Setting up a problem in linprog

This is a follow up to my previous question posted here

I want to define an absolute error loss function of the following form

% Loss Function (absolute error loss)
%% input curves
scale =  1.5;
x1 = [0,4,6,10,15,20]*scale;
y1 =  [18,17.5,13,12,8,10];
x2 = [0,10.5,28]*scale;
y2= [18.2,10.6,10.3];
%% y2 is the target function and y1 is the function to be shifted
f = y1;
g = interp1(x2,y2,x1);
%% linprog inputs
% x = linprog(f,A,b)
% i = 1:length(x1)
A = [fi - 1; -fi -1];
x = [a ui];
b = [gi -gi];
% obj = sum_i u_i (function to minimize)
%% solve system

I am not sure how to set this up and solve this sytem in MATLAB's linprog

Could someone please help?

Answer 1

[Torsten]

I think the whole procedure only makes sense if x1 is a subset of x2, but here we go:

scale =  1.5;
x1 = [0,4,6,10,15,20]*scale;
y1 =  [18,17.5,13,12,8,10];
x2 = [0,10.5,28]*scale;
y2= [18.2,10.6,10.3];
%% y2 is the target function and y1 is the function to be shifted
f = y1;
g = interp1(x2,y2,x1);

ff = [0,ones(size(x1))];
A = [f.',-eye(numel(x1));-f.',-eye(numel(x1))];
b = [g.',-g.'];
sol = linprog(ff,A,b);

Optimal solution found.

a = sol(1)

a = 1.0111

% Compare with least-squares solution
a2 = sum(f.*g)/sum(f.^2)

a2 = 0.9895

And the next step is

min: max_i ( abs( a*f_i - g_i ) )

? :-)

[Deepa Maheshvare]

@Torsten,

Thanks a lot, could you please explain a bit on ff = [0,ones(size(x1))]. I could understand how A is defined but the values defined for `ff` is not clear to me.

I think objective function is ff^Tx and x = [a u1 u2 u3 u4 u5 u6 u1 u2 u3 u4 u5 u6]. I also fail to understand why ff is [0, ones(size(x1))] and not [0 ones(2*size(x1))]. Is x just [a u1 u2 u3 u4 u5 u6] ?

Then ff^T*x yields sum_i ui.

min: max_i ( abs( a*f_i - g_i ) )

sorry, could not understand what you are pointing to.

[Torsten]

Is x just [a u1 u2 u3 u4 u5 u6] ?

Yes.

Then ff^T*x yields sum_i ui.

Yes.

min: max_i ( abs( a*f_i - g_i ) )

sorry, could not understand what you are pointing to.

Besides the two norms you used until now, you can try as a third approach to find "a" that minimizes the maximal deviation of a*f from g:

min: max_i ( abs (a*f_i - g_i) )

Was just a joke :-)

[Deepa Maheshvare]

:-)