1. Max Flow Application: Precedence Relations
Updated 4/2/08

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
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 collectively contain (cover) 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}}
Three-Chain Solution: {{F1, F4},{F2},{F3}} F1 F2 F3 F4

8. Max Flow Formulation of Minimum Chain Covering Problem
a  e, a  f, d  e,
d  c, d  b, f  c a e c d b

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

10. Representing Chains with Flows
If i  j and elements i and j are consecutive elements in the same chain, then let xsi = xij' = xj't = 1
If element i starts a chain, then let xi't = 0
If element i ends a chain, then let xsi = 0
If element i is itself a chain, then let xsi = xi't = 0
Example Covers
6 chains: {{a}, {b}, {c}, {d}, {e}, {f}}
4 chains: {{a, f, c}, {b}, {d}, {e}}
3 chains: {{a, f, c}, {b}, {d, e}}

11. Cover 1: Flow Representationa a' b b' c c' s t d d' e e'
# chains = 6
v = 0 f f'

12. Cover 2: Flow Representation1 b b' c c' 1 1 s t d d' e e' 1 1 1
# chains = 4
v = 2 f f'

13. Cover 3: Flow Representation1 b b' c c' 1 1 s t 1 d d' 1 1 e e' 1 1 1
# chains = 3
v = 3 f f'

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

15. A Feasible Flow v = 2 a a' 1 b b' c c' 1 1 s t d d' e e' 1 1 1 f f'

16. Finding a Chain 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

17. Observation
There is a one-to-one correspondence between feasible flows and chain coverings
The number of chains in a cover plus the value of the flow is equal to the number of elements in the set
Cover 1: = 6
Cover 2: = 6
Cover 3: = 6
# chains + v = |B|
# chains = |B| - v
Since |B| is fixed, maximizing v minimizes the number of chains

18. Interpretation of Min Cuts
Observe that S = {s, 1, 2, …, |B|}, T = N \{S} is a cut with finite capacity.
Arcs of the form (i, j') cannot be in a minimum cut.
If xsi = 0 in a maximum flow then
Node i will be reachable from the source in the residual network
Node j' will also be reachable from the source if i  j

19. Interpretation of Min Cuts
Arcs of the form (s, i) and (i', t) each contribute 1 unit to u[S, T].
Minimizing u[S, T] is equivalent to finding the largest subset A of B such that i can be in S and i' can be in T for all i in A
Since the capacity of a minimum cut is finite, it follows that the set of elements in A mutually incomparable

20. Maximum Flow v = 3 a a' 1 b b' c c' 1 1 s t 1 d d' 1 1 e e' 1 1 1 f f'

21. Residual Network  a a' 1 b b' c c' s t d d' e e' f f'

22. Residual Network  a a' 1 b b' c c' s t d d' e e' f f'

23. Residual Network  a a' 1 b b' c c' s t d d' e e' f f'

24. Residual Network  a a' 1 b b' c c' s t d d' e e' f f'

25. Mutually Incomparable Elementsf
f is not  e, e is not  f,
e is not  b, b is not  e,
f is not  b, b is not  b a e c
b, e, and f form an anti-chain d b

26. Dilworth’s Theorem
Given a set B and a precedence relation, the size of the largest anti-chain is equal to the size of the minimum chain covering
Proof follows from the max-flow min-cut theorem