本题摘自《最优化方法及其Matlab程序设计 第一版》 123(130/271)

【例22】考虑优化问题 m i n f ( x ) = − 2 x 1 2 − x 1 2   , s . t . x 1 2 + x 2 2 − 2 = 0   , − x 1 + x 2 ≥ 0   , x 1 ≥ 0   , x 2 ≥ 0. \mathrm{min} f(x) = -2x^2_1-x^2_1 \ , \\ \mathrm{s.t.} \quad x^2_1+x^2_2-2=0 \ ,\\ -x_1+x_2 \ge 0 \ , \\ x_1 \ge 0 \ , x_2 \ge 0. minf(x)=−2x12​−x12​ ,s.t.x12​+x22​−2=0 ,−x1​+x2​≥0 ,x1​≥0 ,x2​≥0. 试验证 x ∗ = ( 1 , 1 ) T x^*=(1,1)^\mathrm{T} x∗=(1,1)T为 K T KT KT点，并求出问题的 K T KT KT对。

【解】记 f ( x ) = − 2 x 1 2 − x 1 2 , h ( x ) = x 1 2 + x 2 2 − 2 , g 1 ( x ) = − x 1 + x 2 , g 2 ( x ) = x 1 , g 3 ( x ) = x 2 . \begin{aligned} f(x) &= -2x^2_1-x^2_1,\\ h(x) &= x^2_1+x^2_2-2,\\ g_1(x) &= -x_1+x_2, \\ g_2(x) &= x_1, \\ g_3(x) &= x_2. \end{aligned} f(x)h(x)g1​(x)g2​(x)g3​(x)​=−2x12​−x12​,=x12​+x22​−2,=−x1​+x2​,=x1​,=x2​.​ 求梯度，得到 ▽ f ( x ) = [ − 4 x 1 − 2 x 2 ] , ▽ h ( x ) = [ 2 x 1 2 x 2 ] , ▽ g 1 ( x ) = [ − 1 1 ] , ▽ g 2 ( x ) = [ 1 0 ] , ▽ g 3 ( x ) = [ 0 1 ] . \begin{aligned} \bigtriangledown f(x) &= \begin{bmatrix} -4x_1 \\ -2x_2 \end{bmatrix}, \quad \bigtriangledown h(x) = \begin{bmatrix} 2x_1 \\ 2x_2 \end{bmatrix}, \\ \bigtriangledown g_1(x) &= \begin{bmatrix} -1 \\ 1 \end{bmatrix}, \quad \bigtriangledown g_2(x) = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \quad \bigtriangledown g_3(x) = \begin{bmatrix} 0 \\ 1 \end{bmatrix}. \end{aligned} ▽f(x)▽g1​(x)​=[−4x1​−2x2​​],▽h(x)=[2x1​2x2​​],=[−11​],▽g2​(x)=[10​],▽g3​(x)=[01​].​ 把 x ∗ = ( 1 , 1 ) T x^*=(1,1)^\mathrm{T} x∗=(1,1)T代入上面5个式子，由 K T KT KT条件有，

注意：这里一定注意要加负号！！！！

{ − 4 − 2 μ + ω 1 − ω 2 = 0 − 2 − 2 μ − ω 1 + ω 3 = 0 (1) \left\{\begin{matrix} -4-2\mu+\omega_1-\omega_2 = 0 \\ -2-2\mu-\omega_1+\omega_3 = 0 \end{matrix}\right. \tag{1} {−4−2μ+ω1​−ω2​=0−2−2μ−ω1​+ω3​=0​(1) 因为 { ω 2 ▽ g 2 ( x ˉ ) = 0 ω 3 ▽ g 3 ( x ˉ ) = 0 ⇒ { ω 2 ∗ = 0 ω 3 ∗ = 0 \left\{\begin{matrix} \omega_2\bigtriangledown g_2(\bar{x}) = 0 \\ \omega_3\bigtriangledown g_3(\bar{x}) = 0 \end{matrix}\right. \Rightarrow \left\{\begin{matrix} \omega^*_2 = 0 \\ \omega^*_3 = 0 \end{matrix}\right. {ω2​▽g2​(xˉ)=0ω3​▽g3​(xˉ)=0​⇒{ω2∗​=0ω3∗​=0​ 所以 ( 1 ) (1) (1)变为 { − 4 − 2 μ + ω 1 = 0 − 2 − 2 μ − ω 1 = 0 (2) \left\{\begin{matrix} -4-2\mu+\omega_1 = 0 \\ -2-2\mu-\omega_1 = 0 \end{matrix}\right. \tag{2} {−4−2μ+ω1​=0−2−2μ−ω1​=0​(2) 求解 ( 2 ) (2) (2)式，得到 { μ ∗ = − 1.5 ω 1 ∗ = 1 \left\{\begin{matrix} \begin{aligned} \mu^* &= -1.5 \\ \omega^*_1 &= 1 \end{aligned} \end{matrix}\right. {μ∗ω1∗​​=−1.5=1​​ 这表明 x ∗ x^* x∗ 是 K T KT KT 点， ( x ∗ , ( μ ∗ , ω ∗ ) ) (x^*,(\mu^*,\omega^*)) (x∗,(μ∗,ω∗))是 K T KT KT 对，其中， μ ∗ = − 1.5 \mu^*=-1.5 μ∗=−1.5， ω ∗ = ( 1 , 0 , 0 ) T \omega^*=(1,0,0)^\mathrm{T} ω∗=(1,0,0)T。