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NetworkNetwork
ModelsModels
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, dij
– Find the path of minimum total distance that connects node 1 to node n.
2
FAIRWAY VAN LINES
Determine the shortest route from Seattle to El Paso over the following network of highways.
3
4
8
9
7
11
12
1
43
6
2
10
5
Salt Lake City
El Paso
Seattle
Boise
Portland
Butte
Cheyenne
Bakersfield Las Vegas
Albuquerque
Tucson
Phoenix
599
691497180
432 345440
554621
420
280
432
403
314
893
500
290
116
268
577
CommentIn case some arcs are bi-directional, create two directed arcs in two opposite directions, betweenthe same two nodes.
FAIRWAY VAN LINES – The Linear Programming Model
Decision variables
5
X ij
10 if a truck travels on the highway from city i to city j otherwise
Objective = Minimize dijXij
6
6
2
Salt Lake City
1
3 4
Seattle
Boise
Portland
599
497180
432 345
Butte
[The number of highways traveled out of Seattle (the start node)] = 1X12 + X13 + X14 = 1
In a similar manner:[The number of highways traveled into El Paso (terminal node)] = 1X9,12 + X10,12 + X11,12 = 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):X14 + X34 = X46.
Subject to the following constraints:
Non-negativity constraints
FAIRWAY VAN LINES – spreadsheet
7
SOLUTIONTOTAL
DISTANCE= 1731
NODE NAME NODE # FROM TO DISTANCE FROM TO FLOW
Seattle 1 1 2 599 1 2
Butte 2 1 3 180 1 3
Portland 3 1 4 497 1 4 1
Boise 4 2 5 691 2 5
Cheyenne 5 2 6 420 2 6
Salt Lake City 6 3 4 432 3 4
Bakersfield 7 3 7 893 3 7
Las Vegas 8 4 6 345 4 6 1
Albuquerque 9 5 6 440 5 6
Phoenix 10 5 9 554 5 9
Tucson 11 6 8 432 6 8
El Paso 12 6 9 621 6 9 1
7 8 280 7 8
7 10 500 7 10
8 9 577 8 9
8 10 290 8 10
9 12 268 9 12 1
10 11 116 10 11
10 12 403 10 12
11 12 314 11 12
NODE INPUT ARC INPUT
FAIRWAY VAN LINES – The Network Model
The Dijkstra’s 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
An illustration of the Dijkstra’s algorithm
9
(See mathematical formulation of the algorithm in Supplement CD 5).
1056
8
9
7
11
12
1
43
6
2
10
5
Salt Lake City
El Paso
Seattle
Boise
Portland
Butte
Cheyenne
Bakersfield Las Vegas
Albuquerque
Tucson
Phoenix
599
691497180
432 345440
554621
420
280
432
403
314
893
500
290
116
268
577
SEAT.SEAT.
BUT599
POR
180
497BOI
599
180
497POR.POR.
BOI432
Bakersville602
+
+
=
=
612
782
BOI
BOIBOISE.BOISE.
345SLC
+ =
842
BUTTEBUTTE
SLC
420
CHY.691
+
+
=
=
1119
1290
SLC.
SLCSLC.
BAKERSVILLEBAKERSVILLE
… and so on until the whole network is covered.
Dijkstra’s 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
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 Cij for flow on the arc from node i to node j, and capacities Cji for the opposite direction.
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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
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.
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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
UNITED CHEMICAL COMPANY- Network Presentation
16
• Data
17
1 7
2
4
6
5
3
Chemical Drum Safe Tub
10
8
10
61
12
1 4
4 2
2 8
3
3
7
2
Maximum flow from 2 to 4 is 8
Maximum flow from 6 to 3 is 4
UNITED CHEMICAL COMPANY – The Linear Programming Model
• Decision variablesXij - 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 X12 + X13
18
UNITED CHEMICAL COMPANY – The Linear Programming Model
• Constraints– The constraint on each intermediate node is:
Flow out from the node – flow into the node = 0Node 2: X23 +X24 + X26 - X12 - X32 = 0Node 3: X32 +X35 + X36 - X13 - X23 - X63 = 0Node 4: X46 + X47 - X24 - X64 = 0Node 5: X56 + X57 - X35 - X65 = 0Node 6: X63 +X64 +X65 + X67 - X26 - X36 - X46 -X56 = 0
1970
• Data
1 7
2
4
6
5
3
Chemical Drum Safe Tub
10
8
10
61
12
1 4
4 2
2 8
3
3
7
2
Maximum flow from 2 to 4 is 8
Maximum flow from 6 to 3 is 4
2
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 Xij 0
20
UNITED CHEMICAL COMPANY – Spreadsheet
21
SOLUTIONMAXIMUM
FLOW= 17
NODE NAME NODE # FROM TO CAPACITY FROM TO FLOW
SOURCE Chemical Drum 1 1 2 10 1 2 9
SINK Safe Tub 7 1 3 10 1 3 8
Area 2 2 2 3 1 2 3
Area 3 3 2 4 8 2 4 7
Area 4 4 2 6 6 2 6 2
Area 5 5 3 2 1 3 2
Area 6 6 3 5 12 3 5 8
3 6 4 3 6
4 6 3 4 6
4 7 7 4 7 7
5 6 2 5 6
5 7 8 5 7 8
6 3 4 6 3
6 5 2 6 5
6 7 2 6 7 2
ARC INPUTNODE INPUT
8
2
7
9
8
7
8
1 7
42
53
6
2
100/17 = 5,88 minutos
para retirar todo o produto
100.000 galões
17.000 galões/minuto
The role of cuts in a maximum flow networkThe value of the maximum flow = the sum of the capacities of the minimum cut.
22
1
2
3
10
10
8
4 8
2
7
7
5
6
4
3
3
2
12
6
2
This is not a minimal cut
This is a minimal cut,and the flow = 17 is maximal
4
1
1Rede 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
The Traveling Salesman Problem
23
• Problem definition
– There are m nodes.– Unit cost Cij is associated with utilizing arc (i,j)– Find the cycle that minimizes the total cost required to
visit all the nodes exactly once.
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
The Traveling Salesman Problem
25
• ComplexityBoth 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 6.000.000.000.000.000.000.000.000.000.000.000.000 restrições lineares !!
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
To office2 3 4 Home
F Office 1 25 50 50 30r Office 2 40 40 45o Office 3 35 65m Office 4 80
To office2 3 4 Home
F Office 1 25 50 50 30r Office 2 40 40 45o Office 3 35 65m Office 4 80
FEMA traveling salesman Network representation
27
28
30
25
40
35
80
6545
50
50
40
Home
1
2 3
4
Tempo de viagem
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.
FEMA – full enumeration
Possible cyclesCycle Total Cost
1. H-O1-O2-O3-O4-H 210 2. H-O1-O2-O4-O3-H 195 3. H-O1-O3-O2-O3-H 240 4. H-O1-O3-O4-O2-H 200 5. H-O1-O4-O2-O3-H 225 6. H-O1-O4-O3-O2-H 200 7. H-O2-O3-O1-O4-H 265 8. H-O2-O1-O3-O4-H 235 9. H-O2-O4-O1-O3-H 25010. H-O2-O1-O4-O3-H 22011. H-O3-O1-O2-O4-H 26012. H-O3-O1-O2-O4-H 260
30
Minimum
For this problem we have
(5-1)! / 2 = 12 cycles. Symmetrical problemsneed to enumerate only (m-1)! / 2 cycles.
FEMA – optimal solution
31
30
25
40
35
806545
5050
40
Home
1
2 3
4
Tempo total ótimo = 195 minutos
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.
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Variáveis de Decisão: Xij = número de vezes que o arco do nó i ao nó j é usado no ciclo. Assim,
Xij = 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.
FEMA – The Assignment problem approach
• Data
33
To office1 2 3 4 Home
F Office 1 1000000 25 50 50 30r Office 2 25 1000000 40 40 45o Office 3 50 40 1000000 35 65m Office 4 50 40 35 1000000 80
Home 30 45 65 80 1000000
To office1 2 3 4 Home
F Office 1 1000000 25 50 50 30r Office 2 25 1000000 40 40 45o Office 3 50 40 1000000 35 65m Office 4 50 40 35 1000000 80
Home 30 45 65 80 1000000
FEMA – The assignment solution
34
30
25
40
35
806545
5050
40
Home
1
2 3
4
O3 – O4 – O3Sub-tour
O1 – O2 – H – O1Sub-tour
FEMA – The assignment solution• We can prevent the situation of sub-tours by adding
certain constraints.
Exemplo: Seja o nó 5 = Home (H), assimX52 + X21 + X15 2 evita que o sub-tour H-O2-O1-H seja
utilizado.X12 + X23 + X34 + X41 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
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
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
METROPOLITAN TRANSIT – Network presentation
• The network describes:– feasible lines that have been drafted– minimum possible cost for taxpayers per line.
38
39
5
2 6
4
7
81
3
West Side
North Side University
BusinessDistrict
East SideShoppingCenter
South Side
City Center
33
50
30
55
34
28
32
35
39
45
38
43
44
41
3736
40
NETWORK PRESENTATION
40
5
2 6
4
7
81
3
West Side
North Side University
BusinessDistrict
East SideShoppingCenter
South Side
City Center
33
50
30
55
34
28
32
35
39
45
38
43
44
41
3736
40
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
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
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ShoppingCenter
Loop
2 6
4
7
8
West Side
North Side
University
BusinessDistrict
East Side
South Side
City Center
50
30
55
34
28
32
35
39
45
38
43
44
41
3736
40
Continuar até conectar todos os nós
Aplicação do Alg 2
5
LoopLoopLoopLoopLoopL
oop
LoopLo
op LoopLoop Loop LoopLoopLoopLoop
1
3
33
28
32
30
33
44
ShoppingCenter
4
7
8
West Side
North Side
University
BusinessDistrict
East Side
South Side
City Center
30
28
32
35
38
3736Total Cost = $236 million
OPTIMAL SOLUTION
53
1
62
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