In: Statistics and Probability
A company has 4 zones(north, south, east, west) and 4 supervisors available for assignment. It is estimated that a typical supervisor operating in each zone would bring in the following annual turnover: north:126000, south 105000, east 84000, west 63000. The 4 supervisors are also considered to differ in supervisory competencies. It is estimated that working under the same conditions, the company's yearly turnover would be proportionately as follows: supervisor I:7, supervisor II: 5, supervisor III:5, supervisor IV: 4. How should be the company allocate the best supervisor to the various zones in order to maximize its annual turnover. Use Hungarian method.
Here, a company has 4 zones (north,south,east,west) and 4 supervisors available for assignment.
Estimated annual turover for each zone is as follows :
north : 126000 , south : 105000 , east : 84000 , west : 63000
Also, it is estimated that, the company's early turnover would be proportionaly to supervisor as follows:
supervisor I : 7 , supervisor II : 5 , supervisor III : 5 , supervisor IV : 4
So, by using above information we can draw assignment matrix as follows:
supervisor | north | south | east | west |
I | 42000 | 35000 | 28000 | 21000 |
II | 30000 | 25000 | 20000 | 15000 |
III | 30000 | 25000 | 20000 | 15000 |
IV | 24000 | 20000 | 16000 | 12000 |
Above values are calculated as -
For ex; annual turnover of northzone = 126000
Also, we know that the company's early turnover is proportional to supervisor.
i.e. supervisor I : 7 , supervisor II : 5 , supervisor III : 5 , supervisor IV : 4
Now, let x be the proportional multiplier constant .
i.e. 7x + 5x + 5x + 4x = 126000 : this implies that x = 6000
Now, for supervisor I, annual turnover for north zone is 7*6000 i.e 42000.
similarly, we can calculate all other values.
Now, we have to use hungarian method .
Here, the problem is of maximization type and convert it into minimization by subtracting it from maximum value 42000.
supervisor | north | south | east | west |
I | 0 | 7000 | 14000 | 21000 |
II | 12000 | 17000 | 22000 | 27000 |
III | 12000 | 17000 | 22000 | 27000 |
IV | 18000 | 22000 | 26000 | 30000 |
Here, the given problem is balanced.
step 1- Find out each row minimum and subtract it from that row.
supervisor | north | south | east | west |
I | 0 | 7000 | 14000 | 21000 |
II | 0 | 5000 | 10000 | 15000 |
III | 0 | 5000 | 10000 | 15000 |
IV | 0 | 4000 | 8000 | 12000 |
step 2 - find out each column minimum and subtact it from that column.
supervisor | north | south | east | west |
I | 0 | 3000 | 6000 | 3000 |
II | 0 | 1000 | 2000 | 3000 |
III | 0 | 1000 | 2000 | 3000 |
IV | 0 | 0 | 0 | 0 |
step 3 - Make assignment in the opportunity cost
(1) Rowwise cell (I,north) is assigned, so columnwise cell (II,north),(III,north),(IV,north) crossed off(color).
(2) Columnwise cell (IV,south) is assigned, so rowwise cell (IV,east),(IV,west) crossed off(color).
Rowwise & columnwise assignment shown in table :
supervisor | north | south | east | west |
I | [0] | 3000 | 6000 | 3000 |
II | 0 | 1000 | 2000 | 3000 |
III | 0 | 1000 | 2000 | 3000 |
IV | 0 | [0] | 0 | 0 |
Step 4 - Number of assignments = 2, number of rows = 4 which is not equal, so solution is not optimal.
step 5 - Draw a set of horizontal and vertical lines to cover all the 0.( we use color here to cover all 0)
supervisor | north | south | east | west |
I | [0] | 3000 | 6000 | 3000 |
II | 0 | 1000 | 2000 | 3000 |
III | 0 | 1000 | 2000 | 3000 |
IV | 0 | [0] | 0 | 0 |
step 6 - Develop the new revised opportunity cost table by selecting the smallest element, among the cells not colored (say k= 1000)
subtract k= 1000 from every element in the cell not colored, add k = 1000 to every element in the intersection cell of two lines
supervisor | north | south | east | west |
I | 0 | 2000 | 5000 | 2000 |
II | 0 | 0 | 1000 | 2000 |
III | 0 | 0 | 1000 | 2000 |
IV | 1000 | 0 | 0 | 0 |
Again repeat above steps until no. of assignments = no. of rows
step 3 - Make assignment in the opportunity cost
(1) Rowwise cell (I,north) is assigned, so columnwise cell (II,north),(III,north),crossed off(color).
(2) Rowwise cell (II,south) is assigned, so columnwise cell (III,south),(IV,south),crossed off(color).
(3) Columnwise cell (IV,east) is assigned. so rowwise cell (IV,west) crossed off (color).
Rowwise & columnwise assignment shown in table :
supervisor | north | south | east | west |
I | [0] | 2000 | 5000 | 2000 |
II | 0 | [0] | 1000 | 2000 |
III | 0 | 0 | 1000 | 2000 |
IV | 1000 | 0 | [0] | 0 |
Step 4 - Number of assignments = 3, number of rows = 4 which is not equal, so solution is not optimal.
step 5 - Draw a set of horizontal and vertical lines to cover all the 0.( we use color here to cover all 0)
supervisor | north | south | east | west |
I | [0] | 2000 | 5000 | 2000 |
II | 0 | [0] | 1000 | 2000 |
III | 0 | 0 | 1000 | 2000 |
IV | 1000 | 0 | [0] | 0 |
step 6 - Develop the new revised opportunity cost table by selecting the smallest element, among the cells not colored (say k= 1000)
subtract k= 1000 from every element in the cell not colored, add k = 1000 to every element in the intersection cell of two lines
supervisor | north | south | east | west |
I | 0 | 2000 | 4000 | 1000 |
II | 0 | 0 | 0 | 1000 |
III | 0 | 0 | 0 | 1000 |
IV | 2000 | 1000 | 0 | 0 |
step 3 - Make assignment in the opportunity cost
(1) Rowwise cell (I,north) is assigned, so columnwise cell (II,north),(III,north),crossed off(color).
(2) Rowwise cell (II,south) is assigned, so columnwise cell (III,south) crossed off(color).
(3) Columnwise cell (IV,east) is assigned. so rowwise cell (IV,west) crossed off (color).
(4) Columnwise cell (IV,west) is assigned.
Rowwise & columnwise assignment shown in table :
supervisor | north | south | east | west |
I | [0] | 2000 | 4000 | 1000 |
II | 0 | [0] | 0 | 1000 |
III | 0 | 0 | [0] | 1000 |
IV | 2000 | 1000 | 0 | [0] |
Step 4 - Number of assignments = 4, number of rows = 4 which is equal, so solution is optimal.
optimal assignments are
supervisor | north | south | east | west |
I | [0] | |||
II | [0] | |||
III | [0] | |||
IV | [0] |
optimal solution is
zone | supervisor | cost |
north | I | 42000 |
south | II | 25000 |
east | III | 20000 |
west | Iv | 12000 |
Total | 99000 |