\[F\]
factories and \[W\]
warehouses, and the number of used routes from factories to warehouses in a solution is less than \[F+W-1\]
, then this solution is degenerate.We resolve the degeneracy by allocating a small shipment
\[x\]
to an unused route. We calculate the change in cost, and if it is negative, We maximise \[x\]
such that no entries are negative and start again.Example: A transportation problem has cost structure and trial solution below.
Key cost/units
Source | |||||
\[F_1\] |
\[F_2\] |
\[F_3\] |
Demand | ||
\[W_2\] |
0.90/0 | 1.00/5 | 1.00/0 | 5 | |
Destination | \[W_2\] |
1.00/20 | 1.40/0 | 0.80/0 | 20 |
\[W_3\] |
1.30/0 | 1.00/10 | 0.80/10 | 20 | |
20 | 15 | 10 | 45 |
\[x\]
be transported using previously unused route \[F_1W_1\]
. So that demand and supply constraints are satisfied, and the table has no negative for the quantity transported along each route, we MUST have the table below.Source | |||||
\[F_1\] |
\[F_2\] |
\[F_3\] |
Demand | ||
\[W_2\] |
0.90/x | 1.00/5-x | 1.00/0 | 5 | |
Destination | \[W_2\] |
1.00/20-x | 1.40/x | 0.80/0 | 20 |
\[W_3\] |
1.30/0 | 1.00/10 | 0.80/10 | 20 | |
20 | 15 | 10 | 45 |
\[0.90x-1.00x+1.00x-1.40x=0.30x\]
. This is an increase in cost. Evaluating the other unused routes, looking for a decrease in costs results in the final solution below.Source | |||||
\[F_1\] |
\[F_2\] |
\[F_3\] |
Demand | ||
\[W_2\] |
0.90/0 | 1.00/0-x | 1.00/0 | 5 | |
Destination | \[W_2\] |
1.00/20-x | 1.40/09 | 0.80/0 | 20 |
\[W_3\] |
1.30/0 | 1.00/10 | 0.80/10 | 20 | |
20 | 15 | 10 | 45 |