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© Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Pivoting Slide 1 The Simplex Method Pivoting, an Animation

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© Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Pivoting Slide 2 Use the down pointer key to move forward in the animation and the up pointer key to backtrack. Assume the Following Initial Basic Feasible Tableau for a Maximization Problem.

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© Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Pivoting Slide 3 Step 1. Identify the Entering Variable. Step 1, What is the Entering Variable? Rule 1, In a Maximization, the Entering Variable is the one with the most +ve (Cj-Zj) value! So, x2 is the Entering Variable as 20 is the most +ve (Cj-Zj) value! The x2 Column is the “Pivot Column” for this Pivot!

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© Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Pivoting Slide 4 Step 2. Identify the Leaving Variable. Step 2, What is the Leaving Variable? Rule 2, the Leaving Variable is the one with the least +ve Ratio Test result! Take the Bi value in each row and divide it by the value in the row in the Pivot Column. Top row B1 = 200. Value in the pivot column is 2, so the Ratio Test value is 200/2 or 100. Middle row B2 = 72. Value in the pivot column is 0.6, so the Ratio Test value is 72/0.6 or 120. Bottom row B3 = 180. Value in the pivot column is 0.6, so the Ratio Test value is 180/0.6 or 300. So, the minimum Ratio Test value is 100 and S1 is the Leaving Variable.

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© Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Pivoting Slide 5 Naming of Parts The x2 Column is the “Pivot Column”, the S1 Row is the “Pivot Row” and the orange box is the “Pivot Element”.

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© Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Pivoting Slide 6 The Pivot Row We need the Pivot Row to have a 1 in the Pivot Column. To achieve this we will divide all the values in the Pivot Row (except the Cj on the left)! by the value in the Pivot Element.

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© Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Pivoting Slide 7 The Second Row We need the Second Row to have a 0 in the Pivot Column.

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© Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Pivoting Slide 8 The Second Row Multiply the Pivot Row (somewhere off the Tableau!!) by the value in the Pivot Column and Row 2. AND then subtract the result from row 2.

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© Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Pivoting Slide 9 The Third Row We need the Third Row to have a 0 in the Pivot Column.

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© Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Pivoting Slide 10 The Third Row Multiply the Pivot Row (somewhere off the Tableau!!) by the value in the Pivot Column and Row 3. AND then subtract the result from row 3.

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© Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Pivoting Slide 11 Resetting and Recalculating Replace S1 with x2.Update the Cj col. Calculate the Zj row.And the Cj-Zj.

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© Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Pivoting Slide 12 The New Entering and Leaving Variables x1’s (Cj-Zj) value is the largest +ve value so x1 is the Entering Variable and s2 is the leaving variable because its Ratio Test value is the min +ve value.

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© Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Pivoting Slide 13 Running The Next Pivot Should Get You to Here!! The optimal solution is x1 = 63.158, x2 = 77.895, s3 = 38.526, s2 = 0, s1 = 0 and Z = 2,189.5.

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© Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Pivoting Slide 14 Pivoting In a Minimization The only difference is in the choice of the Entering Variable, which involves selecting the variable with the most negative (Cj-Zj) value! In all other ways the procedure is the same.

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© Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Pivoting Slide 15 The End Use the ESC key to exit!

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LINEAR PROGRAMMINGExample 1 MaximiseI = x + 0.8y subject tox + y 1000 2x + y 1500 3x + 2y 2400 Initial solution: I = 0 at (0, 0)

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