拉格朗日松弛（一）——理论及算法

背景

Marshall L.Fisher在1981年发表在《Management Science》上的《The Lagrangian Relaxation Method for Solving Integer Programming Problems》¹在2004年被评为“《Management Science》首个50年中十大最具影响力的主题之一”。

在中国知网中，以“拉格朗日松弛”作为关键词在“篇关摘”选项下检索，并对结果进行计量可视化分析可得到如下的图表。

从发文量这一指标可以看到，近10年涉及这一方法的论文显著增多。

从主要主题分布、次要主题分布中可以看到，在理论层面，该方法与动态规划、整数规划、凸优化和启发式算法等知识都有较为紧密的联系。在应用层面，该方法在机组组合、资源分配、调度、供应链和设施选址等领域都有应用。

拉格朗日松弛——理论

理论和算法部分内容参考资料²进行介绍，主要介绍拉格朗日松弛的思想、重要结论和拉格朗日松弛算法，结论的证明和更多内容可参考该书。除此之外，本文也吸纳了网上其他相关资料，帮助读者理解。

松弛的意思即为放松约束，对于一个标准化为求最小值的优化问题，放松约束会使得有可能得到目标函数值更小的解，换言之，松弛可以求得原问题的一个下界，这为评价其他算法的有效性提供了一种途径，也为原问题的求解提供了更多信息。

该方法常用在整数规划和混合整数规划中。

松弛方法主要有以下四种：线性规划松弛（将整数约束松弛至实数约束）、对偶规划松弛（求解对偶规划，根据弱对偶定理，求 $\text{The dual problem of max}$ provides to find$\text{A lower bound on the original problem of min}$ ), surrogate relaxation (reducing the number of constraints by combining multiple constraints, such as summation), and Lagrangian relaxation. This blog post introduces Lagrangian relaxation.

The basic principle of the Lagrangian relaxation method : absorb the constraints that make the problem difficult into the objective function, and keep the objective function linear, so that the transformed problem can be solved in polynomial time or although it cannot be solved in polynomial time, due to the scale ³ is small and can be solved quickly , thus providing help for the solution of the original problem.

As shown in Figure ⁴ above (the thin line red box should be " $\ge$ " and "$\le$ ", the thick red box should be "lower", "lower bound" means the lower bound), assuming that the constraints of a problem can be divided into two parts, that is,$A$ and$D$（$D_1$、$D_2$、$D_3$), their difference is $D$ is only related to a certain part of the variables, while$A$ is related to all variables, then$A$ is the above-mentioned "constraints that make the problem difficult" (also known as complex constraints). Lagrangian relaxation is to put this complex constraint into the objective function. About$Why\; \lambda$ is required to be non-negative, please refer to⁵. ^{Figure 6}belowcan help to understand.

Further, the objective function value $z_{LR}$is the original problem objective function value $A\; lower\; bound\; of\; z$ , since it needs to be as close as possible to the objective function value of the original problem, it can be used for the relaxed problem$z_{LR}$About $Find\; the\; maximum\; value\; of\; \lambda$ (constitute a dual problem, Dual Problem), and get the best value of all lower bounds$z_{LD}$(The red box in Figure ⁴ below should be $z_{LD}$).

The use of Lagrangian relaxation is as ^follows6 .

The above is the basic theory of Lagrangian relaxation, and some common deformations are introduced below.

The first is relaxation of equality constraints . This is exactly the same as "the dual variable corresponding to the equality constraint in the linear programming dual problem has no sign restriction" and "the coefficient corresponding to the equality constraint in the Kuhn-Tucker condition (KKT condition) has no sign restriction". The equality constraint relaxation corresponds to The coefficients of are also unsigned (subtraction of two non-negative coefficients yields a single unsigned coefficient).

The second is the Lagrangian decomposition . Looking back at question ⁴ above , we might as well set $n=3$ , consistent with the schematic diagram below. Then the xx

in the relaxed problem objective function$x$ according to$x_1$、$x_2$and $x_3$ 分解可以等价变换为如下的问题。
$$\begin{array}{ll}{z}_{LR}\left(l\right)=\text{minimize}& c_1^T{x}_1+c_2^T{x}_2+c_3^T{x}_3+l^T(A_1{x}_1+A_2{x}_2+A_3{x}_3-b)\\ \text{subject to}& D_1^T{x}_1\le e_1\\ & D_2^T{x}_2\le e_2\\ & D_3^T{x}_3\le e_3\\ & x_1,x_2,x_3\in {\mathbb{Z}}_{+}^n\end{array}$$Further, the above problem can be according to $x_1$、$x_2$and $x_3$Decomposed into three sub-problems, since the objective function and constraints are all about $x_1$、$x_2$and $x_3$separable, so this step is an equivalent transformation.

子问题一：
$$\begin{array}{ll}{z}_{LR1}\left(l\right)=\text{minimize}& c_1^T{x}_1+l^T(A_1{x}_1-b)\\ \text{subject to}& D_1^T{x}_1\le e_1\\ & x_1\in {\mathbb{Z}}_{+}^n\end{array}$$子问题二：
$$\begin{array}{ll}{z}_{LR2}(\lambda )=\text{minimize}& c_2^T{x}_2+{\lambda }^T{A}_2{x}_2\\ \text{subject to}& D_2^T{x}_2\le e_2\\ & x_2\in {\mathbb{Z}}_{+}^n\end{array}$$子问题三：
$$\begin{array}{ll}{z}_{LR3}(\lambda )=\text{minimize}& c_3^T{x}_3+{\lambda }^T{A}_3{x}_3\\ \text{subject to}& D_3^T{x}_3\le e_3\\ & x_3\in {\mathbb{Z}}_{+}^n\end{array}$$The objective function value of the decomposed sub-problem has the following relationship with the original problem objective function value.
$$z_{LR1}\left(l\right)+z_{LR2}\left(l\right)+z_{LR}\left(l\right)=z_{LR}\left(l\right)\le The\; z-$$ dual problem is as follows.
$$z_{LD}=\text{max}{z}_{LR1}\left(l\right)+z_{LR2}\left(l\right)+z_{LR}\left(\lambda \right)$$ It can be seen thaton the basis of Lagrangian relaxation, Lagrangian decomposition splits the original problem into sub-problems without complex constraints, so that the original problem can be solved by solving easier sub-problems approximate.

Lagrangian relaxation - algorithm

The Lagrangian relaxation algorithm consists of two parts: on the one hand, it provides the lower bound of the original problem (solving $z_{LD}$), on the other hand, it evolves into a Lagrangian relaxation heuristic algorithm.

subgradient optimization algorithm

A subgradient is very similar to a gradient, a subgradient is an extension of the gradient when the function is not differentiable at some point. The subgradient is introduced because $z_{LR}\left(\lambda \right)$ is a function about$The\; piecewise\; linear\; function\; of\; \lambda$ , which will be explained later.

According to the numbers in ^{2 , the relevant definitions and theorems are introduced as follows, and the detailed proof can be referred to.}

Definition 7.4.1 Function $g:{\mathbb{R}}^m\to \mathbb{R}$ 满足 $$\begin{array}{l}g(\alpha x^1+(1-a)x^2)\ge \alpha g(x^1)+(1-a)g(x^2),\forall x^1,x^2\in {\mathbb{R}}^m,0\le a\le 1\end{array}$$Then say $g\left(x\right)$ is a concave function.

Theorem 7.4.1 If the problem after Lagrangian relaxation ( $z_{LR}$) set of feasible solutions $Q$ is a set of finite integer points, then the objective function$$\begin{array}{l}{z}_{LR}\left(l\right)=\text{min}c{}^{T}x+l^T(Ax-b)\end{array}$$is a concave function. (proved by definition)

Theorem 7.4.2 Differentiable function g ( x ) g(x)The necessary and sufficient condition for$g$$($$x$$)$$\forall x^{\ast }\in {\mathbb{R}}^m$，存在$s=(s_1,s_2,\dots ,s_m{)}^T\in R^m$使得 $$\begin{array}{l}g(x^{\ast })+s^T(x-x^{\ast })\ge g(x),\forall x\in {\mathbb{R}}^m.\end{array}$$(analogous to the first-order conditional understanding of convex functions)

^{Figure 1} below is a schematic diagram of the conclusion of Theorem 7.4.1 ( uu in the figure$u$ means$\lambda$ ), with$The\; variation\; of\; \lambda$ ,$z_{LR}\left(\lambda \right)$ at different points$x$（$x^1$、$x^2$、$x^3$、$x^4$ ) reaches the minimum value, and when$When\; x$ is constant,$z_{LR}\left(\lambda \right)$ about$\lambda$ is a linear function, so in summary,$z_{LR}\left(\lambda \right)$ about$\lambda$ is a piecewise linear function, which is also a concave function. Each straight line represents the minimum value of$Point\; x$ remains the same.

Definition 7.4.2 If the function $g:{\mathbb{R}}^m\to R$ is a concave function, and at the point$x^{\ast }\in R^{At\; m}$ , the vector$s\in R^m$ 满足 $$\begin{array}{l}g(x^{\ast })+s^T(x-x^{\ast })\ge g(x),\forall x\in {\mathbb{R}}^m\end{array}$$Then say $s\in R^m$ is the function$g\left(x\right)$ at$x^{}$A subgradient at${}^{\ast \; .}$$g\left(x\right)$ at$x^{}$The set of all subgradients at${}^{\ast }$$\partial g(x^{\ast })$. (Intuitively, in the case of a one-variable function, the subgradient can be a function at$x^{}$Any value between the left and right derivatives at${}^{\ast \; .}$)

Theorem 7.4.3If $g\left(x\right)$ is a concave function,$x^{\ast }$ 为 max  { g ( x ) ∣ x ∈   R m } \text{max}\space\{g(x)|x\in\ \Bbb{R^m}\} max { g(x)∣x∈ RThe sufficient and necessary condition for the optimal solution of${}^m$$\}$$0\in \partial g(x^{\ast })$。

subgradient optimization algorithm

1. STEP1: Choose an initial Lagrangian multiplier $l^1$ , juxtapose$t=1$.
2. STEP2: λ t \lambda^t for this iteration$l^t$ , solve the minimization problem in the inner layer of the dual problem, and start from$\partial z_{LR}\left(l^t\right)$choose a subgradient$s^t$ ; if$s^t=0$ , then$l^t$ reaches the optimal solution and stops the calculation; otherwise$l^{t+1}=\text{max}\{min{}^t+i_t{s}^t,0\}$ , set$t=t+1$ , repeat STEP2.

It is impossible to iterate infinitely many times in practical applications, so $i_t$The determination method and the condition under which the algorithm stops.

• $i_t$Determination method
  The purpose of the actual calculation is to obtain an acceptable lower bound as soon as possible, so a heuristic method is often used for determination. There are mainly two types of methods.
  The first type of method is to make $i_t$Decreases at an exponential rate with fewer iterations. The update formula is as follows.
  $$i_t=i_0{r}^t,0<r<1$$ The main idea of ​​the second type of method is to use the difference between the upper and lower bounds of the dual problem to correct$i_t$speed of change. The update formula is as follows.
  $$i_t=\frac{z_{UP}(t)-z_{LB}(t)}{||s^t|{|}^2}{b}_t$$where $0\le b_t\le 2$ , generally take$b_0=2$。当 $z_{LR}\left(\lambda \right)$ rises,$b_t$Invariant, when $z_{LR}\left(\lambda \right)$ When there is no change in a given number of steps, then$b_t$ 减半。$z_{UP}\left(t\right)$ is an upper bound of the optimal objective value of the original problem (and thus the upper bound of the dual problem), which can be determined or estimated by a feasible solution. $z_{UP}\left(t\right)$ can be changed with$The\; change\; of\; t$ is gradually corrected. $z_{LB}(t)$ 是 $z_{LR}\left(l^t\right)$, generally take$z_{LB}(t)=z_{LR}\left(l^t\right)$, and sometimes only take a fixed value for simplicity of calculation.
• Conditions for Algorithm Stop
  There are four main conditions for algorithm stop.
  The first is that the number of iterations does not exceed $T.$ _ This is the simplest principle, and it is easy to control the complexity of the calculation, but the quality of the solution cannot be guaranteed.
  The second is$||s^t||\le ϵ$ ($ϵ$ is a small positive number), this is the ideal state$s^t=$An approximation of$0\; .$
  The third is$z_{UP}(t)=z_{LB}\left(t\right)$ , which indicates that the optimal solution to the original problem has been found,$z=z_{UP}\left(t\right)=z_{LB}\left(t\right)$ .
  The fourth type is$l^t$ 或 $z_{LR}\left(l^t\right)$does not change more than a given value within the specified number of steps, at this time it can be considered that the target value will not change, so the calculation is stopped.
  In a specific application, one of the above stop conditions can be used, or a combination of them can be used.

Lagrangian relaxation heuristic algorithm

^{There is such a conclusion 4} in the content mentioned above (the red box should be $z_{LD}$): Comparing the original problem on the left and the dual problem on the right, it can be found that the feasible region is expanded when solving the dual problem , so the better solution we get when solving the dual problem (given by the subgradient optimization algorithm) may not be The feasible solution of the original problem. At this time, the heuristic method is usually used to modify the solution to the feasible solution of the original problem according to the characteristics of the problem. The whole process constitutes the Lagrangian relaxation heuristic algorithm.

References

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1. Fisher, M.L. 1981. The Lagrangian Relaxation Method for Solving Integer Programming Problems. Management Science. 27(1) 1-18. ↩︎ ↩︎

2. Xing Wenxun, Xie Jinxing. Modern Optimal Calculation Methods (Second Edition) [M]. Tsinghua University Press, 2005:210-242. ↩︎ ↩︎

3. Wang Yuan. Analysis of Integer Programming by Lagrangian Relaxation (with Python Code Example) ↩︎

4. TransNET. Lagrangian relaxation problem ↩︎ ↩︎ ↩︎ ↩︎

5. I would like to ask a question, why is the multiplier of the Lagrange multiplier method greater than 0? ↩︎

6. Lagrange relaxation PPT ↩︎ ↩︎