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Concepts from the example problems

Concepts from the example problems. ( F / P, i, N)  Problem 1 ( P / F, i, N)  Problem 2 ( F / A, i, N)  Problem 3 ( A / F, i, N)  Problem 4 ( P / A, i, N)  Problem 5 ( A / P, i, N)  Problem 6 ( P / G, i, N)  Problem 7 ( A / G, i, N)  Problem 8. DIAGRAM:.

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Concepts from the example problems

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  1. Concepts from the example problems ( F / P, i, N)  Problem 1 ( P / F, i, N)  Problem 2 ( F / A, i, N)  Problem 3 ( A / F, i, N)  Problem 4 ( P / A, i, N)  Problem 5 ( A / P, i, N)  Problem 6 ( P / G, i, N)  Problem 7 ( A / G, i, N)  Problem 8

  2. DIAGRAM: $2 000 0 P = F4(P/F,i,n) = 2 000(P|F,12%,4) = 2 000(0.6355) = $1 271 n=4 1 2 3 P? Example 1 How much would you have to deposit today to have $2000 in 4 years if you can get a 12% interest rate compounded annually? GIVEN: F4 = $2 000 i = 12% FIND P:

  3. Different Ways of Looking at P/F • From previous example, if you can earn 12% compounded annually, you need to deposit $1271 to have $2000 in 4 years. • You are indifferent between $1271 today and $2000 in 4 years, assuming you can earn a return on your money of 12%. • $1271 today and $2000 in 4 years are equivalent, if you can earn 12% on your money. • The present worth of $2000 in 4 years is $1271 (i = 12% cpd annually). • If you could get 13% on your money, would you rather have the $1271 today, or $2000 in 4 years

  4. Example 2 Tuition costs are expected to inflate at the rate of 8% per year. The first year’s tuition is due one year from now and will be $10,000. To cover tuition cost for 4 years, a fund is to be set up today in an account that will earn interest at the rate of 5% per year, compounded annually. How much must be deposited into the fund today in order to pay the 4 years of tuition expenses?

  5. P ? 0 1 2 3 n A1 P = A1(P/A,g,i,n) Present Given Gradient (Geometric) • g is the geometric gradient over the time period • (time period: Time 0 to Time n, 1st flow at Time 1) • P is the present value of the flow at Time 0 • (n periods in the past) • i is the effective interest rate for each period Note: cash flow starts with A1 at Time 1, increases by constant g% g = %

  6. Example 2 Tuition costs are expected to inflate at the rate of 8% per year. The first year’s tuition is due one year from now and will be $10,000. To cover tuition cost for 4 years, a fund is to be set up today in an account that will earn interest at the rate of 5% per year, compounded annually. How much must be deposited into the fund today in order to pay the 4 years of tuition expenses?

  7. Example 2 - Concept If your rich Aunt Edna wanted to put a sum of money in the bank today to pay for your next four years of tuition, that sum would be $39,759 assuming 5% return on investment and tuition that begins at $10,000 increasing by 8% per year. This problem assumes tuition is due at the end of the year.

  8. Complex Cash Flows Complex Cash Flows – Break apart (or separate) complex cash flows into component cash flows in order to use the standard formulas. Remember: You can only combine cash flows if they occur at the same point in time. (This is like building with LEGOs!)

  9. Problem 3 A construction firm is considering the purchase of an air compressor. The compressor has the following expected end of year maintenance costs: Year 1 $800 Year 2 $800 Year 3 $900 Year 4 $1000 Year 5 $1100 Year 6 $1200 Year 7 $1300 Year 8 $1400 What is the present equivalent maintenance cost if the interest rate is 12% per year compounded annually?

  10. P = PA + PG + PF = A(P/A,i,n) + G(P/G,i,n) + F(P/F,i,n) = $700(P/A,12%,8) + $100(P/G,12%,8)+ $100(P/F,12%,1) = $700(4.9676) + $100(14.4715)+ $100(0.8929) = $5014 DIAGRAM: P ? PA ? 1 2 3 4 1 2 3 4 n=8 n=8 PF ? $700 0 0 $700 n=1 PG ? $100 $100 0 $200 $100 $700 $300 4 1 2 3 n=8 0 $100 $200 $700 $300 Problem 3 – Alt Soln 1 GIVEN: MAINT COST1-8 PER DIAGRAM i = 12%/YR, CPD ANNUALLY FIND P: NOTE: CAN BREAK INTO 3 CASH FLOWS: ANNUAL, LINEAR GRADIENT, AND FUTURE

  11. PPG ? 0 1 PG ? P = PA + PG(PPG) = A(P/A,i,n) + G(P/G,i,n-1)(P/F,i,1) = $800(P/A,12%,8) + $100(P/G,12%,7)(P/F,12%,1) = $800(4.9676) + $100(11.6443)(0.8929) = $5014 DIAGRAM: PA ? 1 2 3 4 n=8 P ? 0 1 2 3 4 n=8 $800 0 PG ? $800 0 1 2 3 n=7 $200 $100 $600 $200 $100 $600 Problem 3 – Alt Soln 2 GIVEN: MAINT COST1-8 PER DIAGRAM i = 12%/YR, CPD ANNUALLY FIND P: NOTE: PG MUST BE OFFSET ONE YEAR – SO BRING THE OFFSET YEAR BACK TO TIME ZERO

  12. DIAGRAM: $7 000 4 17 P17 = A(F/A,i,n) = A(P/A,i,n) = A(F/A,8%,14) = 7 000(P/A,8%,4) = A(24.2149) = 7 000(3.3121)  A = $957 21 yrs 18 0 A ? Problem 4 A young couple has decided to make advance plans for financing their 3 year old daughter’s college education. Money can be deposited at 8% per year, compounded annually. What annual deposit on each birthday, from the 4th to the 17th (inclusive), must be made to provide $7,000 on each birthday from the 18th to the 21st (inclusive)? GIVEN: WITHDRAWALS18-21 = $7 000 i = r = 8%/YR, CPD YEARLY FIND A4-17: STRATEGY: CAN BREAK INTO 2 CASH FLOWS, SO PICK A CONVENIENT POINT IN TIME AND SET DEPOSITS EQUAL TO WITHDRAWALS…

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