Special Cases in Simplex Method

Special Cases in Simplex Method

• 1.
  Special Cases inSimplex Method Divyansh Verma SAU/AM(M)/2014/14 South Asian University Email : itsmedv91@gmail.com 4/18/2015 1
• 2.
  Contents • Simplex Method •Simplex Table • Special Cases of Simplex Method – Degeneracy – Alternative Optima – Unbounded Solution – Infeasible Solution • References 4/18/2015 2
• 3.
  Simplex Method Step 1: Determine a starting basic feasible solution. Step 2 : Determine the entering basic variable by selecting the non-basic variable with the most negative value (in case of maximization) or with the most positive (in case of minimization) in the Z-row. Stop if there is no entering variable, the last solution is optimal. Else, go to Step 3. Step 3 : Select a leaving variable using the feasibility condition Step 4 : Determine the new basic solution by using the appropriate Gauss-Jordan computations. Repeat Step 2. 4/18/2015 3
• 4.
  Simplex Table Basic Variable X1 X2S1 S2 S3 Solution Min Ratio Z C1 C2 C3 C4 C5 z - S1 a11 a12 a13 a14 a15 b11 b11/a1j S2 a21 a22 a23 a24 a25 b21 b21/a2j S3 a31 a32 a33 a34 a35 b31 b31/a3j Non-Basic Variable Coefficient of Objective Function, decides Entering Variable Decides the Leaving Variable Objective Function Value Right Hand Side Value of Constraints Z-row S1-row S2-row S3-row A format of Simplex Table for some given LPP, here Xj are Decision Variables, Sj are Slack Variables, aij is the coefficient of Xj corresponding to ith constraint. 4/18/2015 4
• 5.
  Special Cases inSimplex Special Cases that arise in the use of Simplex Method : 1. Degeneracy 2. Alternative Optima 3. Unbounded Solution 4. Infeasible Solution 4/18/2015 5
• 6.
  Degeneracy A solution ofthe problem is said to be degenerate solution if the value of at least one basic variable becomes zero. In the simplex table, a tie for the minimum ratio occurs which is broken arbitrarily. It can cause the solution to cycle indefinitely. Graphically, it occurs due to redundant constraint (i.e. a constraint that can be removed from system without changing the feasible solution). 4/18/2015 6
• 7.
  Degeneracy For example, considerthe problem Max z = 3x1 + 9x2 ; Subject to : x1 + 4x2 ≤ 8 ; x1 + 2x2 ≤ 4 ; x1, x2 ≥ 0 4/18/2015 7 Basic Variable X1 X2 S1 S2 Solution Min Ratio Z0 -3 -9 0 0 0 - S1 1 4 1 0 8 2 S2 1 2 0 1 4 2 Z2 -3/4 0 9/4 0 18 - X2 1/4 1 1/4 0 2 8 S2 1/2 0 -1/2 1 0 0 Z3 0 0 3/2 3/2 18 - X2 0 1 1/2 -1/2 2 - X1 1 0 -1 2 0 - Tie in Min Ratio (Degeneracy) X2 enters S1 leaves X1 enters S2 leaves S2 becomes zero in next iteration because of degeneracy Optimum Solution : X1=0, X2=2, Max Z=18
• 8.
  Alternative Optima If thez-row value for one or more non-basic variable is zero in the optimal tableau, alternate optimal solution exists. Graphically, it happens when the objective function is parallel to a non-redundant binding constraint (i.e. a constraint that is satisfied as an equation at the optimal solution). The zero coefficient of non-basic variable xi indicates that it can be made basic, altering the value of basic variable without changing the value of z. 4/18/2015 8
• 9.
  Alternative Optima 4/18/2015 9 Forexample, consider the problem Max z = 2x1 + 4x2 ; Subject to : x1 + 2x2 ≤ 5 ; x1 + x2 ≤ 4 ; x1, x2 ≥ 0 Basic Variable X1 X2 S1 S2 Solution Min Ratio Z0 -2 -4 0 0 0 - S1 1 2 1 0 5 5/2 S2 1 1 0 1 4 4/1 Z2 0 0 2 0 10 - X2 1/2 1 1/2 0 5/2 5 S2 1/2 0 -1/2 1 3/2 3 Z3 0 0 2 0 10 - X2 0 1 1 -1 1 - X1 1 0 -1 2 3 - Optimum Solution, but coefficient of non-basic variable X1 is 0, indicates that it can be made basic. X2 enters S1 leaves X1 enters S2 leaves Alternate Optimum Solution X1 , X1 are basic variables. Optimum Solution : X1=3, X2=1 Max Z=10
• 10.
  Unbounded Solution When determiningthe leaving variable of any tableau, if there is no positive minimum ratio or all entries of pivot column are negative or zero. 4/18/2015 10 For example, consider the problem Max z = 2x1 + x2 ; Subject to : - x1 + x2 ≤ 10 ; -2x1 ≤ 40 ; x1, x2 ≥ 0 Basic Variable X1 X2 S1 S2 Solution Min Ratio Z0 -2 -1 0 0 0 - S1 -1 1 1 0 10 - S2 -2 0 0 1 40 - X1 is the entering variable but all the constraint coefficients under X1 are ≤ 0 , meaning that X1 can be increased indefinitely, which gives unbounded solution.
• 11.
  Infeasible Solution If atleast one artificial variable is positive in the optimum iteration, then the LPP has no feasible solution. This situation obviously does not occur when all the constraints are type ≤ with non-negative right-hand side because all slack variables provide an obvious feasible solution. So, this situation occurs in Big-M Method & Two-Phase Method in which artificial variable are used. 4/18/2015 11
• 12.
  Infeasible Solution 4/18/2015 12 Forexample, consider the problem Max z = 3x1 + 2x2 ; Subject to : 2x1 + x2 ≤ 2 ; 3x1 + 4x2 ≥ 12 ; x1, x2 ≥ 0 Basic Variable X1 X2 S1 S2 R1 Solution Min Ratio Z0 -3 -2 0 0 100 0 - S1 2 1 1 0 0 2 - R1 3 4 0 -1 1 12 - Z0 -303 -402 100 0 0 -1200 - S1 2 1 1 0 0 2 2 R1 3 4 0 -1 1 12 3 Z1 501 0 402 100 0 -396 - X2 2 1 1 0 0 2 - R1 -5 0 -4 -1 1 4 - X2 enters S1 leaves Pseudo Optimum Solution, but artificial variable has a positive value.
• 13.
  References 4/18/2015 13 • OperationsResearch - An introduction (9th edition), Hamdy A. Taha, 2013 • http://www.lokminglui.com/lpch4.pdf • https://www.utdallas.edu/~metin/Or6201/specialcases.pdf
• 14.
  4/18/2015 14 THANK YOU