2

3. Consider the following stochastic optimization problem:Minimize E{ Ax2 _ log(B + y) } Subject to 10x + y 2 10 , Sx + 4y 2 20, 3x +Ty 221, x + 12y 212_Probability...

Question

3. Consider the following stochastic optimization problem:Minimize E{ Ax2 _ log(B + y) } Subject to 10x + y 2 10 , Sx + 4y 2 20, 3x +Ty 221, x + 12y 212_ProbabilityProbability0.3 0.4 0.2 0.10.2 0.4 0.41.5 2.5where A and B are random parameters with the given distributions on the right side_(a) Rewrite this optimization problem in a form that can be solved in CVX(b) Use CVX and solve this optimization problem:

3. Consider the following stochastic optimization problem: Minimize E{ Ax2 _ log(B + y) } Subject to 10x + y 2 10 , Sx + 4y 2 20, 3x +Ty 221, x + 12y 212_ Probability Probability 0.3 0.4 0.2 0.1 0.2 0.4 0.4 1.5 2.5 where A and B are random parameters with the given distributions on the right side_ (a) Rewrite this optimization problem in a form that can be solved in CVX (b) Use CVX and solve this optimization problem:



Answers

Consider the problem of minimizing $$ f(x, y)=x^{2}+2 y^{2} $$ subject to the constraint $g(x, y)=y e^{x^{2}}-1=0$. (A) Solve the constraint equation for $y$, and then substitute into $f(x, y)$ to obtain a function $h(x)$ of the single variable $x$. Solve the original minimization problem by minimizing $h$ (round answers to three decimal places). (B) Confirm your answer by the method of Lagrange multipliers.

Again they give us a constrained maximum are extremely, extremely problem. But now we have this which is basically a sequence of nested ellipses. If you know, if F is a constant, this would be in the lips and then here's our constraint which kind of looks ugly. But if we sell for so for why just get Y equals either the minus X square. So we can obviously plug that back in here and we get F of X. Y equals and to either their minus two, X squared plus X squared um cigarette. And that's just a function of X. So we can take the derivative of it, Set that equal to zero. And we find that we get to solutions actually equals zero and X equals plus or minus three solutions X equals zero and actually plus or minus squared of natural log of two. Yeah. So when actually is cereal, we can just plug that back into here and we get that the function is to and when x is closer minus squared in natural law, go to we get one half plus the natural log of two. So this would be maximum in this would be a minimum. So we have maximum here and minimums here. So if we use the ground multipliers, you can just set up our function. I called K. Take the partial subject the X perspective why? Inspector lambda? You have to make 10. And again this is pretty non linear. But we can, you know, do start doing some eliminations, you know? So for Salem right here, I'm not sure exactly how I did this one. And then so I'll put that into here and then probably solve for why. Yeah, plug everything back into here. And then we get a function of X alone. And we can figure out where that has zeros. It turns out that we have a zero win one X zero. Okay, so when actually zero, we can see this is zero. Um This is lambda and then we have why this is one minus Y. So that why has to be one financial. We got one here and then we could then finally under every one or two. But we can see that that is this point here, right? X zero. This is one by is one. So the other solution is again we can find that it's plus or minus squared of um land the school a square root of lateral log too. We plug that back into here and we just get y equals two in each case. So we can see that we have the same the same solutions and we'll have the same values for the function. Uh Yeah, I think I have anything else really just say about them anyway, you get the same result. Which way is easier, depends on, you know? Um Yeah, depend on the problem and how comfortable you are with uh algebra and calculus.


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