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### Chapter 18

School: MIT
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OPTION A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 B C D E F G H I J THE TO EXPAND Year CF of single machine 0 -1000 Discount rate for machine cash flows Riskless discount rate NPV of single machine 12% 6% -67.48 Number of machines bought next year Option value of single machine purchased in one more year NPV of total project...

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OPTION A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 B C D E F G H I J THE TO EXPAND Year CF of single machine 0 -1000 Discount rate for machine cash flows Riskless discount rate NPV of single machine 12% 6% -67.48 Number of machines bought next year Option value of single machine purchased in one more year NPV of total project 5 0.80 <-- =B24 -63.48 <-- =B8+B10*B11 Black-Scholes Option Pricing Formula S X r T Sigma d1 d2 N(d1) N(d2) Option value = BS call price 1 220 2 300 3 400 4 200 5 150 932.52 PV of machine CFs 1000.00 Exercise price = Machine cost 6.00% Risk-free rate of interest 1 Time to maturity of option (in years) 40% <-- Volatility -0.9818 <-- (LN(S/X)+(r+0.5*sigma^2)*T)/(sigma*SQRT(T)) -0.9918 <-- d1 - sigma*SQRT(T) 0.1631 <--- Uses formula NormSDist(d1) 0.1606 <--- Uses formula NormSDist(d2) 0.80 <-- S*N(d1)-X*exp(-r*T)*N(d2) Data Table -63.48 <-- Table header: =B12 1% -63.48 10% -63.48 700 Project Value as Funct ion of Sigma 20% -63.48 600 30% -63.48 500 40% -63.48 50% -63.48 400 60% -63.48 300 70% -63.48 2 00 100 0 0% 10% 2 0% 30% 40% Sigma 50% 60% 70% 80% A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 B C D E F G H I J K L 0.0953 <-- =E4^2 0.1959 <-- =E4*E5 0.1959 <-- =E4*E5 0.4028 <-- =E5^2 PRICING AN ABANDONMENT OPTION Market data Expected market return Sigma of market return Risk-free rate State prices qu qd 12% 30% 6% 0.3087 0.6347 ### ### One-period "up" and "down" of market Up 1.521962 <-- =EXP(B4+B5), note that a valid alternative is "up" = EXP(B5) Down 0.83527 <-- =EXP(B4-B5), note that a valid alternative is "down" = EXP(-B5) Project cash flows State-dependent present value factors 150 100 0.3087 80 -50 1 80 -50 0.6347 -60 State-by-state present value =C16*I16 14.2981 <-- =E15*K15 30.8740 15.6755 <-- =E17*K17 -50 15.6755 <-- =E19*K19 -31.7328 -24.1673 <-- =E21*K21 Net present value -29.38 <-- =SUM(A24:E30) Cash flows with abandonment Present value with abandonment 150 14.29808 100 30.87402 80 -50 15.67551 -50 0 0 0 0 0 0 Present value with abandonment 10.85 M A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 B C D E F G H I J K L 0.0953 <-- =E4^2 0.1959 <-- =E4*E5 0.1959 <-- =E4*E5 0.4028 <-- =E5^2 PRICING AN ABANDONMENT OPTION Market data Expected market return Sigma of market return Risk-free rate State prices qu qd 12% 30% 6% 0.3087 0.6347 ### ### One-period "up" and "down" of market Up 1.521962 <-- =EXP(B4+B5), note that a valid alternative is "up" = EXP(B5) Down 0.83527 <-- =EXP(B4-B5), note that a valid alternative is "down" = EXP(-B5) Project cash flows State-dependent present value factors 150 100 0.3087 80 -50 1 80 -50 0.6347 -60 State-by-state present value =C16*I16 14.2981 <-- =E15*K15 30.8740 15.6755 <-- =E17*K17 -50 15.6755 <-- =E19*K19 -31.7328 -24.1673 <-- =E21*K21 Net present value -29.38 <-- =SUM(A24:E30) Cash flows with abandonment Present value with abandonment 150 14.29808 100 30.87402 80 -50 15.67551 -50 0 -10 0 -6.34656 0 0 Present value with abandonment 4.50 M A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 B C D E F G H I ALTERNATIVE METHODS OF CALCULATING STATE PRICES Method 1: Uses the risk-free rate, expected market return and sigma of market return Market data Expected market return Sigma of market return Risk-free rate state prices qu qd 12% 30% 6% 0.3087 <-- =(1+B8-B12)/((1+B8)*(B11-B12)) 0.6347 <-- =(B11-1-B8)/((1+B8)*(B11-B12)) One-period "up" and "down" of market up 1.521962 <-- =EXP(B6+B7) down 0.83527 <-- =EXP(B6-B7) Method 3: Uses the risk-free rate and sigma of market return, but not expected market return Market data Sigma of market return Risk-free rate state prices qu qd 30% 6% 0.4944 <-- =(1+B20-B25)/((1+B20)*(B24-B25)) 0.4490 <-- =(B24-1-B20)/((1+B20)*(B24-B25)) One-period "up" and "down" of market up 1.349859 <-- =EXP(B19) down 0.740818 <-- =EXP(-B19) Method 2: Matching state prices to the cost of capital Project cost of capital Risk-free rate 22% <-- This is the discount rate for the project if it has no options 6% Project cash flows 150 100 80 -50 80 -50 -60 =AVERAGE(C35:C39) Project expected cash flows: Assumes equal state probabilities Year 0 1 2 Expected CF -50 25 62.5 Project NPV 12.48 <-- =NPV(B30,C45:D45)+A37 State prices qu qd =AVERAGE(E34:E40) 0.4241 <-- =1/(1+B31)-B51 0.5193 <-- Determined by Solver =C35*B50 Project discounting 26.9797 state-by-state <-- =E34*B50^2 42.4105 17.6187 <-- =E36*B50*B51 -50 17.6187 <-- =E38*B50*B51 -25.9646 -16.1798 <-- =E40*B51^2 =C39*B51 State-by-state NPV 12.48 <-- =SUM(D54:H60) Target cell (0.00) <-- =B66-B46 A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 B C D E F G H I J K L M N O P Q R S ABANDONMENT VALUE--DETAILS OF YEAR 1 CALCULATION Project cash flows Year Cash flow Risk-adjusted discount rate Valuing the year-1 abandonment put Value of project, end year 1 Abandonment value Time to option maturity (years) Risk-free rate Sigma Put value 0 -750 1 100 2 200 3 300 4 400 12% The project's cost of capital 702.44 Like asset value in put formula 300 Like strike price in put formula 3 6% 50% ### <-- =putoption(B10,B11,B12,B13,B14) Black-Scholes Put Option Pricing Formula S -100.00 PV of CFs, discounted at risk-adjusted discount rate X -100.00 Exercise price = Abandonment cost r 6.00% Risk-free rate of interest T 5 Time to maturity of option (in years) Sigma 80% <-- Volatility d1 1.0621 <-- (LN(S/X)+(r+0.5*sigma^2)*T)/(sigma*SQRT(T)) d2 -0.7267 <-- d1 - sigma*SQRT(T) N(d1) 0.8559 <--- Uses formula NormSDist(d1) N(d2) 0.2337 <--- Uses formula NormSDist(d2) 56 Option value = BS call price -68.28 <-- S*N(d1)-X*exp(-r*T)*N(d2) 57 State prices 58 qu 59 Up 122.6% 0.240829 qd 60 Down -55.1% 0.702567 61 Mean 200 300 400 62 63 64 Cash flows 4409.27 65 1485.91 66 445.11 890.22 67 68 -750 300.00 89.87 179.73 69 60.57 70 36.29 71 72 73 Year 1 2 3 74 75 143.1 <-- =SUMPRODUCT(A30:I37,K30:S37) 76 Project present value 77 78 Cash flows with abandonment 100 79 Terminal value on abandonment 80 81 82 4409.27 1485.91 83 445.11 890.22 84 80.00% 300.00 85 86 89.87 179.73 60.57 87 136.29 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 500 12266.27 State dependent present-value factors 0.00 0.01 2476.52 0.06 0.24 500.00 1 0.34 0.70 100.95 0.04 0.12 0.17 0.36 0.49 0.33 0.35 20.38 4 12266.27 2476.52 500.00 100.95 100.00 0.24 A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 B C D E F G H I J K L M N O P Q R S PRICING AN ABANDONMENT OPTION AS A SERIES OF PUTS Project cash flows Year Cash flow 0 -750 Risk-adjusted discount rate 12% The project's cost of capital NPV without options 1 100 2 200 3 300 4 400 -33.53 <-- =NPV(B7,C5:F5)+B5 Sigma Risk-free rate Abandonment value 50% 6% 300 Project can be abandoned at end of any year for this amount NPV of cash flows at RADR Value of abandonment option Adjusted present value -33.53 <-- =B9 ### <-- =NPV(B12,C20:F20) ### End-year value of remaining cash flows Put option value 702.44 ### 586.73 ### =NPV(\$B\$7,E5:\$F\$5) 357.14 ### 0.00 ### =putoption(D19,\$B\$13,\$F\$4-D4+1,\$B\$12,\$B\$11) the parameter are: Asset value -> D19 Exercise price -> B13 Time to exercise -> \$F\$4-D4 Risk-free rate -> B12 Volatility (sigma) -> B11 Black-Scholes Put Option Pricing Formula S -100.00 PV of CFs, discounted at risk-adjusted discount rate X -100.00 Exercise price = Abandonment cost r 6.00% Risk-free rate of interest T 5 Time to maturity of option (in years) Sigma 80% <-- Volatility d1 1.0621 <-- (LN(S/X)+(r+0.5*sigma^2)*T)/(sigma*SQRT(T)) d2 -0.7267 <-- d1 - sigma*SQRT(T) N(d1) 0.8559 <--- Uses formula NormSDist(d1) N(d2) 0.2337 <--- Uses formula NormSDist(d2) Option value = BS call price -68.28 <-- S*N(d1)-X*exp(-r*T)*N(d2) Up Down State prices qu 0.240829 qd 0.7025672 122.6% -55.1% Mean 200 300 400 Cash flows 500 12266.27 State dependent present-value factors 0.00 4409.27 1485.91 445.11 -750 0.01 2476.52 300.00 89.87 0.24 500.00 179.73 60.57 100.95 Project present value 2 3 4 143.1 <-- =SUMPRODUCT(A30:I37,K30:S37) Cash flows with abandonment Terminal value on abandonment 100 12266.27 4409.27 1485.91 445.11 80.00% 2476.52 890.22 300.00 89.87 500.00 179.73 60.57 0.04 0.34 0.17 0.36 0.49 0.33 0.35 20.38 1 1 0.12 0.70 36.29 Year 0.06 890.22 100.95 136.29 100.00 0.24
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CSE 21: Homework 3October 12, 2009Problem 1In how many ways can 6 people be assigned to 4 nonempty teams?Problem 2An urn contains 5 red marbles and 6 white marbles.(a) How many ways can 4 marbles be drawn?(b) What if we must have 2 red marbles and
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Name:Student ID:CSE 21Midterm #2 SolutionsNovember 17, 2009Each problem is worth 20 points. Show your work, especially on decision trees. Also, makesure you write legibly so that I have a chance of being able to read your solutions! Additionalscrat
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CSE21 FA11Homework #3 Solutions (10/15/11)3.1. Prove by induction thatnk2 =k=1n(n + 1)(2n + 1)6Sketch of proof. Step 1. We need to verify the base case for n = 1, i.e.1k=1=k=11 (1 + 1) (2 + 1).6Step 2. Proof by induction. We assume that n=1
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CSE21 FA11Homework #4 (10/17/11)4.1.Sketch of proof. Recall that we showed in class that n=1 k = n(n2+1) for allk22n 1. We will prove that by induction n=1 k 3 = n (n4+1) . We assume thatknn2 (n+1)23for some xed n 1. Note thatk=1 k =4n+1n
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CSE21 FA11Homework #5 (10/17/11)5.1Solution. (a) The probability that no Ace appears on any of the drawsis ( 48 )10 . The probability that at least one King appears in 10 draws is521 ( 48 )10 . The probability that at least 2 Queens appear in the 10
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CSE21 FA11Homework #6 (10/31/11)6.1. Five boys and three girls are throwing Frisbees. Each boy has oneFrisbee and throws it to a random girl. What is the probability that each ofthe girls gets at least one of thrown Frisbees?6.2. A box contains 20 li
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CSE21 FA11Homework #6 (10/31/11)6.1. Five boys and three girls are throwing Frisbees. Each boy has oneFrisbee and throws it to a random girl. What is the probability that each ofthe girls gets at least one of thrown Frisbees?Answer: S (5, 3) 3!35=
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CSE21 FA11Homework #7 (11/7/11)7.1. A bin contains 4 red balls, 5 white balls and 6 blue balls. A randomsubset S of 4 balls is removed (without replacement). Consider the following3 events:(1) E1 : S has exactly 2 red balls.;(2) E2 : S has balls of
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CSE21 FA11Homework #7 Solutions7.1. A bin contains 4 red balls, 5 white balls and 6 blue balls. A randomsubset S of 4 balls is removed (without replacement). Consider the following3 events:(1) E1 : S has exactly 2 red balls.;(2) E2 : S has balls of
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CSE21 FA11Homework #8 (11/14/11)8.1. An urn contains r red marbles and b blue marbles. A random marbleM1 is drawn out. If M1 is red, then it is put back into the urn along withc more blue marbles. On the other hand, if M1 is blue, then it is put back
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CSE21 FA11Homework #8 (11/17/11)8.1Solution. Draw the decision tree. Then compute:(a)P r(M2 = red) =r+drbr+.r+b+cr+b r+b+dr+b(b)P r(M2 = red|M1 = red) =P r(M1 = blue|M2 = blue) =r.r+b+cbbr+b r+d+brb+cbb+ r+b r+d+br+d r+b+c8.2S
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CSE 21Midterm #1October 20, 20111. (a) How many rearrangements using all the letters of the wordSCHWARZENEGGER are there?Answer:14!3!2!2!= 36324288001.(b) How many rearrangements using only 13 of the letters of the wordSCHWARZENEGGER are there?
UCSD - CSE - 21
UCSD - CSE - 21
CSE 21 FA 11Practice Final Exam1. Dene the recurrence a(n + 2) = 4a(n + 1) a(n), , n 0, with a(0) = 1, a(1) = 2.(a) What is the value of a(5)?(b) Find an explicit closed form solution for a(n);(c) Prove by induction that the expression in (b) is vali
FAU - PHI - 2010
NelsonRobert NelsonProfessor Embree04/29/11PHI2010605ArtThere is a philosophy to art. There is a broad spectrum of what would be considered art anddifferent forms of art, even some that disprove another as art. Some questions that would beasked i
FAU - PHI - 2010
NelsonRobert NelsonProfessor EmbreePHY20103/18/2011AgnosticismAgnosticism is set between atheism and religion. It is the view that shows uncertainty betweenbelieving that there can be a god or gods, without leaning on any particular side. It is cer
FAU - PHI - 2010
NelsonRobert NelsonProfessor EmbreePHI201004/1/11EthicsEthics is made up of the principles that we live by. It is something we can argue that is eitherinnate or learned, which is similar to the argument on knowledge and reason. It can even beconsi
FAU - PHI - 2010
NelsonRobert NelsonProfessor EmbreePHY20102/18/11KnowledgeKnowledge can be factual or theoretical. It is the source for our understanding ofanything. It is what we are certain to be true or what we have solid reason to believe in withoutdirect pro