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Engineering Economic Analysis

Engineering Economic Analysis. Chapter 4  More Interest Formulas. Effective Interest Rate. 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

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Engineering Economic Analysis

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  1. Engineering Economic Analysis Chapter 4  More Interest Formulas rd

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

  3. 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% rd

  4. 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% rd

  5. 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 rd

  6. Monthly Payments • What is the monthly payment for a 5-year car loan of $35,000 at 6% compounded monthly? • Find the amount of the principal reduction of the 25th payment. • After making the 50th monthly payment, you decide to pay off the loan what a check for _________. • Find the interest on the 35th payment. • a) A = $35,000(A/P, ½ %, 60) = $676.65. • PR25 = 676.65(P/F, ½ %, 60 – 25 + 1) = $565.44. • B50 = 676.65(P/A, ½ %, 10) = $6584.08 • I35 = B34 * i = 676.65(P/A, ½ %, 26) * ½ % = $82.29 rd

  7. 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 rd

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

  9. 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 rd

  10. 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 = 1214 121 + 0.10 * 121 = 100(1 + 0.10)3 = 133… … … … … … … n 100(1 + 0.10)n – 1 • An = A1(1 + g)n-1 • Each An must be brought back to year 0 to find the present worth. rd

  11. Geometric Gradient An = An-1(1+ g) => An = A1(1 + g)n– 1 Then P = A1(1 + g)n-1(1 + i)-n P = = rd

  12. 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 rd

  13. 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 rd

  14. Geometric Gradient Example You want to accumulate $1M 20 years from now by depositing $A1 at year 1 and increasing the deposit by 6% each year for 20 years. Find A1 if the bank pays 8% interest compounded annually. F = A1(P/A1, g = 6%, i = 8%, n = 20)(F/P, i = 8%, 20) 1M = A1{[1 – 1.0620*1.08-20]/0.02}(1.08)20 A1 = $13,756.85 = (/ 1e6 (FGP (PGGG 1 6 8 20) 8 20) P = rd

  15. 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) = $1417.63Quarterly: 1000(F/P, 1.76%, 20) = $1417.63Semi-annually 1000(F/P, 3.55%, 10) = $1417.63Annually 1000(F/P, 7.23%, 5) = $1417.63Biennially 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 + im)nm = 1000(1 + ia)n rd

  16. 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 rd

  17. Continuous Compounding • In the effective interest formula let m = rp and the formula • becomes • ieff = (1 + r/m)m - 1 e • ieff = (1 + 1/p)rp = • = er as p   • (F/P, r%, N) = erN for continuous compounding rd

  18. 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, e0.005 -1, 60) = $6979.70 • The monthly continuous compounding rate e0.005 -1 = 0.501 rd

  19. Problem 4-36 • $12K is borrowed at 4% per annum and is to repaid in 5 payments. After the 2nd 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 rd

  20. 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) rd

  21. 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 4 5 • 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% rd

  22. 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. rd

  23. Gradient • Find the equivalent sum at year 7 for the following cash flow at 7% compound interest per year. • n 1 2 3 4 5 6 7 • Cf 1000 2000 3000 4050 5000 • F7 = [(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 rd

  24. 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) => im = 2.92% => APR = 34.04% • Iaeff = 41.25% rd

  25. 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 rd

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

  27. 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%. • 5000(A/P, 7%, 5) – 1000(A/F, 7%, 5) = 1219.45 – 173.89 = $1045.56 • (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 rd

  28. Review rd

  29. Change in Rate • You borrow $20,000 at 7% compounded monthly over 48 months. After making the 24th 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. • Am = 20K(A/P, 7/12 %, 48) = $478.92 • B24 = 478.92(P/A, 7/12 %, 24) = $10,696.84 • Aq = 10,698.84(A/P, 1.76%, 8) = $1445.17 rd

  30. Exact Rate of Return • Find the exact rate of return for the following cash flow. • n 0 1 2cf -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 rd

  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 10th unit is ________. • The learning curve rate in percent is ________. • The slope of the learning curve is ________. • The time to make the 13th unit is __________. rd

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

  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) $19,588 b) $20,614 c) $21,547 d) $22,389 • $100K • 6 7 8 9 10 • (AGF (PGF 100E3 10 1) 10 4) 19588.25 rd

  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) $3709 b) $4027 c) $4352 d) 4580 • 2 = (1.03)q => q = 23.35 quarters or 70.35 months • 2K(1.01)70.35 = $4027.50 rd

  35. Measuring Investments • Present Worth (PW) • Annual Worth (AW) • Future Worth • Internal Rate of Return • External Rate of Return • Benefits/Costs ratio • Payback Period • Capitalized Worth rd

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