Solution of Derivative

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Unformatted text preview: invest. If gt < 5%, you should have invested. Therefore we need to solve gT = 5% for the optimal time T to invest: .02 1− 1.5 X(1.02)T = .05 =⇒ T = 24.727 years. (4) The NPV (today’s NPV) is X (1.02)24.727 − 1.5 = .29926. 1.0524.727 (5) The harder way to do this problem is to maximize the log of the NPV directly. X (1.02)T − 1.5 . ln (NP V ) = ln 1.05T (6) which is a calculus exercise (set the derivative equal to zero). The answer (NP V = .29931 and T = 25.224) will be slightly off due to using simple interest rates. Question 17.4. See Table One. 118 Chapter 17 Real Options Table One (Problem 17.4) Widget Year (t) NPV of Price NPV at t 0 Policy 27.00 27.00 1 0.55 29.20 27.81 2 0.57 31.49 28.56 3 0.59 33.87 29.26 4 0.62 36.34 29.90 5 0.64 38.92 30.49 6 0.67 41.59 31.04 7 0.70 44.38 31.54 8 0.72 47.27 32.00 9 0.75 50.28 32.41 10 0.78 53.41 32.79 11 0.81 56.67 33.13 12 0.85 60.06 33.44 13 0.88 63.58 33.72 14 0.92 67.24 33.96 15 0.95 71.05 34.18 16 0.99 75.01 34.36 17 1.03 79.13 34.53 18 1.07 83.42 34.66 19 1.11 87.88 34.78 20 1.16 92.51 34.87 21 1.21 97.33 34.94 22 1.25 102.35 34.99 23 1.30 107.56 35.02 24 1.36 112.98 35.03 25 1.41 118.62 35.03 26 1.47 124.49 35.01 27 1.52 130.59 34.98 28 1.59 136.93 34.93 29 1.65 143.53 34.87 30 1.72 150.39 34.80 31 1.78 157.52 34.71 32 1.86 164.94 34.62 33 1.93 172.66 34.51 34 2.01 180.69 34.39 35 2.09 189.03 34.27 36 2.17 197.72 34.14 37 2.26 206.74 34.00 38 2.35 216.13 33.85 39 2.44 225.90 33.69 40 2.54 236.06 33.53 41 2.64 246.62 33.36 42 2.75 257.60 33.19 43 2.86 269.03 33.01 44 2.97 280.91 32.83 45 3.09 293.26 32.64 46 3.21 306.12 32.45 47 3.34 319.48 32.25 48 3.47 333.38 32.05 49 3.61 347.83 31.85 50 3.76 362.87 31.64 51 3.91 119 Part 4 Financial Engineering and Applications Question 17.6. The strike price (net excavation cost) will be 50m − 20m = 30m. a) No change, 100m. b) 100 − 30e−.5 = 81.804. c) C = 79.66 and H ∗ = 300. a) Each period, the expected price of a widget is .25/2 + 2.25/2 = \$1.25. b) The expected cash ﬂow each period is .25 − 1 2.25 − 1 + = .25 2 2 (7) which will have an NPV of .25/.05 − 10 = −5. It will never be optimal to produce. c) If we can only produce when the widget price is 2.25 we can have expected cash ﬂows of 1.25/2 = .625 (i.e. the second term in the above equation). This will give use an NP V = .625/.05 − 10 = 2.5. This NPV is the same hence we do not have to consider delaying the project. d) If the widget price is either \$.10 or \$2.40 with equal probability, then the expected widget price remains \$1.25, but expected cash ﬂow is .50 (0) + .50(\$2.50 − \$1) = \$.75 which increases our NPV to 5. Note that if the variance of the widget price is greater, expected cash ﬂow is greater. In effect, producing only when it is proﬁtable amounts to having a call option each period, with the strike price being the marginal cost of production. Increases in volatility raise the value of this call. Question 17.8. The discount rate is r + β (rm − r) = .05 + .5 (.08) = .09. The current value of the future cash ﬂows will be 8/.09 = 88.889. The static NPV is negative at 88.89 − 100 = −11.11. Using the binomial model, we need a dividend yield. As in the widget problem, the dividend yield will be the cash ﬂow divided by the PV of the cash ﬂows which is just the discount rate .09. For our options analysis we must use r = ln (1.05) and δ = ln (1.04). Using a 3-step forward tree, see Table Two, we see we would invest in the project in two years in the up-up node; this gives a value of 12.45. With over 100 steps the project has a value around 12.34. With perpetual investment rights we could use the perpetual call real option CallPerpetual (8/.09, 100, .35, ln (1.05) , ln (1.09)) = {19.64, 199.28} . (8) The project is worth 19.64; i.e. the right to invest after three years add approx. 50% to the project value. 120 Chapter 17 Real Options Table Two (Problem 17.10b) Time (yrs) 1 2 3 227.0623 166.1037 121.510371 PV(CF) 88.88889 V of Proj. 12.45509 28.8305216 60.3402646 1.97712275 66.10367 82.48464 5.021933 40.96066 0 American Call Strike = 100 Vol = 35.00%; r = 4.88% Exp = 3 years; Div = 8.62% u = 1.367; d = 0.679 Risk-neutral prob of up = 0.413 Forward tree 127.0623 112.7558 12.75582 55.99288 0 27.80524 0 Question 17.10. The residual value of the land effectively lowers the extraction costs (i.e. the strike price) by R = 1. We must now solve, as in the examples, St = ln (1.05) 12.60 = 15.674 ln (1.04) (9) It will take t years where 15 (1.009615)t = 15.674 for a solution of t = 4.5932. The value of the land is the NPV which will be 15.674 − 12.60 = 2.4568. 1.054.5932 (10) = 2.4568. This higher than extracting now which would give us 2.40. Note the NPV (as of today) for extracting in T years is 1 1.05T 15 12.60 1.05T − 13.60 + 1 = − 15 T T 1.04 1.04 1.05T and we could maximize this directly. 121 (11) Part 4 Financial Engineering and Applications Question 17.12. If shutdown and start-up were costless, we would...
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## This document was uploaded on 03/11/2014 for the course FIN 402 at FPT University.

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