HW1 Solution

# HW1 Solution - Î-1 Î Î 2 1 2 Î = 1-1 1 1 2 1 2 Â 1 = 2 3 Î...

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MIE375H1F - Financial Engineering - HW1 Solutions Nick Yeung September 21, 2010 Question 2.1 Assuming that the current year is 2010 (or 1998, the year the textbook was written), n = 2010 - 1776 = 234 or n = 1998 - 1776 = 222 2.1a PV = \$1(1 + r ) n = \$1(1 + 0 . 033) 234 = \$1992 . 85 or PV = \$1(1 + r ) n = \$1(1 + 0 . 033) 222 = \$1349 . 81 Both are approximately \$1000 in value. 2.1b PV = \$1(1 + r ) n = \$1(1 + 0 . 066) 234 = \$3 , 127 , 417 . 76 or PV = \$1(1 + r ) n = \$1(1 + 0 . 066) 222 = \$1 , 452 , 444 . 15 Both are approximately \$1,000,000 in value. 1

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Question 2.3 2.3a 1 + r eff = (1 + r m ) m r eff = (1 + 0 . 03 12 ) 12 - 1 = 0 . 0304 = 3 . 04% 2.3b 1 + r eff = (1 + r m ) m r eff = (1 + 0 . 18 12 ) 12 - 1 = 0 . 1956 = 19 . 56% 2.3c 1 + r eff = (1 + r m ) m r eff = (1 + 0 . 18 4 ) 4 - 1 = 0 . 1925 = 19 . 25% Question 2.4 f ( λ ) = - 1 + λ + λ 2 f 0 ( λ ) = 1 + 2 λ f ( λ ) f 0 ( λ ) = - 1 + λ + λ 2 1 + 2 λ λ n = λ n - 1 - f ( λ n - 1 ) f 0 ( λ n - 1 ) = λ n - 1 - - 1 + λ n - 1 + λ 2 n - 1 1 + 2 λ n - 1 2
Using Newton’s Method and starting at λ 0 = 1: λ 1 =
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Unformatted text preview: Î»--1 + Î» + Î» 2 1 + 2 Î» = 1--1 + 1 + 1 2 1 + 2 Â· 1 = 2 3 Î» 2 = Î» 1--1 + Î» 1 + Î» 2 1 1 + 2 Î» 1 = 2 3--1 + 2 3 + ( 2 3 ) 2 1 + 2 Â· 2 3 = 13 21 Î» 3 = Î» 2--1 + Î» 2 + Î» 2 2 1 + 2 Î» 2 = 13 21--1 + 13 21 + ( 13 21 ) 2 1 + 2 Â· 13 21 = 610 987 Î» 4 = Î» 3--1 + Î» 3 + Î» 2 3 1 + 2 Î» 3 = 610 987--1 + 610 987 + ( 610 987 ) 2 1 + 2 Â· 610 987 = 0 . 6180 Note: The roots of the quadratic-1 + Î» + Î» 2 are-1 Â± âˆš 5 2 which are approximately equal to 0.618 or -1.618. Question 2.5 PV = 19 X i =0 \$500 , 000 (1 + 0 . 1) i = (\$500 , 000) 1-( 1 1 . 1 ) 20 1-1 1 . 1 = \$4 , 682 , 460 . 05 3...
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HW1 Solution - Î-1 Î Î 2 1 2 Î = 1-1 1 1 2 1 2 Â 1 = 2 3 Î...

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