Chapter 01 - Measurement of Interest

Section 1.1 - Introduction

Definition: Interest is the compensation that a borrower of capital

pays to a lender of capital for its use.

Definition: The principal is the amount of money initially borrowed.

This money accumulates over time. The difference between the
initial amount and the amount returned at the end of the period is
called interest.

Section 1.2 - Basics

Definition: The accumulation function, a(t), describes the

accumulated value at time t of initial investment of 1.

1-1
Properties of the accumulation function:
(1) a(0) = 1
(2) a(t) is generally increasing function of time.
(3) If interest accrues continuously then a(t) will be a continuous
function.

Definition: The amount function, A(t), gives the accumulated value

of an initial investment of k at time t, i.e.
A(t) = ka(t).
It follows that:
A(0) = k ,
and the interest earned during the nth period from the date of
investment is:

In = A(n) − A(n − 1) for n = 1, 2, · · · .

1-2
Section 1.3 - Rates of Interest

Definition: The effective rate of interest, i, is the amount that 1

invested at the beginning of the period will earn during the period
when the interest is paid at the end of the period.
That is,

i = a(1) − a(0) or
(1 + i) = a(1).

The quantity i is always a decimal value even though it is often

expressed as a percent, i.e.
6% interest i = .06
Note that
A(1) − A(0) I1
i = a(1) − a(0) = =
A(0) A(0)

1-3
The effective rate of interest is the interest earned in the period
divided by the principal at the beginning of the period.
Thus by extension,
A(n) − A(n − 1) In
in = =
A(n − 1) A(n − 1)
is the effective rate of interest during the nth period from the date of
investment. Also,

a(n) − a(n − 1)
in = for n = 1, 2, · · · .
a(n − 1)
Example: Consider the accumulation function

a(t) = (.05)t 2 + 1.
Here a(0) = 1 and
i1 = a(1) − a(0) = .05 + 1 − 1 = .05.
is the effective interest rate for period one.
1-4
For the nth period, the effective interest rate is:
a(n) − a(n − 1)
in =
a(n − 1)
(.05)(n)2 + 1 − [(.05)(n − 1)2 + 1]
=
(.05)(n − 1)2 + 1
(.10)n − (.05)
=
(.05)(n − 1)2 + 1

n in
1 .05
2 .143
3 .208

If $100 is invested, how much interest will be earned over three

periods?
Ans = 100[a(3) − a(0)] = 100[(.05)(3)2 + 1 − 1] = $45.

1-5
Section 1.4 - Simple Interest

Under simple interest,

a(t) = 1 + (n − 1)i for n−1≤t <n and n = 1, 2, 3, · · ·

The effective rate of interest for the nth period is

a(n) − a(n − 1)
in =
a(n − 1)
(1 + in) − (1 + i(n − 1))
=
1 + i(n − 1)
i
= .
1 + i(n − 1)
This is a decreasing function of n.
Why is it decreasing?

1-6
Answer: The same amount of interest is credited to the account for
each period. Since the amount in the account is increasing, the
effective rate is
i constant
=
a(n − 1) increasing function

which will be a decreasing function. Simple interest can be extended

to partial periods by letting

a(t) = 1 + it for all t ≥ 0.

This extension is justified when interest will be paid for partial

periods and accumulation increments have an additive nature.

1-7
Additive accumulation increments means:

[a(t + s) − a(0)] = [a(t) − a(0)] + [a(s) − a(0)]

or

a(t + s) = a(t) + a(s) − 1 since a(0) = 1

In this case,
h a(t + s) − a(t) i
a0 (t) = lims→0
s
h a(t) + a(s) − 1 − a(t) i
= lims→0
s
h a(s) − a(0) i
= lims→0
s
= a0 (0)
Thus the derivative is the same for all t, i.e. a(t) is a straight line and

a(t) = 1 + a0 (0)t = 1 + it for all t ≥ 0.

1-8
 2
a(t)
1
0

0 1 2 3 4
Interest Periods

1-9
Example: An account is receiving 6% simple interest. If $100 is
invested, how long must it remain invested until the account contains
$200?

200 = 100(1 + (.06)t)

2 = 1 + (.06)t
.06t = 1
1
t = or 16 23 periods.
.06
In ten periods this account contains

A(10) = 100(1 + (.06)(10)) = 160 dollars

1-10
Section 1.5 - Compound Interest

With compound interest, all interest earned in previous periods is

reinvested to earn interest in subsequent periods.

a(1) = (1 + i)a(0) = (1 + i)
a(2) = (1 + i)a(1) = (1 + i)2
a(3) = (1 + i)a(2) = (1 + i)3 etc.

Extending this to include partial periods produces

a(t) = (1 + i)t for all t ≥ 0.

We will see shortly that this accumulation function is also justified in

terms of a multiplicative property of the accumulation function,

a(t + s) = a(t)a(s) for all t ≥ 0 and s ≥ 0.

1-11
The effective interest rate for the nth period is
a(n) − a(n − 1)
in =
a(n − 1)
(1 + i)n − (1 + i)n−1
=
(1 + i)n−1
= 1+i −1
= i.

Thus the effective interest rate is the same in every period.

For Simple Interest, the accumulation increment

a(t + s) − a(t) does not depend on t

For Compound Interest, the relative growth

a(t + s) − a(t)
does not depend on t
a(t)

1-12
 Accumulation Functions
Compound
2
Simple
a(t)
1
0

0 1 2
Interest Periods

Note that a(0) = 1 and a(1) = 1 + i, for both simple and compound
interest accumulation functions.
When 0 < t < 1, asi (t) > aci (t), and when t > 1, asi (t) < aci (t).
1-13
Aside:
This accumulation function is also justified in terms of a multiplicative
property of the accumulation function,
a(t + s) = a(t)a(s) for all t ≥ 0 and s ≥ 0.
This property says that the accumulation at the end of t + s periods
is the same as if the accumulation over t periods was reinvested for
s periods under the same original plan.

h a(t + s) − a(t) i
a0 (t) = lims→0
s
h a(s) − a(0) i
= a(t)lims→0
s
= a(t)a0 (0)
In other words, the relative growth is always the same, or
d
a0 (t)
ln a(t) = = a0 (0) does not depend on t.
dt a(t)
1-14
It follows that
Z t Z t
d
a0 (0)dr


ln a(r ) dr =
0 dr 0
0
ln a(r ) |t0 = a (0)t



0
ln a(t) − ln(1) = a (0)t.

Note also that

ln a(1) = ln(1 + i) = a0 (0)1

and thus

ln a(t) = ln(1 + i) t = ln (1 + i)t

or

a(t) = (1 + i)t .

End Aside
1-15
Example: An account is receiving 6% compound interest. How long
must $100 be invested for the account to contain $200?

200 = 100(1 + .06)t

2 = (1 + .06)t
ln(2) = t ln(1.06)
ln(2)
t = = 11.896 periods.
ln(1.06)
In 10 periods, the account value is:

A(10) = 100(1.06)10 = 179.08 dollars.

1-16
Example:
A 35-year-old investor deposits $25,000 in an account earning 5%
compound interest until retirement at age 65. What is the value of
the account at retirement?

A(30) = 25000(1.05)30 = 108, 048.55 dollars.

Note also that between ages 35 and 45, the account grows by:

A(10) − A(0) = 25000[(1.05)10 − 1] = 15, 722.36 dollars.

Between ages 45 and 55, it grows by:

A(20)−A(10) = 25000[(1.05)20 −(1.05)10 ] = 25, 610.08 dollars.

Between ages 55 and 65, it grows by

A(30)−A(20) = 25000[(1.05)30 −(1.05)20 ] = 41, 716.11 dollars.

Because interest from earlier years also earns interest in later years,
the fund grows most rapidly toward the end.
Example:

A depositor puts $10,000 in a bank account. It earns an effective

annual interest of i during the first year and an effective annual of
(i − .05), the second year. After two years the account has a balance
of $12,093.75. What would the account contain after 3 years if the
annual effective rate is (i + .09) for each of the three years?

1-18
Section 1.6 - Present Values
An asset of one grows to
a(1) = (1 + i)1
at the end of one period. So (1 + i) is called the accumulation factor.
What would you need to invest today to achieve a value of one after
one interest period?
a(1) = 1 = (1 + i)( what value?)
Solving this equation for the unknown value yields
1
ν= .
(1 + i)
The value ν is called the discount factor. If you are going to receive
an asset of k at the end of one period, its present value is
k
kν = .
(1 + i)
1-19
When set to be in correspondence with one another, accumulation
and discounting are reciprocal processes, i.e. the discount function
and the accumulation function satisfy

a(t)·d(t) = 1,

that is,
1
d(t) = = [a(t)]−1 .
a(t)

So under simple interest

a(t) = (1 + it)
1
d(t) = .
(1 + it)

1-20
Under compound interest

a(t) = (1 + i)t
d(t) = (1 + i)−t = νt

The properties of a discount function are:

1
I d(0) = a(0) =1
1 1
I d(1) = a(1) = (1+i) =ν
I d(t) is typically a decreasing function of t.

1-21
Exercise 1-17:

Two sets of grandparents of a newborn agreed to invest immediately

to fund $20,000 per year for four years of college at age 18 for the
child. Grandparents A agreed to cover the first two years of college
with grandparents B covering the last two years. The account has an
effective rate of 6% per annum. What should each of the
grandparents contribute to the fund today to meet this need?

Grandparents A:

20000ν 18 +20000ν 19 = 7, 006.88+6, 610.26 = 13, 617.14 dollars

Grandparents B:

20000ν 20 +20000ν 21 = 6, 236.09+5, 883.11 = 12, 119.20 dollars

1-22
 100
Payment
50
0

0 1 2
Time

Which would you rather have:

Case A: $100 Today? Or
Case B: Receive $100 two years from today?
1-23
The answer is clear - you would prefer to have the money NOW!
Money in your possession today is more valuable to you (today) than
the same amount of money received in the future.
So what value should you place on this $100 you will receive in two
years? You value it as though it is the amount that you would need to
invest today to have $100 in two years. Therefore its value is
 2
2 1
$100ν = $100
1+i

So this future payment is viewed relative to some interest rate i.

Time Value of Money:
The present value of a future payment is discounted according to
some interest rate i.
In order to be compared, all payments must be viewed from the
same time point and therefore must be accumulated or discounted
to that time point via an interest rate i.

1-24
Section 1.7 - Effective Rate of Discount

The effective rate of interest, i, satisfies:

A(0) + iA(0) = A(1) or

A(1) − A(0) a(1) − a(0)

i= = .
A(0) a(0)

Likewise, the effective rate of discount, d, satisfies:

A(1) − dA(1) = A(0) or

A(1) − A(0) a(1) − a(0)

d= = .
A(1) a(1)

1-25
By extension, during the nth period, the effective rate of discount is:

A(n) − A(n − 1) a(n) − a(n − 1)

dn = =
A(n) a(n)

When do the two rates i and d correspond the the same movement
of assets in opposite directions (equivalent in this sense)?

(1 + i)A(0) = A(1) and

(1 − d)A(1) = A(0) or
(1 + i)(1 − d) = 1.
Solving this equation for d yields:

i
d= = iν
(1 + i)

1-26
and solving it for i yields:

d
i= .
(1 − d)

This shows that

1
1−d = = ν,
(1 + i)

d =1−ν or
ν + d = 1.
In addition,

d = iν = i(1 − d) or

i − d = id > 0.

1-27
For compound discount,

d(t) = ν t = (1 − d)t .

Clearly

d(t)a(t) = ν t (1 + i)t = 1 or

1
d(t) = = [a(t)]−1 .
a(t)
For simple discount:

1
d(t) = (1 − dt) for 0 < t < .
d

1-28
Exercise 01-25 If i and d are equivalent rates of simple interest and
simple discount over t periods, show that i − d = idt.

1-29
Exercise: Give a faculty member a 100i% raise and require each to
pay 100i% into a retirement account. What is the faculty member’s
new salary? What should the raise be to make it even out?

1-30
Section 1.8 - Nominal Rates
Interest is quoted in terms of an annual rate, but frequently is
compounded over shorter intervals. For example, an 8% interest
rate when compounded quarterly means 2% percent interest is
added to the principal at the end of each quarter thus the effective
annual rate i is determined by:
 .08 4
(1 + i) = 1 + or i = .0824.
4
which is larger than 8%.
In this example, 8% is the nominal annual rate (APR) and 8.24% is
the effective annual rate (APY).
In general, suppose a nominal annual rate of i (m) is compounded
over m equal segments of the year. Then the effective annual
interest rate i is determined by
!m
i (m)  i (m) m
(1 + i) = 1 + or i = 1+ − 1.
m m
1-31
Moreover, we can find the nominal annual rate which achieves a
fixed effective annual rate via
1
i (m) = m (1 + i) m − 1 .
 

The accumulation function in this setting is

 i (m) mt
a(t) = 1 + .
m

Suppose i (m) does not depend on m, that is the nominal level stays
the same, no matter how many times the account is compounded.
Call this value i (1) . In calculus we learned that

 i (1) m (1)
lim 1+ =ei .
m→∞ m

1-32
As m → ∞, the account is said to be compounded continuously.
When m → ∞ and i (m) = i (1) for all m (i.e. the nominal rate does not
change as m changes), the effective annual rate is
(1)
i =ei −1 and i (1) = ln(1 + i)

where i (1) is the nominal annual interest rate.

Example: Compound a fixed 5% nominal rate (i (1) = .05 for all m).

Period m i (effective rate)

Annually 1 .05
Semi-annually 2 .050625
Quarterly 4 .050945
Monthly 12 .051162
Daily 365 .051267
Continuously →∞ .051271

1-33
In a similar fashion, discounting can also be applied m times each
period. If d (m) denotes the nominal discount rate in a complete
period, the effective discount rate for a complete period when
convertible m times per period is:
 d (m) m
d =1− 1−
m
or we can solve the relationship for
1  1 
d (m) = m 1 − (1 − d) m = m 1 − ν m .
 

The discount function is

 d (m) mt
d(t) = 1 − for all t > 0.
m
When the nominal discount rate is the same for all m, i.e.
d (m) = d (1) , and m → ∞ (the discounts are convertible
continuously), the effective rate of discount per period is
(1)
d = 1 − e −d .
1-34
Also note that for the effective rates of interest and discount to
match, that is
a(t)d(t) = 1
then
 i (m) mt  d (p) −pt
1+ = 1− .
m p
even when interest is compounded mthly per period and discount is
convertible pthly per period.
Example: Find the accumulated value of $500 at the end of three
years
(a) if the nominal rate of interest is 5% convertible semiannually.

 .05 6
500 1 + = 579.85
2
(b) if the nominal rate of discount is 7% convertible every two years.
500
= 626.93
(1 − .14)1.5
1-35
Exercise 1-23:
Find the present value of $5000 to be paid at the end of 25 months
at a discount of 8% convertible quarterly.
(a) Assume compound discount throughout:
8 31 periods with 2% discount per period

25 25
5000ν 3 = 5000(1 − .02) 3
= 4225.27

(b) Assume compound discount for whole periods and simple

discount for fractional periods.

 
8 .02
5000(1 − .02) 1 − = 4225.46
3

1-36
Exercise 1-29: Given i (m) = .1844144 and d (m) = .1802608 are
nominal annual rates producing effective annual rates, find m.

1-37
Section 1.9 - Force of Interest (Discount)

Earlier we described the effective rate of interest for the nth period to
be
A(n) − A(n − 1) Change in Amount during the period
= .
A(n − 1) Amount entering the period

representing the relative growth in the amount during that period. As

we ignore the discrete period boundaries and focus on the amount
function A(t) at all t > 0, it is natural to describe the relative growth
in this amount function at time t with
A0 (t)
,
A(t)

the instantaneous rate of change in the amount function relative to

its current size.

1-38
Definition: The force of interest δt at time t > 0 is

A(t + h) − A(t) A0 (t) a0 (t)

δt = lim = = .
h→0 hA(t) A(t) a(t)

Examples
Simple Interest:

i
a(t) = 1 + it δt = (decreasing in t)
1 + it
Compound Interest:

a(t) = (1 + i)t δt = ln(1 + i) (constant in t)

Earlier Example:

(.1)t
a(t) = (.05)t 2 + 1 δt =
(.05)t 2 + 1

1-39
Note that
A0 (t) d d
δt = = ln(A(t)) = ln(a(t)).
A(t) dt dt
Replacing t with r and integrating yields
Z t t
Rt d
δr dr = 0 dr ln(A(r ))dr = ln(A(r ))

0 0
 
A(t)
= ln(A(t)) − ln(A(0)) = ln A(0) .

Thus
Rt
δr dr A(t) a(t)
e 0 = = = a(t).
A(0) a(0)
Also
Z t Rt t
A(r )δr dr = A0 (r )dr = A(r )
0 
0
0
= A(t) − A(0).
1-40
A Constant Force of Interest means

δt = δ for all t > 0

In this case
Z t t
δr dr = δr  = δt

0 0

and

a(t) = eδt .

Since

a(1) = 1 + i = eδ ,

δ = ln(1 + i),
where i is the effective rate of interest and δ is the nominal rate of
interest when interest is compounded continuously.
1-41
So when a problem is described as having a constant force of
interest,

a(t) = eδt = (1 + i)t and δ = ln(1 + i)

where i is the effective interest rate, δ is the nominal interest rate

and there is continuous compounding of interest.

The force of discount is defined as

d 0 (t)
δt0 = − ,
d(t)

where the minus sign is included because d 0 (t) is typically negative.

1-42
Because d(t) = [a(t)]−1 ,
d
− dt [a(t)]−1 [a(t)]−2 a0 (t)
δt0 = [a(t)]−1
= [a(t)]−1
,
a0 (t)
= a(t) = δt .

so we dropped the prime, because these two are the same quantity.
Also note that
Rt
d(t) = [a(t)]−1 = e− 0 δr dr
.

A constant force of discount means the same thing as a constant

force of interest, i.e.

d(t) = [a(t)]−1 = e−δt

where δ = ln(1 + i) = − ln(1 − d), for i the effective rate of interest

and d the effective rate of discount.

1-43
Example: What is the account balance after two years when a $500
deposit is made in an account with δt = (.02) + (.03)t 2 ?

Z 2 2
[.02 + .03t 2 ]dt = [(.02)t + (.01)t 3 ] = .04 + .08 = .12

0 0

A(2) = 500e.12 = 563.75

1-44
Exercise 1-46: You are given δt = 2/(t − 1) for 2 ≤ t ≤ 10. For any
one-year interval between n and n + 1, with n = 2, · · · , 9, calculate
the equivalent d (2) .

Z n+1
2 n+1  n 
dt = 2 ln(t − 1) = 2 ln

n (t − 1) n n−1
 
n
−2 ln n−1  d (2) 2
e = 1−
2
 d (2)  n − 1 2
1− = or d (2) =
2 n n

1-45
Exercise: A fund earns interest at a force of interest of δt = kt.
A deposit of 100 at time 0 will grow to 250 at the end of five years.
Find k.

1-46
Section 1.10 - Varying Interest (Discount)
For many investments the interest rates vary from one period to the
next, where the periods are of fixed and equal length. The
compound accumulation function over n periods is then

n
Y
a(n) = (1 + i1 )(1 + i2 ) · · · (1 + in ) = (1 + ij ),
j=1

where ij is the effective interest rate for the j th interest period.

In similar fashion, if the discount rates vary from period to period, the
discount function over n periods is

n
Y
d(n) = (1 − d1 )(1 − d2 ) · · · (1 − dn ) = (1 − dj )
j=1

where dj is the effective rate of discount during the j th period.

1-47
Example: A bank recently advertised a variable interest rate CD that
earns 1.6% APR for the first six months, 1.8% for the second six
months, 2.0% for the third and 2.2% for the fourth. What is the
effective annual interest rate for the CD?

.016 .018 .020 .022

a(2) = (1 + 2 )(1 + 2 )(1 + 2 )(1 + 2 )
= 1.03854

Thus we set

1.03854 = (1 + i)2 and get i = .019089.

1-48
Exercise: At a certain interest rate the present values of the
following are equal:
(a) 200 at the end of 5 years and 500 at the end of 10 years
(b) 400.94 at the end of 5 years.
At the same interest rate, 100 invested now and 120 invested at the
end of 5 years accumulates to P at the end of 10 years. Find P.

1-49
Exercise: An investor puts 100 into fund X and 100 into fund Y. Fund
Y earns compound interest at the annual rate of j > 0 and fund X
earns simple interest at the annual rate of (1.05)j. At the end of 2
years, the amount in fund Y is equal to the amount in fund X.
Calculate the amount in fund Y at the end of 5 years.

1-50
Appendix to Chapter 01 - Time Value of Money Principles

(1) Establish a vision point (some time t = t0 ) at which values of

payments are to be evaluated.
(a) accumulate all earlier payments to the vision point
(b) discount all future payments to the vision point
(2) Only combine payments after they are all projected to the same
(a common) vision point
(3) The value of a payment stream at the time t = t0 is the sum of all
the (projected to t0 ) payment values
(4) Two or more investments (payment streams) are only
comparable when they are all projected to one common vision point.

The two most common vision points are:

PV = Present Value = (vision point is today, i.e. t = 0)
FV = Future Value = (vision point is the end of the last interest
period, i.e. t = n)
1-51