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**Chapter 4 More Interest Formulas**

EGN 3615 ENGINEERING ECONOMICS WITH SOCIAL AND GLOBAL IMPLICATIONS 1

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**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

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**Uniform Series Compound Amount Factor**

F1+F2+F3+F4 =F 1 2 3 4 A 1 2 3 4 A 1 2 3 4 F1 A 1 2 3 4 F2 A 1 2 3 4 F3 A=F4 1 2 3 4

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**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]

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**Uniform Series Compound Amount Factor**

For n periods with interest (per period), F = F1 + F2 + F3 + â€¦ + Fn-1 + Fn = A(1+i)n-1 + A(1+i)n-2 + A(1+i)n-3 + â€¦ + A(1+i) + A = A[(1+i)n-1 + (1+i)n-1 + (1+i)n-3 + â€¦+ (1+i) + 1] A 1 2 3 4 A A n-1 n

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**Uniform Series Compound Amount Factor**

Notation i = interest rate per period n = total # of periods

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**Uniform Series Formulas (Compare to slide 25)**

(1) Uniform series compound amount: Given A, i, & n, find F F = A{[(1+i)n â€“ 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 â€“ 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)

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**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? 1 2 3 4 5 $100 F= 552.6 i=0.05

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**QUESTION CONTINUES (USING INTEREST TABLE)**

1 2 3 4 5 $100 F= 552.6 i=0.05

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**QUESTION CONTINUES(SPREADSHEET)**

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

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**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= 580.2 F1 1 2 3 4 5 $100 i=5%

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**QUESTION CONTINUES(INTEREST TABLE)**

F2= 580.2 F1 1 2 3 4 5 $100 i=5%

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**QUESTION CONTINUES(SPREADSHEET)**

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**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? 1 2 3 4 5 $100 F= ? i=6.5%

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INTERPOLATION 6.0 7.0 6.5 5.6371 5.7507 0.5 X 1 0.1136 Interpolation

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**Uniform Series Sinking Fund Factor**

1 2 3 4 5 F=Given i=Given A=? n=Given A= Equal Annual Dollar Payments F= Future Some of Money i = Interest Rate Per Period n= Number of Interest Periods

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**Uniform Series Sinking Fund Factor**

Notation

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**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=$12,000 1 2 3 4 5 i=5% A =$2172 n=5

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**QUESTION CONTINUES(INTEREST TABLE)**

F=$12,000 1 2 3 4 5 i=5% A = $2172 n=5

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**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? 1 2 3 4 5 F=$12,000 i=5% A = $1764 n=6

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**QUESTION CONTINUES(INTEREST TABLE)**

$1764 1 2 3 4 5 F=$12,000 i=5% A = $1764 n=6

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**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? 1 2 3 4 5 F=$9000 i=6% A = $796.23 6 P=$2,500

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**Uniform Series Capital Recovery Factor**

1 2 3 4 5 P=Given i=Given A=? n P= Present Sum of Money A= Equal Annual Dollar Payments i = Interest Rate n= Number of Interest Periods

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**Uniform Series Capital Recovery Factor**

Notation

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**Uniform Series Formulas (Compare to slide 7)**

(1) Uniform series present worth: Given A, i, & n, find P P = A{[(1+i)n â€“ 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 â€“ 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)

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**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? 1 2 3 4 5 A =26,380 P= $100,000 i=10%

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**Example CONTINUES(INTEREST TABLE)**

1 2 3 4 5 A =26,380 P= $100,000 i=10%

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**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

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**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 manâ€™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

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**EXAMPLE CONTINUES F = 245,868 i=10% A=$2500 i=10% A =$32,332**

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

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**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 1 2 3 4 5 P=? i=Given A=Given n=Given

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**Uniform Series Present Worth Factor**

Notation

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**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? 1 2 3 4 5 A=26,380 P= 100,001 i=10%

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**EXAMPLE CONTINUES (USING INTEREST TABLE)**

1 2 3 4 5 A=26,380 P= 100,001 i=10%

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**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, ) = $100,000.95 See spreadsheet

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**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? 1 2 3 4 5 A=500 6 7 i=5% i=8%

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

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**EXAMPLE CONTINUES (Using MS EXCEL)**

P1 = PV(0.08, 3, -500)(1+0.05)â€“4 = ( )(0.8227) = $1, P2 = PV(0.05, 4, -500) = $1,772.98

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

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**Arithmetic Gradient Present Worth Factor**

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

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**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? 1 2 3 4 5 G=$30 P= $247.11 i=5% $30 $60 $90 $120

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**QUESTION CONTINUES (INTEREST TABLE)**

1 2 3 4 5 G=$30 P= $247.11 i=5% $30 $60 $90 $120

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**Arithmetic Gradient Present Worth Factor**

Question: If interest rate is 8%, what is the present worth of the following sums? + 400 1 2 3 4 5 50 100 150 200 450 500 550 600 = 400 400 400 400

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

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**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

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**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%) 1 2 3 4 5 1000 1500 2000 2500 3000 = 3000 3000 3000 3000 3000 1 2 3 4 5 + 1 2 3 4 5 500 1000 1500 2000

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**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. 1 2 3 4 5 P=? i=5% $100 $110 $121 $133 6 7 8 9 10 g=10% ? ? ? ? ?

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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)

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**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,

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**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)? 1 2 3 4 5 P=? i=5% $100 $110 $121 $133 6 7 8 9 10 g=10% ? ? ? ? ?

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**Geometric Series Present Worth Factor**

1 2 3 4 5 P= $ i=5% $100 $110 $121 $133 6 7 8 9 10 g=10% $146 $161 $195 $177 $214 $236

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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

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**Problem 4-7 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.

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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.

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Problem 4-9 Solution (a) Let x = the toll/vehicle. Then F = $25,000,000 i = 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

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

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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) â€“ B(P/A, 0.12, 2) â€“ B(P/F, 0.12, 3) = â€“ 1.758B Letting PW = 0 yields B = $1, For any cash flow diagram, if PW = 0, then its worth at anytime = 0!

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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 â€“ 27622)/(35512 â€“ 27622)] = 12.14%

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**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 â€“ 1]} (4-5) capital recovery (F=0) A = P{[i(1+i)n]/[(1+i)n â€“ 1]} (4-6) or combined (P â‰ 0, and F â‰ 0) rate(n, A, P, F, type, guess) returns i, given n, A, P, and F

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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 â€“ 1]/[i(1+i)n]} (4-7) or combined (A â‰ 0, and F â‰ 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 â€“ 1]/i} (4-4) or combined (A â‰ 0, and P â‰ 0)

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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 ia, given r and m. effective annual interest rate ia = (1+r/m)m â€“ 1 (3-7) nominal(ia, m) returns r, given ia and m. nominal annual interest rate r = m[(ia â€“ 1)1/m + 1]

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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 ia = (F/P)1/n â€“ 1 = (300/1.5)1/20 â€“ 1 = % ia = rate(20, 0, 1.5, â€“300) = % ia = rate(20, 0, -1.5, 300) = %

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**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

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**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 F(X1) = mX1 + b F(X2) = mX2 + b Subtract 2 from 1: F(X1)-F(X2) = m (X1-X2) ïƒ¨ m = (F(X1)-F(X2))/(X1-X2) From 1 we get b = (F(X1) - mX1)

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**Interpolation-2 F(X1)-F(X2) = m (X1-X2) ïƒ¨ 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)

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**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

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