9 votes

"Partial" Lagrangian Dual in LP

This is called Lagrangian relaxation, no matter what subset of constraints you choose to dualize.
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  • 22.2k
8 votes
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"Partial" Lagrangian Dual in LP

Based on the mentioned references, suppose the primal problem is: \begin{align} \begin{array}{cl} \underset{}{\text{minimize}} & c x \\ \text{subject to} & Ax = a \\ & Dx \leq e \\ & x ...
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  • 5,833
6 votes
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Simple nonlinear programming using convexity analysis and KKT

You should use the KKT conditions (turning the sign restrictions on $x$ into constraints), but it turns out they will not affect the results. First, let me point out that $\partial L/\partial x_1$ is ...
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6 votes

Applicability of Lagrange Multipliers in the analysis of large-scale MILPs?

For an application to very large set covering problems you can see e.g. here This approach can be extended (somehow) to general MILPs and allows one to quickly find a “core” set of variables defining ...
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6 votes

Applicability of Lagrange Multipliers in the analysis of large-scale MILPs?

I am not aware of specific algorithmic methods for MIPs that use Lagrangian multipliers directly. However, as for the interpretation of the solution of a MIP: probably one of the nicer applications I ...
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  • 3,034
5 votes
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Simplex Multiplier

It is explained in this link as: Simplex multipliers are essentially the shadow prices associated with a particular basic solution. Those are the multiples of their initial system of equations such ...
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  • 8,290
5 votes

Applicability of Lagrange Multipliers in the analysis of large-scale MILPs?

Lagrangian relaxation is extremely common as an algorithmic method to solve facility location problems. Because many "minisum"-type problems ($p$-median, uncapacitated facility location problem, etc.) ...
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5 votes
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Augmented Lagrangian Function for Semidefinite Programming Problems

My way of reading it is $\langle X, \mathcal{A}^*(y)+S-C\rangle = \langle X, \mathcal{A}^*(y)-C\rangle + \langle X,S \rangle$. The first term is your standard inner product between dual variable and ...
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4 votes

Augmented Lagrangian Function for Semidefinite Programming Problems

There's a good discussion of this in Convex Optimization by Stephen Boyd and Lieven Vandenberghe. See section 5.9. With an ordinary scalar inequality constraint: $f_{i}(x) \leq 0$, you'll have a term ...
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3 votes
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Why is the Lagrange Multiplier not equal the Shadow Price (Excel solver, Matlab linprog, Gurobi)?

If you look at the "allowable decrease" in the RHS of the highlighted constraint, it's zero. A number of the binding constraints have either allowable increase or allowable decrease zero. ...
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3 votes
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Method of Multipliers: Why is the next iterate always dual feasible?

By dual feasibility, all that the authors mean is that $(x^{k+1}, \lambda^{k+1})$ in Equation (1) satisfies the stationarity equation in the KKT optimality conditions for the problem shown at the top ...
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  • 1,311
2 votes

Recovering Primal Solution from Dual solution

A proof that $x^*$ is optimal can be achieved by a Taylor expansion around $x^*$ at optimality. Observe that $$f(x^*+p) = f(x^*) + p^\top f'(x^*) + \frac12p^\top f''(x^*)p$$ Note that at a local ...
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2 votes

KKT conditions analysis for binary constraints

Binary (Boolean) values are integer values. Therefore, optimization problems with boolean constraints are either integer programming or mixed integer programing (MIP). Generally, there is no easy ...
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1 vote

How to find the optimal solution of a convex program given all KKT points?

This sounds like parametric programming to me. In short, you can calculate a set of regions, typically called "critical regions", within which the same set of constraints is active. This in ...
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  • 3,034
1 vote

Linear Relaxation of Boolean Constraint for Solving Integer Linear Program Using KKT

Minimize $x^2$ where $1 \le x \le 2$. \begin{aligned} \min_{x} \quad & f(x)\\ \textrm{s.t.} \quad & h_{1}(x) \le 0\\ &h_{2}(x) \le 0 \\ \end{aligned} where \begin{align} f(x) &= x^...
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