# Interest Formulas – Equal Payment Series

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Interest Formulas – Equal Payment Series
Engineering Economy

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

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

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

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

Example: Handling Time Shifts in a Uniform Series
First deposit occurs at n = 0 i = 6% \$5,000 \$5, \$5, \$5,000 \$5,000

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

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

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

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

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

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’

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

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

Example: Early Savings Plan – 8% interest
44 Option 1: Early Savings Plan \$2,000 ? ? Option 2: Deferred Savings Plan 44 \$2,000

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

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

At What Interest Rate These Two Options Would be Equivalent?

Using Excel’s Goal Seek Function

Result

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

Interest Formulas (Gradient Series)

Gradient-series present –worth factor P

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

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 =?

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 =?

Method 2:

Example: Supper Lottery
\$3.44 million Cash Option Annual Payment Option \$357,000 G = \$7,000 \$196,000 \$189,000 \$175,000

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

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

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

Alternate Way of Calculating P

Unconventional Equivalence Calculations
EGN3613 Ch2 Part IV

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

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

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

Solution

Cash Flows with Missing Payments
\$100 i = 10% Missing payment

Solution P = ? i = 10% Add this cash flow to offset the change \$100
\$100 Pretend that we have the 10th payment i = 10%

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

Equivalence Relationship

Unconventional Regularity in Cash Flow Pattern
\$10,000 i = 10% 1 C C C C C C C Payment is made every other year

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

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

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

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

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

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