Cal2 ba dinh_hai_slides_ch1

CALCULUS 2 (BA)
Assoc. Prof. Nguyen Dinh
Dr. Nguyen Ngoc Hai
Department of mathematics
INTERNATIONAL UNIVERSITY, VNU-HCM
February 25, 2014
Assoc. Prof. Nguyen Dinh Dr. Nguyen Ngoc Hai CALCULUS 2 (BA)

References
Main textbook:
M. L. Lial, R. N. Greenwell, N. P. Ritchey Calculus
with Applications, 10ed. Pearson, Boston, 2012.
Other textbooks:
L. D. Hoﬀmann, G. L. Bradley, Calculus, Brief 10ed.
McGraw-Hill, Boston, 2010.

Chapter 1 . Mathematics of Finance
Contents
1. Compound Interest
2. Continuous Money Flow: Total money ﬂow,
present value, accumulated amount of money,
continuous deposits.
3. Annuities
4. Amortizations and Sinking Funds

Simple and compound interest
• If you borrow money you have to pay interest on
it. If you invest money in a deposit account you
expect to earn interest on it.
Interest can be interpreted as money paid for the
use of money.
• The original amount borrowed or invested is called
the principal, denoted by P.

The rate of interest r is the amount charged for
the use of the principal for a given length of time,
usually on a yearly (or per annum, abbreviated
p.a.) basis, given either as a percentage (p per
cent) or as a decimal r, i.e.
r =
p
100
.
The total amount received after (investing) a period
of time is called accumulated value. The
accumulated value after t year is A(t).

Simple interest.
Simple interest is interest computed on the
principal for the entire period it is borrowed (or
invested). It is assumed that this interest is not
reinvested together with the original capital.
The principal P with the rate r and time t gives the
simple interest
I(t) = Prt,
and hence,
A(t) = P + Prt = P(1 + rt).

Example 1.1 How much interest will be earned
on $ 4, 000 invested for a year at 0.5%?
Solution We write
0.5% =
0.5
100
= 0.005
and get
0.5% of $ 4, 000 = 0.005 × $ 4, 000 = $ 20.

Compound interest
Compound interest.
Compound interest is interest which is added to
the original investment every time it accrues. The
interest added in one time period will itself earn
interest in the following time period. The total
value of an investment will therefore grow over time.

Compound interest
Suppose that we invest P dollars at interest rate r,
expressed as a decimal and compounded annually.
The amount A1 in the account at the end of the
ﬁrst year is
A1 = P + Pr = P(1 + r).
Going into the second year, we have A1 dollars, so
by the end of the second year, we will have the
amount A2 given by
A2 = A1 + A1r = A1(1 + r) = P(1 + r)2
.

Compound interest
Going into the third year, we have A2 dollars, so by
the end of the third year, we will have the amount
A3 = A2 + A2r = A2(1 + r) = P(1 + r)3
,
and so on.
Theorem 1.1 If an amount P is invested at
interest rate r, expressed as a decimal and
compounded annually, in t years it will grow to the
amount A given by
A = P(1 + r)t
(1)
A is called the compound amount.

Compound interest
Example 1.2 You estimate that you will need
$8, 000 in 3 years’ time to buy a new car. You have
$7, 000 which you can put into a ﬁxed interest
building society account earning 4.5%. Will you
have enough to buy the car?
Solution You need to work out the ﬁnal value of
your savings to see whether it will be greater than $
8, 000. Using Equation 1 with P = 7, 000,
r = 0.045, and t = 3,
A = 7, 000(1+0.045)3
= 7, 000(1.141166) = $7, 988.16.
So the answer is “almost”.

Compound interest
Example 1.3 What principal is required now so
that after 6 years at a rate of interest of 5 per cent
p.a. the ﬁnal amount is 20, 000 EUR?
Solution It follows from Equation 1 that
P =
A
(1 + i)t
.
Substituting t = 6, i = 0.05 and A = 20, 000, we
obtain
P =
A
(1 + i)6
=
20, 000
(1 + 0.05)6
= 14, 924.31 EUR
i.e., the principal required now is equal to 14, 924.31
EUR.

Compound interest
Part year investment
Often, we do not have an annual period for interest
payments, i.e. compounding takes place several
times per year, e.g.,
• semi-annually–there are 2 interest payments
per year, namely after every six months;
• quarterly–there are 4 payments per year,
namely one after every three months;
• monthly–there are 12 payments, namely one
per month;
• daily–compounding takes place 360 times per
year.

Compound interest
If the annual interest rate is r and there are n
interest payments per year, the rate of interest per
payment period is equal to j = r/n, and the number
of interest payments within a period of t years is
equal to tn.
Denoting by A(t) the amount at the end of t years
with n interest payments per year, formula (1)
changes into
A(t) = P (1 + j)nt

Compound interest
Theorem 1.2 If a principal P is invested at
interest rate r, expressed as a decimal and
compounded n times a year, in t years it will grow
to an amount A(t) given by
A(t) = P 1 +
r
n
nt
(2)

Compound interest
Example 1.4 An amount of money of $ 9000 is
invested at a rate 6% annual interest compounded
semi-annually for 4 years. How much interest will be
earned?
Solution In this case r = 0.06, n = 2, and t = 4,
so the compound amount is
A(t) = P 1 +
r
n
tn
= 9000 1 +
0.06
2
(2)(4)
= 11, 400.93.
The interest amount is: $ 11,400.93 - $ 9,000 =
$2,400.93.

Compound interest
Remark The compound amount increases with
the increasing of the number of times paid in a year.

Continuous compounding
As the frequency n with which interest is
compounded increases, the corresponding amount
A(t) also increases. Hence, a bank that compounds
interest frequently may attract more customers than
one that oﬀers the same interest rate but
compounds interest less often.
Question: What happens to the amount at the end
of t years as the compounding frequency n increases
without bound?

In mathematical terms, this question is equivalent
to asking what happens to the expression
A(t) = P 1 +
r
n
nt
as n → ∞?
Note.
lim
x→∞
1 +
1
x
x
= e ≈ 2.718281828.

Theorem 1.3 Suppose a principal P is invested at
interest rate r and the accumulated value in the
account after t years is A(t). If interest is
compounded continuously, then
A(t) = Pert
(3)

Example 1.5 Suppose $ 1, 000 is invested at an
annual interest rate of 6%. Compute the ﬁnal
amount after 10 years if the interest is compounded
(a) Quarterly (b) Monthly (c) Daily
(d) Continuously.
Solution (a) To compute the balance after 10
years if the interest is compounded quarterly, use
the formula (2) with t = 10, P = 1000, r = 0.06,
and n = 4:
A(10) = 1, 000 1 +
0.06
4
4·10
≈ $ 1, 814.02.

(b) This time, take t = 10, P = 1, 000, r = 0.06,
and n = 12 to get
A(10) = 1, 000 1 +
0.06
12
120
≈ $ 1, 819.40.
(c) Take t = 10, P = 1, 000, r = 0.06, and n = 365
to obtain
A(10) = 1, 000 1 +
0.06
365
3,650
≈ $ 1, 822.03.

(d) For continuously compounded interest use the
formula (3) with t = 10, P = 1000, r = 0.06:
A(10) = 1, 000e0.6
= $ 1, 822.12.
This value, $ 1, 822.12, is an upper bound for the
possible balance. No matter how often interest is
compounded, $ 1, 000 invested at an annual interest
rate of 6% cannot grow to more than $ 1, 822.12 in
10 years.
When interest is compounded quarterly in (a), the
value of the investment after ten years is
$ 1, 822.12 − $ 1, 814.02 = $8.1 less compared to
continuous compounding.

2. Present Value
In many situations, it is useful to know how much
money P must be invested at a ﬁxed compound
interest rate in order to obtain a desired
accumulated (future) value A over a given period of
time t.
This investment P is called the present value of
the amount A to be received in t years.
It is the amount of money needed now so that after
depositing this amount for a period of t years at a
per annum rate of interest of r, the amount of an
annuity A results.

2. Present Value
The present value of A in t years invested at the
annual rate r compounded n times per year is
given by
P = A 1 +
r
n
−nt

2. Present Value
If interest is compounded continuously at the
same rate, the present value in t years is given
by
P = Ae−rt
This means that after t units of time, with the rate
of interest r compounded continuously, if you want
to get a compound amount A, the present value you
must deposit is P = Ae−rt
.

2. Present Value
Example 2.1 How much money needs to be
invested now in order to accumulate a ﬁnal sum of
$ 5, 000 in 4 years’ time at an annual rate of
interest of 7% if interest is compounded:
(a) Quarterly (b) Continuously.
Solution The required future value is
A = $ 5, 000 in t = 4 years with r = 0.07.
(a) If the compounding is quarterly, then n = 4 and
the present value is
P = 5, 000 1 +
0.07
4
−(4)(4)
= $ 3, 788.08.

2. Present Value
(b) For continuous compounding, the present value
is
P = 5, 000e−(0.07)(4)
= $ 3, 778.92.

3. Money flow
Total money flow (total income)
Definition 3.1 If f (t) is the rate of money flow,
then the total money flow over the time interval
from t = 0 to t = T is given by
T
0
f (t)dt
This total money flow does not take into account
the interest the money could earn after it is
received. It is simply the total income.

3. Money flow
An amount of money that can be deposited today
at a specified interest rate to yield a given sum in
the future is called the present value of this future
sum. The future sum may be called the future value
or final amount.
To find the present value of a continuous money
flow with interest compounded continuously, let
f (t) represent the rate of the continuous flow.

3. Money flow
The time axis from 0 to T is divided into n
subintervals, each of width ∆t = T/n. The amount
of money that flows during any interval of time is
approximated by f (ti)∆t, which (approximately)
gives the amount of money flow over that
subinterval.

3. Money ﬂow

3. Money flow
Earlier, we saw that the present value P of an
amount A compounded continuously for t years at a
rate of interest r is P = Ae−rt
.
Question: Given a continuous money flow with
interest compounded continuously and with f (t) is
its rate (of change) for T years [t is time variable,
t ∈ [0, T]]. How can we find the present value of
the mentioned continuous money flow?

3. Money flow
Present value of money flow
Letting ti represent the time and replacing A with
f (ti)∆t, the present value of the money flow over
the rth subinterval is approximately equal to
f (ti)∆te−rt
.
The total present value is approximately equal to
the sum n
i=1
f (ti)∆te−rt
.

Pi = [f (ti)∆t]e−rti
, P ≈
n−1
i=0
[f (ti)∆t]e−rti
.

Theorem 3.1
If f (t) is the rate of a continuous money ﬂow at an
interest rate r (at time t) for T years, then the
present value is
P(T) =
T
0
f (t)e−rtdt.

Accumulate amount of money flow at time t
Theorem 3.2 (Accumulate amount of
money flow at time t)
If f (t) is the rate of a continuous money flow at an
interest rate r at time t, the amount of the flow at
time T is
A(T) = erT
T
0
f (t)e−rt dt.

Continuous money flow
Example 3.1 If money flowing continuously at a
constant rate of $ 2000 per year over 5 years at
12% interest compounded continuously, find the
following:
(a) The total amount of the flow over 5-year period.
(b) The accumulate amount compounded
continuously at T = 5.
(c) The total interest earned
(d) The present value with interest.

Solution (sketch)
By assumption: f (t) = 2000, T = 5, and r = 0.12.
(a) The total amount of the ﬂow over 5-year period
is:
5
0
2000dt = 10, 000 (dollars).
(b) The accumulate amount compounded
continuously at t = 5 is:
A = erT
T
0
f (x)e−rt
dt = e5(0.12)
5
0
(2000)e(−0.12)t
dt
= 13, 701.98 (dollars).

Solution (sketch)
(c) The total interest earned is:
13, 701.98 − 10, 000 = 3, 701.98 (dollars).
(d) The present value with interest is:
P =
5
0
f (t)e−rt
dt =
5
0
2000e−0.12t
dt
= 7, 519.81 (dollars).
Answer to the last question: (note P = Ae−rT
).

Comments (on the previous example)
• If f (t) (dollars per unit of time) is the rate of a
continuous money ﬂow at an interest rate r (at time
t) from the time t = a to t = b. Then the present
value of the ﬂow at time t = a is
P =
b
a
f (t)er(a−t)
dt.

Comments (on money ﬂow)
Imagine: A company with a high volume of sales
receives money almost continuously. For purpose of
calculation, it is convenient to assume that the
company literally does receive money continuously.
In such a case, we have a function f (t) that
represents the rate at which money is being received
by the company at the time t.

Capital value
The capital value of an asset (property) is
sometimes deﬁned as the present value of all future
net earning of the asset. If f (t) give the annual rate
at which earnings are produced by an asset at time
t, then the present value formula gives the capital
value as ∞
0
f (t)e−rt
dt
where r is the annual rate of interest.

Capital value
Example 3.2 Suppose income from a rental
property is generated at the annual rate of $ 4000
per year. Find the capital value of this property at
an interest rate 10% computed continuously.
This is a continuous income stream with a
(constant) rate of ﬂow of $4000 per year, that is,
f (t) = 4000. Also, r = 0.1.

4. Continuous Deposits
The capital value is given by
∞
0
4000e−rx
dx = lim
b→∞
b
0
4000e−0.1t
dt
= lim
b→∞
4000
−0.1
e−0.1t
b
0
= lim
b→∞
−40, 000e−0.1b
+ 40, 000
= 40, 000.
The capital value is $40,000.

The accumulate amount A(t) of some amount
money, say P, invested at an annual interest rate r,
compounded continuously, grows according to the
diﬀerential equation
dA
dt
= rA
where t is time (in years).

Suppose regular deposits are made to the account
at frequent intervals at a rate of D dollars per year.
For simplicity, assume these deposits to be
continuous. The diﬀerential equation for the growth
of the account then becomes
dA
dt
= rA + D

Example 4.1 When Michel was born, his
grandfather arranged to deposit $ 5000 in an
account for him at 8% annual interest compounded
continuously. Grandfather plans to add to the
account“continuously” at the rate of $ 1000 a year.
How much will be in the account when Michel is 18?

Since r = 0.08 and D = 1000, the diﬀerential
equation is
dA
dt
= 0.08A + 1000.
Separating the variables and integrating both sides
of the above equation, we get
1
0.08A + 1000
dA = dt,
1
0.08
ln(0.08A + 1000) = t + C,
A = −12500 +
M
0.08
e0.08t
.

Continuous Deposits
By the assumption, A(0) = 5000, we get
A(0) = 5000 = −12500 +
M
0.08
e(0.08)(0)
,
which gives M = 1400, and hence,
A = −12500 + 17500e0.08t
.
When Michel is 18, the amount in the account is
A = ... = 61, 362.18.

5. Annuity
Think about it.
Suppose that $ 1500 is deposit at the end of each
year for the next of the six years in an account
paying 8% per year, compounded annually. How
much is in the count after 6 year?
Such a sequence of equal payments made at equal
period of time is called an annuity

Annuity
To find the amount of this annuity, look at each of
the $ 1500 payments separately.
• The first payment will produce a compound
amount of
1500(1 + 0.08)5
= 1500(1.08)5
at the end of 6 years (note that the money is
deposit at the end of the first year).

Annuity
• The total amount of the annuity is:
1500(1.08)5
+ 1500(1.08)4
+ 1500(1.08)3
+
1500(1.08)2
+ 1500(1.08)1
+ 1500
= 1500
1.086
− 1
1.08 − 1
≈ 11, 003.89($).

Annuity
Amount of annuity
The amount S of an annuity of payments of R
dollars each, made at the end of each period for n
consecutive interest periods at a rate of interest i
per period, is given by
S = R
(i + 1)n
− 1
i
.
How can we get this formula?

Recall: Sum of ﬁrst n terms of geometric sequence
A geometric sequence is a sequence of numbers:
a1, a2, · · · , an, ...., where a, r ∈ R are given, and
an = arn−1
, for all n.
Sn = a1 + a2 + · · · + an
= a + ar + ar2
+ · · · + arn−1
=
a(rn
− 1)
r − 1
.
(How to prove this formula?)

Annuity
Example 5.1 Suppose 1000 dollars is deposited
at the end of each 6-month period for 5 years in an
account paying 6 (percent) per year compounded
semi-annually. Find the amount of the annuity.
• Interest for semi-annually: 0.06/2 = 0.03.
• In 5 years, there are 5 × 2 = 10 semiannual
periods.
• By the formula of the amount of annuity, we get
S = 1000
(1.03)10
− 1
0.03
= 11, 463.88,
i.e., S = 11, 463.88 dollars.

Present value of annuity
Recall
The amount S of an annuity of payments of R
per period, is given by
S = R
(i + 1)n
− 1
i
.
We now suppose that we want to ﬁnd the lump sum
P that must be deposit today at a rate of interest i
per period in order to produce the same amount S
after n periods. The sum P is then called the
present value of the mentioned annuity.
How can we ﬁnd P?Assoc. Prof. Nguyen Dinh Dr. Nguyen Ngoc Hai CALCULUS 2 (BA)

With the assumption, P dollars deposited today will
amount to P(1 + i)n
after n periods and this is
equal to S. We get
P(1 + i)n
= S = R
(i + 1)n
− 1
i
.
This gives (how?)
P = R
1 − (1 + i)−n
i
.

The present value P of an annuity of payments of R
per period is given by:
P = R
1 − (1 + i)−n
i
.

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