# Is there a feasible computation?

I have to find the values of $$p_1$$ and $$p_2$$ which minimizes the expression $$G$$. However, when I took the partial differentiation of $$G$$ with respect to $$p_1$$ and $$p_2$$, the equations I obtained were complicated. And I am unable to solve the variable values for the two equations in 16 parameters. I ran the program using Solve for at least 12 hours for no result. Is there a way to find the symbolic solution of this problem using Mathematica?

R1P = Simplify[(2*mu1 - lambda1P*p1 + c1^2*lambda1P*p1 - lambda2P*p2 +
c1^2*lambda2P*p2)/(2*mu1*(mu1 - lambda1P*p1 - lambda2P*p2))];

R2P = Simplify[(-lambda1P + c2^2*lambda1P - lambda2P +
c2^2*lambda2P + 2*mu2 + lambda1P*p1 - c2^2*lambda1P*p1 +
lambda2P*p2 - c2^2*lambda2P*p2)/(2*
mu2*(-lambda1P - lambda2P + mu2 + lambda1P*p1 + lambda2P*p2))];

R1NP = Simplify[(2 mu1 - lambda1P p1 + c1^2 lambda1P p1 -
lambda1P p1 + c1^2 lambda1P p1 - lambda2P p2 + c1^2 lambda2P p2 -
lambda2P p2 + c1^2 lambda2P p2 + 2 lambda1P mu1 p1 R1P +
2 lambda2P mu1 p2 R1P)/(2 mu1 (mu1 - lambda1P p1 - lambda1P p1 -
lambda2P p2 - lambda2P p2))];

R2NP = Simplify[(-lambda1P + c2^2 lambda1P - lambda1P +
c2^2 lambda1P - lambda2P + c2^2 lambda2P - lambda2P +
c2^2 lambda2P + 2 mu2 + lambda1P p1 - c2^2 lambda1P p1 +
lambda1P p1 - c2^2 lambda1P p1 + lambda2P p2 - c2^2 lambda2P p2 +
lambda2P p2 - c2^2 lambda2P p2 + 2 lambda1P mu2 R2P +
2 lambda2P mu2 R2P - 2 lambda1P mu2 p1 R2P -
2 lambda2P mu2 p2 R2P)/(2 mu2 (-lambda1P - lambda1P - lambda2P -
lambda2P + mu2 + lambda1P p1 + lambda1P p1 + lambda2P p2 +
lambda2P p2))];

G = W1*(R1P + R2P) + W2*(R1NP + R2NP) +
W3*(p1*delta11*(lambda1P/(lambda1P + lambda2P)) + (1 - p1)*
delta12*(lambda1P/(lambda1P + lambda2P)) +
p2*delta21*(lambda2P/(lambda1P + lambda2P)) + (1 - p2)*
delta22*(lambda2P/(lambda1P + lambda2P))) +
W4*(p1*delta11*(lambda1P/(lambda1P + lambda2P)) + (1 - p1)*
delta12*(lambda1P/(lambda1P + lambda2P)) +
p2*delta21*(lambda2P/(lambda1P + lambda2P)) + (1 - p2)*
delta22*(lambda2P/(lambda1P + lambda2P)));
$$$$

• I doubt you would be happy with such a result. If you clear denominators and remove extraneous factors, you have two parametrized polynomials with leaf counts in the tens of thousands. Eliminating a variable, e.g. with Resultant`, is likely to give a univariate polynomial far too large to work with in any sensible manner. Assuming such elimination terminates in reasonable time. – Daniel Lichtblau Nov 14 at 1:27