Biological Effective Dose (BED)

Question

Valentina Vasic 2023-2-20

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1915565-biological-effective-dose-bed

评论： Valentina Vasic 2023-2-21

Hello, everyone,

Does anyone know how to implement the biological effective dose (BED) calculation? The calculation contains a double integral which I do not know how to calculate.

(#) are number and (v) are vectors.

I am trying to use the cumtrapz.

%%  BED
AlphaBeta = 2.6;
nu = 0.3571;
t_w = [0:time_h(end)/length(time_h):time_h(end)]; %??
t_w = t_w';
t_w(end) =[];
Int1 = TAC.*exp(-nu.*(t_w-time_h));
Int2 = TAC.*S.*cumtrapz(time_h,Int1);
G = 2./(max(Dose_Gy3)).^2.*cumtrapz(time_h,Int2);
BED = max(Dose_Gy3).*(1+max(G)./AlphaBeta.*max(Dose_Gy3));

The first internal integral must be made from 0 to t, where t is the time vector. I do not know how to interpret the first integral (dw).

Best regards,

Valentina

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Answer 1

Torsten 2023-2-20

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https://ww2.mathworks.cn/matlabcentral/answers/1915565-biological-effective-dose-bed#answer_1175685

编辑：Torsten 2023-2-20

在 MATLAB Online 中打开

Assuming you have two arrays (time_h and TAC) of the same length:

AlphaBeta = 2.6;
mu = 0.3571;
S = 1;
int_inner = cumtrapz(time_h,TAC.*exp(mu*time_h));
int_outer = trapz(time_h,TAC.*exp(-mu*time_h).*int_inner);
G = 2./AUC^2*int_outer;
BED = S*AUC.*(1+G./AlphaBeta.*S*AUC);

For example:

format long
mu = 0.3571;;
Ddot = @(t)t.^2;
fun = @(t,omega)Ddot(t).*Ddot(omega).*exp(-mu*(t-omega));
value = integral2(fun,0,1,0,@(t)t)
value = 
   0.051556597835785
time_h = 0:0.01:1;
TAC = Ddot(time_h);
int_inner = cumtrapz(time_h,TAC.*exp(mu*time_h));
int_outer = trapz(time_h,TAC.*exp(-mu*time_h).*int_inner)
int_outer = 
   0.051573876925114