Chapter 1 Assignment Problems (DS) (1).pptx

2
Objectives
After completion of this lesson you will be able to:
• formulate the assignment problem
• know Hungarian method to find proper assignment
• employ Hungarian method to find proper assignment

3
The Assignment Problem
The Assignment Problem: Suppose we have n resources to which we want
to assign to n tasks on a one-to-one basis. Suppose also that we know the cost
of assigning a given resource to a given task. We wish to find an optimal
assignment–one which minimizes total cost.

4
Mathematical Model of Assignment Problem
The Mathematical Model: Let 𝐶𝑖,𝑗 be the cost of assigning the 𝑖𝑡ℎresource to
the 𝑗𝑡ℎ task. We define the cost matrix to be the n × n matrix
C=
𝐶1,1 𝐶1,2 … … . 𝐶1,𝑛
𝐶2,1 𝐶2,2 … … 𝐶2,𝑛
… . .
𝐶𝑛,1
… . .
𝐶𝑛,1
… . 𝐶1,1
… … 𝐶𝑛,𝑛
An assignment is a set of n entry positions in the cost matrix, no two of which
lie in the same row or column. The sum of the n entries of an assignment is its cost. An
assignment with the smallest possible cost is called an optimal assignment.

5
Mathematical Model of Assignment Problem
The Mathematical Model:
Mathematically, we can express the problem as follows:
To Minimize z(cost)= 𝒊=𝟏
𝒏
𝒋=𝟏
𝒏
𝒄𝒊𝒋𝒙𝒊𝒋 ……… {i=1,2,3…….n , j=1,2,3…. n}
Where 𝒙𝒊𝒋 =
1 ; 𝑖𝑓 𝑖𝑡ℎ 𝑝𝑒𝑟𝑠𝑜𝑛 𝑖𝑠 𝑎𝑠𝑠𝑖𝑔𝑛𝑒𝑑 𝑡𝑜 𝑗𝑡ℎ 𝑤𝑜𝑟𝑘
0 ; 𝑖𝑓 𝑖𝑡ℎ 𝑝𝑒𝑟𝑠𝑜𝑛 𝑖𝑠 𝑛𝑜𝑡 𝑎𝑠𝑠𝑖𝑔𝑛𝑒𝑑 𝑡𝑜 𝑗𝑡ℎ 𝑤𝑜𝑟𝑘
With the restrictions
i) 𝒊=𝟏
𝒏
𝒙𝒊𝒋=1; j=1,2,….n, i.e. 𝑖𝑡ℎ
𝑝𝑒𝑟𝑠𝑜𝑛 will do only one work
ii) 𝒋=𝟏
𝒏
𝒙𝒊𝒋=1; i=1,2,….n, i.e. 𝑗𝑡ℎ
𝑤𝑜𝑟𝑘 will be done only by one person.

6
The Hungarian Method: The following algorithm applies the above theorem to a given n × n cost matrix to find
an optimal assignment.
Step 1. Subtract the smallest entry in each row from all the entries of its row.
Step 2. Subtract the smallest entry in each column from all the entries of its column.
Step 3. Draw lines through appropriate rows and columns so that all the zero entries of the cost
matrix are covered and the minimum number of such lines is used.
Step 4. Test for Optimality: (i) If the minimum number of covering lines is n, an optimal assignment
of zeros is possible and we are finished. (ii) If the minimum number of covering lines is less than n,
an optimal assignment of zeros is not yet possible. In that case, proceed to Step 5.
Step 5. Determine the smallest entry not covered by any line. Subtract this entry from each uncovered
row, and then add it to each covered column. Return to Step 3.
The Hungarian Method

7
Flowchart to solve Assignment Problem
Start
Prepare the Assignment Table
Is it a
balance
Problem ?
Add Dummy rows or columns
It is a
Maximization
Problem ?
Convert it into a minimization
problem by subtracting all the
elements from the largest element
No
No
Yes
Yes

8
Flowchart to solve Assignment Problem
Yes
Obtain the reduced cost table. For this:
1) Subtract the Minimum element in each row from all the
elements of that row and then
2) Subtract the Minimum element in each column from all the
elements of that row and then
Does the number of
lines draw equal the
order of the matrix ?
Convert it Subtract the
smallest uncovered element
from all the uncovered
elements. Add it to the
elements that lies at the
intersection of two lines.
Keep the remaining
elements unchanged in the
revised cost table.
No
Draw minimum number of lines to cover all the zeroes in the table
Optimum Solution is obtained
Stop

9
We must determine how jobs should be assigned to machines to minimize setup
times, which are given below:
Example 1
Job 1 Job 2 Job 3 Job 4
Machine 1 14 12 15 15
Machine 2 21 18 18 22
Machine 3 14 17 12 14
Machine 4 6 5 3 6

10
Step 1: Find the minimum element in each row of the cost matrix. Form a new matrix by subtracting this
cost from each row i.e. subtract 12 from 1st row, 18 from 2nd row, 12 from 3rd row and 3 from 4th row
respectively
Example 1
Machine 1 14 12 15 15
Machine 2 21 18 18 22
Machine 3 14 17 12 14
Machine 4 6 5 3 6
Machine 1 2 0 3 3
Machine 2 3 0 0 4
Machine 3 2 5 0 2
Machine 4 3 2 0 3
Row Reduction

11
Step 2: Find the minimum element in each column of the cost matrix. Form a new matrix by subtracting
this cost from each column i.e. subtract 2 from 1st col, 0 from 2nd col, 0 from 3rd col and 2 from 4th col
respectively
Example 1
Machine 1 2 0 3 3
Machine 2 3 0 0 4
Machine 3 2 5 0 2
Machine 4 3 2 0 3
Machine 1 0 0 3 1
Machine 2 1 0 0 2
Machine 3 0 5 0 0
Machine 4 1 2 0 1
Column
Reduction

12
Step 3: Cover all the zeroes of the
matrix with the minimum number
of horizontal or vertical lines.
Example 1
Machine 1 0 0 3 1
Machine 2 1 0 0 2
Machine 3 0 5 0 0
Machine 4 1 2 0 1
1
2
3
4
Step 4: Since the minimal number of lines is 4, an optimal assignment of zeroes is
possible.
No. of Rows / Columns = No of Lines
4 = 4

13
Step 5 : Once the No. of Rows / Columns = No of Lines conditions are satisfied Find out the optimum
solution
Process to Find out Optimum Solution: Select the row or column which contain exactly one zero
If you found single zero in row then cut all the zeroes in their column and If you found single zero in
Column then cut all the zeroes in row.
In the given example it can be seen that there is single 0 in the row 4th and col 4th . If you select the 0
from row 4th then cancel all zeroes in the 3rd col.
Example 1
Machine 1 0 0 3 1
Machine 2 1 0 0 2
Machine 3 0 5 0 0
Machine 4 1 2 1
0
Machine 1 0 0 3 1
Machine 2 1 0 0 2
Machine 3 0 5 0 0
Machine 4 1 2 1
0

14
Similarly you can perform the operation for Remaining
Matrix
Example 1
Machine 1 0 3 1
Machine 2 1 0 2
Machine 3 0 5 0
Machine 4 1 2 1
0
0
0
0
Step 6: Once we perform the allocation final assignment
is as follows
Note: For Time Kindly check Original Matrix in given
question
Machine Jobs Time
Machine 1 1 14
Machine 2 2 18
Machine 3 4 14
Machine 4 3 03
Total Time 49 Hrs

15
Solve the following assignment problem for minimization
Example 2
Job 1 Job 2 Job 3 Job 4 Job 5
Workers A 8 8 8 11 12
B 3 9 18 13 6
C 10 7 2 2 2
D 7 11 9 7 12
E 7 9 10 4 12

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Step 1: Find the minimum element in each row of the cost matrix. Form a new matrix by subtracting this cost from each
row i.e. subtract 8 from 1st row, 3 from 2nd row, 2 from 3rd row, 7 from 4th row and 4 from 5th row respectively
Example 2
Row Reduction
A 8 8 8 11 12
B 3 9 18 13 6
C 10 7 2 2 2
D 7 11 9 7 12
E 7 9 10 4 12
A 0 0 0 3 4
B 0 6 15 10 3
C 8 5 0 0 0
D 0 4 2 0 5
E 3 5 6 0 8

17
Step 2: Find the minimum element in each column of the cost matrix. Form a new matrix by subtracting this cost from
each column i.e. subtract 0 from 1st col, 0 from 2nd col, 0 from 3rd col, 0 from 4th col and 0 from 5th respectively
Example 2
Col Reduction
A 0 0 0 3 4
B 0 6 15 10 3
C 8 5 0 0 0
D 0 4 2 0 5
E 3 5 6 0 8
A 0 0 0 3 4
B 0 6 15 10 3
C 8 5 0 0 0
D 0 4 2 0 5
E 3 5 6 0 8

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Step 3: Cover all the zeroes of the
matrix with the minimum number
of horizontal or vertical lines.
Example 2
1
2
3
4
Step 4: Since the minimal number of lines is 4, and total number of rows/ columns are
5 hence we need to perform improvement.
No. of Rows / Columns ≠ No of Lines
5 ≠ 4
( Note: Kindly check the difference in Ex. 1 and Ex.2)
A 0 0 0 3 4
B 0 6 15 10 3
C 8 5 0 0 0
D 0 4 2 0 5
E 3 5 6 0 8

19
Note: The numbers which are covered with
the red lines are called as covered element
and remaining are called as uncovered
elements
Step 5: Process for Improvement: Select
the smallest no from all uncovered element.
Subtract this smallest no from all uncovered
elements and add only at intersection of two
line are happened and keep all other covered
element as it is.
Example 2
1
2
3
4
A 0 0 0 3 4
B 0 6 15 10 3
C 8 5 0 0 0
D 0 4 2 0 5
E 3 5 6 0 8
In the above example 2 can be subtracted
from all uncovered elements and add only at
cell A-Job1, A-job4, C-Job1 and C-Job 4.

20
Improvement 1:
Example 2
1
2
3
4
A 0 0 0 3 4
B 0 6 15 10 3
C 8 5 0 0 0
D 0 4 2 0 5
E 3 5 6 0 8
cell A-Job1, A-job4, C-Job1 and C-Job 4.
A 2 0 0 5 4
B 0 4 13 10 1
C 10 5 0 2 0
D 0 2 0 0 3
E 3 3 4 0 6
After
Improvement
Note: After Improvement start the repeat process with cover zeroes

21
Improvement 1:
Example 2
1
2 3
4
Since the minimal number of lines is 5, an optimal
assignment of zeroes is possible.
5 = 5
A 2 0 0 5 4
B 0 4 13 10 1
C 10 5 0 2 0
D 0 2 0 0 3
E 3 3 4 0 6
5

22
solution
Example 2
A 2 0 5 4
B 4 13 10 1
C 10 5 0 2
D 0 2 0 3
E 3 3 4 6
0
0
0
0
0
Step 7: Once we perform the allocation final
assignment is as follows
Note: For Time Kindly check Original Matrix in
given question
Worker Jobs Time
A 2 05
B 1 03
C 5 02
D 3 09
E 4 04
Total Time 23 Hrs

23
Examples For Practice
Solve the following assignment problem for minimization
Machines
1 2 3 4 5
Jobs A 8 8 8 11 12
B 4 5 6 3 4
C 12 11 10 9 8
D 18 21 18 17 15
E 10 11 10 8 12
Job Machines
A B C D E
1
2
3
4
5
30 37 40 28 40
40 24 27 21 36
40 32 33 30 35
25 38 40 36 36
29 62 41 34 39
Ans: 45 Hrs
Note: You have to perform Improvement till the following condition has to be satisfied
No of Rows/ Columns = No. of Lines
Ans: 149 Hrs

24
1) Unbalanced Problem
2) Multiple Optimum Solution
3) Maximization Problem
4) Prohibited Assignment Problem ( Restricted Problem)
Special Cases in Assignment Problem

25
1) Unbalanced Problem:
When the number of rows is not equal to the number of columns it is called as an
unbalanced assignment problem. Here we add required numbers of dummy rows or
columns with all its element as 0, to the matrix so as to make it square matrix (i.e.
balanced).
Eg:
1 2 3 4 5 6
A 12 10 15 22 18 8
B 10 18 25 15 16 12
C 11 10 3 8 5 9
D 6 14 10 13 13 12
E 8 12 11 7 13 10
1 2 3 4 5 6
A 12 10 15 22 18 8
B 10 18 25 15 16 12
C 11 10 3 8 5 9
D 6 14 10 13 13 12
E 8 12 11 7 13 10
F 0 0 0 0 0 0
As you can seen in the above example total number of rows are 5 and columns are 6 hence its is unbalanced
problem. We can balance the problem after adding dummy row i.e. F with all its elements as ‘0’
After Balance

26
Let consider the previous example after balancing the problem matrix look like this
Example on Unbalanced Problem
1 2 3 4 5 6
A 12 10 15 22 18 8
B 10 18 25 15 16 12
C 11 10 3 8 5 9
D 6 14 10 13 13 12
E 8 12 11 7 13 10
1 2 3 4 5 6
A 12 10 15 22 18 8
B 10 18 25 15 16 12
C 11 10 3 8 5 9
D 6 14 10 13 13 12
E 8 12 11 7 13 10
F 0 0 0 0 0 0
Note: Before Solving any problem in assignment make sure that it has to be balance i.e.
No of Rows = No of Columns if it is not then balance it.
After Balance

27
1 2 3 4 5 6
A 12 10 15 22 18 8
B 10 18 25 15 16 12
C 11 10 3 8 5 9
D 6 14 10 13 13 12
E 8 12 11 7 13 10
F 0 0 0 0 0 0
Row Subtraction
Step 1: Find the minimum element in each row of the cost matrix. Form a new matrix by subtracting this cost from
each row i.e. subtract 8 from 1st row, 10 from 2nd row, 3 from 3rd row, 6 from 4th row ,7 from 5th row and 0 from th
6th row respectively
1 2 3 4 5 6
A 4 2 7 14 10 0
B 0 8 15 5 6 2
C 8 7 0 5 2 6
D 0 8 4 7 7 5
E 1 5 4 0 6 3
F 0 0 0 0 0 0

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Column Subtraction
Step 2: Find the minimum element in each column of the cost matrix. Form a new matrix by subtracting this cost
from each column. As you can seen in the matrix each column has smallest element is ‘0’
1 2 3 4 5 6
A 4 2 7 14 10 0
B 0 8 15 5 6 2
C 8 7 0 5 2 6
D 0 8 4 7 7 5
E 1 5 4 0 6 3
F 0 0 0 0 0 0
1 2 3 4 5 6
A 4 2 7 14 10 0
B 0 8 15 5 6 2
C 8 7 0 5 2 6
D 0 8 4 7 7 5
E 1 5 4 0 6 3
F 0 0 0 0 0 0

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Step 3: Cover all the zeroes of the matrix with the minimum number of horizontal or vertical lines.
1 2 3 4 5 6
A 4 2 7 14 10 0
B 0 8 15 5 6 2
C 8 7 0 5 2 6
D 0 8 4 7 7 5
E 1 5 4 0 6 3
F 0 0 0 0 0 0 1
2
3
4
Step 4: Since the minimal number of
lines is 5, and total number of rows/
columns are 6 hence we need to
perform improvement.
6 ≠ 5
5

30
1 2 3 4 5 6
A 4 2 7 14 10 0
B 0 8 15 5 6 2
C 8 7 0 5 2 6
D 0 8 4 7 7 5
E 1 5 4 0 6 3
F 0 0 0 0 0 0 1
2
3
4
element as it is.
In the given example 2 can be subtracted
cell E1,E3,E6 and F1,F3,F6. Keep remaining
covered element as it is
5

31
1 2 3 4 5 6
A 4 2 7 14 10 0
B 0 8 15 5 6 2
C 8 7 0 5 2 6
D 0 8 4 7 7 5
E 1 5 4 0 6 3
F 0 0 0 0 0 0 1
2
3
4 5
Improvement 1:
1 2 3 4 5 6
A 4 0 7 12 8 0
B 0 6 15 3 4 2
C 8 5 0 3 0 6
D 0 6 4 5 5 5
E 3 5 6 0 6 5
F 2 0 2 0 0 2
After Improvement

32
1
2
3
4
5
Improvement 1:
1 2 3 4 5 6
A 4 0 7 12 8 0
B 0 6 15 3 4 2
C 8 5 0 3 0 6
D 0 6 4 5 5 5
E 3 5 6 0 6 5
F 2 0 2 0 0 2
perform improvement number 2.
6 ≠ 5

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1
2
3
4
5
Improvement 2:
1 2 3 4 5 6
A 4 0 7 12 8 0
B 0 6 15 3 4 2
C 8 5 0 3 0 6
D 0 6 4 5 5 5
E 3 5 6 0 6 5
F 2 0 2 0 0 2
element as it is.
cell A1,A4,C1,C4,F2 and F4. Keep
remaining covered element as it is

34
1
2
3
4
5
Improvement 2:
1 2 3 4 5 6
A 4 0 7 12 8 0
B 0 6 15 3 4 2
C 8 5 0 3 0 6
D 0 6 4 5 5 5
E 3 5 6 0 6 5
F 2 0 2 0 0 2
1 2 3 4 5 6
A 6 0 7 14 8 0
B 0 4 13 3 2 0
C 10 5 0 5 0 6
D 0 4 2 5 3 3
E 3 3 4 0 4 3
F 4 0 2 0 0 2
After Improvement

35
1
2
3
4
5
Improvement 2:
1 2 3 4 5 6
A 6 0 7 14 8 0
B 0 4 13 3 2 0
C 10 5 0 5 0 6
D 0 4 2 5 3 3
E 3 3 4 0 4 3
F 4 0 2 0 0 2
6
6 = 6

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Step 8 : Once the No. of Rows / Columns = No of Lines conditions are satisfied Find out the
optimum solution
1 2 3 4 5 6
A 6 0 7 14 8 0
B 0 4 13 3 2 0
C 10 5 0 5 0 6
D 0 4 2 5 3 3
E 3 3 4 0 4 3
F 4 0 2 0 0 2
0
0
0
0
0
0
Step 9: Once we perform the allocation final assignment is
as follows
question
Worker Jobs Time
A 2 10
B 6 12
C 3 03
D 1 06
E 4 13
F 5 00
Total Time 38 Hrs

37
solution
Example 2
A 2 0 5 4
B 4 13 10 1
C 10 5 0 2
D 0 2 0 3
E 3 3 4 6
0
0
0
0
0
given question
Worker Jobs Time
A 2 05
B 1 03
C 5 02
D 3 09
E 4 04
Total Time 23 Hrs

38
Solve the following assignment problem ?
J
O
B
S
Workers
M1 M2 M3 M4
A 17 23 27 31
B 7 12 16 18
C 9 14 18 21
J
O
B
S
Workers
M1 M2 M3 M4
A 17 23 27 31
B 7 12 16 18
C 9 14 18 21
D 0 0 0 0
After Balance

39
Step 1: Row Subtraction
J
O
B
S
Workers
M1 M2 M3 M4
A 17 23 27 31
B 7 12 16 18
C 9 14 18 21
D 0 0 0 0
After Row Subtraction
J
O
B
S
Workers
M1 M2 M3 M4
A 0 0 10 14
B 0 5 9 11
C 0 5 9 12
D 0 0 0 0

40
Step 1: Column Subtraction
After Column Subtraction
J
O
B
S
Workers
M1 M2 M3 M4
A 0 0 10 14
B 0 5 9 11
C 0 5 9 12
D 0 0 0 0
J
O
B
S
Workers
M1 M2 M3 M4
A 0 0 10 14
B 0 5 9 11
C 0 5 9 12
D 0 0 0 0

41
Step 3 : Cover zeroes
No of Rows ≠ 𝑵𝒐 𝒐𝒇 𝑳𝒊𝒏𝒆𝒔
4 ≠ 3
J
O
B
S
Workers
M1 M2 M3 M4
A 0 0 10 14
B 0 5 9 11
C 0 5 9 12
D 0 0 0 0 1
2
3

42
Improvement No. 1
After Improvement
J
O
B
S
Workers
M1 M2 M3 M4
A 0 0 10 14
B 0 5 9 11
C 0 5 9 12
D 0 0 0 0 1
2
3
J
O
B
S
Workers
M1 M2 M3 M4
A 5 0 10 14
B 0 0 4 6
C 0 0 4 7
D 0 0 0 0

43
Improvement No. 1
No of Rows ≠ 𝑵𝒐 𝒐𝒇 𝑳𝒊𝒏𝒆𝒔
4 ≠ 3
1
2
3
J
O
B
S
Workers
M1 M2 M3 M4
A 5 0 10 14
B 0 0 4 6
C 0 0 4 7
D 0 0 0 0

44
Improvement No. 2
After Improvement
1
2
3
J
O
B
S
Workers
M1 M2 M3 M4
A 5 0 10 14
B 0 0 4 6
C 0 0 4 7
D 0 0 0 0
J
O
B
S
Workers
M1 M2 M3 M4
A 5 0 6 10
B 0 0 0 2
C 0 0 0 3
D 4 4 0 0

45
Improvement No. 2
No. of Rows= No of line
4 = 4
2
1
3
J
O
B
S
Workers
M1 M2 M3 M4
A 5 0 6 10
B 0 0 0 2
C 0 0 0 3
D 4 4 0 0 4

46
Optimum Solution
J
O
B
S
Workers
M1 M2 M3 M4
A 5 0 6 10
B 0 0 0 2
C 0 0 0 3
D 4 4 0 0
0
0
0
0
0
0
Worker Jobs Time
A M2 23
B M1 07
C M3 18
D M4 0
Total Time 48 Hrs
Solution 1
Worker Jobs Time
A M2 23
B M3 16
C M1 09
D M4 00
Total Time 48 Hrs
Solution 2

47
2) Multiple Optimum Solution
After making the assignment (ie. after making the zeroes with square ) to the single unmarked in
all possible rows and columns, it is found that the two or more rows or columns still contain more than one
unmarked zeroes then the problem has multiple optimal solution. To get alternate solutions:
i) Select the row or column containing maximum number of unmarked zeroes (after making assignments
to the single unmarked zeroes.)
ii) Select one of the zeros and mark it with a square ( ).
iii) Cancel all other zeros in its row as well as column.
iv) Proceed further in usual manner to make other assignments.
v) Repeat the procedure by making assignment to each of the zeroes in the row or column selected in (i)
above separately to get the alternate solutions.
vi) All these alternate solutions gives the same optimal value.

48
As we can find that the given problem is un balanced ( No of Rows ≠ 𝑁𝑜 𝑜𝑓 𝐶𝑜𝑙𝑢𝑚𝑛𝑠) hence here we add
dummy row i.e. D with all its element as ‘0’.
Example On Multiple Optimum Solution
Machine
M1 M2 M3 M4
Jobs
A 17 23 27 31
B 7 12 16 18
C 9 14 18 21
Machine
M1 M2 M3 M4
Jobs
A 17 23 27 31
B 7 12 16 18
C 9 14 18 21
D 0 0 0 0

49
Step 1: Find the minimum element in each row of the cost matrix. Form a new matrix by subtracting this cost from
each row i.e. subtract 17 from 1st row, 7 from 2nd row, 9 from 3rd row and 0 from 4th row respectively
Machine
M1 M2 M3 M4
Jobs
A 17 23 27 31
B 7 12 16 18
C 9 14 18 21
D 0 0 0 0 Machine
M1 M2 M3 M4
Jobs
A 0 6 10 14
B 0 5 9 11
C 0 5 9 12
D 0 0 0 0
Row Reduction

50
each column i.e. subtract 0 from 1st col, 0 from 2nd col, 0 from 3rd col, and 0 from 4th respectively
Machine
M1 M2 M3 M4
Jobs
A 0 6 10 14
B 0 5 9 11
C 0 5 9 12
D 0 0 0 0 Machine
M1 M2 M3 M4
Jobs
A 0 6 10 14
B 0 5 9 11
C 0 5 9 12
D 0 0 0 0
Column Reduction

51
Machine
M1 M2 M3 M4
Jobs
A 0 6 10 14
B 0 5 9 11
C 0 5 9 12
D 0 0 0 0
1
2
Step 4: Since the minimal number of lines is 5,
and total number of rows/ columns are 6 hence
we need to perform improvement number 1.
4 ≠ 𝟐

52
Machine
M1 M2 M3 M4
Jobs
A 0 6 10 14
B 0 5 9 11
C 0 5 9 12
D 0 0 0 0 1
2
Step 5: Process for Improvement: Select the
smallest no from all uncovered element. Subtract
this smallest no from all uncovered elements and
add only at intersection of two line are happened
and keep all other covered element as it is.
cell D-M1

53
Machine
M1 M2 M3 M4
Jobs
A 0 6 10 14
B 0 5 9 11
C 0 5 9 12
D 0 0 0 0 1
2
Step 5: Process for Improvement: Select the
smallest no from all uncovered element. Subtract
this smallest no from all uncovered elements and
add only at intersection of two line are happened
and keep all other covered element as it is.
cell D-M1

54
Machine
M1 M2 M3 M4
Jobs
A 0 6 10 14
B 0 5 9 11
C 0 5 9 12
D 0 0 0 0 1
2
Improvement No 1
Machine
M1 M2 M3 M4
Jobs
A 0 1 5 9
B 0 0 4 6
C 0 0 4 7
D 5 0 0 0
Improvement No 1

55
1
2
Improvement No 1
Machine
M1 M2 M3 M4
Jobs
A 0 1 5 9
B 0 0 4 6
C 0 0 4 7
D 5 0 0 0
3
perform improvement number 2.
4 ≠ 𝟑

56
1
2
Improvement No 2
Machine
M1 M2 M3 M4
Jobs
A 0 1 5 9
B 0 0 4 6
C 0 0 4 7
D 5 0 0 0
3
element as it is.
cell D-M1 and D-M2. Keep remaining
covered element as it is

57
1
2
Improvement No 2
Machine
M1 M2 M3 M4
Jobs
A 0 1 5 9
B 0 0 4 6
C 0 0 4 7
D 5 0 0 0
3
Machine
M1 M2 M3 M4
Jobs
A 0 1 1 5
B 0 0 0 2
C 0 0 0 3
D 9 4 0 0
Improvement No 2

58
1
2
Improvement No 2
3
Machine
M1 M2 M3 M4
Jobs
A 0 1 1 5
B 0 0 0 2
C 0 0 0 3
D 9 4 0 0 4
4 = 4

59
Machine
M1 M2 M3 M4
Jobs
A 0 1 1 5
B 0 0 0 2
C 0 0 0 3
D 9 4 0 0
solution
0
0
As you can seen in above matrix we did not find the single ‘0’ either in row or column.
Such type of problem has multiple solution (i.e. More than one solution)
Note: Multiple Optimum Solution problem can be identified only at the time of final allocation

60
Machine
M1 M2 M3 M4
Jobs
A 0 1 1 5
B 0 0 0 2
C 0 0 0 3
D 9 4 0 0
Here Job B can be assigned to either Machine M2 or M3. If you assign Job B to Machine M2 then
Job C can be assigned to M3 or you can assign Job B to Machine M3 and Job C can be assigned to M2.
This can be represented in and
0
0
0
0
0
0

61
Machine
M1 M2 M3 M4
Jobs
A 0 1 1 5
B 0 0 0 2
C 0 0 0 3
D 9 4 0 0
0
0
0
0
0
0
Step 7: Once we perform the allocation final assignment is as follows
Jobs Machine Time
A M1 17
B M2 12
C M3 18
D M4 00
Total Time 47 Hrs
Solution No-1
Jobs Machine Time
A M1 17
B M3 16
C M2 14
D M4 00
Total Time 47 Hrs
Solution No-2

62
Examples For Practice
Q. Solve the following assignment problem
for minimization
Ans: 55 Hrs
Note: You have to perform Improvement till the following condition has to be satisfied
No of Rows/ Columns = No. of Lines
Ans: 32 Hrs
Worker
A B C D E
Jobs
I 16 13 17 19 20
II 14 12 13 16 17
III 14 11 12 17 18
IV 5 5 8 8 11
V 5 3 8 8 10
Q. In a particular plant there are 4 machines to be
installed. There are 5 vacant places available. The
costs of installation of machines at different vacant
places are given in the following table. Find the
optimum assignment.
Places
Machines A B C D E
M1 9 11 15 10 11
M2 12 9 10 15 9
M3 10 13 14 11 7
M4 14 8 12 7 8

63
3) Maximization Problem
Hungarian method can be used for maximization problem as follows:
converting it into equivalent minimization problem as follows:
i) Convert the given profit matrix into relative loss matrix. By subtracting all its element from the largest
element including it.
ii) Add a dummy row or column to it ( with 0 unit profit for its cells) if necessary.
iii) Locate the largest per unit profit figure in the table and subtract all profit figure ( including itself ) from it
to get an equivalent relative loss matrix.
iv) Solve it further as a normal Hungarian method to get optimum assignment
v) To find the total maximum profit consider the original profit elements for the respective assignment.
Maximization Keyword: Profit, Sales, Production, Revenue
Minimization Keyword:- Loss, Time, Space, Expenditure, Cost, Defects

64
Ques. The data given in the table refer to production in certain units: Solve the following
assignment problem
Example on Maximization Problem
OPE
RAT
OR
MACHINES
A B C D
1 10 5 7 8
2 11 4 9 10
3 8 4 9 7
4 7 5 6 4
5 8 9 7 5
As we can find out that the given problem is not balanced. Hence here we add dummy column i.e E with all its
element as ‘0’
Note: Before Solving any problem
in assignment make sure that it has
to be balance i.e.
No of Rows = No of Columns if it
is not then balance it.

65
Step 1: Balance the given problem
Note: Before Solving any problem in assignment make sure that it has to be balance i.e.
No of Rows = No of Columns if it is not then balance it.
After Balance
A B C D
1 10 5 7 8
2 11 4 9 10
3 8 4 9 7
4 7 5 6 4
5 8 9 7 5
A B C D E
1 10 5 7 8 0
2 11 4 9 10 0
3 8 4 9 7 0
4 7 5 6 4 0
5 8 9 7 5 0

66
Step 2: To solve the maximization problem first it has to be converted in to minimization by subtracting all its
element from the largest element.
In given problem the largest element in matrix is 11 hence we can subtract all the element from 11
Once the problem is convert into minimization remaining process is same as we have done in earlier examples
Maximization to
Minimization
A B C D E
1 1 6 4 3 11
2 0 7 2 1 11
3 3 7 2 4 11
4 4 6 5 7 11
5 3 2 4 6 11
A B C D E
1 10 5 7 8 0
2 11 4 9 10 0
3 8 4 9 7 0
4 7 5 6 4 0
5 8 9 7 5 0

67
Step 3: Find the minimum element in each row of the cost matrix. Form a new matrix by subtracting this cost from each row
i.e. subtract 1 from 1st row, 0 from 2nd row, 2 from 3rd row, 4 from 4th row and 2 from 5th row respectively.
A B C D E
1 1 6 4 3 11
2 0 7 2 1 11
3 3 7 2 4 11
4 4 6 5 7 11
5 3 2 4 6 11
A B C D E
1 0 5 3 2 10
2 0 7 2 1 11
3 1 5 0 2 9
4 0 2 1 3 7
5 1 0 2 4 9
Row Subtraction

68
Step 4: Find the minimum element in each column of the cost matrix. Form a new matrix by subtracting this cost from each
column. i.e. subtract 0 from 1st col, 0 from 2nd col, 0 from 3rd col, 1 from 4th col and 7 from 5th row respectively.
A B C D E
1 0 5 3 2 10
2 0 7 2 1 11
3 1 5 0 2 9
4 0 2 1 3 7
5 1 0 2 4 9
Column Subtraction
A B C D E
1 0 5 3 1 3
2 0 7 2 0 4
3 1 5 0 1 2
4 0 2 1 2 0
5 1 0 2 3 2

69
Step 5: Cover all the zeroes of the matrix with the minimum number of horizontal or vertical lines
1 3
2
A B C D E
1 0 5 3 1 3
2 0 7 2 0 4
3 1 5 0 1 2
4 0 2 1 2 0
5 1 0 2 3 2
4
5
Step 6: Since the minimal number of lines is 4,
and total number of rows/ columns are 5 hence
an optimal assignment of zeroes is possible.
5 = 5

70
solution
0
0
A B C D E
1 0 5 3 1 3
2 0 7 2 4
3 1 5 0 1 2
4 0 2 1 2 0
5 1 2 3 2
0
0
0
given question
Operator Machine Production
1 A 10
2 D 10
3 C 09
4 E 00
5 B 09
Total Time 38

71
Q.1 The following table gives the profit of
assignment in Rupees. Also give the optimal
profit.
Examples For Practice on Maximization
Ans: 280
Jobs
J1 J2 J3 J4 J5
M1 50 60 40 30 45
M2 35 55 45 55 40
M3 40 45 50 35 35
M4 60 40 55 40 30
M5 45 35 45 50 55
Q.2 The marketing director of a multi unit company
is faced with a problem of assigning 5 senior
managers to 6 zones. From past experience he
knows that the efficiency percentage judged by
sales, operating cost etc.. depends on manager zone
combination. The efficiency of different managers is
given below
M
A
N
A
G
E
R
ZONES
1 2 3 4 5 6
A 73 91 87 82 78 80
B 81 85 69 76 74 85
C 75 72 83 84 78 91
D 93 96 86 91 83 82
E 90 91 79 89 69 76
Find out which zone will be managed by a junior
manager due to non-availability of a senior manager

72
4) Prohibited Assignment Problem:
• A prohibited assignment problem is a problem which contains one or more constraints. It is also know as
restricted assignment problem. Suppose in case 𝑖𝑡ℎ
person is restricted to perform 𝑗𝑡ℎ
then constrained
assignment occur in the cell (i,j) of the cost matrix.
This are indicated by putting a dash (-) or cross (x) at the positions. To solve the problem:
i) For minimization problem we assume ( +M or + ∞ ) for the prohibited positions and solve further as usual.
ii) For maximization problem we assume ( -M or - ∞ ) for the prohibited positions and solve further as usual.
This are indicated by : -, X, M, ∞
Eg.
Clerk
s
Jobs
A B C D
1 4 7 8 6
2 * 8 7 4
3 3 * 8 3
4 6 6 4 2
As it can be seen in the given example clerk 2 is
not able to perform job A and clerk 3 is not able to
perform job B hence there is no point to assign Job
A and Job to Clerk 1 and 2 respectively that’s why
it can be denoted as *

73
Step 1: Find the minimum element in each row of the cost matrix. Form a new matrix by subtracting this cost from each
row i.e. subtract 4 from 1st row, 4 from 2nd row, 3 from 3rd row and 2 from 4th row respectively.
Example on Prohibited Roots
Clerks Jobs
A B C D
1 4 7 8 6
2 * 8 7 4
3 3 * 8 3
4 6 6 4 2
Row Reduction
Clerks Jobs
A B C D
1 0 3 4 2
2 * 4 3 0
3 0 * 5 0
4 4 4 2 0

74
each column i.e. subtract 0 from 1st col, 3 from 2nd col, 2 from 3rd col and 0 from 4th col respectively
Column Reduction
Clerks Jobs
A B C D
1 0 3 4 2
2 * 4 3 0
3 0 * 5 0
4 4 4 2 0 Clerks Jobs
A B C D
1 0 0 2 2
2 * 1 1 0
3 0 * 3 0
4 4 1 0 0

75
Clerks Jobs
A B C D
1 0 0 2 2
2 * 1 1 0
3 0 * 3 0
4 4 1 0 0
1
2
3
4
perform improvement.
4 = 4

76
Step 8 : Once the No. of Rows / Columns = No of Lines conditions are satisfied Find out the optimum solution
Clerks Jobs
A B C D
1 0 0 2 2
2 * 1 1 0
3 0 * 3 0
4 4 1 0 0
0
Step 9: Once we perform the allocation final assignment is
as follows
question
Clerks Jobs Time
1 B 07
2 D 04
3 A 03
4 C 04
Total Time 18 Hrs
0
0
0

77
Q.1 Solve the following assignment problem
Examples For Practice on Prohibited Roots
Ans: 66
PER
SON
JOBS
J1 J2 J3 J4 J5
P1 27 18 * 20 21
P2 31 24 21 12 17
P3 20 17 20 * 16
P4 20 28 20 16 27
Q.2 Solve the following assignment problem
Task
Machines
X Y X W P
A 19 21 25 20 21
B 27 24 * 25 24
C * 24 27 24 20
D 22 16 20 15 16

78
Extra Problems for Practice
Q.1 Solve the following assignment problem:
1 2 3 4 5
A 4 6 10 5 6
B 7 4 8 5 4
C 12 6 9 6 2
D 9 3 7 2 3
E 6 5 5 3 8
Q.2 Five jobs are to be assigned to five persons
A,B,C,D, E, The time taken (in Minutes) by each of
them on each job is given below. Workout the
optimal assignment as the minimum total time
taken.
Jobs
1 2 3 4 5
Perso
n
A 16 13 17 19 20
B 14 12 13 16 17
C 14 11 12 17 18
D 5 5 8 8 11
E 3 3 8 8 10
Q.3 Solve the following assignment problem and
obtain minimum cost that job can be performed
1 2 3 4 5
A 25 18 32 20 21
B 34 25 21 12 17
C 20 17 20 32 16
D 20 28 20 16 27

79
• Assigning teaching fellows to time slots
• Assigning airplanes to flights
• Assigning project members to tasks
• Determining positions on a team
• Assigning brides to grooms (once called the marriage problem)
• Machine allocation for optimum space utilization
Applications of AP

80
1. A job assignment problem is unbalanced when
A) Each worker can perform only one job B) A worker can not perform all the jobs but can do only some of the jobs
C) The number of jobs and the number of workers are the same D) The number of jobs is not same as the number of workers
Multiple Choice Questions
2. In case multiple zeroes are obtained in all rows and columns
A) No solution is possible for the problem B) A unique solution is exist for the problem
C) The problem has infeasible solutions D) The problem has multiple solutions
3. Balancing of an unbalanced assignment problem involve
A) The introduction of dummy column B) The introduction of dummy row
C) The introduction of dummy row or a dummy column D) All the above
4. In assignment problem at what condition optimum solution is occurred
A) No of rows / columns = no of lines B) No of rows = no of columns
C) No of rows / columns ≠ no of lines D) No of the above

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