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Math Help - Linear programming: simplex method

  1. #1
    Mar 2012

    Linear programming: simplex method

    Trying to solve a task here using the simplex method.
    A company is producing three types of fertilizer. Their chemical composition is as follows (per ton): Type I has 0.1 tonnes of nitrates, 0.1 tonnes of phosphates, 0.2 tonnes of carbonates and 0.6 tonnes of filler; Type II has 0.1 tonnes nitrates, 0.2 tonnes phosphates, 0.1 tonnes carbonates and 0.6 tonnes filler; type III has 0.2 tonnes nitrates, 0.1 tonnes phosphates, 0.1 tonnes carbonates and 0.6 tonnes filler.
    In a month, the plant can make use of 1200 tonnes of nitrates (costing $1500), 2000 tonnes phosphates ($600), 2200 tonnes of carbonates ($1200) and an unlimited supply (!) of filler ($100).
    The fertilizers are sold at a per ton price of $830 (I), $810 (II) and $810 (III). Production costs, except raw material, are estimated at $110 in the case of all three fertilizers.
    Please help me put together a linear programming task to maximise profit. If I could just get the contraints and the optimization formula right, then I could get on with solving the simplex tableu itself.

    Assuming the variables are x1 - type I production, x2 - type II production, x3 - type III production.
    Constraints (non-canonical): 0.1x1 + 0.1x2 + 0.2x3 <= 1200 (for nitrates)
    0.1x1 + 0.2x2 + 0.1x3 <= 2000 (phosphates)
    0.2x1 + 0.1x2 + 0.1x3 <= 2200 (carbonates)
    0.6x1 + 0.6x2 + 0.6x3 <= ...(?)... (filler)

    Some help would be greatly appreciated!
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  2. #2
    Mar 2012

    Re: Linear programming: simplex method

    Quote Originally Posted by JacobE View Post
    0.6x1 + 0.6x2 + 0.6x3 <= ...(?)... (filler)
    One thing that I was wondering: is it even necessary to form such a contraint as its really referring to an unlimited item. Hence there actually is no constraint to be formed?

    As for the optimization formula (maximise profit), I got it like this: F = 210x1 + 250x2 + 160x3. That should represent the profit made per ton of fertilizer sold. This could be totally wrong, but I thought cost of materials per ton of fertilizer can be found by (using Type I fertilizer as an example) 0.1*1500$ + 0.1 * 600$ + 0.2 * 1200 + 0.6 * 100$ = $510. By substracting that and production costs ($110) from the selling price of $830, you'd make a profit of $210 per ton of type I fertilizer sold. Do the same for type II and type III fertilizers and you get $250 and $160 respectively. Is this correct?
    Last edited by JacobE; February 17th 2013 at 03:48 AM.
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