148 Views

Download Presentation
##### EMIS 8374

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -

**EMIS 8374**Network Flow Models updated 29 January 2008**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**Elements of the MCNFP**• Defined on a network G = (N,A) • N is a set of nnodes: {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**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)**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**Graphical Network Flow Formulation**(cij, lij, uij) i j arc (i,j) b(j) b(i)**Example MCNFP**-2 (3, 2, 5) (4, 1,3) 2 5 -3 1 4 (1, 0, 2) (2, 0,2) (4, 0, 3) 3 0**Requirements for a Feasible Flow**• Flow on all arcs is within the allowable bounds: lij xijuijfor 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**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**Example Feasible Solution**-2 (3, 2,5) (4, 1,3) 2 5 3 5 -3 1 4 (1, 0,2) 0 0 0 (4, 0,3) (2, 0,2) 3 Cost = 15 + 12 = 27 0 Arc flows shown in blue.**Optimal Solution for Example MCNFP**-2 (3, 2,5) (4, 1,3) 2 3 1 5 -3 1 4 (1, 0,2) 0 2 2 (4, 0,3) (2, 0,2) 3 Cost = 25 0 Arc flows shown in blue.**Graphical Network Flow Formulation**(cij, uij) i j arc (i,j) b(j) b(i) lij= 0 for all arcs.**-1**A -1 C -1 W -1 Supply Nodes Demand Nodes +4 I F +1 +2 G D S -3 Dummy Node**A**C W Supply Nodes Demand Nodes (13, 1) +4 I -1 (35, 1) +1 F -1 (0,1) (42, 1) +2 G -1 (0,4) (0,2) (9, 1) -3 D S -1 Dummy Node**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**Shortest Path Example 1**5 10 1 2 3 s t 7 1 7 4 1,(1,2),2,(2,3),3 Length = 15 1,(1,2),2,(2,4),4,(4,3) Length = 13 1,(1,4),4,(4,3),3 Length = 14**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**Shortest Path as MCNFP: Graphical Formulation**0 1 2 3 (5,0,1) (10,0,1) 1 -1 (7,0,1) (7,0,1) (1,0,1) 4 0**Shortest Path as MCNFP: Graphical Solution**0 1 2 3 (5,0,1) (10,0,1) 1 -1 1 0 (7,0,1) (7,0,1) 1 (1,0,1) 1 0 4 0 Arc flows shown in blue.**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?**Graphical Network Flow Formulation**(cij) i j arc (i,j) b(j) b(i) lij = 0, uij=1**Formulation as Shortest Path**0 0 9 2 4 4 8 s t 5 1 6 4 3 10 1 6 5 -1 2 3 5 0 0**“Greedy” Solution**0 0 9 2 4 4 8 s t 5 1 6 4 3 10 1 6 5 -1 2 3 5 0 0 x13 = x23 = x35 = x35 = 1, xij = 0 for all other arcs. Objective function value = 19.**Shortest Path: Optimal Solution**0 0 9 2 4 4 8 s t 5 1 6 4 3 10 1 6 5 -1 2 3 5 0 0 x13 = x35 = x56 = 1, xij = 0 for all other arcs. Objective function value = 17.**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**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.**Graphical Network Flow Formulation**(uij) i j arc (i,j) b(j) b(i) lij = 0**Network Representation**2 D C 4 5 s t SF NY 4 6 7 5 H A**MCNF Formulation of Maximum Flow Problems**• Let arc cost = 0 for all arcs • Add an arc from t to s • Give this arc a cost of –1 and infinite capacity • All nodes are transshipment nodes • Circulation Problem**Max Flow Formulation as MCNFP**(0,0,2) D C (0,0,4) (0,0,5) SF NY (0,0,4) (0,0,7) (0,0,6) (0,0,5) H A (-1,0,)**MCNFP Solution**(0,0,2) D C (0,0,4) (0,0,5) 2 2 4 SF NY (0,0,4) 2 (0,0,7) (0,0,6) 5 (0,0,5) H A 7 5 (-1,0,) 9**NSC Example**• Max production per month = 4,000 tons • Inventory holding cost = $120/ton/month • Initial inventory = 1,000 tons • Final inventory = 1,500 tons**P1**D1 4000 -2400 P2 D2 4000 -2200 P3 D3 4000 -2700 P4 D4 4000 -2500 D0 I1 I0 I3 I4 I2 1000 -5700 -1500 Network Flow Formulation**P1**D1 P2 D2 P3 D3 P4 D4 I2 I3 I1 Network Flow Formulation: Arc Costs All other arc costs are 0 I0 120 7400 7520 120 7500 7620 120 7600 7720 120 7800 7920 I4 D0**P1**D1 4000 -2400 P2 D2 4000 -2200 P3 D3 4000 -2700 P4 D4 4000 -2500 I1 I2 I4 d0 I3 -5700 -1500 Network Flow Solution lij = 0 and uij = for all arcs I0 100 900 2300 1700 900 2200 1800 2700 2700 1300 2500 1500 4000