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Network Models (2). Tran Van Hoai Faculty of Computer Science & Engineering HCMC University of Technology. Ballston Electric Assembly line - Inspection. Products transported from Assembly line to Inspection are for Quality Control. - PowerPoint PPT Presentation
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Tran Van Hoai 1
Network Models (2)
Tran Van HoaiFaculty of Computer Science & Engineering
HCMC University of Technology
2010-2011
Tran Van Hoai 2
Products transported from Assembly line to
Inspection are for Quality Control
Ballston ElectricAssembly line - Inspection
2010-2011
Which assembly line should be assigned to which inspection area (to minimize a given objective)?
Tran Van Hoai 3
Time to transport a unit of product from assembly line to inspection
Inspection AreaA B C D E
AssemblyLine
1 10 4 6 10 122 11 7 7 9 143 13 8 8 14 154 14 16 16 17 175 19 11 17 20 19
2010-2011
Time difference mainly comes from the distance difference among pair <assembly line-inspection area>
Tran Van Hoai 4
MIN 10X11 + 4X12 + … + 20X54 + 19X55
S.T.
X11 + X12 + X13 + X14 + X15 = 1 (Assemble line 1 is assigned)
X21 + X22 + X23 + X24 + X25 = 1 (Assemble line 2 is assigned)
X31 + X32 + X33 + X34 + X35 = 1 (Assemble line 3 is assigned)
X41 + X42 + X43 + X44 + X45 = 1 (Assemble line 4 is assigned)
X51 + X52 + X53 + X54 + X55 = 1 (Assemble line 5 is assigned)
X11 + X21 + X31 + X41 + X51 = 1 (Inspection area 1 is assigned)
X12 + X22 + X32 + X42 + X52 = 1 (Inspection area 2 is assigned)
X13 + X23 + X33 + X43 + X53 = 1 (Inspection area 3 is assigned)
X14 + X24 + X34 + X44 + X54 = 1 (Inspection area 4 is assigned)
X15 + X25 + X35 + X45 + X55 = 1 (Inspection area 5 is assigned)
All Xij’s ≥ 0
Define a set of decision variables Xij, which mean1 if Line i is assigned to Inspection j
Xij =0 otherwise
Formulation
2010-2011
Tran Van Hoai 5
Assignment networksDefinition
• Can be solved by– Enumeration– LP– Transportation model– Dynamic programming– Branch-and-bound– Hungarian algorithm
2010-2011
- m workers are to be assigned to m jobs- Unit cost Cij for worker i performing job jGoal: to minimize total cost of assignment
Tran Van Hoai 6
Advanced issues (1)
• Number of workers ≥ number of jobs– Worker constraint changed from “=” to “≤”
• A worker can perform ≥ 1 jobs• Minimization changed to maximization• Additional constraints– Line 1 cannot be assigned to Inspection B
• X12=0
– If Line 1 is assigned to Inspection B, then Line 2 must be assigned to Inspection D
2010-2011
Tran Van Hoai 7
Advanced issues (2)
• Additional constraints– If Line 1 is assigned to Inspection B, then Line 2
must be assigned to Inspection D• X12 ≤ X24
– At least one in Lines {1,3,4} is assigned to Inspection E• X15+X35+X45=1
2010-2011
Tran Van Hoai 8
Generalized assignment model
2010-2011
njmix
njx
miwxw
ts
xp
ij
m
iij
i
n
jijij
m
i
n
jijij
,,1 ,,,1 ,1,0
,,1 ,1
,,1 ,
..
max
1
1
1 1
Tran Van Hoai 9
Marriage service
2010-2011
ASSIGNMENTEach man only assigned to one woman
Pair assignment requires a operational cost
GOAL: to find a match having minimum total cost
Tran Van Hoai 10
Shortest path problem
2010-2011
1 2
3
4
5
6
87
100
151
42
67
150
75
22
8952
2571
11
- n nodes, a starting node (source), an ending node (destination)
- Arcs connecting adjacent nodes with non-negative distances dij
GOAL: to find a shortest path from source to destination
Tran Van Hoai 11
Solution methods
• Dijkstra– Only works for non-negative arc weight
• Bellman-Ford– To find shortest path from a source to all other
nodes– Works with non-negative arc weight (provided that
there is no negative weighted cycle)• Floyd–Warshall– To find shortest path between all node pairs
2010-2011
Not easy to include additional constraints(for non-IT users)
Tran Van Hoai 12
LP-based approach
2010-2011
Define a set of decision variables Xij, which mean1 if arc ij is utilized
Xij =0 otherwise
MIN ∑dijXij
S.T. <number of out-arcs used> = 1 (for source)
<number of in-arcs used> - <number of out-arcs used> = 0
(for intermediate nodes)
-<number of in-arcs used> = -1 (for destination)
Xij = 0 or 1
Tran Van Hoai 13
MIN 100X12 + 151X13 + … + 52X78
S.T. X12 + X13 = 1-X12 + X25 + X26 = 0-X13 + X34 + X35 = 0-X34 + X45 + X47 = 0-X25 - X35 - X45 + X57 = 0-X26 + X67 + X68 = 0-X47 - X57 - X67 + X78 = 0-X68 - X78 = -1
Xij = 0 or 1
2010-2011
Tran Van Hoai 14
Shortest path
2010-2011
1 2
3
4
5
6
87
100
151
42
67
150
75
22
8952
2571
9
Tran Van Hoai 15
Additional constraints
• Not go through node 5X25 = X35 = X45 = X57 = 0Don’t have to remove
node 5• Node 4 must be on the
pathX45+X47= 1
2010-2011
1 2
3
4
5
6
87
100
15142
67
15075
2289 52
2571
9
1 2
3
4
5
6
87
100151
42
67
15075
2289 52
2571
9
Tran Van Hoai 16
Additional constraints
• If node 7 on path, then node 3 must on pathX35 + X34 – X78 ≥ 0
• …
2010-2011
1 2
3
4
5
6
87
100
15142
67
15075
2289 52
2571
9
Tran Van Hoai 17
Maximal flow model
2010-2011
- one source node, generating flows- one terminal node, depositing flows- flow in = flow out on intermediate nodes- capacity Cij on arc from i to j
GOAL: to find maximum flow out of source to terminal, without exceeding arc capacities
Tran Van Hoai 18
Maximal flow problem
2010-2011
1 2
3
4
5
6
87
10
15
5
6
12
7
2
85
47
3
Tran Van Hoai 19
LP-based approach
• Xij: flow from node i to node j (if arc ij exists)
2010-2011
Tran Van Hoai 20
MAX X12 + X13
S.T. -X12 + X25 + X26 = 0-X13 + X34 + X35 = 0-X34 + X45 + X47 = 0-X25 - X35 - X45 + X57 = 0-X26 + X67 + X68 = 0-X47 - X57 - X67 + X78 = 0
Xij = 0 or 1X12≤1, X13≤2, X25≤3, X26≤4, X34≤5, X35≤6, X45≤7,
X47≤8,X57≤9,X67≤10,X68≤11,X78≤12
2010-2011
Tran Van Hoai 21
Maximal flow problem
2010-2011
1 2
3
4
5
6
87
10
15
5
6
12
7
2
85
47
3
7
5 7
5
57
5
Tran Van Hoai 22
Cuts in maximal flow problem
2010-2011
1 2
3
4
5
6
87
10
15
5
6
12
7
2
85
47
3
7
5 7
5
57
5
CUT(all flow from 1 → 8 must cross CUT)
Maximal flow (12) ≤ C25 + C26 + C35 + C45 + C47 (34)
Sum of arc capacities on the cut provides upper bound for maximal flow
Tran Van Hoai 23
Max flow/Min cut theorem
1. The value of max flow = the sum of capacities of min cut
2. The flow of all arcs on min cut will be at their upper bound
2010-2011
Tran Van Hoai 24
Traveling salesman network
• NP-Hard (cannot be solved in polynomial time)
• Connectivity network model
2010-2011
- m nodes- unit cost Cij utilizing arc from i to jGOAL: to find a minimum cost tour (cycle) visiting all nodes (not twice)
Tran Van Hoai 25
Solution methods
• Enumerating all possible tour (cycle)(m-1)! tours for m nodes in symmetric TSP
• LP-based approach
2010-2011
Define a set of decision variables Xij, which mean1 if arc ij is utilized
Xij =0 otherwise
Tran Van Hoai 26
Federal Emergency Managament Agency
2010-2011
12
3
4
H
3050
35
45
65
80
40
50
2540
Tran Van Hoai 27
Assignment constraints• Sum of arcs used out of each node is 1
X11 + X12 + X13 + X14 + X15 = 1
• Sum of arcs used into each node is 1X11 + X21 + X31 + X41 + X51 = 1
2010-2011
12
3
4
H3050
35
45
65
80
40
50
2540INVALID
SOLUTION
Need constraints to remove subtours
Tran Van Hoai 28
Subtour constraints
• One-node subtour constraintsX11, X22, X33, X44, X55 ≤ 0
• Two-node subtour constraintsX12 + X21 ≤ 1, …
• Three-node subtour constraintsX12 + X23 + X31 ≤ 2, …
• Four-node subtour constraintsX12 + X23 + X34 + X41 ≤ 3, …
2010-2011
Tran Van Hoai 29
Vehicle routing problem
• Generalized TSP
2010-2011
Tran Van Hoai 30
Minimum spanning tree network
• Read textbook
2010-2011