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Unformatted text preview: possible but as we do, the current basic variables (s1, s2, s3, s4) will change value. Thus, increasing x1 may cause a basic variable to become negaWve. RATIO From row 1 we see that s1 = 48 – 8x1. Since s1 ≥ 0, x1 ≤ 48 / 8 = 6 From row 2, we see that s2 = 20 – 4x1. Since s2 ≥ 0, x1 ≤ 20 / 4 = 5 From row 3, we see that s3 = 8 – 2x1. Since s3 ≥ 0, x1 ≤ 8 / 2 =4 From row 4, we see that s4 = 5. For any x1, s4 will always be ≥ 0 This means to keep all the basic variables nonnegaWve, the largest we can make x1 is min {6, 5, 4} = 4. 4.3 – The Simplex Algorithm (max LPs) The RaCo Test When entering a variable into the basis, compute the raWo: rhs of row / coeﬃcient of entering variable in row For every constraint in which the entering variable has a posiWve coeﬃcient. The constraint with the smallest ra1o is called the winner of the raWo test. The smallest raWo is the largest value of the entering variable that will keep all the current basic variables nonnegaWve. Make the entering variable x1 a basic variable in row 3 since this row (constraint) was the winner of the raWo test (8/2 =4). 4.3 – The Simplex Algorithm (max LPs) To make x1 a basic variable in row 3, we use elementary row operaWons (ero s) to make x1 have a coeﬃcient of 1 in row 3 and a coeﬃcient of 0 in all other rows. This procedure is called pivoCng on row 3; and row 3 is called the pivot row. The ﬁnal result is that x1 replaces s3 as the basic variable for row 3. Step 5 ‐ The Gauss‐Jordan method using ero s and simplex tableaus shown on the next slide makes x1 a basic variable. 4.3 – The Simplex Algorithm (max LPs) ero
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This note was uploaded on 09/20/2011 for the course ENG 300 taught by Professor Albin during the Fall '11 term at Rutgers.
 Fall '11
 Albin
 Ode

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