IEOR 269, Spring 2010 Integer Programming and Combinatorial ...

IEOR269 notes, Prof. Hochbaum, 2010 5
3.2 Piecewise linear cost function
Consider the following problem:
n∑
min f i (x i )
i=1
where n ∈ N, and for i ∈ {1, . . . , n}, f i is a piecewise linear function on k i successive intervals.
Interval j for function f i has length w j i , and f i has slope c j i
on this interval, (i, j) ∈ {1, . . . , n} ×
{1, . . . , k i }. For i ∈ {1, . . . , n}, the intervals for function f i are successive in the sense that the
supremum of interval j for function f i is the infimum of interval j +1 for function f i , j ∈ {1, . . . , k i −
1}. The infimum of interval 1 is 0, and the value of f i at 0 is fi 0 , for i ∈ {1, . . . , n}. The notations
are outlined in Figure 2.
f i (x i )
✻
f 0 i
c ✁ ✁❅ 1 ❅
c 2 i
i
✁ ❅
✁ ✁ ❅✏ ✏✏ c 3 i ❍ ❍
✁
❇
c
❇
k i
i
❇
❇
w 1 i w 2 i w 3 i w k i
i
✲ x i
Figure 2: Piecewise linear objective function
We define the variable δ j i , (i, j) ∈ {1, . . . , n} × {1, . . . , k i}, to be the length of interval j at which f i
is estimated. We have x i = ∑ k i
j=1 δj i .
The objective function of this problem can be rewritten ∑ (
n
i=1
fi 0 + ∑ )
k i
j=1 δj i cj i
. To guarantee
that the set of δ j i
define an interval, we introduce the binary variable:
{
y j i = 1 if δ j i > 0
0 otherwise
and we formulate the problem as:
⎛
n∑ ∑k i
min ⎝f i 0 +
i=1
j=1
δ j i cj i
⎞
⎠
subject to w j i yj+1 i
≤ δ j i ≤ wj i yj i
for (i, j) ∈ {1, . . . , n} × {1, . . . , k i } (5)
y j i ∈ {0, 1}, for (i, j) ∈ {1, . . . , n} × {1, . . . , k i}