To make it easier to solve the problem, the original maximization problem in (5) can be transformed into a minimization problem as [17]
subject to:
(6)
The first constraint in (5) is relaxed in such a way that it is a real number on the interval of 0[1]. Furthermore, we define as the transmission power used by user k on subchannel s. The case corresponds to an unused subchannel for user k. The most important property of the objective function in (6) is that it is convex. The proof of convexity is given in Appendix.
Letting λ ≥ 0, η
s
≥ 0, ξ
k
,
s
≥ 0 and μ
k
,
s
≥ 0 be the Lagrange multipliers associated with the given constraints, the Lagrangian dual of (6) can be formulated as
(7)
The optimal solution must satisfy the Karush-Kuhn-Tucker (KKT) conditions [18], which can be given as follows:
(8)
(9)
(10)
(11)
(12)
(13)
From (8), we define
(14)
where is the logarithmic function and is the rest function of the first term in (8). From (12) and (13), if subchannel s is allocated to user k, i.e., α
k,s
= 1, then ξ
k
,
s
= 0 and μ
k
,
s
≥ 0. On the other hand, if subchannel s is not allocated to user k, i.e., α
k
,
s
< 1, then ξ
k
,
s
= 0 and μ
k
,
s
= 0. Thus, we can write
(15)
From (11) and (15), it can be concluded that η
s
is a constant for subchannel s of all users and subchannel s can be allocated to the user u(s), who has the maximum Ψ
k
,
s
on that subchannel, i.e.,
(16)
The objective in (16) is equivalent to finding the maximum . Hence, considering (2), we can conclude that
(17)
Note that the condition in (17) corresponds to selecting the user with the maximum weighted rate for subchannel s and given the transmit power levels.
Similarly, from (9) and (10), we may obtain the well-known water-filling solution as
(18)
where λ' is a constant which is a function of λ and can be obtained through substituting (18) into (10) which yields
(19)
where Ω
k
(|Ω
k
| ≤ S) is the set of subchannels assigned to user k.