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TRANSPORTATION SIMPLEX METHOD: A SPECIAL-PURPOSE SOLUTION PROCEDURE 297
Table 7.14 MODI Evaluation of Each Cell in Solution
v j
u i 3 2 –1 0
3 2 7 6
0 3 500 1 500 8 6
7 5 2 3
3 1 2 500 2 000 1 500
2 5 4 5
–1 2 500 4 6 6
Table 7.16 shows the initial feasible solution obtained using the minimum cost
method for a transportation problem involving m ¼ 3 origins and n ¼ 3 destina-
tions. To use the MODI method for this problem, we must have m + n 1 ¼ 3+
3 1 ¼ 5 occupied cells. Since the initial feasible solution has only four occupied
cells, the solution is degenerate.
Suppose that we try to use the MODI method to calculate row and column
indexes to begin phase II for this problem. Setting u 1 ¼ 0 and calculating the
column indexes for each occupied cell in row 1, we obtain v 1 ¼ 3 and v 2 ¼ 6 (see
Table 7.16). Continuing, we then calculate the row indexes for all occupied cells in
columns 1 and 2. Doing so gives u 2 ¼ 5 6 ¼ 1. At this point, we cannot calculate
any more row and column indexes because no cells in columns 1 and 2 of row 3 and
no cells in rows 1 or 2 of column 3 are occupied.
To calculate all the row and column indexes when fewer than m + n 1cells
areoccupied,wemustcreateoneor more‘artificially’ occupied cells with a flow of
zero. In Table 7.16 we must create one artificially occupied cell to have five
occupied cells. Any currently unoccupied cell can be made an artificially occupied
cell if doing so makes it possible to calculate the remaining row and column
indexes. For instance, treating the cell in row 2 and column 3 of Table 7.16 as an
Table 7.15 Optimal Solution to the Foster Electronics Transportation Problem
Route
From To Units Shipped Cost per Unit, E Total Cost, E
Czech Republic Boston 3 500 3 10 500
Czech Republic Dubai 1 500 2 3 000
Brazil Dubai 2 500 5 12 500
Brazil Singapore 2 000 2 4 000
Brazil London 1 500 3 4 500
China Boston 2 500 2 5 000
39 500
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