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

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:

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(0.00246) = $15.27

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?

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(23.5213) = $32 930

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)

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

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) = 50 000(A/P,8.5%,10) $50 000 1 2 3 n=10 A ? DIAGRAM:

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

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

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(10.2737) + 50(89.0280) = 2 568.43 + 4 451.40 = $7 020 P ? 1 2 3 n=30 $250 DIAGRAM: $100 $50 $1 450

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) = 10 000(F/A,8%,42) – 100(F/G,8%,42)

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)