Interior-point method for LP

Authors: Tomas Lopez Lauterio, Rohit Thakur and Sunil Shenoy
Steward: Dr. Fengqi You and Akshay Ajagekar

Introduction

Linear programming problems seeks to optimize linear functions given linear constraints. There are several applications of linear programming including inventory control, production scheduling, transportation optimization and efficient manufacturing processes. Simplex method has been a very popular method to solve these linear programming problems and has served these industries well for a long time. But over the past 40 years, there have been significant number of advances in different algorithms that can be used for solving these types of problems in more efficient ways, especially where the problems become very large scale in terms of variables and constraints. In early 1980s Karmarkar (1984) published a paper introducing interior point methods to solve linear-programming problems. A simple way to look at differences between simplex method and interior point method is that a simplex method moves along the edges of a polytope towards a vertex having a lower value of the cost function, whereas an interior point method begins its iterations inside the polytope and moves towards the lowest cost vertex without regard for edges. This approach reduces the number of iterations needed to reach that vertex, thereby reducing computational time needed to solve the problem.

Lagrange Function

Before getting too deep into description of Interior point method, there are a few concepts that are helpful to understand. First key concept to understand is related to Lagrange function. Lagrange function incorporates the constraints into a modified objective function in such a way that a constrained minimizer (x^*) is connected to an unconstrained minimizer {x^*, λ^*} for the augmented objective function L(x,λ), where the augmentation is achieved with 'p' Lagrange multipliers.
To illustrate this point, if we consider a simple an optimization problem: minimize f(x) subject to: A·x = b, where A ε R^pxn is assumed to have a full row rank Lagrange function can be laid out as:

$L(x,\lambda )=f(x)-\sum _{i=1}^{p}\lambda _{i}\cdot a_{i}(x))$
where, 'λ' introduced in this equation is called Lagrange Multiplier.

Newton's Method

Another key concept to understand is regarding solving linear and non-linear equations using Newton's methods. Assume you have an unconstrained minimization problem in the form:
minimize g(x) , where g(x) is a real valued function with n variables.
A local minimum for this problem will satisfy the following system of equations:
$\left[{\frac {\partial g(x)}{\partial x_{1}}}.....{\frac {\partial g(x)}{\partial x_{n}}}\right]^{T}=\left[0...0\right]$

The Newton's iteration looks like:
$x^{k+1}=x^{k}-\left[\nabla ^{2}g(x^{k})\right]^{-1}\cdot \nabla g(x^{k})$

Theory and algorithm

We first start forming a primal-dual pair of linear programs and use the "Lagrangian function" and "Barrier function" methods to convert the constrained problems into unconstrained problems. These unconstrained problems are then solved using Newton's method as shown above.

Problem Formulation

Visualization of Central Path method in Interior point

Consider a combination of primal-dual problem below:
(Primal Problem formulation)
→ minimize $c^{T}x$ Subject to: $Ax=b$ and $x\geq 0$ .......................................(1)
(Dual Problem formulation)
→ maximize $b^{T}y$ Subject to: $A^{T}y+\lambda =c$ and </math> \lambda \geq 0 </math> ...................(2)
'λ' vector introduced represents the slack variables.
Now we use the "Logarithmic Barrier" function and form 2 Lagrangian equations for primal and dual forms mentioned above:
Lagrange function for Primal : $L_{p}(x,y)=c^{T}\cdot x+\mu \cdot \sum _{j=1}^{n}log(x_{j})-y^{T}\cdot (Ax-b)$ ......................................(3)
Lagrange function for Dual  : $L_{d}(x,y,\lambda )=b^{T}\cdot y+\mu \cdot \sum _{j=1}^{n}log(\lambda _{j})-x^{T}\cdot (A^{T}y-\lambda -c)$ .........(4)

Taking the partial derivatives of L_p and L_d with respect to variables 'x', 'λ' , 'y', and forcing these terms to zero, we get the following equations:
$Ax=b$ and $x\geq 0$ ..........................................(5)
$A^{T}y+\lambda =c$ and $\lambda \geq 0$ .......................(6)
$x_{j}\cdot \lambda _{j}=\mu$ for j= 1,2,.......n .....................(7)

where, μ is strictly positive scaler parameter. For each μ > 0, the vectors in the set {x(μ), y(μ), λ(μ)} satisfying equations (5), (6) and (7), can we viewed as set of points in Rⁿ, R^p, Rⁿ such that when 'μ' varies, the corresponding points form a set of trajectories called "Central Path". THe central path lies in the "Interior" of the feasible regions.

Let us define the following:
X = Diagonal ($x_{1}^{0},....x_{n}^{0}$)
$\lambda$ = Diagonal ($\lambda _{1}^{0},....\lambda _{n}^{0}$ )
e^T = (1 .....1) as vector of all 1's.
Using these newly defined terms, the equation (7) can be written as:
$X\cdot \lambda \cdot e=\mu \cdot e$

Iterations using Newton's Method

Now we employ the Newton's iterative method to solve the following equations:
$Ax-b=0$ ...................................................(8)
$A^{T}y+\lambda =c$ ........................................(9)
$X\cdot \lambda \cdot e-\mu \cdot e=0$ ..................... (10)
Suppose we start with definition of starting point that lies within feasible region as (x⁰, y⁰, λ ⁰) such that x⁰>0 and λ ⁰ >0. Also let us define 2 residual vectors for both the primal and dual equations:
$\delta _{p}=b-A\cdot x^{0}$ .................................(11)
$\delta _{d}=c-A^{0}\cdot y^{0}-\lambda ^{0}$ ...............(12)

Applying Newton's Method to solve equations (8) - (12) we get:
${\begin{bmatrix}A&0&0\\0&A^{T}&1\\\lambda &0&X\end{bmatrix}}\cdot {\begin{bmatrix}\delta _{x}\\\delta _{y}\\\delta _{\lambda }\end{bmatrix}}={\begin{bmatrix}\delta _{p}\\\delta _{d}\\\mu \cdot e-X\cdot \lambda \cdot e\end{bmatrix}}$
So a single iteration of Newton's method involves the following equations. For each iteration, we solve for the next value of x^k+1, y^k+1 and λ^k+1:
$(A\lambda ^{-1}XA^{T})\delta _{y}=b-\mu A\lambda ^{-1}+A\lambda ^{-1}X\delta _{d}$ ...................(13)
$\delta _{\lambda }=\delta _{d}\cdot A^{T}\delta _{y}$ ...................................................(14)
$\delta _{x}=\lambda ^{-1}\left[\mu \cdot e-X\lambda e-\lambda \delta _{z}\right]$ .................(15)
$\alpha _{p}=min\left\{{\frac {-x_{j}}{\delta _{x_{j}}}}\right\}$ for $\delta x_{j}<0$ ..... (16)
$\alpha _{d}=min\left\{{\frac {-\lambda _{j}}{\delta _{\lambda _{j}}}}\right\}$ for $\delta \lambda _{j}<0$ ..... (17)

The value of the the following variables for next iteration (k+1) is determined by:
$x^{k+1}=x^{k}+\alpha _{p}\cdot \delta _{x}$
$y^{k+1}=y^{k}+\alpha _{d}\cdot \delta _{y}$
$\lambda ^{k+1}=\lambda ^{k}+\alpha _{d}\cdot \delta _{\lambda }$

The quantities α^p and α^d are positive with 0 ≤ α_p, α_d ≤ 1.
After each iteration of Newton's method, we assess the duality gap that is given by the expression below and compare it against a small value ε
${\frac {c^{T}x^{k}-b^{T}y^{k}}{1+\left|b^{T}y^{k}\right|}}\leq \varepsilon$
The value of ε can be chosen to be something small 10^-6, which essentially is the permissible duality gap for the problem.