The motivation for removing small items is that, when all items are large, the number of items in each bin must be small, so the number of possible configurations is (relatively) small. Initially, weWe pick some constant ''<math>g\in(0,1)</math>'', and remove from the original instance <math>I</math> all items whose size is lesssmaller than ''<math>g\cdot B</math>''. Let <math>J</math> be the reducedresulting instance. We pack <math>J</math> and get a packing with some <math>b_J</math> bins.
Now, we add the small items into the existing bins in an arbitrary order, as long as there is room. When there is no more room in the existing bins, we open a new bin (as in [[next-fit bin packing]]). Let <math>b_I</math> be the number of bins in the resultingfinal allocationpacking. Then: <blockquote><math>b_I \leq \max(b_J, (1+2 g)\cdot OPT(I) + 1)</math>.</blockquote>''Proof''. If no new bins are opened, then the number of bins remains <math>b_J</math>. If a new bin is opened, then all bins except maybe the last one contain a total size of at least <math>B - g\cdot B</math>, so the total instance size is at least <math>(1-g)\cdot B\cdot (b_I-1)</math>. So the optimal solution needs at least <math>(1-g)\cdot (b_I-1)</math> bins. Therefore, <math>b_I \leq OPT/(1-g) + 1 = (1+g+g^2+\ldots) OPT+1 \leq (1+2g) OPT+1</math>.
