INTEGER PROGRAMMING: THE BRANCH AND BOUNT METHOD
I have taken this example from Taha.
It is quite common for variables to be allowed to take only integer values. For example,
we cannot buy part of a car, or allocate half of a child to a school. This is quite easy to
formulate, all we need to do is to say that the relevant variable needs to take an integer
value. But it is not easy to solve. There is no simple algorithm to do Integer
Programming (IP). There are various methods, such as Branch and Bound (B&B),
which we will study in here. These days IP is to be found everywhere. Many
combinatorial optimisation problems, such as organising a timetable for a university, are
formulated as IP. We consider two kinds of IP problems: all integer problems, where all
the variables take integer values, and Mixed IP, where some variables must take integer
values, while others can take any value.
The B&B method is based on a very simple idea: by introducing extra constraints things
can only get worse.
Take the following problem (Taha, example 9.3-1)
Our first step is to treat it as if it was an ordinary LP problem. By removing the integer
constraints we are giving freedom to the problem, and we obtain the maximum possible
value that x0 can take.
Basic X0X1X2S1S2Sol Ratio
X01 -5 -4 0 0 0
S10 1 1 1 0 5 5/1
S20 10 6 0 1 45 45/10 ←
↑
X01 0 -1 0 1/2 45/2
S10 0 2/5 1 -1/10 1/2 9/5 ←
X10 1 3/5 0 1/10 9/2 31/5
↑
X01 0 0 5/2 1/4 95/4
X20 0 1 5/2 -1/4 5/4
X10 1 0 -3/2 1/4 15/4
The solution is x0 = 95/4 = 23.5, x2 = 5/4 = 1.25, x1 = 15/4 = 3.75.
The decision variables do not satisfy the integer condition. Now we “branch” on one of
the variables. It does not matter which one, so we choose x1. X1 cannot be 3.75. In
fact, it cannot be any number between 3 and 4, as such numbers do not satisfy the
integer condition. This leads us to create two new problems, one with and another one
with .
We try the first option first. We need to solve:
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