Optimization of weighted sum

Question

De Ar 2022-3-28

1
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1682449-optimization-of-weighted-sum

评论： De Ar 2022-3-28

采纳的回答： Matt J

Optimization.png

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!

3 个评论
显示 1更早的评论隐藏 1更早的评论

De Ar 2022-3-28

编辑：De Ar 2022-3-28

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 2022-3-28

编辑：Torsten 2022-3-28

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.

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Matt J 2022-3-28

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1682449-optimization-of-weighted-sum#answer_928704

编辑：Matt J 2022-3-28

在 MATLAB Online 中打开

Dose.mat

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')

1 个评论
显示 -1更早的评论隐藏 -1更早的评论

De Ar 2022-3-28

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!

请先登录，再进行评论。

Answer 2

John D'Errico 2022-3-28

2
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1682449-optimization-of-weighted-sum#answer_928729

编辑：John D'Errico 2022-3-28

在 MATLAB Online 中打开

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)