assignment poblems

1

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.

2

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.

4

As the numbers of rows are equal to columns, we have a balanced assignment table and

move on to step number 2.



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.



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.


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.




6

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.


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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

8

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.



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


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.




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



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


other zeros in that respective column for it to be crossed out. There are no zeros to be

crossed out.


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

13

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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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.




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


zeros in the table.

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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.





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.


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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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


other zeros in that respective column for it to be crossed out. There are two zeros which can

be crossed out.


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:

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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)

Portfolio

Management

(PM)

Private

Mutual Fund

(PMF)

Equity

Research (ER) 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).



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Days Leasing (L)

Portfolio

Management

(PM)

Private

Mutual Fund

(PMF)

Equity

Research (ER) 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


zeros in the table.

Days Leasing (L)

Portfolio

Management

(PM)

Private

Mutual Fund

(PMF)

Equity

Research (ER) 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.




20

Days Leasing (L)

Portfolio

Management

(PM)

Private

Mutual Fund

(PMF)

Equity

Research (ER) 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.





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.


Days Leasing (L)

Portfolio

Management

(PM)

Private

Mutual Fund

(PMF)

Equity

Research (ER) 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:

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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.

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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



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


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

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number of lines are less than the rows and columns, we move on to step number 5 for

modification.




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



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


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







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.


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 nomeans 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.



31


201 202 203 204

101 735 705 495 0

102 855 825 615 0

103 0 30 240 75

104 240 210 0 1005




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.


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.





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


As the number of lines is less than the rows and columns, we again use step number 5 for

modification


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


201 202 203 204

101 80 0 0 0

102 200 140 120 0

103 0 0 400 730

104 240 180 0 1500




201 202 203 204

101 80 0 0 0

102 200 140 120 0

103 0 0 400 730

104 240 180 0 1500








row.


34


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.





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



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


zeros in the table.

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










row.


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.

assignment poblems

Education