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The Minimum Cost Network Flow (MCNF) Problem

Learn about MCNF modeling in IEOR, special cases, transportation, assignment, max flow, min cut problems, LP formulations with examples.

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The Minimum Cost Network Flow (MCNF) Problem

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  1. 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

  2. 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 < 0 => node i is a demand node with a demand of -bi • bi = 0 => node i is a transshipment node • bi > 0 => 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)

  3. 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)

  4. 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 i=6,7,8,9 • Arcs from each supply node to each demand node • cij = swimmer i’s time in event j • uij = 1, only one swimmer can be in any event • lij = 0

  5. John -1 1 Paul -1 1 George -1 1 Ringo 1 -1 5.88 Freestyle 6.91 9.30 Butterfly 9,10 8.46 Backstroke 7.36 5.27 Breaststorke

  6. 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 <= Xij <= 1 for all (i,j)

  7. 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.

  8. 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 -> Suez -> Suez Canal -> 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

  9. 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 can’t go through the Suez Canal • Pipeline from Suez to Alexandria has a capacity of 10 million barrels of oil • Formulate as a MCNF Problem

  10. Network Formulation R PS M S D A N

  11. 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

  12. Arcs Lower bound = 0 for all arcs Upper bound = infinity for all arcs except (S,A) = 10 million (S,PS) = 14 million

  13. 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 <= 14000000 ! At most 70% through canal XSA <= 10000000 ! Pipeline capacity end

  14. 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

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