Engineering Economy

1. Engineering Economy Chapter 4: The Time Value of Money Lecture 5

2. Sometimes cash flows change by a constant amount each period. We can model these situations as a uniform gradient of cash flows. The table below shows such a gradient.

3. Gradient Formulas Arithmetic gradient series – Starts at base amount and increases by constant gradient G in years 2 through n Base Amount A P = A(P/A,i%,n) + G(P/G,i%.n) This total present worth of base and gradient

4. A1+n-1G A1+n-2G A1+2G A1+G 0 1 2 3 n-1 N Linear Gradient Example • Assume the following: This represents a positive, increasing arithmetic gradient

5. Arithmetic Gradients • P/G factor formula for gradient only • A/G factor for annual equivalent of gradient only • P/G and A/G factors are in the tables at the rear of text

6. $8500 $8000 $7500 $7000 $6500 0 1 2 3 4 5 6 7 8 $6000 $5500 $5000 Arithmetic Gradients Example:Estimated annual revenue is $5,000 increasing by $500 per year starting next year. Find P and A equivalent over an 8-year period at i = 10%. Gradient, G = $500 Base A = $5000

7. Arithmetic Gradient Factors • The “G” amount is the constant arithmetic change from one time period to the next. • The “G” amount may be positive or negative. • The present worth point is always one time period to the left of the first cash flow in the series or, • Two periods to the left of the first gradient cash (G) flow.

8. 0 1 2 3 4 5 6 7 Present Worth Point… $700 $600 $500 $400 $300 $200 $100 X The Present Worth Point of the Gradient

9. Present Worth: Linear Gradient • The present worth of a linear gradient is the present worth of the two components: • 1. The Present Worth of the Gradient Component and, • 2. The Present Worth of the Base Annuity flow • Requires 2 separate calculations.

10. Present Worth: Gradient Component • Three contiguous counties in Florida have agreed to pool tax resources already designated for county-maintained bridge refurbishment. At a recent meeting, the county engineers estimated that a total of $500,000 will be deposited at the end of next year into an account for the repair of old and safety-questionable bridges throughout the three-county area. Further, they estimate that the deposits will increase by $100,000 per year for only 9 years thereafter, then cease. • Determine the equivalent (a) present worth and (b) annual series amounts if county funds earn interest at a rate of 5% per year.

12. Example Determine the equivalent annual series amounts if county funds earn interest at a rate of 5% per year.

13. Equations for P/G and A/G

14. The annual equivalent of this series of cash flows can be found by considering an annuity portion of the cash flows and a gradient portion.

15. Uniform Gradient Series Example: An engineer is planning for a 15 year retirement. In order to supplement his pension and offset the anticipated effects of inflation and increased taxes, he intends to withdraw $5,000 at the end of the first year, and to increase the withdrawal by $1,000 at the end of each successive year. How much money must the engineer have in this account at the start of his retirement, if the money earns 6% per year, compounded annually? Want to Find: P Given: A1, G, i, and n $19000 $18000 $8000 $7000 $6000 $5000 T = 0 1 2 3 4 14 15 P

16. Uniform Gradient Series Example: AT = (A1) + G(A/G,i,n) A2 = G(A/G,i,n) = $1000 (A/G,6%,15) = $1000 (5.926) = $5926 AT = $5000 + $5926 = $10,926 P = AT (P/A,i,n) = $10,926 (P/A,6%,15) = $10,926 (9.7123) = $106,120

18. A = Given 4 2 3 0 1 5 PA = ? Shifted Uniform Series A shifted uniform series starts at a time other than period 1 The cash flow diagram below is an example of a shifted series Series starts in period 2, not period 1 FA = ? Shifted series usually require the use of multiple factors Remember: When using P/A or A/P factor, PA is always one year ahead of first A When using F/A or A/F factor, FA is in sameyear as last A

19. The present worth of the cash flow shown below at i = 10% is: (a) $25,304 (b) $29,562 (c) $34,462 (d) $37,908 P0 = ? i = 10% 0 1 2 3 4 5 6 A = $10,000 Example Using P/A Factor: Shifted Uniform Series P1 = ? Actual year Series year 0 1 2 3 4 5 Solution: (1) Use P/A factor with n = 5 (for 5 arrows) to get P1in year 1 (2) Use P/F factor with n = 1 to move P1 back for P0in year 0 P0 = P1(P/F,10%,1) = A(P/A,10%,5)(P/F,10%,1) = 10,000(3.7908)(0.9091) = $34,462 Answer is (c)

20. Example Using F/A Factor: Shifted Uniform Series How much money would be available in year 10 if $8000 is deposited each year in years 3 through 10 at an interest rate of 10% per year? Cash flow diagram is: FA = ? i = 10% Actual year 0 1 2 3 4 5 6 7 8 9 10 Series year 0 1 2 3 4 5 6 7 8 A = $8000 Re-number diagram to determine n = 8 (number of arrows) Solution: FA = 8000(F/A,10%,8) = 8000(11.4359) = $91,487

21. Shifted Series and Random Single Amounts For cash flows that include uniform series and randomly placed single amounts: Uniform series procedures are applied to the series amounts Single amount formulas are applied to the one-time cash flows The resulting values are then combined per the problem statement The following slides illustrate the procedure

22. Example: Series and Random Single Amounts Find the present worth in year 0 for the cash flows shown using an interest rate of 10% per year. PT = ? i = 10% 0 1 2 3 4 5 6 7 8 9 10 A = $5000 $2000 PT = ? i = 10% Actual year 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 Series year A = $5000 $2000 Solution: First, re-number cash flow diagram to get n for uniform series: n = 8

23. Example: Series and Random Single Amounts PA PT = ? i = 10% Actual year 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 Series year A = $5000 $2000 Use P/A to get PA in year 2: PA = 5000(P/A,10%,8) = 5000(5.3349) = $26,675 Move PA back to year 0 using P/F: P0 = 26,675(P/F,10%,2) = 26,675(0.8264) = $22,044 Move $2000 single amount back to year 0: P2000 = 2000(P/F,10%,8) = 2000(0.4665) = $933 Now, add P0 and P2000 to get PT: PT = 22,044 + 933 = $22,977

24. Example Worked a Different Way (Using F/A instead of P/A for uniform series) The same re-numbered diagram from the previous slide is used PT = ? FA = ? i = 10% 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 A = $5000 $2000 Solution: Use F/A to get FA in actual year 10: FA = 5000(F/A,10%,8) = 5000(11.4359) = $57,180 Move FA back to year 0 using P/F: P0 = 57,180(P/F,10%,10) = 57,180(0.3855) = $22,043 Move $2000 single amount back to year 0: P2000 = 2000(P/F,10%,8) = 2000(0.4665) = $933 Same as before Now, add two P values to get PT: PT = 22,043 + 933 = $22,976 As shown, there are usually multiple ways to work equivalency problems

25. Example: Series and Random Amounts Convert the cash flows shown below (black arrows) into an equivalent annual worth A in years 1 through 8 (red arrows) at i = 10% per year. A = ? i = 10% 0 1 2 3 4 5 6 7 8 0 1 2 3 4 5 A = $3000 $1000 Approaches: 1. Convert all cash flows into P in year 0 and use A/P with n = 8 2. Find F in year 8 and use A/F with n = 8 Solution: Solve for F: F = 3000(F/A,10%,5) + 1000(F/P,10%,1) = 3000(6.1051) + 1000(1.1000) = $19,415 Find A: A = 19,415(A/F,10%,8) = 19,415(0.08744) = $1698

26. Shifted gradient begins at a time other than between periods 1 and 2 To find equivalent A series, find PT at actual time 0 and apply (A/P,i,n) Must use multiple factors to find PT in actual year 0 Present worth PG is located 2 periods before gradient starts Shifted Arithmetic Gradients

27. Example: Shifted Arithmetic Gradient John Deere expects the cost of a tractor part to increase by $5 per year beginning 4 years from now. If the cost in years 1-3 is $60, determine the present worth in year 0 of the cost through year 10 at an interest rate of 12% per year. i = 12% PT = ? Actual years 1 10 3 0 2 4 5 Gradient years 0 1 2 3 8 60 60 60 65 70 G = 5 95 First find P2 for G = $5 and base amount ($60) in actual year 2 Solution: P2 = 60(P/A,12%,8) + 5(P/G,12%,8) = $370.41 P0 = P2(P/F,12%,2) = $295.29 Next, move P2 back to year 0 Next, find PA for the $60 amounts of years 1 and 2 PA = 60(P/A,12%,2) = $101.41 Finally, add P0 and PA to get PT in year 0 PT= P0 + PA = $396.70

28. Example: Shifted Arithmetic Gradient