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Let’s work some problems…

DIAGRAM:. F 17 ?. 0. 1. 2. 3. F 17 = P(F/P,i,n) = 1 600(F/P,12%,17) = 1 600(6.8660) = $10 986. n=17. $1 600. Let’s work some problems….

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Let’s work some problems…

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  1. DIAGRAM: F17? 0 1 2 3 F17 = P(F/P,i,n) = 1 600(F/P,12%,17) = 1 600(6.8660) = $10 986 n=17 $1 600 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:

  2. DIAGRAM: $10 986 0 1 2 3 n=17 $1 600 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.

  3. DIAGRAM: $6 200 0 P = F53(P/F,i,n) = F53(1+ i)–n = 6 200(1+ .12)–53 = 6 200(0.00246) = $15.27 1 2 3 n=53 P? 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:

  4. F50= P50 0 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 1 2 3 1 2 3 P50 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 51 52 n=53 P?

  5. DIAGRAM: F10? F10 = A(F/A,i,n) = 1 400(F/A,18%,10) = 1 400(23.5213) = $32 930 0 9 1 2 3 n=10 $1 400 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?

  6. DIAGRAM: $90 000 0 1 2 3 n=72 A ? 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? A = F72(A/F,i,n) = F72(A/F,6.3%,72)

  7. DIAGRAM: P ? P = A(P/A,i,n) = 1 000(P/A,8%,9) = 1 000(6.2469) = $6 247 1 2 3 n=9 0 $1 000 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?

  8. DIAGRAM: $50 000 1 2 3 n=10 0 A ? 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)

  9. DIAGRAM: P ? 1 2 3 n=30 0 $250 $50 $100 $1 450 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?

  10. DIAGRAM: PA ? 1 2 3 P ? n=30 1 2 3 n=30 0 $250 0 $250 $50 PG ? $100 1 2 3 n=30 $1 450 $50 0 $100 $1 450 Example 7 This can be broken into an Annual flow and a Linear Gradient flow

  11. DIAGRAM: P ? 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 1 2 3 n=30 0 $250 $50 $100 $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.

  12. FA ? FG ? 1 2 3 1 2 3 n=42 n=42 0 0 $200 $100 $10 000 $4 100 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? F42 = FA–FG = A(F/A,i,n) – G(F/G,i,n) = 10 000(F/A,8%,42) – 100(F/G,8%,42)

  13. F42 = A(F/A,i,n) – G(F/G,i,n) 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%.

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