You would have noticed that the initial basic feasible solution obtained by various methods provide different starting solution.
Optimality Checking Of The
Initial Basic Feasible Solution: Modi [Modified Distribution Method] Method
You would have noticed that the initial basic
feasible solution obtained by various methods provide different starting
solution. Once you obtain the initial solution by any of the method, the next
step is to check the optimality of the solution obtained and if not, improve
the solution and obtain the optimal solution. Two popularly known methods are ‘Stepping
Stone Method’ and ‘MODI Method or u-v method.
In the stepping stone method, we have to draw as
many closed paths as equal to the unoccupied cells for their evaluation. To the
contrary, in MODI method, only closed path for the unoccupied cell with highest
opportunity cost is drawn. Here, we will discuss only the MODI Method, which is
more scientific and approaches the optimal solution faster.
Modi
[Modified Distribution Method] Method - Step-by-step procedure
Step1
Determine an initial basic feasible solution using
any one of the three methods:
North West Corner Rule Matrix Minimum Method Vogel Approximation Method Step2 Each row, assign one ‘dual’ variable, say- u1, u2,
u3…; for each column, assign on dual variable – say, v1, v2, v3… Now using the basic cells [which are assigned
through any one of the three methods], and the transportation costs of those
basic cells -Cij, we will determine the values of these u_{i} and v_{j}.
Determine
the values of dual variables, ui and vj, using u_{i} + v_{j} =
Cij Since the net evaluation is zero for all basic cells, it follows that z_{ij}
- c_{ij} = u_{i} +v _{j} - C_{ij} , for all
basic cells (i, j). So we can make use of this relation to find the values of u_{i}
and v_{j} Step3 Compute the opportunity cost, for those cells,
which are not allocated for any goods to be transported [known as non-basic
cells], using (u_{i} + v_{j}) - Cij Step4 1. Check the
sign of each opportunity cost. 2. If the
opportunity costs of all these unoccupied cells / non-basic cells are either
negative or zero, the given solution is the optimal solution.
3. On the
other hand, if one or more unoccupied cell has positive entry / opportunity
cost, it is an indication that the given solution is not an optimal solution;
it can be improved and further savings in transportation cost are possible. Step5 1. Select
the unoccupied cell with the highest positive opportunity cost as the cell to
be included in the next solution.
2. This cell
has been left out / missed out by the initial solution method.
3. If this
cell is allocated, it will bring down the overall transportation cost Step6 Draw a closed path or loop for the unoccupied cell
selected in the previous step. A loop in a transportation table is a collection
of basic cells and the cell, which is to be converted as basic cell. It is
formed in such a way that, it has only even number cells in any row or column. 1. After
identifying the entering variable X_{rs}, form a loop; this loop starts
at the non-basic cell (r, s) connecting only basic cells.
2. Assign
alternate plus and minus signs at the unoccupied cells on the corner points of
the closed path with a plus sign at the cell being evaluated 3. Determine
the maximum number of units that should be shipped to this unoccupied cell. 4. The
smallest value with a negative position on the closed path indicates the number
of units that can be shipped to the entering cell.
5. Now, add this quantity to all the cells on
the corner points of the closed path marked with plus signs, and subtract it from those cells marked
with minus signs. 6. In this
way, an unoccupied cell becomes an occupied cell. 7. Other
basic cells, the quantity allocated are modified, in such a way that without
affecting the row availability and column / market requirements 8. Such a
closed path exists and is unique for any non-degenerate solution. Please
note that the right angle turn in this path is permitted only at occupied cells
and at the original unoccupied cell. Step7 Repeat the whole procedure until
an optimal solution is obtained. ---------------------- To
understand the MODI Method, let us consider the solution obtained by Least Cost
Method. Step1
Step2 For each row, we assign u_{i}
and for each column, we will assign v_{j} Let us
assume any one of the ui or vj as zero; preferably a row or column having
maximum number of basic cells Since there is a tie, we break it
arbitrarily. Let us assume, U1= 0 We know that U1+V1= 8 [ui+vj= Cij
for all the basic cells] These findings are
incorporated in the following table
Step3 Let us compute the opportunity cost, for those
cells, which are not allocated for any goods to be transported [known as
non-basic cells], using (u_{i} + v_{j}) - Cij For example, cell (1, 3) Net evaluation for cell (1, 3) = ui+vj-Cij
= u1+v3-C13 = [0 + 12-10] = 2
Net evaluation for cell (1, 4) =
ui+vj-Cij
= u1+v4-C14 = [0 + 1-9] = -8 Similarly for all other
cells, the net evaluation / opportunity cost is computed and placed in left
hand top corner of each non-basic cell; the table is placed below;
Step4 Since all the net evaluation / opportunity costs are not negative, it is
an indication of the solution is not in the optimum stage; it can be improved Among the positive entries, the maximum positive appears in the cell (1,
3), which is 2. So this cell to be converted as a basic cell. Step5 1. After
identifying the entering variable X_{rs}, form a loop; this loop starts
at the non-basic cell (r, s) connecting only basic cells. 2. Assign
alternate plus and minus signs at the unoccupied cells on the corner points of
the closed path with a plus sign at the cell being evaluated 3. Determine
the maximum number of units that should be shipped to this unoccupied cell. 4. The
smallest value with a negative position on the closed path indicates the number
of units that can be shipped to the entering cell. 5. Now, add
this quantity to all the cells on the corner points of the closed path marked
with plus signs, and subtract it from those cells marked with minus signs. 6. In this
way, an unoccupied cell becomes an occupied cell. Other basic cells, the
quantity allocated are modified, in such a way that without affecting the row
availability and column / market requirements. The loop formed is shown below;
In
the loop, between the cells with negative sign, lowest quantity allocated is
15, in the cell (1, 1); it should be shifted to the cell (1, 3), where there is
a positive sign.
Thus,
wherever there is a positive sign, the quantity is added; in the negative
signed cell, it is subtracted. Also, when we move the quantities, we should not
move to 3 cells in a row or column; the movement of goods will be always in
terms of even numbers. You may notice in the above table, while moving the
goods, we skipped the basic cell (1, 2) from the loop. Other allocations remain
as such; we will now compute the cost of this allocation. Now the allocations are: | |
X12 = 20 | X23= 5 |
X13 = 15 | X33 = 10 |
X21 = 45 | X34 = 30 |
The total transportation cost =
(20 X 6)
+ (15 X 10) + (45 X 9) + (5 X 13) + (10 X 16) + (30 X 5) =
120 + 150
+ 405 + 65 + 160 + 150 =
1050 You may note that Least Cost method which has an initial cost of
allocation as 1080, reduced by Rupees 30 by MODI / u-v method. Once again, we check, whether the solution is optimum or not; we start
again to compute the ui and vj for each row and column;
We assume
V3 = 0, since it has 3 basic cells in the column and repeat the same process. Now we will compute the Net
evaluation / opportunity cost for the non-basic cells and try to form the loop,
if there are positive opportunity cost in any of the cells.
The maximum positive appears in the cell (3, 2); it
should be converted as basic cell and the loop is shown in the above table. The
modified distribution is shown below and the cost is computed.
You may note that the cost of
allocation is =(10X 6) + (25X 10) + (45 X 9) +
(5 X 13) + (10 X 9) + (30 X 5) =
60 + 250
+ 405 + 65 + 90 + 150 =
1020 Thus, the total cost is reduced by modifying the
distribution / allocations made earlier. Now once again, we will check whether
the solution is optimum or not. Table-6 Finding ui / vj and checking optimality of the solution
Since, the net evaluation / opportunity cost is
negative for all the non-basic cells, it is indication that the solution is in
the optimum stage. Variations in the Transportation Models
Unbalanced Transportation Models: If the demand is
not equal to supply, the transportation problems are known as unbalanced
transportation problems. You have to add a dummy row / column with zero as
transportation cost with the shortfall (difference between supply and demand)
as the quantity available in the origin or marketing requirements. Then, in the usual manner, using any one of the
initial solution methods will be applied and initial solution will be obtained.
Tags : Operations Management - Transportation / Assignment & Inventory Management
Last 30 days 1983 views