fractional small jobs of priority bags and we need to ensure that the height increase through this rounding is bound by $O(\epsilon)$. We will still see that the number of integral variables is bound by constants dependent on $\epsilon$.

Before we look at the running time necessary to solve this MILP and the number of variables we want to proof, that this MILP will yield a solution, when there exists a solution for our instance.

\[lem:milp\] Given a modified instance $I$ of machine scheduling with bag-constraints. If there is a solution of makespan $T$ for $I$ then the MILP will have a valid solution.

Now that we know that our MILP will have a valid solution the question remains how fast we can compute this solution. We want to apply the result of Kannan [@kannan], which is an improvement on the result of Lenstra [@lenstra]. The time to solve the MILP with these approaches depends strongly on the number of integral variables. Therefore we will look at this number in the following lemma.

\[lem:runtime\] Given an instance $I$ for machine scheduling with bag-constraints. We can solve the respective MILP with a running time of $2^{O(z*log(z))}*poly(n)$ with\
$z = 2^{O(\frac{1}{\epsilon^{k+1}}*log(\frac{1}{\epsilon^{2k+11}}*log^3(\frac{1}{\epsilon})))}$.

With 