Sparsity pattern

mining

Namemining
Downloadmining.mps.gz
Set Membership Challenge
Problem StatusOpen
Problem FeasibilityFeasible
Originator/ContributorK. Eurek
Rows661133
Cols348921
Num. non-zeros in A3844879
Num. non-zeros in c348921
Rows/Cols1.8947928041
Integers
Binaries348920
Continuous1
min nonzero |Aij|0.6859025
max |Aij|27044.12
min nonzero |cj|0.03020538
max |cj|948653700
Integer ObjectiveUnknown
LP Objective-949724584.696851
Aggregation
Variable Bound347953
Set partitioning
Set packing
Set covering
Cardinality
Equality Knapsacks
Bin packing
Invariant Knapsack313101
Knapsacks
Integer Knapsack
Mixed 0/178
General Cons.
References

Unspecified mining application.

Mark Zuckerberg suggests the following tightening of one class of constraints:
The variables in this problem (except for the last one, x348921, which is fixed to 1 and does not appear in any constraint) are arranged in groups of 20, and there are three classes of precedence constraints (or more accurately, implication constraints):
1) For each variable index i=...,348920, if i mod 20 > 0 then there is a constraint x_i <= x_{i+1} (so x1 <= x2 <= ... <= x20, x21<=x22 <= ... <= x40, etc).
2) If i mod 20 = and i>20 then there is a constraint x_i <= x_{i-20}, (so x1 >= x21 >= x41 >=...)
3) and for all other i there is a constraint x_i <=_{i-1} + x_{i-20} (so x22 <= x21 + x2, x23 <= x22 + x3,..., x40 <= x39 + x20, x42 <= x41 + x22,..., x60 <= x59 + x40,...).
Graphically, we can represent the variables as a table with height twenty and width 348920 / 20, with indices 1,...,20 going down the left-most column, then 21,...,40 going down the next column to the right, etc. There is then a precedence constraint (1) from each variable in a column to the one below it, a precedence constraint (2) from the topmost entry in each column to its neighbor on the left, and an "OR" precedence constraint (3) from each entry that isn't on top of a column, requiring that if it has value 1 then either the one above it or the one to its left must have value 1 as well.
But in fact this OR precedence (3) can be replaced with a simple precedence constraint from each variable to the one at its left. To see this note that in any column, the first type of precedence constraint will require that a feasible solution must be a contiguous sequence of 1's from some point in the column until the bottom. Consider column c > 1. The topmost 1 in this column, by precedence constraints (3) (or if the topmost is at the top of the column, then by precedence constraints (2)) implies that the variable to its left (in column c-1) is also 1, which implies that the contiguous sequence of 1's in column c-1 must be at least as high as the sequence in column c. Thus for every 1 in column c, the entry to its left in column c-1 must also be 1.


Last Update February 28, 2017 by Gerald Gamrath
© 2017 by Konrad-Zuse-Zentrum für Informationstechnik Berlin (ZIB)
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