Economics Dynamics Problems 126

Economics Dynamics Problems 126 - suppose the solution...

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110 Economic Dynamics Although software programmes can readily solve for the internal rate of return, there is a problem in the choice of r . n ± t = 0 B t C t (1 + r ) t is a polynomial with the highest power of n , and so theoretically there are n possible roots to this equation. Of course, we can rule out negative values and complex values. For example, the choice problem for Bramwell plc involves r 10 as the highest term and so there are ten possible solutions to the equation 40000 + 7500 ² 1 (1 + r ) 10 r ³ 5000 (1 + r ) 5 = 0 Eight solutions, however, are complex and another is negative. This leaves only one positive real-valued solution, namely r = 0 . 1172 or r = 11 . 72%. Since such a return is well above the typical market interest rate, then the investment should be undertaken. The point is, however, that multiple positive real-valued solutions are possible. 3.8 Solving second-order difference equations 3.8.1 Homogeneous Consider the following general second-order linear homogeneous equation y n + 2 = ay n + 1 + by n (3.19) Similar to the solution for a ±rst-order linear homogeneous equation, we can
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Unformatted text preview: suppose the solution takes the form y n = c 1 r n + c 2 s n for some constants r and s and where c 1 and c 2 depend on the initial conditions y and y 1 . If this indeed is correct, then c 1 r n + 2 + c 2 s n + 2 = a ( c 1 r n + 1 + c 2 s n + 1 ) + b ( c 1 r n + c 2 s n ) Re-arranging and factorising, we obtain c 1 r n ( r 2 − ar − b ) + c 2 s n ( s 2 − as − b ) = So long as r and s are chosen to be the solution values to the general quadratic equation x 2 − ax − b = i.e. x = r and x = s , where r ±= s , then y n = c 1 r n + c 2 s n is a solution to the dynamic system. This quadratic equation is referred to as the characteristic equation of the dynamical system. If r > s , then we call y 1 = c 1 r n the dominant solution and r the dominant characteristic root. Furthermore, given we have obtained the solution values r and s , and given the initial conditions, y and y 1 , then we can solve for the two unknown coef±cients,...
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