CE 203
More Interest Formulas
(EEA Chap 4)
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Example of a Uniform Series CFD
Cost Flow Diagram
F
P
A
1
A
2
A
3
A
4
A
5
A
6
0
n = 6 for this example; i must “match”
all amounts (arbitrarily) shown as positive
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Uniform Series Formulas: Conventions
– “A” is a payment that occurs at the
end of each of a series of time periods
– “P” occurs one payment period before
the first “A”
– “F” occurs at the same time as the last
“A” and “n” periods after “P”
– “i” is the interest rate per period
– “n” is the number of periods
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Uniform Series Compound Amount Factor
(e.g., many investment programs)
F
[
(1 + i)n - 1
i
] = A (F/A, i, n)
A
A
A
F=A
1
2
3
A
4
A
5
A
6
0
(see text p. 87 for derivation of formula)
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Example 1: Series Compound Amount
You deposit $1000 per year for 30 years in an
account that earns 10%, compounded yearly.
How much could you withdraw from the
account at the end of 30 years?
F = 1000 (F/A, 0.1, 30) = A [(1 + i)n - 1 ]/i
= $1000 [(1 + 0.1)30 – 1] / 0.1
= $164,494.02
Or from text p. 578, F = 1000 (164.494)
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Example 2: Series Compound Amount
You deposit $100 per month in a Roth IRA from
age 25 to retirement at age 65 at 8%. How
much would you have in the account at
retirement?
F = 100 (F/A, 0.08/12, 40 x 12)
= $100 [(1 + 0.00667)480 – 1] / 0.00667
= $349,505.19
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Uniform Series Sinking Fund
If the equation
F=A
[
(1 + i)n - 1
i
]
= A (F/A, i, n)
i
(1 + i)n - 1
]
= F (A/F, i, n)
is solved for A
A=F
[
Used to determine how much money (A) needs to be
saved per period to obtain a given future amount (F)
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Example 1: Series Sinking Fund
Your company plans to purchase a new instrument
in 2 years at a guaranteed price of $60,000.
Assuming your bank will pay 6% interest,
compounded monthly, how much should your
company put aside each month in order to
purchase the instrument?
A = 60,000 (A/F, 0.005, 24)
= $60,000 (.005) / [(1 + 0.005)24 – 1]
= $2,359.24 (per month)
Note: $60,000/24 = $2,500
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Uniform Series Capital Recovery
If the equations
A=F
[
i
(1 + i)n - 1
]
and
F = P (1 + i)n
are combined (and solved for A)
A=P
[
i (1 + i)n
(1 + i)n - 1
]
= P (A/P, i, n)
Used to determine how much money (A) needs to be
paid per period to repay or recover an initial amount
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Example 1: Series Capital Recovery
You just purchased a home for $290,000.
Your agreement is 10% down and a 20year mortgage at 7%. What are your
monthly payments going to be?
P = .9 (290,000) = $261,000
i = .07/12 = .00583
n = 240
A = $261,000
[
= $2022.90
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0.00583 (1 + 0.00583)240
(1 + 0.00583)240 - 1
]
(Total payments = $485,496.86 + $29,000)
Uniform Series Present Worth
If the equation
A=P
[
i (1 + i)n
(1 + i)n - 1
]
= P (A/P, i, n)
is solved for P (today’s value or Present Worth)
P=A
[
(1 + i)n - 1
i (1 + i)n
]
= A (P/A, i, n)
Used to calculate the Present Worth of a series of
regular and equal future payments
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Example 1: Series Present Worth
Jill just inherited a sum of money and has decided to
put some of it into a savings account that can be used
to pay her utility bills over the next three years. If her
monthly utility bills are a uniform $180 per month, how
much should she put into the savings account if it
earns 6% per year, compounded monthly?
P = $180
[
(1 + 0.005)36 - 1
0.005 (1 + 0.005)36
]
= $ 5916.78
(or P = 180 (P/A, .005, 36) = 180 (32.871) = $5916.78)
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Note: 36 x $180 = $6480
In-class Example 2 (unknown i)
A bank offers to pay $16,000 in 10 years for a
deposit of $100 per month for that 10-year time
period. A second bank promises a rate of return
of 4.5% for the same payment plan. What interest
rate is the first bank offering? Which is the better
offer? (Assume monthly compounding for both.)
Also can use EXCEL functions FV (trial and error) or IRATE
F = FV (rate, nper, pmt, fv, type) = FV (trial, 120, -100,,)
i = RATE (nper, pmt, pv, fv, type) = RATE (120, -100,,16000,,)
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This knowledge may be useful for your future spreadsheet homework
In-class Example 3
Dave just bought a new car that came with an all
inclusive (i.e., all maintenance included) 5-year
warranty. Starting in year 6, Dave estimates that
maintenance will be $800/year. He plans to keep
the car for 8 years. Assuming he can earn 10%
(compounded yearly) on his money, how much
should he deposit now in order to cover the
maintenance costs in years 6, 7, and 8?
Answer: Calculate the “future present value” in Y5 = $1991
and convert to the present value in Y0 = $1236
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