QF1100 Week 3

▶ Principal of Equivalent

▶ Internal Rate of Return (IRR)

▶ Bond Yields

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Future value vs Present value

Present Value of a cash flow stream = Present payment amount

Consider the entire stream as being replaced by a single flow at the

initial time:

In fact,
F V =x0 (1 + r)n + x1 (1 + r)n−1 + · · · + xn
 
n x1 x2 xn
=(1 + r) × x0 + + + ··· +
1 + r (1 + r)2 (1 + r)n
=(1 + r)n × P V.
Therefore,
FV
PV = .
(1 + r)n
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Example

Given a cash flow stream C = (−30, 20, −40, 50) and an effective
annual rate of 3%.
Find the future value and the present value of C.

Solution.
The future value is

F V =(−30) × (1 + 3%)3 + 20 × (1 + 3%)2 + (−40) × (1 + 3%) + 50

= − 2.763

and the present value is

20 −40 50
P V = (−30) + + + = −2.529.
1 + 3% (1 + 3%)2 (1 + 3%)3

Remark.
F V = P V × (1 + 3%)3 .

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Example: Automobile purchase
▶ Car A costs $20,000, Maintenance cost of $1,000 per year, a
useful mileage life 4 years;
▶ Car B costs $30,000, Maintenance cost of $2,000 per year, a
useful mileage life 6 years;
Assumption: No salvage value, interest rate is 10%, a planning period
of 12 years.

Question: Which car should you buy?

Strategy: Select the car that has the lower present value.

Cash flow steams:

CA = (2 × 104 , 103 , 103 , 103 , 2 × 104 , 103 , 103 , 103 , 2 × 104 , 103 , 103 , 103 , 0)

CB = (3 × 104 , 2 × 103 , 2 × 103 , 2 × 103 , 2 × 103 , 2 × 103 ,

3 × 104 , 2 × 103 , 2 × 103 , 2 × 103 , 2 × 103 , 2 × 103 , 0)
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Solution
Recall that
n n
X xm − xn+1 X 1 − xn+1
xk = and xk = .
1−x 1−x
k=m k=0

The present value of CA is

3
4 3
X 1
P VA =2 × 10 + 10 ×
(1 + 10%)k
k=1
3
1 X 1
+ 2 × 104 × 4
+ 103 ×
(1 + 10%) (1 + 10%)4+k
k=1
3
1 X 1
+ 2 × 104 × 8
+ 103 ×
(1 + 10%) (1 + 10%)8+k
k=1
" 3
#  
4 3
X 1 1 1
= 2 × 10 + 10 × × 1+ +
(1 + 10%)k (1 + 10%)4 (1 + 10%)8
k=1
=$48, 336.
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Solution

The present value of CB is

5
X 1
P VB =3 × 104 + 2 × 103 ×
(1 + 10%)k
k=1
5
1 X 1
+ 3 × 104 × 6
+ 2 × 103
×
(1 + 10%) (1 + 10%)6+k
k=1
" 5
#  
4 3
X 1 1
= 3 × 10 + 2 × 10 × × 1+
(1 + 10%)k (1 + 10%)6
k=1
=$58, 795.

Therefore, car A should be selected because its cost has the lower
present value.

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Frequent and Continuous Compounding
Let r(p) be a nominal interest rate and consider a cash flow stream
(x0 , x1 , . . . , xn ), a total of n periods (each period is of length p1 ).
Then
n n
X r(p) n−k X xk
FV = xk (1 + ) and PV = r (p) k
.
p (1 +
k=0 k=0 p )

Let r(∞) be compounded continuously. The cash flow occur at times

t0 , t1 , . . . , tn and denote the cash flow at time tk by x(tk ).

We have
n n
X (∞) X (∞)
(tn −tk )
FV = x(tk )er and PV = x(tk )e−r tk
.
k=0 k=0

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Principal of Equivalent
Exercise. Let r = 10%. If we change the stream (1, 0, 0) into the
stream (0, 0, 1.21), will we get the different payments after 2 years?
Hint: 1.12 = 1.21.

Two cash flows streams (x0 , x1 , . . . , xn ) and (y0 , y1 , . . . , yn ) are

equivalent streams if they have the same present value.

Example. The streams (1, 0, 0) and (0, 0, 1.21) are equivalent.

Given a cash flow (x0 , x1 , . . . , xn ) associated with an investment (a

series of deposits and withdrawals), the equation
n
X xk
PV = =0
(1 + r)k
k=0

is known as the equation of value. That is, the streams

(x0 , x1 , . . . , xn ) and (0, 0, . . . , 0) are equivalent.
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Internal Rate of Return
Any non-negative root, r of the equation of value is called the yield,
or internal rate of return (IRR), of the cash flow stream
(x0 , x1 , . . . , xn ).
Remark.
▶ The internal rate of return is defined without reference to a
prevailing interest rate and determined entirely by the cash flows
of the stream.
▶ This is the rate that an ideal bank would have to apply to
generate the given stream from an initial balance of zero.
▶ If let c = 1+r 1
, the equation of value is a polynomial equation in c
of degree n, which has at most n roots.
▶ The essence of investment is selection from a number of
alternative cash flow streams.

Example. Find all the IRRs of an investment project with cash flow
stream (−100, 0, 50, 0, 150).

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Solution
Solution. Assume r to be the IRR. We have
50 150
P V = −100 + + = 0.
(1 + r)2 (1 + r)4
1 1
Let c = 1+r (or (1+r)2 ). Then c satisfies a polynomial equation

−100 + 50c2 + 150c4 = 0.

Solve for c and we get

r r
√ 2 2
c = ± −1, c=− , and c = .
3 3
q
3
Then r = 1/c − 1 = 2 − 1 = 22.47%.

Remark that we only choose the non-negative roots r i.e.,

1
r ≥ 0 ⇐⇒ 0 < c = 1+r ≤ 1.

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IRR
Consider the cash flow: (−100, 50, 100). The IRR is 28.08%.
Consider another two cash flows: (100, 0, 0) and (0, 50, 100)

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Exercise
Find the IRR of the cash flow stream (−2, 1, 1, 1) to 4 significant
figures.
1. Assume r to be the IRR. Find the present value of the cash flow
stream.

2. Solve for r.

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Bond Yields
▶ m: the number of coupon payments per year
▶ n: the total number of coupond payments
▶ t: any time in the lifetime of the bond
▶ P (t): the price (present value) of the bond at time t
▶ The nominal yield of the bond is the nominal internal rate of
return compounded m times per annum of holding the bond from
time t to maturity.

The (nominal) yield of the bond λ(t)% at time t satisfies

n−tm
F X F × c%/m
P (t) =  n−tm +  i .
λ(t)%
1+ m
i=1 1 + λ(t)%
m

We will always assume that the face value and the redemption value of
the bond are the same, unless we explicitly say so. 13/21
Bond Yields
1−r n
Pn 1
Recall that i=1 ri = r × 1−r . Let r = .
(1+ λ(t)%
m )

1
 
1−  n−tm 
1 + λ(t)%

 1 c% 1 m

P (t) = F   n−tm + × ×
 
1
m 1−

 1+
 λ(t)% 1 + λ(t)%
m (1+ λ(t)%
m )

m 

1
 
1−  n−tm 
 λ(t)%
 1 c% 1 + m

= F  +
 
n−tm  
m

λ(t)% λ(t)%
 1+
 1+ m −1 
m 

  
1 c%  1
= F  n−tm + 1 −  n−tm  .
 
λ(t)% λ(t)% λ(t)%
1+ m 1+ m

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Bond Yields

    
1 c  1
P (t) = F 1 + −1 +  n−tm  + 1 − 
   
n−tm 
λ(t)% λ(t) λ(t)%
1+ m 1+ m
 
 
c 1
=F +F × − 1 1 − 
 
n−tm 
λ(t) λ(t)%
1+ m
 
 
c − λ(t) 1
=F +F × 1 −  n−tm  .
 
λ(t) λ(t)%
1+ m

A bond is said to be priced at time t

1. at a premium if P (t) > F (if and only if c > λ(t)),
2. at par if P (t) = F (if and only if c = λ(t)),
3. at a discount if P (t) < F (if and only if c < λ(t)).
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Example

Suppose that a 10-year bond has a face value of $100 and a coupon
rate is 6% paid semi-anually. Suppose that at the end of the third
year, the nominal yield is 4%.
Calculate the price of the bond at the end of the third year.

Exercise: Find F , m, n (the number of total payments), t, c%, and

λ(t)% first.

Solution. This bond is at a premium because c% > λ(3)%.

" #
6−4 1
P (3) =100 + 100 × × 1− 4% 20−3×2
4 (1 + 2 )
=$112.11

Note that P (3) > F .

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Price-yield relation
At a time t, consider P as a function of λ. Then P is a decreasing
function of the yield λ.

Remark
1. The above graph of the price-yield curve is decreasing and convex
∂2P
(because ∂P
∂λ < 0 and ∂λ2 > 0).
2. The maximum possible bond price is attained when the yield is
zero, in which case the bond price is simply the sum of all coupon
payments and the redemption value. That is,
c% × (n − tm)
lim P = F + F × .
λ→0 m
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Example: SSB Yield
A bond with face value F makes m coupon payments of Ci /m (where
Ci = F × ci %) in the i-th year and there are k-years in total. Let P
be the current price of the bond.
Consider its cash flow stream:

(−P, C1 /m, . . . , C1 /m, C2 /m, . . . , C2 /m, . . . , Ck /m, . . . , Ck /m + F ).

The the nominal rate λ is so-called Yield To Maturity of a bond.

k 
F X Ci /m Ci /m
P = km
+ + + ···
(1 + λ/m) i=1
(1 + λ/m)(i−1)m+1 (1 + λ/m)(i−1)m+2

Ci /m
+
(1 + λ/m)(i−1)m+m

That is, the yield to maturity (YTM) is the interest rate at which
the present value of the stream of payments (consisting of the coupon
payments and the final face-value) is exactly equal to the current
price.
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Singapore Saving Bonds
Find the yield to maturity (YTM) of the following bonds.

The cash flow is the following: a(0) = −1000

6m 12m 18m 24m 30m 36m 42m 48m 54m 60m
13.15 13.15 13.55 13.55 13.55 13.55 13.55 13.55 13.55 13.55
66m 72m 78m 84m 90m 96m 102m 108m 114m 120m
13.95 13.95 14.30 14.30 14.70 14.70 15.0 15.0 15.2 15.2 + 103
The equation of valuation:
1000 13.15 13.15 13.55 13.55 13.55
1000 = λ 20
+ λ
+ λ 2
+ λ 3
+ λ 4
+ ··· + λ 10
(1 + 2
) (1 + 2
) (1 + 2
) (1 + 2
) (1 + 2
) (1 + 2
)
13.95 13.95 15.2 15.2
+ + + ··· + +
(1 + λ/2)11 (1 + λ/2)12 (1 + λ/2)19 (1 + λ/2)20

Then λ = 2.8%.

Question: What is the YTM if we redeem the bond after 2 years? cf.
the average return in the above table.
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Pricing a bond

To price the bonds, we make the following simplifying assumptions:

▶ Interest rates are constant over the lifetime of the bond. This is
reasonable if economic conditions are stable.
▶ The yield at any point in time has to equal the interest rates.
This is reasonable if the bond does not have significant default or
liquidity risks (for example, US treasury bonds). Then the price
of the bond is equal to its present value.

The price of the bond in time t is the present value by replacing

λ(t) with the current interest rates.
 
 
c − λ(t)  1
P (t) = F + F × 1 − 

n−tm 
λ(t) λ(t)%
1+ m

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Example

A 10-year bond with face value $100 pays coupons annually at 4%.
Given that the price of the bond at t = 0 is $90, find the price of the
bond at t = 4. Give your answer to 3 significant figures.

Exercise: Find F , m, and n. How about c% and the current interest?

Solution Assume that the current interest rate is r%.

The price of the bond at t = 0 is:
4−r
P (0) = 100 + 100 × × (1 − (1 + r%)−10 ) = 90.
r
Solve for r = 5.31%. Then the price of the bond at t = 4 is:
4−r
P (0) = 100 + 100 × × (1 − (1 + r%)−(10−4) ) = $93.40.
r

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