HW3Solutions - ACF329 Interest Theory Fall 2010 University...

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Unformatted text preview: ACF329 Interest Theory Fall 2010 University of Texas at Austin HW 3 - Solutions Instructor: Milica 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 Suzannes 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 Bills 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 . 15 (1 . 06) 8 + 1 . 06 8- 1 . 06 = 20 . 4638 P. On the other hand, Seths 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 . 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 . 9 20 / . 1111 = 11 . 1893 Q....
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This note was uploaded on 12/05/2011 for the course M 341 taught by Professor Hietmann during the Spring '08 term at University of Texas at Austin.

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HW3Solutions - ACF329 Interest Theory Fall 2010 University...

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