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Simplex Method MSci331—Week 3~4 1. Simplex Algorithm Consider the following LP, solve using Simplex: 2.

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Presentation on theme: "Simplex Method MSci331—Week 3~4 1. Simplex Algorithm Consider the following LP, solve using Simplex: 2."— Presentation transcript:

1 Simplex Method MSci331—Week 3~4 1

2 Simplex Algorithm Consider the following LP, solve using Simplex: 2

3 Step 1: Preparing the LP LP Model Is the LP model in normal form? LP in a standard form All constraints are “=“ All RHS >0 All variables>0 If there are = or > Add an artificial variables to these constraints Write Row 0 Move all variables in the objective function equation to LHS. Keep all constants in the RHS A min problem can be treated as a –MAX problem Obtain an initial BFS If the original LP is not in normal form apply the Big M method to obtain the initial BFS. 3

4 Step 2: Express the LP in a tableau form ZX1X2S1S2S3RHSRatio Row Row Row Row

5 Step 3: Obtain the initial basic feasible solution (if available) ZX1X2S1S2S3RHSRatio Row Row Row Row a) Set n-m variables equal to 0 These n-m variables the NBV b) Check if the remaining m variables satisfy the condition of BV = If yes, the initial feasible basic solution (bfs) is readily a available = else, carry on some ERO to obtain the initial bfs

6 Step 4: Apply the Simplex Algorithm ZX1X2S1S2S3RHSRatio Row Row /2 Row /1 Row /1 6 a) Is the initial bfs optimal? (Will bringing a NBV improve the value of Z?) b) If yes, which variable from the set of NBV to bring into the set of BV? - The entering NBV defines the pivot column c) Which variable from the set of BV has to become NBV? - The exiting BV defines the pivot row Pivot cell

7 Summary of Simplex Algorithm for Papa Louis m ≠0 Set: n-m =0 1 7 BFS (intial) BFS (1) BFS (2) BFS (3) The optimal solution is x 1 =20, x 2 =60The optimal value is Z=180 The BFS at optimality x 1 =20, x 2 =60, s 3 =20

8 Geometric Interpretation of Simplex Algorithm 8

9 Class activity Consider the following LP: 9 This is a maximizing LP, in normal form. So an initial BFS exists.

10 Class activity 10

11 Class activity 11 Zx1x1 x2x2 s1s1 s2s2 s3s3 s4s4 RHS

12 Class activity 12 Zx1x1 x2x2 s1s1 s2s2 s3s3 s4s4 RHS /1 6/1 2/2* --- Make this coefficient equal 1 and pivot all other rows relative to it

13 Class activity 13 Zx1x1 x2x2 s1s1 s2s2 s3s3 s4s4 RHS /2001/ / / /205 01/200 15

14 Class activity 14 Zx1x1 x2x2 s1s1 s2s2 s3s3 s4s4 RHS /2001/ / / /205 01/ /2.5* 5/ /0.5 Make this coefficient equal 1 and pivot all other rows relative to it

15 Class activity 15 Zx1x1 x2x2 s1s1 s2s2 s3s3 s4s4 RHS / /50-1/506/ / / /2001/ / / /

16 Example: LP model with Minimization Objective Solve the following LP model: Initial Tableau 16

17 Example: LP model with Minimization Objective Iteration 0 Iteration 1 Optimality test: 17

18 > Constraint x1x1x1x1 x 2 x 2 Constraint 1 Constraint 2 Constraint 3 Z New feasible region Constraint 4 18

19 Equality Constraint x1x1x1x1 x 2 x 2 Constraint 1 Constraint 2 Constraint 3 Z New feasible region 19

20 The Problem of Finding an Initial Feasible BV An LP Model Standard Form Cannot find an initial basic variable that is feasible. 20

21 Example: Solve Using the Big M Method Write in standard form 21

22 Example: Solve Using the Big M Method Adding artificial variables 22

23 Example: Solve Using the Big M Method Put in tableau form 23

24 Example: Solve Using the Big M Method Eliminating a2 from row 0 by operations: new Row 0 = old Row 0 -M*old Row 2 24

25 Example: Solve Using the Big M Method Eliminating a3 from the new row 0 by operations: new Row=old Row-M*old Row 3 25

26 Example: Solve Using the Big M Method The initial basic variables are s1=25, a2=12, and a3=0. Now ready to proceed for the simplex algorithm. The initial Tableau 26

27 Example: Solve Using the Big M Method Using EROs change the column of x1 into a unity vector. Iteration 1 27

28 Example: Solve Using the Big M Method Using EROs change the column of z into a unity vector. Iteration 2 Students to try more iterations. The solution is infeasible. See the attached solution. 28

29 Special case 1: Alternative Optima. 29 See Notes on this slide (below) for more information

30 Special case 1: Alternative Optima 30

31 Special case 2: Unbounded LPs 0 1 s1s1 x1x1 31 See Notes on this slide (below) for more information

32 Special Case 3: Degeneracy 32

33 Special Case 5: Degeneracy Iteration 0 Iteration 1 Iteration 2 33

34 Special Case 5: Degeneracy Degeneracy reveals from practical standpoint that the model has at least one redundant constraint. 34


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