Hello all,
In this moment, I am understanding the mechanism about of the minimization of
cost with curves cost and load benefit. For this reason, I have posed a problem
with generators and dispatchable loads in a two nodes system with the following
equations:
f1 = 0.05*p(1)^2+100*p(1; in node 1 (generator 1); limits 0<p(1)<100
f2 = -0.05*p(2)^2+100*p(2); in node 1(dispatchable load 1); limits
-100<p(1)<0
f3 = -0.06*p(3)^2+120*p(3); in node 2 (generator 2); limits 0<p(1)<100
f4 = 0.06*p(4)^2+120*p(4) ; in node 2 (dispatchable load 2); limits
-100<p(1)<0
With this conditions, I run DC OPF and AC OPF and found the following results
of generation and distpachable load:
DC OPF AC OPF
generator 1 89.0265 90.8313
dispatchable load 1 -89.0265 -0.0000
generator 2 92.4779 0.0000
dispatchable load 2 92.4779 -90.8238
To compare these results, I used the Matlab’s Toolbox, specifically fmimcon, to
optimize these curves and the results are:
x =
90.9091 0.0000 -0.0000 -90.9091
fval =
-909.0909
Why are they so different the DC and AC results?.
PD: The problem is designed to optimized the curves f1 and f4, because the
curves f2 and f3 are the reflecting of the curves f1 and f4 (this case is only
to understand the mechanism). The problem is attached.
Thanks so much for the help.
Santiago Ch.
function mpc = mer
%% MATPOWER Case Format : Version 2
mpc.version = '2';
%% system MVA base
mpc.baseMVA = 100;
%% bus data
% bus_i type Pd Qd Gs Bs area Vm Va
baseKV zone Vmax Vmin
mpc.bus = [
1 3 0 0 0 0 1 1 0
135 1 1.05 0.95;
2 2 0 0 0 0 1 1 0
135 1 1.05 0.95;
];
%% generator data
% bus Pg Qg Qmax Qmin Vg mBase status Pmax
Pmin
mpc.gen = [
1 30 0 60 -15 1 100 1 100
0;
1 -30 0 0 0 1 100 1 0 -100;
2 30 0 60 -15 1 100 1 100
0;
2 -30 0 0 0 1 100 1 0 -100;
];
%% branch data
% fbus tbus r x b rateA rateB rateC ratio
angle status
mpc.branch = [
1 2 0.0001 0.0001 0.0 130 130 130 0
0 1;
];
%%----- OPF Data -----%%
%% generator cost data
% 1 startup shutdown n x1 y1 ... xn
yn
% 2 startup shutdown n c(n-1) ... c0
mpc.gencost = [
2 0 0 3 0.05 100 0;
2 0 0 3 -0.05 100 0;
2 0 0 3 -0.06 120 0;
2 0 0 3 0.06 120 0;
];
