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1 Network Models

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The Shortest Path Model For a given network find the path of minimum distance, time, or cost from a starting point, the start node, to a destination, the terminal node. Problem definition – There are n nodes, beginning with start node 1 and ending with terminal node n. – Bi-directional arcs connect connected nodes i and j with nonnegative distances, d ij – Find the path of minimum total distance that connects node 1 to node n. 2

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FAIRWAY VAN LINES Determine the shortest route from Seattle to El Paso over the following network of highways. 3

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Salt Lake City El Paso Seattle Boise Portland Butte Cheyenne BakersfieldLas Vegas Albuquerque Tucson Phoenix Comment In case some arcs are bi-directional, create two directed arcs in two opposite directions, between the same two nodes.

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FAIRWAY VAN LINES – The Linear Programming Model Decision variables 5 Objective = Minimize d ij X ij

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6 6 2 Salt Lake City 1 34 Seattle Boise Portland Butte [The number of highways traveled out of Seattle (the start node)] = 1 X 12 + X 13 + X 14 = 1 In a similar manner: [The number of highways traveled into El Paso (terminal node)] = 1 X 9,12 + X 10,12 + X 11,12 = 1 [The number of highways used to travel into a city] = [The number of highways traveled leaving the city]. For example, in Boise (node 4): X 14 + X 34 = X 46. Subject to the following constraints: Non-negativity constraints

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FAIRWAY VAN LINES – spreadsheet 7

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FAIRWAY VAN LINES – The Network Model The Dijkstras algorithm: – Find the shortest distance from the START to each other node, in the order of the closest nodes to the START. – Once the shortest route to the m closest node is determined, the (m+1) closest can be easily determined. – This algorithm finds the shortest route from the start to all the nodes in the network. 8

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An illustration of the Dijkstras algorithm 9 (See mathematical formulation of the algorithm in Supplement CD 5).

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10 SEAT. BUT 599 POR BOI POR. BOI 432 Bakersville = = BOI BOISE. 345 SLC + = 842 BUTTE SLC 420 CHY = = SLC. SLC SLC. BAKERSVILLE … and so on until the whole network is covered.

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Dijkstras algorithm - continued When all the network is covered, the shortest route from START to every other node can be identified. Trace the path that leads to each node by backtracking from each node toward node START. 11

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The Maximal Flow Problem Problem definition – There is a source node (labeled 1), from which the network flow emanates. – There is a terminal node (labeled n), into which all network flow is eventually deposited. – There are n - 2 intermediate nodes (labeled 2, 3,…,n-1), where the node inflow is equal to the node outflow. – There are capacities C ij for flow on the arc from node i to node j, and capacities C ji for the opposite direction. 12

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The Maximal Flow Problem Objective The objective is to find the maximum total flow out of node 1 that can flow into node n without exceeding the capacities on the arcs. 13

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UNITED CHEMICAL COMPANY United Chemical produces pesticides and lawn care products. Poisonous chemicals needed for the production process are held in a huge drum. A network of pipes and valves regulates the chemical flow from the drum to different production areas. The safety division must plan a procedure to empty the drum as fast as possible into a safety tub in the disposal area, using a network of pipes and valves. 14

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UNITED CHEMICAL COMPANY 15 The plan must determine: which valves to open and shut What is the estimated time for total discharge

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UNITED CHEMICAL COMPANY- Network Presentation 16

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Data Chemical Drum Safe Tub Maximum flow from 2 to 4 is 8 Maximum flow from 6 to 3 is 4

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UNITED CHEMICAL COMPANY – The Linear Programming Model Decision variables X ij - the flow from node i to node j on the arc that connects these two nodes Objective function – Maximize the flow out of node 1: Max X 12 + X 13 18

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UNITED CHEMICAL COMPANY – The Linear Programming Model Constraints – The constraint on each intermediate node is: Flow out from the node – flow into the node = 0 Node 2: X23 +X24 + X26 - X12 - X32= 0 Node 3: X32 +X35 + X36 - X13 - X23 - X63 = 0 Node 4: X46 + X47 - X24 - X64 = 0 Node 5: X56 + X57 - X35 - X65 = 0 Node 6: X63 +X64 +X65 + X67 - X26 - X36 - X46 -X56=

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UNITED CHEMICAL COMPANY – The Linear Programming Model Constraints – continued – Flow cannot exceed arc capacities X12 10; X13 10; X23 1; X24 8; X26 6; X32 1; X35 15; X36 4; X46 3; X47 7; X56 2; X57 8; X63 4; X64 3; X65 2; X67 2; – All X ij 0 20

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UNITED CHEMICAL COMPANY – Spreadsheet /17 = 5,88 minutos para retirar todo o produto galões galões/minuto

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The role of cuts in a maximum flow network The value of the maximum flow = the sum of the capacities of the minimum cut This is not a minimal cut This is a minimal cut, and the flow = 17 is maximal Rede com as capacidades nos arcos Corte: linha sobre os arcos da rede que separa nó fonte do nó destino Capacidade do Corte: soma das capacidades dos arcos cortados pela linha do corte Cap = 30 Cap = 17

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The Traveling Salesman Problem 23 Problem definition –There are m nodes. –Unit cost C ij is associated with utilizing arc (i,j) –Find the cycle that minimizes the total cost required to visit all the nodes exactly once.

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The Traveling Salesman Problem Importance: – Variety of scheduling application can be solved as a traveling salesmen problem. – Examples : Ordering drill position on a drill press. School bus routing. Military bombing sorties. – The problem has theoretical importance because it represents a class of difficult problems known as NP-hard problems. 24

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The Traveling Salesman Problem 25 Complexity Both writing the mathematical model and solving it are cumbersome (a problem with 20 cities requires over 500,000 linear constraints.) 50 cidades requerem 500 trilhões de restrições lineares 120 cidades requerem restrições lineares !!

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THE FEDERAL EMERGENCY MANAGEMENT AGENCY A visit must be made to four local offices of FEMA, going out from and returning to the same main office in Northridge, Southern California. Data (simétrico) Travel time (min) between offices 26

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FEMA traveling salesman Network representation 27

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Home Tempo de viagem

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FEMA - Traveling Salesman 29 Solution approaches –Enumeration of all possible cycles. This results in (m-1)! cycles to enumerate. Se a rede for simétrica serão (m – 1)!/2 ciclos. Only small problems can be solved with this approach. –A combination of the Assignment problem and the Branch and Bound technique. Problem with up to m=20 nodes can be efficiently solved with this approach.

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FEMA – full enumeration Possible cycles CycleTotal Cost 1. H-O1-O2-O3-O4-H H-O1-O2-O4-O3-H H-O1-O3-O2-O3-H H-O1-O3-O4-O2-H H-O1-O4-O2-O3-H H-O1-O4-O3-O2-H H-O2-O3-O1-O4-H H-O2-O1-O3-O4-H H-O2-O4-O1-O3-H H-O2-O1-O4-O3-H H-O3-O1-O2-O4-H H-O3-O1-O2-O4-H Minimum For this problem we have (5-1)! / 2 = 12 cycles. Symmetrical problems need to enumerate only (m-1)! / 2 cycles.

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FEMA – optimal solution Home Tempo total ótimo = 195 minutos

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FEMA – The Assignment problem approach Refer to the nodes designated From as Workers nodes. Refer to nodes designated by To as Jobs nodes. Assign Workers to Jobs at minimum cost. 32 Variáveis de Decisão: X ij = número de vezes que o arco do nó i ao nó j é usado no ciclo. Assim, X ij = 1 (se o arco está no ciclo) ou 0 (arco não está no ciclo). Restrições no Modelo da Designação: soma dos arcos que saem de um nó tipo Worker = 1; soma dos arcos que chegam em em nó tipo Job = 1.

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FEMA – The Assignment problem approach Data 33

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FEMA – The assignment solution Home O3 – O4 – O3 Sub-tour O1 – O2 – H – O1 Sub-tour

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FEMA – The assignment solution We can prevent the situation of sub-tours by adding certain constraints. Exemplo: Seja o nó 5 = Home (H), assim X 52 + X 21 + X 15 2 evita que o sub-tour H-O2-O1-H seja utilizado. X 12 + X 23 + X 34 + X 41 3 evita que o sub-tour O1-O2-O3-O4- O1 seja utilizado. This makes the problem extremely large. Another approach, that combines the Assignment model with the Branch and Bound solution methodology can efficiently solve problems up to m = 20 nodes. 35

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The Minimal Spanning Tree This problem arises when all the nodes of a given network must be connected to one another, without any loop. The minimal spanning tree approach is appropriate for problems for which redundancy is expensive, or the flow along the arcs is considered instantaneous. 36

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THE METROPOLITAN TRANSIT DISTRICT The City of Vancouver is planning the development of a new light rail transportation system. The system should link 8 residential and commercial centers. The Metropolitan Transit District needs to select the set of lines that will connect all the centers at a minimum total cost. 37

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METROPOLITAN TRANSIT – Network presentation The network describes: – feasible lines that have been drafted – minimum possible cost for taxpayers per line. 38

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West Side North Side University Business District East Side Shopping Center South Side City Center

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NETWORK PRESENTATION West Side North Side University Business District East Side Shopping Center South Side City Center

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THE METROPOLITAN TRANSIT DISTRICT Solution - a network approach (See Supplement CD 5) The algorithm that solves this problem is a very trivial greedy procedure. Two versions of the algorithm are described. The algorithm – version 1: – Start by selecting the smallest arc, and adding it to a set of selected arcs (currently contains only the first arc). – At each iteration, add the next smallest arc to the set of selected arcs, unless it forms a cycle. – Finish when all nodes are connected. 41

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THE METROPOLITAN TRANSIT DISTRICT The algorithm – version 2: – Start by selecting the smallest arc creating the set of connected arcs. – At each iteration add the smallest unselected arc that has a connection to the connected set, but do not create a cycle. – Finish when all nodes are connected See demonstration of version 2 next 42

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43 Shopping Center Loop West Side North Side University Business District East Side South Side City Center Continuar até conectar todos os nós Aplicação do Alg 2 5 Loop

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44 Shopping Center West Side North Side University Business District East Side South Side City Center Total Cost = $236 million OPTIMAL SOLUTION

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