Assignment problem using Hungarian method-1 calculator

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Solution

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1. A computer centre has 3expert programmers. The centre wants 3 application programmes to be developed. The head of thecomputer centre, after studying carefully the programmes to be developed, estimates the computer time in minutes required by the experts for the application programmes as follows.

		Programmers
		A	B	C
Programmes	1	6	3	5
	2	5	9	2
	3	5	7	8

2. An airline company has drawn up a new flight schedule involving five flights. To assist in allocating five pilots to the flights, it has asked them to state their preference scores by giving each flight a number out of 10. The higher the number, the greater is the preference. Certain of these flights are unsuitable to some pilots owing to domestic reasons. These have been marked with a -.

		Flight Number
		I	II	III	IV	V
Pilot	A	8	2	-	5	4
	B	10	9	2	8	4
	C	5	4	9	6	-
	D	3	6	2	8	7
	E	5	6	10	4	3

What should be the allocation of the pilots to flights in order to meet as many preferences as possible?

Example

1. Find Solution of Assignment problem using Hungarian method (MIN case)
_Work\^Job I II III IV V
A 10 5 13 15 16
B 3 9 18 13 6
C 10 7 2 2 2
D 7 11 9 7 12
E 7 9 10 4 12

Solution:
The number of rows = 5 and columns = 5

	`I`	`II`	`III`	`IV`	`V`
`A`	10	5	13	15	16
`B`	3	9	18	13	6
`C`	10	7	2	2	2
`D`	7	11	9	7	12
`E`	7	9	10	4	12

Here given problem is balanced.

Step-1: Find out the each row minimum element and subtract it from that row

	`I`	`II`	`III`	`IV`	`V`
`A`	5 `5=10-5`	0 `0=5-5`	8 `8=13-5`	10 `10=15-5`	11 `11=16-5`	(-5) Minimum element of `1^(st)` row
`B`	0 `0=3-3`	6 `6=9-3`	15 `15=18-3`	10 `10=13-3`	3 `3=6-3`	(-3) Minimum element of `2^(nd)` row
`C`	8 `8=10-2`	5 `5=7-2`	0 `0=2-2`	0 `0=2-2`	0 `0=2-2`	(-2) Minimum element of `3^(rd)` row
`D`	0 `0=7-7`	4 `4=11-7`	2 `2=9-7`	0 `0=7-7`	5 `5=12-7`	(-7) Minimum element of `4^(th)` row
`E`	3 `3=7-4`	5 `5=9-4`	6 `6=10-4`	0 `0=4-4`	8 `8=12-4`	(-4) Minimum element of `5^(th)` row

Step-2: Find out the each column minimum element and subtract it from that column.

	`I`	`II`	`III`	`IV`	`V`
`A`	5 `5=5-0`	0 `0=0-0`	8 `8=8-0`	10 `10=10-0`	11 `11=11-0`
`B`	0 `0=0-0`	6 `6=6-0`	15 `15=15-0`	10 `10=10-0`	3 `3=3-0`
`C`	8 `8=8-0`	5 `5=5-0`	0 `0=0-0`	0 `0=0-0`	0 `0=0-0`
`D`	0 `0=0-0`	4 `4=4-0`	2 `2=2-0`	0 `0=0-0`	5 `5=5-0`
`E`	3 `3=3-0`	5 `5=5-0`	6 `6=6-0`	0 `0=0-0`	8 `8=8-0`
	(-0) Minimum element of `1^(st)` column	(-0) Minimum element of `2^(nd)` column	(-0) Minimum element of `3^(rd)` column	(-0) Minimum element of `4^(th)` column	(-0) Minimum element of `5^(th)` column

Step-3: Make assignment in the opporunity cost table

a. Identify rows with exactly one unmarked 0. Make an assignmment to this single 0 by make a square ( [0] ) around it and cross off all other 0 in the same column.

b. Identify columns with exactly one unmarked 0. Make an assignmment to this single 0 by make a square ( [0] ) around it and cross off all other 0 in the same rows.

c. If a row and/or column has two or more unmarked 0 and one cannot be chosen by inspection, then choose the cell arbitarily.

d. Continue this process until all 0 in rows/columns are either assigned or cross off(

Step-3: Make assignment in the opporunity cost table
(1) Rowwise cell `(A,II)` is assigned

(2) Rowwise cell `(B,I)` is assigned, so columnwise cell `(D,I)` crossed off.

(3) Rowwise cell `(D,IV)` is assigned, so columnwise cell `(C,IV)`,`(E,IV)` crossed off.

(4) Columnwise cell `(C,III)` is assigned, so rowwise cell `(C,V)` crossed off.

Rowwise & columnwise assignment shown in table

	`I`	`II`	`III`	`IV`	`V`
`A`	5	[0] (1) Rowwise cell `(A,II)` is assigned	8	10	11
`B`	[0] (2) Rowwise cell `(B,I)` is assigned so columnwise cell `(D,I)` crossed off.	6	15	10	3
`C`	8	5	[0] (4) Columnwise cell `(C,III)` is assigned so rowwise cell `(C,V)` crossed off.	0 Columnwise `(C,IV)` crossed off because (3) Rowwise cell `(D,IV)` is assigned	0 Rowwise `(C,V)` crossed off because (4) Columnwise cell `(C,III)` is assigned
`D`	0 Columnwise `(D,I)` crossed off because (2) Rowwise cell `(B,I)` is assigned	4	2	[0] (3) Rowwise cell `(D,IV)` is assigned so columnwise cell `(C,IV)`,`(E,IV)` crossed off.	5
`E`	3	5	6	0 Columnwise `(E,IV)` crossed off because (3) Rowwise cell `(D,IV)` is assigned	8

Step-4: Number of assignments = 4, number of rows = 5
Which is not equal, so solution is not optimal.

Step-5: Draw a set of horizontal and vertical lines to cover all the 0

a. Tick(✓) mark all the rows in which no assigned 0.

b. Examine Tick(✓) marked rows, If any 0 cell occurs in that row, then tick(✓) mark that column.

c. Examine Tick(✓) marked columns, If any assigned 0 exists in that columns, then tick(✓) mark that row.

d. Repeat this process until no more rows or columns can be marked.

e. Draw a straight line for each unmarked rows and marked columns.

f. If the number of lines is equal to the number of rows then the current solution is the optimal, otherwise goto step-6

Step-5: Cover the 0 with minimum number of lines
(1) Mark(✓) row `E` since it has no assignment

(2) Mark(✓) column `IV` since row `E` has 0 in this column

(3) Mark(✓) row `D` since column `IV` has an assignment in this row `D`.

(4) Mark(✓) column `I` since row `D` has 0 in this column

(5) Mark(✓) row `B` since column `I` has an assignment in this row `B`.

(6) Since no other rows or columns can be marked, therefore draw straight lines through the unmarked rows `A,C` and marked columns `I,IV`

Tick mark not allocated rows and allocated columns

	`I`	`II`	`III`	`IV`	`V`
`A`	5	[0]	8	10	11
`B`	[0]	6	15	10	3	✓(5) (5) Mark(✓) row `B` since column `I` has an assignment in this row `B`.
`C`	8	5	[0]	0	0
`D`	0	4	2	[0]	5	✓(3) (3) Mark(✓) row `D` since column `IV` has an assignment in this row `D`.
`E`	3	5	6	0	8	✓(1) (1) Mark(✓) row `E` since it has no assignment
	✓ (4) (4) Mark(✓) column `I` since row `D` has 0 in this column			✓ (2) (2) Mark(✓) column `IV` since row `E` has 0 in this column

Step-6: Develop the new revised opportunity cost table

a. Select the minimum element, say k, from the cells not covered by any line,

b. Subtract k from each element not covered by a line.

c. Add k to each intersection element of two lines.

Step-6: Develop the new revised table by selecting the smallest element, among the cells not covered by any line (say k = 2)
Subtract k = 2 from every element in the cell not covered by a line.
Add k = 2 to every elment in the intersection cell of two lines.

	`I`	`II`	`III`	`IV`	`V`
`A`	7 `7=5+2` intersection cell of two lines	0 cell covered by a line	8 cell covered by a line	12 `12=10+2` intersection cell of two lines	11 cell covered by a line
`B`	0 cell covered by a line	4 `4=6-2` cell not covered by a line	13 `13=15-2` cell not covered by a line	10 cell covered by a line	1 `1=3-2` cell not covered by a line
`C`	10 `10=8+2` intersection cell of two lines	5 cell covered by a line	0 cell covered by a line	2 `2=0+2` intersection cell of two lines	0 cell covered by a line
`D`	0 cell covered by a line	2 `2=4-2` cell not covered by a line	0 `0=2-2` cell not covered by a line	0 cell covered by a line	3 `3=5-2` cell not covered by a line
`E`	3 cell covered by a line	3 `3=5-2` cell not covered by a line	4 `4=6-2` cell not covered by a line	0 cell covered by a line	6 `6=8-2` cell not covered by a line

Repeat steps 3 to 6 until an optimal solution is obtained.

Step-3: Make assignment in the opporunity cost table
(1) Rowwise cell `(A,II)` is assigned

(2) Rowwise cell `(B,I)` is assigned, so columnwise cell `(D,I)` crossed off.

(3) Rowwise cell `(E,IV)` is assigned, so columnwise cell `(D,IV)` crossed off.

(4) Columnwise cell `(C,V)` is assigned, so rowwise cell `(C,III)` crossed off.

(5) Rowwise cell `(D,III)` is assigned

Rowwise & columnwise assignment shown in table

	`I`	`II`	`III`	`IV`	`V`
`A`	7	[0] (1) Rowwise cell `(A,II)` is assigned	8	12	11
`B`	[0] (2) Rowwise cell `(B,I)` is assigned so columnwise cell `(D,I)` crossed off.	4	13	10	1
`C`	10	5	0 Rowwise `(C,III)` crossed off because (4) Columnwise cell `(C,V)` is assigned	2	[0] (4) Columnwise cell `(C,V)` is assigned so rowwise cell `(C,III)` crossed off.
`D`	0 Columnwise `(D,I)` crossed off because (2) Rowwise cell `(B,I)` is assigned	2	[0] (5) Rowwise cell `(D,III)` is assigned	0 Columnwise `(D,IV)` crossed off because (3) Rowwise cell `(E,IV)` is assigned	3
`E`	3	3	4	[0] (3) Rowwise cell `(E,IV)` is assigned so columnwise cell `(D,IV)` crossed off.	6

Step-4: Number of assignments = 5, number of rows = 5
Which is equal, so solution is optimal

Optimal assignments are

	`I`	`II`	`III`	`IV`	`V`
`A`	7 Original cost 10	[0] Original cost 5	8 Original cost 13	12 Original cost 15	11 Original cost 16
`B`	[0] Original cost 3	4 Original cost 9	13 Original cost 18	10 Original cost 13	1 Original cost 6
`C`	10 Original cost 10	5 Original cost 7	0 Original cost 2	2 Original cost 2	[0] Original cost 2
`D`	0 Original cost 7	2 Original cost 11	[0] Original cost 9	0 Original cost 7	3 Original cost 12
`E`	3 Original cost 7	3 Original cost 9	4 Original cost 10	[0] Original cost 4	6 Original cost 12

Optimal solution is

Work	Job	Cost
`A`	`II`	5
`B`	`I`	3
`C`	`V`	2
`D`	`III`	9
`E`	`IV`	4
	Total	23