 Greedy two-bin scheduling bound

Put each successive item into a currently lighter bin. Before receiving weight $w_j$, that bin contains at most half the total weight already assigned, hence at most $(s-w_j)/2$, where $s$ is the overall total. Apply this to the last item in a finally heaviest bin to get $\mathrm{ALG}\le s/2+w_{\max }/2$. Since $\mathrm{OPT}\ge \max (s/2,w_{\max })$, the approximation ratio is at most $3/2$. Ordered weights $1,1,2$ give loads $3,1$ rather than the optimal $2,2$, proving the bound is sharp. Its relative-error guarantee is $\epsilon =1/2$.