How to maximize system of linear equations?

6 次查看（过去 30 天）
Clarisha Nijman
Clarisha Nijman 2018-11-26
评论： Clarisha Nijman 2018-12-1
Hello,
Given:
A=[4 3; -1 7; 5 9; 2 4];
x=[x1;x2];
b=[b1; b2; b3; b4];
How can I maximize the linear system of equations: Ax=b?
  4 个评论
显示 2更早的评论
Matt J
Matt J 2018-11-26
编辑：Matt J 2018-11-26
associated with the maximum possible value of b1, b2, b3 and b4
So the idea is to make all b1..b4 as large as possible? Then clearly x=0 and y=Inf are optimal in the example you've shown. They result in b1=b2=b3=b4=Inf.
More generally, though, you cannot simultaneously maximize the right hand side elements b, because they co-depend on the same variables.
Clarisha Nijman
Clarisha Nijman 2018-11-26
Okay, I understand, maybe this example is too simple. In my case I have to minimize the maximal entry of a linear combination of 20 vectors with each 1001 entries. So I was wondering how to find the maximal/optimal entry of such a linear combination of vectors. A linear combination of vectors is a vector; in the simplified example I started with, it is the b-vector on the right hand side. The left hand side is the linear combination. So I need a code/matlab function that can help me to find the maximal entry.
min max (ax+by+cz+......+dv)
where x, y, z and v are vectors and a, b, c and d non-negative scalars. Minmization has as argument the scalars a,b,,c,d. And for simplicity we minimize the maximal entry of the linear combination.
My question is how to find this maximal entry. Is there a matlab function/code I can use?

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Matt J
Matt J 2018-11-26
编辑：Matt J 2018-11-27
You can use fminimax, e.g.,
fminimax(@(x) A*x, x0)
  6 个评论
显示 4更早的评论
Matt J
Matt J 2018-11-30
编辑：Matt J 2018-11-30
Runs fine for me. Here is my complete implementation.
C=[ 0.0038 0.0038 0.0038 0.0038
0.0037 0.0037 0.0037 0.0037
0.0036 0.0036 0.0036 0.0036
0.0034 0.0034 0.0035 0.0035
0.0033 0.0033 0.0034 0.0034
0.0032 0.0032 0.0033 0.0033
0.0031 0.0031 0.0032 0.0033
0.0029 0.0029 0.0031 0.0031
0.0028 0.0028 0.0029 0.0028
0.0027 0.0027 0.0024 0.0023];
A1=C(:,2:end);
A2=-C(:,2:end);
Aleq=[A1;A2];
%The less equal RHS
bleq=[C(:,1) -C(:,1)];
Aeq=ones(3,1);
beq=1;
lb=zeros(3,1);
ub=[];
%initial guess
x0=0.1*rand(3,1);
%Defining the objective function
Cs=C(:,2:4);
[OptArgum, optimalVal] = fminimax(@(x) Cs*x,x0,Aleq,bleq(:),Aeq.',beq,lb,ub);
Clarisha Nijman
Clarisha Nijman 2018-12-1
Ok I see,
Two questions more;
Can you give me some explanation about the first argument of the fminmax code: @(x) Cs*x
Can you give me some explanation about the output? Should I change the tolerance of choose a larger rand initial guess? The output says:
Converged to an infeasible point.
fminimax stopped because the predicted change in the objective function
is less than the default value of the function tolerance but constraints
are not satisfied to within the default value of the constraint tolerance.
<stopping criteria details>
Optimization stopped because the predicted change in the objective function, 0.000000e+00,
is less than options.FunctionTolerance = 1.000000e-06, but the maximum constraint violation,
1.093094e+00, exceeds options.ConstraintTolerance = 1.000000e-06.
Optimization Metric Options
abs(steplength*directional derivative) = 0.00e+00 FunctionTolerance = 1e-06 (default)
max(constraint violation) = 1.09e+00 ConstraintTolerance = 1e-06 (default)

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Solver Outputs and Iterative Display 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by