written 7.8 years ago by
teamques10
★ 67k
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•
modified 7.8 years ago
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Here one column is unbalanced, so we add fifth column with zero values.
60 |
20 |
40 |
82 |
0 |
57 |
35 |
44 |
92 |
0 |
20 |
30 |
35 |
70 |
0 |
45 |
25 |
64 |
100 |
0 |
38 |
45 |
52 |
85 |
0 |
This problem is of maximization, so covering it to minimization type by subtracting biggest value from table from each value.
40 |
80 |
60 |
18 |
100 |
43 |
65 |
56 |
8 |
100 |
80 |
70 |
65 |
30 |
100 |
55 |
75 |
36 |
0 |
100 |
62 |
55 |
48 |
15 |
100 |
Applying Row operation
22 |
62 |
42 |
0 |
82 |
35 |
57 |
48 |
0 |
92 |
50 |
40 |
35 |
0 |
70 |
55 |
75 |
36 |
0 |
100 |
47 |
40 |
33 |
0 |
85 |
Applying column operation as well as assigning jobs
No. of allocation not equal to no. of rows, so this is not feasible solution, so moving towards optimality.
Here we will draw four lines to cover all zero’s and crossed zero’s. Minimum uncovered value is 3. Subtracting 3 from all uncovered values & adding ‘3’ at the junction where two lines cross each other.
No. of allocations are not equal to no. of rows. Here we can’t draw four lines which will cover all zero’s and crossed zero’s. Therefore assigning in same matrix by priority rule Right-Lest & Top-Bottom.
No. of allocation = No. of rows
It is a feasible solution. No match for city ‘C’, since fifth column was dummy column.
Cities |
Matches |
A |
1 |
B |
4 |
C |
- |
D |
3 |
E |
2 |