Why do I get Error "Out of Memory caused by failure in user supplied fitness function" in Problem-Based Optimization

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Bastian
Bastian 2023-4-3
评论： Bastian 2023-4-5
Hello,
I am trying to minimize costs of running a heat pump taking spot market prices as reference, but I receive following error:
"Error using optim.problemdef.OptimizationProblem/solve
Out of memory.
Error in Optimierung_WP_Zeitbegrenzung_v2 (line 102)
[sol,fval,exitflag,output] = solve(problem,initialPoint,'Options',options);
Caused by:
Failure in user-supplied fitness function evaluation. Cannot continue.
Failure in initial user-supplied fitness function evaluation. GA cannot continue."
The household I am considering uses a rooftop PV. Therefore my costs do not only depend on the spot market prices, but also on the amount of electricity I can use from the PV and the electricity injected into the grid. The heat demand is given. The optimization should take place for every minute of the day for a whole year. Simplified assumptions to develop a solid code that can be adjusted later on are: Heat pump has to run 15min when started and needs to rest 15min after switched off, therefore start or stop time is assumed to be at each quarter hour. It can run with max power or 2/3 of max power. To test the code I run it over 3 days (4320 minutes), however, the optimization takes place stepwise in 6 hour setps to reduce calculation amount. This is my code
% Input data
x_Ergebnis = zeros(72,1); % variable to save on/off times of Heat pump after each day (15 minute steps)
y_Ergebnis = zeros(4320,1); % variable to save times of grid injection or grid consumption after each day
P_HP_Ergebnis = zeros(72,1); % variable to save power level decision after each day
Leistung_HP_Ergebnis = zeros(4320,1); % variable to save power level in kW after each day
Zustand_HP_Ergebnis = zeros(4320,1); % variable to save on/off times of heat pump after each day (1 minute steps)
s_Ergebnis = zeros(4320,1); % variable to save storage level after each day
storage_capacity = 13.25; % thermal energy storage capacity
s_Ergebnis(1)=0.8*storage_capacity; % initial heat SOC
for i = 1:12
Summenprofil = load('Summenprofil.mat');
COP = 4; % COP of the heat pump
P_max = 92/COP; % maximum thermal power of the heat pump in kW
P_max_house = ones(360,1).*86; % maximum total power consumption in kW
storage_capacity = 13.25; % thermal energy storage capacity
c_pv_self = ones(360,1).*10; % self-consumed PV electricity price in ct/kWh
c_pv_feed = ones(360,1).*7; % fed-in PV electricity price in ct/kWh
diff = optimexpr(360,1); % variable describing load difference
Self_Consumption_Netzbezug = optimexpr(360,1); % Self-Consumption variable when Grid consumption
Self_Consumption_Netzeinspeisung = optimexpr(360,1); % Self-Consumption varible when Grid injection
Grid_Injection = optimexpr(360,1); % amount of grid injection
Grid_Consumption = optimexpr(360,1); % amount of grid consumption
Leistung_gesamt = optimexpr(360,1); % total power measured from the household
s = optimexpr(360,1); % variable describing SOC of heat storage
z = optimexpr(360,1); % binary variable that is 1 when there is overproduction of PV
Zustand_HP = optimexpr(360,1); % binary variable (heat pump switched on/off)
Leistung_HP = optimexpr(360,1); % determines power of heat pump
m = i*360;
l = 1+(i-1)*360;
q = i*24;
r = 1+(i-1)*24;
PV_prod = Summenprofil.Summenprofil11.PV(l:m)./1000;
Heat_demand = Summenprofil.Summenprofil11.Wrmebedarf(l:m)./1000;
House_demand = Summenprofil.Summenprofil11.Haushalt3(l:m)./1000;
c_grid = Summenprofil.Summenprofil11.Kosten(l:m); % grid electricity price in ct/kWh
% Create optimization variables
x = optimvar("x",24,"Type","integer","LowerBound",0,"UpperBound",1); % heat pump switched on/off
P_HP = optimvar("P_HP",24,"Type","integer", "LowerBound",0,"UpperBound",1); % power level of heat pump
y = optimvar("y",360,"Type","integer","LowerBound",0,"UpperBound",1); % grid injection (yes or no)
% Set initial starting point for the solver
initialPoint.x = ones(size(x));
initialPoint.P_HP = ones(size(P_HP));
initialPoint.y = ones(size(y));
for o = 1:24
if o == 1
Zustand_HP(1:15) = x(1);
else
Zustand_HP((o-1)*15+1:o*15) = x(o); % Zustand_HP = x but with Zustand_HP describing every minute instead of 15 minute time slots
end
end
for t = 1:24
if t == 1
Leistung_HP(1:15) = P_HP(1)*(1/3)*P_max+(2/3)*P_max;
else
Leistung_HP((t-1)*15+1:t*15) = P_HP(t)*(1/3)*P_max+(2/3)*P_max;
end
end
% Determine values of the optimization expressions for every minute
for j = 1:360
diff(j) = (House_demand(j)/60)+(Zustand_HP(j)*Leistung_HP(j)/60)-PV_prod(j)/60; % determines load difference in every minute
Grid_Consumption(j) = diff(j); % positive "diff" means Grid Consumption
Self_Consumption_Netzbezug(j) = PV_prod(j)/60;
Self_Consumption_Netzeinspeisung(j) = (House_demand(j)/60)+(Zustand_HP(j)*Leistung_HP(j)/60);
Grid_Injection(j) = -diff(j); % negative "diff" means Grid Injection
z(j) = 1-y(j); % z always opposite of y
Leistung_gesamt(j) = Zustand_HP(j)*Leistung_HP(j)-PV_prod(j)+House_demand(j); % total power in each minute
% Calculating the SOC
if j == 1
s(1) = s_Ergebnis(l) + (Zustand_HP(1)*Leistung_HP(1) - Heat_demand(1))/60;
else
s(j) = s(j-1) + (Zustand_HP(j)*Leistung_HP(j) - Heat_demand(j))/60;
end
end
% Create problem
problem = optimproblem;
% Define problem objective
problem.Objective = sum(((c_grid./100).*Grid_Consumption.*y)... % cost of grid consumption
+ ((c_pv_self./100).*(Self_Consumption_Netzbezug.*y+Self_Consumption_Netzeinspeisung.*z))... % cost for Self-Consumption
- ((c_pv_feed./100).*Grid_Injection.*z));
problem.Constraints.constraint1 = s >= 0; % Minimum SOC
problem.Constraints.constraint2 = s <= storage_capacity; % Max SOC
problem.Constraints.constraint3 = diff.*(1-2.*y) <= 0; % y = 1 when Grid Consumption and no Grid Injection
problem.Constraints.constraint4 = Leistung_gesamt <= P_max_house; % total household power needs to be below maximum
% Display problem information
show(problem);
options = optimoptions(problem,'Display','iter');
% Solve problem
[sol,fval,exitflag,output] = solve(problem,initialPoint,'Options',options);
% Display results
sol;
exitflag;
fval;
% save the solutions
x_Ergebnis(r:q) = sol.x(1:24);
y_Ergebnis(l:m) = sol.y(1:360);
P_HP_Ergebnis(r:q) = sol.P_HP(1:24);
for p = 1:24
if p == 1
Zustand_HP_Ergebnis(l:l+14) = sol.x(1);
Leistung_HP_Ergebnis(l:l+14) = sol.P_HP(1)*(1/3)*P_max+(2/3)*P_max;
else
Zustand_HP_Ergebnis((p-1)*15+l:p*15+l-1) = sol.x(p);
Leistung_HP_Ergebnis((p-1)*15+l:p*15+l-1) = sol.P_HP(p)*(1/3)*P_max+(2/3)*P_max;
end
end
for k = 1:360
if k == 1 && i == 1
s_Ergebnis(l+k-1) = s_Ergebnis(1) + (Zustand_HP_Ergebnis(k)*Leistung_HP_Ergebnis(k) - Heat_demand(k))/60;
else
s_Ergebnis(l+k-1) = s_Ergebnis(l+k-2) + (Zustand_HP_Ergebnis(l+k-1)*Leistung_HP_Ergebnis(k) - Heat_demand(k))/60; % Speichervariable befüllen
end
end
clearvars -except Summenprofil s x_Ergebnis y_Ergebnis P_HP_Ergebnis s_Ergebnis l m Zustand_HP_Ergebnis Leistung_HP_Ergebnis
if l == 1
s_Ergebnis(l) = s_Ergebnis(1);
else
s_Ergebnis(m+1) = s_Ergebnis(m); % Set current SOC to starting SOC for next time interval
end
end
Used solver:
Solving problem using ga.
Single objective optimization:
408 Variable(s)
408 Integer variable(s)
1440 Nonlinear inequality constraint(s)
Options:
CreationFcn: @gacreationuniformint
CrossoverFcn: @crossoverlaplace
SelectionFcn: @selectiontournament
MutationFcn: @mutationpower
I would really appreciate if you could help me. Thank you in advance!
  11 个评论
显示 9更早的评论
Torsten
Torsten 2023-4-5
编辑：Torsten 2023-4-5
However, I do not get why this error hasn't occured before, because I thought changing continuous P_HP(360,1) to integer P_HP(24,1) would be easier to solve.
No, integer-constraint problems are far more difficult to solve than problems with continuous variables. That's why I asked about the solver MATLAB used before. I didn't look at your system in detail, but you should try to see whether it's possible to formulate your objective function and constraints linearly. It will save a lot of time because MATLAB will be able to either use linprog or intlinprog to solve.
Bastian
Bastian 2023-4-5
I don't see any possibility to do so, especially since I use some constraints to generate some kind of "if-conditions". Eg. I use optimvar y only to determine when variable "diff" is positive or negative.
As I said, my code worked. I tried to extend it with a condition so that the heat pump, when switched on (x=1), has to run for at least 2 time steps (30 mins), but I can't figure out a way to do so, without getting a nonsense solution or error. Do you have any idea to do that, without taking optimvars or optimexpr into "if-conditions"?

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