二次规划的对偶形式(CVX)

文章目录

输入数据
优化代码
计算对偶间隙
原目标
拉格朗日的形式
对偶形式
参考

解析一下 Boyd & Vandenberghe, "Convex Optimization"上的例子，重点在于其对偶形式是怎么得到的
Section 5.2.4: Solves a simple QCQP

输入数据

randn('state',13);
n = 6;
P0 = randn(n); P0 = P0'*P0 + eps*eye(n);
P1 = randn(n); P1 = P1'*P1;
P2 = randn(n); P2 = P2'*P2;
P3 = randn(n); P3 = P3'*P3;
q0 = randn(n,1); q1 = randn(n,1); q2 = randn(n,1); q3 = randn(n,1);
r0 = randn(1); r1 = randn(1); r2 = randn(1); r3 = randn(1);

优化代码

cvx_beginvariable x(n)dual variables lam1 lam2 lam3minimize( 0.5*quad_form(x,P0) + q0'*x + r0 )lam1: 0.5*quad_form(x,P1) + q1'*x + r1 <= 0;lam2: 0.5*quad_form(x,P2) + q2'*x + r2 <= 0;lam3: 0.5*quad_form(x,P3) + q3'*x + r3 <= 0;
cvx_end
obj1 = cvx_optval;

计算对偶间隙

P_lam = P0 + lam1*P1 + lam2*P2 + lam3*P3;
q_lam = q0 + lam1*q1 + lam2*q2 + lam3*q3;
r_lam = r0 + lam1*r1 + lam2*r2 + lam3*r3;
obj2 = -0.5*q_lam'*inv(P_lam)*q_lam + r_lam;
disp('The duality gap is equal to ');
disp(obj1-obj2)

原目标

minimize12xTP0x+q0Tx+r012xTP1x+q1Tx+r1<=012xTP2x+q2Tx+r2<=012xTP3x+q3Tx+r3<=0minimize \ \frac{1}{2}x^TP_0x+q_0^Tx+r_0\\ \frac{1}{2}x^TP_1x+q_1^Tx+r_1<=0\\ \frac{1}{2}x^TP_2x+q_2^Tx+r_2<=0\\ \frac{1}{2}x^TP_3x+q_3^Tx+r_3<=0 minimize 21xTP0x+q0Tx+r021xTP1x+q1Tx+r1<=021xTP2x+q2Tx+r2<=021xTP3x+q3Tx+r3<=0
P0∈S++,P1,P2,P3∈S+P_0 \in S_{++},P_1,P_2,P_3 \in S_+P0∈S++,P1,P2,P3∈S+,原目标是严格凸函数，有唯一极小解

拉格朗日的形式

L(x,λ1,λ2,λ3)=12xTP0x+q0Tx+r0+∑i=13λi(12xTPix+qiTx+ri)=12xTP(λ)x+q(λ)Tx+r(λ)P(λ)=P0+∑i=13λiPiq(λ)=q0+∑i=13λiqir(λ)=r0+∑i=13λiriλ⪰0\mathcal{L}(x,\lambda_1,\lambda_2,\lambda_3)=\frac{1}{2}x^TP_0x+q_0^Tx+r_0+\sum_{i=1}^3\lambda_i(\frac{1}{2}x^TP_ix+q_i^Tx+r_i)\\= \frac{1}{2}x^TP(\lambda)x+q(\lambda)^Tx+r(\lambda)\\ P(\lambda)=P_0+\sum_{i=1}^3\lambda_iP_i\\ q(\lambda)=q_0+\sum_{i=1}^3\lambda_iq_i\\ r(\lambda)=r_0+\sum_{i=1}^3\lambda_ir_i\\ \lambda\succeq 0 L(x,λ1,λ2,λ3)=21xTP0x+q0Tx+r0+i=1∑3λi(21xTPix+qiTx+ri)=21xTP(λ)x+q(λ)Tx+r(λ)P(λ)=P0+i=1∑3λiPiq(λ)=q0+i=1∑3λiqir(λ)=r0+i=1∑3λiriλ⪰0

对偶形式

由于对偶变量λ⪰0\lambda\succeq 0λ⪰0,因此，P(λ)∈S++P(\lambda)\in S_{++}P(λ)∈S++
对拉格朗日求导可以得到极小值的位置
L(x,λ)=12xTP(λ)x+q(λ)Tx+r(λ)→P(λ)x+q(λ)=0→x=−P(λ)−1q(λ)\mathcal{L}(x,\lambda)=\frac{1}{2}x^TP(\lambda)x+q(\lambda)^Tx+r(\lambda)\rightarrow\\ P(\lambda)x+q(\lambda)=0\rightarrow x=-P(\lambda)^{-1}q(\lambda)L(x,λ)=21xTP(λ)x+q(λ)Tx+r(λ)→P(λ)x+q(λ)=0→x=−P(λ)−1q(λ)
代入拉格朗日函数，得到对偶形式
g(λ)=infx(L(x,λ))=−12q(λ)TP(λ)−1q(λ)+r(λ)λ⪰0g(\lambda)=\underset{x}{inf}(\mathcal{L}(x,\lambda))=-\frac{1}{2}q(\lambda)^TP(\lambda)^{-1}q(\lambda)+r(\lambda)\\\lambda\succeq 0g(λ)=xinf(L(x,λ))=−21q(λ)TP(λ)−1q(λ)+r(λ)λ⪰0

参考

http://web.cvxr.com/cvx/examples/cvxbook/Ch05_duality/html/qcqp.html