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Simplex Method MSci331—Week 3~4

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Simplex Algorithm Consider the following LP, solve using Simplex:

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**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.

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**Step 2: Express the LP in a tableau form**

Z X1 X2 S1 S2 S3 RHS Ratio Row 0 1 -3 -2 -- Row 1 - 2 100 Row 2 80 Row 3 40

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**Step 3: Obtain the initial basic feasible solution (if available)**

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 Z X1 X2 S1 S2 S3 RHS Ratio Row 0 1 -3 -2 -- Row 1 - 2 100 Row 2 80 Row 3 40

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**Step 4: Apply the Simplex Algorithm**

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 Exits Pivot cell Enters Z X1 X2 S1 S2 S3 RHS Ratio Row 0 1 -3 -2 -- Row 1 - 2 100 100/2 Row 2 80 80/1 Row 3 40 40/1

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**Summary of Simplex Algorithm for Papa Louis**

Set: n-m=0 m≠0 BFS (intial) BFS (1) BFS (2) BFS (3) 1 The optimal solution is x1=20, x2=60 The optimal value is Z=180 The BFS at optimality x1=20, x2=60, s3=20

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**Geometric Interpretation of Simplex Algorithm**

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**Class activity Consider the following LP: **

This is a maximizing LP, in normal form. So an initial BFS exists.

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Class activity

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Class activity Z x1 x2 s1 s2 s3 s4 RHS 1 -3 100 ----- 4 2 6 -1

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**Class activity Z x1 x2 s1 s2 s3 s4 RHS 1 -3 100 ----- 4 2 6 -1 4/1 6/1**

Make this coefficient equal 1 and pivot all other rows relative to it Enters Exits Z x1 x2 s1 s2 s3 s4 RHS 1 -3 100 ----- 4 2 6 -1 4/1 6/1 2/2* ---

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**Class activity Z x1 x2 s1 s2 s3 s4 RHS 1 -3 100 ----- 4 2 6 -3/2 1/2**

100 ----- 4 2 6 -3/2 1/2 -1 -7.5 3/2 103 2.5 1 -1/2 3 3.5 1 -1/2 5 1/2 1 5

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**Class activity Z x1 x2 s1 s2 s3 s4 RHS 1 -3 100 ----- 4 2 6 -3/2 1/2**

Make this coefficient equal 1 and pivot all other rows relative to it Enters Z x1 x2 s1 s2 s3 s4 RHS 1 -3 100 ----- 4 2 6 -3/2 1/2 -1 -7.5 3/2 103 3/2.5* 2.5 1 -1/2 3 3.5 1 -1/2 5 5/3.5 --- 5/0.5 1/2 1 5

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**Class activity Z x1 x2 s1 s2 s3 s4 RHS 1 ----- 3 112 -7.5 3/2 103 1**

3 112 -7.5 3/2 103 1 2/5 -1/5 6/5 -1.4 1 1/5 0.8 3.5 1 -1/2 5 1 3/5 0.8 2.8 1 -3/2 1/2 -1/5 0.6 1 4.4 1/2 1 5

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**Example: LP model with Minimization Objective**

Solve the following LP model: Initial Tableau

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**Example: LP model with Minimization Objective**

Iteration 0 Iteration 1 Optimality test:

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**> Constraint x2 x1 Constraint 1 Constraint 3 Constraint 2**

40 35 30 25 20 15 10 5 Constraint 1 Constraint 3 Z Constraint 2 x1 Constraint 4 New feasible region

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**Equality Constraint x2 x1 Constraint 1 Constraint 3 Constraint 2**

40 35 30 25 20 15 10 5 Constraint 1 Constraint 3 Z Constraint 2 x1 New feasible region

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**The Problem of Finding an Initial Feasible BV**

An LP Model Standard Form Cannot find an initial basic variable that is feasible.

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**Example: Solve Using the Big M Method**

Write in standard form

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**Example: Solve Using the Big M Method**

Adding artificial variables

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**Example: Solve Using the Big M Method**

Put in tableau form

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**Example: Solve Using the Big M Method**

Eliminating a2 from row 0 by operations: new Row 0 = old Row 0 -M*old Row 2

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**Example: Solve Using the Big M Method**

Eliminating a3 from the new row 0 by operations: new Row=old Row-M*old Row 3

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**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

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**Example: Solve Using the Big M Method**

Using EROs change the column of x1 into a unity vector. Iteration 1

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**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.

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**Special case 1: Alternative Optima**

. Observation: At optimality, all nonbasic variables have nonnegative (≥0) coefficients in row 0, hence the bfs is optimal. At optimality, the non-basic variable x2 has a zero coefficient in row 0. If variable x2 enters into the basis, the basis and bfs changes but the objective function value remains the same. Hence we have multiple optima. Conclusions: If there is no nonbasic variable with a zero coefficient in row 0 of the optimal tableau, then the LP has a unique optimal solution. Even if there is a nonbasic variable with a zero coefficient in row 0 of the optimal tableau, it is possible that the LP may not have an alternative optimal solutions. See Notes on this slide (below) for more information

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**Special case 1: Alternative Optima**

Observation: At optimality, all nonbasic variables have nonnegative (≥0) coefficients in row 0, hence the bfs is optimal.

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**Special case 2: Unbounded LPs**

1 s1 x1 Notice after the first iteration: At this stage, x2 is the entering variable (because it has the most negative reduced cost, but there is no ratio to compute, since there is no positive entry in the column of x2. As we start increasing x2, the value of z increases (from Row 0) and the values of the basic variables increase as well (from Rows 1 and 2). There is nothing to stop them going off to infinity. So the problem is unbounded. Conclusions: An unbounded LP for a maximization problem occurs when a nonbasic variable with a negative coefficient in row 0 has a nonpositive coefficient in each constraint. The objective function value improves with each iteration See Notes on this slide (below) for more information

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**Special Case 3: Degeneracy**

Definition: An LP is degenerate if it has at least one bfs in which a basic variable is equal to zero. Notice S1=0 However notice that degeneracy is detected when the ratio tests for row 1 and 2 are equal. This means as X2 enters, both S1 and S2 will reach zero at the same time. Hence we have a choice of selecting S1 or S2 as the exiting BV.

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**Special Case 5: Degeneracy**

Iteration 0 Iteration 1 Definition: An LP is degenerate if it has at least one bfs in which a basic variable is equal to zero. Iteration 2

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**Special Case 5: Degeneracy**

Degeneracy reveals from practical standpoint that the model has at least one redundant constraint.

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