Your understanding is correct. However, with numerical algorithms such as the
primal-dual interior point method used by MATPOWER’s default OPF solver,
depending on the numerical termination criteria for the algorithm, some
non-binding constraints may still have small non-zero values for their
multipliers. MATPOWER attempts to screen these out with a threshold of 1e-4
required to display a constraint, but in this case that threshold is too small.
Decreasing the termination tolerance eliminates the problem.
mpopt = mpoption('opf.ac.solver', 'MIPS', 'mips.comptol', 1e-7);
r = runopf(m,mpopt);
Best regards,
Ray
> On Apr 8, 2016, at 3:50 AM, Deep Kiran <[email protected]> wrote:
>
> Dear Matpower community,
>
> I was doing some interpretation for pattern of Lagrange multiplier correspond
> to Qg limits on case5.m.
>
> I have done some modification on the network which are as follows:
>
> m = loadcase(case5);
> m.bus(:,[3,4]) = m.bus(:,[3,4])*0.65;
> m.bus(:,[4]) = m.bus(:,[4])*-2;
> r = runopf(m);
>
> With the above modification, I get the following result for reactive power
> limit of generator at bus 4.
>
> Gen Bus Reactive Power Limits
> # # Qmin mu Qmin Qg Qmax Qmax mu
> ---- ----- ------- -------- -------- -------- -------
> 4 4 0.001 -150.00 -145.20 150.00 -
>
> From the above result, Qg(=-145.20) is still within its limits [-150,150].
> My understanding of Lagrange multiplier is that it remains zero for the
> inequality constraint that is < or > 0. Its value will be non-zero only if
> the inequality constraint is = 0. Although, the Qmin mu(=0.001) is small,
> however, Qg is 4.8 off from its lower bound.
>
> Am I missing some obvious link here?
>
> Thank you in advance.
>
> Best regards,
> Deep Kiran
> Research Scholar,
> IIT Delhi,
> India
