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HW3Solutions

# HW3Solutions - ACF329 Interest Theory Fall 2010 University...

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ACF329 Interest Theory Fall 2010 University of Texas at Austin HW 3 - Solutions Instructor: Milica ˇ Cudina 3.7.2. We split the annuity-due from the problem into two annuities-due: one with payments of 100 and the other with payments of 300. The latter annuity is one half-year deferred. With i = 0 . 06, the present value of Suzanne’s annuity-due can then be calculated as 100¨ a 20 i + (1 + i ) - 1 / 2 · 300¨ a 20 i = 100¨ a 20 i (1 + 3(1 + i ) - 1 / 2 ) 100 · 12 . 158 · 3 . 914 4 , 758 . 51 . 3.7.4 Let P = 100, i 1 = 0 . 15 and i 2 = 0 . 06. Then, the accumulated value of Bill’s deposits at the end of 13 years equals P ( s 5 i 1 · (1 + i 2 ) 8 + s 8 i 2 ) = P 1 . 15 5 - 1 0 . 15 · (1 . 06) 8 + 1 . 06 8 - 1 0 . 06 = 20 . 4638 P. On the other hand, Seth’s deposits must satisfy 20 . 4638 P = P · s 13 i = P · (1 + i ) 13 - 1 i . Using our calculator, we solve for i , we get i 0 . 0738. 3.7.6 The present value of the annuity X is PV ( X ) = 1000¨ a 12 + 2000 v 12 ¨ a 18 with v = 1 - d = 0 . 9. So, PV ( X ) = 1 . 1111(6458 . 475 + 4320 . 98553) = 1 . 1111 · 10779 . 4605 = 11977 . 0586 .. The present value of the perpetuity Y can be expressed as PV ( Y ) = Qa 20 + 3 Qv 20 a = Q ( a 20 + 3 v 20 a ) = Q (7 . 9064 + 3 · 0 . 9 20 / 0 . 1111 = 11 . 1893 Q. So, Q = 11977 . 0586 / 11 . 1893 1070 . 40285 . Note: The answer in the back of the textbook is different due to rounding errors.

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