Max Flow Application: Precedence Relations

1. Max Flow Application: Precedence Relations

2. Precedence Relations • Given a finite set of elements B we define a precedence relation as relation  between pairs of elements of S such that: • i not i for all i in B • ij implies ji for all i, j in B • i j and jk implies ik for all i, j, k in B

3. Examples of Precedence Relations • Let B be the set of integers. The < relation is a precedence relation on B. • Let B be a set of jobs and let ij mean that job j cannot start until job i is complete.

4. Network Representation of Precedence Relations f a  e, a f, d e, d  c, d b, f c a e c d b

5. Minimum Chain Covering Problem • A chain i1, i2, …, ik is a sequence of elements in B such that i1i2 … ik. • The minimum chain covering problem is to find a minimum number of chains that covers all the elements of B.

6. Example: Aircraft Scheduling • Given a set of flight legs that must be serviced, determine the minimum number of planes required. • Example • Flight 1: SFO -> LAX • Flight 2: LAX -> DFW • Flight 3: OAK -> MSP • Flight 4: LAX -> CVG

7. Aircraft Scheduling Example Four-Chain Solution: {{F1},{F2},{F3},{F4}} Three-Chain Solution: {{F1,F2},{F3},{F4}} F2 F1 F4 F3

8. Max Flow Formulation  a a' 1 1 b b' 1 1 c c' 1 1 s t 1 1 d d' 1 1 e e' 1 1 f f'

9. Finding Chains from a Feasible Flow • Chose i' such that the flow from i' to t = 0. Let k = 1. chain[k] = i. • If the flow from s to i= 0, stop. {chain[1], chain[2], …, chain[k]} is a chain. • If the flow from s to i= 1 then find j' such that the flow from i to j' = 1. Let k = k+1, chain[k] = j. Let i = j. Go to step 2.

10. A Feasible Flow v = 2 a a’ 1 0 b b’ 0 0 c c’ 0 1 1 s t 0 0 d d’ 0 0 e e’ 1 1 1 f f’

11. Step 1: i’ = a’, k=1, chain[1] = a Step 2: xsa=1 Step 3: j’=f’, k=2, chain[2] = f, i=f Step 2: xsf=1 Step 3: j’=c’, k=3, chain[3] = c, i=c Step 2: xsc=0. Chain: a, f, c Finding a Chain

12. 1 0 0 1 1 1 Finding a Feasible Chain a a’ b b’ c c’ s t d d’ e e’ f f’

13. A Cover with 4 chains f a  e, a  f, d  e, d  c, d  b, f  c a Chain 1: a, f, c Chain 2: e Chain 3: d Chain 4: b e c d b

14. Interpretation of Network Flow Solution: Sink Side • Each node i’ such that xi’t = 0 starts a chain. • Note that there are |B| nodes adjacent to the sink. Each with unit capacity. • The number of chains determined by a feasible flow is |B| - v. • Maximizing the flow minimizes the number of chains.

15. Interpretation of Network Flow Solution: Source Side • Each node j such that xsj = 0 ends a chain. • Observe that S = {s, 1, 2, …, |B|}, T = N \{S} is a cut with finite capacity. • Only arcs of the form (s,j) or (i’,t) can be in a minimum cut. • If xsj = 0 in a maximum flow, then node j will be reachable from the source in the residual network. • If Node j is in S, then (s, j) contributes one unit to u[S,T].

16. Interpretation of Network Flow Solution: Source Side • Each node j such that xsj = 0 ends a chain. • The capacity of a minimum cut is equal to the number of source-adjacent nodes that don’t receive flow in a maximum flow. • The capacity of a minimum cut is equal to the number of chains. • Maximizing the flow, minimizes the cut capacity which minimizes the number of chains.

17. A Maximum Flow v = 3 a a’ 1 0 b b’ 0 0 c c’ 0 1 1 s t 0 1 d d’ 1 0 1 e e’ 1 1 1 f f’

18. A Cover with 3 chains f a  e, a  f, d  e, d  c, d  b, f  c a Chain 1: a, f, c Chain 2: d, e Chain 3: b e c d b