\(\newcommand{\W}[1]{ \; #1 \; }\) \(\newcommand{\R}[1]{ {\rm #1} }\) \(\newcommand{\B}[1]{ {\bf #1} }\) \(\newcommand{\D}[2]{ \frac{\partial #1}{\partial #2} }\) \(\newcommand{\DD}[3]{ \frac{\partial^2 #1}{\partial #2 \partial #3} }\) \(\newcommand{\Dpow}[2]{ \frac{\partial^{#1}}{\partial {#2}^{#1}} }\) \(\newcommand{\dpow}[2]{ \frac{ {\rm d}^{#1}}{{\rm d}\, {#2}^{#1}} }\)
lp_box¶
View page sourceabs_normal: Solve a Linear Program With Box Constraints¶
Syntax¶
lp_box
(Prototype¶
template <class Vector>
bool lp_box(
size_t level ,
const Vector& A ,
const Vector& b ,
const Vector& c ,
const Vector& d ,
size_t maxitr ,
Vector& xout )
Source¶
This following is a link to the source code for this example: lp_box.hpp .
Problem¶
We are given \(A \in \B{R}^{m \times n}\), \(b \in \B{R}^m\), \(c \in \B{R}^n\), \(d \in \B{R}^n\), This routine solves the problem
Vector¶
The type Vector is a
simple vector with elements of type double
.
level¶
This value is less that or equal two.
If level == 0 ,
no tracing is printed.
If level >= 1 ,
a trace of the lp_box
operations is printed.
If level >= 2 ,
the objective and primal variables \(x\) are printed
at each simplex_method iteration.
If level == 3 ,
the simplex tableau is printed at each simplex iteration.
A¶
This is a row-major representation of the matrix \(A\) in the problem.
b¶
This is the vector \(b\) in the problem.
c¶
This is the vector \(c\) in the problem.
d¶
This is the vector \(d\) in the problem. If \(d_j\) is infinity, there is no limit for the size of \(x_j\).
maxitr¶
This is the maximum number of newton iterations to try before giving up on convergence.
xout¶
This argument has size is n and the input value of its elements does no matter. Upon return it is the primal variables \(x\) corresponding to the problem solution.
ok¶
If the return value ok is true, an optimal solution was found.
Example¶
The file lp_box.cpp contains an example and test of
lp_box
.