The Minimum Cost Network Flow (MCNF) Problem

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The Minimum Cost Network Flow (MCNF) Problem

• Extremely useful model in IEOR
• Important Special Cases of the MCNF Problem
• Transportation and Assignment Problems
• Maximum Flow Problem
• Minimum Cut Problem
• Shortest Path Problem
• Network Structure
• BFS for MCNF LPs always have integer values !!!
• Problems can be formulated graphically

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General Form of the MCNF Problem

• Defined on a network N (V,A)
• V is a set of vertices (nodes)
• Each node i has an assoicated value bi
• bi lt 0 gt node i is a demand node with a demand
  of -bi
• bi 0 gt node i is a transshipment node
• bi gt 0 gt node i is a supply node with a supply
  of bi
• A is a set of arcs
• 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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General Form of the MCNF Problem Continued

• A flow is feasible if
• Flow on all arcs is within the allowable bounds
• Flow is balanced (conserved)
• total flow going out of node i - total flow
  coming into node i bi
• We want to find a minimum cost feasible flow
• LP Formulation
• Let xij be the units of flow on arc (i,j)

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Example Medley Relay Team

• Supply Nodes 1, 2, 3 and 4 for John, Paul, George
  and Ringo
• Each has a supply of one swimmer, so bi 1
• Demand Nodes 5, 6, 7 and 8 for Freestyle,
  Butterfly,
• Backstroke and Breaststroke
• Each even needs a swimmer, so bi -1 for
  i6,7,8,9
• Arcs from each supply node to each demand node
• cij swimmer is time in event j
• uij 1, only one swimmer can be in any event
• lij 0

5
5.88
Freestyle
6.91
9.30
Butterfly
9,10
8.46
Backstroke
7.36
5.27
Breaststorke
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LP Formulation

• Let xij 1 if swimmer i swims event j and 0
  otherwise

MIN 5.88 X15 6.91 X16 9.1 X17 7.36 X18
9.3 X25 6.53 X26 7.62 X27 3.28 X28
8.46 X35 4.16 X36 2.62 X37 6.23 X38
5.27 X45 7.01 X46 2.47 X47 7.56 X48
SUBJECT TO 2) X15 X16 X17 X18
1 ! John swimmer 1 3) X25 X26
X27 X28 1 ! Paul swimmer 2 4)
X35 X36 X37 X38 1 ! George swimmer
3 5) X45 X46 X47 X48 1 !
Ringo swimmer 3 6) - X15 - X25 - X35 -
X45 - 1 ! Freestyle event 1 7) -
X16 - X26 - X36 - X46 - 1 ! Butterfly event
2 8) - X17 - X27 - X37 - X47 - 1 !
Backstroke event 3 9) - X18 - X28 -
X38 - X48 - 1 ! Breaststroke event 4
0 lt Xij lt 1 for all (i,j)
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LP Solution

• X15 X28 X36 X47 1 all other variables 0
• John swims Freestyle, Paul swims Breaststroke,
  George swims Butterfly and Ring swims Backstroke.
• The total time is 15.79 minutes.
• Observe that the LP solution is integer valued.

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Example 2 (From Bazarra and Jarvis)

• Transport 20,000,000 barrels of oil from Dhahran,
  Saudi Arabia to Rotterdam (4 Mil.), Marseilles
  (12 Mil) and Naples (4 Mil.) in Europe
• Routes
• Ship oil around Africa to
• Rotterdam 1.20/barrel
• Marseilles 1.40/barrel
• Naples 1.40/barrel
• Dhahran -gt Suez -gt Suez Canal -gt Port Said
• 0.30/barrel from Dhahran to Suez
• 0.20/barrel through Suez Canal
• Port Said to
• Rotterdam 0.25
• Marseilles 0.20
• Naples 0.15

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Example 2 Continued

• Dhahran to Suez then pipeline to Alexandria
• 0.15/barrel through pipeline
• Alexandria to
• Rotterdam 0.22
• Marseilles 0.20
• Naples 0.15
• 30 of oil in Dhahran shipped in large tankers
  that cant go through the Suez Canal
• Pipeline from Suez to Alexandria has a capacity
  of 10 million barrels of oil
• Formulate as a MCNF Problem

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Network Formulation
R
PS
M
S
D
A
N
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Network Formulation Continued

• Supply Node
• Dharhan has a supply of 20 M
• Demand Nodes
• Rotterdam, Marseilles and Naples have demands of
  4 M, 12 M and 4M, respectively
• Transshipment Nodes
• Suez, Alexandria and Port Said

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Arcs
Lower bound 0 for all arcs Upper bound
infinity for all arcs except (S,A) 10
million (S,PS) 14 million
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LP Formulation
minimize 1.2 XDR 1.4 XDM 1.4 XDN 0.3 XDS
0.2 XSPS 0.15 XSA 0.25 XPSR 0.2 XPSM
0.15 XPSN 0.25 XAR 0.2 XAM 0.15 XAN subject
to XDR XDM XDN XDS 20000000 ! D XSPS
XSA - XDS 0 ! S XPSR XPSM XPSN - XSPS
0 ! PS XAR XAM XAN - XSA 0 ! PA -XPSR -
XAR - XDR - 4000000 ! R -XPSM - XAM - XDM -
12000000 ! M -XPSN - XAN - XDN - 4000000 !
N XSPS lt 14000000 ! At most 70 through
canal XSA lt 10000000 ! Pipeline capacity end
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Solution
OBJECTIVE FUNCTION VALUE Cost 13,500,000
VARIABLE VALUE XDS 20,000,000 XSPS
10,000,000 XSA 10,000,000 XPSR
4,000,000 XPSM 6,000,000 XAM
6,000,000 XAN 4,000,000