How to estimate the hardness of SIS instances?
The value $\delta$ characterizes, how short a vector you can expect to find using an algorithm (typically used in the context of lattice reduction).
In particular, for a vector $\mathbf{v} \in \Lambda$ (where $\Lambda$ is a lattice), the associated $\delta$ (often also denoted by $\delta_0$) is defined to be such that $\| \mathbf{v} \| = \delta^n \det(\Lambda)^{1/n}$.
For those don't know how to take the determinant of lattices, see page 4 of this introduction lecture in the course &quot;Lattices in Computer Science&quot; given by Oded Regev.
This was introduced in GN08 in the context for lattice reduction, because it was observed that the $\delta$ for output vectors returned by lattice reduction converges (for growing $n$) to a constant for every type and parametrization of reduction.
Generally, reduction algorithms are parameterized and allow for a trade-off between output quality (i.e. size of $\delta$) and running time. The exact trade-off is still a subject to research, especially as reduction algorithms continue to improve, but a good summary of the state-of-the-art is given in APS15 (Section 3). Roughly speaking, translating a SIS instance to a $\delta$ required to break it, is relatively easy, but translating the $\delta$ to a running time is not as straight-forward. The reason is that the behavior of lattice reduction, the most efficient approximation algorithms to date, is not very well understood and the community has not converged on a model yet.
One issue with lattice reduction is that it uses an exact SVP solver (in lower dimension). In order to instantiate these solvers there are different algorithms, some behaving better in practice and others being asymptotically more efficient. This makes it hard to analyze exactly, how long it will take to obtain a short vector of quality $\delta$.