ACTSC 221
Version 1.0
c Faculty of Mathematics, University of Waterloo
April 7, 2022
Contents
1 Introduction to Interest
1.1 Working with Interest . .
1.2 Compound Interest . . . .
1.3 Nominal Rates of Interest
1.4 Varying Rates of Interest
1.5 Dated Values . . . . . . .
1.6 Unknown Rate and Time
1.7 Inflation . . . . . . . . . .
1.8 Rates of Discount . . . . .
1.9 T-Bills . . . . . . . . . . .
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3
3
6
9
11
13
16
18
21
24
2 Annuities
2.1 Accumulated Value of Annuities . . .
2.2 Present Value of Annuities and Loans
2.3 Calculating the Number of Payments .
2.4 Calculating the Rate . . . . . . . . . .
2.5 Different Focal Dates . . . . . . . . . .
2.6 General Annuities . . . . . . . . . . .
2.7 Perpetuities . . . . . . . . . . . . . . .
2.8 Increasing Annuities . . . . . . . . . .
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26
26
29
32
34
35
41
42
44
3 Loans and Debts
3.1 Repayment of Debts
3.2 Refinancing Debts .
3.3 Mortgages . . . . . .
3.4 Sinking Funds . . . .
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51
51
53
55
58
4 Bonds
4.1 4a. Bond Basics . . . . . . .
4.2 Prices between Coupon Dates
4.3 Bond Amortization . . . . . .
4.4 Yield to Maturity . . . . . . .
4.5 Other Bonds . . . . . . . . .
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60
60
64
67
70
71
5 Investment Analysis
5.1 Net Present Value . . . . . . . .
5.2 Internal Rate of Return . . . . .
5.3 Stocks . . . . . . . . . . . . . . .
5.4 Dollar Weighted Rate of Return .
5.5 Time-Weighted Rate of Return .
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74
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75
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84
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2
Chapter 1
Introduction to Interest
1.1
1.1.1
Working with Interest
Interest and Amount Functions
Every day, investors lend money to borrowers for all sorts of needs. Individuals borrow
money to finance the purchase of a house or a car, or just to take a vacation. Companies
borrow money to invest in new production plants, and governments borrow money to build
roads and bridges.
The lenders (or investors) expect a return (or a profit) on their loan, and those returns (or
profits) are called interest. The study of interest is what this course is about.
Let’s consider a simple example. Suppose we invest (or lend) $100 now and in 1 year, the
borrower will repay $110. In this example, we have earned $110 - $100 = $10 of interest.
Generally speaking, the longer we invest money, the more interest we expect to earn. Consequently, we can define a function A(t) which describes the amount of money our investment
is worth at time t.
Specifically, call the function A(t) the amount function, which represents how much our
initial principal is worth a time t ≥ 0. We think of time t = 0 as today, so A(0) equals our
initial principal, or the amount we invest today.
As in the first example, we have
A(0)
A(1)
= $100
= $110
(This is the initial principal amount)
(This is our final value)
The interest earned is simply the difference
Formula 1
I1 = A(1) − A(0) = $10.
In general, we will define In to be the interest earned between time n − 1 and time n (i.e.
the interest earned over the nth time interval). So In = A(n) − A(n − 1). This is explained
graphically in the following timeline.
3
4
Chapter 1
Introduction to Interest
Figure 1.1.1
As we will see, it will be more useful to work with rates of interest, rather than with actual
dollar amounts. We will do this next.
1.1.2
Rates of Interest
As in the first example, we earned $10 of interest on an initial investment of $100. In this
$10
case, we would say we earned
= 10% interest. So, our initial investment grew at a
$100
rate of 10%.
More formally, we define
Definition 1.1.1
Effective Rate of Interest
The effective rate of interest is the amount of interest earned (or paid) during the period
divided by the initial principal amount, assuming the interest is received (or paid) at the
end of the period.
Example 1
Sally invested $500 for 1 year and earned $25 interest. What was the effective rate of
interest on her deposit?
Solution: The effective rate of interest is simply
$25
= 5%
$500
Generalizing this to the nth period, between time (n − 1) and n, we have that in , which is
the effective rate of interest earned over the nth period is given by:
in =
A(n) − A(n − 1)
In
Interest
=
=
A(n − 1)
A(n − 1)
Amount at the Start
Section 1.1
Working with Interest
1.1.3
5
Simple Interest
The easiest way for interest to accumulate is if we assume that the interest earned is a linear
function of time, and this is called Simple Interest.
More precisely, the interest earned after t years is given by the formula
I =P ×r×t
where P is the initial principal, r the annual rate of simple interest, and t the time in years.
Note that if we hold the investment for twice as long, we make twice the interest.
If we let S be the accumulated value of P , (also called the future value, or the maturity
value of P ) then
S = P + I = P + P × r × t = P × (1 + r × t).
Example 2
A loan of $10,000 is taken out at 5% simple interest. How much will the interest charge be
after
(a) 2 years?
(b) 6 months?
Solution: In each case the interest amount is given by I = P × r × t. Here, this gives:
(a) I = P rt = ($10, 000) × 5% × 2 = $1000.
(b) I = P rt = ($10, 000) × 5% × ( 12 ) = $250.
Once we know how interest is earned as time passes, we can define the amount function
explicitly. Let’s do this for the data in the last example.
Example 3
A loan of $10,000 is taken out at 5% simple interest. Find the amount function, as a
function of time t where time is expressed in years.
Solution: Note that the initial principal is A(0) = $10, 000 and the interest earned over
the time period up to time t is I = A (0) × r × t = A (0) × 5 % × t. Therefore, the total
accumulated value, which equals the principal plus interest is given by
A(t)
= A(0) + A(0) × r × t
= A(0)(1 + rt)
= $10, 000(1 + 5% × t)
Notice that after 2 years, the loan will have accumulated to
A(2) = $10, 000(1 + 5% × 2) = $10, 000(1.1) = $11, 000
which equals the initial principal plus $1,000 interest, which exactly corresponds to the
answer from before
6
Chapter 1
Introduction to Interest
This last example leads us to the general formula for the amount function of an initial
principal P invested for t periods under a period interest rate of r is given by
Formula 2
A (t) = P (1 + rt)
1.1.4
Day Count Conventions
If time is given in actual number of days, there are two conventions to assess the actual
interest charge, so called exact interest and ordinary interest.
In exact interest, we assume a year has 365 days, thus
I = Pr ×
total number of days
.
365
In ordinary interest (also called the Banker’s Rule) we assume a year has 360 days, thus
I = Pr ×
total number of days
.
360
In Canada (and this course) the practice is to use exact interest. The 360 day count
convention is much more common in the United States.
Example 4
A deposit of $2,000 is placed in a bank account earning 7.5% simple interest. How much
will be in the account after 30 days?
Solution: Since we know the number of days (as opposed to knowing the time in years)
we have
S
1.2
1.2.1
= P (1 + rt)
= $2, 000 1 + 7.5% ×
= $2, 012.33
30
365
Compound Interest
Accumulated Values under Compound Interest
By far the most common way for interest to accumulate is by compound interest, where
interest can be earned on interest.
For example, consider an investment of $100 earning 10% interest annually. Under compound interest, we assume the first interest payment is received at the end of the first year,
and reinvested for the second year. Thus, we can earn interest in the second year on the
interest we earned in the first year.
This is easiest to see with a time line.
Section 1.2
Compound Interest
7
Figure 1.2.2
So, after 1 year, our investment has grown to $100(1 + 10%) = $110. That amount will then
be invested for 1 more year at 10%. Over the second year, that amount will the grow to
$110(1 + 10%) = $121.
If we put all this together we see that
$121
= $110(1 + 10%)
= $100(1 + 10%)(1 + 10%)
= $100(1 + 10%)2
Note that $100 accumulating at 10% simple interest would amount to only $120 after 2
years. The difference comes from the interest on the interest. Specifically, the $10 earned
in year one is reinvested in year 2. That $10 makes $10 × 10% = $1 in additional interest.
That $1 is the difference between the simple interest amount of $120 and the compound
interest amount of $121.
Generalizing the above, we have the formula for the amount function for an initial principal
invested for n years at compound interest r is given by
Formula 3
A(t) = P (1 + r)n .
Example 5
Calculate the accumulated value of $2,000 invested at a compound interest rate of 6% per
year at times t = 5 years, t = 10 years and t = 1 month.
Solution:
(a) A(5) = $2, 000(1 + 6%)5 = $2, 676.45
(b) A(10) = $2, 000(1 + 6%)10 = $3, 581.70
(c) A(1/12) = $2, 000(1 + 6%)1/12 = $2, 009.74
8
Chapter 1
1.2.2
Introduction to Interest
Present Values
In each of the previous examples, we computed the accumulated, or future value of the
principal. In many cases, it is the initial value, or present value we wish to find, given a
final value.
An example will make this more clear.
Example 6
Suppose Interest rates are r. How much would you need to invest today to have $1 in 1
year?
Solution: Let the initial value be P . Then the final accumuated value satisfie
A = P (1 + r) = $1.
Solving for P yields
P =
1
.
1+r
This process, where we bring cash flows back in time, is called discounting the values.
The idea that multiplying by 1 + r carries values into the future, and dividing by 1 + r
carries values back to the present can be summarized as follows:
Figure 1.2.3
Since dividing the value by (1 + r) discounts back 1 period, it is not hard to believe that
dividing by (1 + r)n is needed to discount values back n periods (assuming compound
interest). This compares naturally with multiplying by (1 + r) for each period when we
accumulate values into the future.
Therefore, we get the general formula for the present value (PV) of an amount A(t) received
in n periods in the future is given by
Section 1.3
Nominal Rates of Interest
9
Formula 4
A(0) = P V =
Example 7
A(t)
.
(1 + r)n
How much would we need to invest today, if we wished have $10,000 at the end of 10 years?
Assume interest is compounded annually at 5%
Solution: We have
PV
=
$10, 000
(1 + 5%)10
= $6, 139.13
1.3
Nominal Rates of Interest
So far, our examples of compound interest assume that the interest is received and reinvested
at the end of each year.
In many cases, the actual frequency of interest payments may be more often than annually.
For example, many bank accounts pay interest at the end of each month, so the interest
received is reinvested monthly.
In such an instance, we would say that the interest on the account compounds monthly.
In this section, we will cover all types of compounding periods.
Interest rates are typically quoted on an annual basis, even if the compound frequency is
not annual.
We will use the notation i(m) to mean a nominal, or stated annual rate compounded m times
per year. So, i(2) = 10% means that the nominal rate is 10%, compounded semi-annually,
or twice a year.
Thus $1 will grow by 5% in the first 6 months, when the interest will be received. Then
that new principal will grow by another 5% in the next six months.
10
Chapter 1
Introduction to Interest
Figure 1.3.4
For example, if we invest $1 for a year at i(2) = 10% at the end of the year, the accumulated
value will be
10% 2
A = $1 1 +
= $1.1025.
2
This equation makes sense since there are 2 compounding periods (each 6 months long),
and we are receiving half of the interest each period.
In general, $1 invested at i(m) will receive interest at a rate of
(received at the end of each 1/mth of a year.)
i(m)
m ,
m times per year
Therefore, an initial principal, P , will accumulate to a final value, A, when invested at i(m)
for n periods to
Formula 5
A=P
i(m)
1+
m
!n
.
Note that the principal is invested in for n periods, not years. This makes sense, since each
(m)
period, the principle is earning only i m .
Example 8
Suppose $100 is invested for 5 years. Find the final amount assuming: i(1) = 10%, i(2) =
10%, i(12) = 10%, i(365) = 10%
(m) n
Solution: In each case, the accumulated value is given by P 1 + i m
, where n is the
number of total periods.
Section 1.4
Varying Rates of Interest
11
(a) A = $100(1 + 10%)5 = $161.05
2×5
(b) A = $100 1 + 10%
= $162.89. (Note the exponent. Since there are 2 periods per
2
year – given semi-annual compounding – and the investment lasts 5 years, we have 2 × 5
total periods.)
12×5
(c) A = $100 1 + 10%
= $164.53.
12
(d) A = $100 1 +
1.3.1
10%
365
365×5
= $164.86.
Equivalent Rates
The previous example shows for a given nominal rate, we have that i(2) 6= i(12) . As a result,
it is not immediately obvious which generates a higher accumulated value, i(12) = 10% or
i(1) = 11%. We need to be able to compare rates on an equivalent basis.
Definition 1.3.1
Equivalent Rates:
Two rates are called equivalent if a given amount of principal invested for the same length
of time at either rate produces the same accumulated values.
For example, let’s compute the annual rate of interest i(1) that is equivalent to i(4) = 10%.
Assuming $1 in principal invested, we have
i(4)
1+
4
!4
10% 4
= 1+
= 1 + i(1) .
4
Notice that the far left hand side of this expression is the accumulated value of $1 compounded quarterly at i(4) , and the far right hand side of this expression is the accumulated
value of $1 compounded annually at i(1) . These accumulated values are equal, since that is
what it means for the rates to be equivalent.
Solving gives i(1) = 10.381289%. Therefore, investing at 10% compounded quarterly produces the same accumulated value as investing at 10.381289% compounded annually.
This annually compounded rate that is equivalent to the given nominal rate is called the effective annual rate (EAR). Thus 10% compounded quarterly is equivalent to an effective
annual rate of 10.381289%
1.4
1.4.1
Varying Rates of Interest
Varying Rates
In this section, we will consider cases where the interest rate varies over the life of the
investment.
Here is an example.
12
Chapter 1
Example 9
Introduction to Interest
Suppose $200 in invested in an account that pays i(1) = 5%, for the first 2 years, followed
by i(1) = 8% for the next 3 years. Find the final accumulated value.
Solution: Drawing a timeline of the values helps make this a bit more clear.
Figure 1.4.5
Now we can compoute A(2) by
A(2) = $200(1 + 5%)2
and also we have
A(5) = A(2)(1 + 8%)3 .
Putting these together gives
A(5) = $200(1 + 5%)2 (1 + 8%)3 = $277.77.
Equivalent Rates
As an aside, suppose instead this account paid an fixed annual rate of i. What is this rate
such that the account accumulates to the same amount?
Again, we take a look at the time line.
Figure 1.4.6
To answer this, we solve:
Section 1.5
Dated Values
13
$200(1 + i)5 = $277.77 =⇒ i = 6.789849%.
Therefore, investing at 5% for 2 years followed by 8% for 3 years is equivalent to investing
at 6.789849% for 5 years.
1.5
Dated Values
Suppose you are given the choice between receiving $100 in one year or $110 in one year.
Which would you prefer? The answer should be obvious: $110 is better than $100, and
therefore we would rather receive the larger amount of money.
Now instead suppose the choice is between receiving $100 in one year or $110 in two years.
Now which you prefer? The answer here is not so obvious since the amounts of money are
not received at the same point in time. We will see that we need to know both how much
money we are receiving and when we are receiving it before we can decide which option we
prefer.
Consequently, all our calculations involve dated values.
Going back to the previous example, suppose interest rates are 5%. Now can we determine
which option is better? The answer is yes, and to do this we just compute the dated value
(or time value) of both options at a common point in time.
So, how can we convert these options to cash flows that are happening at a common point
in time? There are actually several ways to do this.
Let’s call the option of receiving $100 in one year option A, and the option of receiving the
$110 in two years option B.
Our first solution will be to notice that if we take the $100 that we receive at time 1 and
reinvest it for one year at 5%, then at the end of year two that $100 will have accumulated
to
$100(1 + 5%)1 = $105.00
Essentially, we have just shown that receiving $100 in one year is equivalent to receiving
$105 in two years. They are equivalent because I can convert the $100 into the $105 simply
by investing for one year.
Figure 1.5.7
14
Chapter 1
Introduction to Interest
Consequently, we are now choosing between option A, which is equivalent to $105 in two
years, and option B which is equal to $110 in two years. Since these options are now taken
at the same point in time we can compare them directly, and since $110 is better than $105,
we accept option B – $110.
Let’s do this problem again but at a different point in time. More specifically we want to
convert the $110 at time to to an equivalent amount of time 1. How do we do that?
The solution will be to borrow an amount of money at time one that will accumulate to
$110 a time 2. To compute that amount we just discount the $110 back one period. Let’s
call this value at time one V1 . Now we have
V1 =
$110
= $104.76.
(1 + 5%)1
So, if we borrow $104.76 at time 1, we will have to pay back $110 at time 2, which we will do
with the proceeds from option B. Thus, receiving $110 at time 2 is equivalent to receiving
$104.76 at time 1 since I can convert receving $110 at time 2 by borrowing $104.76 at time
1.
Figure 1.5.8
So, in this case, we are now chosing between option A, which is $100 in 1 year, or $104.76 in
1 year, which is equivalent to option B. Clearly $104.76 is better, so once again we choose
option B.
In our first solution, we decided to compute the relvant values at time 2, and the second
solution we decided to compute the relevent values at time 1. The date we select to compute
the values is often called the focal point, or focal date.
It turns out that no matter what focal point we chose to compute the associated time values,
if a given option is better than another option at one focal point, it will be better at any
focal point.
Although it is arbitrary, sometimes one focal date makes more sense than another, as we
will see with examples. A common focal date to select is in fact now, or time 0. In this
case, the time value is called the present value.
Let’s do this problem for the thrid time by calculating the present value of both options .
We have
Section 1.5
Dated Values
15
P V$100 =
$100
= $95.24
(1 + 5%)1
P V$110 =
$110
= $99.77.
(1 + 5%)2
and
Since option B has a PV of $99.77 which is higher than the PV of option A which is only
$95.24, we select option B. Consequently, receiving $110 in two years has a higher present
value than receiving $100 in one year.
Figure 1.5.9
We will be doing a large number of similar type calculations throughtout this course. Indeed,
this is probably the single most important concept we will face.
1.5.1
Example 10
Equivalent Amounts
We say that 2 different values X and Y , where Y is received n periods later under a period
interest rate of i are equivalent if
Y = X(1 + i)n
or equivalently X = Y (1 + i)−n .
You should recognize that the way we deal with cash flows occuring at different points in
time is to compute equivalent cash flows that occur at the same time. Once the cash flows
are all occuring at the same point in time, it is easy to chose among the various options.
We will find that a large number of the problems we face boil down to computing equivalent
amounts.
16
Chapter 1
Example 11
Introduction to Interest
A person owes $100 in one year and additional $500 is due in five years. What single
payment (a) now, (b) in three years, will satisfy these obligations. Assume i(1) = 10%.
Solution: We are being asked to find an amount either (a) today, or (b) in three years
which is equivalent to the obligations we must pay.
Drawing a timeline is always the first step in tackling these types of problems. We let X be
the equivalent amount today that satisfies the obligations, and Y be the equivalent amount
in 3 years that satisfies the obligations.
Figure 1.5.10
So,
X=
$500
$100
+
= $401.37.
1
(1 + 10%)
(1 + 10%)5
Since we know the time value at time zero, it is easy to find Y , the equivalent value at time
3.
Y = X(1 + 10%)3 = $534.22
Alternatively, we can compute the value Y directly by looking at the time line. We carry
the $100 forward 2 periods, and the $500 back 2 periods, yielding
Y = $100(1 + 10%)2 +
1.6
1.6.1
$500
= $534.22.
(1 + 10%)2
Unknown Rate and Time
Unknown Rate
Recall our fundamental formula
FV = PV
i(m)
1+
m
!n
,
Section 1.6
Unknown Rate and Time
17
where F V is the future value, P V is the present value, i(m) is the annual interest rate
compounded m times per year, and n is the number of periods the money is invested.
We can use this formula to solve for the rate, if we are given the number of periods, and
both the present and future values.
Example 12
Find the rate i(2) such that $100 will grow to $1000 in 10 years.
Solution: Here, the future value is $1000, the present value is $100, and number of periods
is 20. Why 20 and not 10? Since we are looking for an interest rate that is compounded
semi-annually (or twice per year) there are a total of 10 × 2 = 20 periods in 10 years.
Therefore, we need to solve
i(2)
1000 = 100 1 +
2
!20
Dividing by 100 and taking roots gives
i(2)
1+
2
!
=
1000
100
1
20
= 1.122018.
Solving for i(2) gives i(2) ' 24.4%.
Unknown Time
We can solve for unknown time in a similar way.
Example 13
How long will it take for $100 to grow to $1000 if i(4) = 10%.
Solution: In this example, we need to solve for the number of periods, n, in the equation,
10% n
1000 = 100 1 +
.
4
Dividing by 100 and taking logs, gives
10%
1000
n × ln 1 +
= ln
.
4
100
Thus n = 93.249958 periods. Note this is periods, not years. Since we are solving for an
interest rate that compounds quarterly (or 4 times per year) we need to divide the number
of periods by 4 in order to determine the number of years.
Therefore, the answer is years is T = n/4 = 23.3124896 years = 23 years, 3 months, 23 day
18
Chapter 1
1.6.2
Introduction to Interest
Doubling Time
A common question people ask is how long will it take for money to double at a given rate?
If we assume we invest $1, we need to determine the number of periods for this to grow to
$2.
To answer that, we need to solve
2 = (1 + i)n =⇒ n =
So, it takes
ln 2
.
ln(1 + i)
ln 2
periods for money to double when invested at a period rate of i.
ln(1 + i)
A common approximation for this is called the “Rule of 72” which states that the doubling
time n is approximately
n'
72
.
i
So if i = 10%, then n ' 7.2. This compares with the more accurate answer given by the
ln 2
precise formula n =
' 7.2725.
ln 1.1
Remark: Sometimes you will see the “Rule of 70” which states that the doubling time n
70
is approximately n '
, so the only difference is the 70 in the numerator instead of 72.
i
Both are just approximations, so either can be used. The Rule of 72 seems to work better
for interest rates over 5%, and the rule of 70 seems to work better for interest rates below
5%. In any case, these are just approximations to the precise value
n=
1.7
1.7.1
ln 2
ln(1 + i)
Inflation
Exponential Growth
In many ways, problems in compound interest are just examples of general problems involving exponential growth.
The solution to the question “How long does it take money to double at 10%?” is the same
as the solution to the question “How long does it take for a population to double if the
growth rate is 10%?”. In both cases, the quantity under discussion is growing at a rate of
10% per year.
1.7.2
Inflation
Inflation is another example of exponential growth.
Section 1.7
Inflation
Example 14
19
If a loaf of bread costs $2.50 in 2014, how much will it cost in 20 years if inflation is 2% per
year?
Solution: So, in this example, the price of bread in increasing at a rate of 2% per year for
20 years. This looks very much like our future value problems, and the solution is exactly
the same. We solve
$2.50(1 + 2%)20 = 3.71.
1.7.3
Real Rate of Return
An interesting problem is to determine the future value of an investment, after we consider
the effects of inflation.
Consider the following. Suppose we have $100 now. This means we have the ability to
purchase $100 worth of good and services right now.
Now suppose we can invest our money at 10% for 1 year, but we expect the rate of inflation
to be 6%. So at the end of the year, we will have $110, but the original basket of goods now
costs $106. So, by investing and deferring our purchasing decisions, we really only have $4
extra, not $10.
Definition 1.7.1
The Real Rate of Return is defined to be the growth in purchasing power available after
we consider the effects of inflation. This is distinct from the nominal rate of interest,
which is the interest rate that does not adjust for inflation. When people speak of interest
rates on a day-to-day basis, they are really talking about nominal rates.
Suppose the inflation rate is r, and suppose we have $1 now. After 1 year has passed, that
$1
same $1 can now only buy
worth of goods.
1+r
Now, if I can invest the $1 at a nominal rate i, the at the end of the year the $1 has grown
to 1 + i. Therefore, I can buy
1+i
1+r
worth of goods.
We define the real rate of interest to be this growth rate in purchasing power. More
precisely
Formula 6
1 + ireal =
A little algebra yields
1+i
.
1+r
20
Chapter 1
Formula 7
ireal =
Example 15
Introduction to Interest
i−r
.
1+r
Joe invests at 8% interest; however, Joe expects inflation to be 4%. What is his real rate
of return?
Solution: Simply we calculate
ireal =
8% − 4%
= 3.85%.
1 + 4%
So, Joe is able to purchase 3.84% more goods at the end of the year than at the start of the
year.
1.7.4
Taxes
In many cases, investors must pay taxes on the interest they earn. The taxes are paid
usually at a fixed rate of the interest earned.
Suppose you invest at a rate of 10%, but that you must pay taxes at a rate of 40% (i.e.
40% of the interest earned you must pay in taxes).
Thus, you only get to keep 1 − 40% = 60% of the interest. So, at a 10% interest rate, you
only make money at a rate of 60% × 10% = 6%.
Therefore we get that the after tax interest rate is given by
Formula 8
iafter tax = i(1 − T )
where i is the nominal interest rate and T is the tax rate.
1.7.5
Taxes and Inflation
Recall that thet the real rate of return, given an interest rate of i and an inflation rate of r
is given by:
Formula 9
ireal =
i−r
.
1+r
Also recall that given a before tax interest rate of i and a tax rate of T , the after tax interest
rate is given by:
Section 1.8
Rates of Discount
21
Formula 10
iafter tax = i(1 − T )
We can now combine both the effects of taxes and inflation to calculate a real after tax
interest rate which is given by the formula:
Formula 11
ireal after tax =
Example 16
i(1 − T ) − r
1+r
Joe can invest his money at 5% for one year. However, Joe must pay taxes at a rate of
45%, and inflation is 2%. What is his real after tax return?
Solution: His real after tax return is simply
ireal after tax =
5%(1 − 45%) − 2%
= 0.74%.
1 + 2%
This last example shows how the effects of taxes and inflation can significantly reduce the
return an investor will actually realize from an investment.
1.8
1.8.1
Rates of Discount
Rate of Discount
In some unusual cases, the interest is paid at the start of the loan, and we say that it is
being paid in advance, as opposed to being paid at the end of the loan, in which case we
say the interest is being paid in arrears. it is far more common for interest to be paid in
arrears, but in this section we will consider the case where it is paid in advance.
For example, suppose you borrow $100 at a rate of 10%. Naturally, you would expect to
pay interest of $100 × 10% = $10. The big difference with discount is that this interest
will then be paid “up-front”, meaning it must be paid now, when the loan is advanced.
(This is why we say it is paid in advance.) Therefore, the bank will advance to you only
$100 − $10 = $90. You will then be required to repay the loan amount of $100 at the end
of the year.
22
Chapter 1
Introduction to Interest
Figure 1.8.11
When interest is paid this way, we call it a rate of discount and it be denoted by the
letter d.
So, if the loan principal is P , the amount you will receive today is then
Amt Received = Principal - Interest = P − P × d = P (1 − d)
Example 17
Suppose a 1 year discount loan with a principal value of $1000 is made at a rate of discount
of 5%. how much money will be advanced on the loan today?
Solution: The interest amount is 5% × $1000 = $50. This amount will be charged today,
so you will receive today the remainder, which is $1000 − $50 = $950.
Alternatively, we can compute the amount directly by the formula
Amt = $1000 × (1 − 5%) = $950.
The effective rate of discount over period n, denoted dn , is the ratio of the cost of the
loan (or the amount of interest) to the amount at the end of the year. Thus
Formula 12
dn =
A(n) − A(n − 1)
.
A(n)
Recall the effective rate of interest is given by
Formula 13
in =
A(n) − A(n − 1)
.
A(n − 1)
So, in summary discount is paid at the beginning of the year based on the balance at the
end of the year; while interest is paid at the end of the year, based on the balance at the
beginning of the year.
Section 1.8
Rates of Discount
23
Using the data from the last example, the effective rate of discount over the first period is
given by
d1 =
$1000 − $950
= 5%
$1000
as expected (since we assumed a rate of discount of 5% in the question).
Now, the effective rate of interest is then
i1 =
$1000 − $950
= 5.263%
$950
So we see that rates of discount and rates of interest are indeed different. Put another way,
we just showed that borrowing $1000 at a rate of discount of 5% is equivalent to borrowing
$950 at a rate of interest of 5.263%.
1.8.2
Converting rates
Under compounding interest, we have that the future value after n periods given an initial
principal P growing at an interest rate i is given by A(n) = P × (1 + i)n . Substituting this
into the expression for the rate of discount yields
dn
=
Interest Amount Charged
.
Amount at end of Period
=
A(n) − A(n − 1)
A(n)
P (1 + i)n − P (1 + i)(n−1)
P (1 + i)n
i
=
1+i
=
Using some algebra to solve for i, we get
Formula 14
i=
d
.
1−d
The last expression makes sense since you will be charged an interest amount of d on an
amount actually borrowed of 1 − d.
Example 18
A discounted loan is offered at a rate of discount of 15%. What is the equivalent rate of
interest?
Solution Set
i=
d
15%
=
= 17.647%.
1−d
1 − 15%
24
Chapter 1
1.8.3
Introduction to Interest
Future Values
Recall again that A(n) = A(0)(1 + i)n where we accumulate at a compound interest rate of
i. What is the corresponding formula for the future value given rate of discount?
Recall that for a rate of discount d
A(0) = A(1)(1 − d),
thus
A(1) =
A(0)
.
1−d
Extending to 2 periods
A(2) =
A(1)
A(0)
=
.
1−d
(1 − d)2
Thus in general,
Formula 15
A(t) =
A(0)
.
(1 − d)t
In the language of present and future values, the previous expression can be summarized as
FV =
PV
⇐⇒ P V = F V (1 − d)t .
(1 − d)t
This contrasts with regular interest wish can be summarized as
F V = P V (1 + i)t ⇐⇒ P V =
FV
.
(1 + i)t
Finally, it is worth remembering that rates of interest are far more common than rates of
discount, and therefore the last expressions we will be using far more often.
1.9
T-Bills
Treasury Bills are short term notes issued by governments. Essentially, governments use
T-bills as a way to borrow money from the public over a short time frame. Currently, T-bills
make up a sizable percentage of all government debt outstanding.
T-bills are an example of a money market instrument. The so-called money markets
consist of short term loans issued by governments, bank and corporations. By “short term”,
we mean a loan that is due in less than 1 year. Other examples of money market instruments
include
Section 1.9
T-Bills
25
• Commercial paper
• Bankers Acceptance Notes
• Certificates of Deposit.
We will focus on T-Bills here. T-bills are issued for various maturities that are less than 1
year but 91 day bills are the most common.
T-bills pay do not pay interest in the conventional way. Instead, they are issued at a
discount to the face value (or maturity value) and the difference is essentially interest. For
example, a T-bill might have a maturity value of $1000 (meaning the government will pay
the holder $1000 on the maturity date), but the T-bill would be issued to the public for a
lesser amount, say $975. So, a purchaser of the T-bill could buy it for $975 and then redeem
it later for $1000. The difference is essentially the interest.
1.9.1
Canadian T-Bills
In Canada, the price of a T-Bill can be computed by discounting the face value using simple
interest and dividing the exact number of days by 365.
Example 19
Compute the price of a 91 day T-bill if the rate is 5%. Assume a face value of $1,000.
Solution: To find the value, we compute
P =
1.9.2
$1000
= 987.69
91
(1 + 365
5%)
US T-Bills
US T-Bill conventions are a bit strange. Instead of using simple interest, US T-bills use
simple discount conventions. More specifically, the price is computed by discounting the
face value using simple discount and dividing the exact number of days by 360. The basic
idea is that if a 1 year T-bill has a face value of $1000 and it is trading with a rate of 5%,
then its price is trading at a 5% discount to face value, which is $1000(1 − 5%) = $950 .
For periods of less than 1 year, we use simple discount to compute the price.
Example 20
Compute the price of a 91 day US T-bill if the rate is 5%. Assume a face value of $1,000.
Solution: (You may wish to first review the conventions for rates of discount.)
To find the value, we compute
91
P = $1000 1 −
5% = 987.36
360
Chapter 2
Annuities
2.1
2.1.1
Accumulated Value of Annuities
Simple Annuities
An annuity is a stream of payments made at regular intervals. Typically, the payment
amounts are all equal; however, later we will consider examples where the payments amounts
may not all be equal.
The simplest example is a stream of equal cash flows of $R each of which occurs at the end
of each period for n consecutive periods.
Figure 2.1.1
A practical example of an annuity would be the payments on a loan. Often, loan payments
are made in equal amounts at regular intervals until the loan is fully paid off.
We will be interested in calculating the Present and Future Values of various annuities.
2.1.2
Geometric Progressions
It turns out that we will need to recall a high school fact about the sum of a geometric
progression in order to value annuities.
Recall that a geometric progression is a finite sum of terms, where the ratio between consecutive terms is constant. For example, the sum S
S = a + ar + ar2 + · · · + arn .
26
Section 2.1
Accumulated Value of Annuities
27
is a geometric progression with first term a and common ratio r (since if you take any term,
and divide it by its preceding term, you always get r).
The trick to evaluating the sum of such a progression is to multiply both sides by the
common ratio r to obtain
rS = ar + ar2 + ar3 + · · · + arn+1 .
Now if r > 1 we subtract the first equation from the second to get
rS − S = arn+1 − a.
Solving for S yields.
S=
arn+1 − a
.
r−1
If r < 1 then the natural trick is to subtract the second equation from the first to get
S − rS = a − arn+1 .
Again, solving for S gives
S=
a − arn+1
.
1−r
It turns out that these expressions are equal (as long as r 6= 1). However, one is more
natural than the other, depending on the size of r. If r > 1 is seems more convienent to use
the first expression, and when r < 1, the second expression is a bit more natural – as this
keeps both the numerator and denominator positive.
2.1.3
Accumulated Values of Simple Annuities
Now we are ready to compute the accumulated value of an ordinary simple annuity.
What makes an annuity ordinary is that the payments are due at the end of each period.
What makes it simple is that the interest compounding period corresponds to the payment
frequency.
Specifically, consider an annuity of n payments of $1 occuring at the end of each year, where
interest rates are i compounded annually.
Figure 2.1.2
28
Chapter 2
Annuities
We want to find the total accumulated value (or future value) of this annuity at time n. To
do that, we need to find the future values of each cash flow, then add them up.
The future values of each cash flow are as follows:
Figure 2.1.3
The total accumulated value can be found by adding up all the individual accumulated
values.
S
= 1(1 + i)n−1 + 1(1 + i)n−2 + · · · + 1)
= 1 + 1(1 + i) + · · · + 1(1 + i)n−1
(reverse order of the terms)
This is a geometric progression with n terms, a = 1 and r = (1 + i). Thus the sum is
S=
(1 + i)n−1+1 − 1
(1 + i)n − 1
=
.
1+i−1
i
This sum gets special notation
Formula 1
sn i =
(1 + i)n − 1
.
i
which is read “s n angle i” and is called the accumulation factor.
So, in summary, the accumulated value of an ordinary simple annuity with n payments of
$1 computed on the date of the last payment is given by s n i
Example 1
Suppose you deposit $100 into an account at the end of each year earning interest at
i(1) = 10%. How much money have you accumulated after the 15th deposit?
Solution Depositing $100 a year is clearly 100 times as much as depositing $1 per year,
thus we have
A.V.
= $100 × s 15 10%
= $100 ×
(1 + 10%)15 − 1
10%
= $3, 177.25
Section 2.2
Present Value of Annuities and Loans
2.2
2.2.1
29
Present Value of Annuities and Loans
Discounting Annuities
In this section, we will look at discounting the cash flows in an annuity in order to compute
the present value of the payments. Recall the timeline of an annuity of n payments of R
dollars beginning 1 period from today.
Figure 2.2.4
The present value of these payments is found by adding up the present values of each of the
cash flows taken one at a time, as we see below.
Figure 2.2.5
So, to compute the present value of the annuity, we just add up the terms to get
P.V. =
R
R
R
+
+ ··· +
.
(1 + i) (1 + i)2
(1 + i)n
This sum is a geometric progression of n terms with a =
R
(1+i)
and common ratio r =
Therefore, using the formula for the sum of a geometric progression, we have
1
1+i .
30
Chapter 2
PV
=
Annuities
a − arn
1−r
R
1
R
−
(1 + i) (1 + i) (1 + i)n
=
1
1−
(1 + i)
=R×
1 − (1 + i)−n
1+i−1
=R×
1 − (1 + i)−n
i
In particular, the present value of an annuity with equal payments of $1 is equal to
P.V. =
1 − (1 + i)−n
.
i
an i =
1 − (1 + i)−n
.
i
This sum gets special notation
Formula 2
which is read “a n angle i” and is called the discount factor.
So, in summary, the present value of an annuity of n payments of $1 paid at the end of
each period with a period interest rate of i is given by a n i .
Here are a couple of examples.
Example 2
What is the present value of a stream of payments of $100 paid at the end of each year for
10 years? Assume i = 10 %.
Solution: The present value is given by the formula
PV
= $100 × a 10 10%
%)
= $100 × 1−(1+10
10
= $614.46
Example 3
−10
Suppose you wish to buy an annuity that pays $1000 per month for the next 10 years. How
much would this cost today if i(12) = 6%?
Solution: The amount of money we would agree to pay for this annuity is the present
value of the annuity. Now, that present value is given by
Section 2.2
Present Value of Annuities and Loans
31
P V = $1000 × a n i
The common mistake that students make is to assume n = 10 since it is a 10 year annuity,
and that i = 6%, since that is what we are told interest rates are. However, this is incorrect.
The parameter n is the total number of payments. Since the payments are made monthly
for 10 years, there are a total of 12 × 10 = 120 payments. Also, the interest rate i is the rate
per payment period. Since the payment period is monthly, we need a monthly rate. We are
given that interest rates are 6% compounded monthly, and that equates to a monthly rate
of 6%
12 = 0.5%
So now we have that the present value of $1,000 per month for 12 × 10 periods at a period
(12)
rate of i 12 = 6%
12 is
PV
= $1000 × a 12×10 6%
12
−120
= $100 × 1−(1+0.5%)
0.5%
= $90, 073.45
2.2.2
Loans
A major category of problems where we will use the discount factor is in the calculation of
loan payments.
Example 4
I just purchased a car for $30,000. I am going to make monthly payments for the next 5
years. What will my monthly payment be if i(12) = 4%?
Solution: We have 2 choices: buy the car now for $30,000 or borrow the money and make
monthly payments. Financially, these are equivalent choices and hence they should have
the same time value. The easiest point in time to work with is today, so that is our focal
date. The present value of the purchase of $30,000 is simply $30,000 since that payment
must be made today. Therefore, the present value of the loan payments is also $30,000. Let
my payment be denoted by R. Then
30, 000 = R × a 12×5 4% .
12
Therefore
R
=
30000
a 12×5 4%
12
= $552.50.
Again, note that the number of payments is 12 × 5 and not just 5, since the payments
are monthly. Also, the interest rate we use the period rate, which is again monthly, not
annually.
An interesting follow-up question to the last example is “How much total interest did I pay
on my loan?”
32
Chapter 2
Annuities
Note that the total payments on the loan are
$552.50 × 12 × 5 = $33, 150.
So, either I can pay $30,000 for the car today, or I can borrow the money and make loan
payments that total $33,150. Subtracting the original $30,000 that I borrowed from the
total loan payments, means the total interest payments made as part of the loan are
$33, 150 − $30, 000 = $3, 150.
In summary, if I borrow the money, I will make payments totalling $33,150, $30,000 of
which are principal payments and the remainder of $3,150 are interest charges.
2.3
2.3.1
Calculating the Number of Payments
Calculating Number of Payments
Let’s begin with a simple example. Suppose I purchase a car for $20,000 and agree to make
monthly payments of $500. How many payments will I have to make if i(12) = 12%?
Figure 2.3.6
Since we could either buy the car now for $20,000 or make the monthly payments, these
must have the same time value. The easiest focal date is today at time zero, in which case
we need to equate the present values of both options. The present value of the purchase
option is just $20,000 (since we need to pay that amount now at time zero to buy the
car), and the present value of the monthly payments is given by $500 × a n 12% , where n is
12
the number of months (or periods) and our period interest rate (or monthly rate) is
Equating these gives
$20, 000 = $500 × a n 12% .
12
Now,
$20, 000
= $500 × a n 12%
12
= $500 ×
1 − (1 + 1%)−n
1%
12%
12
.
Section 2.3
Calculating the Number of Payments
33
Therefore
$20, 000
× 1% = 1 − (1 + 1%)−n
$500
Simplifying a bit gives
(1 + 1%)−n = 0.6
Taking logs and solving shows
n=−
ln 0.6
' 51.3 months
ln(1 + 1%)
So, I need to make just over 51 payments. But how does this work exactly? 51 payments
is not enough, but 52 is too many.
There are 2 options: The first is to make a slightly larger final payment in order to fully
reply the loan (called the balloon payment). The second (and more common) is to make
a smaller final concluding payment one period later (called the drop payment).
Let’s see if we can determine how big these payments need to be.
Here is the time line of cash flows. Let’s first assume we make our last payment slightly
larger in order to satisfy the loan.
Figure 2.3.7
Now, the present value of this stream of cash flows must total the cost of the car. This
present value can be seen as the sum of the present value of the 51 regular payments, plus
the present value of the additional “top up” payment at time 51. This gives
$20, 000 = $500 × a 51 1% + X(1 + 1%)−51 .
Solving gives X = $167.66.
Therefore, the final balloon payment is the regular payment of $500 plus the extra payment
X, giving
$500 + $167.66 = $667.66.
Now let’s assume we make a final drop payment one period later, which is at time 52.
34
Chapter 2
Annuities
Figure 2.3.8
Again setting the present value of this stream of cash flows to the cost of the car gives
$20, 000 = $500 × a 51 1% + X(1 + 1%)−52 .
In this case, solving for X gives X = $169.33.
Therefore, the final drop payment one period later at time 52 is $169.33
2.4
2.4.1
Calculating the Rate
Calculating the Interest Rate
Let’s begin with an example.
Suppose I purchased a car valued at $20,000. To finance this purchase, I’ve agreed to pay
$500 per month for 5 years. What rate of interest am I being charged on the loan?
As we have seen before, to solve this, we set the present value of the monthly payments to
equal the cost of the car, which is $20,000.
Figure 2.4.9
Notice that this is a 5 year loan and the payments are monthly, therefore there are a total
of 5 × 12 = 60 payments. If we let i(12) be the annual interest rate (compounded monthly)
(12)
that we are trying to find, we have the period (or monthly rate) is i 12 . So, we set
$20, 000 = $500 × a
Therefore, we must solve for i(12) where
(12)
60 i 12
.
Section 2.5
Different Focal Dates
35
1−
i(12)
1+
12
$20, 000 = $500 ×
!−60
i(12)
12
.
Unfortunately, there are no algorithms to solve this.
Instead, we must use numerical tools such as bisection methods, or Newton’s-type methods
to approximate the solution.
Financial calculators have built in solvers, and other tools like Excel also have tools to solve
such expressions. Using Excel, I found the interest rate to be about
i(12) = 17.27%.
Generally speaking, when solving for an interest rate, we will state the rate accurate to 2
decimal places. In practice, rates are often stated to 6 decimal places of accuracy, but to
keep things a bit simpler, we will work to 2 places of accuracy.
2.5
2.5.1
Different Focal Dates
Deferred Annuities
A deferred annuity is like a regular annuity, expect the first payment is delayed by a
certain number of periods.
Assume today is time 0, and we have an annuity of n payments of $1 with the first payment occurring m + 1 periods in the future. We want to calculate the present value and
accumulated value of this annuity.
Figure 2.5.10
Note that if we choose time m to be our focal date (so let’s initially assume that today is
on date m ), then looking that the stream of cash flows at time m, we have just an ordinary
simple annuity. The present value of that annuity is given by a n i .
To find the present value today (at time 0) we need to discount this back m periods. Thus
P.V. =
an i
= a n i × (1 + i)−m
(1 + i)m
36
Chapter 2
Annuities
Now for the accumulated value. If we choose time n + m as our focal date, then as of
that date, the annuity just looks like a regular ordinary simple annuity, and hence its
accumulated value is given by
A.V. = s n i .
Example 5
Sally wishes to purchase an annuity for her retirement. Sally is 30 years old today, and she
wants the first payment to occur in 35 years when she turns 65. She would like the annuity
to make 20 equal annual installments of $20,000. How much will the annuity cost today, if
i(1) = 10%?
Solution: As usual, we draw a time line of the cash flows. Here, the time refers to Sally’s
age, which is currently 30. The first payment will occur when she is 65. (Note the dollar
amounts in the diagram below are in thousands of dollars)
Figure 2.5.11
Thus P.V. = $20, 000 × (1 + 10%)(30−64) × a 20 10% = $6, 664.85
Notice that the annuity is deferred 34 periods, not 35. Many students erroneously think
that since she is 30 years old now and that the first payment is 35 years later when she
is 65, it follows that the annuity is deferred 35 periods. This is wrong, since in a regular
ordinary simple annuity, the first payment is assumed to be 1 period after the focal date.
So in that sense, 1 period is already accounted for, so the annuity is then deferred only 34
periods, not 35.
This is an excellent example of how drawing careful time lines are an important skill when
computing these values. The majority of the time students take to answer questions is often
spent in the creation of a careful time line.
2.5.2
Annuities Due
So far, all our previous examples involved annuities where the payments are made at the
end of the period. An annuity due is an annuity where the payments are made at the
start of each period
Section 2.5
Different Focal Dates
37
Figure 2.5.12
Notice that there is no cash flow at the end of the last period, so there are still a total of n
cash flows, but that they are now all at the start of each period.
The present value of this annuity is calculated at time 0, which is the time that the first
cash flow is made.
There are 2 main ways to find the present and accumulated values of these types of annuities:
the first is to resort back to our first principles and recognize the values as a geometric
progression. The second is to use our existing formulas to solve for these amounts.
Let’s calculate the present value first. Actuaries use the notation ä n i to denote the present
value of an annuity due with n payments. We will calculate this by making the clever
observation that an annuity due consists of a regular ordinary simple annuity (for which we
know how to find the present value) plus one extra payment.
Specifically, if we ignore the first payment on day 1, we are left with a regular annuity, but
with n − 1 payments, as shown in the timeline below.
Figure 2.5.13
Therefore, ä n i is equivalent to the present value of a regular annuity of n − 1 payments,
plus the $1 at time 0.
Therefore
Formula 3
ä n i = a n−1 i + 1
38
Chapter 2
Annuities
Accumulated values work in a similar way. Actuaries use the notation s̈ n i to denote the
accumulated value of an annuity due (at time n) with n payments.
Recall that the accumulated value of a regular annuity calculated on the date the last
payment is made is given by s n i . In this case, the date the last payment is made is time
n − 1. Therefore, the accumulated value of the annuity due at time n − 1 must be s n i . To
find the accumulated value one period later (at time n) we multiply by (1 + i ), as shown
below
Figure 2.5.14
Thus,
Formula 4
s̈ n i = s n i (1 + i).
Students can easily get confused that the accumulated value of the annuity at time n − 1 is
s n i . Why is there not an n − 1 in that expression? Remember that the n parameter in the
expression s n i refers to the number of payments, not to the time at which the time value is
calculated. Since there are n payments, the parameter value is n.
Example 6
John deposits $1000 per year for 10 years. Assume he begins making deposits today, and
each deposit occurs at the begining of each year. How much will he have accumulated by
the end of year 10 if i(1) = 10%?
Solution: As usual, we begin by drawing a timeline. Note that there are a total of 10
payments, with the first payment today. Also note that we want the accumulated value at
the end of year 10, which is 1 period after the last payment.
Figure 2.5.15
Section 2.5
Different Focal Dates
39
The accumulated value is then
$1000 × s̈ 10 10% = $1000 s 10 10% · (1 + 10%) = $17, 531.17
2.5.3
Final Comments
It is very easy for students to confuse all these types of annuities. It is important to keep
in mind that all of these annuities are really the same thing. An annuity is simply a stream
of cash flows equally spaced in time. The only difference between an regular annuity, an
annuity due and a deferred annuity is when the first cash flow occurs relative to today. But
ultimately, as an object, all of them are exactly the same thing – a stream of equal cash
flows occurring at regular intervals.
The timeline below describes a typical annuity. We can compute the time value of this
annuity at any focal date that we choose. For example, if we compute the time value at
point A, the time value equals the present value of a deferred annuity. If we compute the
time value at point D, the time value equals becomes the accumulated value of a regular
annuity. But ultimately the “annuity” is the same.
Figure 2.5.16
This annuity here starts with a cash flow at time C, and has its last cash flow at time D.
However, we may be interested in finding the equivalent time value at various focal dates.
I find the easiest one to consider first is at time B. If this is our focal date, then the annuity
is just an ordinary simple annuity and its time value is given by
B = an i
At time A, to find the equivalent time value, we just discount B back an appropriate number
of periods, so here
A = (1 + i)−m × B =
an i
(1 + i)m
At time C, the annuity is in fact an annuity due, so it is given by the formula
40
Chapter 2
Annuities
C = ä n i = (1 + i) × a n i
Notice that the right hand side of the expression is also the value B carried forward 1 period.
Recall from before that
ä n i = a n−1 i + 1
It is left as an exercise to show that these are equal.
Not that anyone would compute it this way, but notice that since C is equivalent to A and
occurs m + 1 periods later, it follows that C equals A carried forward m + 1 periods, so we
have
C
= (1 + i )m+1 × A
= (1 + i )m+1 × (1 + i )−m × B
= (1 + i ) × B
= (1 + i) × a n i
(from the above formula for A)
(from the above formula for B)
This last equation is exactly what we already computed for B. This is a good illustration
that your initial choice for a focal date is essentially arbitrary.
Since the last payment occurs at time D, if we choose that as our focal date, then we have
the accumulated value of an ordinary annuity and hence
D = sn i
At point E, we are computing the accumulated value of an annuity due, which is alternatively
just the value D carried forward 1 period. Hence
E = s̈ n i = (1 + i) × D = (1 + i) × s n i
Finally, for point F, it is the accumulated value several periods after the last payment. This
can be found by carrying forward any of the previous amounts an appropriate number of
periods. In particular, it is D carried forward k + 1 periods, yielding
F = (1 + i)k+1 × s n i
Personally, I rarely think about an “annuity due”. I tend to use either point B or D as my
initial focal date, and then move that result either forward or backwards in time as needed.
This way, I typically work with the formulas a n i and s n i only.
Here is another example that shows how we can choose any initial focal date when working
to value an annuity.
Section 2.6
General Annuities
Example 7
41
Consider an annuity that makes 10 equal payments of one dollar at the end of each year.
Let’s compute the time value of this annuity on the date that the seventh cash flow is made.
Let’s assume interest rates are 10%.
Solution: As usual, our first step is to draw a time line.
Figure 2.5.17
We wish to find the time value at time 7, which is denoted here by X . We will do this 3
ways:
First, at time 7, we can recognize this annuity as a sum of 2 annuities: the accumulated
value of an annuity of 7 payments (shown in blue) plus the present value of an annuity of
3 payments (shown in red). Therefore
X = $1 × s 7 10% + $1 × a 3 10% = $9.487 + $2.487 = $11.97
Alternatively, we can find the present value at time zero (denoted by A) and carry that
forward by 7 periods. Therefore
X = (1 + 10%)7 × A = (1 + 10%)7 × a 10 10% = 1.949 × 6.145 = $11.97
Finally, we could compute the accumulated value of this annuity at the time of the last
payment at time 10 (denoted by B) and bring it back 3 periods. This gives
X = (1 + 10%)−3 × B = (1 + 10%)−3 × s 10 10% = 0.751 × 15.937 = $11.97
All of these results are the same!!
So, we can see that the initial choice of focal date does not matter. In practice, we choose
our initial focal date to be whichever date is easiest to work with.
2.6
General Annuities
Recall that the simple annuities we have been considering so far all consist of a stream
of equal payments made at intervals which correspond to the compound frequency of the
interest rates. So if we had an annuity of monthly payments, we were given an interest rate
that compounded monthly. This way it was easy to compute the period (or monthly) rate
which is needed in order to use the annuity formulas.
42
Chapter 2
Annuities
We will now begin to generalize these assumptions. First, we will consider cases where the
compounding period of the interest rate does not match the timing of the cash flows.
Let’s begin with an example.
Example 8
John deposits $100 at the start of each year into an account paying i(4) = 5%. How much
is in the account immediately after the 10th deposit?
Solution: We cannot just apply our known formulas to this question, since the deposits are
made annually, but the interest rate is compounded quarterly – so they don’t match. The
solution is to change the interest rate to an equivalent rate that has the same compounding
period as the deposits. Therefore, we need a rate that compounds annually that is equivalent
to the given rate which compounds quarterly. Thus we solve
5% 4
1+i= 1+
=⇒ i = 5.0945337%.
4
Therefore, the rate i = 5.0945337 % compounded annually is equivalent to 5% compounded
quarterly. Now we have
AV
= $100 × s 10 i
= $100 × 12.63353
= $1, 263.35
Here is another similar example
Example 9
Sally is making car payments of $200 per month for 5 years. If the interest rate on the loan
is i(1) = 10%, what was the initial loan amount?
Solution: In this case, the rate is compounded annually, but the payments are made
monthly. So, we first need to find the equivalent monthly compounded rate (matching the
payment frequency). Here we get
1 + 10% =
i(12)
1+
12
!12
=⇒ i(12) = 9.5689685%.
The loan amount is then the present value of these payments, which is given by:
PV
= $200 × a
(12)
5×12 i 12
= $200 × 47.5385
= $9, 507.70
2.7
Perpetuities
Perpetuities are annuities that continuing paying forever. In other words, we can consider
a stream of payments R each period, with no final payment.
Section 2.7
Perpetuities
43
Figure 2.7.18
So, the first payment starts one period from today, and the payments continue each period
forever.
Recall that the present value of an annuity paying $1 per period for n periods with an
interest rate of i is a n i .
Therefore, the present value of a perpetuity paying $1 forever is just the limit of this as
n → ∞.
Formula 5
a ∞ i = lim a n i .
n→∞
Now
a∞ i
= limn→∞ a n i
−n
= limn→∞ 1−(1+i)
i
1−
=
=
lim (1 + i)−n
n→∞
i
1
i
Therefore, we have the simple formula for the present value of a perpetuity of $1 is given
by
Formula 6
1
a∞ i = .
i
Here is an example
Example 10
Find the present value of a perpetuity that pays $10 at the end of each year, assuming
i = 10%.
Solution: Using our formula, the solution is then
P.V. = $10 × a ∞ 10% = $10 ×
1
= $100.
10%
44
Chapter 2
2.7.1
Annuities
General Perpetuities
Deferred perpetuities and perpetuities where the interest rate compounding period differs
from the payment frequency work just as we saw in the previous section. Here is an example.
Example 11
What is the present value of a perpetuity that pays $250 per month, with the first payment
2 years from today? Assume i(1) = 11%.
Solution: There are 2 things going on in this example. First, the perpetuity is deferred
(it does not start paying in the immediate next period) and secondly, the interest rate
compounds annually, which does not correspond to the payment frequency.
Let’s adjust the interest rate first. We have
(1 + 11%) =
i(12)
1+
12
!12
=⇒ i(12) = 10.4815%.
Let’s look at the time line of the payments, where the times indicated are in months.
Figure 2.7.19
If we choose month 23 as our focal date (which is one period before the first cash flow),
then the present value of the perpetuity is given by
P V23 = $250 × a
(12)
∞ i 12
In order to find the present value today, at time zero, we need to discount this back 23
periods, yielding:
P V0 = $250 × a
2.8
(12)
∞ i 12
×
i(12)
1+
12
!−23
= $23, 433.02.
Increasing Annuities
Consider an annuity that pays annually with the first payment of $1 in one year and each
subsequent payment is increased by a factor (1 + g). So, the payments are growing at a rate
Section 2.8
Increasing Annuities
45
of g.
Below we have the time line of these cash flows
Figure 2.8.20
The most common application of this is when the growth rate is the rate of inflation.
How would we find the present value of this? The idea is to find the present value of each
cash flow separately and then add them up, as shown below
Figure 2.8.21
Therefore, the present value is the sum of these individual present value terms
PV =
1(1 + g)
1(1 + g)n−1
1
+
+
·
·
·
+
.
1+i
(1 + i)2
(1 + i)n
Notice that this is a geometric progression with a =
1
1+i
and common ratio r =
1+g
1+i .
Recalling the formula for the sum of a geometric progression and doing some simplification
yields
46
Chapter 2
PV
=a×
1 − rn
1−r
Annuities

1+g n

1 − 1 + i
1

=
·

1+g 
1+i
1−
1+i

1+g n
1−
1+i
=
(1 + i) − (1 + g)
1+g
1−
1+i
=
i−g
n
where
(i 6= g)
Therefore, the present value of an annuity with first payment of $1 in 1 period growing at
a rate of g under an interest rate i is
Formula 7
1+g
1−
1+i
PV =
i−g
n
where
(i 6= g)
As an aside, I do not recommend that students memorize the formuals in this section. It is
better to spend time understanding the concepts of growing annuities. You can make make
a formula page to help you solve actual numerical examples.
Here is an example of a growing annuity problem.
Example 12
You wish to purchase a 20 year annuity that will grow at the rate of inflation which is
expected to be 2%. Assume payments are made at the end of each year, and the first
payment of $10,000 will be at the end of this year. What is the value of this annuity if
i(1) = 10%?
Solution: Using our formula for a growing annuity, we have
1 + 2% 20
1−
1 + 10%
P V = $10, 000 ×
10% − 2%
= $97, 390.42
2.8.1
Arithmetic Growth
In our last examples, the payments were growing in geometric progression. Now we will
consider cases where the growth is in arithmetic progression.
Section 2.8
Increasing Annuities
47
Consider a stream of n payments of $1 increasing by $1 each period. To find the present
value, we will do the same things that we did before: we can take the present value of each
cash flow and then add them up, as shown in the time line below
Figure 2.8.22
So, the present value is the sum of these terms.
Therefore,
PV =
1
2
n−1
n
+
+ ··· +
+
2
n−1
1 + i (1 + i)
(1 + i)
(1 + i)n
The trick to evaluating this is to multiply both sides by 1 + i to get
(1 + i)P V = 1 +
2
3
n
+
+ ··· +
1 + i (1 + i)2
(1 + i)n−1
Subtracting the first equation from the second gives
iP V
=1+
2−1
3−2
n − (n − 1)
n
+
+
−
1+i
(1 + i)2
(1 + i)n−1
(1 + i)n
=
1+
= ä n i −
1
1
1
+
+
1 + i (1 + i)2 (1 + i)n−1
−
n
(1 + i)n
n
(1 + i)n
where the last line come from the observation that the first collection of terms is the sum
of an annuity due with n payments.
This yields,
PV =
ä n i −
n
(1+i)n
i
=
ä n i − n(1 + i)−n
.
i
Actuaries use the notation (Ia) n i to denote the present value of this increasing annuity.
Therefore we have
Formula 8
(Ia) n i =
ä n i − n(1 + i)−n
i
48
Chapter 2
Annuities
which is the present value of an annuity with n payments starting at $1 and incresing by
$1 each period.
We can use this formula to value more general increasing annuities, as shown in the following
example.
Example 13
Consider an increasing annuity that pays $1000 at the end of this year. At the end of each
subsequent year, the annuity will pay an additional $100 for the next 10 years. The final
payment will be $2000 at the end of year 11. What is the present value of this annuity?
Assume i(1) = 10%.
Solution: As usual, a time line can be helpful.
Figure 2.8.23
The idea is to split the cash flows into the 2 parts: an initial constant annuity plus the
increasing part.
Figure 2.8.24
So, the top annuity in blue is just an ordinary simple annuity of $1000, starting at time 1.
Its present value at time 0 is given by
P V0 = $1000 × a 11 10%
The lower annuity in red is an annuity of $100 growing by $100 each period, all starting at
time 2. Remembering that the annuity formulas provide the present value 1 period before
the first cash flow, we have the present value of this annuity at time 1 is given by
P V1 = $100 × (Ia) 10 10%
Therefore, the total present value is the sum of the first term, plus the second term discounted back 1 period, yeilding
Section 2.8
Increasing Annuities
PV
49
= $1000 × a 11 10% + $100 × (Ia) 10 10% × (1 + 10%)−1
ä 10 10% − 10(1 + 10%)−10
= $6, 495.06 + $100 ×
× (1 + 10%)−1
10%
= $6, 495.06 + $2, 639.63
= $9, 134.69
There is actually an alternative way to decompose the cash flows in the last example.
Suppose instead we do the following
Figure 2.8.25
In this decomposition, we recognize the cash flows as the sum of a regular ordinary annuity
of $900 starting at time 1, plus an increasing annuity of $100 also starting at time 1. This
yields:
PV
= $900 × a 11 10% + $100 × (Ia) 11 10%
ä 11 10% − 11(1 + 10%)−11
= $5, 845.55 + $100 ×
10%
= $6, 495.06 + $3, 289.13
= $9, 134.69
which is the same value as before
2.8.2
Accumulated Values
The previous examples all involved the computation of the present value of increasing
annuities.
Let’s now consider the accumulated value. Again let’s consider the stream of n payments of
$1 increasing by $1 each period. We let (Is) n i denote the accumulated value of this annuity
on the date of the last payment.
Figure 2.8.26
50
Chapter 2
Annuities
We could develop a formula for solve for this value from first principles; however, there is
an easier way.
In fact we can see that the accumulated value is just the present value moved forward n
periods.
More precisely,
(Is) n i = (Ia) n i × (1 + i)n
=
ä n i − n(1 + i)−n
× (1 + i)n
i
ä n i × (1 + i)n − n
i
s̈ n i − n
=
i
=
=
s n+1 i − (n + 1)
i
These simplifications use the fact that ä n i × (1 + i)n = s̈ n i and that s̈ n i = s n+1 i − 1 . See
if you prove that these are in fact true.
So, in summary, the accumulated value of an annuity of n payments starting at $1 and
increasing by $1 each period is
Formula 9
(Is) n i =
s n+1 i − (n + 1)
i
Again, it is not recommend that students try to memorize these formuals. It is understanding the ideas behind growing annuities that is more valuable.
Chapter 3
Loans and Debts
3.1
3.1.1
Repayment of Debts
Debts
In this section, we will explore some additional practical issues surrounding the payment of
debts.
When making payments on a loan, the payments are essentially of 2 parts: the first part
pays the interest charges on the loan, and the second part of the payment is used to reduce
the principal. In this case was say that the principal balance is being amortized over the
life of the loan.
In many ways, these problems are quite similar to what we have already seen. Let’s do an
example.
Example 1
Suppose John borrows $100,000 at an interest rate of i(12) = 12%, and he pays off the loan
over 10 years. What are the monthly payments?
Solution: This looks familiar. Note that the monthly interest rate here is
the monthly payment is R, we have
i(12)
12
= 1%. If
$100, 000 = R × a 120 1% .
Solving gives R = $1, 434.709484. Clearly, however, we cannot make a payment of fractions
of 1 cent.
Therefore, we typically round the payment up, say to the nearest cent. Thus, John actually
pays $1, 434.71.
In this case, John is slightly overpaying on the loan. Consequently, the final payment will be
lower than the regular payment. So, John will make only 119 regular payments of $1,434.71
and one final payment that is lower.
More precisely, let X be the size of the final payment. To solve for X we note that the value
of the loan is the present value of the payments, which consist of 119 payments of $1,434.71
plus the one extra payment of X at the end of the loan at time 120. Therefore we have
51
52
Chapter 3
Loans and Debts
$100, 000 = $1, 434.71 × a 119 1% + X × (1 + 1%)−120 .
Solving for X gives the final payment of X = $1, 434.59 (rounding off to the nearest cent).
So this final payment is indeed slightly lower.
3.1.2
Amortization Schedules
The way we record the interest charges, the receipt of payments and the adjusted loan
balance is to construct an amortization schedule. The amortization schedule is a table
of values showing the remaining balance on the loan and records the new interest charges
and payments made on the loan.
Let’s build the amortization schedule for John
Payment Number
1
2
3
..
.
Opening Balance
$100,000.00
99,565.29
99,126.23
Interest
$1,000.00
996.65
Payment
$1,434.71
1,434.71
1,434.71
Closing balance
$99,565.29
99,126.23
98,681.79
119
120
2,826.83
1,420.39
28.27
14.20
1,434.71
1,434.59
1,420.39
0
Table 3.1
How was this schedule built? Here are the key points:
• Each row corrresponds to a payment period on the loan. Since there are 120 payments,
there are 120 rows in this amortization table.
• The opening balance of one row (which corresponds to the balance at the beginning
of the period) is the closing balance of the previous row (which corresponds to the
balance at the end of the previous period)
• The interest charge on the loan during a period is simply the opening balance times
the interest rate, which here is 1%.
• The payments are fixed at $1,434,71, except for the last payment that we already
determined was slightly lower at $1,434.59
• The closing balance is the opening balance plus the interest charged less the payments
made.
3.1.3
General Amortization Schedules
Given a general amortization schedule (and by general we mean without doing any rounding of fractional cents), we can actually determine the values in each row without needing
to construct the entire table.
For example, let A be the initial principal balance, and let R be the periodic payment,
assuming n payments under an interest rate i. The we know
Section 3.2
Refinancing Debts
53
A = Ra n i .
Now let Ak be the remaining principal balance after k payments are made. What is Ak ?
Ak is the present value of the remaining payments, therefore
Ak = Ra n−k i .
The interest component paid as part of the k th payment is given by taking the amount still
owing on the loan, Ak times the interest rate. Therefore we have
Ik
= i × Ak−1
1 − (1 + i)−(n−k+1)
=i×R×
i
= R × 1 − (1 + i)−(n−k+1)
We also know that the principal payment paid as part of the k th payment is the difference
between the total payment less the interest component. This is given by
Pk
= R − Ik
= R − R × 1 − (1 + i)−(n−k+1)
= R(1 + i)−(n−k+1)
It is not recommended that students memorize these formulas. Instead it is better to learn
the concepts, and either compute the amounts explicitly, or use a formula sheet.
3.2
3.2.1
Refinancing Debts
Debt Refinancing
It many instances, a borrower needs to refinance a loan at new terms before the original
amortization period is complete. In this section, we will look at how to calculate the new
payment amounts.
This is easiest to see with an example.
Example 2
John borrowed $20,000 to purchase a new car at an interest rate of i(12) = 6% amortized
over 5 years, paid monthly. Unfortunately, after making 20 payments on time, John fell ill
and was forced to miss 5 consecutive payments. The bank agreed to allow him to restart
making payments, but they would not agree to extend the original amortization period
(thus the loan must be fully paid under the new terms on the same date it would have been
under the original terms). What was his original monthly payment, and what will it be
under the new terms?
Solution: Calculating the original payments is straightforward:
54
Chapter 3
Loans and Debts
$20, 000 = R × a 5×12 6%
12
Solving yields R = $386.6560. Rounding this up to the nearest penny, gives R = $386.66.
Although it is not actually necessary, let’s calculate the final irregular payment, X. Since
the loan is monthly for 5 years, there are a total of 5 × 12 = 60 payments, but only 59 of
which are regular payments of $386.66. The final payment of X is at time 60. Taking the
present value of these payments yields
$20, 000 = $386.66 × a 5×12−1 6%
12
6%
+X 1+
12
−(5×12)
.
Therefore X = $386.38.
In order to calculate the new terms, we need to know the outstanding balance when the
loan is renegotiated, which will be after 25 months. However, John only made 20 payments,
so first we need to find the balance at the time of the last payment. Here is a time line of
John’s payments. The value of the new payments is denoted by R∗
Figure 3.2.1
The balance at time 20, B20 , is the accumulated value of the loan, less the accumulated
value of the payments. This is given by
6% 20
− $386.66 × s 20 6%
B20 = $20, 000 1 +
12
12
= $13, 986.13
Carrying this forward 5 periods yields the balance at time 25, B25 ,
6% 5
B25 = B20 1 +
12 6% 5
= $13, 986.13 × 1 +
12
= $14, 339.29
Taking our focal date to be at time 25, the new payments look like a regular loan with a
loan amount of $14,339.29. This loan must be paid off in 35 periods to ensure it is full paid
Section 3.3
Mortgages
55
at time 60 (which is what the bank has agreed to provide). Now, the new payments then
satisfy
$14, 339.29 = R∗ × a 35 6% .
12
Solving gives R∗ = $447.61.
The solution was just used is called the retrospective method in that we look at the past
payments in order to calculate the outstanding balance.
An alternative method, called the prospective method instead looks at the normal payments to come in the future in order to find the outstanding balance.
Let’s redo the previous question, but using the prospective method.
Here is the original time line of payments we expected John to make. (Note the final small
payment of $386.38 at time 60 we calculated earlier due to rounding.)
Figure 3.2.2
So B20 must equal the present value of the future loan payments. Note that there are
39 more regular payments of $386.66 plus the one final payment of $386.38 in 40 periods.
Therefore,
6% −40
B20 = $386.66 × a 39 6% + $386.38 1 +
12
12
= $13, 986.13
This is exactly the same as the value we found before. The rest of the solution is identical.
In summary, when calculating a balance on a loan at some point in the middle of a loan,
we can either look backwards at the payments already made, or we can look forwards to
the payments yet to come.
3.3
Mortgages
A mortgage is a special type of loan that is secured by real estate. When an individual
needs to borrow money in order to buy a home, the easiest way is to get a mortgage loan
from a bank. Should the borrow not be able to repay the loan, the bank can then seize the
56
Chapter 3
Loans and Debts
home and sell it in order to recover as much of the loan as possible. There are special laws
governing mortgages, but we de not need to get into the legal details here.
Essentially, mortgages are really just a special case of general loans and annuities. However,
given their importance in the marketplace, they deserve a section of their own.
In Canada, mortgage payments are typically made monthly, but the interest rate the banks
quote on mortgages is often a semi-annual rate. Thus, we need to convert the interest rate
before we do the final computations.
Let’s begin with an example, but first an important definition.
Definition 3.3.1
The amortization (or amortization period ) is the length of time it would take to pay
off the entire mortgage loan assuming no change in interest rates and no missed payments.
So, if a mortgage has an amortization of 20 years, it means that the borrower will need to
make payments for 20 years in order to fully repay the loan.
Example 3
Fred purchases a house and obtains a $200,000 mortgage. What are the monthly payments
if the amortization period is 25 years and the interest rate is i(2) = 6%?
Solution: First we convert the rate to a monthly rate to match the frequency of payments.
So,
6%
1+
2
2
=
i(12)
1+
12
!12
=⇒ i(12) = 5.9263464%.
Now for the payments. Since the initial loan amount is $200,000, this must equal the present
value of the payments. If we let P M T be the payment amount, we have
$200, 000 = P M T × a
(12)
25×12 i 12
.
Thus P M T = $1, 279.61.
Notice that there are 12 × 25 periods since payments are made monthly for 25 years, which
is the amortization period of the mortgage.
3.3.1
Term of a Mortgages
A mortgage typically has both a term and an amortization period. As we have seem, the
amortization period is the number of periods over which we calculate the payments. The
term is different. Typically, mortgages come with a shorter length of time that the interest
rate is guaranteed. Once the term ends, the mortgage is due. Typically, borrowers then
refinance the mortgage at whatever new interest rate is in effect at that time.
More precisely
Section 3.3
Mortgages
Definition 3.3.2
57
The term of a mortgage is the length of time that the terms and conditions of the mortgage
have legal effect. In most instances, the significance of this is that the interest rate is
essentially fixed for the term of the mortgage. At the end of the term, the remaining
mortgage balance must be repaid in full, or refinanced (typically at a new rate).
Confusing the term and the amortization period is a common source of errors for students
and home buyers.
Let’s do an example
Example 4
Recall our last example where Fred purchased a house and obtained a $200,000 mortgage
at an interest rate of i(2) = 6%. Assume the mortgage has a 5 year term and a 25 year
amortization. Assume that after the term expires, interest rates increase to i(2) = 10%.
What will the new monthly payments be?
Solution: We computed that the initial monthly payment will be $1,279.61. In order to
compute the new monthly payments, we need to know what the outstanding balance will
be after 5 years.
This balance must be the difference between the accumulated value of the loan, minus the
accumulated value of the payments, as shown in the time line below
Figure 3.3.3
So, the balance at time 60, B60 is given by
B60
!60
i(12)
− $1279.61 × s i(12)
= $200, 000 × 1 +
60 12
12
= $268, 783.28 − $89, 109.64
= $179, 673.63
Now that we have the balance, we can compute the new payments. Since we have been
paying the mortgage for 5 years, we would expect the new amortization period to now be
only 20 years, instead of 25. Thus
$179, 673.63 = P M T × a
(12)
20×12 i 12
.
However, interest rates have changed, so we need to compute the new value i(12) .
58
Chapter 3
Loans and Debts
Since the new rate is i(2) = 10%, we have
10% 2
1+
=
2
i(12)
1+
12
!12
=⇒ i(12) = 9.7978%.
Hence the new payment is
PMT
=
$179, 673.63
a
i(12)
20×12
12
= $1, 709.89
So the payments increased by more than $400 per month.
3.3.2
Down Payments
Typically lenders are not willing to lend 100% of the value of a house to a purchaser.
Instead, the purchaser must contribute some money, and this contribution is called the
down payment.
For example, a home buyer may want to purchase a $400,000 home. The mortgage lender,
however, is only willing to lend $300,000. In this case, the buyer is required to make a down
payment of $100,000 in order to purchase the home. In this case, we would say that the
$100, 000
purchaser made a down payment of
= 25%.
$400, 000
The typical size of down payments can vary quite a bit, and the minumum size is often
determined by government regulations, which often change. At the time of writing, the
minimum down payment in Canada on a home worth less than $500,000 was 5%, and on
a home with a value between $500,000 and $1 million, the minimum down payment is 5%
of the first $500,000 and 10% on the remaining value. However, home buyers often make
down payments in excess of this minimum.
3.4
Sinking Funds
In many instances, it is prudent to place funds aside in order to satisfy some future expected
obligation, such as paying the principal amount when a bond comes due, or in order to
purchase an asset.
Such a fund is called a sinking fund.
Here is an example
Example 5
The board of homeowners in a townhouse complex determined that the parking lot will need
to be rebuilt in 10 years. If the lot were rebuilt now, it would cost $100,000. In order to fund
this expense, the board has decided to make monthly deposits into a sinking fund. Assume
Section 3.4
Sinking Funds
59
the deposits are made into an account earning i(12) = 3%. Assume also that inflation over
the next 10 years is expected to be 1% per year. How much should the deposits be?
Solution: We need the accumulated value of the deposits to equal the expected cost of the
repairs. Since the cost of the repairs is $100,000 today and inflation is expected to be 1%,
we have that the expected cost in 10 years is $100, 000 × (1 + 1%)10 . If we let R be the
size of the deposits, we have
R × s 12×10 3% = $100, 000 × (1 + 1%)10 .
12
Solving yields R = $790.48.
3.4.1
Sinking Fund Schedules
In a manner analogous to an amortization schedule, we can construct a sinking fund
schedule to track how the funds accumulate to the desired amount.
Let’s construct the sinking fund schedule for the last example.
Deposit Number
1
2
3
Opening Balance
$0
790.48
1,582.94
Interest
$0
1.98
3.96
Deposit
$790.48
790.48
790.48
Closing Balance
$790.48
1,582.94
2,377.37
119
120
108,377.50
109,398.80
270.84
273.50
790.48
790.48
109,398.80
110.462.78
Table 3.2
How was this table built?
• As before, the opening balance for one period equals the closing balance of the previous
period.
• The interest earned is the opening balance times the interest rate, which is 0.25% per
month here.
• The deposits are fixed at $790.48
• The closing balance is the opening balance plus interest earned plus the deposit.
Note that $100, 000(1 + 1%)10 = $110, 462.21 which is the expected cost of the parking lot.
The difference between the expected cost of the parking lot and the last entry of the table
is rounding error.
Chapter 4
Bonds
4.1
4.1.1
4a. Bond Basics
Bond Features
Most governments and large corporation borrow money in order to fund their operations.
They are able to do this by issuing bonds to the public. Indeed, bonds make up a large part
of all government borrowings. Essentially, buy purchasing a bond, you are lending money
to the issuer.
A major difference between a bond and a simple loan is that a bond is typically marketable
meaning the initial purchaser of the bond is able to sell the bond prior to the maturity of
the loan. Bonds are traded between investors in the so-called bond markets which are
some of the largest and most important markets in finance.
The bond contract (also called the bond indenture) outlines the terms of the loan. The
terms include:
• The face value of the bond, which is typically, the par value of the note. In most
cases, this is the principal amount that is paid when the bond matures.
• The maturity date is the date when the bond matures, and the face value is paid
to the investor. Maturity dates for bonds vary from 30 days to 30 years or more from
the date of issue.
• The coupon rate, which is the percentage of the face value paid annually, essentially
as interest on the bond.
• The coupon frequency refers to how often a coupon payment is made. The most
common frequency is semi-annually, but other payment frequencies are possible.
• Some bonds have embedded options which can allow, for example, the issuer to repurchase the bond early (essentially this is an early repayment option); or options that
allow the holder to redeem the bond early (essentially an early withdrawal option).
Other more complex options also exist as part of bonds.
A simple bond with a fixed coupon rate, a fixed maturity date and no embedded options are
the main focus of the course. Such bonds are sometime called bullet bonds by specialists
in the field.
60
Section 4.1
4a. Bond Basics
61
Bond Pricing
A key computation for us will be to determine the price (or value) of a given bond.
A major principle in finance is the concept that the value of an investment should be equal
to the present value of the cash flows generated by that asset. So, to calculate the value,
we need to determine the cash flows from the bond. In the case of a regular coupon bond,
this is usually straightforward.
Here is a timeline of the cash flows from a typical bond paying a coupon of C with a face
value of F maturing in n periods. In this case, the bond is paying coupons at regular
intervals (essentially these are the interest payment) plus the face value of the bond at
maturity (essentially, this represents the repayment of principal.)
Figure 4.1.1
To compute the present value of these cash flows, we can observe that this stream of cash
flows is really a combination of an annuity of C dollars lasting n periods, followed by a
single cash flow of F dollars at time n. The present value is then
Value = C × a n i +
F
(1 + i)n
Note that the interest rate i used here is the “period rate”. Normally, bonds pay coupons
semi-annually, and therefore each “period” in the time line is half of a year. Therefore, the
period rate is half the annual stated rate. This rate that we use to discount the cash flows
gets a special name in the case of bonds
Definition 4.1.1
The (annualized) rate that we use to discount the cash flows of the bond is called the Yield
to Maturity (YTM) of the bond. Since bonds usually pay semi-annually, the YTM is
quoted with semi-annual compounding.
Therefore, the formula for the value of a coupon bond paying semi-annual coupons is
Formula 1
Value = C × a n Y T M +
2
where YTM is the yield to maturity of the bond.
Here is an example
F
(1 +
Y TM n
2 )
62
Chapter 4
Example 1
Bonds
Consider a 5% coupon Government of Canada 10 year bond with a YTM of 6%. Find the
value of this bond.
Solution: Since Government of Canada bonds pay coupons semi-annually, each coupon
payment (based on a $1000 face value) is given by
C=
1
× 5% × $1000 = $25.
2
Similarly, since the YTM is expressed at an annual nominal rate, the period rate (or six
month rate) is given by
r=
1
× 6% = 3%.
2
The value is therefore:
P
= C × a 20 3% +
F
(1 + 3%)20
= 25 × a 20 3% +
1000
(1 + 3%)20
= 371.936872 + 553.675754
= 925.612626
4.1.2
Bond Quotes
Bonds are normally quoted in the marketplace at a percentage of face value. Therefore, the
market quote for the bond in the last example would be
Quote =
925.612626
× 100 = 92.561263.
1000
Notice that we are expressing the quote to 6 decimal places. This is the convention in the
bond markets, although in this course we will usually use only 2 decimal places. Also, notice
that even though the quote is expressed as a percentage of face value, we do not put a %
symbol at the end. We would say “the bond is trading at 95” and not “the bond is trading
at 95%” when discussing bond quotes. Everyone in the market would know that the quote
of 95 means 95% of face value.
4.1.3
Premium and Discount Bonds
A bond is called a Par Bond (and we say the bond is trading at a par ) if the quoted price
is equal to 100 (called the par value).
A bond is called a Discount Bond (and we say the bond is trading at a discount) if the
quoted price is less than 100.
Section 4.1
4a. Bond Basics
63
A bond is called a Premium Bond (and we say the bond is trading at a premium) if the
quoted price is greater than 100.
In fact, a bond will trade at par exactly when the coupon rate equals the yield to maturity.
To show this, let’s consider a bond with a face value F paying coupons semi-annually with
a yield to maturity of YTM . Let i = Y T2M be the period rate. Since the coupon rate equals
the yield to maturity, we have the coupon amount is C = F × i . Therefore, using the
formula for the price of the bond, we have
P
= C × an Y T M +
2
= Fi ×
F
(1 +
Y TM n
2 )
1 − (1 + i)−n
i
+
F
(1 + i)n
=F
4.1.4
(Using the defn of a n i )
(Expand and simplify)
Alternative Pricing Formula
The last example motivates an alternative bond pricing formula.
Let r be the coupon rate (per period) and i the yield to maturity (per period), then using
our original pricing formula and doing some substitutions and simplifications, we have
P
F
(1 + i)n
= F r × a n i + F (1 + i)−n
= F r × a n i + F (1 − ia n i )
= F + F (r − i)a n i
= C × an i +
(Since the coupon rate is r)
(Using the defn of a n i )
(expand and collect terms)
Therefore, we are lead to the alternate formula for the price of a bond.
Formula 2
P = F + F (r − i)a n i
From this expression, it is easy to see that a bond trades at a premium exactly when r > i
and a discount when r < i.
Let’s do the first example again using this alternative formula
Example 2
Recall the 5% coupon Government of Canada 10 year bond with a YTM of 6%. Find the
value of this bond using the alternative formula.
Solution: The coupon is paid semi-annually, so the semi-annual rate is r = 5%
2 = 2.5%
6%
, and the semi-annual YTM is i = 2 = 3% . A 10 year bond has 10 × 2 = 20 coupon
periods, so therefore we have (based on $1000 face value)
64
Chapter 4
P
Bonds
= F + F (r − i)a n i
= 1000 + 1000(2.5% − 3%)a 20 3%
= 1000 − 74.387374
= 925.612626
just as before.
4.2
Prices between Coupon Dates
4.2.1
Bond Price Between Coupon Dates
Recall our formula for the price of a coupon bond.
P = C × an i +
F
.
(1 + i)n
This formula only works if we are pricing the bond on a coupon date.
How do we find the price between coupon dates? We will address this question in this
section.
The Full Price of a bond (also called the dirty price or invoice price) is essentially the
value of the bond when discounted back to the settlement day.
This is easier to see with some examples.
Assume we are somewhere between time 0 (when the most recent coupon was paid) and
time 1 (when the next coupon payment will be made).
Figure 4.2.2
Let f represent the fraction of a coupon period that “now” is before the next coupon. So,
more precisely
f=
# of days from the last coupon payment to the settlement date
.
# days in coupon period
The full price of this bond at time “now” equals all the cash flows of the bond discounted
back to now. For example, the first cash flow at time 1 neads to be discounted back the
fraction 1 − f of a period. If we assume a period interest rate of i, its present value is
Section 4.2
Prices between Coupon Dates
65
PV =
C
(1 + i)(1−f )
The second cash flow must be discounted back a full period plus the fraction 1 − f of a
period, for a total of 1 + (1 − f ) = 2 − f periods. Therefore, its present value is given by
PV =
C
(1 + i)(2−f )
Continuing this for each of the cash flows and adding them up gives a total present value
now to be
F Pnow =
C
C
C +F
+
+ ···
(1−f
)
(2−f
)
(1 + i)
(1 + i)
(1 + i)(n−f )
This is a rather ugly sum to work with. However, if we factor out (1 + i)f we get
f
F Pnow = (1 + i) ×
C
C
C +F
+
+ ···
1
2
(1 + i)
(1 + i)
(1 + i)n
The second factor should look familiar: it is the price of a coupon bond on a coupon date
with n periods remaining.
In fact, it is the price of the bond, if we were to value it at time 0. The expression above
shows that the price now can be computed by find the price at time zero and moving that
value forward in time by the fraction f of a period. If we let P0 be the price at time 0, we
have the simplier formula for the full price of a bond.
Formula 3
F Pnow = (1 + i)f × P0 .
Figure 4.2.3
Here is an example
Example 3
Find the full price of a 5% bond due June 30, 2022 with a YTM of 4%. Assume the
settlement day is May 18, 2018 and that the bond pays semi-annual coupons.
Solution: We begin by valuing the bond on the previous coupon date, which is December
31, 2017. Call that value P0 . How do we know that the coupon date was December 31,
66
Chapter 4
Bonds
2017? Since the bond pays coupons semi-annually and it matures on June 30, its coupon
dates must be each June 30 and December 31 (six months later).
Note that there are 9 coupon periods from December 31, 2017 to June 30, 2022. Therefore,
the price of the bond on December 31, 2017 is
P0 = 25 × a 9 4% + 2
1000
1+
4%
2
9 .
= 1040.81
To get the full price on the settlement day we need to count the number of days from
December 31, 2017 to May 18, 2018 (there are 138) as well as the total number of days in
the coupon period from December 31, 2017 to June 30, 2018 (there are 181).
Therefore the fraction f is f = 138
181 . Therefore, the full price today is the full price P0
carried forward this fraction of a period. Thus
F P = (1 + 2%)(138/181) × P0 = 1056.64.
4.2.2
Clean Prices
Bonds are usually quoted in the market at their clean price which is the difference between
the full price and the so called accrued interest.
The accrued interest is the amount of the next coupon payment that accrues to the seller.
This is calculated on a simple interest basis.
More precisely
AI =
# of days from the last coupon payment to the settlement date
×C
# days in coupon period
=f ×C
Using the data from the last example we have AI =
138
181
× 25 = 19.06.
The clean price of a bond (also called the quoted price) is the difference between the
full price and the accrued interest. Or simply
CP = F P − AI.
Again, using the data from the last example, the clean price of the bond in the last example
is given by
Section 4.3
Bond Amortization
67
CP = F P − AI
= 1056.644 − 19.060
= 1037.584
It is convention that the clean price is expressed as a presentage of face value, so here the
clean price would be quoted at 103.7584
You may wonder why we bother with Accrued Interest at all, since it is the Full Price that
is first calculated and billed anyway? There are a few reasons:
• The Full Price increases with the passage time, so it is harder to quickly differentiate
a change in price due to time, or due to real market forces. (We will explore this a
bit more later.)
• Accrued Interest is involved in tax calculations and accounting treatment. (We will
not cover these topics in this course.)
Whether we like it or not, the convention in the bond market is that quotes are always
Clean Prices. So, if someone says they are willing to sell you a bond at 99, they mean at
a clean price of 99. You will then have to add the accrued interest in order to find the full
price (which is what you will actually pay.)
4.2.3
Day Count Conventions
Day count conventions are very important and are different in different markets. The
last example assumed an Actual/Actual day count convention. That is to say, when we
computed the fraction f we counted actual number of days in the relavent periods.
This is not the only convention, however. US corporate bonds and money market instruments use a 30/360 convention; meaning each month is assumed to have 30 days, and the
year is assumed to have 360 days. Most international bonds use Actual/Actual, although
there are others, such as Actual/365, which is the convention in the Canadian Bond market.
In this course, we will use the Actual/Actual day count convention
4.3
4.3.1
Bond Amortization
Amortization of Bond Premium and Discount
Recall the alternative bond pricing formula
Formula 4
P = F + F (r − i)a n i ,
68
Chapter 4
Bonds
where r is the (period) coupon rate and i the (period) yield to maturity. (Again, by “period”
we mean that if the bond pays semi-annual coupons then the rates in the above formula
would be a 6 month rate correstponding to the coupon period.)
So, the bond trades at a premium exactly when r > i and a discount when r < i, and
furthermore, the quantity
F (r − i)a n i
is the size of the premium or discount.
Notice that as time passes, the number of periods until the bond matures, n, declines,
hence the size of the premium or discount also declines. This represents what we call the
amortization of premium or discount. So, over time, the size of the premium or
discount is amortized away until it reaches zero at maturity of the bond.
4.3.2
Amortization Tables
We can construct a bond amortization schedule in a manner exactly analogous to a loan
amortization schedule from previous chapters.
As an example, let’s construct the bond amortization schedule for a 10 year bond with a
10% coupon trading with a yield to maturity of 5%. Assume a $1000 face value and the
coupon payments are made semi-annually.
The first step is to find the price of the bond, which is given by
P = 1000 + 1000
10% 5%
−
2
2
a 20 5%
2
= 1389.72906
Now for the schedule.
Coupon Number
1
2
3
..
.
Opening Value
$1,389.73
1,374.47
1,358.83
Interest
$34.74
34.36
33.97
Coupon
$50.00
50.00
50.00
Closing Value
$1,374.47
1,358.83
1,342.81
19
20
1,048.19
1,024.39
26.20
25.61
50.00
50.00
1,024.39
1,000.00
Table 4.1
How was this table constructed? Once we work through this carefully, you will see that
the exact same principles are used here in the bond amortization schedule as we used in
the loan amortization schedules in an earlier chapter. Indeed, remembering that bonds are
really a type of loan is a very helpful concept here.
The opening value of the bond is just the purchase price. Since the bond is earning a yield
to maturity of 5% annually, the six month interest payment is half that, or 2.5% on the
balance. Therefore the first interest amount is $1, 389.73 × 2.5% = 34.74 . Each period,
the bond pays a fixed coupon. Since the coupon rate is 10% annually, the size of the six
Section 4.3
Bond Amortization
69
month coupon is $1, 000 × 5% = $50.00 . Note that the coupon payment is on the face
value of $1,000. It is not calculated on the current balance. The final value is the initial
value plus interest charged, less the coupon payment received. (You can see how thinking
about bonds just as if they were loans is very helpful here. At the beginning of the first
period, the borrower owed $1,389.73 on a loan with an interest rate of 5% annually. The
first interest charge is $34.74 but the borrower made a payment of $50.00. Therefore at the
end of the period, the new loan balance should be the old loan balance plus the interest
charge minus the payments made. This is exactly the calculation in the table.)
This continues for 20 periods. There are 20 periods since this is a 10 year bond with 2
coupons per year for a total of 20 coupons. Notice that the final balance is exactly $1,000
which is the face value of the bond. This is the amount that will be paid back when the
bond expires at time 20.
4.3.3
Graph of Bond Value Over Time
We can capure the effect of this amortization graphically by graphing the value of a bond
as time passes.
Here is a graph of the value of a premium bond as time passes to maturity. When the bond
is initially purchased, it is worth more than par (that is what it means to be a premium
bond), but as time passes, its value falls as the premium is amortized away. The premium
is finally amortized all the way to zero at maturity, at which point the value of the bond
equals its face value or par value.
Figure 4.3.4
A similar effect occurs with discount bonds. Discount bonds are initially valued below
par, but in this case the amount of the discount is amortized away until it reaches zero at
maturity. At that point the bond is again worth the face or par value.
70
Chapter 4
Bonds
Figure 4.3.5
This effect is often called pulled to par, meaning that as time passes, the value of a bond
(either a premium or discount bond) is pulled closer to the par value, which it will reach at
maturity.
4.4
4.4.1
Yield to Maturity
Finding the Yield to Maturity
In all the previous examples, we have provided the yield to maturity and then computed
the bond’s price.
In practice, it is the price that is negotiated between the buyer and seller, with the yield to
maturity being calculated after the price is determined.
Here is a typical example.
Example 4
A 10 year 6% coupon bond is purchased at 105. Compute its yield to maturity.
Solution: To find the yield to maturity, we need to solve for i in the expression
105 = $3 × a 20 i +
100
.
(1 + i)20
Unfortunatley, we cannot solve for i in this expression algebraically. instead, we need to
use numerical methods. I used Excel (see Excel Demontration below) to give the estimate
i ' 5.348%.
Notice that the bond is at a premium so we know the YTM must be less than the coupon
rate. So a number less that 6% was expected.
Section 4.5
Other Bonds
4.4.2
71
Method of Averages
One approximation method that can be useful to estimate the yield to maturity, is the so
called method of averages that says an approximate value for the yield to maturity can
be given by
i'
the average interest payment
.
the average amount invested
Let’s do the last example again, but we will use the method of averages to estimate the
YTM.
Example 5
A 10 year 6% coupon bond is purchased at 105. Estimate the yield to maturity.
Solution: Let’s assume a $100 face value. Since the bond pays a coupon of 6%, we expect
to receive $6 per year for 10 years for a total of $60. However, we purchase the bond at 105
and have it mature at 100, so we lose $5 at maturity. Thus, the total interest received is
$60 − $5 = $55.
Thus, on average, we receive
$55
= $5.50
10
per year.
Now, we initially invested $105 to buy the bond, but it has amortized down to $100 at
maturity. Thus, our average amount invested is
$105 + $100
= $102.50.
2
So, our approximate value for i is
i'
$5.50
= 5.37%.
$102.50
Recall that Excel gave the value 5.348%, so this estimate is reasonably good.
4.5
4.5.1
Other Bonds
Strip Bonds
Bonds that have a coupon rate of zero, are called Zero Coupon Bonds. They are particularly easy to price, since the price is just given by
P =
F
,
(1 + i)n
72
Chapter 4
Bonds
where i is the period YTM and the n is the number of periods from today that the bond
expires.
Note the the above expression is just the present value of the future cash flows, which in
the case of a zero coupon bond, is just the face value at maturity in n periods.
Here is an example.
Example 6
Find the price of a 7 year zero coupon bond trading with a yield of i(2) = 5%. Assume a
face value of $1000.
Solution: To find the price, we solve
$1000
P =
= $707.73.
5% 14
1+
2
Note that since we are given the YTM in semi-annual compounding, we have 7 × 2 = 14
5%
total periods, with a period (or six month) YTM of i =
.
2
The most common way zero coupon bonds arise in practice is when a regular coupon bond
is stripped of its coupons and the components are traded separately. The residual principle
component is then effectively a zero coupon bond, called a strip bond. Indeed the word
“strip” comes from the phrase Separate Trading of Interest and Principal, where we think
of the principal part and the coupons all as separate things.
4.5.2
Callable Bonds
Some bonds have an embedded option in them which allows the issue to ability to repurchase, or call, a bond before maturity. Such bonds are called callable bonds. If we think
of a bond as a loan, then the call feature is really a early repayment option. Essentially the
issuer (the borrower) is able to repay the loan earlier than the original maturity date.
These bonds are difficult to value, since the cash flows are uncertain: we don’t know if or
when the bonds may be called, so a simple present value calculation won’t work. However,
we can do some analysis of such bonds.
In particular, we can compute the yield to call, which is the yield to maturity we would
receive if the bond were to be called.
Example 7
A 10 year 5% coupon bond is issued at 101. If the bond is callable by the issuer after 5
years at 102, find the yield to call.
Solution: In order to find the yield to call, we simply compute the yield assuming the
bond is in fact called at 102. In this case, we need to find i such that
101 = $2.50 × a 10 i +
2
102
.
(1 + 2i )10
Section 4.5
Other Bonds
73
Notice that the numerator in the last term is 102, not 100, since if the bond is called, we
receive 102 on the bond, instead of the usual 100 at maturity. Using Excel, we find
i ' 5.127%.
Notice that if the bond is not called, the yield to maturity satisfies
101 = $2.50 × a 20 i +
2
100
.
(1 + 2i )20
Thus i ' 4.873%.
For bonds that have more than one call date, we can find a yield to call for each date. The
yield to worst is the lowest of all the yield to call amounts and the yield to maturity on
the bond.
So, in the last example, the yield to worst is in fact the yield to maturity of 4.873%
This yield to worst is essentially the lowest yield an investor would expect to receive on the
bond, regardless of if or when it is called.
Chapter 5
Investment Analysis
5.1
Net Present Value
Businesses are routinely deciding whether or not to make an investment in a new project.
This decision making is often called capital budgeting as the process involves determining
which projects are worth investing the firm’s capital in.
A popular method to evaluate projects is with the net present value method, which is
determined as follows:
N P V = −Cost + P V (future cash flows),
where we compute the present values using the firm’s cost of capital.
The idea behind this method is as follows: the final term – PV(future cash flows) – can be
thought of as the financial value of the project (remember, the value of an asset is simply
the PV of the future cash flows generated by that asset). From this we subtract the cost.
A good project is one where its cost is less than its financial value (i.e. we got the project
“cheap”). A bad project is one where the cost is higher than the finncial value. In that
case, we overpaid for the project and distroyed value by investing.
So, in summary, a project should be accepted if N P V > 0 and rejected if N P V < 0 .
Here is an example
Example 1
Suppose a company is considering a new plant. The plant will cost $10 million to build. If
the plant will generate cash flows of $1.7 million per year for 10 years, calculate the NPV
assuming the cost of capital is i(1) = 10%.
Solution We calculate:
N P V = −10, 000, 000 + 1, 700, 000 × a 10 10%
= −10, 000, 000 + 10, 445, 764
= $445, 764.
Since the N P V > 0 we accept the project
74
Section 5.2
Internal Rate of Return
75
Notice that if we compute the NPV of the project in the last example where the cost of
capital is 12% we get:
N P V = −10, 000, 000 + 1, 700, 000 × a 10 12%
= −10, 000, 000 + 9, 605, 379
= −$394, 621.
In this case, the project fails to generate its cost of capital, so we should reject this project.
What do we mean by the cost of capital ? The idea is that investors in a company expect
a return on their investment in the firm. The cost of capital captures that return that
investors demand for investing. Therefore, the company needs to select projects that earn
at least the cost of capital. Positive NPV projects earn more than the cost of capital, while
negative NPV projects earn less.
In practice, analysis use spreadsheets to compute the cash flows associated to a complex
project, then they discount the cash flows to compute the NPV.
5.2
Internal Rate of Return
As we saw in the last section, the NPV of a project is dependent on the cost of capital used.
The internal rate of return of a project is defined to be the cost of capital such that the
NPV of a project is zero.
Let’s do an example:
Example 2
Find the IRR of a project assuming the initial cost is $25,000 and the cash flows are as
follows:
Year
1
2
3
CashFlow
$10,000
$9,000
$8,000
Table 5.1
Solution: We need to find i such that
NPV = 0
Thus
0 = −$25, 000 +
$10, 000
$9, 000
$8, 000
+
+
.
2
(1 + i)
(1 + i)
(1 + i)3
We cannot solve this explicitly in general. Instead, we use numerical tools such as Excel in
order to approximate the solution. (Much like the yield to maturity of a bond.)
Using Excel, I found
76
Chapter 5
Investment Analysis
i ' 4.16%.
Therefore, the IRR for this project is 4.16%
The idea behind this result is that this project returns about 4.16%. If our cost of capital
is more than this, we would reject this project, and if the cost of capital is below this, we
would accept it.
Here is a graph of the NPV of the project as the cost of capital changes. From the graph,
we can see the regions where the NPV is positive and where it is negative.
Figure 5.2.1
5.2.1
Problems with IRR
Although popular, the IRR methodology does have some problems. Consider the project
having an initial cost of $10,000 with cash flows:
Year
1
2
3
Cash Flow
$5,000
$55,000
-$55,000
Table 5.2
Then we can show that the IRR is both 16.3% and 86.8%. What’s happening?
The NPV calculation boils down to solving the polynomial
0 = −C + CF1 x + CF2 x2 + · · · + CFn xn ,
Section 5.2
Internal Rate of Return
where x =
1
1+i
77
and CFi is the cash flow at time i.
Descartes’ Rule of Signs states that the number of positive roots of a polynomial cannot
exceed the number of sign changes of the coefficients. So, a typical project has terms like
− + + + + (where the first cash flow – the cost – is negative, and the future cash flows are
all positive). Such projects have a unique IRR since the cash flow pattern has only 1 sign
change.
However, the project in our last example has cash flows in the pattern − + +− (so the are
2 changes of sign). Consequently, the project has 2 IRRs. Essentially, the polynomial has
multiple positive roots, as shown in the graph of NPV vs cost of capital below
Figure 5.2.2
Here is another example to highlight another problem with the IRR decision rule. This
problem relates to the size and timing of the cash flows of different projects.
Example 3
Consider 2 projects, E and L, with cash flows shown below. Assume both projects cost
$15,000. For each project, calculate (a) the IRR, (b) The NPV if rates are 3%, and (c) the
NPV if rates are 8%.
Year
1
2
3
E
$10,000
5,000
2,000
L
$3,000
5,000
10,000
Table 5.3
Solution: Using Excel, I calculated the IRR for each project and found
IRRE = 8.69%
and IRRL = 8.01.%
78
Chapter 5
Investment Analysis
Since Project E has the higher IRR, it looks like the better project.
Also, at 3% we have
N P VE = $1, 252
and N P VL = $1, 777.
In this case, we see that Project L is better.
Finally, at 8% we have
N P VE = $134
and N P VL = $3.
In this case, Project E is better.
What’s happening? Why does the decision change as the cost of capital change?
Here is a graph of the NPV of the projects as the cost of capital changes.
Figure 5.2.3
At a rate of 6.9%, both projects have the same NPV (called the crossover rate). Above
that rate, Project E is better, and below that rate Project L is better.
So we can see that the IRR rule does not always provide the best measure of which project
we should accept.
Section 5.3
Stocks
79
A subtle problem with the IRR rule is that it assumes that the cash flows can be reinvested
at the IRR, which is typlically not a good assumption. (There is no reason we should expect
another project to be available with the same IRR that we can reinvest the cash flows in.)
The NPV rule assumes cash flows can be reinvested at the cost of capital, which is generally
a better assumption.
5.3
5.3.1
Stocks
Stock Valuation
All of the material we have learnt so far can be applied to stock valuation. In this section,
we will apply our present value results to the valuation of common and preferred shares.
Companies that require money from investors have 2 main choices of funding for long term
projects: they can borrow money, buy obtaining a bank loan or by issuing a bond; or they
can issue stock (or shares) to the public.
When companies issue stock, they are essentially selling a share of the company to investors.
Investors who then purchase stock in a company become part owners in the company.
Obviously, such investors hope the company earns substantial profits which the investors
can then share in.
There are 2 main classes of shares that a company can issue: common shares and preferred shares.
The major differences between the 2 classes are:
• Common shareholders usually get to vote on major corporate decisions. They get
to choose who will serve on the company’s Board of Directors, which is the most
senior management board of the company. By selecting the Board members, common
shareholders can ensure that the company is run according to their wishes. Perferred
shareholders, on the other hand, typically do not get to vote on any corporate matter.
• Preferred shareholders are usually granted a fixed dividend on their shares. So for example, a preferred share may pay a fixed dividend of $1.00 per share paid in quarterly
instalments of $0.25. (It is the convention in North Americal that most dividends
are paid in quarterly instalments. Different global regions have different conventions,
however.) In the case of common shares, the common shareholders are the owners of
the company, so their earnings are essentially the net profits of the company, which
can vary from year to year. Sometimes it is said that the common shareholders earn
the residual cash flow of the firm. We call it residual since the common shareholders
get to keep whatever is left after all the bondholders, other lenders, and preferred
shareholders get paid. In this sense, common shareholders are at the bottom of the
priority list.
Given that preferred shareholders are typically allocated a fixed dividend, a purchaser of a
preferred share looks very much like a purchaser of a bond. If we buy a bond we expect
to receive fixed coupon payments. If we buy preferred share we expect to receive fixed
dividends. In fact these are actually quite similar, and as a result preferred shares are often
called hybrid securities meaning that they have characteristics similar to both bonds and
stocks. Some of the key differences between preferred shares and bonds include:
80
Chapter 5
Investment Analysis
• There can be tax differences between the receipt of a dividend and the receipt of a
coupon payment on a bond.
• If the company runs out of money and cannot pay the coupon payments on a bond,
the company is essentially bankrupt and the bondholders can force a liquidation of
the company in order to recover as much money as possible. A preferred share on the
other hand is not legally a loan. So, should the company run out of money and not
be able to pay the dividend, the preferred shareholders cannot force the company into
bankruptcy or liquidation in order to recover their money.
We will not focus on these advanced tax and legal distinctions in this course. Instead we
will focus primarily on valuation. As we have seen valuation of any asset, including a stock,
boils down to calculating the present value of the future cash flows generated by that asset.
This is the approach we will take here.
5.3.2
Preferred Shares
Preferred shares typically pay a fixed dividend at regular intervals forever. So we can
recognize the cash flows from a perferred share as a perpituity.
Recall the formula for the present value of a perpituity that pays C dollars per period under
a period interest rate i is given by
PV =
C
.
i
This is the key formula to value preferred shares.
Let’s do an example:
Example 4
How much would you pay for a preferred share that pays a dividend of $0.75 each quarter?
Assume you want a return on your investment of i(4) = 10%.
Solution: Since a preferred share is typically a perpetuity, we recall the present value
formula for a perpituity to get
P =
5.3.3
$0.75
10%
4
= $30.
Common Shares
We can model the price of a common share in a similar way. However, the main difference
here is that common shareholders often expect the dividends on a common share to increase
as the company grows larger. Thus, the dividend stream is really a growing perpituity.
Here is an example:
Section 5.3
Stocks
Example 5
81
RBC stock currently pays a dividend of $0.90 per share per year. Dividends historically
increase at about 5% per year, which we expect to continue in perpituity. If investors wish
to earn 14% on RBC stock, what should the stock price be?
Solution: The present value of the cash flows is given by
P =
$0.90(1 + 5%)
$0.90
+ ··· .
+
1 + 14%
(1 + 14%)2
This is a geometric series with first term a =
The sum is then,
S=
$0.90
1 + 5%
and common ratio r =
.
1 + 14%
1 + 10%
a
1−r
$0.90
1
+ 14%
=
1 + 5%
1−
1 + 14%
=
$0.90
(1 + 14%) − (1 + 5%)
=
$0.90
14% − 5%
= $10.00
The solution to the last problem easily generalizes to what we call the dividend discount
model which says that the value of a stock paying an initial dividend of D growing at a
rate g is given by
Formula 1
P =
5.3.4
D
.
r−g
Dividend Yield
In actual practice, the value of a stock is determined by looking at the stock market to see
at what price a given stock is current trading. Since we know what the current dividend
amount on a stock is, we can compute the dividend yield as follows
Dividend Yield =
Annual Dividend Amount
.
Stock Price
82
Chapter 5
Example 6
Investment Analysis
A stock is trading at $40 per share. If the stock pays an annual dividend of $2.00, find the
dividend yield.
Solution: The dividend yield is
Dividend Yield =
$2.00
= 5.00%.
$40.00
Notice that if we assume the stock is a perpituity that pays the current dividend, the the
present value of the dividend stream taken at the dividend yeild gives is the current price.
Using the data from the last example and the formula for the present value of a perpituity,
we have
Price = P V =
$2.00
= $40.00
5%
which is exactly the market price.
5.4
5.4.1
Dollar Weighted Rate of Return
Rates of Return
When money is invested over several periods, it is always interesting to ask what the average
return or average rate of interest on the investment was.
However, as we will see, this is not always a simple question to answer.
In this section, we will consider two different methods of determining an average rate of
return. These methods are called
• Dollar weighted rate of return, and
• Time weighted rate of return.
5.4.2
Dollar Weighted Rates of Return
Consider this example.
Example 7
Suppose you deposit $1,000 into a fund now. After 1 year, you withdraw $100, and after
two years, you deposit another $200. If the fund is worth $1,200 after 3 years, what interest
rate did you earn?
Solution: Let’s draw the time line of the cashflows.
Section 5.4
Dollar Weighted Rate of Return
83
Figure 5.4.4
Let i be the interest rate earned on the investment. We have that the accumulated value
of the various cash flows is $1200. Therefore, the sum of the accumulated values of all the
cash flows must add up to $1,200. In particular we have
1, 200 = 1000(1 + i)3 − 100(1 + i)2 + 200(1 + i).
Typically, we cannot solve for i algebraically, but instead we need numerical tools. Solving
for with Excel yields
i ' 3.23%.
The is called the dollar weighted rate of interest.
5.4.3
An approximation for the dollar weighted rate of interest
We can quickly find an approximate value for i if we make the assumption that interest
accumulated vis simple interest.
Let’s do the last example again, using this simplifying assumption.
Example 8
Suppose you deposit $1,000 into a fund now. After 1 year, you withdraw $100, and after
two years, you deposit another $200. If the fund is worth $1,200 after 3 years, what interest
rate did you earn?
Solution: Here again is the timeline of cash flows, but now will will accumulate the values
using simple interest.
Figure 5.4.5
84
Chapter 5
Investment Analysis
So, in this case we have
1200 = 1000(1 + 3i) − 100(1 + 2i) + 200(1 + i)
This is now a much simplier expression to deal with. Expanding the brackets and simplifying
gives
3000i = 100
Therefore
i ' 3.33%
This compares favourably with the earlier (more precise) caclulation of i ' 3.23%
This approximation works as long as the time periods are not too long, and if the interest
rate is not too high. In these cases, we know that simple and compound interest are
reasonable approximations of each other.
5.5
5.5.1
Time-Weighted Rate of Return
Time Weighted Rates of Return
The second way to weight varying rates is with time weighted rate of interest.
The easiest way to see time weighted interest is to consider the return on a stock. Suppose
a stock is worth $100 per share today. After 1 year, the stock is worth $75 per share. After
two years, it is worth $150 per share.
So, an investor who purchases 1 share initially and held it for two years, has made a total
holding period return of 50%. This is easily annualized as
(1 + r)2 = 1.5
=⇒ r = 22.47%,
where r is the annualized compounded return. This is the time weighted interest rate.
Suppose now, a second investor also purchases one share initially for $100. However, this
investor will purchase another share for $75 after one year. What is this investors return?
Let’s consider the time line of the cash flows for this investor.
Section 5.5
Time-Weighted Rate of Return
85
Figure 5.5.6
This investor’s return satisfies
$300 = $100(1 + r)2 + $75(1 + r).
Solving this shows r = 39.61%.
Notice that this second answer is actually a dollar weighted rate of interest.
So, a time weighted rate of interest captures the performance of the stock (or the overall
market), while the dollar weighted rate of interest captures the performance of a particular
individual, who’s final returns is likely a function of when they added or withdrew cash.
Let’s do a final example to put all this together
Example 9
Suppose an investor begins with a $10,000 investment, and makes deposits and withdrawals
into the fund according to the table below. Suppose the value of the fund at each time is
also as indicated below. Find both the dollar weighted and time weighted rates of interest
for the fund.
Year
0
1
2
3
4
Deposits
Withdrawals
$1,000
$2,000
$3,000
Balance
$10,000
$12,000
$10,500
$7,000
$9,000
Table 5.4
Solution: For dollar weighted rate of interest, we solve for r where
$9, 000 = $10, 000(1 + r)4 + 1, 000(1 + r)3 − 2, 000(1 + r)2 − 3, 000(1 + r).
Solving with Excel yields r ' 7.37%.
For time weighted, we need to understand the performance of the market, independent of
the deposits and withdrawals the investors happened to make. This is easier to tackle on a
time line.
86
Chapter 5
Investment Analysis
Figure 5.5.7
Here, rk denotes the return on the fund over that specific time interval. So, for example,
the initial $10,000 grew at a rate r1 . At the end of the period, that accumulated value plus
the $1,000 additional deposit amounts to the new balance of $12,000. Therefore we have
10, 000(1 + r1 ) + 1, 000 = 12, 000
Solving gives r1 = 10%.
For the next period, we started with $12,000 that grew at a rate r2 . At the end of the
period, we withdrew $2,000 and finished with $10,500. In this case we have
12, 000(1 + r2 ) − 2, 000 = 10, 500
Solving this gives r2 = 4.167%.
Working with similar equations for r3 and r4 gives
10, 500(1 + r3 ) − 3000 = 7, 000 =⇒ r3 = −4.76%
and
7, 000(1 + r4 ) = 9, 000 =⇒ r4 = 28.57%
Remember that these values represent the return on the underlying investment in each of
the 4 years. If we were to invest $1 at these rates for 4 years, our $1 would accumulate to
(1 + r1 )(1 + r2 )(1 + r3 )(1 + r4 ) = 1.40298
If we let r be the average annual rate of return, then we have
(1 + r)4 = 1.40298
Solving gives r ' 8.83%. This is the time weighted rate of interest.
Recall that the dollar weighted rate of interest was only 7.37%.
What this says that the underlying investments earned 8.83%; however, our investor only
earned 7.37% due to the timing of the deposits and withdrawals into fund. What really
Section 5.5
Time-Weighted Rate of Return
87
happened was that our investor was withdrawing money from the investment before the
4th year. The 4th year was the year that the investment had its highest returns, and our
investor missed out on some of this due to the early withdrawals. Oh well. Better luck next
time.