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Math Help - Optimisation involving if statements and ceiling values?

  1. #1
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    Optimisation involving if statements and ceiling values?

    \frac{ln(999/x) } {ln(y/100+1) }+ 1 = N

    How can I minimise N and maximise x given these constraints:

    1. 1 =< x =< 315
    2. 1 =< y =< 110
    3. x+y =< 340
    4. if x >= 150 then y =< 50
    5. if x >= 250 then y =< 20
    6. if y >= 90 then x =< 60
    7. N is a ceiling value, so 3.1 would be the same as 4.

    By minimising N and maximising x I mean that lets say x = 200, y = 105 gives 3.24 for N while x = 210, y = 90 gives 3.43 for N. Since N is a ceiling value, both results are the same however the second set of answers is more desirable as x is 210 as opposed to 200. This particular set of values isn't valid since it ignores some of the restraints but it demonstrates what I mean by minimising N and maximising x.

    I dont know how to solve this (is it even possible?), it seems like a linear programming type problem or lagrangian multipliers but I'm not sure how to solve it. Also how do you make 1/2 look like 1 'on' 2?
    Last edited by Corpsecreate; December 7th 2010 at 09:38 AM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Corpsecreate View Post
    \frac{ln(999/x) } {ln(y/100+1) }+ 1 = N

    How can I minimise N and maximise x given these constraints:...
    Do you mean:

    Maximise $$x such that

    \displaystyle \left\lceil \frac{\ln(999/x) } {\ln(y/100+1) }+ 1 \right\rceil

    is minimised and subject to the constraints ..

    CB
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