#### Transcript The Transportation Problem Initial Basic Feasible Solution Methods

```INTRODUCTION TO OPERATIONS
RESEARCH
The Transportation Problem
Initial Basic Feasible Solution Methods
TRANSPORTATION PROBLEM
Transportation model:
• The transportation problem deals with the
distribution of goods from several points
of supply (sources) to a number of points
of demand (destinations)
• Usually we are given the capacity of goods
at each source and the requirements at
each destination
• Typically the objective is to minimize total
transportation and production costs
TRANSPORTATION PROBLEM
A typical transportation problem contains
Inputs:
Sources with availability
Destinations with requirements
Unit cost of transportation from various sources to
destinations
Objective:
To determine schedule of transportation to minimize
total transportation cost
TRANSPORTATION PROBLEM
There are three basic methods to find the initial
solution for a balanced Transportation Problem
 Northwest
Corner Method
 Least Cost Method
 Vogel’s Method
NORTHWEST CORNER METHOD
1.
2.
3.
4.
Select the northwest corner cell of the transportation
table and allocate as many units as possible equal to
the minimum between availability supply and demand
requirements
Adjust the supply and demand numbers in the
respective rows and columns allocation
If the supply for the first row is exhausted ,then move
down to the first cell in the second row and first column
and go to step 2.
If the demand for the first column is satisfied, then
move horizontally to the next cell in the second column
and first row and go to step 2
NORTHWEST CORNER METHOD
5.
6.
If for any cell, supply equals demand, then the next
allocation can be made in cell either in the next row or
column.
Continue the procedure until the total available quantity
is fully allocated to the cells as required.
NORTHWEST CORNER RULE
TO
FROM
A
B
1
3
SUPPLY
\$5
\$4
\$3
\$8
\$4
\$3
\$7
\$5
100
200
100
\$9
C
DEMAND
2
100
300
200
Total Cost = \$4,200
200
200
100
300
300
700
LEAST COST METHOD
Although the Northwest Corner Rule is the
easiest, it is not the most efficient starting
point because our objective is not included in
the process.


The Northwest Corner Method does not utilize
shipping costs. It can yield an initial bfs easily
but the total shipping cost may be very high.
The least cost method uses shipping costs in
order come up with a bfs that has a lower cost.
LEAST COST METHOD
Steps of Least Cost Method
1.
2.
3.
Select the cell with the minimum cell cost in the tableau
and allocate as much to this cell as possible, but within
the supply and demand constraints.
Select the cell with the next minimum cell-cost and
allocate as much to this cell as possible within the
demand and supply constraints.
Continue the procedure until all of the supply and
demand requirements are satisfied. In a case of tied
minimum cell-costs between two or more cells, the tie
can be broken by selecting the cell that can
accommodate the greater quantity.
NORTHWEST CORNER RULE
TO
FROM
1
\$5
A
DEMAND
3
SUPPLY
\$4
\$3
\$4
\$3
100
\$8
B
C
2
100
\$9
200
\$7
\$5
300
300
200
Total Cost = \$4,100
200
100
300
300
700
VOGEL’S APPROXIMATION METHOD



Vogel’s Approximation Method (VAM) is not as simple
as the northwest corner method, but it provides a
very good initial solution, often one that is the
optimal solution
VAM tackles the problem of finding a good initial
solution by taking into account the costs associated
with each route alternative
To apply VAM, we first compute for each row and
column the penalty faced if we should ship over the
second-best route instead of the least-cost route
VOGEL’S APPROXIMATION METHOD
Using 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.
 Assign the highest possible value to that variable
 Compute new penalties and use the same procedure.
VOGEL’S APPROXIMATION METHOD

Transportation table with VAM row and column
differences shown
TO
FROM
3
0
0
A
B
C
D
100
E
200
\$8
\$9
F
TOTAL REQUIRED
\$5
300
100
100
200
OPPORTUNITY
COSTS
TOTAL
AVAILABLE
\$4
\$3
\$4
\$3
\$7
200
200
\$5
100
1
300
1
300
2
700
VOGEL’S APPROXIMATION METHOD
VAM Step 2. identify the row or column with the greatest
opportunity cost, or difference (column A in this example)
VAM Step 3.Assign as many units as possible to the lowestcost square in the row or column selected
VAM Step 4. Eliminate any row or column that has been
completely satisfied by the assignment just made by placing
Xs in each appropriate square
VAM Step 5. Recompute the cost differences for the
transportation table, omitting rows or columns eliminated in the
previous step
VOGEL’S APPROXIMATION METHOD

VAM assignment with D’s requirements satisfied
TO
FROM
D
31
03
02
A
B
C
100
E
F
TOTAL REQUIRED
300
\$5
\$4
X
OPPORTUNITY
COSTS
TOTAL
AVAILABLE
\$3
X
\$8
\$4
\$3
\$9
\$7
\$5
200
200
100
1
300
1
300
2
700
VOGEL’S APPROXIMATION METHOD
VAM Step 6. Return to step 2 for the rows and columns
remaining and repeat the steps until an initial feasible solution
has been obtained
 In this case column B now has the greatest difference, 3
 We assign 200 units to the lowest-cost square in the
column, EB
 We recompute the differences and find the greatest
difference is now in row E
 We assign 100 units to the lowest-cost square in the
column, EC
VOGEL’S APPROXIMATION METHOD

Second VAM assignment with B’s requirements satisfied
31
03
02
A
B
C
TO
FROM
D
100
\$8
E
\$9
F
TOTAL REQUIRED
\$5
300
\$4
X
200
X
200
OPPORTUNITY
COSTS
TOTAL
AVAILABLE
\$3
X
\$4
\$3
\$7
\$5
200
100
1
300
1
300
2
700
VOGEL’S APPROXIMATION METHOD

Third VAM assignment with E’s requirements satisfied
TO
A
FROM
D
100
E
X
\$5
\$8
\$9
F
TOTAL REQUIRED
B
300
\$4
X
200
\$4
\$3
X
100
\$7
X
200
TOTAL
AVAILABLE
C
\$3
\$5
200
100
300
300
700
VOGEL’S APPROXIMATION METHOD

Final assignments to balance column and row
requirements
TO
A
FROM
D
100
E
X
F
200
TOTAL REQUIRED
B
\$5
\$8
300
\$9
\$4
X
200
\$4
\$7
X
TOTAL
AVAILABLE
C
\$3
X
100
100
200
Total Cost = \$3,900
200
\$3
\$5
100
300
300
700
```