Tutorial03_-_soln

Tutorial03_-_soln - Solution to Problem Set 3 Q1. PV = 100...

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Unformatted text preview: Solution to Problem Set 3 Q1. PV = 100 exp- Z 2 t dt + 100 exp- Z 4 t dt = 100 exp- Z 2 (0 . 05 + 0 . 01 t ) dt + 100 exp- Z 4 (0 . 05 + 0 . 01 t ) dt = 100 exp- ( . 05 t + 0 . 005 t 2 ) 2 t =0 + 100 exp- ( . 05 t + 0 . 005 t 2 ) 4 t =0 = 100 e- . 12 + 100 e- . 28 = 164 . 27. Q2. (i) A simple interest rate of 10% implies an accumulation function a ( t ) = 1 + 0 . 1 t . Thus, d 5 = a (5)- a (4) a (5) = [1 + 0 . 1 5]- [1 + 0 . 1 4] 1 + 0 . 1 5 = . 1 1 . 5 = 6 . 666667% (ii) A simple discount rate of 10% implies an accumulation function a ( t ) = 1 1- . 1 t . Thus, d 5 = a (5)- a (4) a (5) = 1 1- . 1 5- 1 1- . 1 4 1 1- . 1 5 = 1 6 = 16 . 666667% (iii) A constant interest force of 0.1 implies that a ( t ) = e . 1 t . Thus, d 5 = a (5)- a (4) a (5) = e . 1 5- e . 1 4 e . 1 5 = 1- e- . 1 = 9 . 516258%. Q3. By the question, 1 + i = e and 1 + r = e . Thus it follows from the definition of inflation adjusted interest rate that 1 + j...
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This note was uploaded on 11/30/2010 for the course MATH 235/237 taught by Professor Wilkie during the Spring '10 term at Waterloo.

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Tutorial03_-_soln - Solution to Problem Set 3 Q1. PV = 100...

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