For the function
$$ f(x_1,x_2;a_0,b_0)=\\\small\cases{\frac{1}{2}\left[x_1+x_2-x_1x_2+\left(\frac{-1+b_0(1-a_0)}{a_0}x_1+1\right)\left(\frac{-1+b_0(1-a_0)}{a_0}x_2+1\right)\right],\quad 0 \leq x_1\leq a_0,\, 0 \leq x_2\leq a_0\\ \frac{1}{2}\left[x_1+x_2-x_1x_2+\left(\frac{-1+b_0(1-a_0)}{a_0}x_1+1\right)b_0(-x_2+1)\right],\quad\quad\,\,\,\,\,\quad 0 \leq x_1\leq a_0,\, a_0 \leq x_2\leq 1\\ \frac{1}{2}\left[x_1+x_2-x_1x_2+b_0(-x_1+1)\left(\frac{-1+b_0(1-a_0)}{a_0}x_2+1\right)\right],\quad\quad\,\,\,\,\quad a_0 \leq x_1\leq 1,\, 0 \leq x_2\leq a_0\\ \frac{1}{2}\left[x_1+x_2-x_1x_2+b_0(-x_1+1)b_0(-x_2+1)\right],\quad\quad\quad\quad\quad\quad\quad a_0 \leq x_1\leq 1,\, a_0 \leq x_2\leq 1}$$
and
$$f(x_1=x_2;a_0,b_0)=\cases{\frac{1}{2}\left[2x_1-{x_1}^2+\left(1+\frac{x_1(1+b_0(1-a_0))}{a_0}\right)^2\right],\quad 0 \leq x_1 \leq a_0\\\frac{1}{2}\left[2x_1+b_0^2(1-x_1)^2-x_1^2\right],\quad\quad\quad\quad\,\quad a_0 \leq x_1 \leq 1}$$
I would like to solve the following optimization problem:
$$\max_{a_0,b_0\in[0,1]}\left[\min_{{x_1=x_2\in[0,1]}}f(x_1,x_2;a_0,b_0) -\min_{{x_1,x_2\in[0,1]}}f(x_1,x_2;a_0,b_0)\right]$$
Here is my code for the function $f$:
f[x1_, x2_] := Piecewise[{{(1/2)*(x1 + x2 - x1*x2 + ((-1 + b0*(1 - a0)) x1/a0 + 1) ((-1 + b0*(1 - a0)) x2/a0 + 1)), 0 <= x1 <= a0 && 0 <= x2 <= a0}, {(1/2)*(x1 + x2 - x1*x2 + ((-1 + b0*(1 - a0)) x1/a0 + 1)*b0 (-x2 + 1)), 0 <= x1 <= a0 && a0 <= x2 <= 1}, {(1/2)*(x1 + x2 - x1*x2 + ((-1 + b0*(1 - a0)) x2/a0 + 1)*b0 (-x1 + 1)), a0 <= x1 <= 1 && 0 <= x2 <= a0}, {(1/2)*(x1 + x2 - x1*x2 + b0 (-x1 + 1)*b0 (-x2 + 1)), a0 <= x1 <= 1 && a0 <= x2 <= 1}}]
Would it be possible also to solve this analytically with Mathematica?
b0
in your function definition, something seems to be wrong! $\endgroup$ – Ulrich Neumann Aug 3 '18 at 9:460<b0<1
doesn't appear in your Piecewise-function. For numerical optimization your function must return a numerical value! $\endgroup$ – Ulrich Neumann Aug 3 '18 at 10:27f[x1_ ?NumericQ, x2_?NumericQ , a0 _?NumericQ , b0_?NumericQ ]:=...
.The testf[.9, .8, 0.1, .7]
keeps unevaluated. So any numerical programm cannot succeed $\endgroup$ – Ulrich Neumann Aug 3 '18 at 10:47