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6 Pages
Chapter 18
Course: BUSINESS 201, Spring 2011School: MIT
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Word Count: 1108
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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
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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
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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
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B
C
D
E
F
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H
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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
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5
6
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B
C
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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
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B
C
D
E
F
G
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K
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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
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105
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107
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109
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111
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132
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135
136
137
138
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141
142
143
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147
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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
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7
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163
164
165
166
167
168
169
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171
172
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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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cha06369_tn07.qxd2/12/037:01 PMPage 299TECHNICAL NOTE SEVENtechnical notetechnical note sevenP R O C E S S C A PA B I L I T Y A N DS TAT I S T I C A L Q U A L I T YCONTROLAssignable variation definedCommon variation definedVariation around Us
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Universidad VirtualAdministracin de la Cadena de Valor2.4 Realizar reflexin tema SustentabilidadEQUIPO 37Olga Nora Garca Esquivel A00372979Tatiana Vanessa Mndez Martnez A01304717Amilcar Estrada Rodrguez A01304897Octavio Navarro Daz A01307499Alejan
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LOOKING FOR.SOLUTIONSMANUAL ORTEST BANK ?OVER 10,000 TITLESAVALABLE!Student.Plus@Hotmail.Comwww.student-plus.netThe List Below Updated On 24, JULY, 2010How To Order? Email UsEmail us at STUDENT.PLUS@HOTMAIL.COM asking for the teaching or studyi
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Universidad VirtualAdministracin de la Cadena de Valor2.5 Resolver problemas del captulo 6EQUIPO 37Olga Nora Garca Esquivel A00372979Tatiana Vanessa Mndez Martnez A01304717Amilcar Estrada Rodrguez A01304897Octavio Navarro Daz A01307499Alejandro Be
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Universidad VirtualAdministracin de la Cadena de Valor2.4 Realizar reflexin tema SustentabilidadEQUIPO 37Olga Nora Garca Esquivel A00372979Tatiana Vanessa Mndez Martnez A01304717Amilcar Estrada Rodrguez A01304897Octavio Navarro Daz A01307499Alejan
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Universidad VirtualAdministracin de la Cadena de Valor2.4 Realizar reflexin tema SustentabilidadEQUIPO 37Olga Nora Garca Esquivel A00372979Tatiana Vanessa Mndez Martnez A01304717Amilcar Estrada Rodrguez A01304897Octavio Navarro Daz A01307499Alejan
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Colorado Tech - ADMON - mpm
TATIANA V. MENDEZA01304717ADMINISTRACION DE LA CADENA DE VALORPROBLEMA 2Rockness Recycling reacondiciona a estudiantes de administracin agotados. El procesoutiliza una banda transportadora que lleva a cada estudiante por los cinco pasos del procesoe
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Universidad VirtualAdministracin de la Cadena de Valor2.5 Resolver problemas del captulo 6Profesor TitularDr. Alberto Rodrguez RodrguezCampusSedeEnero 19, 2011.Sante FeCoatzacoalcosProblema 2.- Rockness Recycling reacondiciona a estudiantes de a
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Universidad VirtualAdministracin de la Cadena de Valor2.5 Resolver problemas del captulo 6Profesor TitularDr. Alberto Rodrguez RodrguezCampusSedeEnero 19, 2011.Sante FeCoatzacoalcosProblema 2.- Rockness Recycling reacondiciona a estudiantes de a
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Universidad VirtualAdministracin de la Cadena de Valor2.4 Realizar reflexin tema SustentabilidadEQUIPO 37Olga Nora Garca Esquivel A00372979Tatiana Vanessa Mndez Martnez A01304717Amilcar Estrada Rodrguez A01304897Octavio Navarro Daz A01307499Alejan
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Universidad VirtualAdministracin de la Cadena de Valor2.5 Resolver problemas del captulo 6EQUIPO 37Olga Nora Garca Esquivel A00372979Tatiana Vanessa Mndez Martnez A01304717Amilcar Estrada Rodrguez A01304897Octavio Navarro Daz A01307499Alejandro Be
UCSD - CSE - 21
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
UCSD - CSE - 21
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
UCSD - CSE - 21
Name:Student ID:CSE 21Practice Final ExamDecember 2, 20091. 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
UCSD - CSE - 21
Solutions to the Practice Final ExamDecember 2, 20091. Solution:(a) a(0) = 1, a(1) = 2, a(2) = 7, a(3) = 26, a(4) = 97, and a(5) = 362.(b) Notice that the characteristic polynomial of the recurrence relation is x2 4x + 1, which has roots 2 3,nnyiel
UCSD - CSE - 21
CSE 21: Homework Solutions 2October 5, 2009Problem 1Teams A and B play in baseballs world series. Here, the team that rst wins four games wins the series.(a) What is the number of ways the series can occur?(b) What is the number of ways the series ca
UCSD - CSE - 21
CSE 21: Homework Solutions 3October 12, 2009Let S (n, k ) be the Stirling number of the second kind which counts thenumber of ways to partition n labeled elements into k disjoint, nonempty,unlabeled blocks.Problem 1In how many ways can 6 people be a
UCSD - CSE - 21
CSE21 FA11Homework #1 (9/26/11)1.1. An n-digit number is a list of n digits where the rst digit in the listis not zero (where we assume n 1).(a)How many n-digit numbers are there (as a function of n)?(b)How many n-digit numbers contain no 1s?(c)H
UCSD - CSE - 21
CSE21 FA11Homework #1 (9/26/11)1.1. An n-digit number is a list of n digits where the rst digit in the listis not zero (where we assume n 1).(a)How many n-digit numbers are there (as a function of n)?(b)How many n-digit numbers contain no 1s?(c)H
UCSD - CSE - 21
CSE21 FA11Homework #2 (10/3/11)In this homework, we will consider ordinary decks of playing cards whichhave 52 cards, with 13 of each of the four suits (Hearts, Spades, Diamondsand Clubs) with each suit have the 13 ranks (Ace, 2, 3, . . . , Jack, Quee
UCSD - CSE - 21
CSE21 FA11Homework #2 (10/3/11)In this homework, we will consider ordinary decks of playing cards whichhave 52 cards, with 13 of each of the four suits (Hearts, Spades, Diamondsand Clubs) with each suit have the 13 ranks (Ace, 2, 3, . . . , Jack, Quee
UCSD - CSE - 21
CSE21 FA11Homework #3 (10/10/11)3.1.Prove by induction thatnk=1k2 =n(n+1)(2n+1).63.2. It is desired to form three committees from a group of 3 men and 3women.(a) In how many ways can this be done?(b) Suppose that each of the committees must c
UCSD - CSE - 21
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
UCSD - CSE - 21
CSE21 FA11Homework #4 (10/17/11)4.1.Prove by induction thatnk=1k3 = (nk=1k )2 .(Hint: First get an explicit form form the second sum. Then use induction.)4.2. (Note: This problem is fairly tough! Ill be impressed by anyone whocan get this one!
UCSD - CSE - 21
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
UCSD - CSE - 21
CSE21 FA11Homework #5 (10/24/11)5.1. (a) 10 cards are drawn at random one at a time with replacement froman ordinary deck of cards.What is the sample space? What is the probabilitythat no Ace appears on any of the draws? What is the probability that a
UCSD - CSE - 21
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
UCSD - CSE - 21
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
UCSD - CSE - 21
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=
UCSD - CSE - 21
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
UCSD - CSE - 21
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
UCSD - CSE - 21
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
UCSD - CSE - 21
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
UCSD - CSE - 21
CSE21 FA11Homework #9 (11/22/11)9.1Does every connected graph has a spanning tree? Give either a proof or acounterexample.9.2(a) Select a spanning tree of minimum total weight from the following weightedgraph.(b) Find all spanning trees (list thei
UCSD - CSE - 21
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