4 - Linear Inequalities

4 - Linear Inequalities - Math 1b Practical Nonnegative...

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Math 1b Practical — Nonnegative solutions January 10, 2011 We may ask whether a (nonhomogeneous) system of linear equations with real coeffi- cients has a solution in nonnegative real numbers. This question is part of the subject of linear programming and has many applications. The system of equations (1) on the Frst page of the handout on “Basic solutions; pivot operations”, though it has many solutions, has no nonnegative solutions (with x 1 ,x 2 3 4 5 0). 2 x 1 3 x 2 5 x 3 7 x 4 +2 x 5 =7 , x 1 + x 2 5 x 3 + x 4 4 x 5 =3 , x 1 4 x 3 +3 x 4 + x 5 = 4 . That there is no nonnegative solution is is clear from the Frst equation in (4), x 4 = (26 / 19) (15 / 19) x 5 , since if x 5 is nonnegative, then x 4 is strictly negative. We claim that if a system of linear equations over the real numbers has a nonnegative solution, then it has a basic solution which is nonnegative, though a complete proof will not be given in these notes. If the coefficient matrix of the system is r × n , then there are at most ( n r ) choices for the dependent variables and so there can be at most ( n r ) basic solutions. We could, theoretically, go through the basic solutions one by one and check whether any are non- negative. If we have a system of 50 equations in 100 variables, there are potentially ( 100 50 ) basic solutions, so this is not practical. Here is a method (an algorithm) to Fnd a nonnegative basic solution or else a proof that none exist. To fnd a nonnegative solution oF A x = b, or a prooF that none exist: Start with the matrix M = A b . Use pivot operations or individual elementary row operations to make the “ A -part” of the matrix basic. ( ) If the enties in the last column are nonnegative, then the corresponding basic solution is nonnegative and we stop. Otherwise, there is a negative entry in the last column, say in row i . If all other entries in row i are positive, there are no nonnegative solutions of the i -th equation, and we stop. If there is another negative entry in row i , pick one, say in columns j , and pivot on position ( i, j ) to get a new matrix representing an equivalent system of equations. Go to ( ).
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4 - Linear Inequalities - Math 1b Practical Nonnegative...

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