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

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Presentation on theme: "Chapter 4 More Interest Formulas EGN 3615 ENGINEERING ECONOMICS WITH SOCIAL AND GLOBAL IMPLICATIONS 1."— Presentation transcript:

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

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 2

3 Uniform Series Compound Amount Factor 0 F1+F2+F3+F4 = F AAAA A F1 0 A F2 0 A F3 0 A=F

4 Uniform Series Compound Amount Factor 4 That is, for 4 periods, F = F 1 + F 2 + F 3 + F 4 = 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 5 For n periods with interest (per period), F = F 1 + F 2 + F 3 + … + F n-1 + F n = 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 AAAA A n A n-1 F

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

7 Uniform Series Formulas (Compare to slide 25) 7 (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. Butrate(nper, pmt, pv, fv, type, guess)

8 Uniform Series Compound Amount Factor 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? $100 F= i=0.05 8

9 QUESTION CONTINUES (USING INTEREST TABLE) $100 F= i=0.05 9

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

11 Uniform Series Compound Amount Factor 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? F2= 580.2F $100 i=5% 11

12 QUESTION CONTINUES(INTEREST TABLE) F2= 580.2F $100 i=5% 12

13 QUESTION CONTINUES(SPREADSHEET) 13

14 Uniform Series Compound Amount Factor 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? $100 F= ? i=6.5% 14

15 INTERPOLATION X Interpolation 15

16 Uniform Series Sinking Fund Factor A= Equal Annual Dollar Payments F= Future Some of Money i = Interest Rate Per Period n= Number of Interest Periods F=Given i=Given A=? n=Given 16

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

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? F=$12, i=5% A =$2172 n=5 18

19 QUESTION CONTINUES(INTEREST TABLE) F=$12, i=5% A = $2172 n=5 19

20 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, 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? F=$12,000 i=5% A = $1764 n=6 20

21 QUESTION CONTINUES(INTEREST TABLE) F=$12,000 i=5% A = $1764 n=6 21 $1764

22 Uniform Series Sinking Fund Factor 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? F=$9000 i=6% A = $ P=$2,500 22

23 Uniform Series Capital Recovery Factor P= Present Sum of Money A= Equal Annual Dollar Payments i = Interest Rate n= Number of Interest Periods P=Given i=Given A=? n 23

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

25 Uniform Series Formulas (Compare to slide 7) 25 (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. Butrate(nper, pmt, pv, fv, type, guess)

26 Uniform Series Capital Recovery Factor 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? A =26,380 P= $100,000 i=10% 26

27 Example CONTINUES(INTEREST TABLE) A =26,380 P= $100,000 i=10% 27

28 Uniform Series Capital Recovery (MS EXCEL) 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 28

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 54 th birthday (25 deposits). Starting at age 55, 15 equal annual withdrawals are made. How much should each withdrawal be? Step 1: First A will be converted into F. Step2: F will be considered as P. Step3: P will be converted into Second A Solution 29

30 EXAMPLE CONTINUES F = 245,868 i=10% 2324 A=$ i=10% 1415 P= $245,868 A =$32,332 30

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

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

33 Uniform Series Present Worth Factor 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? A=26,380 P= 100,001 i=10% 33

34 EXAMPLE CONTINUES (USING INTEREST TABLE) A=26,380 P= 100,001 i=10% 34

35 Uniform Series Present Worth ( Using MS EXCEL ) 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, ) = $100, See spreadsheet 35

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? A= i=5% i=8% 36

37 EXAMPLE CONTINUES A= i=5% i=8% 37

38 EXAMPLE CONTINUES (Using MS EXCEL) 38 P 1 = PV(0.08, 3, -500)(1+0.05) –4 = ( )(0.8227) = $1, P 2 = 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 A A A A G 2G 3G 4G A A+G A+2G A+3G A+4G = 39

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

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? G=$30 P= $ i=5% 0 $30 $60 $90 $120 41

42 QUESTION CONTINUES (INTEREST TABLE) G=$30 P= $ i=5% 0 $30 $60 $90 $120 42

43 Arithmetic Gradient Present Worth Factor Question: If interest rate is 8%, what is the present worth of the following sums? =

44 QUESTION CONTINUES

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

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%) = 46

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 P=? i=5% $100 $110 $121 $ g=10% ? ? ? ? ? ? 47

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. Uniform growth rate (g) First year maintenance cost 48

49 Geometric Series Present Worth Factor Geometric Series Present Worth Factor When Interest rate equals the growth rate, 49 Given A 1, g, i, & n, find P (4-29) & (4-30)

50 Geometric Series Present Worth Factor Question: What is the present value (P) of a geometric series with $100 at EOY1 (A 1 ), 5% interest rate (i), 10% growth rate (g), and 10 interest periods (n)? P=? i=5% $100 $110 $121 $ g=10% ? ? ? ? ? ? 50

51 Geometric Series Present Worth Factor P= $ i=5% $100 $110 $121 $ g=10% $146 $161 $195 $177 $214 $236 51

52 Time for a Joke! What is Recession? Recession is when your neighbor loses his or her job. What is Depression? Depression is when you lose yours. By Ronald Reagan 52

53 Problem Purchase a car:$3,000 down payment $480 payment for 60 months If interest rate is 12% compounded monthly, at what purchase price P of a car can one buy? Solution i = 12.0%/12 = 1.o% per month, n = 60, and A = $480 P = (P/A, 0.01, 60) = (44.955) = $24,578 Important: P = $ $480(60) = $31,800, if i = 0.

54 Problem 4-9 $25 million is needed in three years. Traffic is estimated at 20 million vehicles per year. At 10% interest, what should be the toll per vehicle? (a)Toll receipts at end of each year in a lump sum. (b)Traffic distributed evenly over 12 months, and toll receipts at end of each month in a lump sum. 54

55 Problem 4-9 Solution (a)Let x = the toll/vehicle. Then F = $25,000,000i = 10%/year, n = 3 years Find A (=20,000,000x). A = F(A/F, 0.1, 3) 20,000,000x = 25,000,000(0.3021) x = $ = $0.38 per vehicle 55

56 Problem 4-9 Solution (b)Let x = the toll/vehicle. Then F = $25Mi = (1/12)10%/month, n = 36 months Find A (=20,000,000x/12). A = F{i/[(1+i) n –1]} 20,000,000x/12 = 25,000,000{(0.1/12)/(1+0.1/12) 36 –1} x = $0.359 = $0.36 per vehicle 56

57 Problem 4-32 If i = 12%, for what value of B is the PW = 0? Solution Consider now = time 1. Then PW = B+800(P/A, 0.12, 3) – B(P/A, 0.12, 2) – B(P/F, 0.12, 3) = – 1.758B Letting PW = 0 yields B = $1, For any cash flow diagram, if PW = 0, then its worth at anytime = 0! 57

58 Problem 4-46 Solution FW = FW [1000(F/A, i, 10)](F/P, i, 4) = By try and error: At i = 12%, LHS = [1000(17.549)](1.574) = $27,622 too low At i = 15%, LHS = [1000(20.304)](1.749) = $35,512 too high Using linear interpolation: i = 12% + 3%[(28000 – 27622)/(35512 – 27622)] = 12.14% 58

59 Use of MS EXCEL pmt(i, n, P, F, type) returns A, given i, n, P, and F sinking fund (P=0) A = F{i/[(1+i) n – 1]}(4-5) capital recovery (F=0) A = P{[i(1+i) n ]/[(1+i) n – 1]} (4-6) or combined (P ≠ 0, and F ≠ 0) rate(n, A, P, F, type, guess) returns i, given n, A, P, and F 59

60 Use of MS EXCEL pv(i, n, A, F, type) returns P, given i, n, A, and F present worth (A=0) P = F/(1+i) n (3-5) series present worth (F=0) P = A{[(1+i) n – 1]/[i(1+i) n ]} (4-7) or combined (A ≠ 0, and F ≠ 0) fv(i, n, A, P, type) returns F, given i, n, A, and P compound amount (A=0)F = P(1+i) n (3-3) series compound amount (P=0)F = A{[(1+i) n – 1]/i} (4-4) or combined (A ≠ 0, and P ≠ 0) 60

61 Use of MS EXCEL nper(i, A, P, F, type) returns n, given i, A, P, and F. If A = 0,n = log(F/P)/log(1+i) single payment If P = 0,n = log(1+Fi/A)/log(1+i)uniform series If F = 0,n = log[A/(A-Pi)]/log(1+i)uniform series effect(r, m) returns i a, given r and m. effective annual interest rate i a = (1+r/m) m – 1 (3-7) nominal(i a, m) returns r, given i a and m. nominal annual interest rate r = m[(i a – 1) 1/m + 1] 61

62 A Real Life Case Mr. Goodman set up a trust fund of $1.5M for his 2 children in It is worth more than $300M today (January 2012). What is the effective annual interest rate? Solution P = $1.5M, F = $300M, n = 20 years i a = (F/P) 1/n – 1 = (300/1.5) 1/20 – 1 = % i a = rate(20, 0, 1.5, –300) = % i a = rate(20, 0, -1.5, 300) = % 62

63 End of Chapter 4 Uniform Series Compound Interest Formulas Uniform Series Compound Amount Factor: F/A Uniform Series Sinking Fund Factor:A/F Uniform Series Capital Recovery Factor:A/P Uniform Series Present Worth Factor:P/A Arithmetic Gradient Geometric Gradient Spreadsheet Solutions 63

64 Interpolation-1 Given: F(X1); F(X2) What is F(X3) where X1 < X3 < X2? Assuming linearity so that a linear equation will do: Basic equation: y = mx + b so 1. F(X1) = mX1 + b 2. F(X2) = mX2 + b Subtract 2 from 1: F(X1)-F(X2) = m (X1-X2)  m = (F(X1)-F(X2))/(X1-X2) From 1 we get b = (F(X1) - mX1) 64

65 Interpolation-2 F(X1)-F(X2) = m (X1-X2)  m = (F(X1)-F(X2))/(X1-X2) From 1 we get b = (F(X1) - mX1) Now F(X3) = m X3 + b = m X3 + F(X1) - mX1 = m (X3 - X1) + F(X1) = (X3 - X1) (F(X1)-F(X2))/(X1-X2) + F(X1) = F(X1) + (F(X1)-F(X2)) (X3 - X1) /(X1-X2) 65

66 Interpolation-3 F(X3) = F(X1) + (F(X1)-F(X2)) (X3 - X1) /(X1-X2) Suppose that the Xs are interest rates, i, and the Fs are the functions (F/A,i,n), then F(i3) = F(i1) + (F(i1)-F(i2)) (i3 - i1) /(i1-i2) Return 66


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