Summary: Let \(X=\{x_i:1\leq i\leq n\}\subset\mathbb N^+\), and \(h\in\mathbb N^+\). The \(h\)-iterated sumset of \(X\), denoted \(hX\), is the set \(\{x_1+x_2+\dots +x_h:x_1,x_2,\dots,x_h\in X\}\), and the \([h]\)-sumset of X, denoted \([h] X\), is the set \(\bigcup_{i = 1}^h i X\). A \([h]\)-sumset cover of \(S\subset\mathbb N^+\) is a set \(X\subset\mathbb N^+\) such that \(S\subseteq [h]X\). In this paper, we focus on the case \(h=2\), and study the \(\mathbf{APX}\)-hard problem of computing a minimum cardinality [2]-sumset cover \(X\) of \(S\) (i.e. computing a minimum cardinality set \(X\subset\mathbb N^+\) such that every element of \(S\) is either an element of \(X\), or the sum of two – non-necessarily distinct – elements of \(X\)). We propose two new algorithmic results. First, we give a fixed-parameter tractable (FPT) algorithm that decides the existence of a [2]-sumset cover of size at most \(k\) of a given set \(S\). Our algorithm runs in \(O(2^{(3\log k-1.4)k}\text{poly}(k))\) time, and thus outperforms the \(O(5^{\frac{k^2(k+3)}{2}}k^2\log (k))\) time FPT result presented in [I. Fagnot et al., Lect. Notes Comput. Sci. 5609, 378–387 (2009; Zbl 1248.68362)]. Second, we show that deciding whether a set \(S\) has a smaller [2]-sumset cover than itself is \(\mathbf{NP}\)-hard.