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Chapter 4 More Interest Formulas EGN 3615

Chapter 4 More Interest Formulas EGN 3615 ENGINEERING ECONOMICS WITH SOCIAL AND GLOBAL IMPLICATIONS. Chapter Contents. Uniform Series Compound Interest Formulas Uniform Series Compound Amount Factor Uniform Series Sinking Fund Factor Uniform Series Capital Recovery Factor

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Chapter 4 More Interest Formulas EGN 3615

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  1. Chapter 4More Interest Formulas EGN 3615 ENGINEERING ECONOMICS WITH SOCIAL AND GLOBAL IMPLICATIONS

  2. Chapter Contents • Uniform Series Compound Interest Formulas • Uniform Series Compound Amount Factor • Uniform Series Sinking Fund Factor • Uniform Series Capital Recovery Factor • Uniform Series Present Worth Factor • Arithmetic Gradient • Geometric Gradient • Nominal Effective Interest • Continuous Compounding

  3. Uniform Series Compound Amount Factor F1+F2+F3+F4 =F A A A A 0 1 2 3 4 0 1 2 3 4 A F1 0 1 2 3 4 A F2 0 1 2 3 4 A F3 0 1 2 3 4 A=F4 0 1 2 3 4

  4. Uniform Series Compound Amount Factor That is, for 4 periods, F = F1 + F2 + F3 + F4 = A(1+i)3 + A(1+i)2 + A(1+i) + A = A[(1+i)3 + (1+i)2 + (1+i) + 1]

  5. Uniform Series Compound Amount Factor A A A A F For n periods with interest (per period), F = F1 + F2 + F3 + … + Fn-1 + Fn = A(1+i)n-1 + A(1+i)n-2 + A(1+i)n-3 + … + A(1+i) + A = A[(1+i)n-1+ (1+i)n-1 + (1+i)n-3 + …+ (1+i) + 1] 0 1 2 3 4 A A n-1 n

  6. Uniform Series Compound Amount Factor i= interest rate per period n = total # of periods Notation Uniform Series Compound Amount Factor

  7. Uniform Series Formulas (Compare to slide 25) (1) Uniform series compound amount: Given A, i, & n, find F F = A{[(1+i)n – 1]/i} = A(F/A, i, n) (4-4) (2) Uniform series sinking fund: Given F, i, & n, find A A = F{i/[(1+i)n – 1]} = F(A/F, i, n) (4-5) (3) Given F, A, & i, find n n = log(1+Fi/A)/log(1+i) (4) Given F, A, & n, find i There is no closed form formula to use. But rate(nper, pmt, pv, fv, type, guess)

  8. Uniform Series Compound Amount Factor F= 552.6 • Question: If starting at EOY1, five annual deposits of $100 each are made in the bank account, how much money will be in the account at EOY5, if interest rate is 5% per year? i=0.05 0 1 2 3 4 5 $100 $100 $100 $100 $100

  9. QUESTION CONTINUES (USING INTEREST TABLE) F= 552.6 i=0.05 0 1 2 3 4 5 $100 $100 $100 $100 $100

  10. QUESTION CONTINUES(SPREADSHEET) Go to XL--Chap 4 extended examples-A1 Use function: FV(rate, nper, pmt, pv, type)

  11. Uniform Series Compound Amount Factor F1 F2= 580.2 • Question: Five annual deposits of $100 each are made into an account starting today. If interest rate is 5%, how much money will be in the account at EOY5? i=5% 0 1 2 3 4 5 $100 $100 $100 $100 $100

  12. QUESTION CONTINUES(INTEREST TABLE) F1 F2= 580.2 i=5% 0 1 2 3 4 5 $100 $100 $100 $100 $100

  13. QUESTION CONTINUES(SPREADSHEET)

  14. Uniform Series Compound Amount Factor F= ? • Question: If starting at EOY1, five annual deposits of $100 each are made in the bank account, how much money will be in the account at EOY5, if interest rate is 6.5% per year? i=6.5% 0 1 2 3 4 5 $100 $100 $100 $100 $100

  15. INTERPOLATION 6.0 5.6371 0.5 X 1 6.5 0.1136 7.0 5.7507 Interpolation

  16. Uniform Series Sinking Fund Factor F=Given i=Given n=Given 0 1 2 3 4 5 A=? A= Equal Annual Dollar Payments F= Future Some of Money i = Interest Rate Per Period n= Number of Interest Periods

  17. Uniform Series Sinking Fund Factor Notation Uniform Series Sinking Fund Factor

  18. Uniform Series Sinking Fund Factor • Question: A family wishes to have $12,000 in a bank account by the EOY 5. to accomplish this goal, five annual deposits starting at the EOF year 1 are to be made into a bank account paying 6% interest. what annual deposit must be made to reach the stated goal? i=5% n=5 0 1 2 3 4 5 A =$2172 F=$12,000

  19. QUESTION CONTINUES(INTEREST TABLE) i=5% n=5 0 1 2 3 4 5 A = $2172 F=$12,000

  20. Uniform Series Sinking Fund Factor F=$12,000 • Question: A family wishes to have $12,000 in a bank account by the EOY 5. to accomplish this goal, six annual deposits starting today are to be made into a bank account paying 5% interest. What annual deposit must be made to reach the stated goal? i=5% n=6 0 1 2 3 4 5 A = $1764

  21. $1764 QUESTION CONTINUES(INTEREST TABLE) F=$12,000 i=5% n=6 0 1 2 3 4 5 A = $1764

  22. Uniform Series Sinking Fund Factor F=$9000 • Example: the current balance of a bank account is $2,500. starting EOY 1 six equal annual deposits are to be made into the account. The goal is to have a balance of $9000 by the EOY 6. if interest rate is 6%, what annual deposit must be made to reach the stated goal? i=6% 0 1 2 3 4 5 6 P=$2,500 A = $796.23

  23. Uniform Series Capital Recovery Factor P=Given i=Given n 0 1 2 3 4 5 A=? P= Present Sum of Money A= Equal Annual Dollar Payments i = Interest Rate n= Number of Interest Periods

  24. Uniform Series Capital Recovery Factor Notation Uniform Series Capital Recovery Factor

  25. Uniform Series Formulas (Compare to slide 7) (1) Uniform series present worth: Given A, i, & n, find P P = A{[(1+i)n – 1]/[i(1+i)n]} = A(P/A, i, n)(4-7) (2) Uniform series capital recovery: Given P, i, & n, find A A = P{[i(1+i)n]/[(1+i)n – 1]} = P(A/P, i, n) (4-6) (3) Given P, A, & i, find n n = log[A/(A-Pi)]/log(1+i) (4) Given P, A, & n, find i (interest/period) There is no closed form formula to use. But rate(nper, pmt, pv, fv, type, guess)

  26. Uniform Series Capital Recovery Factor P= $100,000 • Example: A person borrows $100,000 from a commercial bank. The loan is to be repaid with five equal annual payments. If interest rate is 10%, what should the annual payments be? i=10% 0 1 2 3 4 5 A =26,380

  27. Example CONTINUES(INTEREST TABLE) P= $100,000 i=10% 0 1 2 3 4 5 A =26,380

  28. Use function: PMT(rate, nper, pv, fv, type) rate = interest rate/period nper = # of periods pv = present worth fv = balance at end of period n (blank means 0). type = 1 (payment at beginning of each period) or 0 (payment at end of a period)(blank means 0) See spreadsheet Uniform Series Capital Recovery (MS EXCEL)

  29. Uniform Series Capital Recovery Factor • Example: At age 30, a person begins putting $2,500 a year into account paying 10% interest. The last deposit is made on the man’s 54th birthday (25 deposits). Starting at age 55, 15 equal annual withdrawals are made. How much should each withdrawal be? Solution • Step 1: First A will be converted into F. • Step2: F will be considered as P. • Step3: P will be converted into Second A

  30. EXAMPLE CONTINUES F = 245,868 i=10% 0 1 2 3 21 22 23 24 A=$2500 i=10% 0 1 2 3 12 13 14 15 A =$32,332 P= $245,868

  31. Uniform Series Present Worth Factor A=Given n=Given 0 1 2 3 4 5 i=Given P=? A= Equal Annual Dollar Payments P= Present Sum of Money (at Time 0) i = Interest Rate/Period n= Number of Interest Periods

  32. Uniform Series Present Worth Factor Notation Uniform Series Present Worth Factor

  33. Uniform Series Present Worth Factor A=26,380 • Example: A special bank account is to be set up. Each year, starting at EOY 1, a $26,380 withdrawal is to be made. After five withdrawals the account is to be depleted. if interest rate is 10%, how much money should be deposited today? 0 1 2 3 4 5 i=10% P= 100,001

  34. EXAMPLE CONTINUES (USING INTEREST TABLE) A=26,380 0 1 2 3 4 5 i=10% P= 100,001

  35. Use function: PV(rate, nper, pmt, fv, type) rate = interest rate/period nper = total # of periods (payments) pmt = constant payment/period fv = balance at end of period n (blank means 0) type = 1 or 0 PV(0.1, 5, -26380) = $100,000.95 See spreadsheet Uniform Series Present Worth (Using MS EXCEL)

  36. Uniform Series Present Worth Factor • Example: Eight annual deposits of $500 each are made into a bank account beginning today. Up to EOY 4, the interest rate is 5%. After that, the interest rate is 8%. What is the present worth of these deposits? i=8% i=5% 0 1 2 3 4 5 6 7 A=500

  37. EXAMPLE CONTINUES i=8% i=5% 0 1 2 3 4 5 6 7 A=500

  38. EXAMPLE CONTINUES (Using MS EXCEL) P1 = PV(0.08, 3, -500)(1+0.05)–4 = (1288.55)(0.8227) = $1,060.09 P2 = PV(0.05, 4, -500) = $1,772.98

  39. Arithmetic Gradient • Arithmetic Gradient series (G): each annual amount differs from the previous one by a fixed amount G. A+4G A+3G 0 0 0 1 1 1 2 2 2 3 3 3 4 4 4 5 5 5 A+2G A+G 4G A 3G A A A A A 2G G + = 0

  40. Arithmetic Gradient Present Worth Factor • Given G, i, & n, find P (4-19) Notation Arithmetic Gradient Present Worth Factor

  41. Arithmetic Gradient Present Worth Factor • Question: You has purchased a new car. the following maintenance costs starting at EOY 2 will occur to pay the maintenance of your car for the 5 years. EOY2 $30, EOY3 $60, EOY4 $90, EOY5 $120. If interest rate is 5%, how much money you should deposit into a bank account today? $120 $90 $60 $30 0 i=5% 0 1 2 3 4 5 G=$30 P= $247.11

  42. QUESTION CONTINUES (INTEREST TABLE) $120 $90 $60 $30 0 i=5% 0 1 2 3 4 5 G=$30 P= $247.11

  43. Arithmetic Gradient Present Worth Factor • Question: If interest rate is 8%, what is the present worth of the following sums? 600 400 400 400 400 550 0 0 0 1 1 1 2 2 2 3 3 3 4 4 4 5 5 5 500 450 200 150 400 400 100 50 + = 0

  44. QUESTION CONTINUES 400 400 400 400 400 0 0 1 1 2 2 3 3 4 4 5 5 200 150 100 50 0

  45. Arithmetic Gradient Uniform Series Factor Convert an arithmetic gradient series into a uniform series Given G, i, & n, find A (4-20) Notation Arithmetic Gradient Uniform Series Factor

  46. Arithmetic Gradient Uniform Series Factor • Question: Demand for a new product will decrease as competitors enter the market. What is the equivalent annual amount of the revenue cash flows shown below? (interest 12%) 3000 3000 3000 3000 3000 3000 2500 0 0 0 1 1 1 2 2 2 3 3 3 4 4 4 5 5 5 2000 1500 1000 = + 0 500 1000 1500 2000

  47. Geometric Series Present Worth Factor ? • Geometric series: Each annual amount is a fixed percentage different from the last. In this case, the change is 10%. • We will look at this problem in a few slides. g=10% ? $133 ? $121 $110 ? $100 ? ? 0 1 2 3 4 5 6 7 8 9 10 i=5% P=?

  48. Geometric Gradient • Unlike the Arithmetic Gradient where the amount of period-by-period change is a constant, for the Geometric Gradient, the period-by-period change is a uniform growth rate (g) or percentage rate. First year maintenance cost Uniform growth rate (g)

  49. Geometric Series Present Worth Factor Given A1, g, i, & n, find P (4-29) & (4-30) Geometric Series Present Worth Factor When Interest rate equals the growth rate,

  50. Geometric Series Present Worth Factor ? • Question: What is the present value (P) of a geometric series with $100 at EOY1 (A1), 5% interest rate (i), 10% growth rate (g), and 10 interest periods (n)? g=10% ? $133 ? $121 $110 ? $100 ? ? 0 1 2 3 4 5 6 7 8 9 10 i=5% P=?

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