Engineering Economy

2. Engineering Economy Chapter 4: The Time Value of Money Lecture 6

3. There are interest factors for a series of end-of-period cash flows. How much will you have in 40 years if you save $3,000 each year and your account earns 8% interest each year?

4. Finding A when given F. How much would you need to set aside each year for 25 years, at 10% interest, to have accumulated $1,000,000 at the end of the 25 years?

5. Uniform Series Factors Find F, given A Formula: Notation: F = A(F/A,i%,n) E.P.S.C.A.F. Find A, given F Formula: Notation: A = F(A/F,i%,n) E.P.S.S.F.F.

6. Uniform Series Factors Example: General Contractors, Inc. is establishing a sinking fund to accumulate $60,000 after 7 years. What is the required annual deposit ? Assume i = 5.5% per year. F = $ 60,000 Use A/F factor A = 60,000(A/F,5.5%,7) = 60,000(0.12096) = $7258 per year

7. Finding the present amount from a series of end-of-period cash flows. How much would is needed today to provide an annual amount of $50,000 each year for 20 years, at 9% interest each year?

8. Finding A when given P. If you had $500,000 today in an account earning 10% each year, how much could you withdraw each year for 25 years?

9. Uniform Series Factors • Four factors involve a uniform series A • A is the same amount in consecutive interest periods Find A, given P Formula: Notation: A = P(A/P,i%,n) E.P.S.C.R.F. Find P, given A Formula: Notation: P = A(P/A,i%,n) . E.P.S.P.W.F.

10. Uniform Series Factors Example: An engineer needs $600 each year for 9 years starting next year. If i = 16% per year, find the equivalent amount now P = 600(P/A,16%,9) = 600(4.6065) = $2764

11. Uniform Series Factors • Remember • When using the P/A factor, the present worth P is always located ONE period prior to the first uniform-series amount A • When using the A/P factor, the uniform-series amount A starts ONE period after the present worth P value and continues for n consecutive periods

12. It can be challenging to solve for N or i. • We may know P, A, and i and want to find N. • We may know P, A, and N and want to find i. • These problems present special challenges.

13. Finding N Acme borrowed $100,000 from a local bank, which charges them an interest rate of 7% per year. If Acme pays the bank $8,000 per year, now many years will it take to pay off the loan? So, This can be solved by using the interest tables and interpolation, but we generally resort to a computer solution.

14. Finding i Jill invested $1,000 each year for five years in a local company and sold her interest after five years for $8,000. What annual rate of return did Jill earn? So, Again, this can be solved using the interest tables and interpolation, but we generally resort to a computer solution.

15. 0 1 2 3 4 5 Example: i unknown • Assume one can invest $3000 now in a venture in anticipation of gaining $5,000 in five (5) years. • If these amounts are accurate, what interest rate equates these two cash flows? $5,000 • F = P(1+i)n • (1+i)5 = 5,000/3000 = 1.6667 • (1+i) = 1.66670.20 • i = 1.1076 – 1 = 0.1076 = 10.76% $3,000

16. We need to be able to handle cash flows that do not occur until some time in the future. • Deferred annuities are uniform series that do not begin until some time in the future. • If the annuity is deferred J periods then the first payment (cash flow) begins at the end of period J+1.

17. Finding the value at time 0 of a deferred annuity is a two-step process. • Use (P/A, i%, N-J) find the value of the deferred annuity at the end of period J (where there are N-J cash flows in the annuity). • Use (P/F, i%, J) to find the value of the deferred annuity at time zero.

18. 0 1 2 3 4 5 6 7 8 Shifted Series Consider: A = -$500/year P2 P0 P of this series is at t = 2 (P2 or F2) P2 = $500(P/A,i%,4) or, could refer to as F2 P0 = P2(P/F,i%,2) or, F2 (P/F,i%,2)

19. P0 A = -$500/year 0 1 2 3 4 5 6 7 8 P2 F6 = ?? Shifted Series: P and F Require F6 • F for this series is at t = 6 • F6 = A(F/A,i%,4) where n = 4 time periods forward.

20. Suggested Steps • Draw and correctly label the cash flow diagram that defines the problem • Locate the present and future worth points for each series • Write the time value of money equivalence relationships • Substitute the correct factor values and solve

21. Series with other single cash flows • It is common to find cash flows that are combinations of series and other single cash flows. • Solve for the series present worth values then move to t = 0. • Solve for the PW at t = 0 for the single cash flows. • Add the equivalent PW’s at t = 0.

22. F4 = $300 A = $500 0 1 2 3 4 5 6 7 8 F5 = -$400 Series with Other cash flows • Consider: i = 10% • Find the PW at t = 0 and FW at t = 8 for this cash flow –watch the signs!

23. F4 = $300 A = $500 0 1 2 3 4 5 6 7 8 i = 10% F5 = -$400 The PW Points are: 1 2 3 t = 1 is the PW point for the $500 annuity; “n” = 3

24. Back 4 periods F4 = $300 A = $500 0 1 2 3 4 5 6 7 8 i = 10% Back 5 Periods F5 = -$400 The PW Points are: 1 2 3 t = 1 is the PW point for the two other single cash flows

25. Write the Equivalence Statement P = $500(P/A,10%,3)(P/F,10%,1) + $300(P/F,10%,4) - 400(P/F,10%,5) Substituting the factor values into the equivalence expression and solving….

26. Substitute the Factors and Solve P = $500( 2.4869 )( 0.9090 ) + $300( 0.6830 ) - 400( 0.6209 ) = $1086.84 $1,130.30 $204.90 $248.36

28. Engineering Economy Chapter 4: The Time Value of Money Lecture 7