Given $n$ symmetric positive definite matrices $\mathrm W_1, \mathrm W_2, \dots, \mathrm W_n \in \mathbb R^{m \times m}$ and $s \in \mathbb N$, where $s < n$,

$$\begin{array}{ll} \text{maximize} & \det \left( \displaystyle\sum_{i=1}^n z_i \mathrm W_i \right)\\ \text{subject to} & \displaystyle\sum_{i=1}^n z_i = s\\ & \mathrm z \in \{0,1\}^n\end{array}$$

Since the matrices are positive definite and $z_i \geq 0$,

$$\det \left( \displaystyle\sum_{i=1}^n z_i \mathrm W_i \right) \geq \displaystyle\sum_{i=1}^n \det \left( z_i \mathrm W_i \right) = \displaystyle\sum_{i=1}^n z_i^m \det \left( \mathrm W_i \right) = \displaystyle\sum_{i=1}^n z_i \det \left( \mathrm W_i \right)$$

Let $c_i := \det \left( \mathrm W_i \right)$. The following binary **integer program** (IP)

$$\begin{array}{ll} \text{maximize} & \displaystyle\sum_{i=1}^n c_i z_i\\ \text{subject to} & \displaystyle\sum_{i=1}^n z_i = s\\ & \mathrm z \in \{0,1\}^n\end{array}$$

provides a **lower bound** on the maximum of the original optimization problem. This lower bound may be too loose, however. It would be nice to relax the integrality constraint and obtain an **upper bound** on the maximum, too.