1. A shifted uniform series starts at a time other than year 1 When using P/A or A/P, P is always one year ahead of first A When using F/A or A/F, F is in same year as last A Solution: (1) Use P/A factor with n=5 to get P in year 1: (2) Use P/F factor with n=1 to move P back to year 0: P 1 = 10,000(P/A,10%,5) =10,000(3.7908) = $37,908 P 0 = P 1 (P/F,10%1) = 37,908(0.9091) = $34,462 Answer is (c) P=? A=10,000 0 1 2 3 4 5 6 i=10% Example: 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

2. For A/P factor, find P in yr 0, then annualize with (A/P,12%,7) For A/F factor, find F in yr 7, then annualize with (A/F,12%7) Using A/F: A=[10,000(F/A,12%,6) + 6,000(F/P,12%,2)](A/F,12%,7) Solution: The annual worth in yrs 1-7 can be found using either the A/P or A/F factors =[10,000(8.1152) + 6,000(1.2544)](0.09912) =$8,789.80 Answer is (a) These cash flows require the use of multiple factors A=10,000 i= 12% 0 1 2 3 4 5 6 7 6000 Example: The equivalent annual worth in yrs 1-7 for the cash flow shown below at i=12% per year is: (a) $8790 (b) $9530 (c) $10,330 (d) $11,780

3. Shifted gradients begin at a time other than between periods 1& 2Must use multiple factors to find P in year 0 Arithmetic

4. Example : John Deere expects the cost of a certain tractor part to increase by $5 /yr beginning 4 years from now. If the cost in years 1-3 is $60, the present worth of the cost thru year 10 at an interest rate of 12%/yr is closest to: (a)$101 (b)$295 (c)$370 (d)$397

5. Example : John Deere expects the cost of a certain tractor part to increase by $5 /yr beginning 4 years from now. If the cost in years 1-3 is $60, the present worth of the cost thru year 10 at an interest rate of 12%/yr is closest to: (a)$101 (b)$295 (c)$370 (d)$397 0 1 2 3 10 45 60 65 70 95 The cash flow diagram is as follows: Cash Flow Diagram i=12%

6. First find P 2 for the gradient ($5) and its base amount ($60) in year 2: P 2 = 60(P/A,12%,8) + 5(P/G,12%,8) = $370.41 0 1 2 3 10 45 P=? 60 65 70 95 01238 Example : John Deere expects the cost of a certain tractor part to increase by $5 /yr beginning 4 years from now. If the cost in years 1-3 is $60, the present worth of the cost thru year 10 at an interest rate of 12%/yr is closest to: (a)$101 (b)$295 (c)$370 (d)$397 The cash flow diagram is as follows: i=12%

7. First find P 2 for the base amount ($60) & gradient ($5) in year 2: P 2 = 60(P/A,12%,8) + 5(P/G,12%,8) = $370.41 Next, move P 2 back to year 0: P 0 = P 2 (P/F,12%,2) = $295.29 0 1 2 3 10 45 P=? 60 65 70 95 01238 Example : John Deere expects the cost of a certain tractor part to increase by $5 /yr beginning 4 years from now. If the cost in years 1-3 is $60, the present worth of the cost thru year 10 at an interest rate of 12%/yr is closest to: (a)$101 (b)$295 (c)$370 (d)$397 The cash flow diagram is as follows: i=12%

8. First find P 2 for the base amount ($60) & gradient ($5) in year 2: P 2 = 60(P/A,12%,8) + 5(P/G,12%,8) = $370.41 Next, move P 2 back to year 0: P 0 = P 2 (P/F,12%,2) = $295.29 Next, find P A for the $60 amounts of years 1 & 2: P A = 60(P/A,12%,2) = $101.41 0 1 2 3 10 45 P=? 60 65 70 95 Example : John Deere expects the cost of a certain tractor part to increase by $5 /yr beginning 4 years from now. If the cost in years 1-3 is $60, the present worth of the cost thru year 10 at an interest rate of 12%/yr is closest to: (a)$101 (b)$295 (c)$370 (d)$397 The cash flow diagram is as follows: i=12%

9. First find P 2 for the base amount ($60) & gradient ($5) in year 2: P 2 = 60(P/A,12%,8) + 5(P/G,12%,8) = $370.41 Next, move P 2 back to year 0: P 0 = P 2 (P/F,12%,2) = $295.29 Next, find P A for the $60 amounts of years 1 & 2: P A = 60(P/A,12%,2) = $101.41 Finally, add P 0 & P A to get P T in year 0: A P T = P 0 + P A = $396.70 Answer is (d) 0 1 2 3 10 45 P T = ? 60 65 70 95 Example : John Deere expects the cost of a certain tractor part to increase by $5 /yr beginning 4 years from now. If the cost in years 1-3 is $60, the present worth of the cost thru year 10 at an interest rate of 12%/yr is closest to: (a)$101 (b)$295 (c)$370 (d)$397 The cash flow diagram is as follows: i=12%

10. Shifted gradients begin at a time other than between periods 1& 2 Geometric Equation yields P for all cash flow(i.e. base amt is included)For negative gradient, change signs in front of both g’s P=A{1-[(1+g)/(1+i)] n /(i-g)}

11. For arithmetic gradients, change sign in front of G term from + to - For goemetric gradients, change signs in front of both g’s All other procedures are the same as for positive gradients

12. How much should a company which manufactures corrugated pipe be willing to pay a contractor who claims he has a device which will reduce the company's energy bill by at least $4,000 per year. Assume the company wants to recover its investment in five years at an interest rate of 15% per year. P = 4,000 (P/A, 15%, 5) = 4,000 (3.3522) = $13,408.80 Present Worth Problem with given A

13. An elastomeric roofing material can be installed on a parts warehouse for $10,000. If the company expects to recover its investment in seven years through reduced energy costs, the required annual savings at an interest rate of 20% per year is closest to: A = 10,000 (A/P, 20%, 7) = 10,000 (0.27742) = $2774.20 Answer is (B) Annual Worth Problem with Given Present Worth

14. Pebble Beach Country Club installed a new computer-controlled irrigation system that uses reclaimed sewage for watering the fairways. The cost of the pumps, piping, and controls was $1,100,000. If the club expects to recover its investment in 10 years using an interest rate of 10% per year, the annual savings in water cost must be nearest to: A = 1,100,000 (A/P, 10%, 10) = 1,100,000 (0.16275) = $179,025 Annual Worth Problem with Given Present Worth

15. How much money could a start-up software company afford to borrow now if it promises to repay the loan with five equal year-end payments of $10,000 if the interest rate is 10% per year? Solution: The amount that could be borrowed is the present worth of the $10,000 series: P = 10,000 (P/A, 10%, 5) = 10,000 (3.7908) = $37,908 Present Worth Problem with Given Annual Worth

16. A sum of $30,000 one year from now would be equivalent to how much money now, at an interest rate of 25% per year? Answer: P = 30000 (P/F, 25%, 1) = 30000 (0.8000) = $24,000 Single Payment Present Worth Problem

17. An engineer invested his bonus check of $7,000 in a mutual fund two years ago. If the value of his investment now is $10,000, the rate of return he made on his investment was closest to: Answer: i = $7000(F/P, i,2) = $10000 i = 10000/7000 factor (1.4286) on interest table for F/P at n=2 i = 19.5% between 18% & 20% tables Rate of Return Problem

18. Wendy's International has received a bid of $9,000 to re- coat a parking lot with a water sealer and repaint the parking-space lines. What is the equivalent annual cost of the job if the company expects to recover its investment in four years at an interest rate of 20% per year? Answer: A = 9,000 (A/P, 20%, 4) = 9,000 (0.38629) = $3476.61 Annual Worth Problem with Given Present Worth

19. American Airlines estimates that unpainted airplanes use less fuel (because of less weight) to the extent of $20,000 per year per plane. How much could the company afford to spend now to strip the paint from an airplane if it expects to recover its investment in five years at an interest rate of 20% per year? Answer: P = 20,000 (P/A, 20%, 5) = 20,000 (2.9906) = $59,812 Present Worth Problem with Given Annual Worth