Notice that x2 is the only variable that appears linearly in the second equation. So rewrite the first equation as
Now substitute that into the second equation
This is a quadratic in x1 (or x5), so solve it for x1
eqn = x1^2-4*(1000 - x1 - x3 - x4 - x5)+5*x3^3+x4^4-3.5*x5^2 == 300
eqn =
X1 = solve(eqn, x1)
X1 =
Now you run two different PSO runs, one substituting 1000 - x1 - x3 - x4 - x5 for x2 and the first of those X1 values for x1; the other run substituting 1000 - x1 - x3 - x4 - x5 for x2 and the second of those X1 values for x1. These runs will not require equality or inequality constraints (unless there are more constraints you did not tell us about.)
The runs you would do would be over 3 variables only, x3, x4, x5, with you having substituted for x1 and x2 in your objective function.
Best Answer