Applied Development of a Procedure for Finding Active Points of Linear Constraints

In this paper, we present an iterative method to determine active point of linear constraints. It is based on two basic operations which are addition and permutation of constraints. This procedure generates a finite sequence of points that basis in a new lemma and a new formula direction, the laspoint of sequence constitutes an active point, and this procedure gives also two matrices. The first one is constituted by the active constraints which are linearly independent and the second one is a matrix whose columns are the basis vectors of the kernel of the first matrix.


Introduction
Currently, the domain of optimization is attracting considerable interest from the academic and industrial communities, see, for instances [1][2][3][4][5]. The various existing techniques for solving a given problem and the efficient algorithmic implementations open up many perspectives and diverse applications in different areas [6][7][8].
In most optimization problems, initialization points are necessary and required in the resolution algorithms for performing numerical implementations [6][7][8]11,12]. However, the choice of the initialization points is not general, and the values of these points depend strongly on the adopted technique. Furthermore, these points are considered as active or feasible in the applied method.
In this study, we are interested by the optimization problem of the CSLP type (constraint satisfaction linear problems), where the set of constraints are linear and it is defined by determined the active point x act of a set E such as: A is an mxn data matrix, not necessarily full rank and b is a given as vector IR m The problem to solve is the determination of the active points satisfying all the aforementioned constraints.
If the values of the matrix A and the vector b components are integer numbers, the above problem is discrete and can be solved by using the ellipse method [4]. However, in the case of optimization continuous, this is not studied in literature.
This past has motivated this investigation in the purpose of giving a theoretical and practical method of resolution of this problem.
The method that we propose is based on the construction of an iterative algorithm, such that, from any initial point (feasible or not) one produces another better point x, then one associates to it two matrices A and Z. The lines of A are constituted by the active constraints, which are linearly independent. The columns of matrix Z are constituted of the kernel of matrix A, and are also linearly independent.
This process allows to generate a sequence of points (x k ) k∈IN which converge to the point that one search (feasible or active). It is important to mention that our method can be applied without knowing whether this domain of constraints is empty or not.
In the numerical implementation, we used the scientific environment FOTRAN F90 under windows, and the obtained results were very satisfactory.
The rest of this paper is organized as follows. In Sec. 2, we give some definitions and propositions that are used in this article. In Sec. 3, we present the construction of a die kernel. In Sec. 4, we give description of the active method and its implementation in Sec.5, we form algorithm for the active method. At last, we summarize our results in the last section.

Definitions and Propositions Definition
The constraint This definition leads to the fact that, for any vector v, we can introduce all We then say that the vector v is a regular point of all eligible

Definition
Let D be a domain of constraints in IR n , defined by [5] ; We call all candidates constraints, any set of constraints, among the t i a (x) ≤ b i considered as active constraint of the solution that we search for.

Proposition
A direction d is tangent to x∈X, if and only if there exists a sequence (d n ) of limit d, and a sequence (μ n ) of positive real zero limit, such that x+µ n d n ∈X. [5].

Remarks
Most of the algorithms fail when they have to solve a problem whose constraints are not qualified in the solution. Therefore it is preferable to change the description of the set of constraints before solving the problem.

Proposition (CS contraints satisfaction)
The constraints of D domain are qualified at the point x ∈ D, if the gradients in x of the active constraints [5].

Construction of Kernel Matrix
This section contains the most important results for the kernel numerical calculation of any matrix, especially in the case where we have a lot of matrices of large size, and to avoid repetition of the calculations.
The calculation of a matrices and vector kernel obeys certain rules of compliance.
These results are given in the following:

Lemma
Let v be a vector v ∈R n , which defines a set of constraints as follows: Where α is a real number (4.1) then v⊥∆ i.e. v is orthogonal to ∆.

Proof
By subtraction (1) of (2) it comes: This gives that 1 2 v x x ┴ for each 1 2 , From this lemma, we can construct sets ∆ + and ∆ -that help us to lead the constraint equations in the following two corollaries:

Corollary
Let Δ be a set of constraints of the form: Where v is a vector of ,

Corollary
Consider the same data of corollary 4.2, then

Lemma
Let v be a vector in |R n* , then its kernel is formed by the following

Proof
It suffices to show that the rows of the matrix it follows:

Lemma
Let v 1 and v 2 be a two vectors in R n , the set { } Then the matrix 1 2 1

Proof
The same proof of Lemma (4.4), in the rows of the matrix (v 1 v 2 z 1 …….z im-1 …..z n-1 ) t which are linearly independent vectors.

Corollary
Keeping the same data of Lemma as the kernel of a basis vector { } Then the set form a common basis of the kernel vectors v 1 and v 2 , and verifying is a basis of the kernel vector v 1 .
We will show that from a common basis vectors v 1 and v 2 , knowing It remains to show that set (4.7) is linearly independent.

Description of the Active Method
We focus in this section on a so-called active point approach We construct the iterated x k+1 by the formula The choice of α k and d k ensures that x k+1 approaches the border of the constraints better than x k , and This process is repeated until the stopping test is satisfied.

Initialization
Location of the starting point x 0 : Let E be a set of constraints in a general form (equalities, inequalities, and mixed) be an arbitrary point and x 0 be a point of departure in IR n .
We can distinguish the situation from the point x 0 with respect to E, in one of the following three cases:

Case 2:
The point x 0 is located outside of E.

Case 3:
The point x 0 is located in the boundary of E.

Geometric representation at point x 0
Adding and permutation of Constraints Adding a constraint: Let A k be the matrix of active constraints at x k point of iteration k, stitch-forming iteration k +1, we add in the matrix A k the constraint 1 t k a + resulting from the following two equations: Note that This result is obtained by remplacing the x k point in all constraint of domain E.
Permutation of constraints: Let x k be the point in iteration k, in which two matrices are associated A k , Z k , and i k is the index on constraints that can be added to A k to obtain A k+1 , z k is the column that can be eliminated from the matrix Z k-1 to reach Z k .
Permute the constraint of index i k by another constraint of index i 0 that result of equality: If and only if where the algorithm is moved from iteration k to iteration k+1 we meet the condition 0 Where n k is the number of columns of matrix Z k . and { } , 1, , is a set of columns eliminated on the matrix Z 1 .

Remarks: (1)
The rows of the matrix A k+1 are linearly independent, they are also active at the point x k+1 . (2) From the kernel of the matrix A k , we can easily determine the Z k+1 matrix whose columns form a basis of the kernel of A k+1 .

Direction of displacement
We consider the matrix A k composed of active constraints linearly independent at the point x k and the columns of the matrix Z k form a basis of the kernel of A k . To determine the direction d k , we distinguish two alternatives: If k=0, we pose ) .
. If k ≠ 0, here, we find also two other alternative: If 0 t ik k nk a Z ≠ , the direction d k is resulted by solution of the following linear system: 0 and z k is the column that can be eliminated from the matrix Z k-1 to obtain Z k .
If 0 t ik k nk a Z = , the direction d k is resulted by the solve of the following linear system: Step of displacement Let x k is the point of iteration k, and d k the direction of displacement at the point x k .
After finding the associated constraint of iteration (k+1) which is active at the point xk+1, then We conclude that Where ik is the index of constraint to added in the matrix A k .

Remarks:
The active constraint at x k , is also active at the point x k+1 .

Theorem of convergence
Let (x k ) k ∈ IN be an iterative sequence defined by , where α k-1 is the displacement step along the direction d k-1 , and x 0 is a finit starting point of IR n .
Then the sequence ( ) x ∈ formed by a set of the directions

Proof
It sufficiently to show that the set of direction (dk) k∈IN is linearly independent By recurrence we can write that the scalar product we suppose that are true for all step m, and we proof it for step m+1.
The same way, we calculate the basis of KerA k+1 . e) K=k+1 and return to a).

Remarks:
i) This method determines the active points of a problem (E), without any constraints condition i.e., it does not require to make the linearly independent constraints.
ii) This method can be applied to any set E, defined by linear constraints, and even if it is empty.

Numerical Tests
From a practical point of view, our method has remarkable advantages.
This will be shown by numerical application of this method in different cases that may exist: the number of constraints, the number of variables, and the size of the matrix to be taken.
The obtained results are listed in the following tables:   Table 2 shows the inequality form in Large size.       Table 3 shows the mixed form in large size.

Case 3 Mixed form (Large size)
a -Let m=23 and n=7 such that m i =8, i=1,2 and m 3 =7        From a numerical point of view, it is difficult to take the best starting point in IR n , which helps us to obtain easily the active point that we search.
In a numerical application, it is competent to verify whether the domain of optimization is empty or not. This problem is very easy to solve it by our method.
We easily can know the state of the domain. This has been illustrated in the numerical test-case 3 ( Table 4).
All the above results show the efficiency of this method in the problem of optimization, where the domains are consecutives, and with small size.
For large size, the problem is substantially the same; one has only to do a large amount of calculations. So, the discussion is similar to that of domains of small sizes.
These results are showed in examples of three cases.

Conclusion
After a long scientific research, we have not found any thing on the method that discusses to solve this type of continues problem optimization, and then we have suggested this method with a new formula direction d k .
In this work, we studied theoretically and algorithmically an active method, which determines the extremes of a set defined by linear constraints. This set is in the form of equalities, inequalities or both of them. These m constraints are linear and function of n variables, our results can be givens in the following points: • Starting from any initial point, it generates points belonging to the set E.
• It is possible to construct from the m constraints two matrices, where the lines of the first are linearly independents and actives, and the columns of the second form a basis of the kernel of the first matrix.
• Our method can be applied to matrices of large sizes.
• The active point is determined in at most n+ np iterations.
• The other advantage is the simplification of the computation, because each used constraint appears at most only once.
• Our method can be used in other algorithms of resolution of optimization problems to simplify their initializations and to improve their results.