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Homework Help: Dynamic Programming to maximize profit

  1. Oct 9, 2011 #1
    1. The problem statement, all variables and given/known data
    Trying to maximize the profit of a farmer using dynamic optimization. Each period the farmer has a stock of seeds. He can plant them at a cost c per seed or sell them for p. Every seed that is planted produces [itex]\gamma[/itex] seeds for next period. In period m there is no longer any demand for the seeds.

    Profit function [itex]\pi[/itex] = p(1-[itex]\alpha[/itex]t)[itex]\gamma[/itex]xt - c[itex]\alpha[/itex]t[itex]\gamma[/itex]xt where [itex]\alpha[/itex]t is the proportion of seeds kept to sow at the end of the period.

    We are trying to maximize [itex]\sum[/itex][[itex]\pi[/itex]t / (1+r)t] from t=0 to m-1

    Initial stock of seeds is x0 = 1
    [itex]\gamma[/itex] = 8
    c = 3
    p = 1
    r = 0.1
    m = 3

    2. Relevant equations
    Bellman Optimization.

    3. The attempt at a solution

    Define [itex]\beta[/itex] as 1/(1+r)t
    Motion rule xt+1 = [itex]\gamma[/itex][itex]\alpha[/itex]txy
    We know that there is no demand for the seeds in period 4 so x4 = 0 = [itex]\gamma[/itex][itex]\alpha[/itex]3x3
    This means that x4 = 0

    The value function V3 becomes:
    V3 = p[itex]\gamma[/itex]x3

    @ t = 2

    V2 = max (p(1-[itex]\alpha[/itex]2)[itex]\gamma[/itex]x2 - c[itex]\alpha[/itex]2[itex]\gamma[/itex]x2 + [itex]\beta[/itex]V3

    Which using motion rule gives

    V2 = max (p(1-[itex]\alpha[/itex]2)[itex]\gamma[/itex]x2 - c[itex]\alpha[/itex]2[itex]\gamma[/itex]x2) + [itex]\beta[/itex](p[itex]\gamma[/itex][itex]\alpha[/itex]2[itex]\gamma[/itex]x2)

    Normally at this point I would differentiate and to find the maximum and then recurse the answer back into t=1 but it is a linear function. So I am guessing I need to take some sort of corner soluition but I am not entirely clear how to proceed.

    Any tips would be welcome. Thank you.
  2. jcsd
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