Because the function is convex, (49) is the global minimizer. We substitute (49) into (46). The Lagrange dual problem associated with the inner minimization problem is (51). Each q$_{i}$ is assumed to be positive. µ is implicity assumed to be nonzero because it appears as a denominator in (51). Therefore, the objective function of (51) decreases with increasing λ$_{i}$ . The optimal value of each λ$_{i}$ should thus be zero. (51) reduces to (52). We use the trick presented in Palomar (2009) to address the max-min problem. We replace the inner minimization problem with the dual maximization problem. This substitution is valid if and only if strong duality holds. Then, the overall structure of the problem is max-max instead of max-min. The double max structure can collapse to a more conventional problem with only one maximization operator. We use the criteria in Boyd and Vandenberghe (2004) to determine whether strong duality holds. If the primal problem is convex and Slater’s condition holds, strong duality holds. Slater’s condition holds if there exists a strictly feasible p ∈ relintD . 36 8.4.2 Applying Strong Duality to the Inner Minimization Problem max µ,ν − µ ∑ N i $_{=1}$q$_{i}$ e − 1 − d i + ν µ − µ Ω − ν s.t. µ ≥ 0 (52) max λ ,µ,ν − µ ∑ N i $_{=1}$q$_{i}$ e − 1+ λ i $_{−}$d i $_{−}$ν µ − µ Ω − ν s.t. λ$_{1}$, ..., λ$_{N}$ ≥ 0 µ ≥ 0 (51) g ( λ ,µ, ν ) = N ∑ i =1 ( d$_{i}$ − λ$_{i}$ ) q$_{i}$e − 1+ λ i $_{−}$d i $_{−}$ν µ + N ∑ i =1 ( − µ + λ$_{i}$ − d$_{i}$ − ν ) q$_{i}$e − 1+ λ i $_{−}$d i $_{−}$ν µ − µ Ω + v N ∑ i =1 q$_{i}$e − 1+ λ i $_{−}$d i $_{−}$ν µ − v = N ∑ i =1 ( − µ ) q$_{i}$e − 1+ λ i $_{−}$d i $_{−}$ν µ − µ Ω − ν = − µ N ∑ i =1 q$_{i}$e − 1+ λ i $_{−}$d i $_{−}$ν µ − µ Ω − ν (50)