Question

In: Statistics and Probability

A company has 4 zones(north, south, east, west) and 4 supervisors available for assignment. It is...

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.

Solutions

Expert Solution

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

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