Interest Formulas – Equal Payment Series

Engineering Economy

Equal Payment Series

0 1 2 N

0 1 2 N

A A A

F

P

0 N

Using equal payment series you can find present value or future value

Compound Amount Factor

0 1 2 N 0 1 2 N

A A A

F

A(1+i)N-1

A(1+i)N-2

1 2(1 ) (1 )N NF A i A i A 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)

F Ai

iA F A i N

N

( )

( / , , )

1 1

Example:

Given: A = $5,000, N = 5 years, and i = 6%

Find: F

Solution: F = $5,000(F/A,6%,5) = $28,185.46

0 1 2 3N

F

A

Finding an Annuity Value

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

0 1 2 3N

F

A = ?

A Fi

i

F A F i N

N

( )

( / , , )

1 1

Example: Handling Time Shifts in a Uniform Series

F = ?

0 1 2 3 4 5

$5,000 $5,000 $5,000 $5,000 $5,000

i = 6%

First deposit occurs at n = 0

5 $5,000( / ,6%,5)(1.06)

$29,876.59

F F A

Annuity Due

Excel Solution

=FV(6%,5,5000,0,1)Beginning period

Sinking Fund Factor

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

0 1 2 3N

F

A

1)1( Ni

iFA

The term between the brackets is called the equal-payment-series sinking-fund factor.

Excel Solution

Given: F = $100,000 i = 7% N = 8 years

1)1( Ni

iFA

0

1 2 3 4 5 6 7 8

$100,000

i = 8%

A = ?

Current age: 10 years old

• Find: AUsing the equation:

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

Capital Recovery Factor

1)1( Ni

iFA

NiPF )1(

1)1(

)1(N

N

i

iiPA

If we need to find A, given P,I, and N

Remember that:

Replacing F with its value

Capital Recovery Factor

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

1 2 3N

P

A = ?

0

A Pi i

i

P A P i N

N

N

( )

( )

( / , , )

1

1 1

This factor is called capital recovery factor

P =$21,061.82

0 1 2 3 4 5 6

A A A A A

i = 6%

0 1 2 3 4 5 6

A’ A’ A’ A’ A’

i = 6%

P’ = $21,061.82(F/P, 6%, 1)

Grace period

Example: Deferred Loan Repayment Plan

Two-Step Procedure

' $21,061.82( / ,6%,1)

$22,325.53

$22,325.53( / ,6%,5)

$5,300

P F P

A A P

Adding the first year interest to the principal then calculating the annuity payment

Present Worth of Annuity Series

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

1 2 3N

P = ?

A

0

P Ai

i i

A P A i N

N

N

( )

( )

( / , , )

1 1

1

0 1 2 3 4 5 6 7 8 9 10 11 12

44

Option 2: Deferred Savings Plan

$2,000

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 1 – Early Savings Plan

10

44

$2,000( / ,8%,10)

$28,973

$28,973( / ,8%,34)

$396,645

F F A

F F P

0 1 2 3 4 5 6 7 8 9 10

44

Option 1: Early Savings Plan

$2,000

?

6531Age

Option 2: Deferred Savings Plan

44 $2,000( / ,8%,34)

$317,233

F F A

0 11 12

44

Option 2: Deferred Savings Plan

$2,000

?

At What Interest Rate These Two Options Would be Equivalent?

44

44

Option 1:

$2,000( / , ,10)( / , ,34)

Option 2:

$2,000( / . ,34)

Option 1 = Option 2

$2,000( / , ,10)( / , ,34) $2,000( / . ,34)

Solve for

F F A i F P i

F F Ai

F A i F P i F Ai

i

123456789101112131415161718192021224041424344454647

A B C D E F

Year Option 1 Option 201 (2,000)$ 2 (2,000)$ Interest rate 0.083 (2,000)$ 4 (2,000)$ FV of Option 1 396,645.95$ 5 (2,000)$ 6 (2,000)$ FV of Option 2 317,253.34$ 7 (2,000)$ 8 (2,000)$ Target cell 79,392.61$ 9 (2,000)$ 10 (2,000)$ 11 (2,000)$ 12 (2,000)$ 13 (2,000)$ 14 (2,000)$ 15 (2,000)$ 16 (2,000)$ 17 (2,000)$ 18 (2,000)$ 19 (2,000)$ 37 (2,000)$ 38 (2,000)$ 39 (2,000)$ 40 (2,000)$ 41 (2,000)$ 42 (2,000)$ 43 (2,000)$ 44 (2,000)$

Using Excel’s Goal Seek Function

Result

0 1 2 3 4 5 6 7 8 9 10

44

0 1 2 3 4 5 6 7 8 9 10 11 12

44

Option 1: Early Savings Plan

Option 2: Deferred Savings Plan

$2,000

$2,000

$396,644

$317,253

Interest Formulas(Gradient Series)

Linear Gradient Series

),,/(

1

112

NiGPGP

ii

iNiGP N

N

P

Gradient-series present –worth factor

Gradient Series as a Composite SeriesWe 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

$1,000$1,250 $1,500

$1,750$2,000

1 2 3 4 50

P =?

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?

Example:

Method 1:

$1,000$1,250 $1,500

$1,750$2,000

1 2 3 4 50

P =?

$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

Method 2:

P P A1 000 12%,5

604 80

$1, ( / , )

$3, .

P P G2 12%,5

599 20

$250( / , )

$1, .

P

$3, . $1, .

$5,204

604 08 599 20

0 1 2 3 4 5 6 7 25 26

$3.44 million

0 1 2 3 4 5 6 7 25 26

Cash Option

$175,000$189,000

$357,000

$196,000G = $7,000

Annual Payment Option

Example: Supper Lottery

Equivalent Present Value of Annual Payment Option at 4.5%

[$175,000 $189,000( / , 4.5%,25)

$7,000( / , 4.5%,25)]( / , 4.5%,1)

$3,818,363

P P A

P G P F

The gradient series is delayed by one period

To return the calculations to year zero

Geometric Gradient SeriesGeometric 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

giifi

NA

giifgi

igA

P

NN

,)1(

,)1()1(1

1

1

Geometric-gradient-series present-worth factor

Alternate Way of Calculating P

1

Let '1

( / , ', )(1 )

i gg

g

AP P A g N

g

Unconventional Equivalence CalculationsEGN3613

Ch2 Part IV

$50

$100 $100 $100

$150 $150 $150 $150

$200

Group 1 $50( / ,15%,1)

$43.48

P P F

Group 2 $100( / ,15%,3)( / ,15%,1)

$198.54

P P A P F

Group 3 $150( / ,15%,4)( / ,15%,4)

$244.85

P P A P F

Group 4 $200( / ,15%,9)

$56.85

P P F

$43.48 $198.54 $244.85 $56.85

$543.72

P

0

1 2 3 4 5 6 7 8 9

Composite Cash Flows

Unconventional Equivalence Calculations

Situation:

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

Multiple Interest Rates

$300$500

$400

5% 6% 6% 4% 4%

Find the balance at the end of year 5.

0

12

34 5

F = ?

Solution1:

$300( / ,5%,1) $315

2 :

$315( / ,6%,1) $500 $833.90

3:

$833.90( / ,6%,1) $883.93

4 :

$883.93( / , 4%,1) $400 $1,319.29

5 :

$1,319.29( / , 4%,1) $1,372.06

n

F P

n

F P

n

F P

n

F P

n

F P

Cash Flows with Missing PaymentsP = ?

$100

0

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Missing paymenti = 10%

Solution

P = ?

$100

01 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Pretend that we have the 10th

paymenti = 10%

$100Add this cash flow tooffset the change

Approach

P = ?

$100

01 2 3 4 5 6 7 8 9 10 11 12 13 14 15

i = 10%

$100

Equivalent Cash Inflow = Equivalent Cash Outflow

$100( / ,10%,10) $100( / ,10%,15)

$38.55 $760.61

$722.05

P P F P A

P

P

Equivalence Relationship

Unconventional Regularity in Cash Flow Pattern

$10,000

0

1 2 3 4 5 6 7 8 9 10 11 12 13 14

C C C C C C C

i = 10%

Payment is made every other year

Approach 1: Modify the Original Cash Flows

$10,000

0

1 2 3 4 5 6 7 8 9 10 11 12 13 14

i = 10%

A A A A A A A A A A A A A A

$10,000( / ,10%,14)

$1,357.46

A A P

Relationship Between A and C

$10,000

01 2 3 4 5 6 7 8 9 10 11 12 13 14

i = 10%

A A A A A A A A A A A A A A

$10,000

0

1 2 3 4 5 6 7 8 9 10 11 12 13 14

C C C C C C C

i = 10%

C

A =$1,357.46

A A

i = 10%

$10,000( / ,10%,14)

$1,357.46

( / ,10%,1)

1.1

2.1

2.1($1,357.46)

$2,850.67

A A P

C A F P A

A A

A

Solution

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

0

1 2 3 4 5 6 7 8 9 10 11 12 13 14

C C C C C C C

i = 21%

$10,000( / , 21%,7)

$2,850.67

C A P

1 2 3 4 5 6 7