Differential equation system in Optimization Toolbox

2012 4 30
1 个回答
2 次查看（30 天）

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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.

8 个评论
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Walter Roberson
Walter Roberson 2012-4-30
Crosscheck here: does z>=0 mean z(1) >= 0 and z(2) >= 0 ?
Walter Roberson
Walter Roberson 2012-4-30
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
Artyom 2012-4-30
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
Richard Brown 2012-4-30
I'm also confused - how do the values of z change over time? Is this a continuous or discrete time problem?
Artyom
Artyom 2012-4-30
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
Richard Brown 2012-4-30
Are the z variables constrained to take on integer values?
Artyom
Artyom 2012-4-30
No, it can take any values. Except constraints of course.

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回答（1 个）

Richard Brown
Richard Brown 2012-5-1

0 个投票

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)

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Artyom
Artyom
2012-4-30

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