Applied Cryptology, Cryptographic Protocols, and Computer by Richard A. Demillo

By Richard A. Demillo

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Transformed row 2 from step 3 0 1 -1 + row 3 0 -1 2 0 new row 3 0 0 1 -2 -2 3 0 0 1 3 1 At the end of the second column operation, A has been transformed into A2 and the identity matrix into Bt : 1 0 A, = 0 1 0 0 1 Bx = 1 2 2 -1 0 3 0 3 1 Finally, the sixth and seventh steps convert a33 into 1 and al3 and a23 into 0. The element a 33 is already 1 and does not need further transformation. 26 / 2: ALGEBRAIC METHODS Step 6 Multiply row 3 by — 1 and add it to row 1. row 3 Step 7 -1[0 0 l][-2 3 1] = [0 transformed row 3 0 + row 1 1 0 new row 1 1 0 0 -1 0 -1][2 2 -3 -1 1 1 -1 0 0 3 -4 -3 -1] -1 Add row 3 to row 2.

Xx + X2 - X6 + ΧΊ = 4 (5") The artificial variable ΧΊ produces an initial feasible solution. If two negative slack variables were in the problem, then two artificial variables would be necessary. In short, an artificial variable must be added for every negative slack variable and when the constraints are equalities. The artificial variables are purely statistical and have no real value. Thefirstphase of the process consists of preparing a standard problem that will ensure the artificial variables can be removed from the basis.

Maximize Z: 0') subject to (2') (3') (4') 3. THE TWO-PHASE PROCESS Most linear programming problems do not limit themselves to positive slack variables. If one or more of the slack variables is negative, or if the problem has one or more equality constraints, then we do not have an initial MATHEMATICAL SOLUTION OF LINEAR PROGRAMMING PROBLEMS / 45 feasible solution because the number of basis vectors must equal the number of constraint equations. A basis vector must be a positive one, a condition not met by a negative slack variable or an equality constraint which does not have a slack variable.

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