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sols1_14

# sols1_14 - Mathematical Finance 1(a(1 0.1/2)2 = 1.1025 so...

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Mathematical Finance 1. (a) (1 + 0 . 1 / 2) 2 = 1 . 1025 so the effective rate is 0 . 1025. (b) (1 + 0 . 1 / 4) 4 - 1 0 . 1038. (c) e 0 . 1 - 1 0 . 1052. 2. (a) 6.93 years (b) 28.41 years. Time to reach αD is t = log α/ log(1 + r ). 3. P ( r ) = 0 is equivalent to a quadratic in r . Solving this we find A = 300 has rate of return r ≈ - 0 . 148, A = 500 has r = 0, A = 700 has r 0 . 123. 4. Answer: a = 1426 . 15. 5. The present value of this sequence is - 30 . 75 < 0 so it’s not worth investing. 6. The present values of the first bond are (a) 1746 . 29 (b) 90 . 26 (c) - 618 . 93. The second bond has present value 40 . 94. 7. The rate of return per month is that r that solves the equation 117600 = 600 36 summationdisplay k =1 (1+ r ) - k +120000(1+ r ) - 36 = 600 1 + r 1 - (1 + r ) - 36 r/ (1 + r ) +120000(1+ r ) - 36 Answer: solve with Newton’s method or bisection to get r 0 . 5616%. 8. Group the positive terms and group the negative terms. Factor out exactly the right power of 1+ r (key step). Show that what remains is monotone and takes both positive and negative values. Deduce that P ( r ) = 0 has a unique root. 9. Due to the different interest rates you have to determine the balance after each year.
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