Dear Mark,

Thanks very much for your response. Let me consider the following optimization problem (for a given K):

and M

P_1: \min_{\{x_{mk}\}} ~~~~\sum\limits_{m=1}^{M}\sum\limits_{k=1}^K x_{mk}

s.t.

\dfrac{\left(\sum_{m=1}^M\sqrt{x_{mk}b_{mk}}\right)^2}{\sum_{k^\prime\ne k}^K\sum_{m=1}^M\sum_{n\ne m}^M \sqrt{a_{mk^\prime k}}\sqrt{a_{nk^\prime k}}\sqrt{x_{mk^\prime}}\sqrt{x_{nk^\prime}}+1} \ge 1

\sum_{m=1}^M\sum_{k=1}^K x_{mk} \le 1

x_{mk}\ge 0,

where a and b are positive numbers. This is equivalent to

P_2: \min_{\{x_{mk}\}} ~~~~\sum\limits_{m=1}^{M}\sum\limits_{k=1}^K x_{mk}

s.t.

\dfrac{\left(\sum_{m=1}^M\sqrt{x_{mk}b_{mk}}\right)^2}{\sum_{k^\prime\ne k}^K\left(\sum_{m=1}^M \sqrt{x_{mk^\prime}}\sqrt{a_{mkk^\prime}}\right)^2-\sum_{k^\prime\ne k}^K \sum_{m=1}^M x_{mk^\prime}a_{mkk^\prime}+1} \ge 1

\sum_{m=1}^M\sum_{k=1}^K x_{mk} \ge 1

x_{mk}\le 0.

Now I introduce new slack variables w_{mlk}=\sqrt{x_{mlk}}. This results in the following optimization problem:

P_3: \min_{\{w_{mk}\}} ~~~~\sum\limits_{m=1}^{M}\sum\limits_{k=1}^K \mathbf{w}_{k}

s.t.

\dfrac{\mathbf{w}_k^T \mathbf{F}_k \mathbf{w}_k}{\sum_{k^\prime\ne k}^K \mathbf{w}_{k^\prime}^T \mathbf{G}_{k} \mathbf{w}_{k^\prime}-\sum_{k^\prime\ne k}^K \mathbf{w}_{k^\prime}^T \mathbf{H}_{k} \mathbf{w}_{k^\prime}+1} \ge 1

\sum_{m=1}^M\sum_{k=1}^K w_{mk}^2 \ge 1

w_{mk}\le 0,

where \mathbf{F}_k=[b_{1k}...b_{Mk}]^T [b_{1k}...b_{Mk}], \mathbf{G}_{k^\prime}=[\sqrt{a_{1kk\prime}}...\sqrt{a_{Mkk\prime}}]^T[\sqrt{a_{1kk\prime}}...\sqrt{a_{Mkk\prime}}] and \mathbf{H}_k=\text{diag}[a_{1kk^\prime}... a_{Mkk^\prime}]

Next, I introduce the new variable \mathbf{W}_{k}=\mathbf{w}_{k}\mathbf{w}_{k}^T, which enables us to reformulate Problem P_3 into a standard SDP using semidefinite relaxation. By utilising the identity \mathbf{w}^T\mathbf{R}\mathbf{w}=\text{Tr}[\mathbf{R}\mathbf{w}\mathbf{w}^T]=\text{Tr}[\mathbf{R}\mathbf{W}], Problem P_3 can be rewritten as follows:

P_4: \min_{\{\mathbf{W}_k\}} ~~~~\sum\limits_{k=1}^K \text{Tr}(\mathbf{W}_k)

s.t.

\text{Tr} (\mathbf{F}_k\mathbf{W}_k)-\sum_{k^\prime \ne k}^K\text{Tr}(\mathbf{G}_{k^\prime}\mathbf{W}_{k^\prime})-\sum_{k^\prime\ne k}^K\text{Tr}(\mathbf{H}_{k^\prime} \mathbf{W}_{k^\prime})\ge 1

\sum_{m=1}^M\sum_{k=1}^K w_{mk}^2 \ge 1

\mathbf{W}_{k}=\mathbf{W}_{k}^T

\mathbf{W}_{lk}\succeq 0

\text{rank}[\mathbf{W}_{lk}]=1,

where \mathbf{W}_{lk}\succeq 0 means that \mathbf{W}_{lk} is a positive semidefinite matrix. By relaxing all rank-one constraints in Problem P_4, we arrive at a standard SDP.

Now, I am going to use SOCP to solve problem P_1. For this, I need to make the denominator as ||\mathbf{v}_k||^2. This is what I am not able to do!! Looking at Problem P_2 (where I re-wrote the denominator), I am NOT able to write the denominator as ||\mathbf{v}_k||^2 as follows:

\sum_{k^\prime\ne k}^K\left(\sum_{m=1}^M \sqrt{x_{mk^\prime}}\sqrt{a_{mkk^\prime}}\right)^2-\sum_{k^\prime\ne k}^K \sum_{m=1}^M x_{mk^\prime}a_{mkk^\prime}+1 \ne ||\mathbf{v}_k||^2, because there is a negative term here.

My supervisor believes there must be a way to write the denominator as ||\mathbf{v}_k|| and solve this by SOCP. However, I do not know how to do it.

I would be extremely thankful if you could please guide me, and any help would be highly appreciated. Thanks a lot in advance!