CE 203
More Interest Formulas
(EEA Chap 4)
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Formulas for non-uniform payments
Arithmetic Gradient: payments
increase by a uniform
AMOUNT each payment period
 Geometric Gradient: payments
increase by a uniform RATE
each payment period

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Arithmetic Gradient Series
A+(n-1)G
Amount increases by
A+3G
“G” each period
A+2G
A+G
A
= (can be divided into)
P
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Arithmetic Gradient Series Components
(n-1)G
A
A
A
A
A
3G
2G
G
0
+
P’
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P’’
Arithmetic Gradient Future Worth
The Future Worth of an arithmetic gradient cash
flow is given by
G
F’’ = i
[
(1 + i)n - 1
]
(n-1)G
-n
i
3G
2G
=G
[
(1 + i)n – in - 1
i2
]
(see text p. 99-100 for
derivation of formula)
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G
0
F’’
Arithmetic Gradient Uniform Series Factor
If the arithmetic gradient factor is multiplied by the
sinking fund factor, the result is called the
Arithmetic Gradient Uniform Series Factor:
A=G
[
(1 + i)n – in - 1
i {(1 +
i)n
- 1}
]
= G (A/G, i, n)
(gives the uniform series cash flow payment
equivalent to that of an arithmetic gradient cash flow)
(see text p. 100 for derivation of formula)
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In-class Example 1 (arithmetic series)
Your company just purchased a piece
of equipment. Maintenance costs are
estimated at $1200 for the first year and
are expected to rise by $300 in each of the
subsequent four years. How much should
be set aside in a “maintenance account”
now to cover these costs for the next five
years? Assume payments are made at the
end of each year and an interest rate of 6%.
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Geometric Gradient Series
A1(1+g)4
A1(1+g)3
A1(1+g)2
A1
P
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A1(1+g)
- Amount changes at the uniform RATE, g
- Useful for some types of problems such
as those involving inflation
Present Worth for Geometric Gradient Series
A1(1+g)4
P = A1
[
1 - (1 + g)n (1 + i)-n
i-g
]
A1(1+g)3
= A1 (P/A, g, i, n)
for i ≠ g
A1
P
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A1(1+g)2
A1(1+g)
Can be very complex unless
programmed on a computer
In-class Example 2 (geometric series)
You have just begun you first job as a civil
engineer and decide to participate in the
company’s retirement plan. You decide to
invest the maximum allowed by the plan which
is 6% of your salary. Your company has told
you that you can expect a minimum 4% increase
in salary each year assuming good performance
and typical advancement within the company.
1) Choose a realistic starting salary
2) Assuming you stay with the company, the
company matches your 6% investment in the
retirement plan, expected minimum salary
increases, and an interest rate of 10%, how
much will you have in your retirement account
after 40 years?
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Nominal vs. Effective interest rate
Nominal interest rate, r: annual interest rate
without considering the effect of any
compounding at shorter intervals
so that i (for use in equations) = r/m
where m is number of
compounding periods per year
(this is what we have been doing)
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Nominal vs. Effective interest rate
Effective interest rate, ia: annual interest
rate taking into account the effect of any
compounding at shorter intervals; also
called “yield”
An amount of $1, invested at r%,
compounded m times per year, would
be worth $1(1 + r/m)m and the effective
interest 1(1 + r/m)m - 1
ia = (1 + i)m – 1
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(see text p. 110, 9th Ed for derivation)
Example
A bank pays 6% nominal interest rate.
Calculate the effective interest with
a) monthly, b) daily, c) hourly d) secondly
compounding
ia = (1 + i)m – 1
ia monthly = (1 + .06/12)12 -1 = 6.1678 %
ia daily = (1 + .06/365)365 -1 = 6.183 %
ia hourly = (1 + .06/8760)8760 -1 = 6.1836 %
ia secondly = (1 + .06/31.5M)31.5M -1 = 6.18365 %
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Continuous Compounding
The effective interest rate for continuous
compounding (i.e., as the length of the
compounding period → 0 and the number of
periods → ∞)
ia = er – 1
(see text pp. 116-117 for derivation and equations
involving continuous compounding)
In our previous example, ia = e0.06 – 1 = 6.183655%
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Nominal vs. Effective interest rate
Comparison of nominal and effective interest rates
for various APR values and compounding periods
Nominal
rate
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Effective interest rate, ia, for compounding …
r
Yearly
Semiannually
Monthly
Daily
Continuously
5
5.0000
5.0625
5.1162
5.1268
5.1271
10
10.0000
10.2500
10.4713
10.5156
10.5171
15
15.0000
15.5625
16.0755
16.1798
16.1834
20
20.0000
21.0000
21.9391
22.1336
22.1403
Adjustments may be needed because…




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Cash flow period is not equal to
compounding period
There is cash flow in only some of
periods
Pattern does not exactly fit any of
basic formulas
Etc.
Partial review

Single payment compound amount
F = P(1+i)n = P(F/P,i,n)

Uniform series (sinking fund)
A=F

CCEE
]
= F (A/F, i, n)
Arithmetic gradient series
A=G
ISU
[
i
(1 + i)n - 1
[
(1 + i)n – in - 1
i {(1 + i)n - 1}
]
= G (A/G, i, n)