Hi Dominik,
I understand that you are trying to convert the transition matrix so that one multiplication of the population with the matrix equals one tenth of a day instead of a full day.
To achieve this, you can simply take the transition matrix to the power of 1/10.
Taking the matrix to the power of 1/10 is equivalent to multiplying the matrix with itself 1/10 times, which is what we want to achieve to scale the transition matrix for one tenth of a day instead of a full day.
You can see modified code below and it shows both matrix are equivalent: -
%development rates of each stage
dpreg = 0.172;
dprL1 = 0.217;
dprL2 = 0.313;
dprL3 = 0.222;
dprL4 = 0.135;
dprPu = 0.099;
%relative mortality rates of each stage
rmreg = 0.017;
rmrL1 = 0.060;
rmrL2 = 0.032;
rmrL3 = 0.022;
rmrL4 = 0.020;
rmrPu = 0.020;
rmrAd = 0.027;
B1 = [1-(dpreg+rmreg),0,0,0,0,0,0;
dpreg,1-(dprL1+rmrL1),0,0,0,0,0;
0,dprL1,1-(dprL2+rmrL2),0,0,0,0;
0,0,dprL2,1-(dprL3+rmrL3),0,0,0;
0,0,0,dprL3,1-(dprL4+rmrL4),0,0;
0,0,0,0,dprL4,1-(dprPu+rmrPu),0;
0,0,0,0,0,dprPu,1-rmrAd];
A = [10;10;10;10;10;10;10];
%%this loop can be used to calculate the result for 10 days to check whether it is the same %result, to do so change the 1:1 to 1:10
for i = 1:1
C = B1*A;
A=C;
end
disp(A);
%Converting transition matrix
A = [10;10;10;10;10;10;10];
B2=B1^(0.1);
for i = 1:10
C = B2*A;
A=C;
end
disp(A);
