凸优化计算实例CVX：Bound-constrained least squares

考虑一个最基本的凸优化问题：
最小二乘，也称为线性回归

设y=Ax+b， x ∈ R n x \in R^n x∈Rn, 最小二乘问题是求x使得y的平方和最小化。
首先，让我们举一个无约束的例子：
m=16;
n=8;
A=randn(m,n);
b=randn(m,1);
最小二乘解KaTeX parse error: Double superscript at position 7: x=(A^T^̲A)^-1^A^T^b很容易通过 backslash operator计算得到:
x_ls = A \ b;
这种简单问题没有必要使用CVX工具计算，但为了验证求解结果，我们求解如下（注意目标函数中的二范数最小化与y的平方和最小化是等价的）：
cvx_begin
variable x(n)
minimize( norm(Ax-b) )
cvx_end
然后，考虑一个有约束的例子，这时就不能直接通过backslash operator计算得到最小二乘解。
例如x的下界为l，上界为u：
bnds = randn(n,2);
l = min( bnds, [], 2 );
u = max( bnds, [], 2 );
但可以使用matlab自带的QP solver求解，但需要转化为标准形式：
x_qp = quadprog( 2A’A, -2A’b, [], [], [], [], l, u );
转化过程详见quadprog函数说明。
此时我们使用CVX求解就显得方便多了：
cvx_begin
variable x(n)
minimize( norm(Ax-b) )
subject to
l <= x <= u
cvx_end
顺便说一句，CVX不会通过对目标函数求平方来将这个问题转换为QP求解， 而是将其转换为SOCP。可以验证，求解结果是相同的，并且转换是自动完成的。
总结：CVX工具就是尿不湿。
CVX工具安装使用链接
https://blog.csdn.net/Laynezhao/article/details/112118753

m=16;
n=8;
A=randn(m,n);
b=randn(m,1);

bnds = randn(n,2);
l = min( bnds, [], 2 );
u = max( bnds, [], 2 );

x_qp = quadprog( 2*A'*A, -2*A'*b, [], [], [], [], l, u )

cvx_begin
variable x(n)
minimize( norm(A*x-b) )
subject to
l <= x <= u
cvx_end
display(x,'x')

Minimum found that satisfies the constraints.

Optimization completed because the objective function is non-decreasing in 
feasible directions, to within the default value of the optimality tolerance,
and constraints are satisfied to within the default value of the constraint tolerance.

<stopping criteria details>


x_qp =

    0.5404
   -0.0915
    0.2436
   -0.8536
    1.2815
    0.1250
   -0.8840
   -0.2598

 
Calling SDPT3 4.0: 33 variables, 9 equality constraints
   For improved efficiency, SDPT3 is solving the dual problem.
------------------------------------------------------------

 num. of constraints =  9
 dim. of socp   var  = 17,   num. of socp blk  =  1
 dim. of linear var  = 16
*******************************************************************
   SDPT3: Infeasible path-following algorithms
*******************************************************************
 version  predcorr  gam  expon  scale_data
    NT      1      0.000   1        0    
it pstep dstep pinfeas dinfeas  gap      prim-obj      dual-obj    cputime
-------------------------------------------------------------------
 0|0.000|0.000|1.6e+00|2.8e+00|1.7e+03| 7.161863e+01  0.000000e+00| 0:0:00| chol  1  1 
 1|0.927|0.966|1.1e-01|1.2e-01|1.3e+02| 5.081575e+01 -9.409293e+00| 0:0:00| chol  1  1 
 2|1.000|1.000|1.4e-06|2.7e-03|1.3e+01| 4.300286e+00 -8.185259e+00| 0:0:00| chol  1  1 
 3|0.867|0.771|2.6e-07|8.3e-04|2.4e+00|-5.141461e+00 -7.467459e+00| 0:0:00| chol  1  1 
 4|0.733|1.000|2.7e-07|2.7e-05|1.2e+00|-6.103270e+00 -7.252439e+00| 0:0:00| chol  1  1 
 5|1.000|0.807|1.5e-09|7.4e-06|2.2e-01|-6.889683e+00 -7.105866e+00| 0:0:00| chol  1  1 
 6|0.878|1.000|8.9e-10|2.7e-07|4.8e-02|-7.003738e+00 -7.051811e+00| 0:0:00| chol  1  1 
 7|1.000|0.902|6.9e-10|5.1e-08|4.3e-03|-7.038176e+00 -7.042472e+00| 0:0:00| chol  1  1 
 8|0.968|0.985|2.8e-10|3.6e-09|2.3e-04|-7.040772e+00 -7.041001e+00| 0:0:00| chol  1  1 
 9|0.984|0.988|1.4e-10|1.0e-10|3.5e-06|-7.040942e+00 -7.040945e+00| 0:0:00| chol  1  1 
10|0.958|1.000|6.1e-12|2.8e-11|2.8e-07|-7.040944e+00 -7.040945e+00| 0:0:00| chol  1  1 
11|1.000|1.000|1.9e-12|1.2e-12|1.7e-08|-7.040945e+00 -7.040945e+00| 0:0:00|
  stop: max(relative gap, infeasibilities) < 1.49e-08
-------------------------------------------------------------------
 number of iterations   = 11
 primal objective value = -7.04094466e+00
 dual   objective value = -7.04094468e+00
 gap := trace(XZ)       = 1.71e-08
 relative gap           = 1.14e-09
 actual relative gap    = 1.13e-09
 rel. primal infeas (scaled problem)   = 1.93e-12
 rel. dual     "        "       "      = 1.23e-12
 rel. primal infeas (unscaled problem) = 0.00e+00
 rel. dual     "        "       "      = 0.00e+00
 norm(X), norm(y), norm(Z) = 3.6e+00, 7.4e+00, 1.0e+01
 norm(A), norm(b), norm(C) = 1.3e+01, 2.0e+00, 1.5e+01
 Total CPU time (secs)  = 0.37  
 CPU time per iteration = 0.03  
 termination code       =  0
 DIMACS: 1.9e-12  0.0e+00  2.0e-12  0.0e+00  1.1e-09  1.1e-09
-------------------------------------------------------------------
 
------------------------------------------------------------
Status: Solved
Optimal value (cvx_optval): +7.04094
 

x =

    0.5404
   -0.0915
    0.2436
   -0.8536
    1.2815
    0.1250
   -0.8840
   -0.2598

标签：00,01,CVX,chol,Bound,12,1.000,squares,norm
来源： https://blog.csdn.net/Laynezhao/article/details/113770689