Linear programming with conditional constraints

Hello,

I'm trying to figure out a way to formulate a linear program to solve an optimization problem with the following constraints:

x1 = 10*e1 if e1>=0

x1 = 0,1*e1 if e1<0

x1 and e1 are variables of the problem.

All the other constraints are regular linear expressions.

Is it possible to model with a linear programming solution?

Is it possible to solve using linprog or intlinprog? If so, how can I do this?

Thanks

Answer 1

[Alan Weiss]

I think that you can do this with mixed-integer linear programming. Create auxiliary binary variables y1 and y2. These variables track whether e1 is positive or not. Assume that M is an upper bound on abs(e1). Include the constraints

e1 <= M*y1; % This enforces y1 = 1 whenever e1 > 0
e1 + M*y2 >= 0; % This enforces y2 = 1 whenever e1 < 0
y1 + y2 = 1;

So now you have y1 as the indicator that e1 > 0 and y2 as the indicator that e1 < 0.

I am not sure at this point how to include y1 and y2 into your problem formulation, but maybe you can figure it out from here.

Good luck,

Alan Weiss

MATLAB mathematical toolbox documentation

[NN]

Hi Sir,

Can i kknow if there is a solution to inlcude y1 and y2 in problem

[Alan Weiss]

The documentation section Integer and Logical Modeling contains information on how to do this kind of modeling. In particular, see the section Combine Logical Constraints to Create New Formulas. That section shows how to make where z is binary. You can extend that formulation to where and are constants (10 and 1/10 in your case) and and are the binary variables from the first part of my answer..

Alan Weiss

MATLAB mathematical toolbox documentation

[NN]

Thank you very much

[Matt J]

Blenda Moreira's comment moved here:

It looks like a good way out!

I'll try to formulate from there and I'll let you know if it works.

Thanks a lot

[NN]

I have used this below code

y1V = optimvar('y1V',N,'Type','integer','LowerBound',0,'UpperBound',1);

y2V = optimvar('y2V',N,'Type','integer','LowerBound',0,'UpperBound',1);

MV = optimvar('MV',N,'LowerBound',0,'UpperBound',16.5e3);

prob.Objective = dt*(y1V.*Nb1+y2V.*Nb2)'*PbattV;

prob.constraints.batt1= PbattEVregV <= MV*y1V; % This enforces y1 = 1 whenever PbattEVregV > 0

prob.constraints.batt2= PbattEVregV + MV*y2V >= 0; % This enforces y2 = 1 whenever e1 < 0

prob.constraints.batt3= y1V + y2V == 1;

It gives below comment for dimesnion error , please advice

An error occurred while running the simulation and the simulation was terminated

Caused by:

• Unable to perform assignment because the left and right sides have a different number of elements.

[Alan Weiss]

I see two issues here. One is that your stated objective function is not linear in the optimization variables. Please follow the documentation more closely, looking carefully at how the examples take extraordinary steps to keep the expressions linear.

 dt*(y1V.*Nb1+y2V.*Nb2)'*PbattV; % I'm not sure about this one
 PbattEVregV <= MV*y1V; % This one for sure.
 % MV should not be an optimization variable if you are trying to use
 % a Big-M formulation

For the second issue, the dimension error, I suggest that you use the debugger to investigate.

Alan Weiss

MATLAB mathematical toolbox documentation

[NN]

dt*(y1V.*Nb1+y2V.*Nb2)'*PbattV; % I'm not sure about this one

PbattEVregV <= MV*y1V; % This one for sure.--->this is PbattV(not PbattEVregV)

% MV should not be an optimization variable if you are trying to use

% a Big-M formulation

[NN]

Assume that M is an upper bound on abs(e1).

If M is not an optimization variable , how can i define it ?Please explain

[Alan Weiss]

Give it a value. 1e4, 1e5, whatever you think is good. Larger values can lead to numerical instability, so take the smallest value you can.

Alan Weiss

MATLAB mathematical toolbox documentation

[Matt J]

I am not sure at this point how to include y1 and y2 into your problem formulation, but maybe you can figure it out from here.

Complete the constraints as follows:

e1 <= M*y1; % This enforces y1 = 1 whenever e1 > 0
e1 + M*y2 >= 0; % This enforces y2 = 1 whenever e1 < 0
y1 + y2 = 1;

10*e1<= x1  <= 10*e1 +  100*M*y2   %equivalent to x1=10*e1 when e1>0
0.1*e1<= x1 <= 0.1*e1 + 100*M*y1   %equivalent to x1=0.1*e1 when e1<0

[NN]

Yes, i did, but with the below expression

PbattV = optimvar('PbattV',N,'LowerBound',-16500,'UpperBound',16500);
y1V = optimvar('y1V',N,'Type','integer','LowerBound',0,'UpperBound',1);
y2V = optimvar('y2V',N,'Type','integer','LowerBound',0,'UpperBound',1);
prob.Objective = dt*(y1V.*Nb1+y2V.*Nb2)'*PbattV;

prob.constraints.batt1= PbattV <= 16500*y1V; % This enforces y1 = 1 whenever PbattV > 0
prob.constraints.batt2= PbattV + 16500V*y2V >= 0; % This enforces y2 = 1 whenever PbattV< 0
prob.constraints.batt3= y1V + y2V == 1;

Here Nb1 and Nb2 are two struct arrays of 1x1 with subfields time(288 x1 double ) and signals(1 x 1 struct ).Signal ha s two subfields dimesions (241) and values( 288*241 double)

I am getting the error ,

Unable to perform assignment because the left and right sides have a different number of elements.

Dimension problem i am unable to solve.Please help

[Alan Weiss]

Your objective needs to be a scalar. You might need to rewrite your objective expression as explicit matrix multiplication rather than using a structure. Use the debugger to see what the sizes of your variables are. Ensure that the objective expression is a scalar.

Alan Weiss

MATLAB mathematical toolbox documentation

[NN]

Ok

[NN]

I have used debugger and getting this error for the line

prob.constraints.batt1= PbattV <= 16500*y1V; % This enforces y1 = 1 whenever PbattV > 0
prob.constraints.batt2= PbattV + 16500V*y2V >= 0; % This enforces y2 = 1 whenever PbattV< 0
prob.constraints.batt3= y1V + y2V == 1;

• Unrecognized method, property, or field 'constraints' for class 'optim.problemdef.OptimizationProblem'.

Is there any formatting problem ?

Thanks

Answer 2

[Matt J]

All the other constraints are regular linear expressions.

If that's true then just divide the problem into two subproblems, one with the constraints,

x1 = 10*e1 
e1>=0

and the other with the constraints,

x1 = 0.1*e1 
e1<=0

Whichever of these two linear sub-programs gives you the most optimal objective value is the solution that you accept.

Answer 3

[Matt J]

Asume a bound e1<=M, and introduce additional binary variables b1 and b2 with the constraints.

%(1)
e1 <= M*b1; %This enforces b1 = 1 whenever e1 > 0
e1 + M*b2 >= 0; % This enforces b2 = 1 whenever e1 < 0
b1 + b2 = 1;

10*e1  <=  x1,   x1 <= 10*e1 + 9.9*M*b2    %(2)
0.1*e1 <= x1,   x1 <= 0.1*e1 + 9.9*M*b1   %(3)

When 0<=e1<=M, the above reduces to,

1. b1=1, b2=0,
2. x1=10*e1,
3. 0.1*e<=x1<=0.1*e1 +9.9*M

Note that (2) is what is required and (3) does not conflict with (2) on 0<=e1<=M

Similarly when -M<=e1<=0,

1. b1=0, b2=1,
2. 10*e1 <= x1 <= 10*e1 + 9.9*M
3. x1=0.1*e1

Here, (3) is the required condition and (2) does not conflict with (3) on the interval -M<=e1<=0.