Linear Programming

Linear Systems

Systems of linear equations (key word being equality) can be solved via gaussian elimination. This is relatively easy, since your solution space ends up being a line, a plane, or a hyperplane in higher dimensions.

Let $a$ be a column vector in $\mathbb{R}^d$, and $x$ a column vector of $d$ variables. We can represent a system using the inner product of these two vectors:

$$ \langle a, x \rangle = a^Tx = a_1x_1 + a_2x_2 + \ldots + a_dx_d = \sum_{i=1}^d a_ix_i $$

A hyperplane is the set of points $x$ such that $\langle a, x \rangle = b$ for some $b$. A handspace is the set of points on one side of a hyperplace, ${x : \langle a, x \rangle \geq b}$ or ${x : \langle a, x \rangle \leq b}$.

The intersection of a system of half spaces creates a polytope, which is a convex set. A convex set is a set where the line segment between any two points in the set is also in the set.

Linear Programs

The goal of a linear program is to optimize some objective function subject to a set of constraints, which are also linear functions. For example...

$$ \begin{aligned} &max & 3x_1 - 4x_3\ &s.t. & x_1 + x_2 \le 5\ & & x_3 + x_1 = 4\ & & x_3 - x_2 \ge -5\ & & x_1, x_2, x_3 \ge 0\ \end{aligned} $$

Linear Algebra Review

$$ \langle a, x \rangle = a^Tx = a_1x_1 + a_2x_2 + \ldots + a_dx_d $$

$$ A = \begin{bmatrix} a_1^T \ a_2^T \ \vdots \ a_m^T \end{bmatrix} \Rightarrow Ax = \begin{pmatrix} \langle a_1, x \rangle \ \langle a_2, x \rangle \ \vdots \ \langle a_m, x \rangle \end{pmatrix} $$

$$ Ax \le b \Rightarrow \begin{array}{c} \langle a_1, x \rangle \le b_1 \ \langle a_2, x \rangle \le b_2 \ \vdots \ \langle a_m, x \rangle \le b_m \end{array} $$

Linear Program Standard Form

We can write any linear program in the standard form below.

$$ \begin{array}{cc} max & \langle c, x \rangle \ s.t., & Ax \le b\ ~ & x \ge 0 \end{array} $$

For example, we can transform the following linear program by turning the minimization problem into a maximization, and turning all equalities into two inequalities.

$$ \begin{array}{cc} min & y_1 - 2y_2\ s.t., & y_1 + 2y_2 = 3\ ~ & y_1 - y_2 \ge 1\ ~ & y_1, y_2 \ge 0\ \end{array} $$

To turn a min into a max, we just negate the objective function. The same holds for reversing the direction of a $\geq$. We can turn the above linear program into the following:

$$ \begin{array}{cc} max & -y_1 + 2y_2\ s.t., & y_1 + 2y_2 \le 3\ ~ & -(y_1 + 2y_2) \le -3\ ~ & y_1 + y_2 \le 1\ ~ & y_1, y_2 \ge 0\ \end{array} $$

In LPs where you don't have non-negativity for all variables, you can replace the variables missing this constraint with the difference of two non-negative variables. For example...

$$ \begin{array}{cc} max & y_1\ s.t., & y_1 + y_2 \le 3\ ~ & y_2 \ge 0\ \end{array} $$

We replace $y_1$ with $z_1 - z_1'$, where $z_1, z_1' \ge 0$. The linear program becomes...

$$ \begin{array}{cc} max & z_1 - z_1'\ s.t., & z_1 - z_1' + y_2 \le 3\ ~ & z_1, z_1', y_2 \ge 0\ \end{array} $$

Applications of Linear Programming.

LPs generalize to all sorts of problems, such as 2-person zero-sum games, shortest path, max-flow, matching, multi-commodity flow, MST, min weighted, arborescence, ...

We can solve linear programs in polynomial time, and they turn out to be useful for approximation algorithms.

Components of a Linear Program

• Set of variables
• Bounding constraints on variables
• Are they non-negative?
• Objective function
• Is it a minimization or maximization problem?
• LP constrains, but they need to be linear
• Is it an equality or inequality?

Max-Flow

Given a graph $G = (V, E)$ with source $s$ and sink $t$, for every edge $e$ we have a variable $x_e$ as the flow on the edge $e$.

The bounding constraints would be...

• $x_e \ge 0$ for all $e$ (since the flow is non-negative)
• $x_e \le c(e)$ for all $e$ (capacity constraint)
• $\sum_{e \text{ out of } v} x_e = \sum_{e \text{ into} v} x_e$ for all $v \ne s, t$ (conservation constraint)

And our objective function would be...

• Maximize $\sum_{e \text{ out of } s} x_e$

So we have an overall LP of...

$$ \begin{array}{ccc} max & \sum_{e \text{ out of } s} x_e & ~\ s.t., & \sum_{e \text{ out of } v} x_e = \sum_{e \text{ into} v} x_e & \forall c \ne s, t\ ~ & x_e \le c(e) & \forall e\ ~ & x_e \ge 0 & \forall e \end{array} $$

Note that the flow this gives you is not necessarily an integer max-flow.

Min-Cost Max-Flow

Expanding on the previous example, we can add a cost to each edge $p(e)$ and try to minimize $p(e)$ while achieving some constant flow $f$.

$$ \begin{array}{ccc} max & \sum_{e \in E} p(e) \cdot x_e & ~\ s.t., & \sum_{e \text{ out of } v} x_e = \sum_{e \text{ into} v} x_e & \forall c \ne s, t\ ~ & \sum_{e \text{ out of } s} x_e = f & ~\ ~ & x_e \le c(e) & \forall e\ ~ & x_e \ge 0 & \forall e \end{array} $$

Weighted Vertex Cover

Given a graph $G = (V, E)$, where each vertex has a cost $c_v$, find the minimum cost vertex cover, i.e. $\min \sum_{v \in S} c_v$

We have a variable $x_v$ for each $v$, where $x_v = 1$ if $v \in S$ and $x_v = 0$ otherwise. Then we have the constraint that for every edge $(u, v) \in

$$ \begin{array}{ccc} max & \sum_{e \in E} p(e) \cdot x_e & ~\ s.t., & \sum_{e \text{ out of } v} x_e = \sum_{e \text{ into} v} x_e & \forall c \ne s, t\ ~ & \sum_{e \text{ out of } s} x_e = f & ~\ ~ & x_e \le c(e) & \forall e\ ~ & x_e \ge 0 & \forall e \end{array} $$