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matrixcor(1)

# matrixcor(1) - Solving LP Models and Interpreting Answer...

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1 Solving LP Models and Interpreting Answer McCarl and Spreen Chapter 3 The matrix version of the basic LP problem can be expressed as in the equations below. Max CX s.t. AX < b X > 0 C is a 1xN vector of profit contributions X is a Nx1 vector of decision variables A is a MxN matrix giving resource use by the X 's b is a Mx1 vector of resource endowments .

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2 Solving LP Models and Interpreting the Answer Slack Variables To solve we need to convert the LP inequality system to equalities . We do this by adding slack variables which account for the difference between the resource endowment (b) and the use of resources by the variables (AX). Thus, define S = b - AX as vector of slack variables and redefine the equations as AX + IS = b, where I is an MX M identity matrix and S is a Mx1 vector. The slack variables appear in the objective function with zero coefficients. Thus, we add an 1xM vector of zero's. The resultant augmented LP is MAX CX + OS s.t. AX + IS = b X, S > 0.
3 Solving LP Models and Interpreting the Answer Slack Variables The slack variables appear in the objective function with zero coefficients. Thus, we add an 1xM vector of zero's. The resultant augmented LP is MAX CX + OS s.t. AX + IS = b X, S > 0. Throughout the rest of this section we redefine the X and C vector to contains terms for both the original X's and the slacks. T he new A matrix will contain the original A matrix along with the identity matrix for the slacks . The resultant problem is MAX CX s.t. AX = b X > 0 Note A is now nonsquare with more columns than rows which involves multiple solutions

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4 Solving LP Models and Interpreting the Answer Matrix Solution Mathematicians have found the above can be solved by carefully choosing M variables to be nonzero and setting the rest to zero then inverting . The set of variables chosen for the set of M are called basic . The set of variables outside of those M is called non basic . So we partition the problem MAX C B X B + C NB X NB s.t. BX B + A NB X NB = b X B , X NB > 0. where X B , are the basic variables, and X NB , are the nonbasic variables. So BX B + A NB X NB = b Or BX B + 0 = b Or X B = B -1 b And if we consider non basic X B = B -1 b - B -1 A NB X NB
5 Solving LP Models and Interpreting the Answer Matrix Solution Now we know X B = B -1 b - B -1 A NB X NB And suppose we don’t know if we have the right basis Out objective function is Z = C B X B + C NB X NB Substituting the X B equation yields Z = C B (B -1 b – B -1 A NB X NB ) + C NB X NB or Z = C B B -1 b - C B B -1 A NB X NB + C NB X NB or Z = C B B -1 b - (C B B -1 A NB -C NB )X NB

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6 Solving LP Models and Interpreting the Answer Matrix Solution Now we know Z = C B B -1 b - (C B B -1 A NB -C NB )X NB Or , letting a j denote the jth column of A NB Now note So if we have a basis and want to increase the objective chose to add a variable which has most negative term C B B -1 a j j If all positive we are optimal NB j ) c - a B (C - x Z j j 1 - B j = ( ) j X = NB j j j -1 B -1 B c - a B C - b B C Z
7 Solving LP Models and Interpreting the Answer Matrix Solution Given variable to enter, one must leave And we know X B = B -1 b - B -1 A

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matrixcor(1) - Solving LP Models and Interpreting Answer...

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