Question

我想编写一个Matlab代码，通过考虑利润也可以优化调度发电机以满足电力需求。我使用了下面在matlab网站中的代码。 https://ww2.mathworks.cn/help/optim/examples/optimal-dispatch-of-power-generators.html?requestedDomain=en 但是在此代码中，他们没有考虑需求。我的目标是，有20座发电厂，我想安排这些发电厂以满足需求。并且运行功率水平可以是每个电厂最大功率值和最小功率值之间的任何值。高成本的电厂也应该调度。

name = 'generationcost.xlsx';
poolPrice = xlsread(name,'AD3:AD50'); % Revenue in dollars per MWh in interval
demand = xlsread(name,'AC3:AC52');
month = numel(demand);
e = xlsread(name,'W2:W4')'; %Cost for a unit of fuel
fuelPrice = repmat(e,1,1,2);
nPeriods = length(poolPrice);
nGens = 3; 
genhigh = xlsread(name,'F2:F4');
genlow = xlsread(name,'Q2:Q4');
gen = [genlow(:),genhigh(:),]; %MW

fuel = [50,330;45,325;55,750]; % Fuel consumption for generator
startCost = [10000;10000;10000]; %Cost in dollars to start a generator after it has been off

y = optimvar('y',nPeriods,nGens,{'Low','High'},'Type','integer','LowerBound',0,'UpperBound',1);
z = optimvar('z',nPeriods,nGens,'Type','integer','LowerBound',0,'UpperBound',1);

powercons = y(:,:,'Low') + y(:,:,'High') <= 1;


yFuel = zeros(nPeriods,nGens,2);
yFuel(:,1,1) = fuel(1,1); 
yFuel(:,1,2) = fuel(1,2); 
yFuel(:,2,1) = fuel(2,1); 
yFuel(:,2,2) = fuel(2,2); 
yFuel(:,3,1) = fuel(3,1); 
yFuel(:,3,2) = fuel(3,2); 

fuelUsed = sum(reshape((sum(y.*yFuel)),[3 2])');
%fuelcons = fuelUsed <= totalFuel;

w = optimexpr(nPeriods,nGens); 
idx = 1:(nPeriods-1);
w(idx,:) = y(idx+1,:,'Low') - y(idx,:,'Low') + y(idx+1,:,'High') - 
y(idx,:,'High');
w(nPeriods,:) = y(1,:,'Low') - y(nPeriods,:,'Low') + y(1,:,'High') - 
y(nPeriods,:,'High');
switchcons = w - z <= 0;

generatorlevel  = zeros(size(yFuel));
generatorlevel(:,1,1) = gen(1,1); 
generatorlevel(:,1,2) = gen(1,2);
generatorlevel(:,2,1) = gen(2,1);
generatorlevel(:,2,2) = gen(2,2); 
generatorlevel(:,3,1) = gen(3,1);
generatorlevel(:,3,2) = gen(3,2);

revenue = optimexpr(size(y));
for ii = 1:nPeriods
    revenue(ii,:,:) = poolPrice(ii)*y(ii,:,:).*generatorlevel(ii,:,:);
end

val = optimexpr(size(y));
demands = optimexpr(nPeriods,1);
id = 1:nPeriods;
val(id,:,:)= y(id,:,:).*generatorlevel(id,:,:);
demands(id,:)= sum(sum(val,3),2);
demandcons = demands-demand ;

fuelCost = sum(sum(fuelPrice.*sum(y.*yFuel)));
startingCost = z*startCost;

profit = sum(sum(sum(revenue))) - fuelCost - sum(startingCost);

dispatch = optimproblem('ObjectiveSense','maximize');
dispatch.Objective = profit;

dispatch.Constraints.powercons = powercons;
dispatch.Constraints.demandcons = demands - demand <= 0 ;

options = optimoptions('intlinprog','Display','final');
[dispatchsol,fval,exitflag,output] = solve(dispatch,'options',options);

subplot(5,1,1)
bar(dispatchsol.y(:,1,1)*gen(1,1)+dispatchsol.y(:,1,2)*gen(1,2),.5,'g') 
xlim([.5,48.5])
ylabel('MW')
title('Generator 1 Optimal Schedule','FontWeight','bold')
subplot(5,1,2)
bar(dispatchsol.y(:,2,1)*gen(2,1)+dispatchsol.y(:,2,2)*gen(2,2),.5,'c')
title('Generator 2 Optimal Schedule','FontWeight','bold')
xlim([.5,48.5])
ylabel('MW')
subplot(5,1,3)
bar(dispatchsol.y(:,3,1)*gen(3,1)+dispatchsol.y(:,3,2)*gen(3,2),.5,'c')
title('Generator 3 Optimal Schedule','FontWeight','bold')
xlim([.5,48.5])
ylabel('MW')
subplot(5,1,4)
bar(demand,.5)
xlim([.5,48.5])
title('Daily Demand','FontWeight','bold')
xlabel('Period')
ylabel('MW')

starttimes = find(round(dispatchsol.z) == 1);
[theperiod,thegenerator] = ind2sub(size(dispatchsol.z),starttimes)

此代码不能满足需求。输出需求值比实际需求值小。而且这里仅使用“高”或“低”功率电平。在两次使用的功率值之间。

发电机的优化调度

0 个答案: