This document discusses transportation problems and their formulation as linear programs that can be solved using the simplex method. It provides examples of how to find an initial basic feasible solution using different methods like the Northwest Corner method, Minimum Cost method, and Vogel's method. It also explains how to perform pivots on the transportation tableau when applying the simplex method to solve transportation problems. Key steps include determining an entering variable, finding the associated pivot row, labeling cells as even/odd, and choosing a leaving variable based on the smallest odd cell value.

1. Transportation, Assignment and
Transshipment Problems
Course: MBA
Subject: Quantitative
Techniques
Unit: 1.2

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. 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. 7.1 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. 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. Table 1. Shipping costs, Supply, and Demand for
Powerco Example
From To
City 1 City 2 City 3 City 4 Supply
(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

7. Solution
1. Decision Variable:
Since we have to determine how much
electricity is sent from each plant to each city;
Xij = Amount of electricity produced at plant i
and sent to city j
X14 = Amount of electricity produced at plant 1
and sent to city 4

8. 2. Objective function
Since we want to minimize the total cost of shipping
from plants to cities;
Minimize Z = 8X11+6X12+10X13+9X14
+9X21+12X22+13X23+7X24
+14X31+9X32+16X33+5X34

9. 3. Supply Constraints
Since each supply point has a limited production
capacity;
X11+X12+X13+X14 <= 35
X21+X22+X23+X24 <= 50
X31+X32+X33+X34 <= 40

10. 4. Demand Constraints
Since each supply point has a limited production
capacity;
X11+X21+X31 >= 45
X12+X22+X32 >= 20
X13+X23+X33 >= 30
X14+X24+X34 >= 30

11. 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)

12. LP Formulation of Powerco’s Problem
Min Z = 8X11+6X12+10X13+9X14+9X21+12X22+13X23+7X24
+14X31+9X32+16X33+5X34
S.T.: X11+X12+X13+X14 <= 35 (Supply Constraints)
X21+X22+X23+X24 <= 50
X31+X32+X33+X34 <= 40
X11+X21+X31 >= 45 (Demand Constraints)
X12+X22+X32 >= 20
X13+X23+X33 >= 30
X14+X24+X34 >= 30
Xij >= 0 (i= 1,2,3; j= 1,2,3,4)

13. 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 si
units.
2. A set of n demand points to which the good is
shipped. Demand point j must receive at least di
units of the shipped good.
3. Each unit produced at supply point i and
shipped to demand point j incurs a variable cost
of cij.

14. Xij = number of units shipped from supply point i to
demand point j
),...,2,1;,...,2,1(0
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),...,2,1(..
min
1
1
1 1
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njdX
misXts
Xc
ij
mi
i
jij
nj
j
iij
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i
nj
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
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









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15. 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

16. Balancing a TP if total supply exceeds total
demand
If total supply exceeds total demand, we
can balance the problem by adding
dummy demand point. Since shipments to
the dummy demand point are not real,
they are assigned a cost of zero.

17. Balancing a transportation problem if total
supply is less than total demand
If a transportation problem has a total
supply that is strictly less than total
demand the problem has no feasible
solution. There is no doubt that in such a
case one or more of the demand will be
left unmet. Generally in such situations a
penalty cost is often associated with
unmet demand and as one can guess this
time the total penalty cost is desired to be
minimum

18. 7.2 Finding Basic Feasible Solution
for TP
Unlike other Linear Programming
problems, a balanced TP with m supply
points and n demand points is easier to
solve, although it has m + n equality
constraints. The reason for that is, if a set
of decision variables (xij’s) satisfy all but
one constraint, the values for xij’s will
satisfy that remaining constraint
automatically.

19. Methods to find the bfs for a balanced TP
There are three basic methods:
1. Northwest Corner Method
2. Minimum Cost Method
3. Vogel’s Method

20. 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 x11 as large as
possible (here the limitations for setting x11 to a
larger number, will be the demand of demand
point 1 and the supply of supply point 1. Your
x11 value can not be greater than minimum of
this 2 values).

21. According to the explanations in the previous slide
we can set x11=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

22. 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

23. 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

24. Finally, we will have the following bfs, which is:
x11=3, x12=2, x22=3, x23=2, x24=1, x34=2
3 2 X
3 2 1 X
2 X
X X X X

25. 2. Minimum Cost Method
The Northwest Corner Method dos not utilize
shipping costs. It can yield an initial bfs easily but the
total shipping cost may be very high. The minimum
cost method uses shipping costs in order come up
with a bfs that has a lower cost. To begin the
minimum cost method, first we find the decision
variable with the smallest shipping cost (Xij). Then
assign Xij its largest possible value, which is the
minimum of si and dj

26. After that, as in the Northwest Corner Method we
should cross out row i and column j and reduce the
supply or demand of the non crossed-out row or
column by the value of Xij. Then we will choose the
cell with the minimum cost of shipping from the
cells that do not lie in a crossed-out row or column
and we will repeat the procedure.

27. An example for Minimum Cost Method
Step 1: Select the cell with minimum cost.
2 3 5 6
2 1 3 5
3 8 4 6
5
10
15
12 8 4 6

28. Step 2: Cross-out column 2
2 3 5 6
2 1 3 5
8
3 8 4 6
12 X 4 6
5
2
15

29. Step 3: Find the new cell with minimum shipping cost
and cross-out row 2
2 3 5 6
2 1 3 5
2 8
3 8 4 6
5
X
15
10 X 4 6

30. Step 4: Find the new cell with minimum shipping cost
and cross-out row 1
2 3 5 6
5
2 1 3 5
2 8
3 8 4 6
X
X
15
5 X 4 6

31. Step 5: Find the new cell with minimum shipping cost
and cross-out column 1
2 3 5 6
5
2 1 3 5
2 8
3 8 4 6
5
X
X
10
X X 4 6

32. Step 6: Find the new cell with minimum shipping cost
and cross-out column 3
2 3 5 6
5
2 1 3 5
2 8
3 8 4 6
5 4
X
X
6
X X X 6

33. Step 7: Finally assign 6 to last cell. The bfs is found as:
X11=5, X21=2, X22=8, X31=5, X33=4 and X34=6
2 3 5 6
5
2 1 3 5
2 8
3 8 4 6
5 4 6
X
X
X
X X X X

34. 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.

35. An example for Vogel’s Method
Step 1: Compute the penalties.
Supply Row 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

36. Step 2: Identify the largest penalty and assign the
highest possible value to the variable.
Supply Row 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

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

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

39. Step 5: Finally the bfs is found as X11=0, X12=5, X13=5,
and X21=15
Supply Row Penalty
6 7 8
0 5 5
15 80 78
15
Demand
Column Penalty _ _ _
_
_
X X X
X
X

40. 7.3 The Transportation Simplex
Method
In this section we will explain how the simplex
algorithm is used to solve a transportation problem.

41. 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.

42. 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

43. 7.5. Assignment Problems
Example: Machine co 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.

44. Hungarian Assignment Method (HAM)
• For a typical balanced assignment problem involving a certain number of
persons and an equal number of jobs, and with an objective function of
the minimization type, the method is applied as listed in the following
steps:
1. Locate the smallest cost element in each row of the cost table. Now
subtract this smallest element from each element in that row. As a
result, there shall be at least one zero in each row of this new table,
called the reduced cost table.
2. In the reduced cost table obtained, consider each column and locate
the smallest element in it. Subtract the smallest value from every other
entry in the column. As a consequence of this action, there would be at
least one zero in each of the rows and column of the second reduced
cost table.
3. Draw minimum number of horizontal and vertical line that are required
to cover all the ‘zero’ elements.

45. Conti…
4. Select the smallest uncovered cost element. Subtract these element
from all uncovered elements including itself and add this element
to each value located at the intersection of any two lines. The cost
element through which only one line passes remain unaltered.
5. Repeat step 3 and 4 until an optimal solution is obtained.
6. Give an optimal solution, make the job assignments as indicated by
the zero element.

46. Setup times
(Also called the cost matrix)
Time (Hours)
Job1 Job2 Job3 Job4
Machine 1 14 5 8 7
Machine 2 2 12 6 5
Machine 3 7 8 3 9
Machine 4 2 4 6 10

47. References
• Quantitative Techniques, by CR Kothari, Vikas
publication
• Fundamentals of Statistics by SC Guta
Publisher Sultan Chand
• Quantitative Techniques in management by
N.D. Vohra Publisher: Tata Mcgraw hill