SlideShare a Scribd company logo
1 of 80
Assignment Problems
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
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
Job 1 Job 2 Job 3 Job 4
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
Job 1 Job 2 Job 3 Job 4
Machine 1 2 0 3 3
Machine 2 3 0 0 4
Machine 3 2 5 0 2
Machine 4 3 2 0 3
Job 1 Job 2 Job 3 Job 4
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
Job 1 Job 2 Job 3 Job 4
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
Job 1 Job 2 Job 3 Job 4
Machine 1 0 0 3 1
Machine 2 1 0 0 2
Machine 3 0 5 0 0
Machine 4 1 2 1
0
Job 1 Job 2 Job 3 Job 4
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
Job 1 Job 2 Job 3 Job 4
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
16
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
Job 1 Job 2 Job 3 Job 4 Job 5
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
Job 1 Job 2 Job 3 Job 4 Job 5
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
Job 1 Job 2 Job 3 Job 4 Job 5
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
Job 1 Job 2 Job 3 Job 4 Job 5
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
18
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)
Job 1 Job 2 Job 3 Job 4 Job 5
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
Job 1 Job 2 Job 3 Job 4 Job 5
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
Job 1 Job 2 Job 3 Job 4 Job 5
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.
Job 1 Job 2 Job 3 Job 4 Job 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
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.
No. of Rows / Columns = No of Lines
5 = 5
Job 1 Job 2 Job 3 Job 4 Job 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
Step 6 : Once the No. of Rows / Columns = No of Lines conditions are satisfied Find out the optimum
solution
Example 2
Job 1 Job 2 Job 3 Job 4 Job 5
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:
Special Cases in Assignment 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
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
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
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
28
Example on Unbalanced Problem
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
29
Example on Unbalanced Problem
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.
No. of Rows / Columns ≠ No of Lines
6 ≠ 5
5
30
Example on Unbalanced Problem
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 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.
In the given example 2 can be subtracted
from all uncovered elements and add only at
cell E1,E3,E6 and F1,F3,F6. Keep remaining
covered element as it is
5
31
Example on Unbalanced Problem
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
Note: After Improvement start the repeat process with cover zeroes
32
Example on Unbalanced Problem
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
Step 6: Since the minimal number of
lines is 5, and total number of rows/
columns are 6 hence we need to
perform improvement number 2.
No. of Rows / Columns ≠ No of Lines
6 ≠ 5
33
Example on Unbalanced Problem
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
Step 7: 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.
In the given example 2 can be subtracted
from all uncovered elements and add only at
cell A1,A4,C1,C4,F2 and F4. Keep
remaining covered element as it is
34
Example on Unbalanced Problem
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
Note: After Improvement start the repeat process with cover zeroes
35
Example on Unbalanced Problem
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
Since the minimal number of lines is 6, an optimal
assignment of zeroes is possible.
No. of Rows / Columns = No of Lines
6 = 6
36
Example on Unbalanced Problem
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
Note: For Time Kindly check Original Matrix in given
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
Step 8 : Once the No. of Rows / Columns = No of Lines conditions are satisfied Find out the optimum
solution
Example 2
Job 1 Job 2 Job 3 Job 4 Job 5
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
38
1) Unbalanced Problem
Solve the following assignment problem ?
Special Cases in 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
1) Unbalanced Problem
Step 1: Row Subtraction
Special Cases in 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
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
1) Unbalanced Problem
Step 1: Column Subtraction
Special Cases in Assignment Problem
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
1) Unbalanced Problem
Step 3 : Cover zeroes
Special Cases in Assignment Problem
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
1) Unbalanced Problem
Improvement No. 1
Special Cases in Assignment Problem
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
1) Unbalanced Problem
Improvement No. 1
Special Cases in Assignment Problem
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
1) Unbalanced Problem
Improvement No. 2
Special Cases in Assignment Problem
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
1) Unbalanced Problem
Improvement No. 2
Special Cases in Assignment Problem
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
1) Unbalanced Problem
Optimum Solution
Special Cases in Assignment Problem
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.
Special Cases in Assignment Problem
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
Example On Multiple Optimum Solution
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
Example On Multiple Optimum Solution
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, 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
Example On Multiple Optimum Solution
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
Step 3: Cover all the zeroes of the matrix with the minimum number of horizontal or vertical lines.
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.
No. of Rows / Columns ≠ No of Lines
4 ≠ 𝟐
52
Example On Multiple Optimum Solution
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.
In the above example 5 can be subtracted
from all uncovered elements and add only at
cell D-M1
53
Example On Multiple Optimum Solution
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.
In the above example 5 can be subtracted
from all uncovered elements and add only at
cell D-M1
54
Example On Multiple Optimum Solution
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
Note: After Improvement start the repeat process with cover zeroes
55
Example On Multiple Optimum Solution
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
Step 6: Since the minimal number of
lines is 3, and total number of rows/
columns are 4 hence we need to
perform improvement number 2.
No. of Rows / Columns ≠ No of Lines
4 ≠ 𝟑
56
Example On Multiple Optimum Solution
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
Step 7: 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.
In the given example 4 can be subtracted
from all uncovered elements and add only at
cell D-M1 and D-M2. Keep remaining
covered element as it is
57
Example On Multiple Optimum Solution
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
Note: After Improvement start the repeat process with cover zeroes
58
Example On Multiple Optimum Solution
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
Since the minimal number of lines is 4, an optimal
assignment of zeroes is possible.
No. of Rows / Columns = No of Lines
4 = 4
59
Example On Multiple Optimum Solution
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
Step 6 : Once the No. of Rows / Columns = No of Lines conditions are satisfied Find out the optimum
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
Example On Multiple Optimum Solution
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
Example On Multiple Optimum Solution
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
Special Cases in Assignment Problem
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
Example on Maximization 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
Example on Maximization Problem
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.
Example on Maximization Problem
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.
Example on Maximization Problem
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
Example on Maximization Problem
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.
No. of Rows / Columns = No of Lines
5 = 5
70
Example on Maximization Problem
Step 7 : Once the No. of Rows / Columns = No of Lines conditions are satisfied Find out the optimum
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
Step 8: Once we perform the allocation final
assignment is as follows
Note: For Time Kindly check Original Matrix in
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.
Special Cases in Assignment Problem
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
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, 3 from 2nd col, 2 from 3rd col and 0 from 4th col respectively
Example on Prohibited Roots
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
Step 3: Cover all the zeroes of the matrix with the minimum number of horizontal or vertical lines.
Example on Prohibited Roots
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
Step 4: Since the minimal number of
lines is 4, and total number of rows/
columns are 4 hence we need to
perform improvement.
No. of Rows / Columns = No of Lines
4 = 4
76
Step 8 : Once the No. of Rows / Columns = No of Lines conditions are satisfied Find out the optimum solution
Example on Prohibited Roots
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
Note: For Time Kindly check Original Matrix in given
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

More Related Content

What's hot

Transportation Problem in Operational Research
Transportation Problem in Operational ResearchTransportation Problem in Operational Research
Transportation Problem in Operational Research
Neha Sharma
 
Recruitment & selection process practiced by a multinational organisation a c...
Recruitment & selection process practiced by a multinational organisation a c...Recruitment & selection process practiced by a multinational organisation a c...
Recruitment & selection process practiced by a multinational organisation a c...
Nikita Palkar
 
Contemporary issues in hrm
Contemporary issues in hrmContemporary issues in hrm
Contemporary issues in hrm
prakhar07
 
Operational research on Assignment ppt
Operational research on Assignment pptOperational research on Assignment ppt
Operational research on Assignment ppt
Nirali Solanki
 

What's hot (20)

Assignment model
Assignment modelAssignment model
Assignment model
 
Transportation Problem in Operational Research
Transportation Problem in Operational ResearchTransportation Problem in Operational Research
Transportation Problem in Operational Research
 
9Johnson's Rule.ppt
9Johnson's Rule.ppt9Johnson's Rule.ppt
9Johnson's Rule.ppt
 
Lpp numerical solved
Lpp numerical solvedLpp numerical solved
Lpp numerical solved
 
Recruitment & selection process practiced by a multinational organisation a c...
Recruitment & selection process practiced by a multinational organisation a c...Recruitment & selection process practiced by a multinational organisation a c...
Recruitment & selection process practiced by a multinational organisation a c...
 
INTERNSHIP REPORT ON HR AT FUTURE RETAIL LTD - HYD
INTERNSHIP REPORT ON HR AT FUTURE RETAIL LTD - HYDINTERNSHIP REPORT ON HR AT FUTURE RETAIL LTD - HYD
INTERNSHIP REPORT ON HR AT FUTURE RETAIL LTD - HYD
 
Recruitment And Selection Process Of Assignment / Report
Recruitment And Selection Process Of Assignment / ReportRecruitment And Selection Process Of Assignment / Report
Recruitment And Selection Process Of Assignment / Report
 
Mba iibms case study answers & solutions
Mba iibms case study answers & solutionsMba iibms case study answers & solutions
Mba iibms case study answers & solutions
 
Transportation Method Operation Research
Transportation Method Operation ResearchTransportation Method Operation Research
Transportation Method Operation Research
 
Contemporary issues in hrm
Contemporary issues in hrmContemporary issues in hrm
Contemporary issues in hrm
 
Operational research on Assignment ppt
Operational research on Assignment pptOperational research on Assignment ppt
Operational research on Assignment ppt
 
Assignment Chapter - Q & A Compilation by Niraj Thapa
Assignment Chapter  - Q & A Compilation by Niraj ThapaAssignment Chapter  - Q & A Compilation by Niraj Thapa
Assignment Chapter - Q & A Compilation by Niraj Thapa
 
Assignment problems
Assignment problemsAssignment problems
Assignment problems
 
Vogel's Approximation Method
Vogel's Approximation MethodVogel's Approximation Method
Vogel's Approximation Method
 
Simplex method: Slack, Surplus & Artificial variable
Simplex method:  Slack, Surplus & Artificial variableSimplex method:  Slack, Surplus & Artificial variable
Simplex method: Slack, Surplus & Artificial variable
 
Internship report
Internship report Internship report
Internship report
 
Project Title : A study on Job Satisfaction of Employees at Bright Future Con...
Project Title : A study on Job Satisfaction of Employees at Bright Future Con...Project Title : A study on Job Satisfaction of Employees at Bright Future Con...
Project Title : A study on Job Satisfaction of Employees at Bright Future Con...
 
Recruitment
RecruitmentRecruitment
Recruitment
 
steps of implementation of HRIS
steps of implementation of HRISsteps of implementation of HRIS
steps of implementation of HRIS
 
Assignment method
Assignment methodAssignment method
Assignment method
 

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

qadm-ppt-150918102124-lva1-app6892.pdf
qadm-ppt-150918102124-lva1-app6892.pdfqadm-ppt-150918102124-lva1-app6892.pdf
qadm-ppt-150918102124-lva1-app6892.pdf
Hari31856
 

Similar to Chapter 1 Assignment Problems (DS) (1).pptx (20)

Decision Science.pdf
Decision Science.pdfDecision Science.pdf
Decision Science.pdf
 
Quantitativeanalysisfordecisionmaking 13427543542352-phpapp02-120719222252-ph...
Quantitativeanalysisfordecisionmaking 13427543542352-phpapp02-120719222252-ph...Quantitativeanalysisfordecisionmaking 13427543542352-phpapp02-120719222252-ph...
Quantitativeanalysisfordecisionmaking 13427543542352-phpapp02-120719222252-ph...
 
Hungarian algorithm
Hungarian algorithmHungarian algorithm
Hungarian algorithm
 
Asssignment problem
Asssignment problemAsssignment problem
Asssignment problem
 
A Comparative Analysis Of Assignment Problem
A Comparative Analysis Of Assignment ProblemA Comparative Analysis Of Assignment Problem
A Comparative Analysis Of Assignment Problem
 
Quantitative Analysis For Decision Making
Quantitative Analysis For Decision MakingQuantitative Analysis For Decision Making
Quantitative Analysis For Decision Making
 
A0280115(1)
A0280115(1)A0280115(1)
A0280115(1)
 
Hungarian Assignment Problem
Hungarian Assignment ProblemHungarian Assignment Problem
Hungarian Assignment Problem
 
Assignment problem notes
Assignment problem notesAssignment problem notes
Assignment problem notes
 
Lesson 33: The Assignment Problem
Lesson 33: The Assignment  ProblemLesson 33: The Assignment  Problem
Lesson 33: The Assignment Problem
 
Qa 2
Qa 2Qa 2
Qa 2
 
Matching problem
Matching problemMatching problem
Matching problem
 
A study on solving Assignment Problem
A study on solving Assignment ProblemA study on solving Assignment Problem
A study on solving Assignment Problem
 
qadm-ppt-150918102124-lva1-app6892.pdf
qadm-ppt-150918102124-lva1-app6892.pdfqadm-ppt-150918102124-lva1-app6892.pdf
qadm-ppt-150918102124-lva1-app6892.pdf
 
Chapter 3.Simplex Method hand out last.pdf
Chapter 3.Simplex Method hand out last.pdfChapter 3.Simplex Method hand out last.pdf
Chapter 3.Simplex Method hand out last.pdf
 
Simplex algorithm
Simplex algorithmSimplex algorithm
Simplex algorithm
 
Hungarian Method
Hungarian MethodHungarian Method
Hungarian Method
 
6. assignment problems
6. assignment problems6. assignment problems
6. assignment problems
 
Assignment model
Assignment modelAssignment model
Assignment model
 
Simplex Algorithm
Simplex AlgorithmSimplex Algorithm
Simplex Algorithm
 

More from PriyankaLunavat (9)

Chapter 1 236(A).pptx
Chapter 1 236(A).pptxChapter 1 236(A).pptx
Chapter 1 236(A).pptx
 
Product Life Cycle.pptx
Product Life Cycle.pptxProduct Life Cycle.pptx
Product Life Cycle.pptx
 
TOP 5 .pptx
TOP 5 .pptxTOP 5 .pptx
TOP 5 .pptx
 
BREXIT Unit 1 IBE.pptx
BREXIT Unit 1 IBE.pptxBREXIT Unit 1 IBE.pptx
BREXIT Unit 1 IBE.pptx
 
TOP 5 truck BRANDS.pptx
TOP 5 truck BRANDS.pptxTOP 5 truck BRANDS.pptx
TOP 5 truck BRANDS.pptx
 
perimeter and area.pptx
perimeter and area.pptxperimeter and area.pptx
perimeter and area.pptx
 
Induction program-2.pptx
Induction program-2.pptxInduction program-2.pptx
Induction program-2.pptx
 
DABUR.pptx
DABUR.pptxDABUR.pptx
DABUR.pptx
 
Sahara India Pariwar Scam.pptx
Sahara India Pariwar Scam.pptxSahara India Pariwar Scam.pptx
Sahara India Pariwar Scam.pptx
 

Recently uploaded

Recently uploaded (20)

aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
 
Post Exam Fun(da) Intra UEM General Quiz - Finals.pdf
Post Exam Fun(da) Intra UEM General Quiz - Finals.pdfPost Exam Fun(da) Intra UEM General Quiz - Finals.pdf
Post Exam Fun(da) Intra UEM General Quiz - Finals.pdf
 
[GDSC YCCE] Build with AI Online Presentation
[GDSC YCCE] Build with AI Online Presentation[GDSC YCCE] Build with AI Online Presentation
[GDSC YCCE] Build with AI Online Presentation
 
INU_CAPSTONEDESIGN_비밀번호486_업로드용 발표자료.pdf
INU_CAPSTONEDESIGN_비밀번호486_업로드용 발표자료.pdfINU_CAPSTONEDESIGN_비밀번호486_업로드용 발표자료.pdf
INU_CAPSTONEDESIGN_비밀번호486_업로드용 발표자료.pdf
 
The Art Pastor's Guide to Sabbath | Steve Thomason
The Art Pastor's Guide to Sabbath | Steve ThomasonThe Art Pastor's Guide to Sabbath | Steve Thomason
The Art Pastor's Guide to Sabbath | Steve Thomason
 
50 ĐỀ LUYỆN THI IOE LỚP 9 - NĂM HỌC 2022-2023 (CÓ LINK HÌNH, FILE AUDIO VÀ ĐÁ...
50 ĐỀ LUYỆN THI IOE LỚP 9 - NĂM HỌC 2022-2023 (CÓ LINK HÌNH, FILE AUDIO VÀ ĐÁ...50 ĐỀ LUYỆN THI IOE LỚP 9 - NĂM HỌC 2022-2023 (CÓ LINK HÌNH, FILE AUDIO VÀ ĐÁ...
50 ĐỀ LUYỆN THI IOE LỚP 9 - NĂM HỌC 2022-2023 (CÓ LINK HÌNH, FILE AUDIO VÀ ĐÁ...
 
Matatag-Curriculum and the 21st Century Skills Presentation.pptx
Matatag-Curriculum and the 21st Century Skills Presentation.pptxMatatag-Curriculum and the 21st Century Skills Presentation.pptx
Matatag-Curriculum and the 21st Century Skills Presentation.pptx
 
Incoming and Outgoing Shipments in 2 STEPS Using Odoo 17
Incoming and Outgoing Shipments in 2 STEPS Using Odoo 17Incoming and Outgoing Shipments in 2 STEPS Using Odoo 17
Incoming and Outgoing Shipments in 2 STEPS Using Odoo 17
 
The impact of social media on mental health and well-being has been a topic o...
The impact of social media on mental health and well-being has been a topic o...The impact of social media on mental health and well-being has been a topic o...
The impact of social media on mental health and well-being has been a topic o...
 
UNIT – IV_PCI Complaints: Complaints and evaluation of complaints, Handling o...
UNIT – IV_PCI Complaints: Complaints and evaluation of complaints, Handling o...UNIT – IV_PCI Complaints: Complaints and evaluation of complaints, Handling o...
UNIT – IV_PCI Complaints: Complaints and evaluation of complaints, Handling o...
 
Post Exam Fun(da) Intra UEM General Quiz 2024 - Prelims q&a.pdf
Post Exam Fun(da) Intra UEM General Quiz 2024 - Prelims q&a.pdfPost Exam Fun(da) Intra UEM General Quiz 2024 - Prelims q&a.pdf
Post Exam Fun(da) Intra UEM General Quiz 2024 - Prelims q&a.pdf
 
Salient features of Environment protection Act 1986.pptx
Salient features of Environment protection Act 1986.pptxSalient features of Environment protection Act 1986.pptx
Salient features of Environment protection Act 1986.pptx
 
Basic_QTL_Marker-assisted_Selection_Sourabh.ppt
Basic_QTL_Marker-assisted_Selection_Sourabh.pptBasic_QTL_Marker-assisted_Selection_Sourabh.ppt
Basic_QTL_Marker-assisted_Selection_Sourabh.ppt
 
Benefits and Challenges of Using Open Educational Resources
Benefits and Challenges of Using Open Educational ResourcesBenefits and Challenges of Using Open Educational Resources
Benefits and Challenges of Using Open Educational Resources
 
How to Manage Notification Preferences in the Odoo 17
How to Manage Notification Preferences in the Odoo 17How to Manage Notification Preferences in the Odoo 17
How to Manage Notification Preferences in the Odoo 17
 
“O BEIJO” EM ARTE .
“O BEIJO” EM ARTE                       .“O BEIJO” EM ARTE                       .
“O BEIJO” EM ARTE .
 
Pragya Champions Chalice 2024 Prelims & Finals Q/A set, General Quiz
Pragya Champions Chalice 2024 Prelims & Finals Q/A set, General QuizPragya Champions Chalice 2024 Prelims & Finals Q/A set, General Quiz
Pragya Champions Chalice 2024 Prelims & Finals Q/A set, General Quiz
 
Advances in production technology of Grapes.pdf
Advances in production technology of Grapes.pdfAdvances in production technology of Grapes.pdf
Advances in production technology of Grapes.pdf
 
....................Muslim-Law notes.pdf
....................Muslim-Law notes.pdf....................Muslim-Law notes.pdf
....................Muslim-Law notes.pdf
 
An Overview of the Odoo 17 Discuss App.pptx
An Overview of the Odoo 17 Discuss App.pptxAn Overview of the Odoo 17 Discuss App.pptx
An Overview of the Odoo 17 Discuss App.pptx
 

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

  • 2. 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. 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. 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. 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. 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. 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. 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. 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. 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 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 Job 1 Job 2 Job 3 Job 4 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. 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 Job 1 Job 2 Job 3 Job 4 Machine 1 2 0 3 3 Machine 2 3 0 0 4 Machine 3 2 5 0 2 Machine 4 3 2 0 3 Job 1 Job 2 Job 3 Job 4 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. 12 Step 3: Cover all the zeroes of the matrix with the minimum number of horizontal or vertical lines. Example 1 Job 1 Job 2 Job 3 Job 4 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. 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 Job 1 Job 2 Job 3 Job 4 Machine 1 0 0 3 1 Machine 2 1 0 0 2 Machine 3 0 5 0 0 Machine 4 1 2 1 0 Job 1 Job 2 Job 3 Job 4 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. 14 Similarly you can perform the operation for Remaining Matrix Example 1 Job 1 Job 2 Job 3 Job 4 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. 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
  • 16. 16 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 Job 1 Job 2 Job 3 Job 4 Job 5 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 Job 1 Job 2 Job 3 Job 4 Job 5 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. 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 Job 1 Job 2 Job 3 Job 4 Job 5 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 Job 1 Job 2 Job 3 Job 4 Job 5 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
  • 18. 18 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) Job 1 Job 2 Job 3 Job 4 Job 5 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. 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 Job 1 Job 2 Job 3 Job 4 Job 5 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. 20 Improvement 1: Example 2 1 2 3 4 Job 1 Job 2 Job 3 Job 4 Job 5 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. Job 1 Job 2 Job 3 Job 4 Job 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 After Improvement Note: After Improvement start the repeat process with cover zeroes
  • 21. 21 Improvement 1: Example 2 1 2 3 4 Since the minimal number of lines is 5, an optimal assignment of zeroes is possible. No. of Rows / Columns = No of Lines 5 = 5 Job 1 Job 2 Job 3 Job 4 Job 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. 22 Step 6 : Once the No. of Rows / Columns = No of Lines conditions are satisfied Find out the optimum solution Example 2 Job 1 Job 2 Job 3 Job 4 Job 5 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. 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. 24 1) Unbalanced Problem 2) Multiple Optimum Solution 3) Maximization Problem 4) Prohibited Assignment Problem ( Restricted Problem) Special Cases in Assignment Problem
  • 25. 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: Special Cases in Assignment 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 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. 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. 27 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 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
  • 28. 28 Example on Unbalanced Problem 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
  • 29. 29 Example on Unbalanced Problem 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. No. of Rows / Columns ≠ No of Lines 6 ≠ 5 5
  • 30. 30 Example on Unbalanced Problem 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 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. In the given example 2 can be subtracted from all uncovered elements and add only at cell E1,E3,E6 and F1,F3,F6. Keep remaining covered element as it is 5
  • 31. 31 Example on Unbalanced Problem 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 Note: After Improvement start the repeat process with cover zeroes
  • 32. 32 Example on Unbalanced Problem 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 Step 6: Since the minimal number of lines is 5, and total number of rows/ columns are 6 hence we need to perform improvement number 2. No. of Rows / Columns ≠ No of Lines 6 ≠ 5
  • 33. 33 Example on Unbalanced Problem 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 Step 7: 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. In the given example 2 can be subtracted from all uncovered elements and add only at cell A1,A4,C1,C4,F2 and F4. Keep remaining covered element as it is
  • 34. 34 Example on Unbalanced Problem 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 Note: After Improvement start the repeat process with cover zeroes
  • 35. 35 Example on Unbalanced Problem 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 Since the minimal number of lines is 6, an optimal assignment of zeroes is possible. No. of Rows / Columns = No of Lines 6 = 6
  • 36. 36 Example on Unbalanced Problem 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 Note: For Time Kindly check Original Matrix in given 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. 37 Step 8 : Once the No. of Rows / Columns = No of Lines conditions are satisfied Find out the optimum solution Example 2 Job 1 Job 2 Job 3 Job 4 Job 5 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
  • 38. 38 1) Unbalanced Problem Solve the following assignment problem ? Special Cases in 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. 39 1) Unbalanced Problem Step 1: Row Subtraction Special Cases in 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 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. 40 1) Unbalanced Problem Step 1: Column Subtraction Special Cases in Assignment Problem 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. 41 1) Unbalanced Problem Step 3 : Cover zeroes Special Cases in Assignment Problem 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. 42 1) Unbalanced Problem Improvement No. 1 Special Cases in Assignment Problem 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. 43 1) Unbalanced Problem Improvement No. 1 Special Cases in Assignment Problem 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. 44 1) Unbalanced Problem Improvement No. 2 Special Cases in Assignment Problem 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. 45 1) Unbalanced Problem Improvement No. 2 Special Cases in Assignment Problem 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. 46 1) Unbalanced Problem Optimum Solution Special Cases in Assignment Problem 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. 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. Special Cases in Assignment Problem
  • 48. 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. 49 Example On Multiple Optimum Solution 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. 50 Example On Multiple Optimum Solution 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, 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. 51 Example On Multiple Optimum Solution 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 Step 3: Cover all the zeroes of the matrix with the minimum number of horizontal or vertical lines. 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. No. of Rows / Columns ≠ No of Lines 4 ≠ 𝟐
  • 52. 52 Example On Multiple Optimum Solution 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. In the above example 5 can be subtracted from all uncovered elements and add only at cell D-M1
  • 53. 53 Example On Multiple Optimum Solution 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. In the above example 5 can be subtracted from all uncovered elements and add only at cell D-M1
  • 54. 54 Example On Multiple Optimum Solution 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 Note: After Improvement start the repeat process with cover zeroes
  • 55. 55 Example On Multiple Optimum Solution 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 Step 6: Since the minimal number of lines is 3, and total number of rows/ columns are 4 hence we need to perform improvement number 2. No. of Rows / Columns ≠ No of Lines 4 ≠ 𝟑
  • 56. 56 Example On Multiple Optimum Solution 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 Step 7: 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. In the given example 4 can be subtracted from all uncovered elements and add only at cell D-M1 and D-M2. Keep remaining covered element as it is
  • 57. 57 Example On Multiple Optimum Solution 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 Note: After Improvement start the repeat process with cover zeroes
  • 58. 58 Example On Multiple Optimum Solution 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 Since the minimal number of lines is 4, an optimal assignment of zeroes is possible. No. of Rows / Columns = No of Lines 4 = 4
  • 59. 59 Example On Multiple Optimum Solution 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 Step 6 : Once the No. of Rows / Columns = No of Lines conditions are satisfied Find out the optimum 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. 60 Example On Multiple Optimum Solution 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. 61 Example On Multiple Optimum Solution 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. 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. 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 Special Cases in Assignment Problem
  • 64. 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. 65 Step 1: Balance the given problem Example on Maximization 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. 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 Example on Maximization Problem 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. 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. Example on Maximization Problem 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. 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. Example on Maximization Problem 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. 69 Step 5: Cover all the zeroes of the matrix with the minimum number of horizontal or vertical lines Example on Maximization Problem 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. No. of Rows / Columns = No of Lines 5 = 5
  • 70. 70 Example on Maximization Problem Step 7 : Once the No. of Rows / Columns = No of Lines conditions are satisfied Find out the optimum 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 Step 8: Once we perform the allocation final assignment is as follows Note: For Time Kindly check Original Matrix in given question Operator Machine Production 1 A 10 2 D 10 3 C 09 4 E 00 5 B 09 Total Time 38
  • 71. 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. 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. Special Cases in Assignment Problem 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. 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. 74 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, 3 from 2nd col, 2 from 3rd col and 0 from 4th col respectively Example on Prohibited Roots 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. 75 Step 3: Cover all the zeroes of the matrix with the minimum number of horizontal or vertical lines. Example on Prohibited Roots 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 Step 4: Since the minimal number of lines is 4, and total number of rows/ columns are 4 hence we need to perform improvement. No. of Rows / Columns = No of Lines 4 = 4
  • 76. 76 Step 8 : Once the No. of Rows / Columns = No of Lines conditions are satisfied Find out the optimum solution Example on Prohibited Roots 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 Note: For Time Kindly check Original Matrix in given question Clerks Jobs Time 1 B 07 2 D 04 3 A 03 4 C 04 Total Time 18 Hrs 0 0 0
  • 77. 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. 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. 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. 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