Optimization of weighted sum

I want to optimize the weights of the sum , where z is depth (in cm), . The terms are vectors (representing dose) of size for with already fixed and know entry values. I need to optimize these weights such that in the region , where is a constant dose vector of size (see attached figure), in order to flatten the plateau as much as possible.

Could anyone help me with this problem?

Many thanks in advance for any guidance and consideration!

[Torsten]

By this weighting, you want to achieve that Dconst equals Dtot(z) in the region 5.5 <= z <= 8.5 ?

How should this be possible if Dconst is a constant value over the complete z-range ?

[De Ar]

I attached the figure Optimization_DoseVec.png. I have also attached the dose vector data in Dose.mat.

That is correct! By the figure Optimization.png, I have already tried computing the most optimal set of weights through Jette D & Chen W (2011) (Creating a spread-out Bragg peak in proton beams), but the plateau dose did not work in my case. Was therefore wondering if I should apply a method such as e.g. gradient descent..?

[Torsten]

Yes, but how should it be possible that the weighted sum becomes a constant over a complete interval ?

The Di curves behave similarly, so every weighted sum will behave similarly as the Di's do. The weighted sum can't become a constant.

Answer 1

[Matt J]

You did not include Dconst or the corresponding z, so I eyeballed both from your attached .jpg image.

load Dose.mat
C=[D1,D2,D3,D4,D5,D6,D7,D8];
[m,n]=size(C);

Dconst=0.1*ones(m,1); %approximated from .jpg image
z=linspace(0,8.5,m);  %approximated from .jpg image

bp=5.5<=z & z<=8.5; %Bragg peak interval

w=lsqlin(C(bp,:),Dconst(bp),[],[],ones(1,n),1,zeros(n,1),ones(n,1));

Minimum found that satisfies the constraints.

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

plot(z, [C*w,Dconst]); legend('Dtot','Dconst','location','southeast')

[De Ar]

Awesome - thank you!! I forgot to mention that I am mostly keen on achieving homogeniety in dose in that particular region than reaching the level dose of . However, this solution works!

Answer 2

[John D'Errico]

First, LEARN TO USE ARRAYS!!!!!!!!

Don't create 8 different numbered variables. Why does that make sense, when you can create ONE array, call it Dose?

Dose = [D1,D2,D3,D4,D5,D6,D7,D8];

MATLAB is a MATRIX LANGUAGE. Learn to use it as such. Anyway, in order to use lsqlin, you will need an array.

Next, you don't actually provide Dtot. So is there a way we can help you more? Probably not really, if your goal is to truly achueve that specific dose goal. If your goal is to achieve an aggregate dose that is as large as possible over the entire region, then you might do this:

Daim = 0.12*ones(501,1); % unachievable, but who cares? Aim for the stars.
LB = zeros(1,8);
UB = ones(1,8);
A = ones(1,8); % equality constraint: the sum must be 1
B = 1; 
W = lsqlin(Dose,Daim,[],[],A,B,LB,UB)

Minimum found that satisfies the constraints.

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

<stopping criteria details>
W =
        0.0271916622048387
         0.043945469132034
        0.0555504667354272
        0.0671084337012187
        0.0905029933411581
         0.113513538791469
         0.167454011263868
         0.434733424829987
         
plot(Dose*W)

In advance, I played with some numbers, wondering what the solution would look like. That is roughly what I found by hand, at least if your goal is a curve that looks vaguely like your plotted aim.

If you wanted a higher loading on the left end, you need something with some mass down there. However, by dropping the aim just a bit, you can flatten the result, just a bit.

Daim = 0.1*ones(501,1);
W = lsqlin(Dose,Daim,[],[],ones(1,8),1,zeros(1,8),ones(1,8))

Minimum found that satisfies the constraints.

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

<stopping criteria details>
W =
0.0800640800727342
0.0450974697116315
0.0615545440518655
0.0676444663541099
0.0843898048745855
0.110747769607969
0.135846098000183
0.414655767326922

plot(Dose*W)

This clearly has a flatter plateau, which you sort of indicated was your goal.