I try to do one optimization, I need to consider how many cycle usage. the “cycle” variable is updated by other binary variables, which equal to “if c1 and c2 then cycle = cycle + 1”.current, I got the correct “cycle” value update.
But I also need to reset cycle = 0, when “cycle == 5”, which is used to simulate restart process. I try to solve this question by the big-M method or semi-continuous set method, the cycle value will not update.
“if cycle == 5 then set cycle =0” condition, I try to use the following code:
cl(i-1) >= 4+0.00001-MC3(i); % if cl(i-1) == 5 then C3(i) = 0
cl(i-1) <= 4+M(1-C3(i));
1e-8-MC3(i)<=cl(i)<= 1e-8+MC3(i); % through C3(i) = 0 to set cl(i) = 0
However, after adding this constrain, the cycle value will stop at 4, it seems the whole system stays idle, there is no operating for the whole system.
for day = 1:12
cvx_begin
variables bc(24,1) bd(24,1) Pg(24,1) E(24,1) Bmax(24,1)
variable q(24,1) binary
variable p(24,1) binary
variable C1(24,1) binary
variable C2(24,1) binary
variable C3(24,1) binary
variable C4(24,1) binary
variable C(24,1) binary
variable CC(24,1) binary
variable cl(24,1) integer
minimize sum(bill_rates.*Pg)+ Prmax*max(Pg)/30+...
0.12*sum(emis_rates.*(Pg+(bc*(1-eff)+bd*(1/eff-1))))
subject to
zeros(24,1) <= bc <= 100*q;
zeros(24,1) <= bd <= 100*p;
zeros(24,1) <= (p + q) <= ones(24,1);
l-s+bc == Pg+bd;
E(1) == E_iter;
cl(1) == cl_iter;
p(1)+q(1) == 0;
Bmax(1) == 400*(1-0.05*cl(1));
for i = 2:24
E(i) == E(i-1)+ bc(i)*eff-bd(i)/eff;
E(i) <= Bmax(i-1);
E(i) >= 0;
ec+sum(bc(1:i))*eff-400*cl(i-1) >= 400+1-M*(1-C1(i))
ec+sum(bc(1:i))*eff-400*cl(i-1) <= 400+M*C1(i)
ed+sum(bd(1:i))/eff-400*cl(i-1) >= 400+1-M*(1-C2(i))
ed+sum(bd(1:i))/eff-400*cl(i-1) <= 400+M*C2(i)
C1(i)>= C(i)
C2(i)>= C(i)
C(i)+1 >= C1(i) + C2(i)
cl(i) == cl(i-1) + C(i)
Bmax(i) == 400*(1-0.05*cl(i));
cl(i-1) >= 4+0.00001-M*C3(i);
cl(i-1) <= 4+M*(1-C3(i));
1e-8-M*C3(i)<=cl(i)<= 1e-8+M*C3(i);
end
cvx_end
end