Recall the linear program from LinearProgram
- maximize x1+x2 subject to x1 ≥ 0, x2 ≥ 0, x1+2x2 ≤ 1, 3x1+x2 <= 1.
Any feasible solution (e.g. x=(1/4, 1/4)) to the linear program provides a lower bound on OPT.
How can we get upper bounds? For any feasible solution x,
- x1+x2
- ≤ x1+2x2 since x2 ≥ 0
- ≤ 1 since x1+2x2 ≤ 1 is a constraint.
Idea: try combinations of the constraints.
goal: bound x1+x2 for any feasible x.
know: x1+2x2 ≤ 1 and 3x1+x2 ≤ 1.
Linear combinations of these constraints give (all) other valid constraints:
- a (x1+2x2) + b(3x1+x2) ≤ a*1 + b*1 (provided a,b ≥ 0)
Rewriting:
- (a+3b) x1 + (2a+b) x2 ≤ a+b . (#1)
Want
- x1+x2 ≤ (a+3b) x1 + (2a+b) x2 (#2)
for then (#1) above implies x1+x2 ≤ a+b.
Desired inequality (#2) will be true as long as 1 ≤ a+3b and 1 ≤ (2a+b),
in which case we know any feasible x satisfies x1+x2 ≤ a+b.
To get the best possible upper bound:
- minimize a+b subject to
- a+3*b ≥ 1
- 2*a+b ≥ 1
- a,b ≥ 0
Exercise: determine optimal solution using pen and paper.
(Geometric interpretation of feasible dual solutions as valid half-spaces.)
General form:
- primal: minimize b.y subject to Ay ≥ c, y ≥ 0.
- is dual of
- dual: maximize c.x subject to ATx ≤ b, x ≥ 0
weak duality: cost of any feasible solution to the primal ≥ cost of any feasible solution to the dual
proof: b.y ≥ (ATx)y = xAy = x(Ay) ≥ x.c
strong duality: cost of optimal solution to the primal = cost of optimal solution to the dual
Notes:
- constraint in primal <--> variable in dual
- variable in primal <--> constraint in dual
- dual of dual is primal
Examples:
References:
- Chapter 12 of Approximation Algorithms by Vazirani.
- [Samir Khuller's notes]
- Chvatal