1. 12/26/2015rd1 Engineering Economic Analysis Chapter 4  More Interest Formulas

2. 12/26/2015rd2 Annual Percentage Rate (APR) or r Nominal rate 6% per year is designated ~ i. Effective interest rate ~ i eff i eff = (1 + r/m) m – 1 where m is the number of pay periods APR is 12% compounded monthly i eff = (1 + 0.12/12) 12 – 1 = 12.68% effective yearly rate. APR is 12% compounded monthly; find effective quarterly rate. i eff = (1 + 0.03/03) 3 – 1 = 3.03% effective quarterly rate. Effective Interest Rate

3. 12/26/2015rd3 Effective Interest Rate Annual Percentage Rate (APR) is 12% If compounded monthly, effective monthly rate is 1% effective quarterly rate is 3.03% effective yearly rate is 12.68% If compounded quarterly, effective quarterly rate is 3% effective yearly rate is 12.55%

4. 12/26/2015rd4 Interest Rate A credit card company charges 1.5% interest on the unpaid balance each month. Nominal annual interest rate is _________________. ans. 12 * 1.5% = 18% Effective annual interest rate is ________. ans. (1 + 0.18/12) 12 – 1 = 19.56%

5. 12/26/2015rd5 Equivalence Are equivalent cash flows equivalent at any common point in time? For example, $1000 now at i = 10% for 10 years is equivalent to 1000(1 + 0.10) 10 = $2593.74. Are these 2 cash flows equivalent at time 3.37? Does $1000(1.1) 3.37 = $2593.74(1.1) -6.63 ? Check: $1378.77 = $1378.77 => Yes Are these cash flows equivalent at i = 8%? No, 1000(1.08) 3.37 = $1296.10  2593.74(1.08) -6.63 = $1557.14

6. 12/26/2015rd6 Monthly Payments a)What is the monthly payment for a 5-year car loan of $35,000 at 6% compounded monthly? b)Find the amount of the principal reduction of the 25 th payment. c)After making the 50 th monthly payment, you decide to pay off the loan what a check for _________. d)Find the interest on the 35 th payment. a) A = $35,000(A/P, ½ %, 60) = $676.65. b)PR 25 = 676.65(P/F, ½ %, 60 – 25 + 1) = $565.44. c)B 50 = 676.65(P/A, ½ %, 10) = $6584.08 d)I 35 = B 34 * i = 676.65(P/A, ½ %, 26) * ½ % = $82.29

7. 12/26/2015rd7 Mortgage Find the total interest paid on a $300,000 30-year loan at 6% compounded monthly. A = 300K(A/P, ½%, 360) = $1798.65 Total interest = 360 * $1798.65 - $300K = $347,514

8. 12/26/2015rd8 Arithmetic Gradient 0 1 2 3... n -1 n Gradient begins in Year 2 = A(P/A, i%, n) + G(P/G, i%, n) G 2G (n-1)G P A

9. 12/26/2015rd9 Gradient Example What is the present worth at pay period 0 of the following yearly cash flow at 7% compounded annually: n 0 1 2 3 4 5 cf 1000 1300 1600 1900 2200 2500 PW(7%) = 1000 + 1300(P/A, 7%, 5) + 300(P/G, 7%, 5) = 1000 + 5330.27 + 2294 = $8624.26 (F/P (PGG 1000 300 7 6) 7 1)  $8624.26

10. 12/26/2015rd10 Geometric Gradient $100 grows geometrically by 10% per year. Compute the growth after n years n Cash Flow 1 100 2 100 + 0.10 * 100 = 100(1 + 0.10) 1 = 110 3 110 + 0.10 * 110 = 100(1 + 0.10) 2 = 121 4 121 + 0.10 * 121 = 100(1 + 0.10) 3 = 133 ………………… n 100(1 + 0.10) n – 1 A n = A 1 (1 + g) n-1 Each A n must be brought back to year 0 to find the present worth.

11. 12/26/2015rd11 Geometric Gradient A n = A n-1 (1+ g) => A n = A 1 (1 + g) n – 1 Then P =  A 1 (1 + g) n-1 (1 + i) -n P = =

12. 12/26/2015rd12 Geometric Gradient Example A machine’s first year cost is $1,000 and increases 8% per year thereafter for 15 years. Maintenance funds earn 10% per year compounded annually. How much should be deposited in the maintenance fund to cover costs? P = 1000[1 – (1.08) 15 (1 + 0.10) -15 ]/ (0.10 – 0.08) = $12,030.40 (PGGG 1000 8 10 15)  12030.39

13. 12/26/2015rd13 Problem 4-125 Find the present worth of a cash flow beginning at $10K and increasing at 8% for 4 years at 6%/yr interest. (PGGG-table 10000 8 6 4) n Cash-flow 8% PW-factor 6% PWorth 1 10000.00 0.9434 9433.96 2 10800.00 0.8900 9611.96 3 11664.00 0.8396 9793.32 4 12597.12 0.7921 9978.10 $38,817.54 PW = 10K[(1 – (1.08) 4 )/(1.06) 4 (0.06 – 0.08)] = $38,817.54 (PGGG 10E3 8 6 4)  38817.54

14. 12/26/2015rd14 Geometric Gradient Example You want to accumulate $1M 20 years from now by depositing $A 1 at year 1 and increasing the deposit by 6% each year for 20 years. Find A 1 if the bank pays 8% interest compounded annually. F = A 1 (P/A1, g = 6%, i = 8%, n = 20)(F/P, i = 8%, 20) 1M = A1{[1 – 1.06 20 *1.08 -20 ]/0.02}(1.08) 20 A1 = $13,756.85 = (/ 1e6 (FGP (PGGG 1 6 8 20) 8 20) P =

15. 12/26/2015rd15 Effective Interest Rates $1000 is deposited at 7% compounded monthly. Find the value 5 years from now using monthly, quarterly, semiannually, yearly and bi-yearly effective rates. Monthly: 1000(F/P, 7/12%, 60) = $1417.63 Quarterly: 1000(F/P, 1.76%, 20) = $1417.63 Semi-annually 1000(F/P, 3.55%, 10) = $1417.63 Annually 1000(F/P, 7.23%, 5) = $1417.63 Biennially 1000(F/P, 14.98%, 2.5) = $1417.63. Pentad 1000(F/P, 41.76%, 1) = $1417.63 e.g. Pentad effective rate = [1 + 0.35/60) 60 – 1 = 41.7625% 1000(1 + i m ) nm = 1000(1 + i a ) n

16. 12/26/2015rd16 Relationships of Interest Factors F/P =1/(P/F); A/P = 1/(P/A); F/A = 1(A/F) F/A = 1 + Σ(F/P, i%, n-1) A/P = A/F + i; A/P = P/A – A/F; CRF = (P-S)(A/P. i%, n) + Si P/F * F/A = P/A; P/F = 1 – P/A * i A/F = A/P – i; 7% n F/P P/F A/F A/P F/A P/A A/G P/G 1 1.0700 0.9346 1.0000 1.0700 1.0000 0.9346 0.0000 0.0000 2 1.1449 0.8734 0.4831 0.5531 2.0700 1.8080 0.4831 0.8735 3 1.2250 0.8163 0.3111 0.3811 3.2149 2.6243 0.9549 2.5061 4 1.3108 0.7629 0.2252 0.2952 4.4399 3.3872 1.4155 4.7948 5 1.4026 0.7130 0.1739 0.2439 5.7507 4.1002 1.8650 7.6467

17. 12/26/2015rd17 Continuous Compounding In the effective interest formula let m = rp and the formula becomes i eff = (1 + r/m) m - 1 e i eff = (1 + 1/p) rp = = e r as p   (F/P, r%, N) = e rN for continuous compounding

18. 12/26/2015rd18 Continuous Compounding You deposit $100 per month in a savings account with an APR of 6% per year compounded continuously. How much will accumulate in 5 years? F = 100(F/A, e 0.005 -1, 60) = $6979.70 The monthly continuous compounding rate e 0.005 -1 = 0.501

19. 12/26/2015rd19 Problem 4-36 $12K is borrowed at 4% per annum and is to repaid in 5 payments. After the 2 nd payment, the borrower was given the option of paying off the loan the following year. How much was then due? A = 12K(A/P, 4%, 5) = $2695.53 Balance = 2695.53 + 2695.53(P/A, 4%, 2) = $7779.54

20. 12/26/2015rd20 Problem 4-38 Sold in 2002 for $150K at 20% down payment and 15-year loan at 8% per year. Buyer makes first payment in 2003. How much will be owed after 2009? Loan amount = 150K - 0.2 * 150K = $120K A = 120K(A/P, 8% 15) = $14,019.55 Balance after making 7 payments is B = 14,019.55 (P/A, 8%, 8) = $80,565.27 (loan 120E3 8 15)

21. 12/26/2015rd21 Problem 4-50 A debt of $5K is repaid according to the cash flow below at 8% compound interest. Find X. n 1 2 3 45 cf $500 1000 1500 2000 X [5K – [500(P/A, 8%, 4) + 500(P/G 8% 4)](F/P, 8% 5) = X => X = $1497.08 (List-pgf '(0 500 1000 1500 2000 1497.08) 8)  $5000 (IRR '(-5000 500 1000 1500 2000 1497.06))  8%

22. 12/26/2015rd22 Capitalized Cost You can have 5% interest in perpetuity (forever). You need to generate $10,000 a year for a scholarship fund. How much investment is needed to do so? P = A/i = 10,000/0.05 = $200,000.

23. 12/26/2015rd23 Gradient Find the equivalent sum at year 7 for the following cash flow at 7% compound interest per year. n 1 2 3 4 567 Cf 100020003000 40505000 F 7 = [(1000(P/A, 7%, 5) + 1000(P/G, 7%, 5) + 50(P/F 7%, 4)](F/P, 7%, 7) = $18,924.23 (F/P (+ (PGG 1000 1000 7 5) (P/F 50 7 3)) 7 7)  18928.52

24. 12/26/2015rd24 Shady Deal You borrow $1000 to be repaid in 24 monthly installments. The interest rate is a mere 1.5% per month. Further Amount requested$1000 Credit risk insurance 5 Credit investigation 25 Total $1030 Interest: ($1030)(24)(0.015) = $371 Total owed: $1030 + $371 = $1401 Payment: $1401/24 = $58.50 Find the effective annual interest rate charged. 1000 = 58.50(P/A, i%, 24) => i m = 2.92% => APR = 34.04% I aeff = 41.25%

25. 12/26/2015rd25 Sports Contracts Headline blares Ace Stacey sings 10-year contract for $50 million paid $5M now and $4M for the first 5 years and $5M for the next 5 years. How much is the contract worth to Ace now if the interest rate is 7%? PW = 5M + 4M(P/A, 7%, 5) + 5M(P/A, 7%, 5)(P/F, 7%, 5) = 5M + 16,400,790 + 14,616,921 = $36,017,710

26. 12/26/2015rd26 Capital Recovery (A/P, i%, n) – (A/F, i%, n) = i P F = Salvage P(A/P, i%, n) – S(A/F, i%, n) EUAC = (P - S)(A/P, i%, n) + Si

27. 12/26/2015rd27 Capital Recovery Example A new machine's first cost is $5,000 with a 5-year life and a salvage value of $1000. Compute the annual cost at i = 7%. 1)5000(A/P, 7%, 5) – 1000(A/F, 7%, 5) = 1219.45 – 173.89 = $1045.56 2)(P – S)(A/P, 7%, 5) + Si = 4000(P/A, 7%,5) + 0.07*1000 = 975.56 + 70 = $1045.56 3) (P – F)(A/F, 7%, 5) + Pi = 4000(A/F, 7%, 5) + 5000 * 0.07 = 695.56 + 350 = $1045.56

28. Review 12/26/2015rd28

29. 12/26/2015rd29 Change in Rate You borrow $20,000 at 7% compounded monthly over 48 months. After making the 24 th payment, you negotiate with the bank to pay off the loan in 8 equal quarterly payments. Determine the quarterly payment at the same interest rate. A m = 20K(A/P, 7/12 %, 48) = $478.92 B 24 = 478.92(P/A, 7/12 %, 24) = $10,696.84 A q = 10,698.84(A/P, 1.76%, 8) = $1445.17

30. Exact Rate of Return Find the exact rate of return for the following cash flow. n 0 1 2 cf-1200 900 700 (quadratic -12 9 7)  1.225857 => 22.5857% (list-pgf '(-1200 900 700) 22.585738)  0 (IRR '(-1200 900 700))  22.585738 12/26/2015rd30

31. Wright Learning Curve Unit Hours 1 1000.00 2 600.00 3 445.02 4 360.00 5 305.41 a) The time to make the 10 th unit is ________. b)The learning curve rate in percent is ________. c)The slope of the learning curve is ________. d)The time to make the 13th unit is __________. 12/26/2015rd31

32. Mortgage You borrow $10,000 at 6% compounded monthly for 24 years. Your monthly payment is closest to a) $660b) $550c) $450d) $350 Your principal reduction on the 12 th payment is closest to a) $315b) $415c) $515 d) not given Total interest paid after making 12 th payment is a) $370b) $470c) $570d) $670 12/26/2015rd32

33. Annual Worth You want $100,000 in a fund 10 years from now, the amount to deposit in years 6 through 9 at i = 10% per year is closest to a) $19,588b) $20,614c) $21,547d) $22,389 $100K 6 7 8 9 10 (AGF (PGF 100E3 10 1) 10 4)  19588.25 12/26/2015rd33

34. Doubling Investment If you invest $2,000 at 12% compounded monthly for the same length of time that it takes an investment to double in value at 12% compounded quarterly, you will have a) $3709b) $4027c) $4352d) 4580 2 = (1.03) q => q = 23.35 quarters or 70.35 months 2K(1.01) 70.35 = $4027.50 12/26/2015rd34

35. 12/26/2015rd35 Measuring Investments 1.Present Worth (PW) 2.Annual Worth (AW) 3.Future Worth 4.Internal Rate of Return 5.External Rate of Return 6.Benefits/Costs ratio 7.Payback Period 8.Capitalized Worth