Exam 21: Option Valuation

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Using the binomial pricing model,calculate the price of a two-year put option on Kinston stock with a strike price of $9.

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This problem requires a two period binomial tree.The solution will start by solving the value of the call option for the up and down branches as of year 1 and then solve for the final value of the option at year 0. This problem requires a two period binomial tree.The solution will start by solving the value of the call option for the up and down branches as of year 1 and then solve for the final value of the option at year 0.    Up branch D =    =    = 0 B =    =    = 0 C = SD + B = $13(0)+ (0)= $0 Down Branch D =    =    = -0.6 B =    =    = 6.285714 C = SD + B = $8(-0.6)+ (6.285714)= $1.49 Value at year 0 D =    =    = -0.298 B =    =    = 3.874 C = SD + B = $10(-0.298)+ 3.874 = $0.89
Up branch
D = This problem requires a two period binomial tree.The solution will start by solving the value of the call option for the up and down branches as of year 1 and then solve for the final value of the option at year 0.    Up branch D =    =    = 0 B =    =    = 0 C = SD + B = $13(0)+ (0)= $0 Down Branch D =    =    = -0.6 B =    =    = 6.285714 C = SD + B = $8(-0.6)+ (6.285714)= $1.49 Value at year 0 D =    =    = -0.298 B =    =    = 3.874 C = SD + B = $10(-0.298)+ 3.874 = $0.89
= This problem requires a two period binomial tree.The solution will start by solving the value of the call option for the up and down branches as of year 1 and then solve for the final value of the option at year 0.    Up branch D =    =    = 0 B =    =    = 0 C = SD + B = $13(0)+ (0)= $0 Down Branch D =    =    = -0.6 B =    =    = 6.285714 C = SD + B = $8(-0.6)+ (6.285714)= $1.49 Value at year 0 D =    =    = -0.298 B =    =    = 3.874 C = SD + B = $10(-0.298)+ 3.874 = $0.89
= 0
B = This problem requires a two period binomial tree.The solution will start by solving the value of the call option for the up and down branches as of year 1 and then solve for the final value of the option at year 0.    Up branch D =    =    = 0 B =    =    = 0 C = SD + B = $13(0)+ (0)= $0 Down Branch D =    =    = -0.6 B =    =    = 6.285714 C = SD + B = $8(-0.6)+ (6.285714)= $1.49 Value at year 0 D =    =    = -0.298 B =    =    = 3.874 C = SD + B = $10(-0.298)+ 3.874 = $0.89
= This problem requires a two period binomial tree.The solution will start by solving the value of the call option for the up and down branches as of year 1 and then solve for the final value of the option at year 0.    Up branch D =    =    = 0 B =    =    = 0 C = SD + B = $13(0)+ (0)= $0 Down Branch D =    =    = -0.6 B =    =    = 6.285714 C = SD + B = $8(-0.6)+ (6.285714)= $1.49 Value at year 0 D =    =    = -0.298 B =    =    = 3.874 C = SD + B = $10(-0.298)+ 3.874 = $0.89
= 0
C = SD + B = $13(0)+ (0)= $0
Down Branch
D = This problem requires a two period binomial tree.The solution will start by solving the value of the call option for the up and down branches as of year 1 and then solve for the final value of the option at year 0.    Up branch D =    =    = 0 B =    =    = 0 C = SD + B = $13(0)+ (0)= $0 Down Branch D =    =    = -0.6 B =    =    = 6.285714 C = SD + B = $8(-0.6)+ (6.285714)= $1.49 Value at year 0 D =    =    = -0.298 B =    =    = 3.874 C = SD + B = $10(-0.298)+ 3.874 = $0.89
= This problem requires a two period binomial tree.The solution will start by solving the value of the call option for the up and down branches as of year 1 and then solve for the final value of the option at year 0.    Up branch D =    =    = 0 B =    =    = 0 C = SD + B = $13(0)+ (0)= $0 Down Branch D =    =    = -0.6 B =    =    = 6.285714 C = SD + B = $8(-0.6)+ (6.285714)= $1.49 Value at year 0 D =    =    = -0.298 B =    =    = 3.874 C = SD + B = $10(-0.298)+ 3.874 = $0.89
= -0.6
B = This problem requires a two period binomial tree.The solution will start by solving the value of the call option for the up and down branches as of year 1 and then solve for the final value of the option at year 0.    Up branch D =    =    = 0 B =    =    = 0 C = SD + B = $13(0)+ (0)= $0 Down Branch D =    =    = -0.6 B =    =    = 6.285714 C = SD + B = $8(-0.6)+ (6.285714)= $1.49 Value at year 0 D =    =    = -0.298 B =    =    = 3.874 C = SD + B = $10(-0.298)+ 3.874 = $0.89
= This problem requires a two period binomial tree.The solution will start by solving the value of the call option for the up and down branches as of year 1 and then solve for the final value of the option at year 0.    Up branch D =    =    = 0 B =    =    = 0 C = SD + B = $13(0)+ (0)= $0 Down Branch D =    =    = -0.6 B =    =    = 6.285714 C = SD + B = $8(-0.6)+ (6.285714)= $1.49 Value at year 0 D =    =    = -0.298 B =    =    = 3.874 C = SD + B = $10(-0.298)+ 3.874 = $0.89
= 6.285714
C = SD + B = $8(-0.6)+ (6.285714)= $1.49
Value at year 0
D = This problem requires a two period binomial tree.The solution will start by solving the value of the call option for the up and down branches as of year 1 and then solve for the final value of the option at year 0.    Up branch D =    =    = 0 B =    =    = 0 C = SD + B = $13(0)+ (0)= $0 Down Branch D =    =    = -0.6 B =    =    = 6.285714 C = SD + B = $8(-0.6)+ (6.285714)= $1.49 Value at year 0 D =    =    = -0.298 B =    =    = 3.874 C = SD + B = $10(-0.298)+ 3.874 = $0.89
= This problem requires a two period binomial tree.The solution will start by solving the value of the call option for the up and down branches as of year 1 and then solve for the final value of the option at year 0.    Up branch D =    =    = 0 B =    =    = 0 C = SD + B = $13(0)+ (0)= $0 Down Branch D =    =    = -0.6 B =    =    = 6.285714 C = SD + B = $8(-0.6)+ (6.285714)= $1.49 Value at year 0 D =    =    = -0.298 B =    =    = 3.874 C = SD + B = $10(-0.298)+ 3.874 = $0.89
= -0.298
B = This problem requires a two period binomial tree.The solution will start by solving the value of the call option for the up and down branches as of year 1 and then solve for the final value of the option at year 0.    Up branch D =    =    = 0 B =    =    = 0 C = SD + B = $13(0)+ (0)= $0 Down Branch D =    =    = -0.6 B =    =    = 6.285714 C = SD + B = $8(-0.6)+ (6.285714)= $1.49 Value at year 0 D =    =    = -0.298 B =    =    = 3.874 C = SD + B = $10(-0.298)+ 3.874 = $0.89
= This problem requires a two period binomial tree.The solution will start by solving the value of the call option for the up and down branches as of year 1 and then solve for the final value of the option at year 0.    Up branch D =    =    = 0 B =    =    = 0 C = SD + B = $13(0)+ (0)= $0 Down Branch D =    =    = -0.6 B =    =    = 6.285714 C = SD + B = $8(-0.6)+ (6.285714)= $1.49 Value at year 0 D =    =    = -0.298 B =    =    = 3.874 C = SD + B = $10(-0.298)+ 3.874 = $0.89
= 3.874
C = SD + B = $10(-0.298)+ 3.874 = $0.89

The Black-Scholes value of a one-year call option on Taggart stock with a strike price of $50 is closest to:

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A

Use the information for the question(s)below. The current price of Kinston Corporation stock is $10.In each of the next two years,this stock price can either go up by $3.00 or go down by $2.00.Kinston stock pays no dividends.The one year risk-free interest rate is 5% and will remain constant. -Using the binomial pricing model,calculate the price of a two-year call option on Kinston stock with a strike price of $9.

Free
(Essay)
4.9/5
(80)
Correct Answer:
Verified

This problem requires a two period binomial tree.The solution will start by solving the value of the call option for the up and down branches as of year 1 and then solve for the final value of the option at year 0. This problem requires a two period binomial tree.The solution will start by solving the value of the call option for the up and down branches as of year 1 and then solve for the final value of the option at year 0.    Up branch D =    =    = 1 B =    =    = -8.571429 C = SD + B = $13(1)+ (-8.571429)= $4.43 Down Branch D =    =    = .4 B =    =    = -2.285714 C = SD + B = $8(.4)+ (-2.285714)= $0.91 Value at year 0 D =    =    = .7040 B =    =    = -4.497143 C = SD + B = $10(.704)+ (-4.497143)= $2.54
Up branch
D = This problem requires a two period binomial tree.The solution will start by solving the value of the call option for the up and down branches as of year 1 and then solve for the final value of the option at year 0.    Up branch D =    =    = 1 B =    =    = -8.571429 C = SD + B = $13(1)+ (-8.571429)= $4.43 Down Branch D =    =    = .4 B =    =    = -2.285714 C = SD + B = $8(.4)+ (-2.285714)= $0.91 Value at year 0 D =    =    = .7040 B =    =    = -4.497143 C = SD + B = $10(.704)+ (-4.497143)= $2.54
= This problem requires a two period binomial tree.The solution will start by solving the value of the call option for the up and down branches as of year 1 and then solve for the final value of the option at year 0.    Up branch D =    =    = 1 B =    =    = -8.571429 C = SD + B = $13(1)+ (-8.571429)= $4.43 Down Branch D =    =    = .4 B =    =    = -2.285714 C = SD + B = $8(.4)+ (-2.285714)= $0.91 Value at year 0 D =    =    = .7040 B =    =    = -4.497143 C = SD + B = $10(.704)+ (-4.497143)= $2.54
= 1
B = This problem requires a two period binomial tree.The solution will start by solving the value of the call option for the up and down branches as of year 1 and then solve for the final value of the option at year 0.    Up branch D =    =    = 1 B =    =    = -8.571429 C = SD + B = $13(1)+ (-8.571429)= $4.43 Down Branch D =    =    = .4 B =    =    = -2.285714 C = SD + B = $8(.4)+ (-2.285714)= $0.91 Value at year 0 D =    =    = .7040 B =    =    = -4.497143 C = SD + B = $10(.704)+ (-4.497143)= $2.54
= This problem requires a two period binomial tree.The solution will start by solving the value of the call option for the up and down branches as of year 1 and then solve for the final value of the option at year 0.    Up branch D =    =    = 1 B =    =    = -8.571429 C = SD + B = $13(1)+ (-8.571429)= $4.43 Down Branch D =    =    = .4 B =    =    = -2.285714 C = SD + B = $8(.4)+ (-2.285714)= $0.91 Value at year 0 D =    =    = .7040 B =    =    = -4.497143 C = SD + B = $10(.704)+ (-4.497143)= $2.54
= -8.571429
C = SD + B = $13(1)+ (-8.571429)= $4.43
Down Branch
D = This problem requires a two period binomial tree.The solution will start by solving the value of the call option for the up and down branches as of year 1 and then solve for the final value of the option at year 0.    Up branch D =    =    = 1 B =    =    = -8.571429 C = SD + B = $13(1)+ (-8.571429)= $4.43 Down Branch D =    =    = .4 B =    =    = -2.285714 C = SD + B = $8(.4)+ (-2.285714)= $0.91 Value at year 0 D =    =    = .7040 B =    =    = -4.497143 C = SD + B = $10(.704)+ (-4.497143)= $2.54
= This problem requires a two period binomial tree.The solution will start by solving the value of the call option for the up and down branches as of year 1 and then solve for the final value of the option at year 0.    Up branch D =    =    = 1 B =    =    = -8.571429 C = SD + B = $13(1)+ (-8.571429)= $4.43 Down Branch D =    =    = .4 B =    =    = -2.285714 C = SD + B = $8(.4)+ (-2.285714)= $0.91 Value at year 0 D =    =    = .7040 B =    =    = -4.497143 C = SD + B = $10(.704)+ (-4.497143)= $2.54
= .4
B = This problem requires a two period binomial tree.The solution will start by solving the value of the call option for the up and down branches as of year 1 and then solve for the final value of the option at year 0.    Up branch D =    =    = 1 B =    =    = -8.571429 C = SD + B = $13(1)+ (-8.571429)= $4.43 Down Branch D =    =    = .4 B =    =    = -2.285714 C = SD + B = $8(.4)+ (-2.285714)= $0.91 Value at year 0 D =    =    = .7040 B =    =    = -4.497143 C = SD + B = $10(.704)+ (-4.497143)= $2.54
= This problem requires a two period binomial tree.The solution will start by solving the value of the call option for the up and down branches as of year 1 and then solve for the final value of the option at year 0.    Up branch D =    =    = 1 B =    =    = -8.571429 C = SD + B = $13(1)+ (-8.571429)= $4.43 Down Branch D =    =    = .4 B =    =    = -2.285714 C = SD + B = $8(.4)+ (-2.285714)= $0.91 Value at year 0 D =    =    = .7040 B =    =    = -4.497143 C = SD + B = $10(.704)+ (-4.497143)= $2.54
= -2.285714
C = SD + B = $8(.4)+ (-2.285714)= $0.91
Value at year 0
D = This problem requires a two period binomial tree.The solution will start by solving the value of the call option for the up and down branches as of year 1 and then solve for the final value of the option at year 0.    Up branch D =    =    = 1 B =    =    = -8.571429 C = SD + B = $13(1)+ (-8.571429)= $4.43 Down Branch D =    =    = .4 B =    =    = -2.285714 C = SD + B = $8(.4)+ (-2.285714)= $0.91 Value at year 0 D =    =    = .7040 B =    =    = -4.497143 C = SD + B = $10(.704)+ (-4.497143)= $2.54
= This problem requires a two period binomial tree.The solution will start by solving the value of the call option for the up and down branches as of year 1 and then solve for the final value of the option at year 0.    Up branch D =    =    = 1 B =    =    = -8.571429 C = SD + B = $13(1)+ (-8.571429)= $4.43 Down Branch D =    =    = .4 B =    =    = -2.285714 C = SD + B = $8(.4)+ (-2.285714)= $0.91 Value at year 0 D =    =    = .7040 B =    =    = -4.497143 C = SD + B = $10(.704)+ (-4.497143)= $2.54
= .7040
B = This problem requires a two period binomial tree.The solution will start by solving the value of the call option for the up and down branches as of year 1 and then solve for the final value of the option at year 0.    Up branch D =    =    = 1 B =    =    = -8.571429 C = SD + B = $13(1)+ (-8.571429)= $4.43 Down Branch D =    =    = .4 B =    =    = -2.285714 C = SD + B = $8(.4)+ (-2.285714)= $0.91 Value at year 0 D =    =    = .7040 B =    =    = -4.497143 C = SD + B = $10(.704)+ (-4.497143)= $2.54
= This problem requires a two period binomial tree.The solution will start by solving the value of the call option for the up and down branches as of year 1 and then solve for the final value of the option at year 0.    Up branch D =    =    = 1 B =    =    = -8.571429 C = SD + B = $13(1)+ (-8.571429)= $4.43 Down Branch D =    =    = .4 B =    =    = -2.285714 C = SD + B = $8(.4)+ (-2.285714)= $0.91 Value at year 0 D =    =    = .7040 B =    =    = -4.497143 C = SD + B = $10(.704)+ (-4.497143)= $2.54
= -4.497143
C = SD + B = $10(.704)+ (-4.497143)= $2.54

Which of the following statements is FALSE?

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Which of the following statements is FALSE?

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The Black-Scholes value of a one-year,at-the-money put option on Taggart stock is closest to:

(Multiple Choice)
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Using the binomial pricing model,the calculated price of a one-year put option on KD stock with a strike price of $20 is closest to:

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Using risk neutral probabilities,the calculated price of a one-year put option on KD stock with a strike price of $20 is closest to:

(Multiple Choice)
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Consider the following equation: B = Consider the following equation: B =   In this equation,the term B,represents: In this equation,the term B,represents:

(Multiple Choice)
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The Black-Scholes Δ of a one-year,at-the-money put option on Taggart stock is closest to:

(Multiple Choice)
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Consider the following equation: C = S × N Consider the following equation: C = S × N   - PV(K)× N   In this equation,the term σ represents: - PV(K)× N Consider the following equation: C = S × N   - PV(K)× N   In this equation,the term σ represents: In this equation,the term σ represents:

(Multiple Choice)
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Which of the following is a corporate application of option pricing?

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Which of the following statements is FALSE?

(Multiple Choice)
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Consider the following equation: D = Consider the following equation: D =   In this equation,the term D,represents: In this equation,the term D,represents:

(Multiple Choice)
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Which of the following statements is FALSE?

(Multiple Choice)
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Use the information for the question(s)below. The current price of KD Industries stock is $20.In the next year the stock price will either go up by 20% or go down by 20%.KD pays no dividends.The one year risk-free rate is 5% and will remain constant. -Using the binomial pricing model,the calculated price of a one-year call option on KD stock with a strike price of $20 is closest to:

(Multiple Choice)
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Use the information for the question(s)below. The current price of Kinston Corporation stock is $10.In each of the next two years,this stock price can either go up by $3.00 or go down by $2.00.Kinston stock pays no dividends.The one year risk-free interest rate is 5% and will remain constant. -Using risk neutral probabilities,calculate the price of a two-year call option on Kinston stock with a strike price of $9.

(Essay)
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Use the information for the question(s)below. The current price of KD Industries stock is $20.In the next year the stock price will either go up by 20% or go down by 20%.KD pays no dividends.The one year risk-free rate is 5% and will remain constant. -Assuming the Beta on KD stock is 1.1,the calculated beta for a one-year call option on KD stock with a strike price of $20 is closest to:

(Multiple Choice)
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Which of the following statements is FALSE?

(Multiple Choice)
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Using risk neutral probabilities,calculate the price of a two-year put option on Kinston stock with a strike price of $9.

(Essay)
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(76)
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