Transportation and Logistic Problem part 2

1
Transportation and Logistic Problem (part 2)

• Lecturer
• Agachai Sumalee

2
Last week

• We learnt about the Decision Model
• Mapping between real world and mathematical form
• We introduced the term Linear Programming
• We solved the LP
• Now we will look at the Single Commodity Network
  Flow problem

3
Graph, Networks, and Flows
Network G (N, A)
Node set N 1, 2, 3, 4
In an undirected graph, (i,j) (j,i)
4
Nodes

• Flow enters to and exits from the network through
  nodes
• Three types of Nodes
• Source nodes (origin) automatically have some
  flows which need to be released from that node
• Sink nodes (destination) have some demand need
  to be satisfied
• Transshipment node total inflow total outflow
• Node can have upper or lower capacity
• Normally node can represent storage facility,
  production unit, stage of project, geographical
  area, (anything else???)

5
Links

• Link connect one node to the other nodes
• Flow can only travel along the link
• A link is defined by two nodes, origin node and
  destination node of that link

i
j
Called link i-j

• Each link will have its
• Link Cost (fixed amount or a function)
• Link capacity (maximum amount of flow which can
  pass through that link)

6
Flows

• Flows enter to the network through Source Nodes
  or Origin Nodes
• Flows travel from one node to the others through
  links
• Flow exits from the network through Sink Nodes or
  Destination Nodes
• Flows can represent numerous variables
• In most cases, flows cannot be negative

x1
y1
x2
y2
inflow
outflow
x3
y3
x1x2x3-(y1y2y3)0
7
Network

• So Network is made up from a set of nodes and a
  set of links!
• Some meaning must be given to a network
  constructed (e.g. personal and fleet planning,
  financial application, shipping route, etc)
• Sometime, the network represents a real physical
  system (e.g. road network, production line, rail
  network, etc.)

8
Example of network formulation

• OSI makes low-cost cars at plants in Korat and
  Saraburee.
• Completed cars are shipped by freight-rail to one
  of OSIs two distribution centers in Ayuthaya and
  Chonburi
• Then distributed to customer facilities in
  Bangkok, Rayong, and Samutparkarn
• Out task is to plan OSIs distribution of new
  model Thai-CEO-2004 cars over the next quarter
• Each plant can produce up to 1000 units during
  this period and none are presently stored at
  distribution centres
• The forecasted demand at Bangkok, Rayong, and
  Samutparkarn are 610, 450, and 500 respectively
• Transfers between the distribution centres is
  possible with no costs but limited to 25 units
• Costs of transports between each locations are
  shown in the following table

9
Example of network formulation
10
Rayong
5
Korat
Ayuthaya
1
3
Samutpakarn
6
2
4
BKK
Saraburee
Chonburee
7
11
Converting Graph to Equation

• Firstly, define variable which is link flow.
• As mentioned, each link can be defined by its
  starting node and ending node, so we will also
  use the same system to define variables for link
  flow
• Let Xi,j denoted the amount of link flow travel
  on link i-j (i.e. from node i to node j)
• Also we put the amount of produced cars at each
  factory unit and the demand at each area
• So we can get the symbol for all link flows in
  the network

12
Rayong
5
Korat
Ayuthaya
450
1
3
1000
Samutpakarn
6
500
2
4
1000
BKK
Saraburee
Chonburee
7
610
13

• Now we can continue to put the network structure
  into a mathematical equation
• Mainly, we only need to put the flow conservation
  at each node
• This will also give us the actual connection
  between each links in the network!
• Example (node 4)

X3,4
X4,3
X1,4
(X1,4X2,4X3,4)
-(X4,3X4,5X4,6X4,7) 0
X4,5
4
X4,6
X2,4
X4,7
14

• But for a source of sink node the right-hand side
  value will change to the amount of product and
  demand respectively
• For node 1 X1,3X1,4 -1000
• For node 5 X3,5X4,5 610
• Next we added all flow conservation equation for
  all nodes for the OSIs network

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Node-Link Incidence Matrix

• Believe it or not? There is a clear link between
  the set of flow conservation equations and the
  network structure
• This is represented through Node-Link Incidence
  Matrix
• We can see from the previous example that from
  the flow conservation equation of node 4 all flow
  entering this node will have the positive sign
• On the other hand, all flow exiting from node 4
  will have the negative sign
• We can define a matrix from a network based on
  the same principle

17
Node-Link Incidence Matrix

• Call A Node-Link Incidence Matrix
• The value of am,n (value in the row m and column
  n of A) can be determined by looking at the
  network
• Each row of A represents one Node
• Each column of A represents one link
• If node m is the starting node of link n, then
  am,n is -1
• If node m is the ending node of link n, then am,n
  is 1
• Otherwise, am,n is 0
• From our OSIs network, there are totally 7 nodes
  and 12 links, so our A matrix will have 7 rows
  and 12 columns

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Flow Conservation in Matrix form

• After making the A matrix from the network, we
  can then define the flow conservation for the
  whole network by one single matrix equation

• Where X is a flow vector and B is zero in all row
  except the row associated with source node and
  sink node
• Basically, B is the vector of all values on the
  right hand side of the flow conservation equations

20
Others constraints

• Flow conservation will be added as constraints to
  the Mathematical Program of this problem
• Other constraints include node capacity
  constraint and link capacity constraint
• From OSI example, we need to link capacity
  constraint for link 4,3 and 3,4 which is limited
  to 25 unit of flows

21
Minimum cost flow problem

• Now we have got all constraints needed (further
  constraints can also be added into the program)
• One of the most widely used problem is Minimum
  cost flow problem
• From our OSI, we seek the optimal distribution
  plan to minimize the total cost of transportation
• So the objective of our mathematical program is
  total transport costs which is

22
Total Demand Total Supply

• Before finally formulate the mathematical program
  for this problem, we need to check the overall
  equality of the demand and supply
• Total production unit of two factories are 2000,
  but total demand are only 1560
• This leaves an excess supply of 440
• We need to modify the network to take account of
  this excess supply
• This can be done by adding a dummy sink node
  (node 8) and link 1-8 and 2-8 to the network
• These two links will have no costs and capacities
• Node 8 will have a dummy demand of 440

23
Rayong
5
Korat
Ayuthaya
450
1
3
1000
X1,8
Samutpakarn
8
6
440
500
X2,8
2
4
1000
BKK
Saraburee
Chonburee
7
610
24

• After adding node 8 and link 1-8 and 2-8, we can
  then add the flow conservation equation for node
  8 into the problem
• Also, we need to modify the flow conservation
  equations for node 1 and 2.

25
Complete problem
Is this a Linear Program (LP)???
26
Solution from Excel
Excel
Rayong
5
Korat
Ayuthaya
450
1
3
1000
440
Samutpakarn
8
6
440
500
2
4
1000
BKK
Saraburee
Chonburee
7
610
Link cap infinity Total cost 25355
27
Adding Link Capacity Constraint

• Now, we will start to include some link capacity
• Let assume that all link has the same capacity of
  600 and 650 units
• By adding this link capacity, will we get the new
  solution? Can we observe these results from the
  previous solution?

28
Solution from Excel
Excel
Rayong
5
Korat
Ayuthaya
450
1
3
1000
440
Samutpakarn
8
6
440
500
2
4
1000
BKK
Saraburee
Chonburee
7
610
Link cap 600 Total cost 26215
29
Solution from Excel
Excel
Rayong
5
Korat
Ayuthaya
450
1
3
1000
440
Samutpakarn
8
6
440
500
2
4
1000
BKK
Saraburee
Chonburee
7
610
Link cap 550 Total cost 26865
30
Decision on improved links

• If OSI is actually run the whole freight-rail by
  itself and is the organisation who control the
  capacity on all movement
• OSI ask you to recommend (with a minimum level of
  study budget) the most effective link to be
  improved (by introducing new locomotive and
  rolling stock) to reduce the whole transport cost
• Can you give some quick recommendation based on
  previous results?
• Also, assume that the costs of link capacity
  improvement is 10, 11, 12 unit/one unit of
  capacity, will the recommendation change?

31
Maximum Flow Problem

• A new instance of network flow problem arises
  when
• all links have fixed capacities
• there is only one source node and one sink node
• the main aim is to find out how much flow can be
  delivered through this network without violating
  all link capacities

32
OSI example 2

• OSI is running a freight-rail network between BKK
  and Khon-Kaen (network shown in the figure below

8
2
4
3
7
10
6
3
1
6
7
2
1
1
4
2
10
4
3
2
8
5
12

• Each link has link capacity
• OSI wants to estimate the maximum level of good
  they can transport from BKK to Khon-Kaen in one
  period of time

33
Excel Solution
Excel
34
Transportation Assignment Problem

• Network flow problem without transhipment nodes
  (flows could be interpreted as people, tools,
  materials, water, oil, money, or almost anything
  else
• Examples include
• N persons apply for M jobs
• N persons allocated to M activities
• Assign N pilot to M aircrafts
• Assigns N unit of commodities to M routes
• And many more

35
Transportation problem Network
ci,j
1
A
10
15
2
B
20
10
10
15
3
C
36
Assignment Network
ci,j
1
A
Mr.X
2
B
Mr.Y
Mr.Z
3
C
37
Critical Path in Project Plan

• Nodes represent activities in the project
• Links represent the precedence between different
  activities
• Link costs represent the time to complete the
  activity at the starting node of that link
• Find a path with maximum cost
• Inflow 1 unit

38
3
5
6
8
12
1
6
7
0
1
10
-1
2
9
4
Total duration 38 time units
39
Time-Expanded Network Model

• Network flow can also be extended to many
  applications involving time-expanded formulations
  (flows over time)
• Especially true for inventory management
• Each node can represent different time interval
• Links then reflect either flows between points in
  time or in location

40
OSI example with time-expanded

• OSI produces machinery products upto 15,000
  units/ quarter
• Cost 35 pounds per thousand
• Need to deliver the products to two customer
  areas
• Demand from two locations varies by quarter

• Inventory can be maintained at the plant for 8
  pounds per thousand per quarter

41
-44
source
(35,15)
(35,15)
(35,15)
(35,15)
P,1
P,2
P,3
P,4
(8,inf)
(8,inf)
(8,inf)
(8,inf)
(11,inf)
(11,inf)
(11,inf)
(11,inf)
(17,inf)
(17,inf)
(17,inf)
(17,inf)
(11,inf)
C1,1
C1,2
C1,3
C1,4
C2,1
C2,2
C2,3
C2,4
5
9
2
1
3
14
6
4