# Optimal instance of quadratically constrained program

Consider the following optimization problem. Let $n, m \in \mathbb N$ and $0 < p_1 \leq \ldots \leq p_n ~ (p_i \in \mathbb R)$ be constant. The feasible region is described by a partition $T_1, \ldots, T_m$ of the index set $\lbrace 1, \ldots, n \rbrace$, i.e. $\bigcup_{1 \leq i \leq n} T_i = \lbrace 1, \ldots, n \rbrace$ and $T_i \cap T_j = \emptyset ~ (i \neq j)$ and the following constraints.

$$\forall 1 \leq i \leq n:~ 0 \leq r_i \leq p_i,~ 0 \leq c_i \leq p_i$$ $$r_1 = c_1$$ $$\forall 1 < i \leq m ~ \forall j \in T_i \backslash \lbrace 1 \rbrace:~ r_j = c_j + r_{j-1} + \sum_{1 \leq k < j ~ \wedge ~ k \in T_i} \left(\left\lceil \frac{r_j}{p_k} \right\rceil - w_{i,j,k} \right) c_k$$ $$\text{where} ~ w_{i,j,k} = \left\lceil \frac{r_{j-1}}{p_k} \right\rceil ~ \text{if} ~ j - 1 \in T_i ~ \text{and} ~ w_{i,j,k} = \left\lfloor \frac{r_{j-1}}{p_k} \right\rfloor ~ \text{if} ~ j - 1 \notin T_i$$

The optimization problem is to find the maximum of $\sum_{1 \leq i \leq n} \frac{c_i}{p_i}$ over all partitions.

My conjecture is that the optimum can always be obtained by a partition where for each $T_i$ there is exactly one $c_j > 0 ~ (j \in T_i)$ while $c_k = 0 ~ \forall k \in T_i \backslash \lbrace j \rbrace$. I can show this for various small concrete problem instances but I am lacking a general proof idea to show the conjecture for any such problem. Do you have an idea how to approach the proof? Thanks for your time.

Edit: The problem can be simplified in two ways. First, we can assume $m = 2$ (for $m = 1$ the conjecture holds trivially). Second, we can restrict the feasible region in order to get rid of $\lfloor \cdot \rfloor$ and $\lceil \cdot \rceil$ by replacing the third constraint above by the following one.

$$\forall 1 < i \leq m ~ \forall j \in T_i \backslash \lbrace 1 \rbrace:~ r_j = c_j + r_{j-1} + \sum_{1 \leq k < j ~ \wedge ~ k \in T_i} \left(\frac{r_j - r_{j-1}}{p_k} + 2 \right) c_k$$

Do you have a proof idea for the conjecture in this simpler case?

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