SLIDE 14 162
0 yi 1 yi = 0 or 1 1 1 v(LR
i)
v(LR
i)
Figure 2
taken by a given yi decides alone the fate of all other variables containing the same value of the index i – that usually means that if variable yi is 0, all variables in “its family” are 0, and if it is 1, they are solutions of a subproblem – one may be able to reformulate the problem in terms of the variables yi only. Often, but not always, when this property holds, it is because the Lagrangean problem, after removal of all constraints containing
- nly the yi’s – let us call it (LRPλ), for partial problem – decomposes into
- ne problem (LRP i
λ) for each i, i.e., for each 0-1 variable yi.
The use of this property is based on the following fact. In problem (LRP i
λ), the integer variable yi can be viewed as a parameter, however we
do know that for the mixed-integer problem (LRP i
λ), the feasible values of
that parameter are only 0 and 1, and we can make use of the fact that there are only two possible values for v(LRP i
λ), the value computed for yi = 1,
say vi (= vi · yi for yi = 1), and the value for yi = 0, that is, 0 (= vi · yi for yi = 0), which implies that for all possible values of yi, v(LRP i
λ) = vi · yi.
Hence the name “integer linearization”, as one replaces a piecewise linear function corresponding to 0 ≤ yi ≤ 1 by a line through the points (0, 0) and (1, vi). We will first present an example of the simple decomposable case. Example 6.1. Suppose that (LRλ) is of the form minx{fx + gy | Aixi ≤ piyi, all i, x ∈ X, By ≤ b, yi = 0 or 1, all i}, where there is one set of con- straints Aixi ≤ piyi for each i, the constraints By ≤ b are over y alone, Ai and pi are nonnegative, and X = Πi(Xi). Here xi may be a vector, with possibly some integer part. To solve (LRλ), one can proceed as follows: (i) ignore at first the constraints containing only the yi’s, i.e., By ≤ b;