3

Let's look at an example. Suppose you have four cities, A, X, Y, and Z, and A is at distance $1$ from each of the others, while the distance between any two of X, Y, Z is $100$. Then of course a solution to TSP starting at A is given by A to X to A to Y to A to Z to A – let's write this as AXAYAZA, which has length $6$, and visits A multiple times. Now let'...


2

Without derivatives: $$2+x^2(3-x)=4x^2-x^3+2-x^2=x^2(4-x)+(4-x)(4+x)-14\geq-14.$$ The equality occurs for $x=4$, which says that we got a minimal value.


2

Also, we can use Shur here. Indeed, we need to prove that $$a^2+b^2+c^2+2\sqrt{3abc(a+b+c)}\geq\frac{1}{2}(a+b+c)^2$$ or $$a^2+b^2+c^2+4\sqrt{3abc(a+b+c)}\geq2(ab+ac+bc).$$ By AM-GM $$4\sqrt{3abc(a+b+c)}\geq4\sqrt{3abc\cdot3\sqrt[3]{abc}}=12\sqrt[3]{a^2b^2c^2}>3\sqrt[3]{a^2b^2c^2}$$ and it's enough to prove that: $$a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2}\geq2(...


2

Let $a+2b+c = x $ etc. $$ f(a,b,c)=\frac{x+y+z}{4}-\frac{9(x+y+z)}{4}(\frac{1}{(x)^2}+\frac{1}{(y)^2}+\frac{1}{(z)^2})+\frac{9z}{(x)^2}+\frac{9x}{(y)^2}+\frac{9y}{(z)^2} $$ By the rearrangement inequality, $\frac{9z}{(x)^2}+\frac{9x}{(y)^2}+\frac{9y}{(z)^2}\ge \frac{9x}{(x)^2}+\frac{9y}{(y)^2}+\frac{9z}{(z)^2} $ as the minimum occurs when numerators and ...


2

Local minima of a convex function are global minima. Hence if $[0,1]$ does not contain a local minimum, it must be monotonic on the interval and so the constrained minimum will occur at $0$ or $1$.


2

Hint:At first place you can ignore $y$ it dose not play role in problem P. Secondly: You need only consider cases $\mu , \lambda \geq 0$ and what if $ \mu = 0 $ ? $D$ is an open set and you do not need incorporate it as a sign inequality constraint. Even more the standard definition of Lagrange Duality does not involve any strict inequality.


2

If $A$ is invertible but not positive definite, it has an eigenvalue $\lambda < 0$. If $v$ is the corresponding eigenvector, then $\lim_{c \to \infty} (cv)^T A (cv) + b^T(cv) = \lim_{c \to \infty} \lambda c^2 + c (b^Tv) = -\infty$. If $A$ is not symmetric, you can replace it with $0.5(A+A^T)$ and get the same function values (and same optimum).


2

There is only one critical point but it is indefinite. This is seen for example by eliminating $d$ and studying the Hessian of the resulting unconstrained problem (see below for the details). Therefore there are no local extrema. Your function is unbounded. For example, if we set $c=0$, then the constraint is satisfied by $d=63-a-b$, but this will not ...


2

Given the matrix-valued function $$F=(XAX^T)^{-1}$$ its differential is easy to find. $$\eqalign{ I &= FF^{-1} \cr 0 &= F\,dF^{-1}+dF\,F^{-1} \cr dF &= -F\,dF^{-1}\,F \cr &= -F\,dX\,AX^TF - FXA\,dX^T\,F \cr }$$ A difficult issue is that the matrix-by-matrix gradient, i.e. $\frac{\partial F}{\partial X}$, is not a matrix but a 4th order ...


2

Sure. Take Newton's method on the function $f(x) = \frac{1}{x}$, on the domain $(0, \infty)$. Take a first estimate $x_0 = 1$, and define $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} = x_n - \frac{\frac{1}{x_n}}{-\frac{1}{x_n^2}} = 2x_n.$$ Inductively, we see $x_n = 2^n$. Note that $f(x_n) = \frac{1}{2^n} \to 0$, the infimal value of the function, but the ...


1

If you group all terms with $\xi_i$, you get $(C-\alpha_i-\mu_i)\xi_i$ in the dual objective, and you get $C-\alpha_i-\mu_i=0$ as a dual constraint. The term in the objective vanishes (and you can drop the constraint since you can always choose $\mu_i$ to satisfy it).


1

Let $x=\operatorname{vec}(X)$ and $y=\operatorname{vec}(Y)$. The constraints $A_iXe_i=Be_i$ and $CX^T=0$ can be written as $(e_i^T\otimes A_i)x=\operatorname{vec}(Be_i)$ and $(C\otimes I_m)x=0$ and they can be combined into a single linear equation $$ \underbrace{\pmatrix{ e_1^T\otimes A_1\\ \vdots\\ e_n^T\otimes A_n\\ C\otimes I_m}}_A\ x =\underbrace{\...


1

For the Hamiltonian approach you get $$ I(x,u,\lambda)=\int_0^1(x^2+xu+u^2+λ(2x+u-\dot x))dt=\int_0^1 (H(x,u,λ)-λ\dot x)\,dt $$ The optimal control then minimizes $H(x,u,λ)$ for fixed $u$ and $λ$. As this is a quadratic convex function, the minimum is easy to compute as $x+2u+λ=0\implies u=-\frac{x+λ}2$. The saddle point can be computed by setting the ...


1

The Weierstrass theorem asserts that if you minimize a continuous function over a closed and bounded set in $R^{n}$, then the minimum will be achieved at some point in the set. I read in a paper that if an objective function is convex a, using Weierstrass theorem, we can conclude that the optimal objective function is finite, and the optimal set is ...


1

Numerical recipes by Press, Teukolsky and two others does do the analysis including the effects of machine-epsilon rounding.


1

Real world Simplex codes allow variables to have a lower and upper bound. A non-basic variable can be at lower bound or at upper bound but usually not in between (there are some exceptions to this rule, sometimes these are called superbasic). Solvers typically will tell you in their solution not only "this variable is non-basic", but more precisely "this ...


1

Ok, so if you were talking about anything other than a scalar optimization, I would question it, but in this case I think that works. To formally prove it, your original problem is $$\begin{array}{ll} \min_x &f(x)\\ \mathrm{s.t.} &0 \leq x \leq 1.\end{array} \qquad (1) $$ Define $g^* = \frac{\partial f}{\partial x} (x^*)$ where $x^*$ is the ...


1

The problem is quasi-convex in that it is convex for every fixed $a$ and the feasible set decreases for increasing $a$. Hence it can be solved by bisection in $a$ Example code on bisection to explain strategy https://yalmip.github.io/example/decayrate/ If you use MATLAB/YALMIP, you can do it very easily using built-in functionality (here using mosek as SDP ...


1

By AM-GM and C-S we obtain: $$\sum_{cyc}\left(a+\frac{9}{(a+2b+c)^2}\right)\geq6\sqrt{\sum_{cyc}a\sum_{cyc}\frac{a}{(a+2b+c)^2}}\geq6\sum_{cyc}\frac{a}{a+2b+c}=$$ $$=6\sum_{cyc}\frac{a^2}{a(a+2b+c)}\geq\frac{6(a+b+c)^2}{\sum\limits_{cyc}a(a+2b+c)}=\frac{6(a+b+c)^2}{\sum\limits_{cyc}(a^2+3ab)}\geq\frac{9}{2},$$ where the last ineqyality it's just $\sum\...


1

Use a Quadratic Programming (QP) solver, of which there are many for almost any computing environment.. That will allow you to specify the nonnegativity constraint.If you need $x$ to be strictly positive, you will have to specify small_number and input constraint as $x \ge $ small_number. All QP solvers can handle this problem if the objective function is ...


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