Transportation And Assignment Problems - Operations Research

Transportation And Assignment Problems - Operations Research
Source - https://www.stanfordesp.org/download/6d4d63c3c582620a73c1f62de12ad9bb/E1605_TRANSP_PROBLEM.ppt

1. .
1
Transportation and Assignment
Problems

2. .
2
Applications
Physical analog
of nodes
Physical analog
of arcs
Flow
Communication
systems
phone exchanges,
computers,
transmission
facilities, satellites
Cables, fiber optic
links, microwave
relay links
Voice messages,
Data,
Video transmissions
Hydraulic systems
Pumping stations
Reservoirs, Lakes
Pipelines
Water, Gas, Oil,
Hydraulic fluids
Integrated
computer circuits
Gates, registers,
processors
Wires Electrical current
Mechanical systems Joints
Rods, Beams,
Springs
Heat, Energy
Transportation
systems
Intersections,
Airports,
Rail yards
Highways,
Airline routes
Railbeds
Passengers,
freight,
vehicles,
operators
Applications of Network
Optimization

3. .
3
Description
A transportation problem basically deals with the
problem, which aims to find the best way to fulfill
the demand of n demand points using the
capacities of m supply points. While trying to find
the best way, generally a variable cost of shipping
the product from one supply point to a demand
point or a similar constraint should be taken into
consideration.

4. .
4
Formulating Transportation
Problems
Example 1: Powerco has three electric
power plants that supply the electric needs
of four cities.
•The associated supply of each plant and
demand of each city is given in the table 1.
•The cost of sending 1 million kwh of
electricity from a plant to a city depends on
the distance the electricity must travel.

5. .
5
Transportation tableau
A transportation problem is specified by
the supply, the demand, and the shipping
costs. So the relevant data can be
summarized in a transportation tableau.
The transportation tableau implicitly
expresses the supply and demand
constraints and the shipping cost between
each demand and supply point.

6. .
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Table 1. Shipping costs, Supply, and Demand
for Powerco Example
From To
City 1City 2City 3City 4Supply
(Million kwh)
Plant 1 $8 $6 $10 $9 35
Plant 2 $9 $12 $13 $7 50
Plant 3 $14 $9 $16 $5 40
Demand
(Million kwh)
45 20 30 30
Transportation Tableau

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Solution
1.Decision Variable:
Since we have to determine how much electricity
is sent from each plant to each city;
X
ij = Amount of electricity produced at plant i
and sent to city j
X
14 = Amount of electricity produced at plant 1
and sent to city 4

8. .
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2. Objective function
Since we want to minimize the total cost of shipping
from plants to cities;
Minimize Z = 8X
11+6X
12+10X
13+9X
14
+9X
21+12X
22+13X
23+7X
24
+14X
31+9X
32+16X
33+5X
34

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3. Supply Constraints
Since each supply point has a limited production
capacity;
X
11+X
12+X
13+X
14 <= 35
X
21+X
22+X
23+X
24 <= 50
X
31+X
32+X
33+X
34 <= 40

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4. Demand Constraints
Since each supply point has a limited production
capacity;
X
11+X
21+X
31 >= 45
X
12+X
22+X
32 >= 20
X
13+X
23+X
33 >= 30
X
14+X
24+X
34 >= 30

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5. Sign Constraints
Since a negative amount of electricity can not be
shipped all Xij’s must be non negative;
Xij >= 0 (i= 1,2,3; j= 1,2,3,4)

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LP Formulation of Powerco’s Problem
Min Z = 8X
11+6X
12+10X
13+9X
14+9X
21+12X
22+13X
23+7X
24
+14X
31+9X
32+16X
33+5X
34
S.T.: X
11+X
12+X
13+X
14 <= 35 (Supply Constraints)
X
21+X
22+X
23+X
24 <= 50
X
31+X
32+X
33+X
34 <= 40
X
11+X
21+X
31 >= 45 (Demand Constraints)
X
12+X
22+X
32 >= 20
X
13+X
23+X
33 >= 30
X
14+X
24+X
34 >= 30
Xij >= 0 (i= 1,2,3; j= 1,2,3,4)

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General Description of a Transportation
Problem
1.A set of m supply points from which a good is
shipped. Supply point i can supply at most s
i
units.
2.A set of n demand points to which the good is
shipped. Demand point j must receive at least d
i
units of the shipped good.
3.Each unit produced at supply point i and shipped
to demand point j incurs a variable cost of c
ij.

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X
ij = number of units shipped from supply point i to
demand point j),...,2,1;,...,2,1(0
),...,2,1(
),...,2,1(..
min
1
1
11
njmiX
njdX
misXts
Xc
ij
mi
i
jij
nj
j
iij
mi
i
nj
j
ijij
==
=
=



=
=
=
=
=
=
=
=

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Balanced Transportation Problem
If Total supply equals to total demand, the
problem is said to be a balanced
transportation problem:
=
=
=
=
=
nj
j
j
mi
i
i ds
11

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Methods to find the bfs for a balanced TP
There are two basic methods:
1.Northwest Corner Method
2.Vogel’s Method

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1. Northwest Corner Method
To find the bfs by the NWC method:
Begin in the upper left (northwest) corner of the
transportation tableau and set x
11 as large as
possible (here the limitations for setting x
11 to a
larger number, will be the demand of demand
point 1 and the supply of supply point 1. Your
x
11 value can not be greater than minimum of
this 2 values).

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According to the explanations in the previous slide
we can set x
11=3 (meaning demand of demand
point 1 is satisfied by supply point 1).5
6
2
3 5 2 3 3 2
6
2
X 5 2 3

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After we check the east and south cells, we saw that
we can go east (meaning supply point 1 still has
capacity to fulfill some demand).3 2 X
6
2
X 3 2 3 3 2 X
3 3
2
X X 2 3

20. .
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After applying the same procedure, we saw that we
can go south this time (meaning demand point 2
needs more supply by supply point 2).3 2 X
3 2 1
2
X X X 3 3 2 X
3 2 1 X
2
X X X 2

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Finally, we will have the following bfs, which is:
x
11=3, x
12=2, x
22=3, x
23=2, x
24=1, x
34=23 2 X
3 2 1 X
2 X
X X X X

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3. Vogel’s Method
Begin with computing each row and column a penalty.
The penalty will be equal to the difference between
the two smallest shipping costs in the row or column.
Identify the row or column with the largest penalty.
Find the first basic variable which has the smallest
shipping cost in that row or column. Then assign the
highest possible value to that variable, and cross-out
the row or column as in the previous methods.
Compute new penalties and use the same procedure.

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An example for Vogel’s Method
Step 1: Compute the penalties.SupplyRow Penalty
6 7 8
15 80 78
Demand
Column Penalty 15-6=9 80-7=73 78-8=70
7-6=1
78-15=63
15 5 5
10
15

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Step 2: Identify the largest penalty and assign the
highest possible value to the variable.SupplyRow Penalty
6 7 8
5
15 80 78
Demand
Column Penalty 15-6=9 _ 78-8=70
8-6=2
78-15=63
15 X 5
5
15

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Step 3: Identify the largest penalty and assign the
highest possible value to the variable.SupplyRow Penalty
6 7 8
5 5
15 80 78
Demand
Column Penalty 15-6=9 _ _
_
_
15 X X
0
15

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Step 4: Identify the largest penalty and assign the
highest possible value to the variable.SupplyRow Penalty
6 7 8
0 5 5
15 80 78
Demand
Column Penalty _ _ _
_
_
15 X X
X
15

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Step 5: Finally the bfs is found as X
11=0, X
12=5,
X
13=5, and X
21=15SupplyRow Penalty
6 7 8
0 5 5
15 80 78
15
Demand
Column Penalty _ _ _
_
_
X X X
X
X

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The Transportation Simplex
Method
In this section we will explain how the simplex
algorithm is used to solve a transportation problem.

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How to Pivot a Transportation Problem
Based on the transportation tableau, the following
steps should be performed.
Step 1. Determine (by a criterion to be developed
shortly, for example northwest corner method) the
variable that should enter the basis.
Step 2. Find the loop (it can be shown that there is
only one loop) involving the entering variable and
some of the basic variables.
Step 3. Counting the cells in the loop, label them as
even cells or odd cells.

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Step 4. Find the odd cells whose variable assumes the
smallest value. Call this value θ. The variable
corresponding to this odd cell will leave the basis. To
perform the pivot, decrease the value of each odd cell
by θ and increase the value of each even cell by θ. The
variables that are not in the loop remain unchanged.
The pivot is now complete. If θ=0, the entering
variable will equal 0, and an odd variable that has a
current value of 0 will leave the basis. In this case a
degenerate bfs existed before and will result after the
pivot. If more than one odd cell in the loop equals θ,
you may arbitrarily choose one of these odd cells to
leave the basis; again a degenerate bfs will result

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Assignment Problems
Example: Machineco has four jobs to be completed.
Each machine must be assigned to complete one job.
The time required to setup each machine for completing
each job is shown in the table below. Machinco wants to
minimize the total setup time needed to complete the
four jobs.

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Setup times
(Also called the cost matrix)
Time (Hours)
Job1Job2Job3Job4
Machine 114 5 8 7
Machine 22 12 6 5
Machine 37 8 3 9
Machine 42 4 6 10

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The Model
According to the setup table Machinco’s problem can be
formulated as follows (for i,j=1,2,3,4):10
1
1
1
1
1
1
1
1..
10629387
5612278514min
44342414
43332313
42322212
41312111
44434241
34333231
24232221
14131211
4443424134333231
2423222114131211
==
=+++
=+++
=+++
=+++
=+++
=+++
=+++
=+++
++++++++
+++++++=
ijijorXX
XXXX
XXXX
XXXX
XXXX
XXXX
XXXX
XXXX
XXXXts
XXXXXXXX
XXXXXXXXZ

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For the model on the previous page note that:
X
ij=1 if machine i is assigned to meet the demands of
job j
X
ij=0 if machine i is not assigned to meet the demands
of job j
In general an assignment problem is balanced
transportation problem in which all supplies and
demands are equal to 1.

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The Assignment Problem
In general the LP formulation is given as
Minimize 11
1
1
11
11
0
, , ,
, , ,
or 1,
nn
ij ij
ij
n
ij
j
n
ij
i
ij
cx
x i n
x j n
x ij
==
=
=
=  =
=  =
=



Each supply is 1
Each demand is 1