1. Approximation Algorithms Duality My T. Thai @ UF

2. My T. Thai mythai@cise.ufl.edu 2 Duality  Given a primal problem: P: min c T x subject to Ax ≥ b, x ≥ 0  The dual is: D: max b T y subject to A T y ≤ c, y ≥ 0

3. My T. Thai mythai@cise.ufl.edu 3 An Example

4. My T. Thai mythai@cise.ufl.edu 4 Weak Duality Theorem  Weak duality Theorem: Let x and y be the feasible solutions for P and D respectively, then:  Proof: Follows immediately from the constraints

5. My T. Thai mythai@cise.ufl.edu 5 Weak Duality Theorem  This theorem is very useful  Suppose there is a feasible solution y to D. Then any feasible solution of P has value lower bounded by b T y. This means that if P has a feasible solution, then it has an optimal solution  Reversing argument is also true  Therefore, if both P and D have feasible solutions, then both must have an optimal solution.

6. My T. Thai mythai@cise.ufl.edu 6 Hidden Message ≥ Strong Duality Theorem: If the primal P has an optimal solution x* then the dual D has an optimal solution y* such that: c T x* = b T y*

7. My T. Thai mythai@cise.ufl.edu 7 Complementary Slackness  Theorem: Let x and y be primal and dual feasible solutions respectively. Then x and y are both optimal iff two of the following conditions are satisfied: (A T y – c) j x j = 0 for all j = 1…n (Ax – b) i y i = 0 for all i = 1…m

8. My T. Thai mythai@cise.ufl.edu 8 Proof of Complementary Slackness Proof: As in the proof of the weak duality theorem, we have: c T x ≥(A T y) T x = y T Ax ≥ y T b (1) From the strong duality theorem, we have: (2) (3)

9. My T. Thai mythai@cise.ufl.edu 9 Proof (cont) Note that and We have: x and y optimal  (2) and (3) hold  both sums (4) and (5) are zero  all terms in both sums are zero (?)  Complementary slackness holds (4) (5)

10. My T. Thai mythai@cise.ufl.edu 10 Why do we care?  It’s an easy way to check whether a pair of primal/dual feasible solutions are optimal  Given one optimal solution, complementary slackness makes it easy to find the optimal solution of the dual problem  May provide a simpler way to solve the primal

11. My T. Thai mythai@cise.ufl.edu 11 Some examples  Solve this system:

12. My T. Thai mythai@cise.ufl.edu 12 Min-Max Relations  What is a role of LP-duality  Max-flow and Min-Cut

13. My T. Thai mythai@cise.ufl.edu 13 Max Flow in a Network  Definition: Given a directed graph G=(V,E) with two distinguished nodes, source s and sink t, a positive capacity function c: E → R+, find the maximum amount of flow that can be sent from s to t, subject to: 1.Capacity constraint: for each arc (i,j), the flow sent through (i,j), f ij bounded by its capacity c ij 2.Flow conservation: at each node i, other than s and t, the total flow into i should equal to the total flow out of i

14. My T. Thai mythai@cise.ufl.edu 14 An Example s t 4 34 3 2 3 2 3 2 3 1 5 2 4 2 3 4 1 1 3 2 0 0 1 1 4 3 1 2 0 0 0

15. My T. Thai mythai@cise.ufl.edu 15 Formulate Max Flow as an LP  Capacity constraints: 0 ≤ f ij ≤ c ij for all (i,j)  Conservation constraints:  We have the following:

16. My T. Thai mythai@cise.ufl.edu 16 LP Formulation (cont) s t 4 34 3 2 3 2 3 2 3 1 5 2 4 2 3 4 1 1 3 2 0 0 1 1 4 3 1 2 0 0 0 ∞

17. My T. Thai mythai@cise.ufl.edu 17 LP Formulation (cont)

18. My T. Thai mythai@cise.ufl.edu 18 Min Cut Capacity of any s-t cut is an upper bound on any feasible flow If the capacity of an s-t cut is equal to the value of a maximum flow, then that cut is a minimum cut

19. My T. Thai mythai@cise.ufl.edu 19 Max Flow and Min Cut

20. My T. Thai mythai@cise.ufl.edu 20 Solutions of IP Consider: Let (d*,p*) be the optimal solution to this IP. Then:  p s * = 1 and p t * = 0. So define X = {p i | p i = 1} and X = {p i | p i = 0}. Then we can find the s-t cut  d ij * =1. So for i in X and j in X, define d ij = 1, otherwise d ij = 0.  Then the object function is equal to the minimum s-t cut

21. My T. Thai mythai@cise.ufl.edu 21 LP-relaxation  Relax the integrality constraints of the previous IP, we will obtain the previous dual.

22. My T. Thai mythai@cise.ufl.edu 22 Design Techniques  Many combinatorial optimization problems can be stated as IP  Using LP-relaxation techniques, we obtain LP  The feasible solutions of the LP-relaxation is a factional solution to the original. However, we are interested in finding a near-optimal integral solution:  Rounding Techniques  Primal-dual Schema

23. My T. Thai mythai@cise.ufl.edu 23 Rounding Techniques  Solve the LP and convert the obtained fractional solution to an integral solution:  Deterministic  Probabilistic (randomized rounding)

24. My T. Thai mythai@cise.ufl.edu 24 Primal-Dual Schema  An integral solution of LP-relaxation and a feasible solution to the dual program are constructed iteratively  Any feasible solution of the dual also provides the lower bound of OPT  Comparing the two solutions will establish the approximation guarantee