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K.K.T conditions are both necessary and sufficient. These K.K.T conditions include: 

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along with the complementary slackness conditions η$_{u,}$$_{1}$θ$_{u,}$$_{1}$ = 0 , ϑ$_{u,b}$γ$_{u,b}$ = 0 , ∀ u ∈ U ( b $^{)}$, b ∈ B ′ 1 and (1 − ∑ u ∈U ′ θ$_{u,}$$_{1}$ ) λ = 0 , (1 − ∑ u ∈U ( b ) γ$_{u,b}$ ) β$_{b}$ = 0 , ∀ b ∈ B ′ $_{1}$, and the feasibility ones θ$_{u,}$$_{1}$, γ$_{u,b}$ ∈ [0 , 1] , ∀ u, b , ∑ u ∈U ′ θ$_{u,}$$_{1}$ ≤ 1 , ∑ u ∈U ( b ) γ$_{u,b}$ ≤ 1 , ∀ b . Note that λ, { β$_{b}$ } , { η$_{u,}$$_{1}$ } , { ϑ$_{u,b}$ } are non-negative Lagrangian variables. Manipulating the first two K.K.T conditions in (36) along with the complementary slackness conditions, we see that for any two distinct users u, u ′ ∈ U ( b ) 

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which is a contradiction. Consequently, there can be at-most one user u ∈ U ( b ) in each b ∈ B ′ 1 for which θ$_{u,}$$_{1}$γ$_{u,b}$ > 0 . From the remaining conditions, we can further deduce that 

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Further, an optimal solution must fully use the available resource so that ∑ u ∈U ′ θ$_{u,}$$_{1}$ = 1 and ∑ u ∈U ( b ) γ$_{u,b}$ = 1 , ∀ b ∈ B ′ $_{1}$. Next, for each b ∈ B ′ $_{1}$, let us determine the scalars µ$_{m,b}$, m = 1 , · · · , N$_{b}$ by sorting the set of ratios { R$_{u,}$$_{1}$ $_{R$_{u,b}$}$}$_{u}$$_{∈U}$ ( b ) in the increasing order. Then, define the set S A b = ∪$_{m}$$_{=1}$$_{,}$$_{···}$$_{,N}$$_{b}$ (( m − 1) µ$_{m,b}$, mµ$_{m,b}$ ) along with S B b = ∪$_{m}$$_{=2}$$_{,}$$_{···}$$_{,N}$$_{b}$$_{+1}$ [( m − 1) µ$_{m}$$_{−}$$_{1}$$_{,b}$, ( m − 1) µ$_{m,b}$ ] , which together partition the set of positive real numbers into two non-overlapping parts. Notice that if λ ∈ S A b then the only solution to the K.K.T conditions pertaining to TP b must have exactly one user k ∈ U ( b ) with θ$_{k,}$$_{1}$γ$_{k,b}$ > 0 and this user must be the one which has the m th smallest ratio µ$_{m,b}$ , where m ∈ { 1 , · · · , N$_{b}$ } : λ ∈ (( m − 1) µ$_{m,b}$, mµ$_{m,b}$ ) . Each user u whose corresponding ratio R$_{u,}$$_{1}$ R$_{u,b}$ is smaller than µ$_{m,b}$ is assigned resource only in TP b whereas each user u whose corresponding ratio R$_{u,}$$_{1}$ R$_{u,b}$ is greater than µ$_{m,b}$ is assigned resource only by the Macro. In this case, the total load imposed by users associated to TP b on the Macro is given by N$_{b}$ λ − 1 $_{µ$_{m,b}$}$. Similarly, whenever λ ∈ S B b then the only solution to the K.K.T consitions pertaining to TP b must have an orthogonal split of associated users in that each user associated to TP b is assigned resource either by the Macro or by TP b . In particular, when λ ∈ [( m − 1) µ$_{m}$$_{−}$$_{1}$$_{,b}$, ( m − 1) µ$_{m,b}$ ] for 

