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Unformatted text preview: 1 ) × P] – [N(d 2 ) × PV(EX)] = [0.4002 × 1.7] – [0.3435 × (2/1.12 1 )] = $0.0669 million or $66,900 2 12. The asset value from Practice Question 11 is now reduced by the present value of the rents: PV(rents) = 0.15/1.12 = 0.134 Therefore, the asset value is now (1.7 – 0.134) = 1.566 P = 1.566 EX = 2 σ = 0.15 t = 1.0 r f = 0.12 0.9503 ) 1.0 (0.15 0.8003 t σ d d 0.8003 /2 ) 1.0 (0.15 ) 1.0 )]/(0.15 (2/1.12 log[1.566/ /2 t σ t X)]/ σ log[P/PV(E d 1 2 1 1 − = × − − = − = − = × + × = + = N(d 1 ) = N(0.8003) = 0.2118 N(d 2 ) = N(0.9503) = 0.1710 Call value = [N(d 1 ) × P] – [N(d 2 ) × PV(EX)] = [0.2118 × 1.566] – [0.1710 × (2/1.12 1 )] = $0.0263 million or $26,300...
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This note was uploaded on 05/09/2011 for the course FNAN 522 taught by Professor Wilson during the Spring '11 term at University of Louisiana at Lafayette.
 Spring '11
 Wilson

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