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- 1. Chapter 2 Factors: How Time and Interest Affect Money MS291: Engineering Economy
- 2. Content of the Chapter Single-Payment Compound Amount Factor (SPCAF) Single-Payment Present Worth Factor (SPPWF) Uniform Series Present Worth Factor (USPWF) Capital Recovery Factor (CRF) Uniform Series Compound Amount Factor Sinking Fund Factor (SFF) Arithmetic Gradient Factor Geometric Gradient Series Factor (Optional Topic)
- 3. Simple and Compound Interest Example: $100,000 lent for 3 years at interest rate i = 10% per year. What is repayment after 3 years ? Simple Interest Compound Interest Here Interest, year 1: I1 = 100,000(0.10) = $10,000 Total due, year 1: F1 = 100,000 + 10,000 =$110,000 P=$100,000 n= 3 i= 10% Simple interest = P X n x i Interest = 100,000(3)(0.10) = $30,000 Total due = 100,000 + 30,000 = $130,000 Interest, year 2: I2 = 110,000(0.10) = $11,000 Total due, year 2: F2 = 110,000 + 11,000 = $121,000 Interest, year 3: I3 = 121,000(0.10) = $12,100 Total due, year 3: F3 = 121,000 + 12,100 = $133,100 Simple: $130,000: Compounded: $133,100
- 4. Single Payment Compound Amount Factor (SPCAF) If an amount “P” is invested at time “t=0” the amount accumulated after a year is given as to generalize the process for period “n” we can write as; F1 = P + Pi = P(1 + i) ………. (1) F = P(1+i)n At the end of second year, the accumulated amount F2 is given as; F2 = F1 + F1 i = P(1+i) + P(1+i)i (from Eq. 1) = P + Pi + Pi+ Pi2 = P(1+i)2 …………(2) Similarly; F3 = F2 + F2 i = P(1+i)3 ………..(3) The term “(1+i)n” is known as Single Payment Compound Amount Factor (SPCAF) It is also refer as F/P factor This is a converting factor, when multiplied by “P” yields the future amount “F” of initial amount “P” after “n” years at interest rate “i”
- 5. Simple and Compound Interest Example: $100,000 lent for 3 years at interest rate i = 10% per year. What is repayment after 3 years ? Simple Interest Compound Interest Here Now we have F = P(1+i)n P=$100,000 n= 3 i= 10% Simple interest = P X n x i Interest = 100,000(3)(0.10) = $30,000 Total due = 100,000 + 30,000 = $130,000 P = $100,000 n=3 i=10% So F = 100,000 (1+0.10)3 F= 100,000 (1.331) F = 133100 Simple: $130,000: Compounded: $133,100
- 6. From SPCAF to SPPWF Now we have the formula how to “convert” present amounts into future amount at a given interest rate i.e. F = P(1+i)n What if we are given a future amount (F) and we are asked to calculate present worth (P) ? F = P(1+i)n => P = F [1/(1+i)n] or P = F(1+i)-n
- 7. Single Payment Present Worth Factor (SPPWF) The term “(1+i)-n” is known as Single Payment Present Worth Factor (SPPWF) It is also refer as P/F factor This is a converting factor, when multiplied by “F” yields the present amount “P” of initial amount “F” after “n” years at interest rate “i”
- 8. Example Find the present value of $10,000 to be received 10 years from now at a discount rate of 10% F = $10,000 r = 10% n = 10 P = F (1+i)-n => P = 10,000 (1+0.1)-10 = 10,000 x 0.385 = $385
- 9. A Standard Notation Instead of writing the full formulas of SPCAF and SPPWF for simplicity there is a standard notation This notations includes two cash flows symbols, interest rate and number of periods General form is: (X/Y, i, n) which means “X” represents what is sought, Y is given, i is interest rate and n is number of periods Examples: Name Single-payment compound amount Single-payment present worth Equation with factor formula Notation Standard Notation Equation Find/ Given F = P(F/P, i, n) F/P A = P(1+i)n (F/P, i, n) P = F(1+i)-n (P/F, i, n) P = F(P/F, i, n) P/F
- 10. Using Standard Notation Example What will be the future value of Rs. 100,000 compounded for 17 years at rate of interest 10% ? F= (1+ i)n or F = P(1+0.1)n now writing that in standard notation we have (F/P, i, n) F = P(F/P, i, n) F = 100,000(F/P, 10%, 17) That value you F = 100,000 (5.054) get from “Table” F= 505400
- 11. Single Payments to Annuity Normally, in real world we do not face Single payments mostly instead faces cash flows such as home mortgage payments and monthly insurance payments etc An annuity is an equal annual series of cash flows. It may be equal annual deposits, equal annual withdrawals, equal annual payments, or equal annual receipts. The key is equal, annual cash flows
- 12. Uniform Series Present Worth Factor (USPWF) P=? t = given 1 2 3 n-1 t=0 A = given n
- 13. Capital Recovery Factor (CRF) P = given t = given 1 2 3 A=? n-1 n t=0 Name Uniform Series Present Worth Capital Recovery Equation with factor formula Notation Standard Notation Equation (P/A, i, n) P = A(F/P, i, n) (A/P, i, n) A = P(A/P, i, n)
- 14. Example 1: Uniform Series Present Worth (P/A) A chemical engineer believes that by modifying the structure of a certain water treatment polymer, his company would earn an extra $5000 per year. At an interest rate of 10% per year, how much could the company afford to spend now to just break even over a 5 year project period? The cash flow diagram is as follows: Solution: A = 5000 A = $5000 I = 10% P = A(P/A, i, n) 0 1 2 i =10% P=? 3 4 5 P = 5000(P/A,10%,5) = 5000(3.7908) = $18,954 n=5
- 15. Example 2: Uniform Series Capital Recovery (A/P) A chemical product company is considering investment in cost saving equipment. If the new equipment will cost $220,000 to purchase and install, how much must the company save each year for 3 years in order to justify the investment, if the interest rate is 10% per year? Solution: The cash flow diagram is as follows: A=? P = 220,000 I = 10% n=3 A = P(A/P, i, n) 0 1 P = $220,000 2 i =10% 3 A = 220,000(A/P,10%,3) = 220,000(0.40211) = $88,464
- 16. Uniform Series Compound Amount Factor (USCAF) F=? t = given 0 1 2 n-1 3 A = given n
- 17. Sinking Fund Factor (SFF) t = given 0 1 2 n-1 3 n A=? F = given
- 18. Example 3: Uniform Series Involving F/A An industrial engineer made a modification to a chip manufacturing process that will save her company $10,000 per year. At an interest rate of 8% per year, how much will the savings amount to in 7 years? Solution: The cash flow diagram is: A =10,000 i =8% n =7 F=? i = 8% 0 1 2 3 4 A = $10,000 5 6 7 F = A(F/A, i, n) F = 10,000(F/A,8%,7) = 10,000(8.9228) = $89,228 Practice Example 2.5
- 19. Uniform Series Compound Amount Factor (USCAF) F=? t = given 0 1 2 n-1 3 A = given n
- 20. Sinking Fund Factor (SFF) F = given t = given 0 1 2 n-1 3 A=? n
- 21. Example 3: Uniform Series Involving F/A An industrial engineer made a modification to a chip manufacturing process that will save her company $10,000 per year. At an interest rate of 8% per year, how much will the savings amount to in 7 years? Solution: The cash flow diagram is: A =10,000 i =8% n =7 F=? A = $10,000 0 1 2 i = 8% 3 4 5 6 7 F = A(F/A, i, n) F = 10,000(F/A,8%,7) = 10,000(8.9228) = $89,228 Practice Example 2.5
- 22. Arithmetic Gradient Factors (P/G, A/G) Cash flows that increase or decrease by a constant amount are considered arithmetic gradient cash flows. The amount of increase (or decrease) is called the gradient CFn = base amount + (n-1)G Cash Flow Formula (n-1)G A+(n-1) G A A+3G A+2G A+G A A A A 3G 2G A G = + 0 1 2 3 4 PT = 0 n 1 2 3 4 0 n 0 PA + PG 1 2 3 4 n
- 23. Arithmetic Gradient Factors (P/G, A/G) PT = PA + PG PA = A(P/A, i, n) or Uniform Series Present worth Factor PG = G(P/G, i, n) or Arithmetic Gradient Present Worth Factor Alternatively, PG can also be calculated by following formula PG G i (1 i ) n 1 n i (1 i ) n (1 i ) n
- 24. Solving Arithmetic Gradient related problems Present value of the Arithmetic Gradient series can be calculated as follows: 1. Find the gradient and base 2. Cash flow diagram maybe helpful if u draw it 3. Break the gradient series into a Uniform series and a Gradient Series as shown on next slide 4. The formula for calculating present value of the Arithmetic Gradient series is as follows; PT = PA + PG 5. Calculate PA and PG and use the above formula to get the present value of the Arithmetic Gradient
- 25. Example (Problem 2.25) Profits from recycling paper, cardboard, aluminium, and glass at a liberal arts college have increased at a constant rate of $1100 in each of the last 3 years. If this year’s profit (end of year 1) is expected to be $6000 and the profit trend continues through year 5, (a) what will the profit be at the end of year 5 and (b) what is the present worth of the profit at an interest rate of 8% per year? G = $1100, Base = $6000
- 26. Example (Problem 2.25) (a) what will the profit be at the end of year 5 & (b) what is the present worth of the profit at an interest rate of 8% per year? G = $1100 Base = $6000 $10400 $9300 $8200 1 2 $6000 $1100 + => $7100 $6000 0 0 3 $4400 $3300 $2200 4 5 Find the cash flows as follows: CF = Base + G(n-1)G CF1 = 6000 + 1100(1-1)= 6000 CF2 = 6000 + 1100(2-1)= 7100 CF3 = 6000 + 1100(3-1)= 8200 CF4 = 6000 + 1100(4-1)= 9300 CF5 = 6000 + 1100(5-1)= 10400 1 2 3 4 0 5 P = A(P/A, i, n) P = 6000(P/A, 8%, 5) P = 6000(3.9927) P = 326066 + + + 1 2 3 4 5 G(P/G, i, n) 1100(P/G, 8%, 5) 1100(7.3724)
- 27. Arithmetic Gradient Future Worth The following formula is used for calculating future value of arithmetic gradient given the values of gradient (the term in bracket is called Arithmetic gradient future worth factor) FG 1 (1 i ) n 1 n G i i (n-1)G 3G 2G G 0 Converting Arithmetic Gradient (G) into Uniform Series (A) (to convert G into (A/G, i, n) Formula for conversation from G to (A/G, i, n) 1 n AG G n i (1 i ) 1 Also AT = AA + AG Where AA = P(A/P, i, n) and AG = G(A/G, i, n) F?
- 28. Geometric Gradient Factors (Pg /A) A Geometric gradient is when the periodic payment is increasing (decreasing) by a constant percentage: A1 is the initial cash in Year 1 i = given g = given A1 (1+g)n-1 A1 (1+g)2 Pg is the sum of whole series including A1 A1 (1+g) A1 It maybe noted that A1 is not considered separately in geometric gradients 1 g 1 1 i Pg A ig nA Pg 1 i n for g i 0 Pg f or g i Note: If g is negative, change signs in front of both g values where: A1 = cash flow in period 1 and g = rate of increase 1 2 3 ----- n
- 29. Example: Geometric Gradient Determine the present worth of a geometric gradient series with a cash flow of $50,000 in year 1 and increases of 6% each year through year 8. The interest rate is 10% per year. n 1 g 1 1 i Pg A ig 8 1 0 .6 1 1 0 .10 50 ,000 1 0 .743 50000 0 .10 0 .06 0 .04 0 .257 50 ,000 0 .04 50 ,000 ( 6 .425 ) $ 321 ,250
- 30. Thank You
- 31. Using Factor Tables
- 32. Using Factor Tables

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