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Logistics Routing Plans: Max Flow Problem
Objectives and Agenda:
1. Examples for flow of materials over limited capacity channels
2. Finding maximum flows: Ford-Fulkerson Method
Logistics supply problem: Example 1
What is the maximum power we can supply to Wan Chai for a Light-n-Sound Show?
Lamma
Power Station
WanC
hai
NorthPoint
RepulseBay
Aberdeen
PokFuLam
WesternCentral
HappyValley
30
50
40
2020
2020
10 20
5
40
15 15
4025
1520
Legend:Node: Sub-stationEdge: Power lineWeight: Line capacity
Logistics supply problem: Example 2
Detroit
Kansas City
Minneapolis
San Francisco
Boise
Phoenix
8
14
146
12
10
10
7
17
6
Detroit
Kansas City
Minneapolis
San Francisco
Boise
Phoenix
8
14
146
12
10
10
7
17
6
Legend:nodes: train line junctions;edges: rail line;weights: max no. of compartments/day
Maximum number of compartments per day from Detroit SF ?
Maximum Flow Problem: definitions
D
M
K
S
B
P
8
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146
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10
10
7
17
6
D
M
K
S
B
P
8
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146
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10
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SOURCE:Node with net outflow:Production point
SINK:Node with net inflow;Consumption point
CAPACITY:Maximum flowon an edge
Efficient method to solve such problems: Ford-Fulkerson Method
Ford-Fulkerson Method..
Three fundamental concepts:
1. Flow cancellation
2. Augmentation flow
3. Residual network
Ford-Fulkerson Method: Flow Cancellation
a b
14
6
Net flows:
5/14
3/6a b
2/14
6a b
Additional 7 units from b a ?!14
5/6
a b
Network
Flow: 5 units, a b, 3 units b a
Flow Cancellation: Compute the NET FLOW between a pair of nodes
Ford-Fulkerson Method: Augmenting Path
Augmenting Path: any path from source sink with positive capacity
Examples
8
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146
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10
10
7
17
6
0
0
0
0
00
0
0
D
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K
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B
P
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146
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10
10
7
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6
0
0
0
0
00
0
0
D
M
K
S
B
P
Path Capacity
<D, M, B, S> 8
<D, M, K, P, S> 6
…
Ford-Fulkerson Method: Residual network
6/8
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6/146
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6/10
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6/6
0
0
0
0
00
0
0
D
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K
S
B
P
6/8
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6/146
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6/10
7
17
6/6
0
0
0
0
00
0
0
D
M
K
S
B
P
2
14
812
12
10
4
7
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0
0
6
0
0
06
0
6
D
M
K
S
B
P
2
14
812
12
10
4
7
17
0
0
6
0
0
06
0
6
D
M
K
S
B
P
Given a Network G, with flow, | f |, on path p
Flow cancellation
residualnetwork
Ford-Fulkerson Method..
Initialize the network: zero flow
any augmenting path pin network ?
Apply maximum flow allowed on p
Compute residual network
ResidualnetworkCarries
Max FlowYES
NO
Ford-Fulkerson Method: Initialize
Step 1. Add 0-capacity links to pair ‘one-way’ edges
D
M
K
S
B
P
8
14
146
12
10
10
7
17
6
D
M
K
S
B
P
8
14
146
12
10
10
7
17
6
8
14
146
12
10
10
7
17
6
0
0
0
0
00
0
0
D
M
K
S
B
P
8
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146
12
10
10
7
17
6
0
0
0
0
00
0
0
D
M
K
S
B
P
Step 1
Ford-Fulkerson Method: Find an augmentation path
8
14
146
12
10
10
7
17
6
0
0
0
0
00
0
0
D
M
K
S
B
P
8
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146
12
10
10
7
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6
0
0
0
0
00
0
0
D
M
K
S
B
P
6/8
14
6/146
12
10
6/10
7
17
6/6
0
0
0
0
00
0
0
D
M
K
S
B
P
6/8
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6/146
12
10
6/10
7
17
6/6
0
0
0
0
00
0
0
D
M
K
S
B
P
Flow, f = 6 units
Step 2. Find a positive flow from Source Sink
Ford-Fulkerson Method...
6/8
14
6/146
12
10
6/10
7
17
6/6
0
0
0
0
00
0
0
D
M
K
S
B
P
6/8
14
6/146
12
10
6/10
7
17
6/6
0
0
0
0
00
0
0
D
M
K
S
B
P
2
14
812
12
10
4
7
17
0
0
6
0
0
06
0
6
D
M
K
S
B
P
2
14
812
12
10
4
7
17
0
0
6
0
0
06
0
6
D
M
K
S
B
P
Step 3. Update the residual network due to flow f
Current total flow: 6
Ford-Fulkerson Method….
2
14
812
12
10
4
7
17
0
0
6
0
0
06
0
6
D
M
K
S
B
P
2
14
812
12
10
4
7
17
0
0
6
0
0
06
0
6
D
M
K
S
B
P
Augmentation path: <D, M, B, S>Max flow: 2
Current total flow: 6+2 Residual network
0
14
812
10
4
7
15
0
2
6
0
08
0
6
D
M
K
S
B
P
14
812
10
10
4
7
0
6
0
0
2
0
6
Ford-Fulkerson Method…..
0
14
812
10
4
7
15
0
2
6
0
08
0
6
D
M
K
S
B
P
14
812
10
10
4
7
0
6
0
0
2
0
6
0
182
10
4
7
5
0
12
6
0
08
10
6
D
M
K
S
B
P
4
0
107
0
0
12
Augmentation path: <D, K, M, B, S>Max flow: 10
Current total flow: 6+2+10 Residual network
Ford-Fulkerson Method……
0
182
10
4
7
5
0
12
6
0
08
10
6
D
M
K
S
B
P
4
0
107
0
0
12
0
182
10
0
1
0
16
6
08
14
10
D
M
K
S
B
P
0
0
1034
0
12
Augmentation path: <D, K, P, B, S>Max flow: 4
Current total flow: 6+2+10+4 Residual network
No moreAugmentation paths
DONE
Ford-Fulkerson Method: Proof
Network G, flow f1 residual network Gf1
Network Gf1, flow f2 residual network Gf1, f2
Network G, flow (f1 + f2) residual network Gf1, f2
=> We can solve the problem in stages!
Property 1: We can add augmentation flows
Ford-Fulkerson Method: Proof..
Property 2: Every source-containing bag has same net outflow
D
M
K
S
B
P
6/8
6/14
0/146/6
12/12
6/10
6/10
0/7
6/17
6/6
Example:Compare net flow out ofblue bag and red bag
Network G, flow f, amount: | f |
Each source-containing bag, net outflow = |f|
Why ?
Ford-Fulkerson Method: Proof...
Definition: Outflow capacity of a bag = total capacity of outflows
D
M
K
S
B
P
6/8
6/14
0/146/6
12/12
6/10
6/10
0/7
6/17
6/6
Examples:Outflow capacity of red bag = 8+6+10 = 24Outflow capacity of blue bag = 12+10 = 22
Ford-Fulkerson Method: Proof….
Suppose, at termination, total flow in network = f*
Using f*, we have no augmentation path from source sink
OUT-OF-BAG:Set of nodes with no augmentationpath from source
IN-BAG:Set of nodes with augmentationpath from source
Residual network, Gf*
Source
Sink
0
0
0
0
=> Existence of | f | > |f*| impossible!
Concluding remarks
(a)How to find augmenting paths ?
-- Need to search all possibilities on the network
(b) Classical terminology: The Max-flow Min-cut theorem
(c) Applications:
(i) Transportation Logistics (ships, airlines, trains)
(ii) Design of supply networks (water, sewage, chemical plant, food processing, roads)
next topic: Project management using CPM/PERT
Ford-Fulkerson Method..
Ford-Fulkerson Method..
