A branch-and-bound node can close for three different reasons. It can be integer, infeasible, or unable to beat the incumbent bound. This lesson shows all three cases in one small tree.

highlighted = computed this step

Integer candidate

The y-low child gives (3, 0) with z=9. Why: an integer LP optimum becomes a candidate incumbent.

z=9z=9
Three fathom typesThe full tree exhibits integer, infeasible, and bound fathoming.Three fathom typesy<=0y>=1x<=2x>=3rootLP (3, 1/2)z=10branchedy<=0LP (3, 0)z=9incumbent optimaly>=1LP (5/2, 1)z=19/2branchedx<=2LP (2, 3/2)z=9bound fathomedx>=3LP infeasibleinfeasible

Bound fathom

The x-low leaf has LP bound 9, equal to the incumbent value. Why: a node that cannot beat the incumbent is closed.

boundzinc\text{bound}\le z_{inc}
Three fathom typesThe full tree exhibits integer, infeasible, and bound fathoming.Three fathom typesy<=0y>=1x<=2x>=3rootLP (3, 1/2)z=10branchedy<=0LP (3, 0)z=9incumbent optimaly>=1LP (5/2, 1)z=19/2branchedx<=2LP (2, 3/2)z=9bound fathomedx>=3LP infeasibleinfeasible

Infeasible fathom

The x-high leaf has no feasible LP relaxation. Why: an empty relaxation cannot contain an integer feasible point.

LP infeasible\text{LP infeasible}
Three fathom typesThe full tree exhibits integer, infeasible, and bound fathoming.Three fathom typesy<=0y>=1x<=2x>=3rootLP (3, 1/2)z=10branchedy<=0LP (3, 0)z=9incumbent optimaly>=1LP (5/2, 1)z=19/2branchedx<=2LP (2, 3/2)z=9bound fathomedx>=3LP infeasibleinfeasible

Diagram note

The three leaf statuses are recomputed by solving the LP relaxation at each node. Pixel positions are rounded for layout; every number shown is exact.

integer, infeasible, and bound fathoms close nodes\text{integer, infeasible, and bound fathoms close nodes}
Three fathom typesThe full tree exhibits integer, infeasible, and bound fathoming.Three fathom typesy<=0y>=1x<=2x>=3rootLP (3, 1/2)z=10branchedy<=0LP (3, 0)z=9incumbent optimaly>=1LP (5/2, 1)z=19/2branchedx<=2LP (2, 3/2)z=9bound fathomedx>=3LP infeasibleinfeasible