Engineering Optimization: Theory and Practice, Fourth Edition

102 Classical Optimization Techniques

that is,

x 1 − 05 ≥ 0 (E 7 )

x 1 +x 2 − 001 ≥ 0 (E 8 )

x 1 +x 2 +x 3 − 501 ≥ 0 (E 9 )

λj≤ 0 , j= 1 , 2 , 3

that is,

λ 1 ≤ 0 (E 10 )

λ 2 ≤ 0 (E 11 )

λ 3 ≤ 0 (E 12 )

The solution of Eqs.(E 1 ) ot (E 12 ) an be found in several ways. We proceed to solvec

these equations by first nothing that eitherλ 1 = or 0 x 1 = 0 according to Eq. 5 (E 4 ).

Using this information, we investigate the following cases to identify the optimum

solution of the problem.

Case 1:λ 1 = 0.

Equations(E 1 ) ot (E 3 ) iveg

x 3 = −

λ 3

2

x 2 = − 10 −

λ 2

2

−

λ 3

2

(E 13 )

x 1 = − 20 −

λ 2

2

−

λ 3

2

Substituting Eqs.(E 13 ) n Eqs.i (E 5 ) nda (E 6 ) we obtain,

λ 2 ( 30 − 1 −λ 2 −λ 3 )= 0

λ 3 ( 80 − 1 −λ 2 −^32 λ 3 )= 0 (E 14 )

The four possible solutions of Eqs. (E 14 ) rea

1.λ 2 = 0 ,− 180 −λ 2 −

3

2

λ 3 =. These equations, along with Eqs. 0 (E 13 ) yield,

the solution

λ 2 = 0 , λ 3 = − 120 , x 1 = 04 , x 2 = 05 , x 3 = 06

This solution satisfies Eqs.(E 10 ) ot (E 12 ) ut violates Eqs.b (E 7 ) nda (E 8 ) nda

hence cannot be optimum.

2.λ 3 = 0 ,− 130 −λ 2 −λ 3 =. The solution of these equations leads to 0

λ 2 = − 130 , λ 3 = 0 , x 1 = 54 , x 2 = 55 , x 3 = 0