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Below, we describe each of these steps in turn.
== Step 1. Removing and adding small items ==
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. We pick some constant ''<math>g\in(0,1)</math>'', and remove from the original instance <math>I</math> all items smaller than ''<math>g\cdot B</math>''. Let <math>J</math> be the resulting instance. Note that in <math>J</math>, each bin can contain at most ''<math>1/g</math>'' items. 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 final packing. 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>.
== Step 2. Grouping and un-grouping items ==
The motivation for grouping items is to reduce the number of different item sizes, to reduce the number of constraints in the configuration LP. The general grouping process is:
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== Step 3. Constructing and rounding the LP ==
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''e'' - a fraction in (0,1), such that ''eB'' is the smallest size of an item.
== Step 4. Solving the fractional LP ==
=== The dual LP ===
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