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Unformatted text preview: Chapter 7: Rate of Return Analysis 71 $125 = $10 (P/A, i%, 6) + $10 (P/G, i%, 6) at 12%, $10 (4.111) + $10 (8.930) = $130.4 at 15%, $10 (3.784) + $10 (7.937) = $117.2 i* = 12% + (3%) ((130.4 – 125).(130.4117.2)) = 13.23% 72 The easiest solution is to solve one cycle of the repeating diagram: $120 = $80 (F/P, i%, 1) $120 = $80 (1 + i) (1 + i) = $120/$80 = 1.50 $80 $80 $80 $80 $80 $200 $200 = $80 $80 $80 $20 $12 $12 $10 $20 $30 $40 $50 $60 i* = 0.50 = 50% Alternative Solution: EUAB = EUAC $80 = [$200 (P/F, i%, 2) + $200 (P/F, i%, 4) + $200 (P/F, i%, 6)] (A/P, i%, 6) Try i = 50% $80 = [$200 (0.4444) + $200 (0.1975) + $200 (0.0878)] (0.5481) = $79.99 Therefore i* = 50%. 73 $42.55 = $5 (P/A, i%, 5) + $5 (P/G, i%, 5) Try i = 15% , $5 (3.352) + $5 (5.775) = $45.64 > $42.55 Try i = 20%, $5 (2.991) + $5 (4.906) = $39.49 < $42.55 Rate of Return = 15% + (5%) [($45.64  $42.55)/($45.64  $39.49)] = 17.51% Exact Answer: 17.38% 74 For infinite series: A = Pi EUAC= EUAB $3,810 (i) = $250 + $250 (F/P, i%, 1) (A/F, i%, 2)* Try i = 10% $250 + $250 (1.10) (0.4762) = $381 $3,810 (0.10) = $381 i = 10% *Alternate Equations: $3,810 (i) = $250 + $250 (P/F, i%, 1) (A/P, i%, 2) $3,810 (j) = $500  $250 (A/G, i%, 2) $5 $10 $15 $20 $25 $42.55 75 At Year 0, PW of Cost = PW of Benefits $412 + $5,000 (P/F, i%, 10) = ($1000/i) (P/F, i%, 10) Try i = 15% $412 + $5,000 (0.2472) = ($1,000/0.15) (0.2472) $1,648 = $1,648 ROR = 15% 76 The algebraic sum of the cash flows equals zero. Therefore, the rate of return is 0%. 77 Try i = 5% $1,000 =(?) $300 (3.546) (0.9524) =(?) $1,013.16 Try i = 6% $1,000 =(?) $300 (3.465) (0.9434) =(?) $980.66 Performing Linear Interpolation: i* = 5% + (1%) (($1,013.6  $1,000)/($1,013.6  $980.66)) = 5.4% …………. n = 10 $41 Yr 0 $5,000 P’ A = $1,000 n = ∞ $1,000 A = $300 78 $400 = [$200 (P/A, i%, 4)  $50 (P/G, i%, 4)] (P/F, i%, 1) Try i = 7% [$200 (3.387)  $50 (4.795)] (0.9346) = 409.03 Try i = 8% [$200 (3.312)  $50 (4.650)] (0.9259) = $398.08 i* = 7% + (1%) [($409.03  $400)/($409.03  $398.04)] = 7.82% 79 $100 = $27 (P/A, i%, 10) (P/A, i%, 10) = 3.704 Performing Linear Interpolation: ( P/A , i %, 10) i 4.192 20% 3.571 25% Rate of Return = 20% + (5%) [(4.192 – 3.704)/(4.912 – 3.571)] = 23.9% 710 Year Cash Flow$500 1$100 2 +$300 3 +$300 4 +$400 5 +$500 $500 + $100 (P/F, i%, 1)= $300 (P/A, i%, 2) (P/F, i%, 1) + $400 (P/F, i%, 4) + $500 (P/F, i%, 5) Try i = 30% $500 + $100 (0.7692) = $576.92 $300 (1.361) (0.7692) + $400 (0.6501) + $500 (0.2693)= $588.75 ∆ = 11.83 Try i = 35% $500 + $100 (0.7407) = $574.07 $300 (1.289) (0.7407) + $400 (0.3011) + $500 (0.2230)= $518.37 ∆ = 55.70 Rate of Return = 30% + (5%) [11.83/55.70) = 31.06% Exact Answer: 30.81% 711 Year Cash Flow$223 1$223 2$223 3$223 4$223 5$223 6 +$1,000 7 +$1,000 8 +$1,000 9 +$1,000 10 +$1,000 The rate of return may be computed by any conventional means. On closer inspection one observes that each $223 increases to $1,000 in five years....
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This note was uploaded on 08/13/2010 for the course IME IME 314 taught by Professor Freeman during the Spring '10 term at Cal Poly.
 Spring '10
 Freeman

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