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MIE376 Lec 3 - Revised Simplex.page2

# MIE376 Lec 3 - Revised Simplex.page2 - From the constraint...

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MIE376 Mathematical Programming Lecture Notes Daniel Frances © 2011 2 the vector b remains the same the matrix A will be separated to have the columns of the basic variables in B, and those of the non- basic variables in the matrix N Recall the initial basis is B = {x 3 , x 4 } and N = {x 1 , x 2 } We can then write the problem as Max z = c B T x B + c N T x N s.t. x B ≥0 , x N ≥0, Bx B +Nx N =b where = 0 0 B c = 2 1 N c = 4 3 x x x B = 2 1 x x x N = 1 0 0 1 B = 1 1 3 1 N Note that we are using the symbol B to mean three different but related items, easily recognizable by context. 1. B - as a subscript - is the set of basic variables 2. B – as a set of the basic variables, and 3. B – as a matrix – is the square matrix of columns of A corresponding to basic variables in the order of the B-set. ITERATION 1 Update B -1 Since B = I, the Identity matrix, therefore B -1 =I as well Entering Variable Now we need to set the stage for a “matrix" approach for finding the entering variable.
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Unformatted text preview: From the constraint Bx B + Nx N = b pre-multiplying both sides by B-1 x B = B-1 b - B-1 Nx N . Substituting x B back into the objective z = c B T (B-1 b - B-1 Nx N ) + c N T x N z = c B T B-1 b - (c B T B-1 N - c N T ) x N Note • The term c B T B-1 b is the numerical value of the objective function associated with the B-set. • The term c B T B-1 N-c N T is the vector of coefficients of the non-basic variables in the z-equation. • For maximization problems the entering variable has the most negative coefficient, for minimization problems it has the most positive coefficients. In the first iteration this method will appear like overkill, but we can then the same approach for future iterations. Thus the required expression for identifying the entering variable is...
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