InfSolve

Purpose

Find a constrained minimax solution with the use of any suitable TOMLAB solver.

infSolve solves problems of the type:

${\begin{array}{cccccc}\min \limits _{x}&\max r(x)\\{\mbox{subject to}}&x_{L}&\leq &x&\leq &x_{U}\\&b_{L}&\leq &Ax&\leq &b_{U}\\&c_{L}&\leq &c(x)&\leq &c_{U}\\\end{array}}$

where $x,x_{L},x_{U}\in \mathbb {R} ^{n}$ , $r(x)\in \mathbb {R} ^{N}$ , $c(x),c_{L},c_{U}\in \mathbb {R} ^{m_{1}}$ , $b_{L},b_{U}\in \mathbb {R} ^{m_{2}}$ and $A\in \mathbb {R} ^{m_{2}\times n}$.

Calling Syntax

Result=infSolve(Prob,PriLev)

Inputs

Prob	Problem description structure. Should be created in the cls format. infSolve uses two special fields in Prob:
	SolverInf	Name of solver used to solve the transformed problem. Valid choices are conSolve, nlpSolve, sTrustr and clsSolve. If TOMLAB /SOL is installed: minos, snopt, npopt.
	InfType	1 - constrained formulation (default). 2 - LS penalty approach (experimental).
	The remaining fields of Prob should be defined as required by the selected subsolver.
PriLev	Print level in infSolve. =0 Silent except for error messages. >0 Print summary information about problem transformation. Calls PrintResult with specified PriLev. =2 Standard output from PrintResult (default).

Outputs

Result

Structure with results from optimization. Output depends on the solver used. The fields x_k, r_k, J_k, c_k, cJac, x_0, xState, cState, v_k are transformed back to match the original problem. g_k is calculated as J_kT · r_k. The output in Result.Prob is the result after infSolve transformed the problem, i.e. the altered Prob structure

Description

The minimax problem is solved in infSolve by rewriting the problem as a general constrained optimization problem. One additional variable $z\in \mathbb {R}$, stored as $x_{n+1}$ is added and the problem is rewritten as:

${\begin{array}{cccccc}\min \limits _{x}z\\\\{\mbox{subject to}}&x_{L}&\leq &(x_{1},x_{2},\ldots ,x_{n})^{T}&\leq &x_{U}\\&-\infty &\leq &z&\leq &\infty \\&b_{L}&\leq &Ax&\leq &b_{U}\\&c_{L}&\leq &c(x)&\leq &c_{U}\\&-\infty &\leq &r(x)-ze&\leq &0\\\end{array}}$

where $e\in \mathbb {R} ^{N},\;e(i)=1\ \forall i$.

To handle cases where an element $r_{i}(x)$ in $r(x)$ appears in absolute value: $\min \max |r_{i}(x)|$, expand the problem with extra residuals with the opposite sign: $[r_{i}(x);-r_{i}(x)]$

Examples

minimaxDemo.m.

See Also

clsAssign.