# non linear minimization problem

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Alberto Menichetti
on 10 May 2016

Commented: Alberto Menichetti
on 11 May 2016

Suppose I have

aA + bB + cC = E

Where A,B,C are column vectors with components in R+; I need to find a,b,c in R such that a,b,c >= 1 and E is minimized. What's the "right" name for that kind of problem? Which class of algorithms should I use to solve that problem in matlab?

##### 3 Comments

John D'Errico
on 10 May 2016

I'm intrigued how it is that the requirement that each of a,b,c are all greater than 1 is different from the requirement that they are greater than abs(1).

Must be the new math.

### Accepted Answer

Teja Muppirala
on 11 May 2016

It depends on what you mean by "minimize".

Do you mean the sum of squares, sum(E.^2)? If so then it's a quadratic programming problem and it's very easy to set up.

Do you mean the max(abs(E)) or sum(abs(E))? Then it is linear programming, and it is a bit more tricky (you need to set up the constraint matrices right).

In any case, the following is a sample of how you would do it.

----------------------------------

%% First: I'll make some random data to work with

rng(10)

N = 5; % Say 5-dimensions

A = randn(N,1);

B = randn(N,1);

C = randn(N,1);

ABC = [A B C];

%% Case 1: Minimize the sum of squares of E, also called L-2 norm

ABC = [A B C];

Q = ABC'*ABC;

abc_opt_L2 = quadprog(Q,[],[],[],[],[],ones(3,1))

E_opt = ABC*abc_opt_L2;

%% Case 2: Minimize sum(abs(E)) also called L-1 norm

f = [0 0 0 ones(1,N)];

A_cons = -eye(3+N);

A_cons = [A_cons; ABC -eye(N); -ABC -eye(N)];

b = [-1; -1 ;-1; zeros(N,1); zeros(2*N,1)];

values = linprog(f,A_cons,b,[],[]);

abc_opt_L1 = values(1:3)

E_opt = ABC*abc_opt_L1;

%% Case 3: Minimize max(abs(E)) also called L-infinity norm

f = [0 0 0 1];

A_cons = blkdiag(-eye(3+1));

A_cons = [A_cons; ABC -ones(N,1); -ABC -ones(N,1)];

b = [-1; -1 ;-1; -1; zeros(2*N,1)];

values = linprog(f,A_cons,b,[],[]);

abc_opt_Linf = values(1:3)

E_opt = ABC*abc_opt_Linf;

The code above gives me the following optimal values for a,b,c in each case

abc_opt_L2 =

1.0000

1.0851

1.0802

abc_opt_L1 =

1.0000

1.6379

1.6855

abc_opt_Linf =

1.0000

1.3236

1.0000

### More Answers (3)

John D'Errico
on 10 May 2016

You want to solve for 4 parameters, a,b,c,E. They all enter the problem linearly.

However, since the set of equality constraints

a*A + b*B + c*C - E = 0

will not by solved exactly in general, you have two competing goals, which you have not said how they will be resolved. So if there are multiple elements in those vectors, no single value for {a,b,c,E} exists to satisfy all equations at once.

So, you MIGHT choose to define this as a linear least squares problem subject to bound constraints on a,b,c, but that ignores the goal that E is minimized. (May E be negative?)

This is effectively a homogeneous linear system in 4 unknowns, subject to bound constraints on {a,b,c}, as well as an additional requirement on E.

1. If your vectors have length greater than 4 then there will in general be NO exact solution.

2. If your vectors have length exactly 4, then depending on the vectors, there may be infinitely many solutions, or there will be no solution other than that all the unknowns are 0. Of course, in that case, it would be impossible for a solution to satisfy the requirement that each of {a,b,c}>1.

3. If there are less than 4 elements in each vector, or if the system is rank deficient, then there will usually be infinitely many solutions and one would choose a solution that minimizes E, subject to the bounds.

So, much will depend on the length of your vectors, and the vectors themselves.

Brendan Hamm
on 10 May 2016

##### 0 Comments

Bjorn Gustavsson
on 10 May 2016

Am I correct in interpreting R+ as the N-dimensional positive real number sub-space? And you want to minimize E for a, b, and c >= 1?

If so I suggest you turn off your computer and have a short sit-down with pen and paper, pencil will also do, and look at this problem for the 2-D case.

HTH

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