Assignment Poblems

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Assignment Problems
Key Points
 An assignment problem is a particular case of transportation problem.
 The assignment is to be made on a one-to-one basis (one job to one worker).
 The objective in assignment problem is to assign certain number of resources or facilities to an equal number of tasks or activities.( examples-job to machines or workers, products to factories, salesmen to territories, contracts to bidders, etc) to minimize total cost /time or maximize profit/revenue/efficiency.
 An assignment problem is a balanced when the number of rows is equal to number of columns.
 An unbalanced assignment problem can be balanced by adding dummy rows or columns as the case may be to make rows equal to columns.
 In the dummy row or column all elements are zero.
 An assignment problem involving restrictions in allocations (worker cannot be assigned to a machine as he may not possess the skill to operate the machine) is known as prohibited assignment.
 In case of prohibited assignment problem, we assign cost/time of higher value say M at the prohibited combination.
 An assignment problem can have more than one solution giving the same answer.
 Alternate (Multiple) optimum assignment solution exists, when there are multiple zeros in columns and rows.
 Hungarian method is the most efficient method to solve assignment problems when the objective is minimization.
 In case the objective function is maximization of say profit, we first convert the profit table into an opportunity loss table before we apply the Hungarian method.
 A travelling salesman problem is typical assignment problem with two additional constraints:
 Salesmen should not visit the city twice until he has visited all the cities.

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 No assignment should be made along the diagonal line.
 Assignment technique is of little use to a firm whose facilities are perfect substitutes of each other. For example if there are three identical machines doing the same job, any machine can be assigned for the job and hence one-to-one allocation will be disturbed.
Hungarian Method
 Developed by Hungarian Mathematician D. Konig.
 It works on the principle of reducing the given cost matrix to a matrix of opportunity cost.
 It reduces the given cost/time or opportunity loss matrix to the extent of having one zero in each row and column.
Steps involved in solving Assignment Problems by Hungarian Method Step1-Express the given problem into an n × n cost matrix or table. Step 2- Subtract the minimum element of each row of the matrix from all elements of the respective row. This step is known as Row Minima. Step 3- Subtract the minimum element of each column from all the elements in that respective column. This step is known as Column Minima. Step 4- Draw the least (minimum) number of horizontal and/or vertical lines to cover all zeros in the matrix. If the number of lines drawn are equal to the number of rows or columns, optimum solution has been reached. In such a case proceed to step number 7 for assignment. However if the number of lines are not equal to rows or columns, proceed to step number 5 for modification. Step 5- Select the smallest element not covered by the lines, subtract it from all uncovered elements including itself, add it to the elements which are crossed by two lines and reproduce other elements crossed by one line intact. Step6-Repeat steps 4 and 5 until an optimal solution is reached. Step 7- Proceed to job assignments as follows:
Examine the rows one by one starting with the first row until a row with an exactly one zero is found. Mark the zero by enclosing it in a square indicating assignment of the task to the

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facility. After doing so, cross out all the zeros (if any) in that column as they cannot be used to make other assignments. Examine next the columns for any column having a single zero, starting from the first column. Mark the zero as mentioned above, crossing out the remaining zeros (if any) in that row. Repeat steps (i) and (ii) alternatively until either of the following conditions occur: All the zeros have been marked or crossed, ensuring in the process that each row has a zero marked. This means that optimum solution has been reached. All the zeros cannot be marked /crossed .There are at least two zeros in each row and column which cannot be marked by using the above mentioned step 7-(i) and (ii).This means that more than one solution exists. In such a case we mark any one zero in a row of our choice and cross out the remaining zeros in the row and the column where the zero is marked. Carry out this exercise till we get one zero marked or assigned in each row and there is no further assignment required. Step 8- Summarize the result by having the final assignment table. Illustrations 1. A company has five jobs to be done. The following table shows the cost of assigning each job to each machine. Assign five jobs to the five machines so as to minimize the total cost in INR in ‘000’.
Machine
Job
1
2
3
4
5
M1
5
11
10
12
4
M2
2
4
6
3
5
M3
3
12
14
6
4
M4
6
14
4
11
7
M5
7
9
8
12
5
Solution: Hungarian method is used to obtain optimal solution.

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As the numbers of rows are equal to columns, we have a balanced assignment table and move on to step number 2. Step 2- Subtract the minimum element of each row of the matrix from all elements of the respective row. This step is known as Row Minima.
Machine
Job
1
2
3
4
5
M1
1
7
6
8
0
M2
0
2
4
1
3
M3
0
9
11
3
1
M4
2
10
0
7
3
M5
2
4
3
7
0
Note-In the first row 4 is lowest element and we have subtracted this lowest element from all the elements in that row. The same methodology is used for the other rows. Step 3- Subtract the minimum element of each column from all the elements in that respective column. This step is known as Column Minima.
Machine
Job
1
2
3
4
5
M1
1
5
6
7
0
M2
0
0
4
0
3
M3
0
7
11
2
1
M4
2
8
0
6
3
M5
2
2
3
6
0

5
Note- Column 1, 3 and 5 remain the same, as we have zero as the lowest element in each of these columns. In column 2 and 4 we have 2 and 1 as the lowest elements which we have subtracted from all elements in those respective columns. Step 4- Draw the least (minimum) number of horizontal and/or vertical lines to cover all zeros in the table.
Machine
Job
1
2
3
4
5
M1
1
5
6
7
0
M2
0
0
4
0
3
M3
0
7
11
2
1
M4
2
8
0
6
3
M5
2
2
3
6
0
Note- We see from the above table that only four lines are sufficient to cross all zeros.This is achieved by drawing minimum number of lines (horizontal as well as vertical) with each line crossing out maximum zeros. This is the most important step in the method and there is a chance that students can make a mistake by drawing more lines to cross out all zeros than necessary. In case you draw more lines than rows or columns it is an indication that you have a mistake. Cancel the drawn lines and draw it up fresh. Use pencils to draw the lines. As numbers of lines are not equal to rows or columns optimum solution has not been reached and we move to step number 5 for modification. Step 5- Select the smallest element not covered by the lines, subtract it from all uncovered elements including itself, add it to the elements which are crossed by two lines and reproduce other elements crossed by one line intact.

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Machine
Job
1
2
3
4
5
M1
1
3
4
5
0
M2
2
0
4
0
5
M3
0
5
9
0
1
M4
4
8
0
6
5
M5
2
0
1
4
0
Note- 2 is the smallest non-crossed element we have subtracted from all non-crossed elements including itself and added this smallest element 2 to the elements which are crossed by two lines (for example element 3 in the second row was crossed by two lines and hence 2 was added to the element 3 by which the new element in this row is reading 5) .All other elements which are crossed by one line remain intact (for example digit 2 in the last row remains unchanged as it is crossed by only one line). We again draw minimum number of lines crossing out all zeros in the table as shown below:
Machine
Job
1
2
3
4
5
M1
1
3
4
5
0
M2
2
0
4
0
5
M3
0
5
9
0
1
M4
4
8
0
6
5
M5
2
0
1
4
0
Note- We require five lines as they are minimum number lines which are required to cross out all zeros. As the number of lines is equal to rows or columns optimum solution has been reached and we move to step number 7 for assignment. Step 7- Proceed to job assignments as follows:

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Examine the rows one by one starting with the first row until a row with an exactly one zero is found. Mark the zero by enclosing it in a square indicating assignment of the task to the facility. After doing so, cross out all the zeros (if any) in that column as they cannot be used to make other assignments. Examine next the columns for any column having a single zero, starting from the first column. Mark the zero as mentioned above, crossing out the remaining zeros (if any) in that row. Repeat steps (i) and (ii) alternatively until all the zeros have been either assigned or crossed.
Machine
Job
1
2
3
4
5
M1
1
3
4
5
0
M2
2
0
4
5
M3
0
5
9
0
1
M4
4
8
0
6
5
M5
2
1
4
0
Note- (i) On examining, we find that since in row number one there is a single zero for assignment which we mark it by having a square around it. After doing so we check for any other zeros in that respective column for it to be crossed out. We find that a zero is therein column number five which we cross out. (ii) After doing the row exercise, we try to find a single zero in the column starting from column number one. We find that in column no one itself there is a single zero which is marked by having a square around it. After doing so let us find whether there are other zeros in that respective row .We find that there is one zero in row number three which we cross out. (iii)We again proceed to look for a single zero in a row going row by row. We find there is a single zero in row number two for assignment which we mark it by having a square around it. Again we should not forget to cross out any other zeros in that respective column. There are no zeros in that respective column which is row number three.
0
0
0
0
0

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Proceed to look for a single zero in a column by going column by column. We locate it in the column number four for assignment. We mark it by having a square around it and search for any other zeros in that respective row to be crossed out. There is none to be crossed out in row number four. We continue this row column exercise till all unique zeros is marked and others crossed out. Step 8-Summary of the assignment is given in the table below:
Machine
Job
Cost in INR in ‘’000’’
M1
5
4
M2
4
3
M3
1
3
M4
3
4
M5
2
9
Total
23
Each of four workers A, B, C, D can do each of the four jobs I, II, III, IV. The figures within the matrix given below show the time taken by each worker in minutes to do each job. Assign the jobs to the four workers, only one to each so as to minimize total time to do all the jobs.
Jobs
Workers
A
B
C
D
I
4
3
12
7
II
5
4
7
9
III
3
1
6
2
IV
5
6
9
5
Solution: Hungarian method is used to obtain optimal solution.

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Jobs
Workers
A
B
C
D
I
1
0
9
4
II
1
0
3
5
III
2
0
5
1
IV
0
1
4
0
Note-In the first row 3 is lowest element and we have subtracted this lowest element from all the elements in that row. The same methodology is used for the other rows. Step 3- Subtract the minimum element of each column from all the elements in that respective column. This step is known as Column Minima.
Jobs
Workers
A
B
C
D
I
1
0
6
4
II
1
0
0
5
III
2
0
2
1
IV
0
1
1
0
Step 4- Draw the least (minimum) number of horizontal and/or vertical lines to cover all zeros in the table.

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Note- We see that only three lines are sufficient to cancel all zeros. We have drawn minimum number of horizontal and vertical lines with each line striking off maximum zeros. As the number of lines are less than the rows and columns, we move on to step number 5 for modification. Step 5- Select the smallest element not covered by the lines, subtract it from all uncovered elements including itself, add it to the elements which are crossed by two lines and reproduce other elements crossed by one line intact.
Jobs
Workers
A
B
C
D
I
0
0
5
3
II
1
1
0
5
III
1
0
1
0
IV
0
2
1
0
Note- 1 is the smallest non-crossed element we have subtracted from all non-crossed elements including itself and added this smallest element 1 to the elements which are crossed by two lines (for example element 1 in the last row was crossed by two lines and hence 1 was added to the element 1 by which the new element in this row instead is reading 2) .All other elements which are crossed by one line remain intact (for example digit 1 in the last row remains unchanged as it is crossed by only one line). Again we draw minimum number of horizontal or vertical lines crossing out all zeros.
Jobs
Workers
A
B
C
D
I
1
0
6
4
II
1
0
0
5
III
2
0
2
1
IV
0
1
1
0

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Jobs
Workers
A
B
C
D
I
0
0
5
3
II
1
1
0
5
III
1
0
1
0
IV
0
2
1
0
Note- We require four lines (minimum) to cross out all zeros. As the numbers of lines drawn are equal to rows and columns optimum solution has been reached and we proceed to Step 7 for assignment.
Jobs
Workers
A
B
C
D
I
0
0
5
3
II
1
1
0
5
III
1
0
1
0
IV
0
2
1
0
Note- (i) On examining, we find that since in row number two there is a single zero for assignment which we mark it by having a square around it. After doing so we check for any other zeros in that respective column for it to be crossed out. There are no zeros to be crossed out. (ii) After doing the row exercise, we try to find a single zero in the column starting from column number one. We do not find a single zero in any of the columns (iii)We again proceed to look for a single zero in a row going row by row. We find there is a no single zero in any row. This situation signifies that we have more than one solution to the problem .i.e. Multiple Solutions
0
0
0

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In such a case, we select the remaining zeros to be blocked of our choice ,remembering one think that while doing so ,we ensure that there is only one zero in each row for assignment. When we block the zero arbitrarily, any zeros in that respective column as well as row get crossed out. For solution 1 this is done by marking zero in the first row and first column by which zero in that column as well as row gets crossed out. Now we see that zero in third row and second column is free for assignment and zero in the fourth column in the same row gets crossed out. We have only one zero left in row number four and column number four which we mark. This completes the assignment and we get Solution-1 By doing this exercise, we get two optimum solutions whose assignment is different, but the total minutes to do all the jobs by the workers remain the same. Solution-1
Job
Worker
Time in minutes
I
A
4
II
C
7
III
B
1
IV
D
5
Total
17
Solution-2 Note-In the second solution we have marked zero in the first row second column and crossed zeros in the first row and second column. We continue this exercise as described above to get the solution which is given below:
Jobs
Workers
A
B
C
D
I
0
0
5
3
II
1
1
0
5
III
1
0
1
0
IV
0
2
1
0
0
0
0
0

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Job
Worker
Time in minutes
I
B
3
II
C
7
III
D
2
IV
A
5
Total
17
3. The government solicits five different proposals with the intent of giving one job to each of the companies. The bid amounts in thousands of INR are given below with X denoting no bid submitted as the company does not meet the technical criteria for that job. Find the optimal assignment to companies such that the total cost is minimum? Solution: This prohibited assignment problem and can be solved by Hungarian method as the objective is to minimize total cost.
Company
Proposals(INR in ‘’000’’)
1
2
3
4
5
A
50
85
100
75
80
B
80
85
95
X
90
C
70
80
85
75
80
D
X
90
95
70
85
E
85
80
90
80
90

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Company
1
2
3
4
5
A
0
35
50
25
30
B
0
5
15
M
10
C
0
10
15
5
10
D
M
20
25
0
15
E
5
0
10
0
10
Note- M is so very high that even after subtracting small elements like 50 or 70, M remains unchanged. Step 3- Subtract the minimum element of each column from all the elements in that respective column. This step is known as Column Minima.
Company
1
2
3
4
5
A
0
35
40
25
20
B
0
5
5
M
0
C
0
10
5
5
0
D
M
20
15
0
5
E
5
0
0
0
0

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Note- As minimum number of horizontal and vertical lines are less than number of rows or columns, we move on to Step 5 for modification. Step 5- Select the smallest element not covered by the lines, subtract it from all uncovered elements including itself, add it to the elements which are crossed by two lines and reproduce other elements crossed by one line intact.
Company
1
2
3
4
5
A
30
35
25
20
B
0
0
M
0
C
5
0
5
0
D
M
15
10
0
5
E
0
0
5
5
Note- As the number of lines drawn are equal to rows or columns, optimum solution has been reached and we proceed to Step number 7 for assignment.
Company
1
2
3
4
5
A
0
35
40
25
20
B
0
5
5
M
0
C
0
10
5
5
0
D
M
20
15
0
5
E
5
0
0
0
0
10
0
0
0

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Company
1
2
3
4
5
A
0
30
35
25
20
B
0
0
0
M
0
C
0
5
0
5
0
D
M
15
10
0
5
E
10
0
0
5
5
Note- (i) On examining, we find that since in row number two there is a single zero for assignment which we mark it by having a square around it. After doing so we check for any other zeros in that respective column for it to be crossed out. There are two zeros which can be crossed out. (ii) After doing the row exercise, we try to find a single zero in the column starting from column number one. We do not find a single zero in any of the columns (iii)We again proceed to look for a single zero in a row going row by row. We find that there is one zero in fourth row which we mark .After marking we find that there is no zero in that respective column which can be crossed out. Now there are multiple zeros in row as well as column. This situation signifies that we have more than one solution to the problem .i.e. Multiple Solutions In such a case, we select the remaining zeros to be blocked of our choice ,remembering one think that while doing so ,we ensure that there is only one zero in each row for assignment. When we block the zero arbitrarily, any zeros in that respective column as well as row get crossed out. By doing this exercise, we get two optimum solutions whose assignment is different, but the total proposal cost in INR in “000” remains the same. Solution -1

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Company
Proposals
INR in “000”
A
1
50
B
2
85
C
5
80
D
4
70
E
3
90
Total
375
In a similar way we do the arbitrarily allotment as mentioned above (Student if need be can refer to illustration 2) to get Solution-2.
Company
Proposals
INR in “000”
A
1
50
B
3
95
C
5
80
D
4
70
E
2
80
Total
375
4. Schedule the training seminars in five working days of the week so that the number of students unable to attend is kept to the minimum. The details are as follows:

18
Days
Leasing (L)
Portfolio Management (PM)
Private Mutual Fund (PMF)
Equity Research (ER)
Monday
50
40
60
20
Tuesday
40
30
40
30
Wednesday
60
20
30
20
Thursday
30
30
20
30
Friday
10
20
10
30
Solution: As the number of columns is not equal to the number of rows, this is a case of unbalanced assignment problem. Hence before proceeding ahead with the Hungarian method, we need to ensure that numbers of rows are equal to number of columns and this is done by introducing dummy column having all its elements as zero. The balanced assignment table is given below:
Days
Leasing (L)
Dummy
Monday
50
40
60
20
0
Tuesday
40
30
40
30
0
Wednesday
60
20
30
20
0
Thursday
30
30
20
30
0
Friday
10
20
10
30
0
As each row has a zero as minimum element, we straightaway proceed to Step3 (column minima). Step 3- Subtract the minimum element of each column from all the elements in that respective column. This step is known as Column Minima.

19
Days
Leasing (L)
Dummy
Monday
40
20
50
0
0
Tuesday
30
10
30
10
0
Wednesday
50
0
20
0
0
Thursday
20
10
10
10
0
Friday
0
0
0
10
0
Days
Leasing (L)
Dummy
Monday
40
20
50
0
0
Tuesday
30
10
30
10
0
Wednesday
50
0
20
0
0
Thursday
20
10
10
10
0
Friday
0
0
0
10
0
Note- We see that only four lines are sufficient to cancel all zeros. We have drawn minimum number of horizontal and vertical lines with each line striking off maximum zeros. As the number of lines are less than the rows and columns, we move on to step number 5 for modification. Step 5- Select the smallest element not covered by the lines, subtract it from all uncovered elements including itself, add it to the elements which are crossed by two lines and reproduce other elements crossed by one line intact.

20
Days
Leasing (L)
Dummy
Monday
30
10
40
0
0
Tuesday
20
0
20
10
0
Wednesday
50
0
20
10
10
Thursday
10
0
0
10
0
Friday
0
0
0
20
10
As the number of lines is equal to rows or columns optimum solution has been reached and we move to step number 7 for assignment. Step 7- Proceed to job assignments as follows: Examine the rows one by one starting with the first row until a row with an exactly one zero is found. Mark the zero by enclosing it in a square indicating assignment of the task to the facility. After doing so, cross out all the zeros (if any) in that column as they cannot be used to make other assignments. Examine next the columns for any column having a single zero, starting from the first column. Mark the zero as mentioned above, crossing out the remaining zeros (if any) in that row. Repeat steps (i) and (ii) alternatively until all the zeros have been either assigned or crossed.
Days
Leasing (L)
Dummy
Monday
30
10
40
0
0
Tuesday
20
0
20
10
0
Wednesday
50
0
20
10
10
Thursday
10
0
0
10
0
Friday
0
0
0
20
10
Step 8-Summary of the assignment is given in the table below:

21
Days
Training Seminar
Absenteeism of students
Monday
ER
20
Tuesday
-
-
Wednesday
PM
20
Thursday
PMF
20
Friday
L
10
Tuesday is the day off from the training program. 5. The Marketing Director of the multinational company faced with the problem assigning five senior marketing managers to six zones. From past experience he knows that the efficiency percentage by sales depends on marketing manager-zone combination given in the following table:
Marketing Manager
Zones
1
2
3
4
5
6
A
71
83
85
80
76
78
B
79
83
67
74
72
83
C
73
70
81
82
76
89
D
91
94
84
89
81
80
E
88
89
77
87
67
74
As an advisor to the company, recommend which zone should be manned by junior manager so as to maximize the overall efficiency of the company. Solution: We see two things which are different: The objective function is maximization of the efficiency of the company.
The problem is an unbalanced transportation problem as the number of rows is not equal to number of columns.

22
Note- Hungarian method can only be issued when the objective function is minimization. There is a deviation here as the objective function is maximization. In such a case we first convert the given assignment table (let the table be balanced or unbalanced) to an opportunity loss table by subtracting each and every element of the table from the highest element in the table which in this case is 94.By doing so the opportunity loss table is as following:
Marketing Manager
Zones
1
2
3
4
5
6
A
23
11
9
14
18
16
B
15
11
27
20
22
11
C
21
24
13
12
18
5
D
3
0
10
5
13
14
E
6
5
17
7
27
20
The above opportunity loss table is unbalanced as the number of rows is not equal to number of columns .A dummy row is added with each element in that row being zero. By doing so the balanced opportunity loss table is given as follows:
Marketing Manager
Zones
1
2
3
4
5
6
A
71
83
85
80
76
78
B
79
83
67
74
72
83
C
73
70
81
82
76
89
D
91
94
84
89
81
80
E
88
89
77
87
67
74
Dummy
0
0
0
0
0
0
We now proceed to Step number 2 for row minima.

23
Step 2- Subtract the minimum element of each row of the matrix from all elements of the respective row. This step is known as Row Minima.
Marketing Manager
Zones
1
2
3
4
5
6
A
14
2
0
5
9
7
B
4
0
16
9
11
0
C
16
19
8
7
13
0
D
3
0
10
5
13
14
E
1
0
12
2
22
15
Dummy
0
0
0
0
0
0
As each column has a zero, column minima is not required and we move on to step number 4 Step 4- Draw the least (minimum) number of horizontal and/or vertical lines to cover all zeros in the table.
Marketing Manager
Zones
1
2
3
4
5
6
A
14
2
0
5
9
7
B
4
0
16
9
11
0
C
16
19
8
7
13
0
D
3
0
10
5
13
14
E
1
0
12
2
22
15
Dummy
0
0
0
0
0
0
Note- We see that only four lines are sufficient to cancel all zeros. We have drawn minimum number of horizontal and vertical lines with each line striking off maximum zeros. As the

24
number of lines are less than the rows and columns, we move on to step number 5 for modification. Step 5- Select the smallest element not covered by the lines, subtract it from all uncovered elements including itself, add it to the elements which are crossed by two lines and reproduce other elements crossed by one line intact.
Marketing Manager
Zones
1
2
3
4
5
6
A
14
3
0
5
9
10
B
1
0
13
6
8
0
C
13
19
5
4
10
0
D
0
0
7
2
10
14
E
0
2
11
1
21
17
Dummy
0
3
0
0
0
3
Note- 1 is the smallest non-crossed element we have subtracted from all non-crossed elements including itself and added this smallest element 1 to the elements which are crossed by two lines (for example element 2 in the second column was crossed by two lines and hence 1 was added to the element 2 by which the new element in this column is reading 3) .All other elements which are crossed by one line remain intact (for example digit 14 in the first row remains unchanged as it is crossed by only one line). We again draw minimum number of lines crossing out all zeros in the table as shown below:

25
Marketing Manager
Zones
1
2
3
4
5
6
A
14
3
0
5
9
10
B
1
0
13
6
8
0
C
13
19
5
4
10
0
D
0
0
7
2
10
14
E
0
2
11
1
21
17
Dummy
0
3
0
0
0
3
Note- We again see that only five lines are sufficient to cancel all zeros. We have drawn minimum number of horizontal and vertical lines with each line striking off maximum zeros. As the number of lines is less than the rows and columns, we again use step number 5 for modification.
Marketing Manager
Zones
1
2
3
4
5
6
A
14
3
0
4
8
10
B
1
0
13
5
7
0
C
13
19
5
3
9
0
D
0
0
7
1
9
14
E
0
2
11
0
20
17
Dummy
1
4
1
0
0
4
We again draw minimum number of lines crossing out all zeros in the table as shown below

26
Marketing Manager
Zones
1
2
3
4
5
6
A
14
3
0
4
8
10
B
1
0
13
5
7
0
C
13
19
5
3
9
0
D
0
0
7
1
9
14
E
0
2
11
0
20
17
Dummy
1
4
1
0
0
4
Marketing Manager
Zones
1
2
3
4
5
6
A
14
3
0
4
8
10
B
1
0
13
5
7
0
C
13
19
5
3
9
0
D
0
0
7
1
9
14
E
0
2
11
0
20
17
Dummy
1
4
1
0
0
4

27
Step 8-Summary of the assignment is given in the table below
Marketing Manager
Zone
Efficiency
A
3
85
B
2
83
C
6
89
D
1
91
E
4
87
: Zone 5 is left out, which can be manned by a junior manager to maximize overall efficiency of the company. 6. An airline that operates seven days in a week has time table shown below. Crews must have a minimum layover of five hours between flights. Obtain the pairing of flights that minimize layover time away from home. For any given pair the crew will be based at the city that results in smaller layover.
Delhi-Jaipur
Flight No
Departure
Arrival
101
7 A.M.
8 A.M.
102
8 A.M.
9 A.M.
103
1.30 P.M.
2.30 PM
104
6.30 P.M.
7.30 PM

28
Jaipur-Delhi
Flight No
Departure
Arrival
201
8 A.M.
9.15A.M.
202
8.30 A.M.
9.45 AM
203
12 Noon
1.15PM
204
5.30 P.M.
6.45 PM
For each pair also mention the place where the crew should be based. Solution: The illustration mentions about flight operating from Delhi-Jaipur and Jaipur-Delhi on a daily basis. We need to locate the crew for pair of flights where layover time (idle time) is the lowest. Since the objective is to minimize layover time between flights, we can use the Hungarian method. However before doing so we need to calculate the layover time for each pair of flights from Delhi to Jaipur and back with crew based at Delhi and similarly calculate the layover time for each pair of flights from Jaipur to Delhi and back with crew based at Jaipur. After getting the respective tables containing layover time, we select the lowest layover time out of the two tables for each pair of flights and by which get the lowest layover time table on which we need to carry out the Hungarian method.

29
Table consisting of Layover time in minutes for crew based in Delhi Delhi-Jaipur
Return Flight
Flight Number
Delhi-Jaipur Flights
201
202
203
204
101
*1440
1470
1680
570
102
1380
1410
1620
510
103
1050
1080
1290
1620
104
750
780
990
1320
Note-*101 flight lands at Jaipur at 8 am. In case the crew flying on 101 wants to come back to Delhi by 201 flight which takes off from Jaipur at 8 am, the layover time between these two flights is 24 hours (8am takeoff next day from Jaipur-8am, the time it lands at Jaipur.) equals 24 hours which is 1440 minutes. Let us do one more calculation. Consider 104-202 combination. Note-104 flight lands at Jaipur at 7.30 pm and 202 flight takes off from Jaipur at 8.30 am. Therefore the layover time between this pair of flight will be 8.30 am next day and 7.30 pm earlier day, which is equal to 13 hours and in minutes 13×60= 780 minutes. Note-We have to keep in mind that the minimum layover time has to be five hours. Layover time between pair of flights is equal to the difference between takeoff of the return flight and the landing of the earlier flight. Further we avoid fractions in terms of hours we have taken minutes as the basis for the formation of the above table. Can you do the balance calculations? I am sure you will. Let us do the same exercise of calculating layover time in minutes for crew based in Jaipur for the Jaipur-Delhi sector.

30
Table consisting of Layover time in minutes for crew based in Jaipur Delhi-Jaipur
Jaipur-Delhi
Flight Number
Return Flights
201
202
203
204
101
1305
**1275
1065
735
102
1365
1335
1125
795
103
1695
1665
1455
1125
104
555
525
315
1425
Note-** 202 flight lands at Delhi at 9.45 am and in case the crew on this flight wants to come back by 101 flight which departs from Delhi at 7am,the difference works out to 21 hours plus 15 minutes which is equal to 21×60+15 minutes=1275 minutes. As said before we now get the lowest layover time table from the above two tables for each pair of flights an which is as given below:
Flights
Flight Number
201
202
203
204
101
1305
1275
1065
570
102
1365
1335
1125
510
103
1050
1080
1290
1125
104
555
525
315
1320
 means crew is based at Jaipur and if there is nomeans crew is based at Delhi. As the above table is balanced and least cost table, we can use the Hungarian method and proceed to Step number 2 for Row Minima. Step 2- Subtract the minimum element of each row of the matrix from all elements of the respective row. This step is known as Row Minima.

31
Flights
Flight Number
201
202
203
204
101
735
705
495
0
102
855
825
615
0
103
0
30
240
75
104
240
210
0
1005
Step 3- Subtract the minimum element of each column from all the elements in that respective column. This step is known as Column Minima.
Flights
Flight Number
201
202
203
204
101
735
675
495
0
102
855
795
615
0
103
0
0
240
75
104
240
180
0
1005
Step4-Draw the least (minimum) number of horizontal and/or vertical lines to cover all zeros in the table.
Flights
Flight Number
201
202
203
204
101
735
675
495
0
102
855
795
615
0
103
0
0
240
75
104
240
180
0
1005

32
As the number of lines are less than the rows and columns, we move on to step number 5 for modification. Step 5- Select the smallest element not covered by the lines, subtract it from all uncovered elements including itself, add it to the elements which are crossed by two lines and reproduce other elements crossed by one line intact.
Flights
Flight Number
201
202
203
204
101
240
160
0
0
102
360
300
120
0
103
0
0
240
570
104
240
180
0
1500
Note- We again see that only four lines are sufficient to cancel all zeros. We have drawn minimum number of horizontal and vertical lines with each line striking off maximum zeros. As the number of lines is less than the rows and columns, we again use step number 5 for modification
Flights
Flight Number
201
202
203
204
101
80
0
0
0
102
200
140
120
0
103
0
0
400
730
104
240
180
0
1500
We again draw minimum number of lines crossing out all zeros in the table as shown below

33
Flights
Flight Number
201
202
203
204
101
80
0
0
0
102
200
140
120
0
103
0
0
400
730
104
240
180
0
1500
As the number of lines is equal to rows or columns optimum solution has been reached and we move to step number 7 for assignment.
Flights
Flight Number
201
202
203
204
101
80
0
0
0
102
200
140
120
0
103
0
0
400
730
104
240
180
0
1500
Step 7- Proceed to job assignments as follows: Examine the rows one by one starting with the first row until a row with an exactly one zero is found. Mark the zero by enclosing it in a square indicating assignment of the task to the facility. After doing so, cross out all the zeros (if any) in that column as they cannot be used to make other assignments. Examine next the columns for any column having a single zero, starting from the first column. Mark the zero as mentioned above, crossing out the remaining zeros (if any) in that row. Repeat steps (i) and (ii) alternatively until all the zeros have been either assigned or crossed.

34
Flights
Flight Number
201
202
203
204
101
240
160
0
0
102
360
300
120
0
103
0
0
240
570
104
240
180
0
1500
Step 8-Summary of the assignment is given in the table below
Pair of Flights
Crew based at
Layover Time in minutes
101-202
Jaipur
1275
102-204
Delhi
510
103-201
Delhi
1050
104-203
Jaipur
315
Total
3150
Note-For pair of flight101-202 we see from the least layover time table, the crew is based at Jaipur. Remember the marking and similarly for the other pair of flight we can know where the crew will be based. 7. A travelling salesman has to visit five cities. He wishes to start from a particular city, visit each city once and then return to his starting point. The travelling cost (Rs in 000) of each city from a particular city is given below:

35
From City
To City
A
B
C
D
E
A
M
2
5
7
1
B
6
M
3
8
2
C
8
7
M
4
7
D
12
4
6
M
5
E
1
3
2
8
M
What is the sequence of visit of the salesman to achieve the least cost route? Solution: The given assignment table is a balance assignment table and is regarding travelling salesman. We will use Hungarian method with a difference to satisfy the condition that, the salesman should visit each city before returning to the city where he started from. Note- A travelling salesman problem is typical assignment problem with two additional constraints: Salesmen should not visit the city twice until he has visited all the cities. No assignment should be made along the diagonal line. As the numbers of rows are equal to columns, we have a balanced assignment table and move on to step number 2. Step 2- Subtract the minimum element of each row of the matrix from all elements of the respective row. This step is known as Row Minima.

36
From City
To City
A
B
C
D
E
A
M
1
4
6
0
B
4
M
1
6
0
C
4
3
M
0
3
D
8
0
2
M
1
E
0
2
1
7
M
Step 3- Subtract the minimum element of each column from all the elements in that respective column. This step is known as Column Minima.
From City
To City
A
B
C
D
E
A
M
1
3
6
0
B
4
M
0
6
0
C
4
3
M
0
3
D
8
0
1
M
1
E
0
2
0
7
M

37
From City
To City
A
B
C
D
E
A
M
1
3
6
0
B
4
M
0
6
0
C
4
3
M
0
3
D
8
0
1
M
1
E
0
2
0
7
M
From City
To City
A
B
C
D
E
A
M
1
3
6
0
B
4
M
0
6
0
C
4
3
M
0
3
D
8
0
1
M
1
E
0
2
0
7
M

38
However the solution gives the sequence A-E-A .This does not satisfy the condition that the salesman has to visit each city once before returning to the starting point. Hence we have to look at the next best solution which satisfies the above mentioned condition. This can be obtained by the next (non-zero) minimum element i.e. 1 into the solution. In the above table cost 1 occurs at three different places .Therefore we consider all these three one’s, one by one until the acceptable solution or next best solution is found to meet the above condition. Case1- We select 1 in the cell AB instead of zero assignment in the cell AE and delete row A and column B. After this step we select exclusive zero in the row or column .Except at one place i.e row number four we get exclusive zero in a row or a column. After selecting lowest element 1 in this row, in cell DE we are able to satisfy the condition. The assignment table on the above basis is as follows:
From City
To City
A
B
C
D
E
A
M
1
3
6
0
B
4
M
0
6
0
C
4
3
M
0
3
D
8
0
1
M
1
E
0
2
0
7
M
We have not taken other two cases as we achieved the next best solution satisfying the condition. The table below gives us the total transportation cost:

39
From City
To City
Travelling Cost INR
A
B
2000
B
C
3000
C
D
4000
D
E
5000
E
A
1000
Total
15000
Note-No feasible solution is obtained in case we had selected 1 in cell DC.
CLOSING NOTES ON ASSIGNMENT A variety of assignment problems have been taken to give you a flavour of this topic. Students are advised to make all attempts to solve these illustrations in the CD-ROM on their own, an only when they are stuck up as a last measure refer to the solutions therein. Go step by step to achieve the optimum solution. Notes are provided in the illustrations only for guidance. You need not be a maths expert to solve the illustrations, just believe in your ability and keep a positive approach.
The notes given in the illustrations are for guidance only. The notes and steps mentioned in this book as well as the CD-ROM should be used by the students to get a good insight on the diversity of the illustrations solved. While solving illustrations on their own, students are requested to mention the step in brief before writing the table below the same.

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