CE 203 More Interest Formulas (EEA Chap 4)

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CE 203 More Interest Formulas(EEA Chap 4)
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Example of a Uniform Series CFD
Cost Flow Diagram
F
P
A
A
A
A
A
A
1
2
3
4
5
6
0
n 6 for this example i must matchall
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
F A A (F/A, i, n)
(1 i)n - 1
i
A
A
A
A
A
A
1
2
3
4
5
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 A (F/A, i, n)
(1 i)n - 1
i
is solved for A
A F F (A/F, i, n)
i
(1 i)n - 1
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 and F P (1 i)n
i
(1 i)n - 1
are combined (and solved for A)
A P P (A/P, i, n)
i (1 i)n
(1 i)n - 1
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 20-year 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
(Total payments 485,496.86 29,000)
0.00583 (1 0.00583)240
(1 0.00583)240 - 1
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Uniform Series Present Worth
If the equation
A P P (A/P, i, n)
i (1 i)n
(1 i)n - 1
is solved for P (todays value or Present Worth)
P A A (P/A, i, n)
(1 i)n - 1
i (1 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 5916.78 (or
P 180 (P/A, .005, 36) 180 (32.871) 5916.78)
(1 0.005)36 - 1
0.005 (1 0.005)36
Note 36 x 180 6480
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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,,)
This knowledge may be useful for your future
spreadsheet homework
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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