Wow, its been 2 years of using Matpower and I was unaware of this setting. Thank you Sir.
Best regards, Deep Kiran On 8 April 2016 at 19:37, Ray Zimmerman <[email protected]> wrote: > 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 > > >
