EMIS 8374 Network Flow Models updated 29 January 2008.
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Transcript of EMIS 8374 Network Flow Models updated 29 January 2008.
![Page 1: EMIS 8374 Network Flow Models updated 29 January 2008.](https://reader036.fdocuments.net/reader036/viewer/2022081504/56649f535503460f94c78839/html5/thumbnails/1.jpg)
EMIS 8374
Network Flow Modelsupdated 29 January 2008
![Page 2: EMIS 8374 Network Flow Models updated 29 January 2008.](https://reader036.fdocuments.net/reader036/viewer/2022081504/56649f535503460f94c78839/html5/thumbnails/2.jpg)
The Minimum Cost Network Flow Problem (MCNFP)
• Extremely useful model in OR & EM• Important Special Cases of the MCNFP
– Transportation and Assignment Problems– Maximum Flow Problem– Minimum Cut Problem– Shortest Path Problem
• Network Structure– BFS’s for MCNFP LP’s have integer values !!!– Problems can be formulated graphically
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Elements of the MCNFP
• Defined on a network G = (N,A)
• N is a set of n nodes: {1, 2, …, n}– Each node i has an associated value b(i)
• b(i) < 0 => node i is a demand node with a demand for –b(i) units of some commodity
• b(i) = 0 => node i is a transshipment node
• b(i) > 0 => node i is a supply node with a supply of b(i) units
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Elements of the MNCFP
• A is a set of arcs that carry flow– Decision variable xij determines the units of
flow on arc (i,j)
– The arc (i,j) from node i to node j has• cost cij per unit of flow on arc (i,j)
• upper bound on flow of uij (capacity)
• lower bound on flow of lij (usually 0)
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Example MCNFP
• N = {1, 2, 3, 4}b(1) =5, b(2) = -2, b(3) = 0, b(4) = -3
• A ={(1,2), (1,3), (2,3), (2,4), (3,4)}c12 = 3, c13 = 2, c23 =1, c24 = 4, c34 = 4
l12 = 2, l13 = 0, l23 = 0, l24 = 1, l34 = 0
u12 = 5, u13 = 2, u23 = 2, u24 = 3, u34 = 3
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Graphical Network Flow Formulation
b(j)b(i)
i j(cij, lij, uij)
arc (i,j)
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Example MCNFP
5 1 4(1, 0, 2)
3
2
0
-3
-2
(2, 0,2)
(4, 1,3)
(4, 0, 3)
(3, 2, 5)
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Requirements for a Feasible Flow
• Flow on all arcs is within the allowable bounds: lij xij uij for all arcs (i,j)
• Flow is balanced at all nodes:flow out of node i - flow into node i = b(i)
• MCNFP: find a minimum-cost feasible flow
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LP Formulation of MCNFP
Ajiux
Nibxx
xc
ijijij
ijiij
ijij
Ajij Aijj
Aji
),(
s.t.
min
),( : ),( :
),(
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LP for Example MCNFP
Min 3X12 + 2X13 + X23 + 4X24 + 4X34 s.t. X12 + X13 = 5 {Node 1} X23 + X24 - X12 = -2 {Node 2}
X34 - X13 - X23 = 0 {Node 3} - X24 - X34 = -3 {Node 4}
2 X12 5, 0 X13 2, 0 X23 2, 1 X24 3,
0 X34 3
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Example Feasible Solution
5 1 4(1, 0,2)
3
2
0
-3
-2
(2, 0,2)
(4, 1,3)
(4, 0,3)
(3, 2,5)
5 3
000
Cost = 15 + 12 = 27Arc flows shown in blue.
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Optimal Solution for Example MCNFP
5 1 4(1, 0,2)
3
2
0
-3
-2
(2, 0,2)
(4, 1,3)
(4, 0,3)
(3, 2,5)
3 1
022
Cost = 25Arc flows shown in blue. slide 12
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Balanced Transportation Problems
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Graphical Network Flow Formulation
b(j)b(i)
i j(cij, uij)
arc (i,j)
lij = 0 for all arcs.
slide 14
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DDummy Node -3
C
W
+4
+1
+2
Supply Nodes
I
S
G
Demand Nodes
-1
-1
-1
-1
A
F
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C
W
+4
+1
+2
Supply Nodes
I
S
G
Demand Nodes
-1
-1
-1
-1
A
F
(13, 1)
(35, 1)
(9, 1)
(42, 1)
Dummy Node-3
(0,4)
(0,2)
(0,1)
D
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Shortest Path Problems
• Defined on a network with two special nodes: s and t
• A path from s to t is an alternating sequence of nodes and arcs starting at s and ending at t:
s,(s,v1),v1,(v1,v2),…,(vi,vj),vj,(vj,t),t
• Find a minimum-cost path from s to t
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Shortest Path Example 1
1 2 3
4
5 10
7 71
s t
1,(1,2),2,(2,3),3 Length = 151,(1,2),2,(2,4),4,(4,3) Length = 131,(1,4),4,(4,3),3 Length = 14
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MCNFP Formulation of Shortest Path Problems
• Source node s has a supply of 1
• Sink node t has a demand of 1
• All other nodes are transshipment nodes
• Each arc has capacity 1
• Tracing the unit of flow from s to t gives a path from s to t
slide 19
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Shortest Path as MCNFP: Graphical Formulation
21 -11 3
4
(5,0,1) (10,0,1)
0
0
(7,0,1)(7,0,1)(1,0,1)
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Shortest Path as MCNFP: Graphical Solution
21 -11 3
4
(5,0,1) (10,0,1)
0
0
(7,0,1)(7,0,1)
Arc flows shown in blue.
0
01
1
1
(1,0,1)
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Shortest Path Example 2• In a rural area of Texas, there are six farms connected my
small roads. The distances in miles between the farms are given in the following table.
• What is the minimum distance to get from Farm 1 to Farm 6?
From Farm To Farm Distance 1 2 81 3 102 3 42 4 92 5 53 4 63 5 24 5 34 6 65 6 5
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Graphical Network Flow Formulation
b(j)b(i)
i j
lij = 0, uij=1
arc (i,j)
(cij)
slide 23
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Formulation as Shortest Path
s t
1
2 4
3
9
10
56
6
8 45
5
4
2
3
1 -1
0 0
00
slide 24
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LP Formulation of Shortest Path Problem
10
1
0
0
0
0
1st
min
5646
45352556
34244546
23133534
12252423
1312
5646453534
2524231312
54326594108
xxx
xxxxxxxxxxxxxxxx
xxxxxxxxxxxx
ij
slide 25
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“Greedy” Solution
s t
1
2 4
3
9
10
56
6
8 45
5
4
2
3
1 -1
0 0
00
x13 = x23 = x35 = x35 = 1, xij = 0 for all other arcs.Objective function value = 19. slide 26
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Shortest Path: Optimal Solution
s t
1
2 4
3
9
10
56
6
8 45
5
4
2
3
1 -1
0 0
00
x13 = x35 = x56 = 1, xij = 0 for all other arcs.Objective function value = 17. slide 27
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Maximum Flow Problems
• Defined on a network– Source node s– Sink node t– All other nodes are transshipment Nodes– Arcs have capacities, but no costs
• Maximize the total flow from s to t
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Example: Rerouting Airline Passengers
Due to a mechanical problem, Fly-By-Night Airlines had to cancel flight 162 which is its only non-stop flight from San Francisco to New York.
Formulate a maximum flow problem to reroute as many passengers as possible from San Francisco to New York.
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Data for Fly-by-Night Example
Flight From To Number of seats 160 San Francisco Denver 5 115 San Francisco Houston 6 153 Denver Atlanta 4 102 Denver Chicago 2 170 Houston Atlanta 5 150 Atlanta New York 7 180 Chicago New York 4
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Graphical Network Flow Formulation
b(j)b(i)
i j
lij = 0
arc (i,j)
(uij)
slide 31
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Network Representation
s t
SF
D C
H
2
6
A5
NY
5 4
4
7
slide 32
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LP Formulation
40
70,50,40
20,60,50
0
0
0
0
stmax
,
,,,
,,,
,,
,,,
,,
,,,
,,
,,
xxxxxxx
xxxxx
xxxxx
xxxx
NYC
NYAAHAD
CDHSFDSF
NYANYC
ADAHNYA
CDNYC
DSFADCD
HSFAH
HSFDSF
v
vv
slide 33
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Max Flow: Optimal Solution
v
vv
xxxxx
xxxxx
xxxx
NYANYC
ADAHNYA
CDNYC
DSFADCD
HSFAH
HSFDSF
,,
,,,
,,
,,,
,,
,,
0
0
0
0
stmax
9,2,7,2,2
,5,5,4
NYC,
NYA,D,CAD,
AH,HSF,DSF,
vxxxxxxx
slide 34
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MCNF Formulation of Maximum Flow Problems
1. Let arc cost = 0 for all arcs
2. Add an arc from t to s– Give this arc a cost of –1 and infinite
capacity
3. All nodes are transshipment nodes
4. Circulation Problem
slide 35
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Max Flow Formulation as MCNFP
SF
D C
H
(0,0,2)
(0,0,6)
A(0,0,5)
NY
(0,0,5) (0,0,4)
(0,0,4)
(0,0,7)
(-1,0,)
slide 36
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LP Formulation of MCNFP Representation
0
0
0
0
0
0st
min
,, ,
,,,
,,
,,,
,,
,,,
,
xxxxxx
xxxxx
xxxxx
x
NYASFNY NYC
ADAHNYA
CDNYC
DSFADCD
HSFAH
SFNYHSFDSF
SFNY
SFNY,NYC,
NYA,AH,AD,
D,CHSF,DSF,
0,40,70,50,40
,20,60,50
xxxxxxxx
slide 37
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MCNFP Solution
SF
D C
H
(0,0,2)
(0,0,6)
A(0,0,5)
NY
(0,0,5) (0,0,4)
(0,0,4)
(0,0,7)
9
4
22
5 75
2
(-1,0,)
slide 38
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LP Formulation of MCNFP Representation: Optimal Solution
0
0
0
0
0
0st
min
,, ,
,,,
,,
,,,
,,
,,,
,
xxxxxx
xxxxx
xxxxx
x
NYASFNY NYC
ADAHNYA
CDNYC
DSFADCD
HSFAH
SFNYHSFDSF
SFNY
9,2,7,2,2
,5,5,4
SFNY,NYC,
NYA,D,CAD,
AH,HSF,DSF,
xxxxxxxx
slide 39
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NSC Example
Month Demand Production Cost1 2,400 $7,4002 2,200 $7,5003 2,700 $7,6004 2,500 $7,800
• Max production per month = 4,000 tons• Inventory holding cost = $120/ton/month• Initial inventory = 1,000 tons• Final inventory = 1,500 tons
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Network Flow Formulation
D1
D2
D3
D4
P1
P2
P3
P4
4000
4000
4000
4000
-2400
-2200
-2700
-2500
I4 -1500
I0 1000
D0 -5700
I1
I2
I3
slide 41
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Network Flow Formulation: Arc Costs
D1
D2
D3
D4
P1
P2
P3
P4
I4
I0
D0
I1
I2
I3
slide 42
7400
7500
7600
7800
7520
7620
7720
7920
120
120
120
120
All other arc costs are 0
![Page 43: EMIS 8374 Network Flow Models updated 29 January 2008.](https://reader036.fdocuments.net/reader036/viewer/2022081504/56649f535503460f94c78839/html5/thumbnails/43.jpg)
Network Flow Solution
D1
D2
D3
D4
P1
P2
P3
P4
4000
4000
4000
4000
-2400
-2200
-2700
-2500
I4 -1500
I0 100
d0 -5700
I1
I2
I3
900
1800
25001500
1700
2300 900
2200
13002700
4000slide 43
lij = 0 and uij = for all arcs
2700