cant find the value of 'f' for various values of b(1) and b(2)

function f = objfun1(x)
x(1)=1
x(2)=1
b(1)=x(1)
b(2)=x(2)
%minimize the objective function f for various values of b(1) and b(2)
f= 2.2+256.1777777*b(2)^-2+3.7333334*b(1)^2*b(2)^-2-31.733333*b(2)^-1+4.0*b(1)*b(2)^-1
x=fmincon(@objfun1,[1;1],[0],[0])

Answer 1

[Walter Roberson]

objfun1 = @(b) 2.2+256.1777777*b(2)^(-2)+3.7333334*b(1)^2*b(2)^(-2)-31.733333*b(2)^(-1)+4.0*b(1)*b(2)^(-1);
x = fmincon(objfun1,[1;1],[0],[0]);

However, please check your fmincon() call. You are calling with A=0 and b=0, so you are imposing the condition that A*x <= b which is then 0*x <= 0 which is then 0 <= 0 which is always going to be true. Are you sure you are not trying to code lower bounds?

A = [], b = []; Aeq = [], beq = []; lb = [0,0]; ub = [];
x = fmincon(objfun1, [1;1], A, b, Aeq, beq, lb, ub);

[konica garg]

Error using ==> fmincon at 504
FMINCON cannot continue because user supplied objective function failed with the following
error:
Error using ==> fmincon at 504
FMINCON cannot continue because user supplied objective function

Error in ==> objfun1 at 14
x=fmincon(@objfun1,[1;1],A,B,Aeq,Beq,lb,ub);

these were the errors when i wrote the above code

i want 'lb' should always be greater than 0 and b(1) and b(2) should take any value and from that minimum objective function (f) value can be obtained

[Walter Roberson]

If b(1) and b(2) are non-negative then you can simplify the formula by examination. b(1) only occurs in locations that have positive coefficients multiplied by a power of b(1) multiplied by a power of b(2). b(2) is nonnegative as well, so the powers are going to be non-negative; when you combine that with the positive coefficient, the minimum contribution for those terms is going to occur when b(1) is 0 (or b(2) is infinite.)

You can therefore simplify the expression to an expression in one variable, which is then easier to minimize. For example, differentiate it with respect to b(2), solve for 0; you can get the exact result to within the limitations of the accuracy of your coefficients. The solution is b[1] = 0, b[2] = 4610992496 / 285587145

[Walter Roberson]

I suspect that there was more to the error message, such as the reason for failure.