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**Interest Formulas – Equal Payment Series**

Engineering Economy

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**Equal Payment Series F P**

Using equal payment series you can find present value or future value 1 2 N A A A P N 1 2 N

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**Compound Amount Factor**

A(1+i)N-2 A A A A(1+i)N-1 N 1 2 1 2 N This is a geometric series with the first term as A and the constant r = (1+i) The formula for a geometric series = A (1- r^n)/1-r

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**Equal Payment Series Compound Amount Factor (Future Value of an annuity)**

N A Example: Given: A = $5,000, N = 5 years, and i = 6% Find: F Solution: F = $5,000(F/A,6%,5) = $28,185.46

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**Finding an Annuity Value**

N A = ? Example: Given: F = $5,000, N = 5 years, and i = 7% Find: A Solution: A = $5,000(A/F,7%,5) = $869.50

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**Example: Handling Time Shifts in a Uniform Series**

First deposit occurs at n = 0 i = 6% $5,000 $5, $5, $5,000 $5,000

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Annuity Due Excel Solution Beginning period =FV(6%,5,5000,0,1)

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**Example: College Savings Plan **

Sinking Fund Factor The term between the brackets is called the equal-payment-series sinking-fund factor. F N A Example: College Savings Plan Given: F = $100,000, N = 8 years, and i = 7% Find: A Solution: A = $100,000(A/F,7%,8) = $9,746.78

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**Excel Solution Given: Find: A Using the equation:**

N = 8 years Find: A Using the equation: Using built in Function: =PMT(i,N,pv,fv,type) =PMT(7%,8,0,100000,0) =$9,746.78

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**Capital Recovery Factor**

If we need to find A, given P,I, and N Remember that: Replacing F with its value

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**Capital Recovery Factor**

This factor is called capital recovery factor P N A = ? Example: Paying Off Education Loan Given: P = $21,061.82, N = 5 years, and i = 6% Find: A Solution: A = $21,061.82(A/P,6%,5) = $5,000

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**Example: Deferred Loan Repayment Plan**

i = 6% Grace period A A A A A P’ = $21,061.82(F/P, 6%, 1) i = 6% A’ A’ A’ A’ A’

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Two-Step Procedure Adding the first year interest to the principal then calculating the annuity payment

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**Present Worth of Annuity Series**

N A Example:Powerball Lottery Given: A = $7.92M, N = 25 years, and i = 8% Find: P Solution: P = $7.92M(P/A,8%,25) = $84.54M

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**Example: Early Savings Plan – 8% interest**

44 Option 1: Early Savings Plan $2,000 ? ? Option 2: Deferred Savings Plan 44 $2,000

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**Option 1 – Early Savings Plan**

44 Option 1: Early Savings Plan $2,000 ? Age 31 65

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**Option 2: Deferred Savings Plan**

44 Option 2: Deferred Savings Plan $2,000 ?

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**At What Interest Rate These Two Options Would be Equivalent?**

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**Using Excel’s Goal Seek Function**

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Result

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**Option 1: Early Savings Plan**

$396,644 Option 1: Early Savings Plan 44 $2,000 $317,253 Option 2: Deferred Savings Plan 44 $2,000

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**Interest Formulas (Gradient Series)**

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**Linear Gradient Series**

Gradient-series present –worth factor P

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**Gradient Series as a Composite Series**

We view the cash flows as composites of two series a uniform with a payment amount of A1 and a gradient with a constant amount of G

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Example: $2,000 $1,750 $1,500 $1,250 $1,000 How much do you have to deposit now in a savings account that earns a 12% annual interest, if you want to withdraw the annual series as shown in the figure? P =?

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Method 1: $2,000 $1,750 $1,500 $1,250 $1,000 $1,000(P/F, 12%, 1) = $892.86 $1,250(P/F, 12%, 2) = $996.49 $1,500(P/F, 12%, 3) = $1,067.67 $1,750(P/F, 12%, 4) = $1,112.16 $2,000(P/F, 12%, 5) = $1,134.85 $5,204.03 P =?

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Method 2:

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**Example: Supper Lottery**

$3.44 million Cash Option Annual Payment Option $357,000 G = $7,000 $196,000 $189,000 $175,000

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**Equivalent Present Value of Annual Payment Option at 4.5%**

The gradient series is delayed by one period To return the calculations to year zero

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**Geometric Gradient Series**

Geometric Gradient is a gradient series that is been determined by a fixed rate expressed as a percentage instead of a fixed dollar amount For example the economic problems related to construction cost which involves cash flows that increase or decrease by a constant percentage

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Present Worth Factor Geometric-gradient-series present-worth factor

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**Alternate Way of Calculating P**

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**Unconventional Equivalence Calculations**

EGN3613 Ch2 Part IV

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**Composite Cash Flows $200 $150 $150 $150 $150 $100 $100 $100 $50**

$150 $150 $150 $150 $100 $100 $100 $50

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**Unconventional Equivalence Calculations**

Situation: What value of A would make the two cash flow transactions equivalent if i = 10%?

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**Multiple Interest Rates**

F = ? Find the balance at the end of year 5. 6% 4% 4% 6% 5% 2 4 5 1 3 $400 $300 $500

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Solution

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**Cash Flows with Missing Payments**

$100 i = 10% Missing payment

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**Solution P = ? i = 10% Add this cash flow to offset the change $100**

$100 Pretend that we have the 10th payment i = 10%

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Approach P = ? $100 $100 i = 10% Equivalent Cash Inflow = Equivalent Cash Outflow

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**Equivalence Relationship**

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**Unconventional Regularity in Cash Flow Pattern**

$10,000 i = 10% 1 C C C C C C C Payment is made every other year

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**Approach 1: Modify the Original Cash Flows**

$10,000 i = 10% 1 A A A A A A A A A A A A A A

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**Relationship Between A and C**

$10,000 i = 10% 1 C C C C C C C $10,000 i = 10% 1 A A A A A A A A A A A A A A

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Solution i = 10% C A A A =$1,357.46

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**Approach 2: Modify the Interest Rate**

Idea: Since cash flows occur every other year, let's find out the equivalent compound interest rate that covers the two-year period. How: If interest is compounded 10% annually, the equivalent interest rate for two-year period is 21%. (1+0.10)(1+0.10) = 1.21

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Solution $10,000 i = 21% 1 C C C C C C C

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