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Hungarian Algorithm: 1:for each i 2 [ n ] 2: do u[ i ] A maxj w[ (i, j )] 3: v[ i ]A 0 4:while ( G_(u,v)) has no perfect matching 5: do X A minimum vertex cover of ( G_(u,v)) 6:“ A min u[ i ] + v[ j ] - w[ i, j ] : fi, jg2 (A n X Є (B n X)) 7:for each I 2 A nX 8: do u[ i ] A u[ i ] –“ 9:For each I 2 B \ X 10: do v[ i ] A v[ i ] +“ 11:return perfect matching of ( G_(u,v)) |
Three workers to do Three task at the same time. Apply the Hungarian algorithm the cost matrix is,
TASK_A | TASK_B | TASK_C | |
WORKER_1 | 250 | 450 | 350 |
WORKER_2 | 400 | 400 | 350 |
WORKER_3 | 200 | 500 | 250 |
STEP 1:
Identify minimum in each row (show in ‘green’)
TASK_A | TASK_B | TASK_C | ROW_MINS | |
WORKER_1 | 250 | 450 | 350 | 250 |
WORKER_2 | 400 | 400 | 350 | 350 |
WORKER_3 | 200 | 500 | 250 | 200 |
Substract row mins from each element in their respective rows:
TASK_A | TASK_B | TASK_C | |
WORKER_1 | 0 | 200 | 100 |
WORKER_2 | 50 | 50 | 0 |
WORKER_3 | 200 | 300 | 50 |
COL_MINS | 0 | 50 | 0 |
Identify minimum in each column (show in ‘green’).
Substract column mins from each element from their respective columns:
TASK_A | TASK_B | TASK_C | |
WORKER_1 | 0 | 150 | 100 |
WORKER_2 | 50 | 0 | 0 |
WORKER_3 | 0 | 250 | 50 |
This is reduced cost matrix.
STEP 2:
Find the minimum no of horizontal and vertical lines to cover all 0’s:
TASK_A | TASK_B | TASK_C | |
WORKER_1 | 0 | 150 | 100 |
WORKER_2 | 50 | 0 | 0 |
WORKER_3 | 0 | 250 | 50 |
Since < 3 lines suffice, continue to step2.
Identify min of uncovered elements: uncovered min = 50
Substract min of uncovered matrix elements from the uncovered rows:
TASK_A | TASK_B | TASK_C | |
WORKER_1 | -50 | 150 | 100 |
WORKER_2 | 50 | 0 | 0 |
WORKER_3 | -50 | 250 | 50 |
Add the min of the uncovered matrix elements to the covered columns:
TASK_A | TASK_B | TASK_C | |
WORKER_1 | 0 | 100 | 50 |
WORKER_2 | 100 | 0 | 0 |
WORKER_3 | 0 | 200 | 0 |
STEP 3:
Find the minimum number of horizontal and vertical lines to cover all 0’s:
TASK_A | TASK_B | TASK_C | |
WORKER_1 | 0 | 100 | 50 |
WORKER_2 | 100 | 0 | 0 |
WORKER_3 | 0 | 200 | 0 |
Since it takes 3 linkes to cover all zero’s, STOP: an assignment is possible.
Construct a valid assignment of row to column labels using only matrix zeros:
TASK_A | TASK_B | TASK_C | |
WORKER_1 | 0 | 100 | 50 |
WORKER_2 | 100 | 0 | 0 |
WORKER_3 | 0 | 200 | 0 |