This document provides an overview of the transportation problem and method. It defines the transportation problem as finding the minimum cost of distributing a commodity from supply centers to demand centers. Key aspects covered include: - Developing a transportation model requires supply/demand quantities and unit transportation costs - Finding an initial feasible solution using methods like the Northwest Corner method - Evaluating solutions for optimality using techniques like the stepping stone method - Developing improved solutions by reallocating quantities based on negative cell evaluations - Special considerations like dealing with degenerate solutions or unequal supply/demand

1. THE TRANSPORTATION
PROBLEM
CHAPTER THREE

2. Transportation-2
The Transportation Problem
 The problem of finding the minimum distribution
cost of a given commodity from a group of supply
centers (sources) to a group of receiving centers
(destinations)
 Each source has a certain supply (si)
 Each destination has a certain demand (dj)
 The cost of shipping from a source to a destination
is directly proportional to the number of units
shipped

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Figure 6S–1 Overview of the Transportation Method

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Formulating the Model
 To develop a model of a transportation problem,
it is necessary to have the following information:
 Supply quantity (capacity) of each origin
 Demand quantity of each destination
 Unit transportation cost for each origin-
destination route.

5. Assume Sunshine Construction Materials has
contracted to provide sand for three residential
housing developments projects.
Example
Demand for the sand
generated by the
construction projects is:
Sand can be supplied from
three different areas as
follows:

6. • The manager has estimated the cost per cubic yard to
ship each of the possible routes
Copyright © 2007 The McGraw-Hill Companies. All rights reserved. McGraw-Hill/Irwin 6–8

7. Copyright © 2007 The McGraw-Hill Companies. All rights reserved. McGraw-Hill/Irwin 6–9
Table 1 Transportation Table format

8. Table 2 Transportation Table
Copyright © 2007 The McGraw-Hill Companies. All rights reserved. McGraw-Hill/Irwin 6S–10

9. Problem formulation
Min Z = 4X11 + 2X12 + 8X13 + 5X21 + 1X22 +
9X23 +7X31 + 6X32 + 3X33
X11, …, X33  0
Subject to:

10. Solution To Transportation Problem
Copyright © 2007 The McGraw-Hill Companies. All rights reserved. McGraw-Hill/Irwin 6–12

11. Finding Initial Feasible Solution
• A feasible solution is one in which assignments are
made in such a way that all supply and demand
requirements are satisfied.
• The number of nonzero (occupied) cells should equal
one less than the sum of the number of rows and the
number of columns in a transportation table.
• Methods of finding initial feasible solution:
–The Northwest Corner Method
–An Intuitive Approach/Least Cost Method
–Vogel’s Approximation/ Penalty Method

12. Finding an Initial Feasible Solution:
The Northwest Corner Method
• A systematic approach for developing an
initial feasible solution.
• Simple to use and easy to understand.
• Does not take transportation costs into
account.
• Gets its name because the starting point for
the allocation process is the upper left hand
(northwest) corner of the transportation table.

13. Principles that Guides the
Allocation
 Begin with the upper left hand cell, and
allocate as many units as possible to that cell.
 Remain in a row or column until its supply
or demand is completely exhausted or satisfied

14. Initial Feasible Solution Using Northwest Corner
Method
Total Cost = 50(4) + 50(2) + 100(1) + 100(9) + 200(3)
= Birr 1900

15. Finding an Initial Feasible Solution:
The Intuitive Approach
1. Identify the cell that has the lowest unit cost.
2. Cross out the cells in the row or column that has
been exhausted (or both, if both have been
exhausted), and adjust the remaining row or
column total accordingly.
3. Identify the cell with the lowest cost from the
remaining cells.
4. Repeat steps 2 and 3 until all supply and demand
have been allocated.

16. Initial Feasible Solution Using the Intuitive
Approach
Total Cost = 50(4) + 50(8) + 150(1) + 50(9) + 200(3)
= Birr 1800

17. Vogel’s Approximation Method
(VAM)
• Construct the cost, requirement, and
availability matrix
• Compute a penalty for each row(NLC-LC) and
column(NLC- LC ) then select the highest
penalty.
• Then the row and column with the largest
penalty; allocate and cross out the exhausted
row/column
• Repeat steps 1 to 3 for the reduced table until
the entire capabilities are used to fill the
requirement at different warehouses.

19. Initial Feasible Solution Using Vogel’s
Approximation Method (VAM)
Total Cost: 1x150 + 3x200+4x50 + 8x50 + 9x50 = Birr
1800

20. Evaluating a Solution for Optimality
• The test for optimality for a feasible solution
involves a cost evaluation of empty cells.
• We shall consider two methods for cell
evaluation:
–The Stepping Stone Method
–The MODI Method

21. The Stepping Stone Method
• Involves tracing a series of closed paths in the
transportation table, using one such path for each
empty cell. Rules for tracing Stepping-stone paths:
• All unoccupied cells must be evaluated.
• Except for the cell being evaluated, only add or
subtract in occupied cells.
• A path will consist of only horizontal and vertical
moves, starting and ending with the empty cell that is
being evaluated.
• Alter + and – signs, beginning with a + sign in the
cell being evaluated.

22. Evaluation Path for Cell C-1

23. Empty Cell Evaluation
•Cell B-1: + 2 = 5 – 1 + 2 – 4
•Cell C-1: +10 = 7 – 3 + 9 – 1 + 2 – 4
•Cell A-3: -2 = 8 – 9 + 1 – 2
•Cell C-2: +11 = 6 – 3 + 9 – 1
•The negative value for cell A-3 indicates
an improved solution is possible

24. The MODI method
• Involves the use of index numbers that are established for
the rows and columns. These are based on the unit costs
of the occupied cells.
• The index numbers can be used to obtain the cell
evaluations for empty cells
The cell evaluations for each of the unoccupied cells are
determined using the relationship:

25. Cell Evaluations Using the MODI Method

26. Developing an Improved
Solution
• Developing an improved solution to a transportation
problem requires focusing on the unoccupied cell
that has the largest negative cell evaluation.
• Improving the solution involves reallocating
quantities in the transportation table.
• The stepping-stone path for that cell is used for
determining how many units can be reallocated (both
the magnitude and direction of changes)
• The + signs in the path indicate units to be added,
the – signs indicate units to be subtracted. The limit
on subtraction is the smallest quantity in a negative
position along the cell path.

27. Optimal Solution

28. Summary of the Transportation Method
1. Obtain an initial feasible solution. Use either the NWC
method, the intuitive method, or the VAM. Generally, the
intuitive method and Vogel’s approximation are the
preferred approaches.
2. Evaluate the solution to determine if it is optimal. Use
either the stepping-stone method or MODI. The solution
is not optimal if any unoccupied cell has a negative cell
evaluation.
3. If the solution is not optimal, select the cell that has the
most negative cell evaluation. Obtain an improved
solution using the stepping-stone method.
4. Repeat steps 2 and 3 until no cell evaluations (reduced
costs) are negative. Once you have identified the optimal

29. Special Issues
1. Determining if there are alternate optimal
solutions.
2. Recognizing and handling degeneracy (too
few occupied cells to permit evaluation of a
solution).
3. Avoiding unacceptable or prohibited route
assignments.
4. Dealing with problems in which supply and
demand are not equal.
5. Solving maximization problems.

30. Alternate Optimal Solutions
• The existence of an alternate solution is signaled by
an empty cell’s evaluation equal to zero.

31. Degeneracy
• A solution is degenerate if the number of occupied cells is
less than the number of rows plus the number of columns
minus one.
• The modification is to treat some of the empty cells as
occupied cells. This is accomplished by placing a delta
() in one of the empty cells. w/c is very smallest cost

32. Unacceptable Routes
 Certain origin-destination combinations may be
unacceptable due to weather factors, equipment
breakdowns, labor problems, or skill requirements
that either prohibit, or make undesirable, certain
combinations.
 In order to prevent that route from appearing in the
final solution, the manager could assign a unit cost to
that cell that was large enough to make that route
uneconomical and, hence, prohibit its occurrence.
 One rule of thumb would be to assign a cost that is
10 times the largest cost in the table (or a very big
+M).

33. Maximization
• Transportation-type problems that concern
profits or revenues rather than costs with the
objective to maximize profits rather than to
minimize costs.
• Such problems can be handled by adding one
additional step at the start:
• Identify the cell with the largest profit and
subtract all the other cell profits from that
value.
• Replace the cell profits with the resulting
values.

34. Unequal Supply and Demand
• Situations in which supply and demand are not
equal such that it is necessary to modify the
original problem so that supply and demand are
equalized.
• This is accomplished by adding either a dummy
column or a dummy row; a dummy row is added
if supply is less than demand and a dummy
column is added if demand is less than supply.
• The dummy is assigned unit costs of zero for
each cell, and it is given a supply (if a row) or a
demand (if a column) equal to the difference

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•End of Chapter
three

37. 4 Assignment
Problems

38. Assignment Problems
• Involve the matching or pairing of two sets of items
such as jobs and machines, secretaries and reports,
lawyers and cases, and so forth.
• Have different cost or time requirements for different
pairings.

39. The Hungarian Method
• Provides a simple heuristic that can be used to find
the optimal set of assignments.
• It is based on minimization of opportunity costs that
would result from potential pairings. These are
additional costs that would be incurred if the lowest-
cost assignment is not made

40. Requirements for Use of the Hungarian Method
• Situations in which the Hungarian method can be
used are characterized by the following:
 There needs to be a one-for-one matching of two
sets of items.
 The goal is to minimize costs (or to maximize
profits) or a similar objective
 The costs or profits are known or can be closely
estimated.

41. The Hungarian Method
• Step 1: Locate the smallest cost element in each row of the
cost table. Now subtract this smallest from each element in
that row.
• Step 2: Consider each column and locate the smallest element
in it. Subtract the smallest value from every other entry in the
column.
• Step 3: Draw the minimum number of horizontal and vertical
lines required to cover the entire ‘zero’ elements. If the
number of lines drawn is equal to n (the number of
rows/columns) the solution is optimal
• Step 4: Select the smallest uncovered cost element. Subtract
this element from all uncovered elements including itself and
add this element to each value located at the intersection of
any lines.
• Step 5: Repeat steps 3 and 4 until an optimal solution is
obtained.

42. Example

43. Column Reduction of Opportunity (Row Reduction)
Costs

44. Determine the Minimum Number of Lines Needed to Cover
the Zeros
Further Revision of the Cost Table

45. Optimal Assignments

46. Example
• A production supervisor is considering how he should
assign the four jobs that are performed, to four of the
workers working under him. He want to assign the jobs to
the workers such that the aggregate time to perform the
job in the least. Based on the previous experience, he has
the information on the time taken by the four workers in
performing these jobs, as given in below

47. The final assignments is 1-B, 2-D, 3-C,

48. Special Situations
• Among those situations are the following:
• The number of rows does not equal the number of
columns.
• The problem involves maximization rather than
minimization.
• Certain matches are undesirable or not allowed.
• Multiple optimal solutions exist.

49. Unbalanced Assignment Problems
• In such situations, dummy column(s)/row(s),
whichever is smaller in number, are inserted with
zeros as the cost elements
A-1 B-4 C-3 D-2

50. Constrained/Prohibited/ Assignment
Problems
• It happens sometimes that a worker cannot perform a certain
job or is not to be assigned a particular job. To cope with this
situation, the cost of performing that job by such person is
taken to be extremely large
• Example: Determine the optimal set of pairings given the
following cost table. Note that assignment B-3 is undesirable,
as denoted by the M in that position:

51. Multiple Optimal Solutions
• In some cases, there are multiple optimal solutions to a
problem. This condition can be easily recognized when
making the optimal assignments.
• Example: Given this final assignment table, identify the
optimal solutions:

52. Maximization
• One extra step must be added to the start of the process.
• Identify the largest value in each column and then
subtract all numbers in each column from the column
maximum.
• Example: Let’s consider the following assignment table
where the values are unit profits.
The optimal assignment will be to match A with 2, B
with 1 and C with 3 and the maximum profit is Br. 92 =

53. End of
Chapter Four