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Comp_Fin_HW3

# Comp_Fin_HW3 - Jaime Frade Computational Finance Dr Kopriva...

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Jaime Frade Computational Finance: Dr. Kopriva Homework #3 0. Executive Summary In this assignment there is a need to determine the minimal interest rate, i , at which a person can invest \$1000 per month for 15 years to ensure retirement nest egg of \$1 million. In order to calculate i three root-finding algorithms, bisection method, Newton-Rapshon, and a combination, were programmed in C++. It will be shown that even though the bisection method does provide solutions, it suffers from a slow error linear convergence. On the other hand, using Newton’s method will show fast quadratic convergence, however, highly depends on the choosing initial starting point appropriately. A combination of these methods will combine the speed and convergence stability needed to solve the problem. In order to reassure the person of my calculations and show a comparison of each of the three algorithms, I will test several functions with known roots. Results will be shown. This was done by advice of my collegue. From the calculations, it was found that an monthly interest rate, i , of approximately 1.5796708% under the IEEE standard of single precision. Therefore, to create this “nest egg,” the investor will need to find a fund offering an annual interest rate of 18.95% compounded monthly. Unfortunately, as of today, retirement will be impossible due to investments, such as CDs/savings, only offering rates 3-4%. 1. Statement of Problem Determine the interest rate which will satisfy the following equation: A = P i [(1 + i ) n - 1] , (1) where A amount in the account, P deposit amount at each month, and i interest rate per period. For the “nest egg,” n = 15 years (180 month since compounded monthly), P = 1 , 000 and A = 1 , 000 , 000. To determine i , will find a root using the bisection method, Newton-Rapshon, and a combination of the two. 2. Description of the Mathematics From lecture, the problem is to find a root for a function, f ( i ) : f ( i ) = A - P i [(1 + i ) n - 1] , i = 0 (2) such that it solves f ( i ) = 0. The root exists by the Intermediate value theorem, if f ( i ) is a continuous function on an closed interval, [a,b], and f ( a ) f ( b ) < 0 . Then there an i such that it satisfies, (2), f ( i ) = 0 for some i [ a, b ]. Bisection method will repeatedly check for an i , by making an initial guess such that i = a + b 2 . Newton’s method will use functional values to determine the i which satisfies f ( i ) = 0 . If i is an approximation to ˆ i satisfying f ( ˆ i ) = 0, then by using simplification on the Taylor series expansion on the function (2) there exists a better approximation for ˆ i such that ˆ i = i n - f ( i n ) f ( i n ) , where (3) f ( i ) = Pni (1 + n ) n - 1 - P (1 + i ) n + 1 i 2 (4) 1

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3. Description of the Algorithm Bisection Method If trying to determine a root in a closed interval [a,b], denote AbsTol absolute tolerance and RelTol denote the relative tolerence.
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