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Engineering Economy Annuity Problems document sample
Lecture No. 2 Factors: How Time and Interest Affect Money 1. Single Payment Factors A. F/P, P/F The fundamental factor in engineering economy is the one that determines the amount of money F accumulated after n periods from a single present worth, P, with interest compounded one time per period. For n =1 F1 = P + interest = P + Pin, for n=1 F1 = P + Pi F1 = P(1 + i)1 For n =2 F2 = F1 + interest for F1 F2 = F1 + F1i F2 = (P + Pi) + (P + Pi)i F2 = P + Pi + Pi + Pi2 F2 = P(1 + 2i + i2) F2 = P (1 + i)2 For n=3 F3 = F2 + interest for F2 F3 = P (1 + i)2 + P (1 + i)2i F3 = P (1 + i)3 For n = n Fn = P(1+i)n (1+i)n is called the single-payment compound amount factor, SPCAF or the F/P factor. F = P (1 + i)n, F/P Solving for P: 1 P = F[ ] , P/F (1+i)n 1 [ (1+i)n ] is called the single payment present worth factor, SPPWF, or the P/F factor. Both factors are for single payments. Lecture No. 2, Factors: How Time and Interest Affect Money Page No. 2 P = given i = given 0 1 2 n-2 n-1 n F=? B. Standard Notation Standard notation has been adopted for all factors, which includes: two cash flow symbols, the interest rates and the number of symbols. It is always in the general form: (X/Y, i, n) X = what is sought Y = what is given i = interest rate, % n = number of periods Given: (F/P, 12%, 10) Find: Translate the above, calculate F/P then check the table value, calculate P/F F the future worth is what is being sought P the present worth is that is known 12% is the interest rate 10 is the number of time periods (F/P, 12%, 10) = (1+i)n = (1+.12)10 = 1.1210 (F/P, 12%, 10) = 3.105848 From T17, p.718 (F/P, 12%, 10) = 3.1058 P/F = 1/F/P = 1/3.105848 P/F = .3219732 Table T17, p.718 P/F = .3220 Given: A crankshaft company is trying to decide whether to upgrade now for $150,000 or in 3years at 12% per annum Find: How much will it cost them in 3 years? F = P(F/P, 12%, 3) = 150,000(1.4049) F = $210,735 Lecture No. 2, Factors: How Time and Interest Affect Money Page No. 3 Given: The carmaker Renault will have to pay $75M in 3 years at 15% per annum Find: What is the $75M worth today. P = F(P/F, 15%, 3) = 75M(.6575) P = $49.31M See and review T2-1 p. 51 Examples 2.1 through 2.3 p.52 2. Uniform Series: Present Worth Factor, P/A, and Capital Recovery Factor, A/P P=? i = given 0 1 2 n-2 n-1 n A = given Note that A, the annuity, begins in year 1 i.e. not year 0. P, the present worth, is calculated in year 0. 1 1 1 P = A[ (1+i)1 ] + A[ (1+i)2 ] …. A[ (1+i)n ] 1 1 1 P = A{[ (1+i)1 ] + [ (1+i)2 ] …. [ (1+i)n ] } eq.2-4 Multiply each side by the P/F factor 1/1(+i) P 1 1 1 = A{[ (1+i)2 ] + [ (1+i)3 ] …. [ (1+i)n+1 ] eq.2-5 1+i Subtract the two expressions, eq. 2.5 minus eq. 2.4 1 1 1 1 1 1 A{[ (1+i)2 ] + [ (1+i)3 ] …. [ (1+i)n+1 - [ A{[ (1+i)2 ] + [ (1+i)3 ] …. [ (1+i)n+1 ] ] 1 1 P = A[ (1+i)n+1 - (1+i)1 ] Lecture No. 2, Factors: How Time and Interest Affect Money Page No. 4 -i Multiply both sides by 1/1+i A 1 P = -i [ (1+i)n - 1] Find LCD (1+i)n-1 P=A[ ] i0, P/A i(1+i)n The term in brackets is the is the uniform series present worth factor, USPWF, P/A. P/A is used to calculate the equivalent P value in year 0 for a uniform end of period series of A values beginning at the end of period 1 and extending for n years. If A is sought: i(1+i)n A=P[ ], A/P (1+i)n-1 The term in the brackets is the Capital Recovery Factor, CRF, A/P. The CRF calculates the equivalent uniform annual worth A over n years for a given P in year 0 when the interest rate is i. P is always one period prior to A. Summary of factors: T2-2, p.58 Given: (A/P, 12%, 10) Find: Translate the above, calculate A/P, calculate P/A A uniform series end of period payment P the present worth is that is known 12% is the interest rate 10 is the number of time periods i(1+i)n .12(1+.12)10 .12(3.1058) .3727 (A/P, 12%, 10) = n = = = (1+i) -1 (1+.12)10-1 3.1058-1 2.1058 (A/P, 12%, 10) = .17699 From T17 p.718 (A/P, 12%, 10) = .17698 P/A = 1/F/P = 1/.17699 P/A = 5.6500 From T17 p.718 (P/A, 12%, 10) = 5.6502 Examples: Given: An amusement park intends to pays $55,000 per year for 5 years at 15% per annum interest. Find: Using the same amount of money, how much is that today. P = A(P/A, 15%, 5) = 55,000(3.3522) P = $184,371 Lecture No. 2, Factors: How Time and Interest Affect Money Page No. 5 Given: You won the lottery, $1,000,000 payable over 20 years at 10% interest, $50,000 a year for 20 years. Find: How much are you going to get now? P = A(P/A, 10%, 20) = 50,000(8.5136) P= $425,680. Examples 2.4 p.59 3. Sinking Fund Factor, A/F, and Uniform Series Compound Amount Factor, F/A 1 i(1+i)n Substitute P = F[ (1+i)n ] into A = P [ (1+i)n-1 ] 1 i(1+i)n A = F [ (1+i)n ] [ (1+i)n-1 ] i A=F[ ], A/F (1+i)n-1 The expression in brackets is the sinking fund factor, A/F; it determines the uniform annual series that is equivalent to a given future worth. F = given i = given 0 1 2 n-2 n-1 n A=? (1+i)n-1 F=A[ ], F/A i The term in brackets is called the uniform series compound amount factor, USCAF, F/A. The future amount, F, occurs in the same period as the last A. Summary of factors: T2-3, p.61 Given: (A/F, 12%, 10) Find: Translate the above, calculate A/F, calculate F/A A uniform series end of period payment F the future worth is what is being sought 12% is the interest rate 10 is the number of time periods Lecture No. 2, Factors: How Time and Interest Affect Money Page No. 6 i .12 .12 .12 (A/F, 12%, 10) = A = [ ]= = = (1+i)n-1 (1+.12)10-1 3.10584-1 2.10584 (A/F, 12%, 10) = .056984 From T17 p.718 (A/F, 12%, 10) = .05698 F/A = 1/F/A = 1/.05698 F/A = 17.5487 From T17 p.718 (F/A, 12%, 10) = 17.5487 Given: A moving company wants to buy a new truck for $250,000 in 4 years and interest rates are 10% per year. Find: How much should they set aside each year? A = F(A/F, 10%, 4) = $250,000(.21547) A = $53,868 Examples 2.5 through 2.6 p.61 4. Arithmetic Gradient Factors, P/G, A/G P=? i = given Base Amount 0 1 2 3 n-1 n 1G 2G (n-2)G (n-1)G An arithmetic gradient is a cash flow series that either increases or decreases by a constant amount, the gradient. The cash flow changes by the same amount each period. Each year-end amount is different although the increase is constant. Note well that the first value commences in year 1, the first gradient, increase (or decreases), appears between year 1 and year 2 and is expressed in year 2. Thereafter, year 3, 4, …, n increases (or decreases) by the constant gradient amount. n is the number of periods and begins in year 0 which is 1 period to the left of the first value and two years to the left of the first increased value. The value used in the formulas is therefore n, which expresses itself 2 periods to the left of the first gradient or 1 period to the left of the first value. Lecture No. 2, Factors: How Time and Interest Affect Money Page No. 7 CFn = base amount + (n-1)G Given: A cash flow starts at year 1 at $7000 and increases by $250 each year through the year 5. Find: a. Determine the value of the gradient b. Determine the cash flow in year 5 c. Determine n for the gradient a. Gradient G = $250 b. Cash flow CFn = base amount + (n-1)G = 7000 + (5-1)(250) CF5 = $8000 c. n n=5 The present worth at year 0 of only the gradient is equal to the sum of the present worths of the individual values, where each value is a future amount. P = G(P/F, i, 2) + 2G(P/F, i, 3)…[(n-1)G](P/F, i, n) Eventually yields: (see p.67) G (1+i)n-1 n P= [ n - ] , P/G, P does NOT include the base amount i i(1+i) (1+i)n The above is the general relationship to convert an arithmetic gradient, G, NOT including the base amount, for n years into a present worth at year 0. The equivalent uniform annual series, A, for an arithmetic gradient, G, is found by multiplying the present worth by the A/P expression. G (1+i)n-1 n i(1+i)n A = i [ i(1+i)n - (1+i)n ] x [ (1+i)n-1 ] 1 n A=G[ - ] i (1+i)n-1 The expression in brackets is called the arithmetic gradient uniform series factor, A/G. The F/G factor, arithmetic gradient future worth factor is derived by multiply the P/G and F/P factors. 1 (1+i)n-1 F=G[ -( - n )], F/G, F does NOT include the base amount i i The base amount, PA, the present worth, must be added to the gradient. The base amount in the uniform series, A that begins in year 1 and extends through year n is represented by PA. The gradient, PG, is added to or subtracted from PA to yield the total, PT. PT = PA + PG and PT = PA – PG Lecture No. 2, Factors: How Time and Interest Affect Money Page No. 8 Given: (P/G, 12%, 10) Find: Translate the above, calculate P/G, calculate A/G, Verify the answers using the tables. P the present worth is that is known g the amount of increase or decrease in a series of payments 12% is the interest rate 10 is the number of time periods 1 (1+i)n-1 n 1 (1+.12)10-1 10 (P/G, 12%, 10) = [ n - n ]= [ - ]= i i(1+i) (1+i) .12 .12(1+.12)10 (1+.12)10 3.1058-1 10 8.333[ - ] = 8.333[ -3.2198] = 8.333(5.6012-3.2198) = 8.333(2.4303) .12(3.1058) 3.1058 (P/G, 12%, 10) = 20.25186 From T17 p.718 (P/G, 12%, 10) = 20.2541 1 n 1 10 10 (A/G, 12%, 10) = [ - ]=[ - ] = 8.333 - = 8.3333 – 4.7488 i (1+i)n-1 .12 (1+.12)10-1 3.1058-1 (A/G, 12%, 10) = 3.5848 From T17 p.718 (A/G, 12%, 10) = 3.5847 Given2.28: Exxon has a well that is producing $180,000 at the end of the year 1 a value which is decreasing for the next 6 years, at the end of the 6 th additional year (total 7=6+1) the value of the well will be 0. The going interest is 12% per annum. Find: What is the gradient and present worth. CFn = 0 =base amount + (n-1)G = 180,000 + (7-1)G G = 30,000 PT = PA – PG P = 180,000(P/A, 12%,7) – 30,000(P/G, 12%,7) = 180,000(4.5638) – 30,000(11.6443) = 821,484 – 349,329 P = $472,155 NB: $180,000 is the base amount and is independent of the gradient calculation. The first term, P/A, is predicated on the base amount and the second term is the gradient calculation, P/G, and n=7. Examples 2.10 through 2.6 p.69 5. Geometric Gradient Series Factor P=? g = given i = given 0 1 2 3 n-1 n A1 A1(1+g) A1(1+g)2 A1(1+g)n-1 Lecture No. 2, Factors: How Time and Interest Affect Money Page No. 9 As opposed to the previous analysis in which the gradient increased or decreased by a constant amount, a geometric progression is one in which the increase or decrease is a constant percentage i.e. a geometric increase as opposed to an arithmetic increase. A new term, g, is in order: g = constant rate of change in decimal form, by which amounts increase or decrease from one period to the next. The derivation begins on p.71 and results in the following: Pg = A1(P/A, g, i, n) Note well: Unlike the arithmetic gradient in which the gradient and base amount are calculated SEPARATELY, the geometric gradient formula, Pg, is the total amount and INCLUDES the base amount. Note well: There are no “tables” for geometric gradients. 1+g 1 – ( 1+i )n (P/A, g, i, n) = gi i-g n (P/A, g, i, n) = 1+i g=i It is possible to derive factors for the A and F values; however, it is easier to first determine Pg and then multiply Pg by the A/P factor to yield A or the F/P factor to yield F. There are no direct spreadsheet solutions. See example 2.11, p.73. Note: There are no “tables” for geometric series. Given: (P/A, g, i, n) = (P/A, 7, 12%, 10) g i Find: Translate the above, calculate P/A P the present worth is that is known A uniform series end of period payment g, 7, the amount of increase or decrease in a series of payments 12% is the interest rate 10 is the number of time periods 1+g n 1–( ) 1+i (P/A, g, i, n) = gi i-g 1+.07 10 1.07 10 1–( ) 1–( ) 1+.12 1.12 1 – (.6333) .3666 (P/A, 7%, 12%,10) = = = = .12-.07 .05 .05 .05 (P/A, 7%, 12%,10) = 7.3326 Given2.40: A geometric series has $2000 in year 1 and increases by 10% each year for 7 years at 15% interest per year. Find: P Pg = A1(P/A, g, i, n) Lecture No. 2, Factors: How Time and Interest Affect Money Page No. 10 1+.10 7 1.10 7 1–( ) 1–( ) 1+.15 1.15 1 – (.73259) Pg = A1(P/A, 10%, 15%,7) =$2000 = $2000 = $2000 = $2000 .15-.10 .05 .05 .26741 .05 Pg = $10,696 5. Unknown Interest Rate Single amount, uniform series or a uniform gradient can be found by direct solution; non-uniform series and more complex problems can be solved by trial and error or a numerical method. As opposed to solving directly for a rate, it often easier to solve for the value of the factor and then use the tables to fine the actual rate. Given: $1000 (P/F, i, n) Find: i need $10,000 in 14 years, what interest rate is required. A. Use direct calculation B. Use tables A. Direct Calculation P = F(P/F, i, n) = F(1/(1+i))n 1 1000 = 10,000 (1+i)14 (1+i)14=10000/1000 = 10 1+i = 1.1788 i= .1788 = 17.88% B. Tables P = F(P/F, i, n) 1000 = 10,000(P/F, i, n) (P/F, i, n) = 1000/10000 = .100 Tables starting p.702 Scan the P/F factors at n=14 for .100 16% = .1253 18% = .0985 Given: A soil analysis laboratory can lease space for its equipment and pay now or pay latter. It can pay $72,000 now for 3 years or can pay $30,000 at the end of each of three years. Find: What rate of return does the company make if it pays now? 72,000 = 30,000(P/A, i, 3) (P/A, i, 3) = 2.4000 i = 12.04% by interpolation from the tables, or use of the basic equation, or use of a spreadsheet function A.Tables Go to the tables starting on p. 702, start scanning in the n=3 row and the P/A, present worth, column looking for 2.4000. 12% yields 2.4018 which is too high and 14% yields 2.3216 which is too low, therefore, the correct answer is between 12 and 14% and obviously closer to 12%. Note that the function is decreasing with increasing interest rates. Interpolate: Lecture No. 2, Factors: How Time and Interest Affect Money Page No. 11 The target value is 2.4000 and you can work the 12 side or the 14 side but you must be consistent, consider the 12% side: Difference between target value and 12 %: 2.4018 – 2.4000 = .0018 Total difference between the two available values i.e. 12 and 14%: 2.4018 – 2.3216 = .0802 target value difference % difference = total difference total percentage .0018 % difference = .0802 (14-12) % difference = .04488 Since the 12% was chosen the % difference would be added to 12% Interpolated value = 12 + .0448 Interpolated value = 12. 04% Note well: The 14% value can also be used: Difference between target value and 1 %: 2.4000 – 2.3216 = .0784 Total difference between the two available values i.e. 12 and 14%: 2.4018 – 2.3216 = .0802 % difference = total percentage .0784 % difference = .0802 (14-12) % difference = 1.9551 Since the 14% was chosen, the % difference would be subtracted from to 14% Interpolated value = 14 – 1.9551 Interpolated value = 12. 04% Note well: The function is not linear; therefore the interpolated value is only valid to the first significant figure. A better answer may be 12.0% because the .04 reflects TWO significant figures. The exact value to any degree of accuracy by using the formula. Note well: The signage, + or - , may be determined by simply setting up the ratios properly, however, it is “safer” to figure out the sense of the addition or subtraction by inspection. B.Formula (1+i)n-1 P=A[ ] i0, P/A i(1+i)n (1+i)3-1 72,000 = 30,000 i(1+i)3 3 (1+i) -1 72000 = = 2.4000 i(1+i)3 30000 Trial and error: Assume various values of i, when the resulting value on the left equals 2.4000, the assumed i is the correct one. See examples 2.12, 2.13, p.75. The Excel function Rate(number_years, A,P,F) can also be used. 5. Unknown Number of Years Direct solutions and interpolation of the tables are possible as with i. Given: $1000, 3% interest (P/F, i, n) Find: How many years until it reaches $1500. A. Use direct calculation Lecture No. 2, Factors: How Time and Interest Affect Money Page No. 12 B. Use tables A. Direct Calculation P = F(P/F, i, n) = F(1/(1+i))n 1 1000 = 1500 (1+.03)n (1+.03)n = 1500/1000 = 1.5 n log (1.03) = log 1.5 n = .1761/.01283 n =13.72 periods B. Tables P = F(P/F, i, n) 1000 = 1500(P/F, i, n) (P/F, i, n) = 1000/1500 = .6667 Scan the P/F factors Table 8, p.709 at i=3% for .6667 13 = .6810 14 = .6611 Interpolate to the next significant figure i.e. 13.X% where X is the interpolated value. 13 .6810-.6667 .0143 = = 14 .6810-.6611 .0199 .0143 14-13 = 1% x = .72 .0199 n = 13 + .72 n = 13.72 See examples 2.14, p.78. The Excel function NPER(%, A,P,F) can also be used.