How to solve absolute problem in optimization

2023 6 8
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更新时间：2023 6 8
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Hello, i want to solve a quadratic optimization problem (prob.Objective =sum((PgridV).^2);), in constrains i have a one variable where i need to find the absolute value of (Pb1dc). so i introduce a variable called 'K' . where -K<=Pb1dc<=K;
  • FYI: i have used absoulte value in the following constrains (prob.Constraints.loadBalanceAC=Pb1==Pb1dc-(0.05*K);)
MATLAB solution showing exactely whati want . its working well.
However, when i want to chage the objective function with little modifcation (prob.Objective =sum((PgridV-M).^2);) where M is a reference signal (mean of Pload), when i run the simulation
absolute vaules (K) are not getting exactley what mean for. its showing random values ( K = absolute ((Pb1dc))
clc
clear all
Pload=[0;1;3;4;2;6;9;10;2;4]; % load
N=10;
M=mean(Pload)+zeros(N,1);
Einit1=0.5; % initial energy
E=zeros(N,1);
Emin1 = 0; % mini energy
Emax1 = 3;
dt=1;
prob = optimproblem;
PgridV = optimvar('PgridV',N,'LowerBound',0,'UpperBound',20); % grid power
Pb1= optimvar('Pb1',N,'LowerBound',-1,'UpperBound',1); % ac power
Pb1dc= optimvar('Pb1dc',N,'LowerBound',-1,'UpperBound',1); % dc power
K=optimvar('K',N,'LowerBound',0);% absolute of (Pb1dc)
EbattV1 = optimvar('EbattV1',N,'LowerBound',Emin1,'UpperBound',Emax1); % energy
prob.ObjectiveSense = 'minimize';
% prob.Objective =sum(K.^2);
%prob.Objective =sum((PgridV).^2);
prob.Objective =sum((PgridV-M).^2);
% 1 constrains
prob.Constraints.energyBalance = optimconstr(N);
prob.Constraints.energyBalance(1) = EbattV1(1) == Einit1-Pb1dc(1)*dt; %Its Ploss is constanat
prob.Constraints.energyBalance(2:N-1) = EbattV1(2:N-1) == EbattV1(1:N-2)-Pb1dc(2:N-1)*dt;
prob.Constraints.energyBalance(N) = EbattV1(N) ==Einit1;
% K constrain for Pb1dc modulous
prob.Constraints.kbalance1=optimconstr(N);
prob.Constraints.kbalance1(1:N)=-K(1:N)<=Pb1dc(1:N);
prob.Constraints.kbalance2=optimconstr(N);
prob.Constraints.kbalance2(1:N)=Pb1dc(1:N)<=K(1:N);
% load Balance
prob.Constraints.loadBalance=PgridV+Pb1==Pload;
% loss term
prob.Constraints.loadBalanceAC=Pb1==Pb1dc-(0.05*K);
options = optimoptions(prob.optimoptions,'Display','final');
% options = optimoptions(prob.optimoptions,'Algorithm','interior-point');
[values,fval,exitflag] = solve(prob,'Options',options)
Solving problem using lsqlin. Minimum found that satisfies the constraints. Optimization completed because the objective function is non-decreasing in feasible directions, to within the value of the optimality tolerance, and constraints are satisfied to within the value of the constraint tolerance.
values = struct with fields:
EbattV1: [10×1 double] K: [10×1 double] Pb1: [10×1 double] Pb1dc: [10×1 double] PgridV: [10×1 double]
fval = 57.4575
exitflag =
OptimalSolution
% Parse optmization results
if exitflag <= 0
PgridV = zeros(N,1);
Pb1 = zeros(N,1);
Pb1dc = zeros(N,1);
EbattV1 = zeros(N,1);
K = zeros(N,1);
else
PgridV = values.PgridV ;
Pb1 = values.Pb1;
Pb1dc = values.Pb1dc
EbattV1 = values.EbattV1;
K = values.K
end
Pb1dc = 10×1
-0.6434 -0.6204 -0.6701 0.1224 -0.6886 1.0000 1.0000 1.0000 -0.8624 -0.0018
K = 10×1
7.1327 7.5914 6.5986 4.4490 6.2283 1.0000 1.0000 1.0000 2.7522 1.9640
.

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Torsten
Torsten 2023-6-8
Can you show in your code above exactly what you mean ?
I ran the code with
prob.Objective =sum((PgridV-M).^2);
instead of
prob.Objective =sum((PgridV).^2);
(see above). What is not satisfactory with the results ?
The K values : which are absolute values of (Pb1dc)
Torsten
Torsten 2023-6-8
编辑：Torsten 2023-6-8
What's wrong with the K values listed above ?
They are not defined as the absolute values of Pb1dc in your constraints.
Your constraint is abs(Pb1dc) <= K.

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回答（1 个）

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Could you please help me how can write constrain if i want a variable K, which is abs(Pb1dc)
Torsten
Torsten 2023-6-8
Yes, it's easily seen from the output of your program (see above) that the constraints are satisfied:
abs(Pb1dc) <= K
Torsten
Torsten 2023-6-8
fun = @(x,y) x-abs(y);
prob.Constraints.kbalance1=optimconstr(N);
prob.Constraints.kbalance1(1:N)=fcn2optimexpr(fun,K,Pb1dc)==0;
Torsten
Torsten 2023-6-8
编辑：Torsten 2023-6-8
Didn't you read my answer ? By using "fcn2optimexpr" as done above. But note that this will make your problem nonlinear. Do you really need the absolute values of Pb1dc ? It will make your simulation much more complicated (although "abs" sounds so easy).
I have modified
fun = @(x,y) x-abs(y);
prob.Constraints.kbalance1=optimconstr(N);
prob.Constraints.kbalance1(1:N)=fcn2optimexpr(fun,K,Pb1dc)==0;
with
prob.Constraints.kbalance1=optimconstr(N);
prob.Constraints.kbalance1(1:N)=-K(1:N)<=Pb1dc(1:N);
prob.Constraints.kbalance2=optimconstr(N);
prob.Constraints.kbalance2(1:N)=Pb1dc(1:N)<=K(1:N);..
..
.. i got following error since i never used fmincon
Solving problem using fmincon.
Error using optim.problemdef.OptimizationProblem/solve
SOLVE requires a non-empty initial point structure to solve a nonlinear problem.
Error in modfun_effi_loss (line 48)
[values,~,exitflag] = solve(prob,'Options',options);
Torsten
Torsten 2023-6-8
编辑：Torsten 2023-6-8
Yes, add an initial guess for your solution variables in the call to "solve".
See the second example under
https://uk.mathworks.com/help/optim/ug/optim.problemdef.optimizationproblem.solve.html
on how to do this.
Yes, i need absolute value. because i tried initially with mixed integer , and my optimization problem become MINL, where i am unable to solve the issue with ga, (which i already tried), To overcome mixed integer i used this method.
Torsten
Torsten 2023-6-8
Without using fcn2optimexpr, you could add the constraints
K.^2 - Pb1dc.^2 == 0
with K constrained to be >=0.
Thanks . In my case i have a variables PgridV,Pb1,Pb1dc,K,EbattV1,
did i need to mention initial values for all above five variables with (N,1) like
x0.PgridV=3+zeros(N,1);
x0.Pb1=1+zeros(N,1);
x0.Pb1dc=1+zeros(N,1);
x0.K=1+zeros(N,1);
x0.EbattV1=1.5+zeros(N,1);
and finally at end
[values,~,exitflag] = solve(prob,x0,'Options',options);
is this something correct ?
Torsten
Torsten 2023-6-8
is this something correct ?
Yes. Is "lsqnonlin" chosen as solver ?
Is there any other way i can improve simulation time , since i need to consider N= 96 samples and optimizer taking a lot of time.
Torsten
Torsten 2023-6-8
编辑：Torsten 2023-6-8
Ran in:
Ran in:
clc
clear all
Pload=[0;1;3;4;2;6;9;10;2;4]; % load
N=10;
M=mean(Pload)+zeros(N,1);
Einit1=0.5; % initial energy
E=zeros(N,1);
Emin1 = 0; % mini energy
Emax1 = 3;
dt=1;
prob = optimproblem;
PgridV = optimvar('PgridV',N,'LowerBound',0,'UpperBound',20); % grid power
Pb1= optimvar('Pb1',N,'LowerBound',-1,'UpperBound',1); % ac power
Pb1dc= optimvar('Pb1dc',N,'LowerBound',-1,'UpperBound',1); % dc power
K=optimvar('K',N,'LowerBound',0);% absolute of (Pb1dc)
EbattV1 = optimvar('EbattV1',N,'LowerBound',Emin1,'UpperBound',Emax1); % energy
prob.ObjectiveSense = 'minimize';
% prob.Objective =sum(K.^2);
%prob.Objective =sum((PgridV).^2);
prob.Objective =sum((PgridV-M).^2);
% 1 constrains
prob.Constraints.energyBalance = optimconstr(N);
prob.Constraints.energyBalance(1) = EbattV1(1) == Einit1-Pb1dc(1)*dt; %Its Ploss is constanat
prob.Constraints.energyBalance(2:N-1) = EbattV1(2:N-1) == EbattV1(1:N-2)-Pb1dc(2:N-1)*dt;
prob.Constraints.energyBalance(N) = EbattV1(N) ==Einit1;
% K constrain for Pb1dc modulous
prob.Constraints.kbalance1=optimconstr(N);
prob.Constraints.kbalance1=K.^2-Pb1dc.^2==0;
% load Balance
prob.Constraints.loadBalance=PgridV+Pb1==Pload;
% loss term
prob.Constraints.loadBalanceAC=Pb1==Pb1dc-(0.05*K);
x0.PgridV=3+zeros(N,1);
x0.Pb1=1+zeros(N,1);
x0.Pb1dc=1+zeros(N,1);
x0.K=1+zeros(N,1);
x0.EbattV1=1.5+zeros(N,1);
options = optimoptions(prob.optimoptions,'Display','final');
% options = optimoptions(prob.optimoptions,'Algorithm','interior-point');
[values,fval,exitflag] = solve(prob,x0,'Options',options)
Solving problem using lsqnonlin. Local minimum found that satisfies the constraints. Optimization completed because the objective function is non-decreasing in feasible directions, to within the value of the optimality tolerance, and constraints are satisfied to within the value of the constraint tolerance.
values = struct with fields:
EbattV1: [10×1 double] K: [10×1 double] Pb1: [10×1 double] Pb1dc: [10×1 double] PgridV: [10×1 double]
fval = 58.5791
exitflag =
OptimalSolution
% Parse optmization results
if exitflag <= 0
PgridV = zeros(N,1);
Pb1 = zeros(N,1);
Pb1dc = zeros(N,1);
EbattV1 = zeros(N,1);
K = zeros(N,1);
else
PgridV = values.PgridV ;
Pb1 = values.Pb1;
Pb1dc = values.Pb1dc
EbattV1 = values.EbattV1;
K = values.K
end
Pb1dc = 10×1
-0.9524 -0.9524 -0.3679 0.7251 -0.9524 1.0000 1.0000 1.0000 -0.9524 -0.0952
K = 10×1
0.9524 0.9524 0.3679 0.7251 0.9524 1.0000 1.0000 1.0000 0.9524 0.0952
i am using MATLAB 20222a, is this reason why it is used fimncon rather than lsqnonlin?
Torsten
Torsten 2023-6-8
Yes. Nonlinear constraints in lsqnonlin were introduced recently.

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