How can I solve optimization problem without using Symbolic Math Toolbox?

Is it possible to solve this optimization problem without using Symbolic Math Toolbox??

Thank you.

Answer 1

[Alan Weiss]

Sure, you can use Optimization Toolbox. You also need the constraint that all the q_i are nonnegative, otherwise the problem is unbounded.

The most straightforward way is to use fmincon. Let

fn = @(q)(1 + 0.15*(q/4000).^4);
fun1 = @(q)integral(@(x)(30/100)*fn(x),0,q);
fun2 = @(q)integral(@(x)(25/100)*fn(x),0,q);
fn2 = @(q)(1 + 0.15*(q/6000).^4);
fun3 = @(q)integral(@(x)(30/100)*fn2(x),0,q);
fun = @(x)(fun1(x(1)) + fun2(x(2)) + fun3(x(3)));
Aeq = [1,1,1];
beq = 14000;
x0 = 14000/3*Aeq;
lb = [0,0,0]);
z = fmincon(fun,x0,[],[],Aeq,beq,lb)

Sorry if I got some details wrong, I am not sure that I was able to read the problem correctly (small lettering on my screen).

Alan Weiss

MATLAB mathematical toolbox documentation

Answer 2

[John D'Errico]

Is this your homework? If so, then you need to be doing your homework. If not, then you need to use common sense.

Why not do most of the work by pencil and paper?

Each of those integrals is trivial. The first, integral for example:

30/100*(1+0.15*(q/4000)^4)

The integral is high school calc.

3/10*q + 3/10*3/20*q^5/5*/(4000)^4

With limits of 0 & q1, we get

q1^5*9/2.56e17 + q1*3/10

The other two are quite similar, yielding 5th degree polynomials in q2 and q3. Thus...

q2^5*3/1.024e17 + q2/4

q3^5/1.44e17 + q3*3/10

So you wish to solve the minimization problem

q1^5*9/2.56e17 + q1*3/10 + q2^5*3/1.024e17 + q2/4 + q3^5/1.44e17 + q3*3/10

Subject to the linear constraint:

q1 + q2 + q3 = 14000

So, first, consider that there is no minimum, if these variables can go to -inf. That should be obvious, even with the equality constraint. So assume they are constrained to be non-negative.

The simple answer is to just use fmincon now. But you could also formulate it in terms of a Lagrange multiplier. Either way will work. A good starting guess should be easy, because the solution will surely be close to q1=q2=q3.

Using fmincon, the solution is easily found as:

     3657.5656004969          4856.08772111945          5486.34667838365

NO. I'm not going to do it for you, since this easily could be homework.

Answer 3

[Walter Roberson]

Reduce it by one variable by making the substitution q3 = 14000-q1-q2

Now each of the terms can be integrated independently of the others. Each of them is a simple quartic that can be easily integrated in formula by inspection; then substitute the endpoints. You get a 5th order multinomial in two variables out.

Now you can use any of the minimization techniques, such as fminsearch() or fminunc(). Or you could inspect it and see that it has two terms of the form

positiveconstant * q1^5 + positiveconstant * q2^5

and if you think for a moment you will come up with the point at which the expression will be minimized: q1 = q2 = q3 = -infinity