The Minimum Cost Network Flow Problem Problem Instance: Given a - PowerPoint PPT Presentation

EMIS 8374 [MCNFP Review] 1 The Minimum Cost Network Flow Problem Problem Instance: Given a network G = ( N, A ), with a cost c ij , upper bound u ij , and lower bound ℓ ij associated with each directed arc ( i, j ), and supply of, or demand for, b ( i ) units of some commodity at each node. Supply Nodes: b ( i ) > 0 Transshipment Nodes: b ( i ) = 0 Demand Nodes: b ( i ) < 0

EMIS 8374 [MCNFP Review] 2 Linear Programming Formulation: The problem is that of finding a minimum-cost feasible flow : min � c ij x ij ( i,j ) ∈ A subject to x ij − x ji = b ( i ) ∀ i ∈ N, � � { j :( i,j ) ∈ A } { j :( j,i ) ∈ A } ℓ ij ≤ x ij ≤ u ij ∀ ( i, j ) ∈ A Note the problem is not feasible unless the supplies and demands are balanced (i.e., i ∈ N b ( i ) = 0). �

EMIS 8374 [MCNFP Review] 3 The Circulation Problem: A MCNFP where b ( i ) = 0 for all nodes. The Transportation Problem: A MCNFP where • A bipartite network where N = N 1 ∪ N 2 ( N 1 ∩ N 2 = ∅ ) • N 1 are supply nodes and N 2 are demand nodes • i ∈ N 1 and j ∈ N 2 for all ( i, j ) ∈ A The Assignment Problem: A Transportation Problem where • | N 1 | = | N 2 | • b ( i ) = +1 for all i ∈ N 1 and b ( j ) = − 1 for all j ∈ N 2

EMIS 8374 [MCNFP Review] 4 The Shortest Path Problem: defined on a network with arc costs, but no capacity limits. The objective is to find a path from node s , the source , to node t , the sink , that minimizes the sum of the arc costs along the path. To formulate as MCNFP: • s is a suppy node with b ( s ) = 1. • t is a demand node with b ( t ) = − 1. • All other nodes ( N \{ s, t } ) are transshipment nodes. • ℓ ij = 0 and u ij = 1 for all arcs.

EMIS 8374 [MCNFP Review] 5 LP Formulation of the Shortest Path Problem min � c ij x ij ( i,j ) ∈ A x sj − subject to � � x js = 1 , { j :( s,j ) ∈ A } { j :( j,s ) ∈ A } x ij − x ji = 0 ∀ i ∈ N \ { s, t } , � � { j :( i,j ) ∈ A } { j :( j,i ) ∈ A } x tj − x jt = − 1 , � � { j :( t,j ) ∈ A } { j :( j,t ) ∈ A } 0 ≤ x ij ≤ 1 ∀ ( i, j ) ∈ A.

EMIS 8374 [MCNFP Review] 6 The Maximum Flow Problem: defined on a directed network with capacities, and no costs. In addition two nodes are specified, a source node, s , and sink node, t . The objective is the find the maximum possible flow between the source and sink while satisfying the arc capacities.

EMIS 8374 [MCNFP Review] 7 LP Formulation of the Maximum Flow Problem max v x sj − subject to � � x js = v, { j :( s,j ) ∈ A } { j :( j,s ) ∈ A } x ij − x ji = 0 ∀ i ∈ N \ { s, t } , � � { j :( i,j ) ∈ A } { j :( j,i ) ∈ A } x tj − x jt = − v, � � { j :( t,j ) ∈ A } { j :( j,t ) ∈ A } ℓ ij ≤ x ij ≤ u ij ∀ ( i, j ) ∈ A. The scalar variable v is referred to as the value of the flow vector x = { x ij } .

EMIS 8374 [MCNFP Review] 8 The Max Flow Problem Formulated as MCNFP Convert the problem to an equivalent minimum cost circulation problem as follows: • Let c ij = 0 for all ( i, j ) ∈ A . • Let b ( i ) = 0 for all i ∈ N . • Add an arc from s to t with cost c st = − 1. min − x ts s.t. x ij − x ji = 0 ∀ i ∈ N, � � { j :( i,j ) ∈ A } { j :( j,i ) ∈ A } ℓ ij ≤ x ij ≤ u ij ∀ ( i, j ) ∈ A.

EMIS 8374 [MCNFP Review] 9 The Texas Confectionery Company (TCC) produces three types of candy bars at two different plants. • Houston plant produces Rice Krunchy and Aggie bars. • Austin plant produces Aggie Bars and Longhorn Bars. • There are 160 hours of production time available per month at each plant.

EMIS 8374 [MCNFP Review] 10 Product Production Cost Product Time Demand Name Houston Austin (Minutes) (Units) Rice Krunchy Bar $0.04 - 30 200 Aggie Bar $0.05 $0.06 20 280 Longhorn Bar - $0.06 15 320 (a) Formulate a MCNFP that TCC can solve to determine how to minimize the cost of meeting the demand for its products. Briefly describe what the elements (nodes, arcs, capacities, etc.) of your network model represent. Hint: How many hours of production time are required to meet the demand for each type of candy bar?

EMIS 8374 [MCNFP Review] 11 (a) Nodes: H and A are supply nodes representing the Houston and Austin plants, respectively. These nodes each have a supply of 9600 minutes of production time. RK , AB and LB are demand nodes representing the demand for Rice Krunchy, Aggie and Longhorn Bars, respectively. The demand at RK is 6000 minutes (30 minutes times 200 units), the demands at AB and LB are 5600 and 4800 minutes, respectively. D is a dummy node to absorb the excess supply of minutes.

EMIS 8374 [MCNFP Review] 12 Arcs: The arc costs for ( H, RK ), ( H, AB ), ( A, AB ) and ( A, LB ) are $0 . 04 30 , $0 . 05 20 , $0 . 06 and $0 . 06 15 , respectively. 20 These arcs are uncapacitated. A unit of flow on arc ( i, j ) represents one minute of production time at plant i used for making bar j . The arcs from the supply nodes to the dummy node have zero cost, zero lower bound and an upper bound of 9600.

EMIS 8374 [MCNFP Review] 13 Network for TCC Problem 0.0013 -6000 H RK 9600 0.0025 D -2800 -5600 AB 0.003 0.004 -4800 A LB 9600

EMIS 8374 [MCNFP Review] 14 (b) Suppose that TCC wants to plan two months in advance and believes that it will need 240 Rice Krunchy Bars, 360 Aggie Bars and 480 Longhorn bars at the end of next month. If there is time left over in the current month (after the production for this month’s demand is finished), then some of these bars can be manufactured this month and held in inventory until they are needed. Extend your MCNFP from part (a) to minimize TCC’s cost for meeting the demand for its products over the next two months.

EMIS 8374 [MCNFP Review] 15 (b) Rename nodes RK , AB and LB to RK 1, AB 1 and LB 1 (demand in month 1) and add nodes RK 2, AB 2 and LB 2 to represent the demand in month 2. Add nodes RKI ABI and LBI to represent the surplus production of Rice Krunchy, Aggie and Longhorn bars from month 1 that is held inventory. Rename nodes H and A to H 1 and A 1, respectively, and add nodes H 2 and A 2 to represent production in month two.

EMIS 8374 [MCNFP Review] 16 -7200 9600 9600 RK2 RKI -6000 H2 H1 RK1 -7200 D -400 AB2 -5600 ABI AB1 -7200 -4800 A2 LBI A1 LB1 LB2 9600 9600

EMIS 8374 [MCNFP Review] 17 Arc Cost Arc Cost Arc Cost $0 . 04 $0 . 04 $0 . 04 (H1, RK1) (H1, RKI) (H2, RK2) 30 30 30 $0 . 05 $0 . 05 $0 . 05 (H1, AB1) (H1, ABI) (H2, AB2) 20 20 20 $0 . 06 $0 . 06 $0 . 06 (A1, AB1) (A1, ABI) (A2, AB2) 20 20 20 $0 . 06 $0 . 06 $0 . 06 (A1, LB1) (A1, LBI) (A2, LB2) 15 15 15 All arcs have zero lower bound and can have an infinite upper bound. Arcs not listed above have zero cost. Note that the network can be simplified as shown in the next figure.

EMIS 8374 [MCNFP Review] 18 -7200 9600 9600 RK2 -6000 H2 H1 RK1 -7200 D -400 AB2 -5600 AB1 -7200 -4800 A2 A1 LB1 LB2 9600 9600

EMIS 8374 [MCNFP Review] 19 (c) Candy bars that are stored in inventory must be kept in a special storage facility so that they do not become stale. How would you extend your MCNFP from part (b) to account for inventory holding costs of $0.02 for each Rice Krunchy Bar, $0.01 for each Aggie bar and $0.05 for each Longhorn bar?

EMIS 8374 [MCNFP Review] 20 (c) Add the storage cost to the cost on the arcs from H 1 and A 2 to RK 2, AB 2, and LB 2 Arc Cost Arc Cost $0 . 04+$0 . 02 $0 . 05+$0 . 01 (H1, RK2) (H1, AB2) 30 20 $0 . 06+$0 . 01 $0 . 06+$0 . 05 (A1, AB2) (A1, LB2) 20 15