Duration Gap & Clumping

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Transcript Duration Gap & Clumping

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

ASSETS

Fixed rate (5%) 10 Y Mortgages

Dec. 31, 2008 € m LIABILITIES

100 Fixed Rate (3%) 2 y Notes Shareholders’ Equity

Total 100 Total € m

90 10

100

NII

2009 

II

2009 

IE

2009   5 %  100   3 %  90   5  2 .

7  2 .

3

ASSETS

Cash Fixed rate (5%) 10 Y Mortgages

Total Dec. 31, 2009 € m LIABILITIES

2.3

100 Fixed Rate (3%) 2 y Notes Shareholders’ Equity

102.3 Total € m

90 12.3

102.3

4

follows

ASSETS

Cash Fixed rate (5%) 10 Y Mortgages

Total Dec. 31, 2010 € m LIABILITIES

4.6

100 Fixed Rate (3%) 2 y Notes Shareholders’ Equity

104.6 Total

On the 1/1/2009 the ECB increase the interest rates of 100 bp

€ m

90 14.6

104.6

Nothing changes in the FS of the bank

ROE

2009  2 .

3  23 % 10 

ROE

2010  2 .

3 12 .

3  18 .

7 % 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%

NII

2011 

II

2011 

IE

2011   5 %  100   4 %  90   5  3 .

6  1 .

4

ROE

2011  1 .

4 14 .

6  9 .

59 % 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

MV Mortgage

  

t

9   1  1  5 6 %

t

  1  100 6 %  9  93 .

2

MV Note

    1 92 .

7  4 %   89 .

13

ASSETS

Cash Fixed rate (5%) 10 Y Mortgages

Total Dec. 31, 2009 € m LIBILITIES

2.3

93.2

Fixed Rate (3%) 2 y Notes Shareholders’ Equity

95.5 Total € m

89.13

6.37

95.5

P

/

L

 

MV A

 

MV L

 

MV SE P

/

L

2009   93 .

2  5  100   89 .

13  2 .

7  90    1 .

8  1 .

83   3 .

7 63

follows

Next Year (2010)

MV Mortgage

  

t

8   1  1  5 6 %

t

  1  100 6 %  8  93 .

79 Notes (maturity)  90

ASSETS

Cash Fixed rate (5%) 10 Y Mortgages

Total Dec. 31, 2010 € m LIBILITIES

4,6 93,79 Fixed Rate (3%) 2 y Notes Shareholders’ Equity

98,39 Total € m

90,00 8,39

98,39

P

/

L

2010   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 MVA

   1

D

A i A

  

i A

MVL MVL

   1

D L

i L

  

i L

MVA

 

MVA

  1

D A

i A

  

i A

 

MVA

MD A

 

i A

MVL

 

MVL

  1

D L

i L

  

i L

 

MVL

MV L

 

i L

MVE

 

MVA

 

MVL

  

MVA

MD A

 

i A

  

MVL

MD L

 

i L

 

MVE

  

MVA

MD A

MVL

MD L

  

i

MVE

  

MD A

LEV

MD L

  

i MVA

MVE

  

MD A

LEV

MD L

 

MVA

 

i

10

The Duration gap

MVE

  

MD A

LEV

MD L

 

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 MV A = MV L changes if MD A  MVE is not sensitive to interest rates = MD L .

• If MV A > MV L  MVE is not sensitive to interest rates changes if DG=0, i.e. MD A < MD L . 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 MVE

  

MD A

LEV

MD P

 

MVA

 

i MVE

12

The example again

D Mortgage

   Let’s go back to our bank

t

10   1

t

CF t i t MV

t

9   1

t

 5  1 .

05 

t

100  10   105 1 .

05  10 100  8 .

108

MD Mortgage

  

D Mortgage

     8 .

108 1 .

05  7 .

722

D Note

  

t t

  1

t

CF t

 

t MV

 1  2 .

7 1 .

90  2  1 92 .

.

03 7   2 90  1 .

971

MD Note

  

D Note

     1 .

971 1 .

03  1 .

914 13

follows

DG

 

MD A

LEV

MD L

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

1.

2.

3.

  D uration (and duration gap) changes every instant, when interest rate change, or simply because of the passage of time Immunization policies based on duration gap should be updated continuously D uration (and duration gap linear approximation ) is based on a Impact not estimated precisely The model assumes uniform interest rate changes ( rates  i ) of assets and liabilities interest 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

t

1

t

2

t

3

Coupon payments

time 17

Answer to problem 2: convexity

Rather than proxying % change in value with the first derivative only 

MV A MV A

dMV A di MV A

 

i

…we could add the second term in Taylor (or McLaurin) including second derivative: 

MV A MV A

dMV A di MV A

 

i

d

2

MV A di

2

MV A

 ( 

i

) 2 2

See following slides

18

Answer to problem 2: convexity

Second derivative of

VM

A to

i d

2

MV A di

2 

d di t N

  1 

t

CF t

  

t

 1 

t N

  1 

t

( 

t

 1 )

CF t

  

t

 2   1 1

i

2

t N

  1 (

t

2 

t

)

CF t

 

t

Dividing both terms by

MV

A :

d

2

MV A di

2

MV A

 1 1

i

2

t N

  1 (

t

2 

t

)

CF t

 

MV A t

Modified convexity MC

Convexity, C

19

Answer to problem 2: duration gap and convexity gap

Substituting duration and convexity in the second order expansion 

MV MV A A

 

MD A

 

i

MC A

 (  2

i

) 2 Multiplying both terms by

MV

A : 

MV A

 

MD A

 

i

MV A

MC A

 ( 

i

) 2 2 

MV A

Same for liabilities: 

MV L

 

MD L

 

i

MV L

MC L

 ( 

i

) 2 

MV L

2 The change in market value of the bank’s equity can now be better estimated: 

MV E

  

MD A

L

MD L

  

i

MV A

 

MC A

L

CM L

  ( 

i

) 2 2 

MV A

duration gap convexity gap

20

Duration gap and convexity gap: our example

MC A

MC mortgage

   1  1 5 %  2  

t

8   1 (

t

2 

t

) 5  1  5 % 

t

 ( 81  9 ) 105 100  1  5 %  9 100    69 .

79

CM L

CM CD

  1  1 3 %  2 92 .

7  1  3 %   0 .

97 90 

VM B

 

DG

 1 %  100   69 .

79  0 .

9  0 .

97   ( 1 %) 2 2  100   5 .

93

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: 

i A

 

A

 

i

i P

 

P

 

i

Then substitute in the change of the value of the bank 

MV B

        

MV MD

MV A A A

MV

  

MD MD

A A

A A

  

L

i

  

MV L A A

  

MD L

  

MV L

MV L L

 

L

  

MD L MD L

  

L

i L

   

i

 

MV A

 

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 size of the bank ( MV A ) 22

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

Liabilities

Checking accounts Fixed rate CDs Fixed rate bonds

Value

1000 600 800 1200

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 uniform change of interest rates for different maturities of The Clumping o cash-bucketing model different maturities  a model with independent changes of interest rates at  The model is built upon the repricing gap and the duration gap model were focused on the yield curve).

zero-coupon curve (both the  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 ( coupon rate  coupon paying bond with a price of 100.

r 2,5 ) zero For this maturity we only have a 4.5% (semi-annual)  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 ( r t )) 30

How to estimate zero coupon rates: bootstrapping

1. From prices of

zcb

we extract the corresponding

r

t

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% Ex.

r t

t

100

VM Zt

 1

r

2  2 100  1  4 .

26 % 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

0.5

2.21

2.25

1

2.16

8.55

2.25

1,5

2.12

2.25

2

2.07

2.5

Es.

( 1  2 .

25 4 .

26 31 %) 2

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 0

2.25

0.5

2.21

2.25

1

2.16

8.55

2.25

1.5

2.12

100

r

2 .

5 

r

102 .

25 ( 1 

r

) 2 .

5  91 .

45

102.25

2.25

2

2.07

2.5

= 91.45

r

2 .

5  2 .

5 102 .

25 91 .

45  1  4 .

57 % 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 dates, into a simplified portfolio, based on a limited number q (< p 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  M x N 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

Some simplifying cash-flow mapping techniques

 Analytical principal  Given M securities, “maps” each of them to the “principal” maturity node 

Modified analytical principal

Analytical duration  Given M securities, it maps each of them to its duration

method

 Synthetic principal  Given M securities, it only considers the maturity of principal (computes an average) Does not consider coupons reinvestment risk  Synthetic duration  Given M securities, it only considers the duration (computes an average) 35

An hybrid technique: modified principal

   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 10 7.5

5 2.5

0 0 2.5

5

Time to maturity

Coupon =0% Coupon =2% Coupon =5% Coupon =15% 7.5

10 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 (MD

i

)

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

0,5 0,5 0,75 1,25 1,75 2,25 Clumping: 1 0,75 1,25 1,75 2,25 1 2,5 2,75

dates nodes

2,75

dates

2,5

38

nodes

Clumping

  In the clumping model a large number of cash flows, maturing in curve.

p different dates are reduced to virtual cash flows on q q (with q

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 short term part of the zero coupon curve 39

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 the nodes have: into two virtual cash flows with maturities on n and n+1 (with n < t < n+1 ), we must  The same market value  The same modified duration 40

Mapping in practice

 We have two unknowns and two equations  

MV t

  1

NV t

r t

t

MV n

MV n

 1   1

MV n

r n

1

MV

r n n

  1  1

n

 1  

MD t

MD n MV n MV n

MV n

 1 

MD n

 1

MV n MV n

  1

MV n

 1 

MD n MV n MV t

MD n

 1

MV n

 1

MV t

      

NV n NV n

 1  

MV n MV n

 1  1   1

r n

 

n

r n

 1  

MV t

n

 1   

MD t MD

 

MV t n

 

MD n

 1

MD n

 1     1 

r n

n

 

MD n MD n

 

MD t MD n

 1     1 

r n

 1 

n

 1 41

 

An example

A cash flow with a nominal value of 50,000 € and maturity 3y and 3m.

Zero-coupon IR : 3.55%

Maturity

1 month 2 months 3 months 6 months 9 months 12 months 18 months 2 years

3 years Zero-Coupon Rate

2.80% 2.85% 2.90% 3.00% 3.10% 3.15% 3.25% 3.35%

3.50% 4 years

5 years 7 years

3.70%

3.80% 3.90% 4.00% 10 years 15 years 4.10% 30 years 4.25%

r

3 , 25 

r

3  (

r

4 

r

3 ) ( 3 .

25  3 ) ( 4  3 )  3 .

5 %  ( 3 .

7 %  3 .

5 %) 0 .

25 1  3 .

55 % 42

follows

Market Value and Modified Duration for the real cash flows 

MV t

  1

NV t

r t t

 

MD t

  1

D t

r t

50 , 000 1 .

0355 3 .

25 1 .

0355   3 .

25  44 , 640 .

82  3 .

139 Modified Duration for the two virtual cash flows 

MD n

  1

D n

r n

 

MD n

 1   1

D n

r

 1

n

 1 3 1 .

035   2 .

899 4 1 .

037   3 .

857 43

follows

Market value for the two virtual cash flows 

MV n

 

MV n

 1   44 , 640 .

82  44 , 640 .

82    3 .

139 2 .

899   2 .

899 2 .

899   3 .

857   3 .

857   3 .

139 3 .

857     33 , 464 .

45 11 , 176 .

37 Nominal value for the two virtual cash flows

NV n NV n

 1  

MV n

  1

MV n

 1    1

r n

 

n

r n

 1  37 , 102 .

63

n

 1  12 , 924 .

56 44

follows

Real Cash Flow 3Y Virtual Cash Flow 4Y Virtual Cash Flow

T 3.25

3.00

4.00

NV 50,000.00

37,102.63

12,924.56

MV

44,640.82

33,464.45

11,176.37

r

3.55%

3.50% 3.70% D

3.25

3 4

MD

3.139

2.899

3.857

 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. VM t e VM t+1 are chosen in such a way that: 2 

s

2   

VM VM t s

  2 

t

2   

VM t

 1

s

  

t

2  1  2

VM VM t s VM VM t

 1

s

t

2 ,

t

 1 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 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 from 15 to 20 years beyond 20 years Average maturity (D i ) 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 17.5 years 22.5 years Band 9 10 11 12 1 2 3 4 5 6 7 8 13 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, NP i 50

1 2 3 4 5 6 7 8 9 10 11 12 13 14

The Basel Committee Approach

Band Modified duration MD i = D i /(1+5%) 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 13.01 years Weighting factor MD i  y i (with  y i =2%) 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 % 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%  

NP i

 

NP i

MD i

 

y i

Total risk is computed as the sum of all these  NP i 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 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 yearly compounded 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 0.5 1 1.5 2 2.5 Price 98 96 94 92 99 58