Dear all, I try to calculate lambda in cpf by using extended OPF formulation:
min f(x) + fu(x; z) subject to g(x) = 0 h(x) <= 0 xmin <= x <= xmax l <= A[x; z] <= u zmin <= z <= zmax However, if we want to do cpf, lambda must be considered in Jacobi and Hessian matrix, such as g(x, lambda) = 0 h(x, lambda) <= 0 >From extended OPF formulation, does it mean that user-define variable can only >be used in linear constraint? Then the origin goal is impossible if I don't modify Jacobi and Hessian by myself? Regards, > LarryHo ________________________________ 寄件者： Ray Zimmerman <[email protected]> 收件者： MATPOWER Discussion List <[email protected]> 寄件日期： 2013/8/7 (週三) 2:02 AM 主旨： Re: About using opf to do cpf It may be possible to use the OPF with user-defined variables/constraints to solve the problem you have in mind, but you would need to specify the formulation of your problem in more detail to be sure. There are examples of some simple user-defined extensions to the OPF in the OPF test functions (e.g. t/t_opf_mips.m). -- Ray Zimmerman Senior Research Associate B30 Warren Hall, Cornell University, Ithaca, NY 14853 phone: (607) 255-9645 On Aug 6, 2013, at 7:27 AM, [email protected] wrote: Dear all, > > > I am wondering if it is possible to use opf's user-define variable to >calculate lambda in cpf? > Such as > > > max lambda > F(x, lambda) = 0 > > > And where can I find the example of user-define function? It's not that >helpful by reading the manual. > > > Regards, > LarryHo
