MATLAB: Defining simple optimization constraint: When x1 > 0 then x2 has to equal 0

MATLABmatrixoptimization

Hi there,

I'm working on an optimization problem with 2 decision variables x1 and x2. The problem optimizes the values of x1 and x2 each hour over a 24 hour period to minimize the objective function.

I've already defined 2 constraints as follows (using the code below):

Constraints:
 0 <= x1 <= 3
 0 <= x2 <= 3
N =24;
lb = zeros(2*N,1);     % lower bound is 0
ub = 3*ones(2*N,1);    % upper bound is 3

Now, I'm trying to define a constraint that says that only x1 or x2 can be greater than 0 at any one time. i.e. If x1 is greater than 0 then x2 has to equal 0 and vice versa. In the problem x1 and x2 refer to the energy flow in and out of a battery, therefore only one variable can be positive at a time.

if x1 > 0 then  x2=0
or
if x2 > 0 then x1 = 0

I'm scratching my head but I can't seem to figure out how to define this constraint in matrix form.

I'd welcome any suggestions and thank you for your help.

Sincerely

Best Answer

This is a classical application of Dynamic Programming and that might be the most efficient way for you to go about this problem, see

http://en.wikipedia.org/wiki/Dynamic_programming#Optimal_consumption_and_saving

One thing I don't understand, though, is why you have two separate variables, x1 and x2 for Energy_in and Energy_out? Can't you just have one decision variable x in each hourly period and treat x>0 as energy out and x<=0 as energy in? In other words, can't you just view Energy_out as positive energy consumption and Energy_in as negative energy consumption?

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Hi!

Use arrays to save the results:

% allocate arrays for results
Energyin = zeros(JN-1, 1);
Energyout= zeros(JN-1, 1);
for i=2:JN-1
       SOC(i) = 20 - (at(i)-dt(i))*1.22;
       % save results
       [a, b] = Optimiseenergy(at(i),dt(i+1),SOC(i),20);
  Energyin(i) = a;
Energyout(i) = b;
  end

MATLAB: Please Help Optimization using PSO

%{


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 x5
eqn = x1^2-4*(1000 - x1 - x3 - x4 - x5)+5*x3^3+x4^4-3.5*x5^2 == 300
eqn = 
X1 = solve(eqn, x1)
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.

Best Answer

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