Lagrange multipliers



In mathematical optimization problems, the method of Lagrange multipliers, named after Joseph Louis Lagrange, is a method for finding the extrema of a function of several variables subject to one or more constraints; it is the basic tool in nonlinear constrained optimization.

Simply put, the technique is able to determine where on a particular set of points (such as a circle, sphere, or plane) a particular function is the smallest (or largest).

More formally, Lagrange multipliers compute the stationary points of the constrained function. By Fermat's theorem, extrema occur either at these points, or on the boundary, or at points where the function is not differentiable.

It reduces finding stationary points of a constrained function in n variables with k constraints to finding stationary points of an unconstrained function in n+k variables. The method introduces a new unknown scalar variable (called the Lagrange multiplier) for each constraint, and defines a new function (called the Lagrangian) in terms of the original function, the constraints, and the Lagrange multipliers.

Introduction
Consider a two-dimensional case. Suppose we have a function $$f(x,y)$$ we wish to maximize or minimize subject to the constraint


 * $$g\left( x, y \right) = c,$$

where c is a constant. We can visualize contours of $$f$$ given by


 * $$f \left( x, y \right)=d_n$$

for various values of $$ d_n $$, and the contour of $$ g $$ given by $$ g ( x, y ) = c $$.

Suppose we walk along the contour line with $$ g = c $$. In general the contour lines of $$ f $$ and $$ g $$ may be distinct, so traversing the contour line for $$ g = c $$ could intersect with or cross the contour lines of $$ f $$. This is equivalent to saying that while moving along the contour line for $$ g = c $$ the value of $$f $$ can vary. Only when the contour line for $$ g = c $$ touches contour lines of $$ f $$ tangentially, we do not increase or decrease the value of $$ f $$ - that is, when the contour lines touch but do not cross.

This occurs exactly when the tangential component of the total derivative vanishes: $$df_\parallel = 0$$, which is at the constrained stationary points of $$f$$ (which include the constrained local extrema, assuming $$f$$ is differentiable). Computationally, this is when the gradient of $$f$$ is normal to the constraint(s): when $$\nabla f = \lambda \nabla g$$ for some scalar $$\lambda$$. Note that the constant $$\lambda$$ is required because, even though the directions of both gradient vectors are equal, the magnitudes of the gradient vectors are most likely not equal.

A familiar example can be obtained from weather maps, with their contour lines for temperature and pressure: the constrained extrema will occur where the superposed maps show touching lines (isopleths).

Geometrically we translate the tangency condition to saying that the gradients of $$ f $$ and $$ g $$ are parallel vectors at the maximum, since the gradients are always normal to the contour lines. Thus we want points $$(x,y)$$ where $$\nabla_{x,y} f = \lambda \nabla_{x,y} g$$, and, further, $$g(x,y) = c$$. To incorporate both these conditions into one equation, we introduce an unknown scalar, $$\lambda$$, and solve


 * $$ \nabla_{x,y,\lambda} F \left( x, y, \lambda \right)=0 $$

with
 * $$ F \left( x, y, \lambda \right) = f \left(x, y \right) + \lambda \left(g \left(x, y \right) - c \right), $$

and
 * $$ \nabla_{x,y,\lambda} = \left( \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial \lambda} \right). $$

Justification
As discussed above, we are looking for stationary points of $$f$$ seen while travelling on the level set $$g(x,y)=c$$. This occurs just when the gradient of $$f$$ has no component tangential to the level sets of $$g$$. This condition is equivalent to $$ \nabla_{x,y} f(x,y) = \lambda \nabla_{x,y} g(x,y) $$ for some $$\lambda$$. Stationary points $$(x,y,\lambda)$$ of $$F$$ also satisfy $$g(x,y) = c$$ as can be seen by considering the derivative with respect to $$\lambda$$.

Caveat: extrema versus stationary points
Be aware that the solutions are the stationary points of the Lagrangian $$F$$, and are saddle points: they are not necessarily extrema of $$F$$. $$F$$ is unbounded: given a point $$(x,y)$$ that doesn't lie on the constraint, letting $$\lambda \to \pm \infty$$ makes $$F$$ arbitrarily large or small. However, under certain stronger assumptions, as we shall see below, the strong Lagrangian principle holds, which states that the maxima of $$f$$ maximize the Lagrangian globally.

A more general formulation: The weak Lagrangian principle
Denote the objective function by $$f(\mathbf x)$$ and let the constraints be given by $$g_k(\mathbf x)=0$$, perhaps by moving constants to the left, as in $$h_k(\mathbf x)-c_k=g_k(\mathbf x)$$. The domain of f should be an open set containing all points satisfying the constraints. Furthermore, $$f$$ and the $$g_k$$ must have continuous first partial derivatives and the gradients of the $$g_k$$ must not be zero on the domain. Now, define the Lagrangian, $$\Lambda$$, as


 * $$\Lambda(\mathbf x, \boldsymbol \lambda) = f + \sum_k \lambda_k g_k.$$
 * $$k$$ is an index for variables and functions associated with a particular constraint, $$k$$.
 * $$\boldsymbol\lambda$$ without a subscript indicates the vector with elements $$\mathbf \lambda_k$$, which are taken to be independent variables.

Observe that both the optimization criteria and constraints $$g_k(x)$$ are compactly encoded as stationary points of the Lagrangian:


 * $$\nabla_{\mathbf x} \Lambda = \mathbf{0}$$ if and only if $$\nabla_{\mathbf x} f = - \sum_k \lambda_k \nabla_{\mathbf x} g_k,$$
 * $$\nabla_{\mathbf x}$$ means to take the gradient only with respect to each element in the vector $$\mathbf x$$, instead of all variables.

and


 * $$\nabla_{\mathbf \lambda} \Lambda = \mathbf{0}$$ implies $$g_k = 0.$$

Collectively, the stationary points of the Lagrangian,


 * $$\nabla \Lambda = \mathbf{0}$$,

give a number of unique equations totaling the length of $$\mathbf x$$ plus the length of $$\mathbf \lambda$$. This often makes it possible to solve for every $$x$$ and $$\lambda_k$$, without inverting the $$g_k$$. For this reason, the Lagrange multiplier method can be useful in situations where it is easier to find derivatives of the constraint functions than to invert them.

Interpretation of $$ \lambda_i $$
Often the Lagrange multipliers have an interpretation as some salient quantity of interest. To see why this might be the case, observe that:


 * $$\frac{\partial \Lambda}{\partial {g_k}} = \lambda_k.$$

So, λk is the rate of change of the quantity being optimized as a function of the constraint variable. As examples, in Lagrangian mechanics the equations of motion are derived by finding stationary points of the action, the time integral of the difference between kinetic and potential energy. Thus, the force on a particle due to a scalar potential, $$F=-\nabla V$$, can be interpreted as a Lagrange multiplier determining the change in action (transfer of potential to kinetic energy) following a variation in the particle's constrained trajectory. In economics, the optimal profit to a player is calculated subject to a constrained space of actions, where a Lagrange multiplier is the value of relaxing a given constraint (e.g. through bribery or other means).

The method of Lagrange multipliers is generalized by the Karush-Kuhn-Tucker conditions.

Very simple example


Suppose you wish to maximize $$f(x,y)=x+y$$ subject to the constraint $$x^2+y^2=1$$. The constraint is the unit circle, and the level sets of f are diagonal lines (with slope -1), so one can see graphically that the maximum occurs at $$(\sqrt{2}/2,\sqrt{2}/2)$$ (and the minimum occurs at $$(-\sqrt{2}/2,-\sqrt{2}/2)$$

Formally, set $$g(x,y)=x^2+y^2-1$$, and
 * $$\Lambda(x, y, \lambda) = f(x,y) + \lambda g(x,y) = x+y + \lambda (x^2 + y^2 - 1)$$

Set the derivative $$d\Lambda=0$$, which yields the system of equations:


 * $$\begin{align}

\frac{\partial \Lambda}{\partial x}      &= 1 + 2 \lambda x &&= 0, \qquad \text{(i)} \\ \frac{\partial \Lambda}{\partial y}      &= 1 + 2 \lambda y &&= 0, \qquad \text{(ii)} \\ \frac{\partial \Lambda}{\partial \lambda} &= x^2 + y^2 - 1  &&= 0, \qquad \text{(iii)} \end{align}$$ As always, the $$\partial \lambda$$ equation is the original constraint.

Combining the first two equations yields $$x=y$$ (explicitly, $$x \neq 0$$ (otherwise (i) yields 1 = 0), so one can solve for $$\lambda$$, yielding $$\lambda=-1/(2x)$$, which one can substitute into (ii)).

Substituting into (iii) yields $$2x^2=1$$, so $$x=\pm \sqrt{2}/2$$ and the stationary points are $$(\sqrt{2}/2,\sqrt{2}/2)$$ and $$(-\sqrt{2}/2,-\sqrt{2}/2)$$. Evaluating the objective function f on these yields


 * $$f(\sqrt{2}/2,\sqrt{2}/2)=\sqrt{2}\mbox{ and } f(-\sqrt{2}/2, -\sqrt{2}/2)=-\sqrt{2},$$

thus the maximum is $$\sqrt{2}$$, which is attained at $$(\sqrt{2}/2,\sqrt{2}/2)$$ and the minimum is $$-\sqrt{2}$$, which is attained at $$(-\sqrt{2}/2,-\sqrt{2}/2)$$.

Simple example


Suppose you want to find the maximum values for


 * $$ f(x, y) = x^2 y \, $$

with the condition that the x and y coordinates lie on the circle around the origin with radius √3, that is,


 * $$ x^2 + y^2 = 3. \, $$

As there is just a single condition, we will use only one multiplier, say λ.

Use the constraint to define a function g(x, y):


 * $$g (x, y) = x^2 +y^2 -3. \, $$

The function g is identically zero on the circle of radius √3. So any multiple of g(x, y) may be added to f(x, y) leaving f(x, y) unchanged in the region of interest (above the circle where our original constraint is satisfied). Let


 * $$\Lambda(x, y, \lambda) = f(x,y) + \lambda g(x, y) = x^2y + \lambda (x^2 + y^2 - 3). \, $$

The critical values of $$\Lambda$$ occur when its gradient is zero. The partial derivatives are


 * $$\begin{align}

\frac{\partial \Lambda}{\partial x}      &= 2 x y + 2 \lambda x &&= 0, \qquad \text{(i)} \\ \frac{\partial \Lambda}{\partial y}      &= x^2 + 2 \lambda y   &&= 0, \qquad \text{(ii)} \\ \frac{\partial \Lambda}{\partial \lambda} &= x^2 + y^2 - 3      &&= 0. \qquad \text{(iii)} \end{align}$$

Equation (iii) is just the original constraint. Equation (i) implies $$ x = 0 $$ or λ = &minus;y. In the first case, if $$x=0$$ then we must have $$y = \pm \sqrt{3}$$ by (iii) and then by (ii) λ=0. In the second case, if λ = &minus;y and substituting into equation (ii) we have that,


 * $$x^2 - 2y^2 = 0. \, $$

Then x2 = 2y2. Substituting into equation (iii) and solving for y gives this value of y:


 * $$y = \pm 1. \, $$

Thus there are six critical points:


 * $$ (\sqrt{2},1); \quad (-\sqrt{2},1); \quad (\sqrt{2},-1); \quad (-\sqrt{2},-1); \quad (0,\sqrt{3}); \quad (0,-\sqrt{3}). $$

Evaluating the objective at these points, we find


 * $$ f(\pm\sqrt{2},1) = 2; \quad f(\pm\sqrt{2},-1) = -2; \quad f(0,\pm \sqrt{3})=0. $$

Therefore, the objective function attains a global maximum (with respect to the constraints) at $$(\pm\sqrt{2},1)$$ and a global minimum at $$(\pm\sqrt{2},-1).$$ The point $$(0,\sqrt{3})$$ is a local minimum and $$(0,-\sqrt{3})$$ is a local maximum.

Example: entropy
Suppose we wish to find the discrete probability distribution with maximal information entropy. Then


 * $$f(p_1,p_2,\ldots,p_n) = -\sum_{k=1}^n p_k\log_2 p_k.$$

Of course, the sum of these probabilities equals 1, so our constraint is g(p) = 1 with


 * $$g(p_1,p_2,\ldots,p_n)=\sum_{k=1}^n p_k.$$

We can use Lagrange multipliers to find the point of maximum entropy (depending on the probabilities). For all k from 1 to n, we require that


 * $$\frac{\partial}{\partial p_k}(f+\lambda (g-1))=0,$$

which gives


 * $$\frac{\partial}{\partial p_k}\left(-\sum_{k=1}^n p_k \log_2 p_k + \lambda (\sum_{k=1}^n p_k - 1) \right) = 0.$$

Carrying out the differentiation of these n equations, we get


 * $$-\left(\frac{1}{\ln 2}+\log_2 p_k \right) + \lambda = 0.$$

This shows that all pi are equal (because they depend on λ only). By using the constraint ∑k pk = 1, we find


 * $$p_k = \frac{1}{n}.$$

Hence, the uniform distribution is the distribution with the greatest entropy.

Economics
Constrained optimization plays a central role in economics. For example, the choice problem for a consumer is represented as one of maximizing a utility function subject to a budget constraint. The Lagrange multiplier has an economic interpretation as the shadow price associated with the constraint, in this case the marginal utility of income.

The strong Lagrangian principle: Lagrange duality
Given a convex optimization problem in standard form


 * $$\begin{align}

\text{minimize }   &f_0(x) \\ \text{subject to } &f_i(x) \leq 0,\ i \in \left \{1,\dots,m \right \} \\ &h_i(x) = 0,\ i \in \left \{1,\dots,p \right \} \end{align}$$

with the domain $$\mathcal{D} \subset \mathbb{R}^n$$ having non-empty interior, the Lagrangian function $$L: \mathbb{R}^n \times \mathbb{R}^m \times \mathbb{R}^p \to \mathbb{R} $$ is defined as


 * $$L(x,\lambda,\nu) = f_0(x) + \sum_{i=1}^m \lambda_i f_i(x) + \sum_{i=1}^p \nu_i h_i(x).$$

The vectors $$\lambda$$ and $$\nu$$ are called the dual variables or Lagrange multiplier vectors associated with the problem. The Lagrange dual function $$g:\mathbb{R}^m \times \mathbb{R}^p \to \mathbb{R} $$ is defined as


 * $$g(\lambda,\nu) = \inf_{x\in\mathcal{D}} L(x,\lambda,\nu) = \inf_{x\in\mathcal{D}} \left ( f_0(x) + \sum_{i=1}^m \lambda_i f_i(x) + \sum_{i=1}^p \nu_i h_i(x) \right ).$$

The dual function $$g$$ is concave, even when the initial problem is not convex. The dual function yields lower bounds on the optimal value $$p^*$$ of the initial problem; for any $$\lambda \geq 0 $$ and any $$\nu$$ we have $$g(\lambda,\nu) \leq p^* $$. If a constraint qualification such as Slater's condition holds and the original problem is convex, then we have strong duality, i.e. $$d^* = \max_{\lambda \ge 0, \nu} g(\lambda,\nu) = \inf f_0 = p^*$$.