Uniform Payment Series: Calculating Compound Amount & Present Worth, Slides of Calculus

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Chapter 4

More Interest Formulas

Uniform Series Compound Interest Formulas

• Why?

Many payments are based on a uniform

payment series.

e.g. automobile loans, house payments, and

many other loans.

“A” payments in terms of “F”

Amount A is invested at the end of each year for

four years A A A A

F

F = A(1+i)

3

• A(1+i)

2

• A(1+i) + A

F = A(1+i)

n-

• A(1+i)

n- + ….. + A( 1+i)

2

• A(1 +i) + A …( 1)

Multiply by (1+i)

(1+i)F = A(1+i)

n +A(1+i)

n-

• ...+A(1+i)

3 +A(1+i)

2 +A(1+i)

…( 2)

Subtract eqn (1) from eqn (2) n

4

(1+i)F – F = A(1+i)

n

- A

i F = A[(1 + i )

n

- 1 ]

A^ F A i n

i

i F A

n

, %,

Where A(F/A, i%, n) is called uniform series

compound amount factor

Example 4-

A man deposits $500 in a credit union at the end of

each year for 5 years. The credit union pays 5%

interest, compounded annually.

At the end of 5 years, immediately after the fifth

deposit, how much does the man have in his

account?

A = $500 , i = 5% , n = 5

A A A A

F

A

0 1 2 3 4 5 A A A A A

0 1 2 3 4 5

F

Man’s point of view Credit Union point of view

Formula relating A and P

F^ A F i n

i

i A F n , %, 1 1

  

  



 



P A P i n i

i i A P n

n

^ 

10

A A^ A^ A^ A^ A^ A^ A

P

(^1) n

n

i

i A P i

Where P(A/P, i%, n) is called uniform series capital

recovery factor

Also, we can solve for P in terms of A

 

 

A P A i n i i

i P A n

n

, %, 1

Where A(P/A, i%, n) is called uniform series

present worth factor

Example 4-

• An investor has a purchase contract on some machine tools.
• He will be paid $140 at the end of each month for a 5-year period
• The investor offers to sell you the contract for a $6800 cash today.
• If you otherwise can make 1% per month on your money, would you accept or reject the investor’s offer.

P

A = $

i = 1% n = 60 months

i.e. if you take the contract, you

will be paid $140 per month for a period of 60 months. That means a total of $ over the 5-year period

We need to determine if the contract is worth $

P = A(P/A, 1%, 60) = 140(44.955) = $6293.

… Example 4-

• It is clear that if we pay $6800 for the contract and receive

$140 per month, we will receive less than the 1% per month interest (which is the interest that we can otherwise make). Therefore the offer should be rejected.

• OR, we can say that the $140 per month payment is

equivalent to $6293.7. Therefore if we pay $6800 for a benefit of $140 per month, we will lose money

• OR, we can say that the $6800 will give me more than $

per month if invested at the given interest rate of 1%, which means that investing the $6800 in the purchase contract will be a loss when compared to investing it in in the other investment opportunity (which gives a 1% interest rate)

Therefore, Reject the offer.

Example 4-

• Using 15% interest rate, compute F

100

F

100 100

Example 4-

• Using 15% interest rate, compute P

20

P

30 20

Relationships Between Compound Interest Factors

How?

P = A(1+ i)-1^ + A(1+ i)-2^ + ……+ A( 1 + i)-n

P = A[(1+ i)-1^ + (1+ i)-2^ +……+ ( 1+ i)-n^ ]

P = A[(P/F,i,1)+(P/F,i,2)+...+(P/F,i,n)]

since P = A(P/A, i, n)

We conclude that (P/A, i, n) = (P/F,i,1)+(P/F,i,2)+...+(P/F,i,n)

 

 

n

J

P A i n P F i J 1

( / , , ) ( / , , )





n

J

P A i n P F i J 1

( / , , ) ( / , , )

(^0 1 2 3) n

A A A A …..

P

For Example:

(P/A, 5%, 4) = (P/F, 5%, 1) + (P/F, 5%, 2) + (P/F, 5%, 3) + (P/F, 5%, 4)

Relationships Between Compound Interest Factors



 

  

1 1

( / ,, ) 1 ( / ,, )

n J

F Ain F Pi J







 

1

1

( / , , ) 1 ( / , , )

n

J

F A i n F P i J

0 1 2 3 n

A

F

A A A …..

F = A + A(1+ i) + A(1+ i)^2 + ……+ A( 1+ i)n-

F = A[ 1 + (1+ i) + (1+ i)^2 + ... + (1+ i)n-1^ ]

F = A[ 1 + (F/P,i,1) + (F/P,i,2) + ... + (F/P,i,n-1)]

since F = A(F/A,i,n)

We conclude that (F/A,i,n) = 1 + (F/P,i,1) + (F/P,i,2) + ... + (F/P,i,n-1)