minimum:6.06+0.000579(x1)-2.88(x2)-0.552(x3)-1.1059(x4), maximum:30.3+0.0085(x1)+6.8(x2)-7.31(x3)+3.69(x4), subjected to : 3200≤(x1)≤3800 , 0.40≤(x2)≤0.60 , 1.5≤(x3)≤2 , 12≤(x4)≤16 , optimized values and pareto chart ? steps to follow please.
MATLAB: Formulation of multi objective optimization problem
optimization
Related Solutions
f=[-out1(j0), -out2(j0), -out3(j0), 0,0,0,0,0];
Best wishes
Torsten.
%{
x1+x2+x3+x4+x5 =1000;x1^2-4x2+5x3^3+x4^4-3.5x5^2 = 300%}
Notice that x2 is the only variable that appears linearly in the second equation. So rewrite the first equation as
%{x2 = 1000 - x1 - x3 - x4 - x5%}Now substitute that into the second equation
%{x1^2-4*(1000 - x1 - x3 - x4 - x5)+5x3^3+x4^4-3.5x5^2 = 300%}This is a quadratic in x1 (or x5), so solve it for x1
syms x1 x2 x3 x4 x5eqn = x1^2-4*(1000 - x1 - x3 - x4 - x5)+5*x3^3+x4^4-3.5*x5^2 == 300
X1 = solve(eqn, 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.
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