CO227 Lecture 13: co 227 lec13

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For a basis b with a feasible basic solution x(bar), if cn <= 0 then x(bar) is an optimal solution. Proof: the balue of x(bar) is z(x(bar)) = z(bar) + cnx(bar)n = z(bar) + cn (0) = z(bar) Note: x>=0 for any feasible solution and cn <=0. Nxn <=0 so z(x) <= z(bar) for any feasible solution x. 4x1 x2 + 6x3 +x4 x1 + 2x2 + 3x3 3x4 = 3. 3x1 + 2x2 x3 -4x4 = 4. X1 + 3x2 + x3 2x4 = -1 x>=0. Rewrite in canonical form for basis b = {1,2,3} Bb, new lp in canonical form for basis b = {1,2,3}: Note: the constraints on t are t<= 3/2, -t <=0 , 0t<1/2. Any value of t>=0 will satisfy these constraints, we can increase x4 to infinity. The lp is unbounded, the constraints for t are useless because the entries in the 4 th column are all <=0.

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