Lecture 6 Simplex Method: Artifical Starting Solution and ...

1 Lecture 6 Simplex Method: Artifical Starting Solutionand Some Special CasesSeptember 4, 2009 Lecture 6 Outline: Initial table vs initial Simplex table Artificial start: Two-phase method Special cases: Degeneracy Multiple optimal solutions Infinite optimal value Infeasibility of the problemOperations Research Methods1 Lecture 6 Simplex method: started at a feasible basic solutionIllustrated on the Reddy Mikks problemOriginal LP formulationmaximizez= 5x1+ 4x2subject to6x1+ 4x2 24x1+ 2x2 6x1,x2 0 Standard LP formmaximizez= 5x1+ 4x2subject to6x1+ 4x2+x3= 24x1+ 2x2+x4= 6x1,x2,x3,x4 0 Operations Research Methods2 Lecture 6 Initial table vs Initial Simplex tableBasisx1x2x3x4 RHS Valuesz 5 4 0 0= 0?? 6 4 1 0= 24?? 1 2 0 1= 6 Basis:x3andx4 Basisx1x2x3x4 RHS Valuesz 5 4 0 0= 0x36 4 1 0= 24x41 2 0 1= 6 Initial table can be used as initial Simplex tableOperations Research Methods3 Lecture 6 Basis:x1andx4 Basisx1x2x3x4 RHS Valuesz 5 4 0 00x164 1 024x412 0 16 This table cannot be used as the initial Simplex table!

2 We have to transformthe table (Gauss-Jordan elimination usingx1-column elements)Basisx1x2x3x4 RHS Valuesz0 2356020x11231604x4043 1612 This table is aninitial Simplex table, , the Simplex method can Research Methods4 Lecture 6 Artificial Start: Two-phase method Sometimes, it is not easy to find an initial feasible Solution ( , tochoose initial bases yielding a feasible point) Two-phase method is used in such situations In first phase, afeasibility problemassociated with the LP is solvedby a Simplex method In the second phase, the Solution from the first phase is used to startrunning the Simplex methodOperations Research Methods5 Lecture 6 Two-phase Method: ExampleOriginal LP formulationminimizez= 4x1+x2subject to3x1+x2= 34x1+ 3x2 6x1+ 2x2 4x1,x2 0 Standard LP formminimizez= 4x1+x2subject to3x1+x2= 34x1+ 3x2 x3= 6x1+ 2x2+x4= 4x1,x2,x3,x4 0It is not easy to find a basis yielding a basic feasible Solution !

3 Phase I: we first solve afeasibility problem associated with the LP:introduce new variablesR1andR2minimizer=R1+R2subject to3x1+x2+R1= 34x1+ 3x2 x3+R2= 6x1+ 2x2+x4= 4x1,x2,x3,x4,R1,R2 0 Operations Research Methods6 Lecture 6 Phase I: Initial tableBasisx1x2x3x4R1R2 RHS Valuesr0 0 0 0 1 10R13 1 0 0 1 03R24 3 1 0 0 16x41 2 0 1 0 04 InitialsimplextableBasisx1x2x3x4R1R2 RHS Valuesr7 4 1 0 0 09R13 1 0 0 1 03R24 3 1 0 0 16x41 2 0 1 0 04 Operations Research Methods7 Lecture 6 Applying the Simplex method, we will obtain an optimal solutionBasisx1x2x3x4R1R2 RHS Valuesr0 0 0 0 1 10x11 015035 1535x20 1 350 453565x40 0 1 1 1 11Q: How do we know this is optimal Solution to the feasibility problem?

4 What is the Solution ? What is the optimal value forr?Use the Solution of Phase I to start Phase II (solving the original LP)Basisx1x2x3x4 RHS Valuesz 4 1 0 00x11 015035x20 1 35065x40 0 1 11 Operations Research Methods8 Lecture 6 The initial table of Phase IIBasisx1x2x3x4 RHS Valuesz0 0150185x11 015035x20 1 35065x40 0 1 11We continue with Simplex iterations until we find an optimal Research Methods9 Lecture 6 Special Case: DegeneracyDegeneracy is a term used for abasic feasible Solution having one ormore basic variables at value 0 Original LP formulationmaximizez= 3x1+ 9x2subject tox1+ 4x2 8x1+ 2x2 4x1,x2 0 Standard LP formmaximizez= 3x1+ 9x2subject tox1+ 4x2+x3= 8x1+ 2x2+x4= 4x1,x2,x3,x4 0At the end of the Simplex method, (started with basisx3andx4), we haveBasisx1x2x3x4 RHS Valuesz0 0323218x20 112 122x11 0 1 20 There is another basis corresponding to the same Solution !

5 Operations Research Methods10 Lecture 6 Special Case: Multiple SolutionsOriginal LP formulationmaximizez= 2x1+ 4x2subject tox1+ 2x2 5x1+ 2x2 4x1,x2 0 Standard LP formmaximizez= 2x1+ 4x2subject tox1+ 2x2+x3= 5x1+x2+x4= 4x1,x2,x3,x4 0At the end of the Simplex method, (started with basisx3andx4), we haveBasisx1x2x3x4 RHS Valuesz00 2 010x212112052x4120 12132 There is a nonbasic variable with 0 coefficient in the optimal means we can bring that variable in the basis without changing Research Methods11 Lecture 6 Choosingx1to enter the basis, we perform another Simplex iteration andfind an alternative optimal solutionBasisx1x2x3x4 RHS Valuesz0 0 2 010x20 1 1 11x41 0 1 23 Operations Research Methods12 Lecture 6 Special Case: Unbounded Optimal ValueOriginal LP formulationmaximizez= 2x1+x2subject tox1 x2 102x1 40x1,x2 0 Standard LP formmaximizez= 2x1+x2subject tox1 x2+x3= 102x1+x4= 40x1,x2,x3,x4 0At some iteration of the Simplex method, (in this example, it happenedto be the initial iteration)a nonbasic variable withnegative coefficientcan enter the basis without a bound on its value(maximization)Basisx1x2x3x4 RHS Valuesz 2 10 00x31 11 010x4200 140 This means we can bring that variable in the basis and increase thez-valueto+ (since the variable can be increased to+ ).

6 Operations Research Methods13 Lecture 6 Special Case: InfeasibilityOriginal LP formulationmaximizez= 3x1+ 2x2subject to2x1+x2 23x1+ 4x2 12x1,x2 0 Standard LP formmaximizez= 3x1+ 2x2subject to2x1+x2+x3= 23x1+ 4x2 x4= 12x1,x2,x3,x4 0 None of the basic solutions is feasible!Here, it is not as easy to read off a feasible basis -apply Two phase methodPhase ISolve the feasibility problemminimizer=Rsubject to2x1+x2+x3= 23x1+ 4x2 x4+R= 12x1,x2,x3,x4,R 0 Operations Research Methods14 Lecture 6 Initial table for feasibility problemBasisx1x2x3x4 RRHS Valuesr0 0 0 0 10x32 1 1 0 02R3 4 0 1 112 Initial Simplex table for feasibility problemBasisx1x2x3x4 RRHS Valuesr3 4 0 1 012x32 1 1 0 02R3 4 0 1 112 The table is not a nonbasic variable with positiver-coefficient (minimization)

7 , saychoosex2 Operations Research Methods15 Lecture 6In the next Simplex iteration, we haveBasisx1x2x3x4 RRHS Valuesr 5 0 4 1 04x22 1 1 0 02R3 4 0 1 112 This table is optimal (the feasibility problem involves minimization)Indication of infeasibility:In the optimal (phase I) table, The optimal value is positive The basis Solution contains an artificial variable with a positive valueThe material is in Sections and Research Methods16