Sections 2.1, 2.2, 2.3, 2.4, 2.5
Interest problems generally involve four quantities: principal(s),
investment period length(s), interest rate(s), accumulated value(s).
Two or more amounts of money payable at different points in time
cannot be compared until a common date, called the comparison date, is
established.
An equation of value accumulates or discounts each payment to the
comparison date. (A time diagram can be helpful in setting up an
equation of value.)
With compound interest, an equation of value will produce the same
answer for an unknown value regardless of what comparison date is
selected; however, this is not necessarily true for other patterns of
interest.
In return for a payment of $1200 at the end of 10 years, a lender agrees
to pay $200 immediately, $400 at the end of 6 years, and a final amount
at the end of 15 years. Find the amount of the final payment at the end
of 15 years if the nominal rate of interest is 9% converted semiannually.
Time Diagram:
$200
$400
$X
Lender
Borrower
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
$1200
present
Equation of Value:
v = 1 / (1 + 0.045)
(Time periods are counted in half-years,
since interest is converted semiannually.)
200 + 400v12 +Xv30 = 1200v20
1200v20 – 200 – 400v12
X = —————————
v30
=
$231.11
In return for payments of $5000 at the end of 3 years and $4000 at the
end of 9 years, an investor agrees to pay $1500 immediately and to make
an additional payment at the end of 2 years. Find the amount of the
additional payment if i(4) = 0.08.
$1500 $X
Time Diagram:
Investor
Borrower
0 1 2 3 4 5 6 7 8 9 10
$5000
Equation of Value:
(Time periods are counted in quarter-years,
since interest is converted quarterly.)
v = 1 / (1 + 0.02) 1500 + Xv8 = 5000v12 + 4000v36
$4000
X = $5159.24
Consider the graphs of f(x) = x – 1 and f(x) = lnx .
y=x–1
y
We find that for any x > 0,
y = lnx
(1,0)
x – 1  lnx
x
Now, suppose a1 , a2 , … , an are all positive.
Let T = a1 a2 … an , and let yi = ai / T1/n for each i = 1, 2, …, n.
Then, y1 y2 … yn = 1, and ln(y1 y2 … yn) = 0 . Also, we see that
(y1 – 1) + (y2 – 1) + … + (yn – 1)  ln(y1) + ln(y2) + … + ln(yn) =
ln(y1 y2 … yn) = 0.
We now have that
(y1 – 1) + (y2 – 1) + … + (yn – 1)  0 
y1 + y2 + … + yn  n 
y1 + y2 + … + yn
———————  1
n
a1 + a2 + … + an
———————
 1 
1/n
T n

a1 + a2 + … + an
———————  (a1a2…an)1/n
n
In Section B of Appendix C (page 593), find the definition of the
arithmetic mean and the geometric mean. Then observe that we have
proven that that the arithmetic mean must always be greater than or
equal to the geometric mean (assuming of course only positive values).
Amounts $500, $800, and $1000 are to be paid at respective times 3
years from today, 5 years from today, and 11 years from today. Suppose
we would like to find the number of years t from today when one single
payment of $500 + $800 + $1000 = $2300 would be equivalent to the
individual payments made separately. If the rate of interest per annum is
known to be i, then v = 1 / (1 + i), and the equation of value is
2300vt = 500v3 + 800v5 + 1000v11
In general, suppose amounts s1 , s2 , … , sn are to be paid at respective
times t1 , t2 , … , tn . Also suppose we would like to find the time t when
one single payment of s1 + s2 + … + sn would be equivalent to the
individual payments made separately. If v = 1 / (1 + i) is known, then the
equation of value is
(s1 + s2 + … + sn)vt = s1v t1 + s2v t2 + … + snv tn
(Observe what must be done to find an exact formula for the unknown
value of t.)
(s1 + s2 + … + sn)vt = s1v t1 + s2v t2 + … + snv tn
With the exact t, this is the true present
value for the equivalent single payment.
An approximate value of t can be obtained from the method of equated
time. This method estimates t with
t =
s1 t 1 + s2 t 2 + … + sn t n
—————————
s1 + s2 + … + sn
With the approximation t, the estimated present value for the equivalent
single payment is
(s1 + s2 + … + sn)vt
In order to discover an interesting property about how the exact value of
t and the estimate t compare, let us imagine that we have a list of
numbers where s1 of the numbers are equal to v t1 , s2 of the numbers are
equal to v t2 , etc.
In order to discover an interesting property about how the exact value of
t and the estimate t compare, let us imagine that we have a list of
numbers where s1 of the numbers are equal to v t1 , s2 of the numbers are
equal to v t2 , etc.
t
Then v can be written as v
s1t1 + s2t2 + … + sntn
—————————
s1 + s2 + … + sn
From Section B of Appendix C,
we see that this is the geometric
mean of the list of numbers.
v
s1t1
=
v s2t2 … v sntn
1
——————
s1 + s2 + … + sn
This is the product of
the list of numbers.
and v t can be written as
s1 v t 1 + s2 v t 2 + … + sn v t n
——————————
s1 + s2 + … + sn
From Section B of Appendix C,
we see that this is the arithmetic
mean of the list of numbers.
This is the average of
the list of numbers.
Since the arithmetic mean of positive numbers must always be larger
than the geometric mean, we have that
vt > vt
s1 v t 1 + s2 v t 2 + … + sn v t n
—————————— > v t
s1 + s2 + … + sn
s1v t1 + s2v t2 + … + snv tn > (s1 + s2 + … + sn)vt
We now see that the true present value (on the left) will always be
larger than the estimated present value (on the right). Consequently, the
estimated time t from the method of equated time must be an
overestimation.
Amounts $500, $800, and $1000 are to be paid at respective times 3
years from today, 5 years from today, and 11 years from today. The
effective rate of interest is 4.5% per annum.
(a) Use the method of equated time to find the number of years t from
today when one single payment of $500 + $800 + $1000 = $2300
would be equivalent to the individual payments made separately.
t =
s1 t 1 + s2 t 2 + … + sn t n
(500)(3) + (800)(5) + (1000)(11)
————————— = ————————————— =
s1 + s2 + … + sn
2300
7.174 years
(b) Find the exact number of years t from today when one single
payment of $500 + $800 + $1000 = $2300 would be equivalent to
the individual payments made separately.
With v = 1 / (1 + 0.045), the equation of value is
2300vt = 500v3 + 800v5 + 1000v11
3 + 800v5 + 1000v11
500v
t = 6.917 years
vt = —————————— = 0.73752
2300
Given a particular rate of interest, how long will it take an investment to
double?
By solving (1 + i)n = 2, we find the exact solution
ln(2)
n = ———
ln(1 + i)
0.6931
0.6931
i
Observe that n  ——— = ——— ——— ,
ln(1 + i)
i
ln(1 + i)
0.72
72
and when i = 0.08 in the second factor, we have n  —— = —— .
i
100i
This is called the rule of 72, and it gives amazingly accurate results for
a wide range of interest rates! (See Table 2.1 on page 57 of the text.)
Suppose $2000 is invested at a rate of 7% compounded semiannually.
(a) Find the length of time for the investment to double by using the
exact formula.
t = years  2000(1.035)2t = 4000  t = 10.074 years
(b) Find the length of time for the investment to double by using the
rule of 72.
72/3.5 = 20.571 half years  t = 10.286 years
(c) Find the length of time for the investment to be worth $3000 by
using the exact formula.
t = years  2000(1.035)2t = 3000  t = 5.893 years
Suppose it is the rate of interest i that is unknown in an equation of
value. There are several methods available for solving for i, and
depending on the type of situation, one method may be better suited than
another.
At what interest rate convertible semiannually would $500 accumulate to
$800 in 4 years?
When a single payment is involved, the method that works best is to
solve for i directly in the equation of value (using exponential and
logarithmic functions).
500(1 + j)8 = 800
1 + j = (8/5)1/8
j = (8/5)1/8 – 1 = 0.0605
i(2) = 2j = (2)(0.0605) = 0.0121 = 12.1%
At what effective interest rate would the present value of $1000 at the
end of 3 years plus $2000 at the end of 6 years be equal to $2700?
When multiple payments are involved, the equation of value is
generally a polynomial, and the method to solve for i is one for
obtaining the roots of a polynomial. Sometimes this can be done
using the quadratic formula.
1000v3 + 2000v6 = 2700
2000v6 + 1000v3 – 2700 = 0
20v6 + 10v3 – 27 = 0
We are only interested in v > 0, and the only positive root from the
quadratic formula is v3 = 0.9385 , from which we obtain v = 0.9791.
v = 1 / (1 + i) = 0.9791
i = 0.02145
At what interest rate convertible semiannually would an investment of
$1000 immediately and $2000 three years from now accumulate to
$5000 ten years from now?
If we let j = i(2)/2, then the equation of value is
1000(1 + j)20 + 2000(1 + j)14 – 5000 = 0
When solving the equation of value involves a polynomial whose
roots cannot be obtained easily, an iterative procedure can be used.
The use of a computer generally makes an iterative procedure much
easier to use. Use the Excel file Interest_Solver to find the solution
to this equation of value as follows:
Type the formula =1000*(1+j)^20+2000*(1+j)^14-5000 in cell A5.
Select options Tools > Solver to solve the equation of value as
indicated below (and if necessary, first use options Tools > Add-Ins)
You should find that j = 0.032178 is the solution, from which we have
i = 0.065391.
Section 2.7 of the textbook discusses some of the ambiguous
interpretations of words used in the “real world” versus the precise
meanings used in mathematics.