Linear Programming double inequalities

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Homework Help Overview

The discussion revolves around finding the dual of a linear programming problem involving double inequalities, specifically the form -d ≤ Ax - b ≤ d, with constraints on x being non-negative. Participants are exploring the conversion of this problem into a canonical form suitable for dual formulation.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the challenges of converting the given double inequality into a canonical linear programming form. There are attempts to simplify the process and questions about combining separate matrix inequalities into a single expression.

Discussion Status

Some participants have offered guidance on rewriting the inequalities and have confirmed the validity of proposed approaches. There is an ongoing exploration of different interpretations and methods to combine the inequalities effectively.

Contextual Notes

Participants are working within the constraints of standard linear programming formulations and are questioning the assumptions related to the structure of the inequalities and the definitions of the matrices involved.

zcd
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Homework Statement


Find the dual of
[tex]-d \leq Ax-b \leq d[/tex]
[tex]x \geq 0; c \cdot x = min[/tex]
where A is mxn matrix and [tex]x,d,b \in \mathbb{R}^n[/tex]

Homework Equations


dual of canonical is of the form
maximize [tex]b \cdot y[/tex]
[tex]A^{T}y \leq[/tex]
where [tex]y \in \mathbb{R}^m[/tex]

The Attempt at a Solution


I tried converting it to the canonical LP and then applying transpose to A, but it turned out to be a huge mess; is there a simpler way?
 
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zcd said:

Homework Statement


Find the dual of
[tex]-d \leq Ax-b \leq d[/tex]
[tex]x \geq 0; c \cdot x = min[/tex]
where A is mxn matrix and [tex]x,d,b \in \mathbb{R}^n[/tex]

Homework Equations


dual of canonical is of the form
maximize [tex]b \cdot y[/tex]
[tex]A^{T}y \leq[/tex]
where [tex]y \in \mathbb{R}^m[/tex]

The Attempt at a Solution


I tried converting it to the canonical LP and then applying transpose to A, but it turned out to be a huge mess; is there a simpler way?

Do you know how to write the dual of an LP of the form
[tex]\min \; c \cdot x[/tex]
s.t.
[tex]F x \geq f[/tex]
[tex]x \geq 0 ?[/tex]
Just re-write your LP in that form and use what you already know.

RGV
 
How would I combine two separate matrix inequalities to one?
[tex]Ax \geq b-d[/tex]
[tex]-Ax \geq -b-d[/tex]
 
Last edited:
zcd said:
How would I combine two separate matrix inequalities to one?
[tex]Ax \geq b-d[/tex]
[tex]-Ax \geq -b-d[/tex]

You've just done it. Can't you see what the matrix F is?

RGV
 
My guess would be to make a block matrix on the left with A and -A vertically and a block vector with b-d and -b-d on the right. Could it be this simple?
 
zcd said:
My guess would be to make a block matrix on the left with A and -A vertically and a block vector with b-d and -b-d on the right. Could it be this simple?

Yes.

RGV
 

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