Graph for [2]
Graph for [3]
Graph for [4]
Therefore, since x1 can take on any continuous value ≥ 0 and x2 must be an integer ≥ 0, we
found our solution using branch and bound to be:
z=5.5 where x1 = 1.5, x2 = 1
b) The first step of the solution above (graph 2) is the solution to the relaxed LP of the IP given
above where z=5.6667 with x1 = 4/3 and x2 = 5/3. The solution to our IP and the relaxed LP are
different. The relaxed LP has a slightly higher max since it was not constrained to selecting only
integers. In contrast, our IP solution is a bit lower because x2 was constrained to choosing only
integer values.
0
