Abdulla Alothman 1

Asset Valuation

A Unified Approach

Part1 Valuation of Assets With Deterministic Payoffs

The Time Value of Money

The Theory

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• Zero Coupon Bonds (The Building Blocks)

• The Term Structure of Interest Rates

• Bonds

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VALUATION OF ZERO COUPON BONDS

Zero Coupon Certificates are IOU’s that make a single payment of $1.00 to the holder at maturity.

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Objective: To find the “fair market value” of Z (t,r;T)

T = tnt

( ; )n

r t t

( , ; ) ????n

Z r t t =

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Step 0: Collect necessary market dataThe Term Structure of Interest Rates at time t is observed and noted. We will denote this by:

1 2 3( ) { ( ; ), ( ; ), ( ; ) ( ; )}

nr t r t t r t t r t t r t tº ¼

( ; )r t T

T

ti r(t; ti)

t1 5%t2 6%t3 6.5%t4 6.8%

T = t5 7.3%

t

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Step1: Simplify the task ,price a bond with only one period to maturity first..

T = t1t

1( ; )r t t

1( , ; )Z t r t =

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Step1 cont…

T= t1 t

11

( , ; ) $1.00/ (1 )Z t r t r= +

1( ; )r t t

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1(0,0.05;1) $1.00/ (1 0.05) 0.9524Z = + =

T=t1 t=0

Interest = 5% per period*

Example: Find the fair market price of Z(0, 0.05; 1)

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Step 2: Pricing a Zero Coupon Bond with 2 periods to maturity

T = t2t

2( ; )r t t

2( , ; ) ????Z r t t =

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The idea:

2( , ; ) ????Z r t t =

T = t2t t1

2$1.00/ (1 ( ; ))A r t t= +

A

A

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The idea cont..…

22 2 2

( , ; ) / (1 ( ; )) $1.00/ (1 ( ; ))Z t r t A r t t r t t= + = +

t T = t1

2$1.00/ (1 ( ; ))A r t t= +

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2(0,0.06;2) $1.00/ (1 0.06) 0.8900Z = + =

T=2 t=0

Interest = 6% per period*

Example: Find the fair market price of Z(0, 0.06; 2)

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Final Step: Pricing a Zero Coupon Bond with n periods to maturity

T = tnt

( ; )n

r t t

( , ; ) ????n

Z r t t =

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Key Idea: Pretend you’re at time tn and work backwards to tn-1, then tn-2, tn etc, to get…

( , ; ) $1.00/ (1 )nn

Z t r t r= +

T = tnt t2

We now have a general formula to price Z’s of any maturity.

tn-1t(n-2)t1 ……….

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Example: Find the fair market price of Z(4, 0.08; 8)

T=8t=4

Interest = 8% per period*

4(4,0.08,8) $1.00/ (1 0.08) 0.7350Z = + =

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Example: Let’s price the rest………..

ti Z(t; ti) r(t, ti)

t1 0.95248 5%

t2 0.88996 6%

t3 0.82785 6.5%

t4 0.7686 6.8%

T = t5 0.7031 7.3%

( ; )r t T

T

( ; )Z t T

T

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Summary:1 2 3

( ) { ( ; ), ( ; ), ( ; ) ( ; )}n

r t r t t r t t r t t r t tº ¼

1 2 3( ) { ( ; ), ( ; ), ( ; ) ( ; )}

nZ t Z t t Z t t Z t t Z t tº ¼

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Working Backwards:1 2 3

( ) { ( ; ), ( ; ), ( ; ) ( , )}n

r t r t t r t t r t t r t tº ¼

1 2 3( ) { ( ; ), ( ; ), ( ; ) ( , )}

nZ t Z t t Z t t Z t t Z t tº ¼

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Extracting the term structure from market prices:

1 1 1 12 1/ 2

2 2 2 23 1/ 3

3 3 3 3

1/

1/ (1 ) (1/ ) 1

1/ (1 ) (1/ ) 1

1/ (1 ) (1/ ) 1

1/ (1 ) (1/ ) 1n nn n n n

Z r r Z

Z r r Z

Z r r Z

Z r r Z

= + Þ = -

= + Þ = -

= + Þ = -

= + Þ = -

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1 1 12

2 2 23

3 3 3

0.8689 1/ (1 ) 15.09%

0.7890 1/ (1 ) 12.58%

0.7064 1/ (1 ) 12.28%

Z r r

Z r r

Z r r

= = + Þ =

= = + Þ =

= = + Þ =

Example:

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VALUATION OF BONDS

Are IOU’s making a stream of payments to the holder over time.

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Example:

……….. t1 t2 T = tn t3 t4t

( ; )t

B B t Tº

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Note: Zero Coupons are Special Bonds:

………..

t1 t2 T= tn t3 t4t

Z(t ; T)

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………..

t1 t2 T =tn t3 t4

C CCCC C

t

$1000.00

Objective: To find the “fair market value” of B(t,r,C;T)

( , ; ) ???B t C T

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Step 0: Collect necessary market dataThe Term Structure of Interest Rates at time t, is observed and noted. We denote this by:

1 2 3( ) { ( ; ), ( ; ), ( ; ) ( ; )}

nr t r t t r t t r t t r t tº ¼

( ; )r t T

T

ti r(t; ti)

t1 5%t2 6%t3 6.5%t4 6.8%

T = t5 7.3%

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Step 1: Value the Zero’s

ti r(t, ti) Z(t; ti)

t1 5% 0.95248

t2 6% 0.88996

t3 6.5% 0.82785

t4 6.8% 0.7686

T = t5 7.3% 0.7031

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Step 2: Observe that a Bond is just a portfolio of Zero Coupon Bonds

Item Price Number of Items

Cost

Z1 C C*Z1Z2 C C*Z2

Z3 C C*Z3

………… ……….. ………….Zn 1000+C (C+1000)*Zn

Summing up the values in the third column therefore then gives…..

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1

( , ; ) 1000i n

i ni

B t C n cZ Z=

=

= +å

The Bond’s fair market value

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ti Z(t; ti) B(t,C; ti)

t1 0.9524 C*0.9524+952.48

t2 0.8900 C*(0.9524+0.8900)+889.00= 1.84C+889.96

t3 0.8279 C*(0.9524+0.8900+0.8279) + 827.90 = 2.67C+827.90

t4 0.7686 C*(0.9524+0.8900+0.8279+0.7686) + 768.6 = 3.44C+768.6

T = t5 0.7031 (0.95428+0.8900+0.8279+0.7686 +0.7031)*C+703.1 = 4.14C+703.1

Example:

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3

4

( ,4; ) 4* 2.67 827.85 838.53

( ,5; ) 5* 3.4403 768.6 785.80

B t t

B t t

= + =

= + =

Abdulla Alothman 33

1

1

( , , ; ) 1000 )

F(t,y,t ) 1/ (1 ) 1000/ (1 )

k

k i ki

ki i

ki

P B t C t C Z Z

P C y y

=

=

= = +

= - + + + +

å

å

Given a family of Bonds:

r

Define a function:

F(t,y,t )k

* ( , ; )y y t cT=

*y

Yield to Maturity . Also known as Bond’s IRR

Yield to Maturity (IRR)

EXTRACTING THE PRICE OF ZERO’S FROM TRADED BONDS

The Bootstrapping Technique

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1 1 1

2 2 1 2 2

1 2

1

1 2

( ; ) 1000

( ; ) ( 1000)

( ; ) ( 1000)

Solve for Z in equation 1.

Substitute for Z in equation 2, and solve for Z

Subsitut

n n n n n

B t t C Z

B t t C Z C Z

B t t C Z C Z C Z

= +

= + +

= + + + +

Given Market Prices:

Step 1:

Step 2:

Step n:

L L

K

1 2 1e for Z ,Z ....Z in equation n,and solve for Z

n n-

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( ) { ( ; ) : 0 }i

Z t Z t t t T= £ £( ) { ( ; ) : 0 }i

r t r t t t T= £ £

( ) { ( , ; ) : 0 , 0 }i i

B t B t C t t T C= £ £ £

Summary

( ) { ( , ; ) : 0 , 0 }i i

y t y t C t t T C= £ £ £

Term Structure Zero Prices

Yield Curves

Bond Prices

Applications

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• Annuities

• Perpetuities

• Amortizing Loans

• Forwards

• Swaps

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Flat Term Structure Assumption (Slides 39 – 55) :

( ; )r t T

T

r

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An Annuity:

• Is just a Bond with a face value of Zero

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………..

Example:

t1 t2 T =tn t3 t4

( , , ; )t

A A t C Tº r

t

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……….. t1 t2 T =tn t3 t4

Pricing an Annuity:

( ; ) ( , , ; )A t T A t r C Tº

t

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Step 0: Collect necessary market dataThe Term Structure of Interest Rates at time t, is observed and noted. We will denote this by:( ) { , , }r t r r r rº ¼

( ; )r t T

T

ti r(t, ti)

t1 r%t2 r%t3 r%t4 r%

T = t5 r%

Abdulla Alothman 44

Step 2: An Annuity is just a portfolio of Zero Bonds

Item Price Number of Items

Cost

Z1 C C*Z1Z2 C C*Z2

Z3 C C*Z3

………… ……….. ………….Zn C C*Zn

Summing up the values in the third column then gives…..

Abdulla Alothman 45

1

( , ; ) (1 (1 ) )i n

ni

i

cA t C t n c Z r

r

=-

=

+ = = - +åAn Annuity's fair market value

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Example:

……….. t2 T =t5t3 t4

100 100100100 100

5100(0;5) (1 (1.05) ) 432.95

0.05A -= - =

t t1

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A PERPETUITY:

• Is simply an ANNUITY with an infinite coupon stream

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………..

t1 t2 T = t3 t4

Pricing an Perpetuity P(t,C,r;):

( ; ) ( , , ; )P t T P t r C Tº

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Step 0: Collect necessary market data ……

• The Term Structure of Interest Rates at time t, is observed and noted. We will denote this by:

( ; )r t T

¥

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Step 2: Item Price Number of

ItemsCost

Z1 C C*Z1

Z2 C C*Z2

Z3 C C*Z3

………… ……….. ………….

Zn C C*Zn

…………. ………… ……………

Summing up the values in the third column then gives…..

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The Perpetuity’s fair market value!!

1

( , ; ) (1 (1 ) )i

ii

C CA t C t n c Z r

r r

=¥- ¥

=

+ = = - + =å

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………..

t+1 t+2 T = t+3 t+4

1

50( ,0.06,50) 50 1/ (1.06) 833.33

0.06

ii

i

P t=¥

=

= = =å

t

Example: Find the fair market price of P(t, 0.06,50)

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t1 t2 T =tn t3 t4

Example: Calculating Loan Payments:

1

1 1

So:

/ 1/ (1 ) * / (1 (1 ) )

i n

ii

i n i ni n

ii i

LOAN PMT Z

PMT LOAN Z LOAN r LOAN r r

=

=

= =-

= =

=

= = + = - +

å

å å

……………….

Interest = r% per period*

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t1 t2 T =t5 t3 t4

Example :

51000* 0.05/ (1 (1 0.05) ) 230.97PMT -= - + =

……………….

Interest = 5% per period*

$1000.00

t

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LOAN AMORTIZATION (Optional):

This is just the “Depreciation” of a Loan and has nothing to do with valuation, but rather, with how bonds are accounted for by companies.

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LOAN PMT INTEREST 5% PRINCIPLEPAY DOWN

BALANCE

$1000.00 -$230.9 -$50.00 -$180.90 $819.10

$819.00 -$230.9 -$40.95 -$189.95 $629.05

629.05 -$230.9 -$31.45 -$199.44 $429.71

429.71 -$230.9 -$21.455 -$209.45 $220.26

220.26 -$230.9 -$11.13 -$219.89 $0.00

LOAN AMORTIZATION TABLE

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FORWARDS:

Are contracts which allow the (Buyer / Seller ) to lock in today (time t) the Future (Buying / Selling ) price of an asset at some future date t+i.

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Assumption: We take the following Market Data as an given…. The Term Structure of Interest Rates at time t:

1 2 3( ) { ( ; ), ( ; ), ( ; ) ( ; )}

nr t r t t r t t r t t r t tº ¼

( ; )r t T

T

ti r(t, ti)

t1 5%t2 6%t3 6.5%t4 6.8%

T = t5 7.3%

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Motivating question one:How much do you need to invest at time ti to receive $1.00 at time ti+1?

t T ti ti+1

$1.00

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Analysis…At time t construct the following portfolio:

1. Buy one Zero maturing at time ti+1 . Cost: Zi+1 .

2. Finance your purchase by selling (going short) n =Zi+1/Zi Zero’s maturing at time ti . Revenue: n*Zi= Zi+1 .

,

t = 0 ... ti ti+1

-Zi+1 ... 0.00 $1.00

+n*Zi=Zi+1 ... -n*$1.00(Zero’s mature)

0.00

Net = $0.00 -n=-Zi+1/Zi $1.00

Transaction Cash Flows

Abdulla Alothman 61

t T

Solution:

ti ti+1

How much do you need to invest at time ti to get $1.00 at time ti+1

Answer =Zi+1/Zi

$1.00

Abdulla Alothman 62

Motivating question two:What is the effective one period future interest rate f(t; ti; ti+1)* you’ve locked in for your time ti investment?

1*Definition ( ; ; ) :

i if t t t

+The forward interest rate locked in at time t for an investment starting time ti and maturing at time ti+1

ti+1ti

1( ; ; )

i if t t t

+

1/

i iZ Z

+

$1.00

Abdulla Alothman 63

Analysis:1 1 1

1

11 1

11

1

1 ( ; ; ) $1.00/ ( / ) /

But:

$1.00/ (1 ) , 1

Substituting for , gives:

1+ ( ; , ) (1 ) / (1 )

So:

(1 )( ; , ) 1

(1 )

i i i i i i

ll l

i i

i ii i i i

ii

i i ii

f t t t Z Z Z Z

Z r l i i

Z Z

f t t t r r

rf t t t

r

+ + +

+

++ +

++

+

+ = =

= + = +

= + +

+= -

+

Abdulla Alothman 64

Solution:

t(i+1)ti

1/

i iZ Z

+

$1.001

11

(1 )( ; ; ) 1

(1 )

ii

i i ii

rf t t t

r

++

+

+= -

+

Abdulla Alothman 65

The picture:

……….. t1 t2 T = tn t+i ti+1t

$1.00

11

(1 ( ; ))ii

r t t ++

+

(1 ( ; ))i

ir t t+ 1(1 ( ; ; ))

i if t t t

++´

Abdulla Alothman 66

Generalizations: Using the same arguments as in the previous slides, will, mutatis mutandis, yield the following….

11

11

22 2

22

33 3

33

(1 ( ; ))(1 ( ; ; ))

(1 ( ; ))(1 ( ; ))

(1 ( ; ; ))(1 ( ; ))

(1 ( ; ))(1 ( ; ; ))

(1 ( ; ))

(1 ( ; ))(1 ( ; ; )) ( )

iii

i i ii i

iii

i i ii i

iii

i i ii i

ik i ki

i i ki k

r t tZf t t t

Z r t tr t tZ

f t t tZ r t t

r t tZf t t t

Z r t t

r t tZf t t t

Z

++

++

++

++

++

++

++

+

++ = =

++

+ = =+

++ = =

+

++ = =

K

(1 ( ; ))

k

ii

r t t

+

+

Abdulla Alothman 67

Summary:The forward rate f(t ; ti; ti+1) is the one period rate that would have to prevail at time ti – unknown at time t – for you to be indifferent between the following two investments:

• Investing at time t, for i+1 periods at the rate of r(t ; ti+1) per period.

• Investing at time t for i periods at the rate of r(t,ti) per period , then, when your deposit matures, rolling the investment over for one more period at the , currently unknown, rate of r(ti ; ti+1).

Abdulla Alothman 68

Example:1 2 3

( ) { ( ; ), ( ; ), ( ; ) ( , )}n

r t r t t r t t r t t r t tº ¼

( ; )r t T

T

ti r(t, ti)

t1 5%t2 6%t3 6.5%t4 6.8%

T = t5 7.3%

Abdulla Alothman 69

11/ 1 1/ 11

11

21/ 2 1/ 22

22

31/ 3 1/ 33

33

4

(1 ( ; ))1 ( ; ; ) ( ) { }

(1 ( ; ))(1 ( ; ))

1 ( ; ; ) ( ) { }(1 ( ; ))

(1 ( ; ))1 ( ; ; ) ( ) { }

(1 ( ; ))

1 ( ; ; ) (

iii

i i ii i

iii

i i ii i

iii

i i ii i

ii i

i

r t tZf t t t

Z r t tr t tZ

f t t tZ r t t

r t tZf t t t

Z r t tZ

f t t tZ

++

++

++

++

++

++

+

++ = =

++

+ = =+

++ = =

+

+ =4

1/ 4 1/ 44

45

1/ 5 1/ 555

5

(1 ( ; ))) { }

(1 ( ; ))(1 ( ; ))

1 ( ; ; ) ( ) { }(1 ( ; ))

ii

ii

iii

i i ii i

r t t

r t tr t tZ

f t t tZ r t t

++

++

++

+

+=

++

+ = =+

Example Cont...:

Abdulla Alothman 70

Example cont: The time t +1 forward term structure

1 2 1 3 1 4 1 5( ; 1) { ( ; ; ), ( ; ; ), ( ; ; ), ( ; ; )}f t t f t t t f t t t f t t t f t t t+ º

( ; 1; )f t t T+

5t

t1 f(t, t+1;t+2))

t2 7.025%%t3 7.2634%t4 7.411%t5 7.884%t6 na

1t

Abdulla Alothman 71

Example cont.. The time t +2 forward term structure

2 3 2 4 2 5( ; 2) { ( ; ; ), ( ; ; ), ( ; ; )}f t t f t t t f t t t f t t t+ º

( ; 1; )f t t T+

5t

t2 f(t, t+1;t+2))

t3 7.50%%t4 7.61%t5 8.172%t6 nat7 na

2t

Abdulla Alothman 72

Exercises:1. Verify the time t+1 forward term structure…2. Verify the time t+2 forward term structure3. A client wishes to buy 10 ounces of gold from you

at time t2. Market data at time t are given below:

t t2

Price of Gold (ounce)

$850.00 ?

Cost of Insurance* 5% of Value 5% of Value

(Z1 ,Z2,Z3) (0.95,0.90,0.85) Na

Abdulla Alothman 73

Exercices cont…..

The client is concerned that the price at time t= 2 might be higher than today and wishes to lock in a rate today.

What rate would you quote him and why ?

Abdulla Alothman 74

AnswersQ.3

Action t t2

Buy 10 Gold: -8500.00

Insurance Premium 1:

- 850.00

Loan 9350.00

Payback2 Loan+Interest

-10,388.88

Client Payment 10,388.88

Cash Flow 0.00 0.00

2

1)Insurance costs 2* 0.05* 8500

2)Total to reapay:

8500*(1+2*0.05)/ Z

Abdulla Alothman 75

Exercices cont…..

4. The Current Euro / USD exchange rate is 1.00 /2.00 .

A USD Z1 is selling for 0.95.

A Euro Z1 bond is selling for 0.90.

The client wants to buy from you 100 Euros at time t+1. He would like however, to lock in the rate today. What rate should you quote him?

Abdulla Alothman 76

SWAPS:• Are agreements between two parties, party A and

Party B to exchange a series of future cash flows….

PARTY A PARTY Bt

PARTY A

PARTY A

PARTY B

PARTY B

t+1

t+n

……

. ……

.

*In this course we are only going to looking at a basic interest rate swaps

Abdulla Alothman 77

The Basic Interest Rate Swap

• Are agreements between two parties, party A and Party B to exchange a series of future cash flows….

PARTY A PARTY B

PARTY A

PARTY A

PARTY B

PARTY B

……

.

……

.

1t

2t

nt

1( ; )r t t

1 2( ; )r t t

C

C

C1

( ; )n n

r t t-

1( ; ) :Is the one period rate at time

C : Is some fixed constant to be determinedi i i

r t t t-

Abdulla Alothman 78

Exercise: Given the Market data below, find the value C which makes the both cash streams have identical t0 values.

ti r(t, ti)

t1 5%t2 6%t3 6.5%t4 6.8% t5 7.3%

( ; )r t T

T

Abdulla Alothman 79

Step 1: Price the Zeros ti r(t, ti) Z(t; ti)

t1 5% 0.95248

t2 6% 0.88996

t3 6.5% 0.82785

t4 6.8% 0.7686

T = t5 7.3% 0.7031

Abdulla Alothman 80

Solution: Think about how to generate the random stream....

100 100 100 100100

1100 ( ; )r t t+

1 2100 ( ; )r t t+

2 3100 ( ; )r t t+

3 4100 ( ; )r t t+

4 5100 ( ; )r t t+

100

5100Z

1 1 2 2 3 3 4 4 5{ ( ; ), ( ; ), ( ; ), ( ; ), ( ; )}r t t r t t r t t r t t r t t

5

Solution:

1 0100 0t

V Z= -

Abdulla Alothman 81

Think about how to generate the fixed stream....

100

C

5

1

Solution:

t ii

B C Z=

= å

C C C C

Abdulla Alothman 82

Set both streams equal and solve for C...

5

5

51

5

5

4

Solving for C give

1

s

00

:

C

100

=100(1 ) / 7.

,

1

; )

8

(

%

i

ti

i

i

C Z B t C

Z

Zt V

Z

=

=

= = = -

- =å

å

Abdulla Alothman 83

END OF PART 1