it possible to improve the
precision of an interest-rate risk measure based on
duration gap
B. The convexity gap is a second-order effect
C. The convexity gap is an adjustment needed because
the relationship between the interest rate and the
value of a bond portfolio is linear
D. The convexity gap is the second derivative of the
value function with respect to the interest rate,
divided by a constant which expresses the bond
portfolio’s current value.
26
Questions & Exercises
4. Using the data in the table below
i) compute the bank’s net equity value, duration gap and convexity gap;
ii) based on the duration gap only, estimate the impact of a 50 basis points
increase in the yield curve on the bank’s net value;
iii) based on both duration and convexity gap together, estimate the impact of
a 50 basis points increase in the yield curve on the bank’s net value;
iv) briefly comment the results
Assets
Open credit lines
Floating rate securities
Fixed rate loans
Fixed rate mortgages
Value
1000
600
800
1200
Liabilities
Checking accounts
Fixed rate CDs
Fixed rate bonds
Value
1200
600
1000
Modified
duration
0
0.25
3.00
8.50
Modified
duration
0
0.5
3
Modified
convexity
0
0.1
8.50
45
Modified
convexity
0
0.3
6.7
27
Agenda
 Market value versus historical cost
accounting
 The duration gap model
 The Clumping Model
28
A common problem and a
possible solution
 Repricing gap and duration gap  assumption of
uniform change of interest rates for different
maturities
 The Clumping o cash-bucketing model a model
with independent changes of interest rates at
different maturities
 The model is built upon the zero-coupon curve (both the
repricing gap and the duration gap model were focused
on the yield curve).
 The model works trough the mapping of single cash
flows on a predetermined number of nodes (or
maturities) on the term structure.
29
How to estimate zero coupon
rates: bootstrapping
 For longer maturities we typically have no zero
coupon bonds
 We need to extract them from coupon bonds
 One possibility is through bootstrapping
 Assume we want to estimate the 2.5 (r2,5) zerocoupon rate
 For this maturity we only have a 4.5% (semi-annual)
coupon paying bond with a price of 100.
 For the preceding maturities (t = 0.5; 1; 1.5; 2) we
have zero coupon bonds (from their prices we can get
their yield to maturity (rt))
30
How to estimate zero coupon
rates: bootstrapping
1. From prices of zcb we extract the corresponding rt
Zero Coupon Bond
6 months
1 year
18 months
24 months
Maturity
0.5
1
1.5
2
Price
98
96
94
92
Rate
4.12%
4.17%
4.21%
4.26%
rt  t
100
1
VM Zt
100
2
r

 1  4.26%
Ex. 2
92
2. We use these zero-coupon rates to estimate the present value of the first four
cash flows (coupons) of the 4.5% coupon paying bond
102.25
0
2.25
2.25
2.25
2.25
0.5
1
1,5
2
2.21
2.16
2.12
2.07
8.55
2.5
Es.
2.25
31 2
(1  4.26%)
How to estimate zero coupon
rates: bootstrapping
3. Find the rate that equates the present value of 102.5 to the residual value
of the bond which has not been explained by the PV of the four coupons
102.25
0
100
2.25
2.25
2.25
2.25
0.5
1
1.5
2
2.21
2.16
2.12
2.07
-
2.5
8.55
102.25
r2.5  r
 91.45
2.5
(1  r )
=
r2.5 
2.5
91.45
102.25
 1  4.57%
91.45
32
What is the mapping for?
 The mapping is a procedure to simplify the
representation of the financial position of the bank.
 Mapping is used to transform a portfolio with real
cash flows, associated to an excessive number of p
dates, into a simplified portfolio, based on a limited
number q (<p) of maturity nodes (standard dates).
 After mapping, it’s easier to implement effective risk
management policies
 Goal: reduce all the banks’ cash flows to a small
number of significant nodes (maturities).
33
Cash-flow mapping
 We can get an interest rate curve with different
rates for every individual maturity
 Do I really need to consider MxN nodes?
 No, cash-flow mapping allows to map a portfolio of
assets and liabilities (with a large number of cash
flows associated to a large number of maturities) to a
limited number of maturity nodes
 It represents a special case of mapping
 A methodology to map a portfolio to a limited number of risk
factors: e.g. international equity portfolio to S&P500, Dax
and MIB 30
34
 Analytical duration
 Given M securities, it
Modified
maps each of them to its
analytical
duration
principal
 Synthetic principal
 Given M securities, it
only considers the
maturity of principal
(computes an average)
Does not consider coupons
reinvestment risk
method
 Synthetic duration
 Given M securities, it
only considers the
duration (computes an
average)
Extremely
simplified
 Analytical principal
 Given M securities,
“maps” each of them
to the “principal”
maturity node
Requires M
nodes
Some simplifying cash-flow
mapping techniques
35
An hybrid technique:
modified principal
10
7.5
Modified duration
 Computing analytic duration
for each asset and liability
might be complex
 Using principal is not
precise as it does not
consider the coupons
 However, given the level of
interest rates (e.g. 5% in
the chart), there exists a
relationship between
principal and duration for
bonds with different coupon
level
5
2.5
Coupon =0%
Coupon =2%
Coupon =5%
Coupon =15%
0
0
2.5
5
7.5
10
Time to maturity
36
Modified principal
 To simplify the step from principal to duration
 consider only two cases e.g., < o > 3%)
 Divide principal values in few large maturity buckets
 Assign an average duration to each maturity (“modified
principal”)
Residual Life Bracket (i)
Coupon < 3%
Coupon  3%
Up to 1 month
1 - 3 months
3 - 6 months
6 - 12 months
Up to 1 month
1 - 3 months
3 - 6 months
6 -12 months
Average
modified
duration (MDi)
0.00
0.20
0.40
0.70
37
A more refined technique:
clumping
 The objective is the same:
link real cash flows to a
number q (<p) of “nodes”
 What changes? Rather than
compacting flows into a
single one at a unique
date, each cash flow gets
divided into more nodes
 How to map cash flows?
 Building a new security,
identical to the real cash
flow in terms of market
value and riskiness
0,75
0,5
1,25
1,75
2,25
1
2,75
dates
nodes
2,5
Clumping:
0,75
0,5
1,25
1
1,75
2,25
2,75
2,5
dates
nodes
38
Clumping
 In the clumping model a large number of cash flows,
maturing in p different dates are reduced to q (with q<p)
virtual cash flows on q different dates called “nodes” on the
curve.
 In order to choose the number and the position of the
nodes we have to remember that:
 Changes in short term interest rate are more frequent and larger
than changes in long term interest rates.
 The relationship between volatility and maturity of interest rates is
negative.
 Usually cash flows with short maturities are more frequent that cash
flows with long maturities
 It’s better to have a larger number of nodes on the short39
term part of the zero coupon curve
The nodes
 The choice of the node is also influenced by the
availability of hedging instruments: FRA, futures,
swaps, etc.
 When we divide a real cash flow with maturity in
date t into two virtual cash flows with maturities on
the nodes n and n+1 (with n<t<n+1), we must
have:
 The same market value
 The same modified duration
40
Mapping in practice
 We have two unknowns and two equations
NVt
MVn
MVn 1

MV


MV

MV


n
n 1
n
n 1
 t 1  r t




1

r
1

r
t
n
n 1


MVn
MVn1
MVn
MVn 1

MD

MD

MD

MD

MD
n
n 1
n
n 1
 t
MV

MV
MV

MV
MV
MVt
n
n 1
n
n 1
t




MDt  MDn 1  
n
n




NV

MV
1

r

MV
1

r
 n
n
n
n
 t



MD

MD
n
n 1 






MDn  MDt  
n 1
n 1




NV

MV
1

r

MV
1

r
 n 1
n 1
n 1
n 1
 t

MDn  MDn1 


41
An example
 A cash flow with a
nominal value of 50,000 €
and maturity 3y and 3m.
 Zero-coupon IR: 3.55%
r3, 25
Maturity
Zero-Coupon Rate
1 month
2.80%
2 months
2.85%
3 months
2.90%
6 months
3.00%
9 months
3.10%
12 months
3.15%
18 months
3.25%
2 years
3.35%
3 years
3.50%
4 years
3.70%
5 years
3.80%
7 years
3.90%
10 years
4.00%
15 years
4.10%
30 years
4.25%
(3.25  3)
0.25
 r3  (r4  r3 )
 3.5%  (3.7%  3.5%)
 3.55%
(4  3)
1
42
follows
Market Value and Modified Duration for the real cash flows
NVt
50,000

MVt  1  r t  1.03553.25  44,640.82

t

MD  Dt  3.25  3.139
t

1  rt  1.0355
Modified Duration for the two virtual cash flows
Dn
3

MDn  1  r   1.035  2.899

n

MD  Dn 1  4  3.857
n 1

1  rn1  1.037
43
follows
Market value for the two virtual cash flows

3.139 3.857

MVn  44,640.82 2.899 3.857  33,464.45


MV  44,640.82 2.899 3.139  11,176.37
n 1

2.899 3.857

Nominal value for the two virtual cash flows
n



NV

MV

1

r
 37,102.63
 n
n
n

n 1

 NVn1  MVn1  1  rn1   12,924.56
44
follows
T
NV
Real Cash Flow
3.25
50,000.00
3Y Virtual Cash Flow
3.00
4Y Virtual Cash Flow
4.00
MV
r
D
MD
44,640.82 3.55%
3.25
3.139
37,102.63
33,464.45 3.50%
3
2.899
12,924.56
11,176.37
4
3.857
3.70%
 The sum of the market values of the two virtual flows is
equal to the market value of the real cash flow.
 The market value of the 3Y cash flow is greater than the
MV of the 4Y cash flow. This happens because the real
flow maturity is nearer to 3 than to 4
45
Clumping on the basis of price
volatility
 Another form of clumping centers on the equivalence
between price volatility of the initial flow and the total price
volatility of the two new virtual positions
 This is calculated by taking into account also the
correlations between volatilities associated with price
changes for different maturities. VMt e VMt+1 are chosen in
such a way that:
 VM 
2
 VM

2
VM VM
t
t 1
t
t 1
  t2  
  t21  2
 s2  
 t2,t 1
VM s VM s
 VM s 
 VM s 
 Since this is a quadratic equation, we get two solutions for
  we need to assume that the original position and the
two new virtual positions have the same sign 
0  1
46
Clumping
 After the mapping of all the bank positions
on the nodes it’s possible to:
Evaluate the effect on the market value of the
shareholders’ equity of a change of the interest
rates for certain maturities
Implement interest risk management activities
Implement hedging activities
47
Residual Problems

Assumption of a uniform change of assets and
liabilities’ interest rates.

The model does not consider the effect of a
variation of interest rates on the volume of
financial assets and liabilities
48
The Basel Committee
Approach
 Banks are required to allocate their assets and
liabilities to 14 maturity buckets based on their
residual maturity
 For each bucket, they estimate the difference
between assets and liabilities (long and short
positions, i.e. net position)
 The net position is weighted by a coefficient that
proxies the potential change in value
 The product between the average modified duration
and a 2% change in the interest rate (parallel shift of
the yield curve)
49
The Basel Committee Approach
Time Band
Average
maturity
(Di)
Band
Revocable or sight
Up to 1 month
from 1 to 3 months
from3 to 6 months
from 6 months to 1 year
from 1 year to 2 years
from 2 to 3 years
from 3 to 4 years
from 4 to 5 years
from 5 to 7 years
from 7 to 10 years
from 10 to 15 years
0
0.5 month
2 months
4.5 months
9 months
1.5 years
2.5 years
3.5 years
4.5 years
6 years
8.5 years
12.5 years
1
2
3
4
5
6
7
8
9
10
11
12
from 15 to 20 years
17.5 years
13
beyond 20 years
22.5 years
14
• Banks are required to
allocate their assets
and liabilities to 14
different maturity
bands
• For each maturity
bucket, the net
position must be
calculated (difference
assets and liabilities)
• Net position, NPi
50
The Basel Committee Approach
Band
Modified
duration
MDi = Di
/(1+5%)
Weighting factor
MDi yi (with
yi=2%)
1
2
3
4
5
6
7
8
9
10
11
12
13
0
0.04 years
0.16 years
0.36 years
0.71 years
1.38 years
2.25 years
3.07 years
3.85 years
5.08 years
6.63 years
8.92 years
11.21 years
0.00 %
0.08 %
0.32 %
0.72 %
1.43 %
2.77 %
4.49 %
6.14 %
7.71 %
10.15 %
13.26 %
17.84 %
22.43 %
14
13.01 years
26.03 %
 The net position for each
maturity bucket is
weighted by a risk
coefficient espressing the
potential change in value
 Product between
average modified
duration and y = 2%
NPi   NPi  MDi  yi
 Total risk is computed as
the sum of all these NPi
51
The Basel Committee Approach:
pros
 It’s an economic value approach
 It does not only measure the impact of interest rate
changes on the bank’s income, but also on its equity value
 It considers the independence of interest rate curves for
different currencies:
 The risk indicator has to be computed separately for each
currency abosrbing at least 5% of the bank’s balance sheet
 It considers the link between risk and capital
 The sum of all the risk indicators (in absolute value) related
to the different currencies must be computed as a ratio to
the bank’s regulatory capital
52
The Basel Committee
Approach: cons
 It considers a unique interest rate volatility for both
short and long term rates, while the latter are
empirically less volatile because of a mean
reversion phenomenon
 It allows a full netting among the positions of
different time buckets, implicitly assuming parallel
shifts of the curve
These two drawbacks are overcome by the
generic risk indicator for debt securities in
the market risk capital requirement
53
framework (trading portfolio)
The Basel Committee
Approach: cons
 It’s an economic value approach, but it uses
as inputs the book values of assets and
liabilities
 It treats rather imprecisely
 Amortizing items
 Items with an uncertain rate repricing date
 Customer assets & liabilities with no
precise maturity (e.g. demand deposits)
54
Questions & Exercises
1.
A bank holds a zero-coupon T-Bill with a time to
matuity of 22 months and a face value of one million
euros. The bank wants to map this position to two
given nodes in its zero-rate curve, with a maturity of
18 and 24 months, respectively. The zero coupon
returns associated with those two maturities are 4.2%
and 4.5%.
Find the face values of the two virtual cash flows
associated with the two nodes, based on a clumping
technique that leaves both the market value and the
modified duration of the portfolio unchanged.
55
Questions & Exercises
2. Cash flow bucketing (clumping) for a bond involves …
A) …each individual bond cash flow gets transformed into
an equivalent cash flow with a maturity equal to that of
one of the knots;
B) … the different bond cash flows get converted into one
unique cash flow;
C) … only those cash flows with maturities equal to the
ones of the curve knots are kept while the ones with
different maturity get eliminated through compensation
(“cash-flow netting”);
D) …each individual bond cash flow gets transformed into
one or more equivalent cash flows which are associated
to one or more knots of the term structure.
56
Questions & Exercises
3. Bank X adopts a zero-coupon rate curve (term structure)
with nodes at one month, three months, six months,
one year, two years. The bank hold a security cashing a
coupon of 6 million euros in eight months and another
payment (coupon plus principal) of 106 million euros in
one year and eight months.
Using a clumping technique based on the
correspondence between present values and modified
durations, and assuming that the present term structure
is flat at 5% for all maturities between one month and
two years, indicate what flows the bank must assign to
the three-month, six-month, one-year and two-year
nodes.
57
Questions & Exercises
4. Based on the following market prices and using
the bootstrapping method, compute the yearlycompounded zero-coupon rate for a maturity of
2.5 years
Security
6-month T-bill, zero coupon
12-month T-bill, zero coupon
18-month T-bill, zero coupon
24-month T-bill, zero coupon
30-month T-bond with a 2% coupon every 6 months
Maturity
Price
0.5
1
1.5
2
2.5
98
96
94
92
99
58