Oh, yes, that is exactly what I meant. I did not post on the algorithm on drawing from the prob simplex as that is already solved in one of the links I posted.

Your comment to reformulate the problem in additive format, was very helpful for my work. Could I acknowledge you in my work as Ilya Bogdanov?

Is there a way to ensure that the bump fn introduced does not violate $ f((t+1)x)<f(tx)+f(x)$?

If I use $f(x)=-\log g(e^{-x})$, I would get $g(x)=e^{-f(-\log x)}$. Given the boundary conditions $g(0)=0$ and $g(1)=1$, I would require, the candidate function $f$ to satisfy $f(0)=0$ and $f(\infty)=\infty $. I was thinking of following the steps you recommended by using $f(x)=x^{1/2}$. This satisfies the boundary conditions and $ f((t+1)x)<f(tx)+f(x) $ for all $x \in(0,+\infty)$ and positive integer $t$. I perturb the function at $5,5/2$ and in $(1,1.9)$ Someone mentioned using bump functions to smooth the slope discontinuities. I hope using a bump fn dont violate $ f((t+1)x)<f(tx)+f(x)$.