4 3 2 2 1 2 1 2 21 20 γ γ α α since the minimum of these ratios

# 4 3 2 2 1 2 1 2 21 20 γ γ α α since the minimum

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4 3 2 2 / 1 , 2 / 1 2 , min 11 10 21 20 = = γ γ α α since the minimum of these ratios correspond to 11 10 γ γ , non-basic variables can be removed. Thus we introduce a free variable, u 1 as an additional non-basic variable, defined by 1 1 1 1 4 1 2 2 1 2 1 2 1 x x x f u = = = Note that now the basis has two basic variables x 2 and x 1 (just entered). That is, we have ( ) ( ) 2 1 1 , , x x x and u s x B NB = = . Expressing the basic x B in terms of non-basic x NB , we have, 1 1 4 1 u x = ( ) . 2 1 2 1 8 15 4 2 1 1 3 1 2 s u x x x and + = = The objective function, expressing in terms of x NB is, . 2 3 16 97 4 1 2 1 2 1 8 15 3 4 1 2 2 1 2 1 1 1 u s u s u u f = + + = Now, 2 3 0 0 = = = u NB x s F ; 0 2 1 1 0 0 = = = = u u f u NB x δ δ since 0 1 u f for all x j in x NB and , 0 = u f the current solution is optimal. Hence the optimal basic feasible solution to the given problem is: 16 97 * , 8 15 , 4 1 2 1 = = Z x x Similarly we can find that by Wolfe’s algorithm the optimal point is at (1/4, 15/8). which was introduced by [14]. Thus for the optimal solution for the given QP problem is ( ) 2 1 2 1 2 1 2 2 3 1 1 5 1 2 3 4 8 4 9 7 1 1 5 , , 1 6 4 8 M a x Z x x x a t x x = + = + = = Therefore the solution obtained by graphical solution method, Kuhn-Tucker conditions, Beale’s method and Wolf’s algorithm are same. The computational cost is that by the graphical solution method using MATLAB Programming it will take very short time to determine the plan of action and the solution obtained by graphical method is more effective than any other methods we considered.