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Note: in light of Robert's catch below, I've updated the problem I'm facing to its full prose.

Suppose we have an optimization problem of the form

$\max_{A,p} \sum_{i=1}^n f(A_i^T D) p_i$

s.t.

$Ap=c$

$\sum_{i=1}^n p_i = 1$

$0 \le p_i \le 1$ $i=1,\cdots,n$

$\sum_{i=1}^m A_{ij} = 1$ $i=1, \cdots, n$

$0 \le A_{ij} \le 1$ $i=1,\cdots,n,j=1,\cdots,m$

Where $A \in \mathbb{R}^{m\times n}$, $p \in \mathbb{R}^n$, $f(v) : \mathbb{R}^{d} \mapsto \mathbb{R}$ and is piecewise linear, D is constant and $\in \mathbb{R}^{m \times d}$, and c is constant and $\in \mathbb{R}^m$. As stated, this appears to be a Quadratically Constrained Quadratic Program (which can be seen after expressing the objective in its epigraph form, and then expanding the resulting max constraints into a set of linear inequalities).

But suppose, assuming m > n, it is known that the space of rank-n A contains an optimal solution. If we were to optimize over that restricted space, only a single $p$ satisfies the constraint for each $A$. So in that case it seems unnecessary to optimize over both $A$ and $p$ when a feasible assignment of $A$ determins a single feasible assignment to $p$.

What I'm wondering is whether there is anything to be gained by reformulating the problem to eliminate $p$, something like

$\max_{A \in \{\text{rank} \; \text{n} \;mxn \; \text{matrices}\}} \sum_{i=1}^n f(A_i ^T D)(A^{-1}c)_i$

s.t.

$\sum_{i=1}^n (A^{-1}c)_i = 1$

(other constraints)

What problem class does the latter formulation belong to, and does the transformation admit a more efficient solution? Is there some other way to take advantage of there being a single feasible p for every feasible A according to the assumptions above?

I'm new to optimization, so please excuse my ignorance on the subject.

Made a crucial mistake in the problem formulation; please delete.

Note: in light of Robert's catch below, I've updated the problem I'm facing to its full prose.

Suppose we have an optimization problem of the form

$\max_{A,p} \sum_{i=1}^n f(A_i^T D) p_i$

s.t.

$Ap=c$

$\sum_{i=1}^n p_i = 1$

$0 \le p_i \le 1$ $i=1,\cdots,n$

$\sum_{i=1}^m A_{ij} = 1$ $i=1, \cdots, n$

$0 \le A_{ij} \le 1$ $i=1,\cdots,n,j=1,\cdots,m$

Where $A \in \mathbb{R}^{m\times n}$, $p \in \mathbb{R}^n$, $f(v) : \mathbb{R}^{d} \mapsto \mathbb{R}$ and is piecewise linear, D is constant and $\in \mathbb{R}^{m \times d}$, and c is constant and $\in \mathbb{R}^m$. As stated, this appears to be a Quadratically Constrained Quadratic Program (which can be seen after expressing the objective in its epigraph form, and then expanding the resulting max constraints into a set of linear inequalities).

But suppose, assuming m > n, it is known that the space of rank-n A contains an optimal solution. If we were to optimize over that restricted space, only a single $p$ satisfies the constraint for each $A$. So in that case it seems unnecessary to optimize over both $A$ and $p$ when a feasible assignment of $A$ determins a single feasible assignment to $p$.

What I'm wondering is whether there is anything to be gained by reformulating the problem to eliminate $p$, something like

$\max_{A \in \{\text{full} \; \text{rank} \;nxn \; \text{matrices}\}} \sum_{i=1}^n f(A_i ^T D)(A^{-1}c)_i$

s.t.

$\sum_{i=1}^n (A^{-1}c)_i = 1$

(other constraints)

What problem class does the latter formulation belong to, and does the transformation admit a more efficient solution? Is there some other way to take advantage of there being a single feasible p for every feasible A according to the assumptions above?

I'm new to optimization, so please excuse my ignorance on the subject.

Note: in light of Robert's catch below, I've updated the problem I'm facing to its full prose.

Suppose we have an optimization problem of the form

$\max_{A,p} \sum_{i=1}^n f(A_i^T D) p_i$

s.t.

$Ap=c$

$\sum_{i=1}^n p_i = 1$

$0 \le p_i \le 1$ $i=1,\cdots,n$

$\sum_{i=1}^m A_{ij} = 1$ $i=1, \cdots, n$

$0 \le A_{ij} \le 1$ $i=1,\cdots,n,j=1,\cdots,m$

Where $A \in \mathbb{R}^{m\times n}$, $p \in \mathbb{R}^n$, $f(v) : \mathbb{R}^{d} \mapsto \mathbb{R}$ and is piecewise linear, D is constant and $\in \mathbb{R}^{m \times d}$, and c is constant and $\in \mathbb{R}^m$. As stated, this appears to be a Quadratically Constrained Quadratic Program (which can be seen after expressing the objective in its epigraph form, and then expanding the resulting max constraints into a set of linear inequalities).

But suppose, assuming m > n, it is known that the space of rank-n A contains an optimal solution. If we were to optimize over that restricted space, only a single $p$ satisfies the constraint for each $A$. So in that case it seems unnecessary to optimize over both $A$ and $p$ when a feasible assignment of $A$ determins a single feasible assignment to $p$.

What I'm wondering is whether there is anything to be gained by reformulating the problem to eliminate $p$, something like

$\max_{A \in \{\text{rank} \; \text{n} \;mxn \; \text{matrices}\}} \sum_{i=1}^n f(A_i ^T D)(A^{-1}c)_i$

s.t.

$\sum_{i=1}^n (A^{-1}c)_i = 1$

(other constraints)

What problem class does the latter formulation belong to, and does the transformation admit a more efficient solution? Is there some other way to take advantage of there being a single feasible p for every feasible A according to the assumptions above?

I'm new to optimization, so please excuse my ignorance on the subject.

Note: in light of Robert's catch below, I've updated the problem I'm facing to its full prose.

Suppose we have an optimization problem of the form

$\max_{A,p} \sum_{i=1}^n f(A_i^T D) p_i$

s.t.

$Ap=c$

$\sum_{i=1}^n p_i = 1$

$0 \le p_i \le 1$ $i=1,\cdots,n$

$\sum_{i=1}^m A_{ij} = 1$ $i=1, \cdots, m$

$0 \le A_{ij} \le 1$ $i=1,\cdots,m,j=1,\cdots,n$

Where $A \in \mathbb{R}^{m\times n}$, $p \in \mathbb{R}^n$, $f(v) : \mathbb{R}^{d} \mapsto \mathbb{R}$ and is piecewise linear, D is constant and $\in \mathbb{R}^{m \times d}$, and c is constant and $\in \mathbb{R}^m$. As stated, this appears to be a Quadratically Constrained Quadratic Program (which can be seen after expressing the objective in its epigraph form, and then expanding the resulting max constraints into a set of linear inequalities).

But suppose, assuming m > n, it is known that the space of rank-n A contains an optimal solution. If we were to optimize over that restricted space, only a single $p$ satisfies the constraint for each $A$. So in that case it seems unnecessary to optimize over both $A$ and $p$ when a feasible assignment of $A$ determins a single feasible assignment to $p$.

What I'm wondering is whether there is anything to be gained by reformulating the problem to eliminate $p$, something like

$\max_{A \in \{\text{full} \; \text{rank} \;nxn \; \text{matrices}\}} \sum_{i=1}^n f(A_i ^T D)(A^{-1}c)_i$

s.t.

$\sum_{i=1}^n (A^{-1}c)_i = 1$

(other constraints)

What problem class does the latter formulation belong to, and does the transformation admit a more efficient solution? Is there some other way to take advantage of there being a single feasible p for every feasible A according to the assumptions above?

I'm new to optimization, so please excuse my ignorance on the subject.

Note: in light of Robert's catch below, I've updated the problem I'm facing to its full prose.

Suppose we have an optimization problem of the form

$\max_{A,p} \sum_{i=1}^n f(A_i^T D) p_i$

s.t.

$Ap=c$

$\sum_{i=1}^n p_i = 1$

$0 \le p_i \le 1$ $i=1,\cdots,n$

$\sum_{i=1}^m A_{ij} = 1$ $i=1, \cdots, n$

$0 \le A_{ij} \le 1$ $i=1,\cdots,n,j=1,\cdots,m$

Where $A \in \mathbb{R}^{m\times n}$, $p \in \mathbb{R}^n$, $f(v) : \mathbb{R}^{d} \mapsto \mathbb{R}$ and is piecewise linear, D is constant and $\in \mathbb{R}^{m \times d}$, and c is constant and $\in \mathbb{R}^m$. As stated, this appears to be a Quadratically Constrained Quadratic Program (which can be seen after expressing the objective in its epigraph form, and then expanding the resulting max constraints into a set of linear inequalities).

But suppose, assuming m > n, it is known that the space of rank-n A contains an optimal solution. If we were to optimize over that restricted space, only a single $p$ satisfies the constraint for each $A$. So in that case it seems unnecessary to optimize over both $A$ and $p$ when a feasible assignment of $A$ determins a single feasible assignment to $p$.

What I'm wondering is whether there is anything to be gained by reformulating the problem to eliminate $p$, something like

$\max_{A \in \{\text{full} \; \text{rank} \;nxn \; \text{matrices}\}} \sum_{i=1}^n f(A_i ^T D)(A^{-1}c)_i$

s.t.

$\sum_{i=1}^n (A^{-1}c)_i = 1$

(other constraints)

What problem class does the latter formulation belong to, and does the transformation admit a more efficient solution? Is there some other way to take advantage of there being a single feasible p for every feasible A according to the assumptions above?

I'm new to optimization, so please excuse my ignorance on the subject.