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Lecture 5

Theconstraintwiththesmallestra1oiscalledthewinner

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Unformatted text preview: ve) variable for row 0. Since z appears in row 0 with a coeﬃcient of 1, and z does not appear in any other row, we use z as the basic variable. With this convenWon, the basic feasible soluWon for our iniWal canonical form has: BV = {z, s1, s2, s3, s4} and NBV = {x1, x2, x3 }. For this iniWal bfs, z = 0, s1= 48, s2= 20, s3 = 8, s4 =5, x1= x2 = x3 = 0. As this example indicates, a slack variable can be used as a basic variable if the rhs of the constraint is nonnegaWve. 4.3 – The Simplex Algorithm (max LPs) Step 3 – Determine if the Current BFS is OpWmal Once we have obtained a bfs, we need to determine whether it is opWmal. Determine if there is any way zcan be increased by increasing some nonbasic variable from its current value of zero while holding all other nonbasic variables at their current values of zero. Solving for z in row 0 yields: Z = 60x1 + 30x2 + 20x3 Z ‐ 60x1 ‐ 30x2 ‐ 20x3 = 0 •  For each nonbasic variable, we can use the equaWon above to determine if increasing a nonbasic variable will increase z. •  Increasing any of the nonbasic variables will cause an increase in z. •  Increasing x1 causes the greatest rate of increase in z (x1 has the most negaWve coeﬃcient in row 0). •  If x1 increases from its current value of zero, it will have to become a basic variable. For this reason, x1 is called the entering variable. 4.3 – The Simplex Algorithm (max LPs) Step 4 ‐ We choose the entering variable (in a max problem) to the nonbasic variable with the most negaWve coeﬃcient in row 0 (Wes broken arbitrarily). We desire to make x1 as large as...
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