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k−1
yk( ) = ( +1 r C) k − p ∑ ( +1 r) i (2.7)
i=0

Using the expression for the sum of a geometric series, from the appendix, the
expression for y(k) then reduces to:

 ( +1 r) −  1
k
yk() = ( +1 r C) k − p   (2.8)
 r 
At k = N, the debt is paid off and the bank is owed no further payment;
therefore:

( +
N
 1 r) −  1
( +
N
yN() = 0 = 1 r) C − p   r   (2.9)
From this equation, we can determine the amount of each of the (equal)
payments:
r + r) N
(1
p = C (2.10)
(1 + r) N − 1
Question: What percentage of the first payment is going into retiring the
principal?

In-Class Exercises
Pb. 2.3 Given the principal, the number of periods and the interest rate, use
Eq. (2.10) to write a MATLAB program to find the amount of payment per
period, assuming the payment per period is the same for all periods.

Pb. 2.4 Use the same reasoning as for the amortization problem to write the
difference equation for an individual’s savings plan. Let y(k) be the savings
th
balance on the first day of the k year and u(k) the amount of deposit made
th
in the k year.
Write a MATLAB program to compute y(k) if the sequence u(k) and the inter-
est rate r are given. Specialize to the case where you deposit an amount that
increases by the rate of inflation i. Compute and plot the total value of the
savings as a function of k if the deposit in the first year is $1000, the yearly
interest rate is 6%, and the yearly rate of inflation is 3%. (Hint: For simplicity,
assume that the deposits are made on December 31 of each year, and that the
balance statement is issued on January 1 of each year.)

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