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) 0 1 2 3 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 0 1 2 3 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% 0 1 2 3 4 5 $5,000 $5,000 $5,000 $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 0 1 2 3 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 1 2 3 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% 0 1 2 3 4 5 6 Grace period A A A A A P’ = $21,061.82(F/P, 6%, 1) i = 6% 0 1 2 3 4 5 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 1 2 3 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 0 1 2 3 4 5 6 7 8 9 10 44 Option 1: Early Savings Plan $2,000 ? ? Option 2: Deferred Savings Plan 0 1 2 3 4 5 6 7 8 9 10 11 12 44 $2,000

Option 1 – Early Savings Plan 0 1 2 3 4 5 6 7 8 9 10 44 Option 1: Early Savings Plan $2,000 ? Age 31 65

Option 2: Deferred Savings Plan 0 11 12 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 0 1 2 3 4 5 6 7 8 9 10 44 $2,000 $317,253 Option 2: Deferred Savings Plan 0 1 2 3 4 5 6 7 8 9 10 11 12 44 $2,000

Interest Formulas (Gradient Series)

Linear 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 1 2 3 4 5 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 2 3 4 5 $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 0 1 2 3 4 5 6 7 25 26 Annual Payment Option $357,000 G = $7,000 $196,000 $189,000 $175,000 0 1 2 3 4 5 6 7 25 26

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 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

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 1 2 3 4 5 6 7 8 9

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 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 $100 i = 10% Missing payment

Solution P = ? i = 10% Add this cash flow to offset the change $100 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 $100 Pretend that we have the 10th payment i = 10%

Approach P = ? $100 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 $100 i = 10% Equivalent Cash Inflow = Equivalent Cash Outflow

Equivalence Relationship

Unconventional Regularity in Cash Flow Pattern $10,000 i = 10% 1 2 3 4 5 6 7 8 9 10 11 12 13 14 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 2 3 4 5 6 7 8 9 10 11 12 13 14 A A A A A A A A A A A A A A

Relationship Between A and C $10,000 i = 10% 1 2 3 4 5 6 7 8 9 10 11 12 13 14 C C C C C C C $10,000 i = 10% 1 2 3 4 5 6 7 8 9 10 11 12 13 14 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 2 3 4 5 6 7 1 2 3 4 5 6 7 8 9 10 11 12 13 14 C C C C C C C