A Simple Application of the Attic Method

In the starting point, the Attic method verifies whether any variable has only floor constraints as satisfied constraints. In this problem, both the variables have the same sign in both the objective function and in the satisfied constraints hence there are no roof constraints for them and it is possible to perform a search along X by obtaining xi = l x2 =0 F = —5, and along x 2 by obtaining xi = 0 x 2 = 1 F = —6. 

Having performed the aforementioned searches, the Attic method executes another search with the equality constraints and the artificial constraints as the initial matrix J. Since no equality constraints are involved in this problem, the initial matrix J is the identity matrix. Thus, the initial operating matrix, Fi, that must be factorized, has dimensions 0x0. 

The search along such a direction indicates the point Bauw Xi = 10/9 X2 = 4/3 placed on the constraint  

In correspondence with this point, the function is equal to F = —(50/9) — 8 — 6 and is better than the other points obtained by the previous searches along the axes X and X2- Thus, it is opportune to continue on from the point Xi = 10/9 %2 = 4/3 and the constraint (10.40) is inserted. 

The new matrix J will have this (and only this) active constraint as its first row. The other constraints are artificial constraints even though some of them were active. The second row is still an artificial constraint. Therefore, the operating matrix, Fi, that must be factorized, has dimensions 1x1. If the first row is factorized by using variable no. 1 as the pivot, the artificial constraint will involve variable no. 2. 

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