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I saw this paper,and it said to make the first problem "tractable" ,so we have to modify the first problem to the second problem

First problem:

$\max\limits_{\{\mathbf v\},\rho_k}\min\limits_{\mathbf h_k \in H_k} SINR_k(\{\mathbf v_k\},\rho_k)$

s.t.

$\zeta(1-\rho_k)(\sum\limits_{j=1}^{K}|\mathbf h_k^H \mathbf v_j|^2+\sigma^2_k) \ge e_k,\forall k$

$\sum\limits_{k=1}^{K}tr(\mathbf D_m \mathbf v_k \mathbf v_k^H) \le P_m,\forall m$

$0 \le \rho_k \le 1,\forall k$

the paper said is non-convex due to coupled variables {ρk} and {vk} in both the objective function and the EH constraint, and thus, is difficult to solve efficiently.

Second problem

$\min\limits_{\{\mathbf v\},\rho_k}\max\limits_{1 \le m \le M} \frac{\sum\limits_{k=1}^{K}tr(\mathbf D_m \mathbf v_k \mathbf v_k^H )}{P_m}$

s.t.

$\zeta(1-\rho_k)(\sum\limits_{j=1}^{K}|\mathbf h_k^H \mathbf v_j|^2+\sigma^2_k) \ge e_k,\forall k$

$SINR_k(\{\mathbf v_k\},\rho_k) \ge \Gamma,$

$0 \le \rho_k \le 1,\forall k$

the paper said To make problem (3) tractable, we decompose the problem into a set of the min-max per-DA port power balancing problems, one for each given SINR target Γ > 0 [15]. Using bisection search over Γ

I want to ask that what does that "tractable" mean?nonconvex?hard to calculate?can't be calculate?why can we just want to make the first problem "trackble" ,then modify the objective function from max-min to min-max?

I don't know why will we rewrite the first formula to the second formula,because it seems that the objected functions of two problem are inverse proportion,so the max-min will rewrite to min-max,but i don't know why will they be inverse proportion

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  • $\begingroup$ The relationship between the two problems is not clear. Sometimes such statements mean "we don't know how to solve the original problem, so let's solve this other one instead". $\endgroup$ – copper.hat Jun 27 at 5:29
  • $\begingroup$ @copper.hat yes i know,but i don't know why will we rewrite the first formula to the second formula,because it seems that the objected functions of two problem are inverse proportion,so the max-min will rewrite to min-max,but i don't know why will they be inverse proportion $\endgroup$ – XM551 Jun 27 at 5:41

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