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I want to find an optimal solution for a linear set of equations with more unknowns than equations, ultimately up to 3 equations and some 4-10 variables.

Lets start simple with 2 equations, 3 unknowns

coefficient matrix:

A=[1 1 0; 0 0 1];

b = [10;10];

Solution represent a set of forces which have a certain upper limit, e.g.: ub=[20 20 20];

I do the following:

1. find particular solution:

Fp = A\b;

2. find null solution:

F0 = null(A);

Now all possible solutions are given by:

F = Fp + F0 * const

and the target is to find a constant which gives an optimal solution. Optimal I define as follows:

We are talking about force equilibrium, F represent forces which have a certain maximum (ub) and minimum (lb).

My solution is optimal if the sum of the square of the ratios F/Fmax is minimal:

obj = sum( (F./ub).^2 )

In this simple example it is easy to hand-calculate that this is obtained with F = [5 5 10]

I use optimisation tool fmincon. For the F0 multiplier "const" I add a new variable to my solution and add a column of zeros to A which gives Aeq.

Using the following piece of code:

>>>

%%%Add zeros after A

Aeq(:,numF+1:numX) = 0;

beq = b;

x0=[0; 0; 0; 0];

options = optimset('LargeScale','off');

[x,fval,exitflag,output] = fmincon(@(x) Fobjfun(x, Fp, F0, ub),x0,[],[],Aeq,beq,lb,ub,[],options)

>>>

With Fobjfun:

>>>function obj = Fobjfun(x, Fp, F0, ub)

x(1:3) = Fp(i) + F0(i)*x(4);

obj = sum( (x(1:numF)./ub(1:numF)).^2 );

>>>

This works, it gives the answer: x=[5;5;10;-7.0711] and obj 0.375.

Next:

I lower one constraint:

ub = [4; 20; 20]

Now I get the answer:

>>>

x =

3.9255

6.0745

10.0000

-13.5982

obj =

0.4904

>>>

Verification by simply trying a range of possible C's (x4) shows indeed that x4 shall be -13.5982 and the corresponding obj value is 0.4904. However the corresponding forces are not gives in x(1:3). I have to add a final line which reconstructs forces as is done in the objective funtion. This seems strange to me. x4 does not correspond with x1:x3. If I reconstruct forces with F=Fp + F0 * x4 than a set of forces is given which is differnt from x1-x3 but gives the correct (and indeed optimal) objective value.

Questions:

1. Anyone understands what is going wrong here? Or is it not going wrong?

2. Anyone has better ideas in finding the optimal solution from a set of linear equations with more unknowns than equations?

Any help is greatly appreciated,

Mark

> Dear All,

>

> I want to find an optimal solution for a linear set of equations with more unknowns than equations, ultimately up to 3 equations and some 4-10 variables.

> Lets start simple with 2 equations, 3 unknowns

> coefficient matrix:

> A=[1 1 0; 0 0 1];

> b = [10;10];

>

> Solution represent a set of forces which have a certain upper limit, e.g.: ub=[20 20 20];

>

> I do the following:

> 1. find particular solution:

> Fp = A\b;

> 2. find null solution:

> F0 = null(A);

>

> Now all possible solutions are given by:

> F = Fp + F0 * const

>

> and the target is to find a constant which gives an optimal solution. Optimal I define as follows:

> We are talking about force equilibrium, F represent forces which have a certain maximum (ub) and minimum (lb).

> My solution is optimal if the sum of the square of the ratios F/Fmax is minimal:

> obj = sum( (F./ub).^2 )

> In this simple example it is easy to hand-calculate that this is obtained with F = [5 5 10]

>

> I use optimisation tool fmincon. For the F0 multiplier "const" I add a new variable to my solution and add a column of zeros to A which gives Aeq.

> Using the following piece of code:

> >>>

> %%%Add zeros after A

> Aeq(:,numF+1:numX) = 0;

> beq = b;

> x0=[0; 0; 0; 0];

> options = optimset('LargeScale','off');

> [x,fval,exitflag,output] = fmincon(@(x) Fobjfun(x, Fp, F0, ub),x0,[],[],Aeq,beq,lb,ub,[],options)

> >>>

> With Fobjfun:

> >>>function obj = Fobjfun(x, Fp, F0, ub)

> x(1:3) = Fp(i) + F0(i)*x(4);

> obj = sum( (x(1:numF)./ub(1:numF)).^2 );

> >>>

>

> This works, it gives the answer: x=[5;5;10;-7.0711] and obj 0.375.

> Next:

> I lower one constraint:

> ub = [4; 20; 20]

>

> Now I get the answer:

> >>>

> x =

>

> 3.9255

> 6.0745

> 10.0000

> -13.5982

> obj =

>

> 0.4904

> >>>

> Verification by simply trying a range of possible C's (x4) shows indeed that x4 shall be -13.5982 and the corresponding obj value is 0.4904. However the corresponding forces are not gives in x(1:3). I have to add a final line which reconstructs forces as is done in the objective funtion. This seems strange to me. x4 does not correspond with x1:x3. If I reconstruct forces with F=Fp + F0 * x4 than a set of forces is given which is differnt from x1-x3 but gives the correct (and indeed optimal) objective value.

> Questions:

> 1. Anyone understands what is going wrong here? Or is it not going wrong?

> 2. Anyone has better ideas in finding the optimal solution from a set of linear equations with more unknowns than equations?

>

> Any help is greatly appreciated,

>

> Mark

>

>

Dear All,

I found the solution. The reason I use a multiplier for my null solution is the way I previously "solved" these kind of problems: simply by trying a huge range of possibilties of multipliers. However using decent optimisation technology this is unnecessary. I can enter my coefficient matrix directly as Aeq. There is no need for adding an extra variable for every null space vector. So now my code would be:

>>>

A = [1 1 0 0 0;...

0 0 1 1 1;...

1 -1 0.5 1 -1];

beq = [10;10;10];

ub = [20; 20; 20; 20; 20];

lb = [-20; -20; -20; -20; -20];

Aeq = A;

x0=[0; 0; 0; 0; 0];

options = optimset('LargeScale','off');

[x,fval,exitflag,output] = fmincon(@(x) Fobjfun(x, ub),x0,[],[],Aeq,beq,lb,ub,[],options)

>>>

and the objective function:

>>>

function obj = Fobjfun(x, ub)

obj = sum( (x./ub).^2 );

>>>

This works. However I am still curious about what happens in the previous posted method. Any ideas?

Brgds,

Mark

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