introducing a comparison related to variables in linear programming

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ennes mulla 2023-1-20

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编辑： Matt J 2023-1-20

Hello,

I want to run linear programming to make a trade off between multiple sources of energy. I want to include a battery in the system and size it using linear programming. Therefore, I have the following varaibles:

Pdemand: demands of energy which I have it already as a vector colmun.

X1= A_pv [m^2 ]-maximum total area occupied by the solar photovoltaic system

X2=A_w [m^2 ]-maximum total wind rotor area

.

X8=E_bess (t) [kWh]-energy in the BESS at time t

X9=P_(bess.in) (t) [kW]-power in battery at time t

X10=P_(bess.out) (t) [kW]-power out from battery at time t

I started to encounter problems when setting my first inequality since X9 and X10 are related to X1 and X2 as following

if X1 + X2 > Pdemand

X9 = (X1 +X2) - Pdemand;

X10 = 0;

else

X10 = Pdemand - (X1 + X2);

X9= 0;

Now I want to make the first constrain for linear programming by reporting the coefficient to use it later in linear programming

X1 + X2 +X9 +X10 = demands

I included X9 and X10 here in the equation since one of them needs to be zero at the time the other is presented.

A1 = [CX1 CX2 0 0 0 0 0 0 0 CX9 CX10]

As you can see, I am trying to solve for X9 and X10 but before I need them to go throght the if statment which is not possible till I get the results of linear programming first.

Can someone help me please, I have zero ideas how to solve this. Thanks.

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Torsten 2023-1-20

编辑：Torsten 2023-1-20

You say you have Pdemand as a column vector. So X1 and X2 are also X1(t) and X2(t) ?

It would be helpful to have your problem stated precisely.

ennes mulla 2023-1-20

yes yes. I over simplified the problem to mention less details, But man ignore this details please, how would you introduce if statment to the LP

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Answer 1

Matt J 2023-1-20

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编辑：Matt J 2023-1-20

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The constraints you describe are non-convex, so there is no way to formulate them in a linear way. However, you can modify your problem so that the solution to a linear program is encouraged to satisfy those constraints. Introduce additional variables U,V and the constraints,

U=(X1 +X2) - Pdemand
-V<=U<=V  %implies V>=abs(U)
X9>=0
V>=X9>=U    %(1a)
X10>=0
V>=X10>=-U  %(1b)
X9+X10=V    %(2)

Also, modify your linear objective to have a penalty term,

min. f(X) + largeNumber*V

So, if largeNumber is large enough, the penalty will force V to its lower bound V=abs(U) at the solution. In that case when U>0 then U=V and constraint (1a) implies X9=U=(X1 +X2) - Pdemand. Additionally constraint (2) then implies X10=0.

A similar argument shows that if U<0 and so -U=V, then constraint (1b) implies X10=V=abs(U)= Pdemand-(X1+X2) and X9=0.

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Answer 2

Torsten 2023-1-20

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编辑：Torsten 2023-1-20

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Your constraint says that

X9 = max(0,X1+X2-Pdemand)
X10 = max(0,Pdemand-(X1+X2))

Usually, it suffices to set the constraints

X9  >= 0
X9  >= X1+X2-Pdemand
X10 >= 0
X10 >= Pdemand-(X1+X2)

since you usually won't "oversupply" in an optimization.

If the constraints don't suffice for your application, look up on how to linearize the max-operator independent of the objective following Larry Snyder's answer under

https://or.stackexchange.com/questions/711/how-to-formulate-linearize-a-maximum-function-in-a-constraint

But in case you have to use this formulation, your problem becomes a problem for intlinprog - it can no longer be solved with linprog.