1. Let’s work some problems…
If $1,600 is deposited (today) into an account earning 12% compounded annually, what amount of money will be in the account at the end of 17 years?
GIVEN:
P = $1 600
i = 12%
FIND F17: 1 2 3
n=17
$1 600
F17?
DIAGRAM:
F17 = P(F/P,i,n)
= 1 600(F/P,12%,17)
= 1 600(6.8660) = $10 986

2. Example 1 - Concept
Assuming a 12% per year return on my investment, I am indifferent between having $1,600 today and $10,986 in 17 years. 1 2 3
n=17
$1 600
$10 986
DIAGRAM:

3. Example 2
What is the present value of having $6,200 fifty-three years from now at 12% compounded annually?
GIVEN:
F53 = $6 200
i = 12%
FIND P: 1 2 3
n=53 P?
$6 200
DIAGRAM:
P = F53(P/F,i,n) = F53(1+ i)–n
= 6 200(1+ .12)–53
= 6 200( ) = $15.27

4. Example 2 - Concept
Having $15.27 today is equivalent to having $6,200 in 53 years assuming that I will invest that $15.27 and earn 12% per year on my money.
P50 = F53(P/F,i,n)
= 6 200(P/F,12%,3)
= 6 200(0.7118) = F50
ALTERNATIVE:
$6 200
F50 = P50
P50 3 1 2 1 2 3 51 52
n=53
P = F50(P/F,i,n)
= P50(P/F,12%,50)
= 6 200(0.7118)(P/F,12%,50)
= 6 200(0.7118)(0.0035) = $15.45 P?

5. Example 3
If $1,400 is deposited at the end of each year (every year) for 10 years, what is the accumulated value at the end of 10 years at 18% compounded annually? 1 2 3
n=10
$1 400
DIAGRAM: 9
F10 ?
F10 = A(F/A,i,n)
= 1 400(F/A,18%,10)
= 1 400( ) = $32 930

6. Example 4
What series of equal yearly payments must be made into an account to accumulate $90,000 in 72 years at 6.3% compounded annually?
$90 000 1 2 3
n=72
A ?
DIAGRAM:
A = F72(A/F,i,n) = F72(A/F,6.3%,72)

7. Example 5
What is the present worth of deposits of $1,000 at the end of each of the next 9 years at 8% compounded annually?
P ? 1 2 3
n=9
$1 000
DIAGRAM:
P = A(P/A,i,n)
= 1 000(P/A,8%,9)
= 1 000(6.2469) = $6 247

8. Example 6
What series of equal, annual payments is necessary to repay $50,000 in 10 years at 8.5% compounded annually?
A = P(A/P,i,n)
= (A/P,8.5%,10)
$50 000 1 2 3
n=10
A ?
DIAGRAM:

9. Example 7
What is the present worth of a series of 30 end of the year payments that begin at $250 and increase at the rate of $50 a year if the interest rate is 9% compounded annually?
P ? 1 2 3
n=30
$250
DIAGRAM:
$100
$50
$1 450

10. Example 7
This can be broken into an Annual flow and a Linear Gradient flow
P ? 1 2 3
n=30
$250
DIAGRAM:
$100
$50
$1 450
PA ? 1 2 3
n=30
$250
PG ? 1 2 3
n=30
$100
$50
$1 450

11. Example 7 - Concept
Annual payments that begin at $250 and increase with a linear gradient of $50 each year for 30 years are equivalent to $7,020 today, assuming 9% return on investment.
P = PA + PG = A(P/A,i,n) + G(P/G,i,n)
= 250(P/A,9%,30) + 50(P/G,9%,30)
= 250( ) + 50( )
= = $7 020
P ? 1 2 3
n=30
$250
DIAGRAM:
$100
$50
$1 450

12. Example 8
What is the future worth of a series of 42 deposits that begin at $10,000 and decrease at the rate of $100 a year with 8% interest compounded annually?
FA ? 1 2 3
n=42
$10 000
FG ? 1 2 3
n=42
$200
$100
$4 100
F42 = FA – FG = A(F/A,i,n) – G(F/G,i,n)
= (F/A,8%,42) – 100(F/G,8%,42)

13. Example 8 - Concept
$2,714,631 would be accumulated if yearly payments are made that begin at $10,000 and decrease yearly with a linear gradient of $100, given an interest rate of 8%.
F42 = A(F/A,i,n) – G(F/G,i,n)