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Math Help - More Details on The Optimization Problem

  1. #1
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    More Details on The Optimization Problem

    Hello,

    I need the solution of the following problem:



    max/min f(x_{1},x_{2},...,x_{n})= \frac{x_{1}}{1-x_{n}} subject to x_{1}+x_{2}+...+x_{n}=1.

    where 0<a_{1}\leq x_{1} \leq b_{1}<1, 0<a_{2}\leq x_{2}   \leq b_{2}<1, ...,0<a_{n}\leq x_{n} \leq b_{n}<1 .


    Best Regards.
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  2. #2
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by raed View Post
    Hello,

    I need the solution of the following problem:



    max/min f(x_{1},x_{2},...,x_{n})= \frac{x_{1}}{1-x_{n}} subject to x_{1}+x_{2}+...+x_{n}=1.

    where 0<a_{1}\leq x_{1} \leq b_{1}<1, 0<a_{2}\leq x_{2}   \leq b_{2}<1, ...,0<a_{n}\leq x_{n} \leq b_{n}<1 .


    Best Regards.
    If there are no mistakes f= \frac{x_{1}}{1-x_{n}} seems to be function only of x_{1} and x_{n} and not of x_{2}, x_{3},..., x_{n-1}...

    Kind regards

    \chi \sigma
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  3. #3
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    Quote Originally Posted by chisigma View Post
    If there are no mistakes f= \frac{x_{1}}{1-x_{n}} seems to be function only of x_{1} and x_{n} and not of x_{2}, x_{3},..., x_{n-1}...

    Kind regards

    \chi \sigma
    Thank you very much for your reply, there is no mistake because in my own problem I will maximize/minimize functions such the one above but in each case x_{1} will be replaced by x_{2}, x_{3},..., x_{n-1}

    Best Regards,
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    The function f belongs to \mathcal{C}^1(A) where A is the open set A=\prod_{i=1}^n(a_i,b_i) .

    On the other hand, g(x_1,\ldots,x_n)=x_1+\ldots+x_n-1 also belongs to \mathcal{C}^1(A) and =1" alt="\textrm{rank}(\;\nabla g(x_1,\ldots,x_n)\=1" /> for all x=(x_1,\ldots,x_n)\in A (maximun rank).

    According to the multipliers Lagrange theorem if f has a local maximum or minimum at x_0\in A there exists \lambda\in \mathbb{R} such that x_0 is critical point of

    F(x_1,\ldots,x_n)=f(x_1,\ldots,x_n)+\lambda g(x_1,\ldots,x_n)

    Verify that such a point does not exist. This means that the absolute maximum an minimum value for f is in the boundary \partial (K) of the compact set:

    K=\left(\;\prod_{i=1}^n[a_i,b_i]\;\right)\cap\{(x_1,\ldots,x_n):x_1+\ldots+x_n=1\}

    because f:K\to \mathbb{R} is continuous.
    Last edited by FernandoRevilla; March 8th 2011 at 08:51 AM.
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  5. #5
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    We can write

    <br />
\displaystyle<br />
f=\frac{x_1}{1-x_n}=\frac{x_1}{\sum_1^{n-1}x_i}<br />

    If x_i are independent in their range

    <br />
\displaystyle<br />
max \; f=\frac{max(x_1)}{min(\sum_1^{n-1}x_i)}=\frac{b_1}{\sum_1^{n-1}a_i}<br />

    and in the similar way

    <br />
\displaystyle<br />
min \; f=\frac{min(x_1)}{max(\sum_1^{n-1}x_i)}=\frac{a_1}{\sum_1^{n-1}b_i}<br />
    Last edited by zzzoak; March 8th 2011 at 09:48 AM.
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  6. #6
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    Quote Originally Posted by FernandoRevilla View Post
    The function f belongs to \mathcal{C}^1(A) where A is the open set A=\prod_{i=1}^n(a_i,b_i) .

    On the other hand, g(x_1,\ldots,x_n)=x_1+\ldots+x_n-1 also belongs to \mathcal{C}^1(A) and =1" alt="\textrm{rank}(\;\nabla g(x_1,\ldots,x_n)\=1" /> for all x=(x_1,\ldots,x_n)\in A (maximun rank).

    According to the multipliers Lagrange theorem if f has a local maximum or minimum at x_0\in A there exists \lambda\in \mathbb{R} such that x_0 is critical point of

    F(x_1,\ldots,x_n)=f(x_1,\ldots,x_n)+\lambda g(x_1,\ldots,x_n)

    Verify that such a point does not exist. This means that the absolute maximum an minimum value for f is in the boundary \partial (K) of the compact set:

    K=\left(\;\prod_{i=1}^n[a_i,b_i]\;\right)\cap\{(x_1,\ldots,x_n):x_1+\ldots+x_n=1\}

    because f:K\to \mathbb{R} is continuous.
    Thank you very much for your reply, but the function f belongs to \mathcal{C}^1(A) where A is the closed set A=\prod_{i=1}^n[a_i,b_i] not the open set A=\prod_{i=1}^n(a_i,b_i) .
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  7. #7
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    Quote Originally Posted by zzzoak View Post
    We can write

    <br />
\displaystyle<br />
f=\frac{x_1}{1-x_n}=\frac{x_1}{\sum_1^{n-1}x_i}<br />

    If x_i are independent in their range

    <br />
\displaystyle<br />
max \; f=\frac{max(x_1)}{min(\sum_1^{n-1}x_i)}=\frac{b_1}{\sum_1^{n-1}a_i}<br />

    and in the similar way

    <br />
\displaystyle<br />
min \; f=\frac{min(x_1)}{max(\sum_1^{n-1}x_i)}=\frac{a_1}{\sum_1^{n-1}b_i}<br />
    Thank you very much for reply, accourding to my problem the x_i are dependent in their range.


    Best Regards,
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  8. #8
    Grand Panjandrum
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    Quote Originally Posted by raed View Post
    Hello,

    I need the solution of the following problem:



    max/min f(x_{1},x_{2},...,x_{n})= \frac{x_{1}}{1-x_{n}} subject to x_{1}+x_{2}+...+x_{n}=1.

    where 0<a_{1}\leq x_{1} \leq b_{1}<1, 0<a_{2}\leq x_{2}   \leq b_{2}<1, ...,0<a_{n}\leq x_{n} \leq b_{n}<1 .


    Best Regards.
    There are only two variables $$x_1 and $$x_n, and the function has no calculus like minimum or maximum in the interior of the feasible region.

    As the feasible region is closed there is a minimum/maximum on the boundary and it can be found by examination of the function on the boundary.

    CB
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  9. #9
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by raed View Post
    Thank you very much for your reply, but the function f belongs to \mathcal{C}^1(A) where A is the closed set A=\prod_{i=1}^n[a_i,b_i] not the open set A=\prod_{i=1}^n(a_i,b_i) .

    Look at my previous post, I took the restriction to A=\textrm{Int}(K) to prove that we get the absolute maximum and minimum on the boundary of K.
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  10. #10
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    Quote Originally Posted by FernandoRevilla View Post
    Look at my previous post, I took the restriction to A=\textrm{Int}(K) to prove that we get the absolute maximum and minimum on the boundary of K.
    Dear Prof. Fernando Revilla,

    I understand you but what is the solution of the problem in this case.

    Best Regards,
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  11. #11
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by raed View Post
    I understand you but what is the solution of the problem in this case.

    First of all, you should show some work. Look rule #11 here:

    http://www.mathhelpforum.com/math-he...ng-151424.html


    I propose you the following: consider n=3 (then, you'll analyze what difficulties you can find in the general case).

    (i) Prove that there are no critical points in \textrm{Int}(K) for:

    F(x,y,z)=\dfrac{x}{1-z}+\lambda (x+y+z-1)

    (ii) Find the absolute maximum and minimum of f:\partial (K)\to \mathbb{R}

    " alt="f(x,y,z)=\dfrac{x}{x+y},\quad \partial (K) \equiv \begin{Bmatrix} a_1\leq x \leq b_1\\ a_2 \leq y \leq b_1\\a_3\leq 1-x-y \leq b_3 \end{matrix}\quad (\; [a_i,b_i]\subset (0,1)\ " />

    Draw in the xy plane the set \partial(K) . According to the relations of a_3,b_3 with a_1,b_1,a_2,b_2 , analyze the different kind of boundaries you can find.


    P.S. Have you seen the problem in a book? . In what a context?. Is it homework?.
    Last edited by FernandoRevilla; March 8th 2011 at 11:20 PM.
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  12. #12
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    Dear Prof. Fernando Revilla,

    First of all thank you very much for every thing.
    My problem is neither in a book nor a homework. Nowadays I am working on a paper and I countered this optimization problem. I think if I was able to sove it, then I would get some results concerning my research. Therefore, I first introduced the problem in a general case without determining the function and the constraint.

    Best Regards.
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  13. #13
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by raed View Post
    My problem is neither in a book nor a homework. Nowadays I am working on a paper and I countered this optimization problem. I think if I was able to sove it, then I would get some results concerning my research. Therefore, I first introduced the problem in a general case without determining the function and the constraint.

    Well, in that case and if we have no more information about the parameters a_i,b_i this could be a large, large,...., large problem with a lot of cases to analyze depending on the form of the boundary.
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