Finding integrality gap for maximum weight independent set
Suppose you prove that an upper bound on the integrality gap is $\alpha$.
  The goal is to “create” an instance (weighted graph, or better yet, a family of weighted graphs) where the value of the optimal LP solution is $\alpha$
  times the value of the ILP solution. 

Independent set ILP
$$O_{ILP}=\max\sum_{v\in V}x_{v}\cdot w(v) $$
such that
$$x_{v}+x_{u} \leq 1\ \ \ \forall u,v\in V$$
$$x_{v}\in\{0,1\}  \ \ \forall v\in V$$

The relaxation is straight forward. say the value of the optimal solution to the LP is $O_{LP}.$ Now, you want to upper bound $ O_{LP}/O_{ILP}.$
  To do this, try to consider a rounding scheme that converts a solution for the LP into a feasible solution for the corresponding ILP. One such scheme involves reasoning about the constraints, that is $x_{v}+x_{u}\leq1$
 for every edge in G.
  This implies that at least one of $x_{u}$
  or $x_{v}$
  is atleast $1/2$,
  and therefore all vertices in $v\in V$
  such that $x_{v}&gt;\frac{1}{2}$
  is an independent set in $G$.
  I'll leave it to you to reason about how to compare $O_{LP}$
  and $O_{ILP}$, (i.e. determine the maximum of the ratio $\alpha=\max O_{LP}/O_{ILP}$), and to find an instance where $O_{LP}$ is atleast $\alpha.$