Computer Aided Engineering Design

358 COMPUTER AIDED ENGINEERING DESIGN

∂

∂

⎡

⎣⎢

⎤

⎦⎥

L

x

xx x

1

= 2 ( 1112 – 1) – 2 – = 2 (1 – ) 112 – 1 –

1

2

λλ λ λ = 0

∂

∂

L

x

x

2

= 2 + = 0 21 λ

λ 12 [(xx + 2) – 12 ] = 0

and λ 2 x 1 = 0

Case I (λ 1 = λ 2 = 0): Here, x 2 = 0 and x 1 = 1 for which g 1 (x 1 ,x 2 ) = 1 while g 2 (x 1 ,x 2 ) = –1. Since one
constraint is not satisfied, the solution in infeasible.

Case II (λ 1 ≠ 0, λ 2 = 0): This is equivalent to the solution in Case II of Example 12.9(a). Since
λ 1 ≠ 0, g 1 (x 1 ,x 2 ) is active and satisfied for the three solution points. Evaluating g 2 (x 1 ,x 2 ) at these
points, we get

g 2 (1.36, – 0.13) = –1.36 < 0

g 2 (–1, –1) = 1 > 0

and g 2 (–0.36, –1.86) = 0.36 > 0

which gives the only feasible minimum at (1.36, – 0.13), that is, point A in Figure 12.12. That
g 2 (x 1 ,x 2 ) is inactive at A, the Hessian is still positive definite with eigenvalues 2(1 – 0.27) and 2.

Case III (λ 2 ≠ 0, λ 1 = 0): We have x 1 = 0 and x 2 = 0 from the second and fourth KKT conditions
above. Note that g 2 (x 1 ,x 2 ) is active while g 1 (0, 0) = 2 which is greater than 0 and thus this solution
is not feasible.

Case IV (λ 1 ≠ 0, λ 2 ≠ 0): We have the two constraints active, that is, x 1 = 0 and (x 2 + 2) – x 12 = 0
which implies that x 2 = –2. From the first and second KKT conditions for this problem, computing
the multipliers gives λ 1 = 4 and λ 2 = –2. Since both multipliers are not positive, (0, –2) is not a
minimum for the function.
Combining the Cases I-IV gives (1.36, – 0.13) or point A as the constrained minimum which is
suggested by Figure 12.12 as points B and C do not comply with g 2 (x 1 ,x 2 )≤ 0. Also, the sufficiency
condition is satisfied for point A.
Section 12.3 discussed the classical methods to solve multi-variable optimization problems without
and with equality and inequality constraints. There may be problems wherein we may require handling
both equality and inequality constraints for which the KKT conditions get slightly modified. If f(X)
is to be minimized subject to m inequality constraints gi(X)≤ 0, i = 1,... , m and p equality
constraintshj(X) = 0, j = 1,... , p, then the KKT necessary conditions can be stated as

∇∇ ∇fg hi

m

iij

p

+ ΣΣ=1λβ + =1 jj= 0

λigi = 0, i = 1,... , m

gi≤ 0, i = 1,... , m

hj = 0, j = 1,... , p

and λi≥ 0, i = 1,... , m (12.28a)