Lecture 4

# Lecture 4 - Simplex Method 1 Properties of the Standard...

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1 Simplex Method

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2 Properties of the Standard Form LP The set of constraints has m equations on n variables ( m < n ) If we set n – m variables to be 0, then we have a set of m constraints and m remaining variables, and thus these m variable can be uniquely determined by the constraints. These m variables are called a set of basic variables, and the corresponding solution is called a basic solution. n j x m i b x a x c z j i n j j ij n j j j , , 1 , 0 , , 1 , subject to Maximize 1 1 = = = = = =
3 BFS Definition . For an LP in the standard form, a basic solution to the m constraints is called a basic feasible solution (BFS) if it satisfies the non- negativity constraints. Theorem . For an LP in the standard form: If there is a feasible solution, there is a BFS. If these is an optimal feasible solution, there is an optimal BFS. Remark: Each BFS corresponds to a corner point in the graphic representation of the LP.

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4 Simplex Form of an LP 5 ,..., 1 , 0 20 35 3 5 subject to 3 2 Maximize 5 1 4 2 1 3 2 1 2 1 = = + = + + = + + - + = i x x x x x x x x x x x z i 5 ,..., 1 , 0 20 35 3 5 0 3 2 subject to z Maximize 5 1 4 2 1 3 2 1 2 1 = = + = + + = + + - = - - i x x x x x x x x x x x z i Simplex form: Row 0 Row 1 Row 2 Row 3
5 Conditions for the Simplex Form A simplex form corresponds to a BFS: Each basic variable corresponds to a row, and the value of the basic variable is the right-hand side of the row The value of the objective function is equal to the right- hand side of row 0 Every basic variable appears in one and only one equation, but not row 0 Each basic variable has the coefficient 1 in the equation it appears Each equation has only one basic variable Variable z only appears in row 0 with coefficient 1

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6 Tableau of Simplex Form 5 ,..., 1 , 0 20 35 3 5 0 3 2 subject to z Maximize 5 1 4 2 1 3 2 1 2 1 = = + = + + = + + - = - - i x x x x x x x x x x x z i Coefficients Row Basic z x 1 x 2 x 3 x 4 x 5 RHS 0 Z 1 -2 -3 0 0 0 0 1 0 -1 1 1 0 0 5
7 Is the Current BFS Optimal? The current BFS is optimal if and only if row 0 has no negative coefficient 5 ,..., 1 , 0 20 35 3 5 0 3 2 subject to z Maximize 5 1 4 2 1 3 2 1 2 1 = = + = + + = + + - = - - i x x x x x x x x x x x z i In row 0, x 1 ’s coefficient 2<0, implying that increasing x 1 from 0 to a small positive number will also increase z . So the current BFS is not optimal.

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8 Moving from One Simplex Form to Another An LP can be written in different simplex forms by some linear operations upon the rows A linear operation refers to: row j = row j * non-zero parameter row j = row j + row k * a parameter Steps: Choose a non-basic variable with negative element in row 0: a pivot column • for example, x 2 Choose a basic variable leaving the current basic solution: a pivot row • for example, x 3 (row 1 for x 3 ) So a pivot element is determined For the pivot row, each element is multiplied by a parameter so that the pivot element becomes 1 For other rows, use linear operations to make the element in the pivot column to be 0
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Lecture 4 - Simplex Method 1 Properties of the Standard...

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