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Unformatted text preview: e solve for the values of the n – (n – m) = m variables (the basic variables, or BV) that saWsfy Ax = b. 4.2 – Preview of the Simplex Algorithm Diﬀerent choices of nonbasic variables will lead to diﬀerent basic soluWons. Consider the basic soluWons to the system of 2 equaWons shown to the right: x1 + x2 = 3 ‐ x2 + x3 = ‐1 The number of nonbasic variables = 3 – 2 = 1. Sebng, for example, NBV = {x3} (as shown to the right, then BV = {x1, x2}. We can obtain the values for these basic variables by sebng x3 = 0. Solving we ﬁnd x1 = 2, x2 = 1. Thus, x1 = 2, x2 = 1, and x3 = 0 is a basic soluCon. x1 + x2 = 3 ‐ x2 + 0 = ‐1 • If NBV = {x1} and BV = {x2, x3} are chosen instead, the basic soluWon becomes x1 = 0, x2 = 3, and x3 = 2. [basic feasible soluWon] • If NBV = {x2} and BV = {x1, x3} are chosen instead, the basic soluWon becomes x1 = 3, x2 = 0, and x3 = ‐1. [not basic feasible soluWon] 4.2 – Preview of the Simplex Algorithm Some sets of m variables do not yield a basic soluWon. Consider the linear system shown to the right: If NBV = {x3} and BV = {x1, x2} the corresponding basic soluWon would be: x1 + 2x2 + x3 = 1 2x1 + 4x2 + x3 = 3 x1 + 2x2 = 1 2x1 + 4x2 = 3 Since this system has no soluWon, there is no basic soluWon corresponding to BV = {x1, x2}. Why????? 4.2 – Preview of the Simplex Algorithm Theorem 1 The feasible region for any linear programming problem is a convex set. Also, if an LP has an opWmal soluWon, there must be an extreme point of the feasible region that is opWmal. Theorem 2 For any LP, there is a unique extreme point of the LP’s feasible region co...
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 Fall '11
 Albin
 Ode

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