The Vogel Approximation Method [VAM] is an iterative procedure for computing an initial basic feasible solution of a transportation problem.
Initial
Basic Feasible Solution – Vogel’s Approximation Method [Vam]
The Vogel Approximation Method [VAM] is an
iterative procedure for computing an initial basic feasible solution of a
transportation problem. This method is preferred over the two methods discussed
in the previous sections, because the initial basic feasible solution obtained
by this method is either optimal or very nearer to the optimal solution.
Therefore the amount of time required to arrive at the optimum solution is
greatly reduced.
Vogel’s
Approximation Method [Vam]:– Step-by-step procedure
Step 1
Compute a penalty for each row in the
transportation table. The penalty for a given row is the difference between the
smallest cost and the next
smallest cost in that particular row; write this difference (penalty) along the
side of the table against the corresponding row.
Step 2
Compute a penalty for each column of the
transportation table. Identify the cell having minimum and next to minimum
transportation cost in each column and write the difference (penalty) against
the corresponding column.
Step 3
1. Identify
the row or column with the largest penalty; make a mark in the penalty [circle
the respective penalty for future references]
2. In this
identified row or column, choose the cell, which is having the lowest
transportation cost
3. Allocate
the maximum possible quantity to the cell having lowest cost in that row or
column
• Exhaust
either the supply at a particular source or satisfy demand at a warehouse.
4. If there
is a tie between / among the penalties computed, select that row/column which
has minimum cost.
5. If there
is a tie in the minimum cost also select that row/column which will have
maximum possible assignments. It will considerably reduce computational work.
6. Also, you
have an option of breaking the tie between / among the penalties corresponding
to two or more rows or columns by breaking the tie arbitrarily
Step 4
Reduce
the row supply or the column demand by the amount assigned to the cell.
Step 5 1. If the
row supply is now zero, eliminate the row; 2. If the
column demand is now zero, eliminate the column 3. If both
the row supply and the column demand is zero, eliminate either the row or
column. 4. After
eliminating the row or column, in the left out row or column, enter the balance
as zero [this will help in avoiding degeneracy, which is an issue while solving
the problem for optimality check] 5. In future
iterations, if the row or column with zero is selected, follow the same
procedure, as if you are allocating a quantity; only the difference is, you
will allocate zero as quantity to a cell. This will avoid number of other
steps, which are slightly complicated. Step 6 Recompute the row and column difference for the
reduced transportation table, which has obtained by omitting rows or columns
crossed out in the preceding step. Step 7 Repeat the above procedure until the entire supply
at factories are exhausted to satisfy demand at different warehouses. For our better understanding of the method, we
solve the same Adani Power limited problem which introduced and solved by
North-west corner method as well as Least Cost Methods. 1. The
lowest and next lowest for each row is located and penalty is computed and
marked against the respective rows. 2. Similarly,
the lowest and next lowest is located and penalty is computed for each column
and marked against the respective columns.
3. The
maximum among all the penalties [rows as well as columns together] is marked; • In case
of a tie, you may choose the row or column that is having the lowest
transportation cost cell;
4. That row
or column should be allocated first; the allocation is made to the cell, which
is having lowest transportation cost; in case of tie between cells, you can
break it arbitrarily. In the
above table, it is 4, appearing in the 3rd row; the lowest transportation cost
5 appears in the cell (3, 4). Availability at the plant-3 is 40 units and city-4 requires 30 units; so
we supply the entire 30 units from plant-3; Now the city-4 is eliminated from further consideration and plant-3
availability is reduced to 10. The transportation table is modified with the above changes and placed
below; in that table, again the process is repeated [while doing future
calculations, we will not consider costs in the column-4 anymore] and row/
column penalty-2 is computed.
In the above table, highest penalty is 5, appearing
again in the 3rd row; the lowest transportation cost 9 appears in the cell (3,
2). Availability at the plant-3 is 10 units and city-2
requires 20 units; so we supply the entire 10 units from plant-3; Now the plant-3 is eliminated from further consideration
and city-2 requirement is reduced to 10. The transportation table is modified with the above
changes and placed below; in that table, again the process is repeated [while
doing future calculations, we will not consider costs in the column-4 & Row-3
anymore] and row/ column penalty-3 is computed.
In the above table, highest penalty is 6, appearing again in the 2nd
column; the lowest transportation cost 6 appears in the cell (1, 2). Availability at the plant-1 is 35 units and city-2 requires 10 units; so
we supply the entire 10 units from plant-1; Now the city-2 is eliminated from further consideration and plant-1
requirement is reduced to 25. The transportation table is modified with the above changes and placed
below; in that table, again the process is repeated [while doing future
calculations, we will not consider costs in the column-4, column-2 & Row-3
anymore] and row/ column penalty-4 is computed.
In the above table, highest penalty is 6, appearing
again in the 2nd column; the lowest transportation cost 6 appears in the cell
(1, 2). Availability at the plant-1 is 35 units and city-2
requires 10 units; so we supply the entire 10 units from plant-1; Now the city-2 is eliminated from further consideration
and plant-1 requirement is reduced to 25. Now we left out with only one column, city-3. This
is serviced by plant-1, 25 units and plant-2, by 5 units. Thus, we fulfil the
requirements of all the cities and from the plants. We made the initial allocation
through Vogel’s approximation method.
Now
the allocations are:X21 = 45 | X23 = 5 |
X32 = 10 | X34 = 30 |
The total transportation cost =
(10 X 6)
+ (25 X 10) + (45 X 9) + (5 X 13) + (10 X 9) + (30 X 5) =
60 + 250
+ 405 + 65 + 90 + 150 =
1020 You can note that the total
cost of allocation through the North West corner method is 1180; least cost
method is 1080, whereas through VAM, we get 1020, which is a superior basic feasible solution than a solution
obtained through other methods.
Note: You may
note that the number of basic cells in any transportation problem should be equal to (row+column-1).
Tags : Operations Management - Transportation / Assignment & Inventory Management
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