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How to minimize

  1. Jun 25, 2012 #1
    How to minimize the following:

    Subject to : x/(1-e^(-βx)))- (1/β)≤ T

    I need the find the value of x. All A,B,T and β are constants. The numerator is linear. I don't know if the equation remains linear when the denominator is added. When plotted, the equation is a straight increasing line starting from 0. Any help please?
    Last edited: Jun 25, 2012
  2. jcsd
  3. Jun 25, 2012 #2


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    All linear equations are of the form "y= Ax+ B" so, no, this does not "remain linear when the denominator is added". The "standard method" of finding minimum values is to set the derivative equal to 0. Here, if I have done the calculations correctly, that reduces to [tex](1- x)e^{\beta x}= 1[/tex] which would be best solved graphically. In fact, the graph of this function looks to me like it does not have any local max or min.

    Attached Files:

  4. Jun 25, 2012 #3
    This bit doesn't make sense. Please clarify. (Look at your parentheses.)) :wink:
    Last edited: Jun 25, 2012
  5. Jun 25, 2012 #4
    Thanks HallsofIvy. So let me get this correctly.....for the above optimization, we just take derivative of the objective function? How do we incorporate the constraint in that case?

    In my case derivation of the objective function yields e^(βx)-βx=1-(βA)/B. And then considering 1/β is large, I get x=√(2A/βB). But then I do not consider the constraint at all?
  6. Jun 27, 2012 #5
    I really need this problem solved. Some help please.
  7. Jun 27, 2012 #6
    [tex]f(x) = \frac{A+Bx}{1-e^\textrm{ β x}} - \frac{B}{β}[/tex]
    [tex]f'(x) = \frac{B(1-e^\textrm{ β x}) + βe^\textrm{ β x}(A+Bx)}{(1-e^\textrm{ β x})^{2}}[/tex]
    [tex]f'(x) = 0 = \frac{B-Be^\textrm{ β x} + Aβe^\textrm{ β x}+Bxβe^\textrm{ β x}}{(1-e^\textrm{ β x})^{2}}[/tex]
    [tex]B-Be^\textrm{ β x} + Aβe^\textrm{ β x}+Bxβe^\textrm{ β x} = 0[/tex]
    [tex]Be^\textrm{ -β x} - B + Aβ+Bxβ = 0[/tex]
    [tex]B(e^\textrm{ -β x} + βx) = B - Aβ[/tex]
    [tex]\frac{1}{e^\textrm{ β x}} + βx = 1 - \frac{Aβ}{B}[/tex]
  8. Jun 28, 2012 #7
    Thanks Math man,

    But how do I incorporate the constraint ?
  9. Jun 28, 2012 #8
    Sorry, I forgot about the constraint. You can use Kuhn-Tucker conditions:
    [tex]f(x) = \frac{A+Bx}{1-e^\textrm{ β x}} - \frac{B}{β}[/tex]
    [tex]g(x) = \frac{x}{1-e^\textrm{ β x}} - \frac{1}{β}[/tex]
    [tex]g(x) ≤ T[/tex]
    The Kuhn-Tucker conditions for minimizing single variable functions are:
    [tex]0 = -f'(x) - λg'(x)[/tex]
    [tex]g(x) ≤ T[/tex]
    [tex]λ ≥ 0[/tex]
    [tex]λ(g(x)-T) = 0[/tex]
    So to apply the Kuhn Tucker conditions here:
    [tex]\frac{-B+Be^\textrm{ β x} - Aβe^\textrm{ β x}-Bxβe^\textrm{ β x}}{(1-e^\textrm{ β x})^{2}} = λ\frac{(1-e^\textrm{ β x}) + βxe^\textrm{ β x}}{(1-e^\textrm{ β x})^{2}}[/tex]
    [tex]-B+Be^\textrm{ β x} - Aβe^\textrm{ β x}-Bxβe^\textrm{ β x} = λ-λe^\textrm{ β x} + λβxe^\textrm{ β x}[/tex]
    [tex]e^\textrm{ β x}(B - Aβ-Bxβ+λ - λβx)-B-λ = 0[/tex]
    This can be combined with the remaining conditions:
    [tex]λ(\frac{x}{1-e^\textrm{ β x}}-\frac{1}{β}-T) = 0[/tex]
    [tex]\frac{x}{1-e^\textrm{ β x}} ≤ T + \frac{1}{β}[/tex]
    [tex]λ ≥ 0[/tex]
    Use these to solve for x.
  10. Jun 29, 2012 #9


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    Minimising {(A+Bx)/(1-e-βx)}-(B/β) is the same as minimising f(x) = (A+Bx)/(1-e-βx) [0]
    Ignoring the constraint for the moment, setting the derivative to zero gives me:
    eβx = 1 + βx + Aβ/B [1]
    (Rather different from what HofI got...)
    Whether a solution of the above blows the constraint may well depend on the details of the constants. As oay pointed out, there's a syntax error in your constraint. I assume it's just an extra right parenthesis, so should read:
    x/(1-e-βx) ≤ T + 1/β [2]
    Since the function to be minimised clearly has exactly one minimum between 0 and +∞, if the solution of [1] is disallowed by [2] then the desired answer gives equality in [2]:
    x/(1-e-βx) = T + 1/β [3]
  11. Jun 30, 2012 #10
    Thank you very much!
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