[PDF] [PDF] 1 Overview 2 Basic Feasible Solutions - Harvard SEAS

[PDF] Finding feasible solutions to a LP

basic feasible solution: put the slack variables on the left hand side How- ever, this is not Problem: The artificial variable may allow us to find “solutions” that

A1 LINEAR PROGRAMMING AND OPTIMAL SOLUTIONS A2

A nonnegative vector of variables that satisfies the constraints of (P) is called a feasible solution to the linear programming problem A feasible solution that minimizes the objective function is called an optimal solution

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2 oct 2014 · The corresponding basic feasible solution is x = 0, z = b We use this to initialize the simplex algorithm The simplex method can be one of two 

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Degenerate basic feasible solution: A basic feasible solution where one or more of the basic variables is zero Discrete Variable: A decision variable that can only  

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Recall the definition of a polyhedron, and a basic feasible solution: • P ⊆ Rn is a polyhedron, if it can be expressed as P = {x ∈ Rn : Fx ≥ g} for some matrix F 

[PDF] 1 Overview 2 Basic Feasible Solutions - Harvard SEAS

19 fév 2014 · Any feasible solution in the pyramid only has 3 linearly independent active constraints, but we need at least 4 constraints to represent the pyramid 

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THEOREM: For a feasible linear program in its standard form, the optimum value of the objective over its nonempty feasible region is (a) either unbounded or (b) 

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The feasible region has exactly one corner point, at (0,0); and that this corner point is not optimal This clearly is a disturbing factor for Procedure Solve LP A linear 

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Recall the definition of a basic feasible solution: Definition 1 Let P be a polyhedron defined by linear equality and inequality constraints, and consider x ∗ ∈ Rn

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AM 221: Advanced OptimizationSpring 2014

Prof. Yaron Singer Lecture 7 | February 19th, 20141 Overview In our previous lecture we saw the application of the strong duality theorem to game theory, and then saw how that is applied to learning theory where we showed that weak learning implies strong learning. Today we'll present the simplex method for solving linear programs. We will start with discussing basic solutions and then show how this applies to the simplex algorithm.

2 Basic Feasible Solutions

Denition 1.We say that a constraintaxbisactive(orbinding) at pointxifax=b. Denition 2.A solution inP=fx:Axbgis calledbasic feasibleif it hasnlinearly independent active constraints. Denition 3.A solution inP=fx:Axbgis calleddegenerateif it has more thannlinearly independent active constraints. Example: Degeneracy does not imply redundancy.Consider the pyramid inR3. Any feasible solution in the pyramid only has 3 linearly independent active constraints, but we need at least 4 constraints to represent the pyramid.

2.1 Basic solutions in standard form

We say that an LP is in standard form if we express it as: mincTx s:t: Ax=b x0 Let us assume thatAis amnmatrix. Any linear program can be written in the standard form withmn. Without loss of generality we can assume thatrank(A) =m(ifrank(A)< m, then the system has redundant constraints that can be identied and removed). Pick a set of indices B[n] that correspond tomlinearly independent columns of the matrixA. Now, we can think of the matrixAas the concatenation of two matricesABandANwhereABis themmmatrix 1 whose columns are indexed by the indices inB, andANis them(nm) matrix whose columns are indexed by the indices in [n]nB. Similarly we can think ofxas [xB;xN] in a natural manner.

A= [ABjAN];

x= [xBjxN]: Remark 4.For any basic feasible solutionx, we have a setB[n]ofmindices that correspond to a linearly independent set of columns ofAsuch that:

1.xN= 0

2.xB=A1

Bb. In addition, for any setB[n]ofmindices that correspond to a linearly independent set of columns, ifxB=A1

Bb0then(xB;xN)is basic feasible.

We'll conclude this discussion with an important theorem which will be used in the simplex method. Theorem 5.LetP=fx2Rn:Axbgthenxis an extreme point ofPif and only ifxis a basic feasible solution ofP. The proof follows the same principles as the proofs for extreme points and is left as an exercise in your next problem set.

3 The Simplex Algorithm

From the above discussion, it is clear that in order to nd an optimal solution, it is sucient to search over the basic feasible solutions to nd the optimal one. The Simplex Algorithm, given by

Dantzig, does this search in an organized fashion.Algorithm 1Simplex1:Let (xB;xN) be a basic feasible solution.

2: cT cTcTBA1 BA

3:xB b:=A1

Bb

4:ifc0then

5:STOP and return (xB;xN) as optimal solution.

6:end if

7:Selectj2Nsuch thatcj<0

8:dj A1

BAj

9:ifdj0then

10:STOP; return \LP is unbounded!! YOLO".

11:end if

12:k argminfi2B:dji>0g(bi=dji)

13:B (Bn fkg)[ fjg;N (Nn fjg)[ fkg

14:go to step 1.2

Example.Consider the following linear program.

minx13x2 s.t. 2x1+ 3x26 x1+x21 x 1;x20

Let us rst write this in standard form:

minx13x2 s.t. 2x1+ 3x2x3= 6 x1+x2x4= 1 x

1;x2;x3;x40

Say we start withB=f3;4gandN=f1;2g. It should be clear that the resulting solution (x1= 0;x2= 0;x3= 1;x4= 1) is a basic feasible solution (Verify that the corresponding columns of A are linearly independent). Then, A B=1 0 0 1 ; A N=2 3 1 1

Iteration 1.

cTN=cTNcTBA1 BAN = (1;3)(0;0)1 0 0 1 2 3 1 1 = (1;3) Next, x

B=b=A1

Bbx3 x 4 =b3 b 4 =1 0 0 1 6 1 =6 1

Say we selectj= 2 to enter the basis.

d 2=A1

BA2d23

d 24
=1 0 0 1 3 1 =3 1 Thus, k=argminfb3=d23;b4=d24g=argminf6=3;1=1g= 4 leaves the basis. So, the nextB=f3;2gandN=f1;4gand A B=1 3 0 1 ; A N=2 0 1 1 3

Iteration 2.

cTN=cTNcTBA1 BAN = (1;0)(0;3)13 0 1 2 0 1 1 = (4;3) Next, x

B=b=A1

Bbx3 x 2 =b3 b 2 =13 0 1 6 1 =3 1

We selectj= 1 to enter the basis.

d 1=A1

BA1d13

d 12 =13 0 1 2 1 =5 1 Thus, k=argminfb3=d13g= 3 leaves the basis. So, the nextB=f1;2gandN=f3;4gand A B=2 3 1 1 ; A N=1 0 0 1

Iteration 3.

cTN=cTNcTBA1 BAN = (0;0)(1;3)1=53=5

1=5 2=5

1 0 0 1 = (4=5;3=5) and hence the optimal solution is x

B=b=A1

Bbx1 x 2 =1=53=5

1=5 2=5

6 1 =3=5 8=5

Proposition 6.IfcT=cTcTBA1

BA0then the solution is optimal.

Proof.Consider the dual of the problem:

maxyTb s.t.yTAcT(1) which in the language of basic and nonbasic solutions we can write as: maxyTb s.t.yTABcTB y

TANcTN

4

Observe thatcTcTBA1

BA0 implies thatyT=cTBA1

Bis a feasible solution to the dual problem.

The value of this solution in the dual objective is: c TBA1 Bb: Now observe that for a feasible solutionxwe have thatAx=b, which in the language of basic feasible solutions, is:

Ax=ABxB+ANxN=ABxB

sincexN= 0. So a feasible solution respectsxB=A1

Bband therefore the value in the primal

objective for such a solution is: c TBA1 Bb:

And therefore whencTcTBA1

BA0 we have that the primal and dual objectives have the same

value. From the strong duality theorem this implies that the solution must be optimal.Remarks.What is the running time of the Simplex algorithm? Is it even nite? What if the

algorithm repeatedly cycles over the same setBof basis indices? Observe that the algorithm is

well-dened only after we specify the tie-breaking rules to be used for selecting the indices that leave

and enter the basis (Steps 7 and 12). One way to break ties is by picking the least indices among the possible choices. This rule, due to Robert Bland, provably avoids cycling thereby ensuring that the algorithm is nite. It remains open to design variations of the tie-breaking rules so that the total number of iterations performed by the Simplex algorithm is polynomial in the number of variables. 5quotesdbs_dbs20.pdfusesText_26