MIE376 Lec 3 - Revised Simplex

MIE376 Lec 3 - Revised Simplex - MIE376 Mathematical...

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MIE376 Mathematical Programming Lecture Notes Daniel Frances © 2011 1 Lecture 3 Revised Simplex Method – Matrix Notation Before we showed how to solve problems by simple substitution; now we will deal with algorithms which more closely resemble how commercial codes perform the calculations. The notation introduced here is essential to be able to deal with the more advanced LP concepts of decomposition and column generation to be covered later in this course. Only the second box in the earlier flow chart is changed to: Back to the lathe example: Maximize z = x 1 +2*x 2 + 0*x 3 + 0*x 4 subject to x 1 + 3*x 2 + x 3 = 16 x 1 + x 2 +x4 = 7 x 1 , x 2 , x 3 , x 4 0 In Matrix form Max c T x s.t. Ax=b, x≥0 where = 4 3 2 1 x x x x x = 0 0 2 1 c = 7 16 b = 1 0 1 1 0 1 3 1 A Let’s separate all the elements of the original problem into a basic and a non-basic part the vector x will have the basic variables in a vector x B and the non-basic variables in x N the vector c will have the coefficients of the basic variables in c B and the non-basic ones in c N None None Find an initial basic feasible solution Update the basis Update B -1 Identify the entering and departing variables entering variable? departing variable? Optimal solution Unbounded solution
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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.
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MIE376 Lec 3 - Revised Simplex - MIE376 Mathematical...

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