[Optimization] Convex Optimization Problem
1.1 Basic terminology
\[minimize \quad f_0(x) \\ subject \, to \quad f_i(x) \leq 0, \quad i = 1, ... , m \\ \qquad \qquad \quad \, h_i(x) = 0, \quad i = 1, ... , p\]- Optimization variable := $x$
- Optimization function(or cost function) := $f_0(x)$
- Inequality constraints := $f_i(x) \leq 0$
- Inequality constraint functions := $f_i : R^n \rightarrow R$
- Equality constraints := $h_i(x) = 0$
- Equality constraint functions := $h_i : R^n \rightarrow R$
- Domain of the optimization problem := the set of points for which the objective and all constraint functions are defined
- Feasible point $x \in D$ := a point satisfing the constraints $f_i(x) \leq 0, i = 1, … , m, and \; h_i(x) = 0, i = 1, … , p$
- Feasible problem := the problem in which there exists at least one feasible point and infeasible otherwise
- Feasible set(or constraint set) := the set of all feasible points
- Optimal value := $p*$
- The problem is unbounded below := if there are feasible points $x_k$ with $f_0(x_k) \rightarrow -\infty$, then $p* \rightarrow -\infty$
Optimal and locally optimal points
- Optimal point := $x$ if $x$ is feasible and $f_0(x) = p$
- Optimal set := the set of all optimal points $X_{opt}$
- Optimal value is attained or achieved(and the problem is solvable) : there exists an optimal opint for the optimization problem
- Optimal value is not attained or not achieved : $X_{opt}$ is empty
- $\epsilon$ - suboptimal point : a feasible point $x$ with $f_0(x) \leq p* + \epsilon$
- $\epsilon$ - suboptimal set : the set of all $\epsilon$ - suboptimal points
- Locally optimal : a feasible point $x$ in which there is an $R > 0$ such that
or, in other word, $x$ solve the optimization problem
\[\text{minimize} \;f_0(z) \\ \text{subject to} \;f_i(z) \leq 0, \;\; i = 1, ... ,m \\ h_i(z) = 0, \;\; i = 1, ... ,p \\ ||z - x|| \leq R\]Feasibility problems
- Feasibility problem : a problem in which if the objective function is identically zero, the optimal value is either zero or inifinity
Example 4.1 - We illustrate these definitions with a few simple unconstrained optimization problems with variable $x \in R$, and $\textbf{dom} f_0 = R++$
- $f_0(x) = \dfrac{1}{x} : p* = 0$, but the optimal value is not achieved.
- $f_0(x) = -\log x : p* = \infty$, so this problem is unbounded below.
- $f_0(x) = x \log x : p* = -1/e$, achieved at the unique optimal point $x* = 1/e$
1.2 Expressing problems in standard form
In the standard form problem we adopt the convention that the righthand sied of the inequlity and equlity constraints are zero.
Example 4.2 - Box constraints. Consider the optimization problem \(\text{minimizate} \;f_0(x) \\ \text{subject to} \; i_i \leq x_i \leq u_i, \;\; i = 1, ... , n,\) where $x \in R$ is variable. The constraints are called variable bounds (since they give lower and upper bounds for each $x_i$) or box constraints (since the feasible set is a box)
We can express this problem in standard form as
$$
\text{minimizate} \;f_0(x)
\text{subject to} \; i_i - x_i \leq 0 \;\; i = 1, … , n,
x_i - u_i \leq 0 \;\; i = 1, … , n,
\(There are 2n inequality constraint functions:\) f_i(x) = i_i - x_i \;\; i = 1, … , n \(and\) f_i(x) = x_i - u_i \leq 0 \;\; i = n, … , 2n, $$
Maximization problems
We can solve the maximization problem by minizing the function $-f_0$ subject ot the constraints
- Utility(or satisfaction levbel) : an objective in the maximization problem
Maximization problem
\[\text{maximize} \;\; f_0(x) \\ \text{subject to} \;\; f_i(x) \leq 0, \;\;\; i = 1, ..., m \\ h_i(x) = 0 \;\;\; i = 1, ..., p\] \[p* = \sup\{f_0(x) | f_i(x) \leq 0, i = 1, ..., m, \;\; h_i(x) = 0, i = 1, ..., p\}\]1.3 Equivalent problems
Tow problems are equivalent : from a solution of one, a solution of the other is readily found, and vice versa
1.4 Parameter and oracle problem descriptions
2. Convex optimization
2.1 Convex optimization problems in standard form
\[minimize \quad f_0(x) \\ subject \, to \quad f_i(x) \leq 0, \quad i = 1, ... , m \\ \qquad \qquad \quad \, a_i^T x = b_i, \quad i = 1, ... , p\]Comparing with the general standard form problem, the convex problem hase three additional requirements
- the objective functions must be convex.
- the inequality constraint functions must be convex.
- the equality constraint functions $h_i(x) = a_i^Tx - b_i$ must be affine
The feasible set of a convex optimization problem is convex, since it is the intersection of the domain of the problem which is a convex set, with m sublevel sets and p hyper planes. Thus, in aconvex optimization problem, we minimize a convex objective function over a convex set.
\[D = \cap_{i=0}^m \textbf{dom}f_i\]Concave maximization problems
\[\text{maximizate} \quad f_0(x) \\ \text{subject} \, to \quad f_i(x) \leq 0, \quad i = 1, ... , m \\ \qquad \qquad \quad \, a_i^T x = b_i, \quad i = 1, ... , p\]This concave optimization problem is readily solved by minimizing the convex objective function $-f_0$
Abstract form convex optimization problem
It is important to note a subtlety in our definition of convex optimization problem.
This problem is not a convex optimization problem in standard form since the equality function $h_1$ is not affine, and the inequality constraint function $f_1$ is not convex.
\[\text{minimize} \quad f_0(x) = x_1^2 + x_2^2 \\ \text{subject} \, to \quad f_1(x) = \dfrac{x_1}{1 + x_2^2} \leq 0 \\ \qquad \qquad \quad \, h_1(x) = (x_1 + x_2)^2 = 0\]The problem is readily reformulated as
\[\text{minimize} \quad f_0(x) = x_1^2 + x_2^2 \\ \text{subject} \, to \quad f_1(x) = x_1 \leq 0 \\ \qquad \qquad \quad \, h_1(x) = (x_1 + x_2) = 0\]2.2 Local and global optima
A fundemental property of convex optimization problems is that any locally optimal point is also globally optimal
Pf
Supoose that $x$ is locally optimal for a convex optimization - $x$ is feasible and
\[f_0(x) = \text{inf} \; \{f_0(z) | z \;\text{feasible}, ||z - x||_2 \leq R\}\]for some $R > 0$.
Suppose that $x$ is bot globally optimal - there is a feasible $y$ such that $f_0(y) < f_0(x)$
Evidently $ | y - x | _2 > R$, since otherwise $f_0(x) \leq f_0(y)$. |
Consider the point $z$ given by
\[z = (1 - \theta) x + \theta y, \;\; \theta = \dfrac{R}{2||y - x||_2}.\]Then we have
\[||z - x||_2 = ||(1 - \dfrac{R}{2||y - x||_2}) x + \dfrac{R}{2||y - x||_2} y - x||_2 = \dfrac{y - x}{2||y - x||_2}R = \dfrac{R}{2}\], and by convexity of feasible set, $z$ is feasible.
By convexity of $f_0$ we have
\[f_0(z) = f_0((1-\theta)x + \theta y) \leq (1 - \theta)f_0(x) + \theta f_0(y) \leq (1 - \theta)f_0(x) + \theta f_0(x) = f_0(y)\]which contradicts.
Hence there exists no feasible $y$ with $f_0(y) < f_0(x)$, $x$ is globally optimal.
2.3 An optimality criterion for differentiable $f_0$
Suppose that the objective $f_0$ in a convex optimization problem is differentiable, so that for all $x, y \in \textbf{dom} f_0$
\[f_0(y) \geq f_0(x) + \nabla f_0^T(x)(y - x)\]Let X denote the feasible set, i.e.,
\[X = \{x | f_i(x) \leq 0, i = 1, ..., m, h_i(x) = 0 \}\]x is optimal if and only if $x \in X$ and
\[\nabla f_0(x)^T(y - x) \geq 0 \;\;\text{for all} \; y \in X\]Proof of optimal condition
First suppose $x \in X$ and satisfies
\[\nabla f_0(x)^T(y - x) \geq 0 \;\;\text{for all} \; y \in X\]Then if $y \in X$,
\[f_0(y) - f_0(x) \geq \nabla f_0^T(x)(y - x) \geq 0 \rightarrow f_0(y) \geq f_0(x)\]Conversely, suppose $x$ is optimal, but the conditon
\[\nabla f_0(x)^T(y - x) \geq 0 \;\;\text{for all} \; y \in X\]does not hold, i.e., for some $y \in x$ we have
\[\nabla f_0(x)^T(y - x) < 0\]Consider the point $z(t) = ty + (1-t)x$, where $t \in [0, 1]$ is a parameter.
Since $z(t)$ is on the line segment between $x$ and $y$, and the feasible set is convex, $z(t)$ is feasible.
For small positive $t, f_0(z(t)) \leq f_0(x)$, which will prove that $x$ is not optimal.
Unconstrainted problems
\[x \;\text{is optimal} \quad \Leftrightarrow \quad\nabla f_0(x) = 0\]Suppose this optimality condition, which means that $x \in \textbf{dom} f_0$, and for all feasible $y \quad\nabla f_0(x)(y - x) \geq 0$
Since $f_0$ is differentiable, its domain is open, so all y sufficiently close to $x$ are feasible.
Let us take $y = x - t\nabla f_0(x)$, where $t \in R$ is parameter.
For $r$ small and positive, $y$ is feasible, and so
\[\nabla f_0(x)(y - x) = -t||\nabla f_0(x)||^2 \geq 0\]from which we conclude $f_0(x) = 0$
Example 4.5 Unconstrainted quadratic optimization. Consider the problem of minimizing the quadrtic function
\[f_0(x) = \dfrac{1}{2}x^TPx + q^Tx + r\]where $P \in S_+^n$ (which makes $f_0$) convex.
The necessary and sufficient condition for $x$ to be a minimizer of $f_0$ is
\[\nabla f_0(x) = Px + q = 0\]- If $q \notin R(p)$, then there is no solution. In this case $f_0$ is unbounded below.
- If $P > 0$ (which is the condition for $f_0$ to be strictly convex), then there is a unique minimizer, $x* = - P^{-1}q$
- If
Problems with equality constraints only
Minimization over the nonnegative orthant
2.4 Equivalent convex problems
2.5 Quasiconex optimization
Locally optimal solutions and optimality conditions
Let X denote the feasible set for the quasiconvex optimization problem. x is optimal if
\[x \in X, \;\; \nabla f_0(x)^T(y - x) > 0 \;\text{for all} \; y \in X\{x\}\]Reference
Stephen Boyd - Convex Optimization