where the [mathematical expression not reproducible] is the set of the dual variables, and [rho] > 0.
Firstly the initialization of the auxiliary variables includes: randomly choosing the auxiliary voltages, the auxiliary powers and the dual variables. Significantly, the value of the dual variables must satisfy [[lambda].sup.k.sub.g,h] = [[lambda].sup.k.sub.h,g] for ([c.sub.g], [c.sub.h]) [member of] [[OMEGA].sub.c].
In solving the dual problem, a decomposition and iterative method is proposed to obtain the optimal
dual variables.
[mathematical expression not reproducible]; are
dual variables that correspond to (7), (8), (9), (10), (11), (12), and (13), respectively.
Where [v.sub.i], i = 1, ..., m;[u.sub.r], r = 1, ..., s and [w.sub.f], f = 1, ..., h are all positive
dual variables associated with the first three constraints from the model (1).
, [v.sub.K]) [greater than or equal to] 0 are the vectors of the
dual variables associated with the individual power constraint for the users, individual power constraint for the relays, and individual rate constraint for the users, respectively.
Let w [member of] [R.sup.n] be optimal
dual variables corresponding to the first n constraints of linear program LP(S).
Before showing how this is done, let us give the reader some feel for how the
dual variables "pay" for a primal solution by considering the following simple setting: suppose LP (2) has an optimal solution that is integral, say I [subset or equal to] F and [Phi]: C [right arrow] I.
In the optimal solution, the lease-lines estimate of [p.sub.r] - [p.sub.t] is simply the difference between the
dual variables for sources r and t, while [q.sub.r] - [q.sub.t] is estimated by the difference between the
dual variables for destinations r and t.
where [tau] and [[bar.[tau]].sub.m] are the
dual variables associated with the power constraint at the source and the different relays respectively.
Different users adjust the water level through
dual variables [alpha], [[beta].sub.m], [[mu].sub.k].