Differential equation system in Optimization Toolbox

Hi, everyone.

I'm trying to solve dynamic programming problem. I have differential equation system like that one:

dydt = zeros(2,1); 

dydt(1) = z(1);

dydt(2) = z(2)-z(1);

With constraints:

z>=0;

z(1)+z(2)<=x(2);

z(2)<=x(1);

x(2)>=0;

I must maximize x(2) on period T=5.

How can I optimize DES and find z with Optimization Toolbox?

Thank you.

[Walter Roberson]

Crosscheck here: does z>=0 mean z(1) >= 0 and z(2) >= 0 ?

[Artyom]

Yes.

z(i)>=0; %i=1,2

[Walter Roberson]

Which of the values are unknowns? Are z or x intended to be interpreted as being time-dependent, z1(t) ?

If you must maximize x(2) then why not just say x(2) = infinity ?

[Artyom]

My mistake. System of equation:

x'(1)= z(1); - where x1 - steel capacity

x'(2)= z(2)-z(1); - where x2 - steel stockpile

Value x(1) and x(2) we know.

We take some steel from stockpile (z1) and increase steel capacity at period [0;1]. Than we do it again at period [1;2]. Final period is T=5.

Constraints:

z(1)>=0;z(2)>=0;

z(1)+z(2)<=x(2); - we can't take more steel that we have

z(2)<=x(1); - bottleneck, where capacity greater then z2

x(2)>=0;

Problem: maximize x'(2) at period T=5.

If you have more quastions just ask.

[Richard Brown]

I'm also confused - how do the values of z change over time? Is this a continuous or discrete time problem?

[Artyom]

Ok, we take, for example, z(1)=3; z(2)=3 on period [0;1], z(1)=5;

z(2)=6 on [1;2] and so on. Our goal to maximize x(2) on [0;T]. So we take decision abou z(i) in moments of time 0,1,2,3,4.

[Richard Brown]

Are the z variables constrained to take on integer values?

[Artyom]

No, it can take any values. Except constraints of course.

Answer 1

[Richard Brown]

Your problem is a linear program in the arrays x1, x2, z1, z2. Because the RHS of the ODEs is piecewise constant, the problem can be reformulated as a series of difference equations. You therefore have the following linear equality constraints:

k = 0, ..., 4:

x1[k+1] = x1[k] + z1[k]
x2[k+1] = x2[k] + z2[k] - z1[k]

where x1[0] and x2[0] are known initial conditions

And you have the following linear inequality constraints:

k = 0, ..., 4:

z1[k] >= 0
z2[k] >= 0
z1[k] + z2[k] <= x2[k]
z2[k] <= x1[k]
z2[k] - z1[k] >= -x2[k]

Your cost is also linear:

c(z1, z2, x1, x2) = x2[5]

All of this defines a linear program that you can solve with linprog. All you require is a little bookkeeping to formulate the constraint matrices (and conversion to ones-based indexing)