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The Minimum Cost Network Flow (MCNF) Problem

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 !!!

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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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