Max Flow Application: Precedence Relations
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Max Flow Application: Precedence Relations
description
Max Flow Application: Precedence Relations. 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 - PowerPoint PPT Presentation
Transcript of Max Flow Application: Precedence Relations
Max Flow Application: Precedence Relations
Precedence Relations
• Given a finite set of elements B we define a precedence relation as relation between pairs of elements of S such that:
1. i not i for all i in B2. i j implies j i for all i, j in B3. i j and j k implies i k for all i, j, k in
B
Slide 2
Examples of Precedence Relations
1. Let B be the set of integers. The < relation is a precedence relation on B.
2. Let B be a set of jobs and let i j mean that job j cannot start until job i is complete.
Slide 3
Network Representation of Precedence Relations
a
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a e, a f, d e, d c, d b, f c
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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.
Slide 5
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
Slide 6
Aircraft Scheduling Example
F1
F3
F2
F4
Four-Chain Solution: {{F1},{F2},{F3},{F4}}
Three-Chain Solution: {{F1,F2},{F3},{F4}}
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Max Flow Formulationa
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Finding Chains from a Feasible Flow
1. Chose i' such that the flow from i' to t = 0. Let k = 1. chain[k] = i.
2. If the flow from s to i = 0, stop. {chain[1], chain[2], …, chain[k]} is a chain.
3. 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.
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A Feasible Flow v = 2a
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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
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Finding a Feasible Chaina
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A Cover with 4 chains
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a e, a f, d e, d c, d b, f c
Chain 1: a, f, cChain 2: eChain 3: dChain 4: b
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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.
Slide 14
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].
Slide 15
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.
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A Maximum Flow v = 3a
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s t
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A Cover with 3 chains
a
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a e, a f, d e, d c, d b, f c
Chain 1: a, f, cChain 2: d, eChain 3: b
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