2. ISU CCEE CE 203 EEA Chap 3 Interest and Equivalence

3. ISU CCEE l The use of money has value – Somebody will pay you to use your money – You will pay others to use their money l Interest is “Rent ” for the use of money l When decisions involve “cash flows” over a considerable length of time, economic analysis should include the effects of: – Interest – Inflation Time Value of Money

4. ISU CCEE l $1000 in savings account at 10% for 1 year? l $1000 per year for 30 years at 10%? l For typical “investment plan”, $1000 per year for 30 years at 10% yields $164,494 (see formula for “sinking fund”) Pre-Course EEA Assessment

5. ISU CCEE Would you rather have $100 now or – $100 a year from now? – $110 a year from now? – $120 a year from now? – $150 two years from now? Your time value of money?

6. ISU CCEE l Interest computed only on the original sum of money or “principal” l Total interest earned = I = P x i x n where P = principal or present sum of money i = interest rate per period n = number of periods l Example: $1000 borrowed at 8% for two years, simple interest I = $1000 x.08 x 2 = $160 Simple Interest

7. ISU CCEE l F = P + P x in = P (1 + in) = P [1 +i(period)] l Example: $1000 borrowed at 8% for five years, simple interest F = $1000 (1 +.08(5)) = $1400 Simple Interest Future Value, F, of a Loan, P Simple Interest Future Value, F, of a Loan, P

8. ISU CCEE Compound Interest Compound Interest Interest on original sum + on interest Example $1000 @ 10% per year, F F 1 (first year) = $1000 + $1000(0.1) or $1000 (1 + 0.1) = $1100 F 2 (second year) = $1100 + $1100(0.1) or $1000(1+0.1)(1+0.1) = $1210 or $1000(1+0.1) 2 F 3 = $1000(1+0.1) 3 ; F n = $1000(1+0.1) n General: F n = P(1+i) n

9. ISU CCEE Interest computed on the original sum and any unpaid interest Compound Interest Period Beginning Balance Interest for Period Ending Balance 1PiPP(1 + i) 2 iP(1 + i)P(1 + i) 2 3 iP(1 + i) 2 P(1 + i) 3 nP(1 + i) n-1 iP(1 + i) n-1 P(1 + i) n

10. ISU CCEE l Future Value, F, = P (1 + i) n – P = principal or present sum of money – i = interest rate per period – n = number of periods (years, months, …) l Example: $1000 borrowed at 8% for five years, compound interest, all of principal and interest repaid in 5 years: F = $1000 (1 +.08) 5 = $1469.33 F = P (F/P, i, n) = $1000 (F/P,.08, 5) = $1469.00 (see EEA p. 576) Compound Interest Note: Simple interest yield would be $1400

11. ISU CCEE l Total interest earned = I n = P (1 + i) n - P P = principal or present sum of money i = interest rate n = number of periods l Example: $1000 borrowed at 8% for five years, compound interest I = $1000 (1 +.08) 5 - $1000 = $469.33 Compound Interest

12. ISU CCEE Future value, F, for P = $1000 at 8% Simple vs. Compound Interest PeriodsF/P, simple iF/P, compound i 1$1080 2$1160$1166 3$1240$1260 4$1320$1360 5$1400$1469 10$1800$2159 15$2200$3172 20$2600$4661

13. ISU CCEE Specification of Interest Rate, i 1) “8%” - assumed to mean per year and compounded annually 2) “8% compounded quarterly” - 2% per each 3 months, compounded every 3 months 3) “8% compounded monthly” – 2/3% each month, compounded every month 4) “8%” =.08 in equations NB 1000

14. ISU CCEE In-class Example Would you rather have $5000 today or $35,000 in 25 years if the interest rate was 8% compounded yearly/ monthly… 1) yearly? F = P(F/P,.08, 25) = 5000 (6.848) = $34, 240 2) monthly? F = P (F/P,.006667, 300) = $36,700

15. ISU CCEE Four Ways to Repay a Debt Plan Principal repaid … Interest paid … Trend on interest earned 1 in equal annual installments on unpaid balance declines 2at end of loan on unpaid balance constant 3in equal annual installments declines at increasing rate 4at end of loan at end of loan (compounded) increases at increasing rate

16. ISU CCEE Loan Repayment Plan 1 ($5000, 5 years, 8%) Year Owed at begin- ning Annual Interest Total owed at end Principal Paid Total pay- ment 1$5000$400$5400$1000$1400 2$4000$320$4320$1000$1320 3$3000$240$3240$1000$1240 4$2000$160$2160$1000$1160 5$1000$80$1080$1000$1080 $1200$5000$6200 Bank loans with yearly payments: Principal repaid in equal installments

17. ISU CCEE Loan Repayment Plan 2 ($5000, 5 years, 8%) Year Owed at begin- ning Annual Interest Total owed at end Princi- pal Paid Total pay- ment 1$5000$400$5400$0$400 2$5000$400$5400$0$400 3$5000$400$ 5400$0$400 4$5000$400$ 5400$0$400 5$5000$400$ 5400$5000$5400 $2000$5000$7000 Interest only loans: used in bonds and international loans

18. ISU CCEE Loan Repayment Plan 3 ($5000, 5 years, 8%) Year Owed at begin- ning Annual Interest Total owed at end Princi- pal Paid Total pay- ment 1$5000$400$5400$852$1252 2$4148$331$4479$921$1252 3$3227$258$ 3485$994$1252 4$2233$178$ 2411$1074$1252 5$1159$93$ 1252$1159$1252 $2000$5000$6260 Equal annual installments: Auto/home loans (but usually monthly payments)

19. ISU CCEE Loan Repayment Plan 4 ($5000, 5 years, 8%) Year Owed at begin- ning Annual Interest Total owed at end Princi- pal Paid Total pay- ment 1$5000$400$5400$0 2$5400$432$5832$0 3$5832$467$ 6299$0 4$6299$504$ 6804$0 5$6803$544$ 7347$5000$7347 $2347$5000$7347 Interest and principal repaid at end of loan: Certificates of deposit (CD’s) and IRA’s

20. ISU CCEE Loan repayment plans 1-4 l Are all equivalent to $5000 now in terms of time value of money, l May not be equally attractive to loaner or borrower l Have different cash flow diagrams l “Equivalent in nature but different in structure”

21. ISU CCEE The present sum of money is equivalent to the future sum(s) (from our perspective), if....we are indifferent as to whether we have a quantity of money now or the assurance of some sum (or series of sums) of money in the future (with adequate compensation) EquivalenceEquivalence

22. ISU CCEE l Used to make a meaningful engineering economic analysis l Apply by finding equivalent value at a common time for all alternatives – value now or “Present Worth” – value at some logical future time or “Future Worth” l Assume the same time value of money (interest rate) for all alternatives EquivalenceEquivalence

23. ISU CCEE You borrow $8000 to help pay for senior year at ISU – the bank offers you two repayment plans for paying off the loan in four years: Year Plan 1 payment Plan 2 payment 1$0 2 $3300 3$0$3300 4$10,400$3300 Total$10,400$9900 Which would you choose?

24. ISU CCEE “Technique of Equivalence” requires that we 1)Determine our “time value of money” 2)Determine a single equivalent value at a selected time for Plan 1 3) Determine a single equivalent value at the same selected time for Plan 2 4) Compare the two values Comparing Economic Alternatives

25. ISU CCEE Cash flows can be compared by calculating their total Present Worth (“now value”) or their total Future Worth (at some time in the future) – Future Worth, F = P (1 + i) n – Solving the previous equation for P gives Present Worth, P = F (1 + i) -n » P = some present amount of money » F = some future amount of money » i = interest rate per time period (appropriate time value of money) » n = number of time periods Technique of Equivalence

26. ISU CCEE Technique of Equivalence Year Plan 1 payment Plan 2 payment 1$0 2 $3300 3$0$3300 4$10,400$3300 Total$10,400$9900 To compare payment plans, “move” all amounts to now (Present Worth) or to 4 years hence (Future Worth)

27. ISU CCEE Technique of Equivalence – Present Worth Assume our “time value of money” is 7% APR 0 0 1 1 2 2 3 3 4 4 P 4 = 10,400(1 +.07) -4 = 7,934.11 Total PW = $7,934.11 Or P = $10,400 (P/F,.07, 4) =.7629 (10,400) = $7934.16 $10,400$10,400 Present Worth: Payment Plan 1 P = F(1 + i) -n PW?

28. ISU CCEE Technique of Equivalence – Present Worth Assume our “time value of money” is 7% APR 0 0 1 1 2 2 3 3 4 4 $3300$3300 P2 = 3300(1 +.07) -2 = 2882.35 P3 = 3300(1 +.07) -3 = 2693.78 P4 = 3300(1 +.07) -4 = 2517.55 Total PW = $8093.68 $3300$3300$3300$3300 PW? Present Worth: Payment Plan 1 P = F(1 + i) -n

29. ISU CCEE Technique of Equivalence – Present Worth Which would you choose? 0 0 1 1 2 2 3 3 4 4 $3300$3300$3300$3300$3300$3300 $8093.68 0 0 1 1 2 2 3 3 4 4 $10,400$10,400 $7934.11 Plan 2 Plan 1

30. ISU CCEE Technique of Equivalence – Future Worth Assume our “time value of money” is 7% APR 0 0 1 1 2 2 3 3 4 4 F 4 = 10,400(1 +.07) 0 = 10,400 Total FW = $10,400.00 $10,400$10,400 Future Worth: Payment Plan 1 F = P(1 + i) n

31. ISU CCEE Technique of Equivalence – Future Worth Assume our “time value of money” is 7% APR 0 0 1 1 2 2 3 3 4 4 $3300$3300 F2 = 3300(1 +.07) 2 = 3778.17 F3 = 3300(1 +.07) 1 = 3531.00 F4 = 3300(1 +.07) 0 = 3300.00 Total FW = $10609.17 Again, larger than Plan 1 $3300$3300$3300$3300 Future Worth: Payment Plan 2 P = F(1 + i) -n FW?

32. ISU CCEE In-class Example You deposit $3000 in an account that earns 5%, compounded daily. How much could you withdraw from the account at the end of two years? F = 3000 (F/P,.05/365, 730) = $3,315.49 How much could you withdraw if the compounding were monthly? F = 3000 (F/P,.05/12, 24) = $3,314.82

33. ISU CCEE In-class Example You are considering two designs for a wastewater treatment facility. Plan 1: Costs $1,700,000 to construct and will have to be replaced every 20 years. Plan 2: Costs $2,100,000 to construct and will have to be replaced every 30 years. Which is the better of the two designs? Assume 7% APR and neglect inflation and operating, maintenance and disposal costs. Sketch CFD’s for each plan and compare.

34. ISU CCEE How to compare two schemes with different lives l One plant lasts 20y and the other 30y l Could we make a comparison over 20y? l Could we make a comparison over 30y? l What would be a decent period over which to compare? The smallest common denominator, i.e. in this case 60y.

35. ISU CCEE Technique of Equivalence – Present Worth Based on dollars only, which is the better plan? 0 0 10 $1.7 M $2.254 M (if provision is made for final replacement, $ 2.283M) Plan 1 20 30 40 50 60 $1.7 M repeats P = $ 1.7 + $ 1.7 (1.07) -20 + $ 1.7 (1.07) -40 [ + $ 1.7 (1.07) -60 ] * [$1.7 M]* This term will only apply when it is a requirement to replace at the end, not normally and it has been excluded from this analysis. You could get to 60 years without a final replacement.

36. ISU CCEE Technique of Equivalence – Present Worth Based on dollars only, which is the better plan? 0 0 10 $2.1 M $2.376 M Plan 2 20 30 40 50 60 $2.1 M repeats Other option was $2.254 M Not required = P = 2.1 + 2.1(1.07) -30 + the 60y replacement, which does not apply