1. Solve the assignment problem for the optimal solution

Time taken by the workers.

workers JOB

A B C D

1 45 40 51 67

2 57 42 63 55

3 49 52 48 64

4 41 45 60 55

The solution of this problem in step wise manner..

Step 1 Row wise reduction so matrix will be

workers JOB

A B C D

1 5 0 11 27

2 15 0 21 13

3 1 4 0 16

4 0 4 19 14

Step 2 Column wise reduction

workers JOB

A B C D

1 5 0 11 14

2 15 0 21 0

3 1 4 0 3

4 0 4 19 1

Step 3: Draw the line in such a way so that minimum no. of line should cover all

zero.

So

workers JOB

A B C D

1 5 0 11 14

2 15 0 21 0

3 1 4 0 3

4 0 4 19 1

As we observed that the no. of lines and no. of columns or rows are equal, so that

we allocate the work.

workers JOB

A B C D

1 5 0 11 14

2 15 0 21 0

3 1 4 0 3

4 0 4 19 1

Assignment is made so that

1 B, 2 D, 3 C, 4 1 and time taken is (40+55+48+41) = 184 mins.

2. Using the optimal cost matrix, determine the optimal job assignment and cost of

assignment.

Machinist JOB

1 2 3 4 5

A 10 3 3 2 8

B 9 7 8 2 7

C 7 5 6 2 4

D 3 5 8 2 4

E 9 10 9 6 10

Step 1 Reduction of matrix

Machinist JOB

1 2 3 4 5

A 8 1 1 0 6

B 7 5 6 0 5

C 5 3 4 0 2

D 1 3 6 0 2

E 3 4 3 0 4

Step 2 : Column reduction

Machinist JOB

1 2 3 4 5

A 7 0 0 0 4

B 6 4 5 0 3

C 4 2 3 0 0

D 0 2 5 0 0

E 2 3 2 0 2

Draw the lines such a way so that Minimum no. of line should cover all the zeros.

Machinist JOB

1 2 3 4 5

A 7 0 0 0 4

B 6 4 5 0 3

C 4 2 3 0 0

D 0 2 5 0 0

E 2 3 2 0 2

Since the no. of lines covering all the zeros are less than the no. of columns/rows, so again

we have do the reduction by 2 as minimum value is 2 in matrix.

Machinist JOB

1 2 3 4 5

A 7 0 0 2 6

B 4 2 3 0 3

C 2 0 1 0 0

D 0 2 5 2 2

E 0 1 0 0 2

Machinist JOB

1 2 3 4 5

A 7 0 0 2 6

B 4 2 3 0 3

C 2 0 1 0 0

D 0 2 5 2 2

E 0 1 0 0 2

A 2, B 4, C5, D 1, E3

So that the cost which will occur is

A = 3

B = 2

C = 4

D = 3

E = 9

Which equal to 21.

3. Using the optimal cost matrix, determine the optimal job assignment and cost of

assignment.

Machinist JOB

1 2 3 4 5

A 10 3 3 2 8

B 9 7 8 2 7

C 7 5 6 2 4

D 3 5 8 2 4

E 9 10 9 6 10

Step 1 Reduction of matrix

Machinist JOB

1 2 3 4 5

A 8 1 1 0 6

B 7 5 6 0 5

C 5 3 4 0 2

D 1 3 6 0 2

E 3 4 3 0 4

Step 2 : Column reduction

Machinist JOB

1 2 3 4 5

A 7 0 0 0 4

B 6 4 5 0 3

C 4 2 3 0 0

D 0 2 5 0 0

E 2 3 2 0 2

Draw the lines such a way so that Minimum no. of line should cover all the zeros.

Machinist JOB

1 2 3 4 5

A 7 0 0 0 4

B 6 4 5 0 3

C 4 2 3 0 0

D 0 2 5 0 0

E 2 3 2 0 2

Since the no. of lines covering all the zeros are less than the no. of columns/rows, so again

we have do the reduction by 2 as minimum value is 2 in matrix.

Machinist JOB

1 2 3 4 5

A 7 0 0 2 6

B 4 2 3 0 3

C 2 0 1 0 0

D 0 2 5 2 2

E 0 1 0 0 2

Machinist JOB

1 2 3 4 5

A 7 0 0 2 6

B 4 2 3 0 3

C 2 0 1 0 0

D 0 2 5 2 2

E 0 1 0 0 2

A 2, B 4, C5, D 1, E3

So that the cost which will occur is

A = 3

B = 2

C = 4

D = 3

E = 9

Which equal to 21.

4.

JObs Machines

M1 M2 M3 M4

W1 40 50 60 65

W2 30 38 46 48

W3 25 33 41 43

W4 39 45 51 59

How would the task to be assigned to minimize the cost?

Answer: Step 1 Row wise reduction

Jobs Machines

M1 M2 M3 M4

W1 0 10 20 15

W2 0 8 16 18

W3 0 8 16 18

W4 0 6 12 20

Step 2 Column Wise Reduction

Jobs Machines

M1 M2 M3 M4

W1 0 4 8 7

W2 0 2 4 0

W3 0 2 4 0

W4 0 0 0 2

Jobs Machines

M1 M2 M3 M4

W1 0 4 8 7

W2 0 2 4 0

W3 0 2 4 0

W4 0 0 0 2

Step 3 In three line all zeros are covered so that we have again reduce the matrix by 2.

Jobs Machines

M1 M2 M3 M4

W1 0 2 4 7

W2 0 0 0 0

W3 0 0 0 0

W4 4 0 0 4

All zeros are covered by four lines

Jobs Machines

M1 M2 M3 M4

W1 0 2 4 7

W2 0 0 0 0

W3 0 0 0 0

W4 2 0 0 4

So work assign in this way

W1—M1, W2---M4, W3—M3, W4—M2.

40+48+41+45 = 174.

5. The college department has the problem of providing the teachers for all courses

offered by his department at the highest possible level of educational quality. He has

got the three professors and one teaching assistant (TA). Four courses must be offered

and after appropriate introspection and evaluation he has arrived at the following

relative ratings (100= best rating) regarding the ability of each instructor to teach the

four courses respectively.

Course 1 Course 2 Course 3 Course 4

Prof. 1 70 50 70 80

Prof. 2 30 70 60 80

Prof. 3 30 40 50 70

TA 40 20 40 50

How should be the assign his staff to the course to maximize educational quality of the

department.

Answer :

Step 1: We have to maximize the problem so reduce it by 70.

Course 1 Course 2 Course 3 Course 4

Prof. 1 10 30 10 0

Prof. 2 50 10 20 0

Prof. 3 50 40 30 10

TA 40 60 40 30

Step 2 : Reduce row wise

Course 1 Course 2 Course 3 Course 4

Prof. 1 10 30 10 0

Prof. 2 50 10 20 0

Prof. 3 40 30 20 0

TA 10 30 10 0

Step 3 : Reduce column wise. And cover all the zero by minimum line

Course 1 Course 2 Course 3 Course 4

Prof. 1 0 20 0 0

Prof. 2 40 0 10 0

Prof. 3 30 20 10 0

TA 0 20 0 0

So the no. of line is 4 which is equal to no. of column or rows.

Course 1 Course 2 Course 3 Course 4

Prof. 1 0 20 0 0

Prof. 2 40 0 10 0

Prof. 3 30 20 10 0

TA 0 20 0 0

So the course offer by

Prof. 1 Course 1 70

Prof. 2 Course 2 70

Prof. 3 Course 4 70

TA Course 3 40

Total 250

6. Captain MS Dhoni has to allot five Middle batting position to five batsmen. The

average run scored by each batsman at these position is as follows

BATSMAN BATTING POSITION

I II III IV V

VIRENDRA 40 40 35 25 50

GAUTAM 42 30 16 25 27

SURESH 50 48 40 60 50

YUVARAJ 20 19 20 18 25

SACHIN 58 60 59 55 53

Find out the proper batting assignement.

Answer:

Step 1 Reduction of matrix.

BATSMAN BATTING POSITION

I II III IV V

VIRENDRA 20 20 25 35 10

GAUTAM 18 30 44 35 33

SURESH 10 22 20 0 10

YUVARAJ 40 41 40 42 35

SACHIN 02 00 01 05 02

Step 2 reduction row wise

BATSMAN BATTING POSITION

I II III IV V

VIRENDRA 10 10 15 25 10

GAUTAM 0 12 26 17 15

SURESH 10 22 20 0 10

YUVARAJ 5 6 5 7 0

SACHIN 02 00 01 05 02

Step 4 Reduction Column Wise

BATSMAN BATTING POSITION

I II III IV V

VIRENDRA 10 10 14 25 10

GAUTAM 0 12 25 17 15

SURESH 10 22 19 0 10

YUVARAJ 5 6 4 7 0

SACHIN 02 00 0 05 02

Step 5 Reduce by 10

BATSMAN BATTING POSITION

I II III IV V

VIRENDRA 0 0 4 15 0

GAUTAM 0 12 25 17 15

SURESH 10 22 19 0 10

YUVARAJ 5 6 4 7 0

SACHIN 02 00 0 05 02

So the batting order is like this

Gautam—virendra—Sachin—Suresh—yuvaraj runs are 40+42+59+60+25=226

7. Darda oil mills has four plants each of which can manufacture anyone of these four

product. The manufacturing cost differs from plant to plant so the sales revenue.

Sales

Plants

Products

I II III IV

A 70 88 69 82

B 80 90 71 94

C 75 87 73 80

D 78 85 74 89

revenue

Plants

Products

I II III IV

A 59 70 55 71

B 65 73 55 79

C 62 72 59 68

D 65 74 58 76

Reduction to profit matrix

Plants

Products

I II III IV

A 11 18 14 11

B 15 17 16 15

C 13 15 14 12

D 13 11 16 13

Plants

Products

I II III IV

A 7 0 4 7

B 3 1 2 3

C 5 3 4 6

D 5 7 2 5

Column wise reduction

Plants

Products

I II III IV

A 4 0 2 4

B 0 1 0 0

C 2 3 2 3

D 2 7 0 2

repeat

Plants

Products

I II III IV

A 3 0 2 3

B 0 2 1 0

C 1 3 2 2

D 1 7 0 1

Plants

Products

I II III IV

A 2 0 2 2

B 0 3 2 0

C 0 3 2 1

D 0 7 0 0

Plants

Products

I II III IV

A 2 0 2 2

B 0 3 2 0

C 0 3 2 1

D 0 7 0 0

A= II, B=IV, C=I,D=III ==18+15+13+16 = 61

8. The planning engineering has the four jobs and four machine for the allocation

the job-wise allocation is given below.

JObs

Machines

W X Y Z

A 1200 1440 480 1080

B 720 600 360 960

C 240 360 480 600

D 720 840 360 1200

STEP : This is simple assignment problem ( minimization and unrestricted)

After reducing by row wise and column wise we get answer …2280Rs.

With Assignment A to Z, B to X, C to W and D to Y.

9. The engg. Workshop has five lathes and five operators. The workshop is involved in

the producing of standard part which is required in large quantity. Since the operator

have the varying skills and machines alos have the different efficiency the output are

varying. The wekly output has given below.

Operators

Lathes

1 2 3 4 5

O1 30 33 37 42 46

O2 29 33 39 44 50

O3 33 38 45 49 44

O4 31 34 41 57 52

O5 32 38 41 46 51

This is simple matrix of (maximization, unrestricted allocation)

So first we have to make it in minimization form by subtracting from 52.

The by row and column reduction and following the procedure.

We get the answer.209 units. O1—1, O2—4, O3—3, O4—5 AND O5—2.

10. Captain MS Dhoni has to allot five Middle batting position to five batsmen. The

average run scored by each batsman at these position is as follows

BATSMAN BATTING POSITION

I II III IV V

VIRENDRA 35 36 35 25 40

GAUTAM 52 61 14 24 27

SURESH 50 47 42 13 41

YUVARAJ 19 25 40 47 52

SACHIN 60 72 84 56 65

Find out the proper batting assignment.

11. Assign three jobs on three machines for following cost matrix:

Jobs Machines

M1 M2 M3

J1

J2

J3

Rs. 14

Rs.11

Rs.20

Rs.12

Rs.17

Rs. 8

Rs.16

Rs.21

Rs. 7

Solution

Step 1: Subtract minimum entry in each column from all the entries on that column. This is a job opportunity cost matrix.

Job-opportunity cost matrix

Step 2: Subtract minimum entry in each row of job-opportunity cost matrix from all the entries of that row. This is a total opportunity cost matrix:

Total opportunity cost matrix

Step 3: Check for Optimality

Draw minimum number of horizontal and vertical lines to cover all zeros. This can be done in 2 lines, which is one less than the number of rows (which is 3). Thus, the solution is yet not final. Hence, go

JOBS/MACHINES M1 M2 M3

J1 0 1 6

J2 0 9 14

J3 9 0 0

to step 4.

Check for optimality

Step 4: From the uncovered entries, the minimum is 1. Thus, subtract 1 from all entries, which are uncovered. Add one at junction of lines, i.e., at J3-M1. We get the following matrix now, as the received opportunity cost matrix.

JOBS/MACHINES M1 M2 M3

J1 0 0 5

J2 0 8 13

J3 10 0 0

Revised total opportunity cost matrix

Now, go to step 3 to check the optimality. We can cover all zeros of above matrix by at least three lines (which is also equal to number of rows). Hence, above solution may be used for optimal assignment.

Step 5: Assignment Scheme

Refer revised total opportunity cost matrix. Row J2 has only one zero at M1 column. Hence, assign J2 to M1. Remove row J2 and column M1.

Column M3 of remainder matrix has one zero at J3 row. Assign J3 to M3. The last assignment is remainder job J1 to M2. Thus, the final assignment is:

Job Machine Cost

J1

J2

J3

M2

M1

M3

Rs.12

Rs.11

Rs. 7

Total Cost Rs.30

12. Four persons A,B,C and D are to be assigned four jobs I, II, III and IV. The cost matrix is given as under, find the proper assignment.

A B C D

I 8 10 17 9

II 3 8 5 6

III 10 12 11 9

IV 6 13 9 7

Solution : In order to find the proper assignment we apply the Hungarian algorithm as follows: I (A) Row reduction

A B C D

I 0 2 9 1

II 0 5 2 3

III 1 3 2 0

IV 0 7 3 1

I (B) Column reduction

A B C D

I 0 0 7 1

II 0 3 0 3

III 1 1 0 0

IV 0 5 1 1

II(A) and (B) Zero assignment

A B C D

I 0 0 7 1

II 0 3 0 3

III 1 1 0 0

IV 0 5 1 1

In this way all the zero’s are either crossed out or assigned. Also total assigned zero’s = 4 (i.e., number of rows or columns). Thus, the assignment is optimal. From the table we get I to B; II to C; III to D and IV to A.

13. There are five machines and five jobs are to be assigned and the associated cost matrix is as follows. Find the proper assignment. Machines

I II III IV V

A 6 12 3 11 15

B 4 2 7 1 10

C 8 11 10 7 11

D 16 19 12 23 21

E 9 5 7 6 10

Solution: In order to find the proper assignment, we apply the Hungarian method as follows: IA (Row reduction) Machines

I II III IV V

A 3 9 0 8 12

B 3 1 6 0 9

C 1 4 3 0 4

D 4 7 0 11 9

E 4 0 2 1 5

IB (Column reduction) Machines

I II III IV V

A 2 9 0 8 8

B 2 1 6 0 5

C 0 4 3 0 0

D 3 7 0 11 5

E 3 0 2 1 1

II (Zero assignment) Machines

I II III IV V

A 0** 7 0 6 6

B 2 1 8 0** 5

C 0 4 5 0 0**

D 1 5 0** 9 3

E 3 0** 4 1 1

Thus, we have got five assignments as required by the problem. The assignment is as follows: A -- I, B-- IV, C -- V, D -- III and E -- II. This assignment holds for table given in step IV but from theorem 1 it also holds for the original cost matrix. Thus from the cost matrix the minimum cost = 6+1+11+12+5=Rs.35.

14. A firm produces four products. There are four operators who are capable of producing

any of these products. The firm records 8 hours of day and allows 30 minutes for

lunch. The processing time in minutes and the profit for each product are given below:

Operator Product

A B C D

1 15 9 10 6

2 10 6 9 6

3 25 15 15 9

4 15 9 10 10

Profit per unit 8 6 5 4

Find the optimal assignment of products to operators?

Ans: Step 1

8 hours working = 480 minutes

(-)Lunch time = 030 minutes

Available hours = 450 minutes

No. of products each operator can produce can be found out by dividing 450 by the given processing

time.

For eg: Operator 1 can produce 450/15=30 units of product A at a profit rate of Rs. 8,implies a total

profit of Rs.240

Profit Matrix

Operator Product

A B C D

1 240 300 225 300

2 360 450 250 300

3 144 180 150 200

4 240 300 225 180

To solve the problem it has to be converted into a minimisation problem

450 is subtracted from each value:

Operator Product

A B C D

1 210 150 225 150

2 90 0 200 150

3 306 270 300 250

4 210 150 225 270

We use Hungarian method to solve the Matrix, we get the following table:

Step 2 Row-wise reduction,

Operator Product

A B C D

1 60 0 75 0

2 90 0 200 150

3 56 20 50 0

4 60 0 75 120

Step 3 Column-wise reduction

Operator Product

A B C D

1 4 0 25 0

2 34 0 150 150

3 0 20 0 0

4 4 0 25 120

Step 4 Since the minimum no. of lines to cover all zeros is 4 , further reduction is done by 4

Operator Product

A B C D

1 0 0 21 (0)

2 30 (0) 146 150

3 0 20 (0) 0

4 (0) 0 21 120

Thus the optimal assignment is:

Operator Product Profit

1 D 300

2 B 450

3 C 150

4 A 240

Total 1,140

15. ABC airline, operating 7 days a week has given the following time table. The crews

must have a minimum lay-over of 5 hours between flights. Obtain the pairing of flights

that minimises lay-over time away from home. For any given pairing the crew will be

based at the city that results the smallest lay-over.

Hyderabad-Delhi Delhi-Hyderabad

Flight No. Departure Arrival Flight No. Departure Arrival

A1 6AM 8AM B1 8AM 10AM

A2 8AM 10AM B2 9AM 11AM

A3 2PM 4PM B3 2PM 4PM

A4 8PM 10PM B4 7PM 9PM

We assume that all the crew is based at Hyderabad. Using this assumption we can obtain the lay-

over times for various combinations of flights. To illustrate, the flight A1 which starts from

Hyderabad at 6AM, reaches Delhi at 8AM. If it has to return as B1, the schedule time for which is

8PM , then it can do so after 24 hrs , since it has a lay-over time of 5 hours.

Similarly, lay-over times for other flights can be obtained,

Lay-over Time-Crew at Hyderabad

Flight B1 B2 B3 B4

A1 24 25 6 11

A2 22 23 28 9

A3 16 17 22 27

A4 10 11 16 21

When crew is based at Delhi,

Lay-over Time-Crew at Delhi

Flight B1 B2 B3 B4

A1 20 19 14 9

A2 22 21 16 11

A3 28 27 22 17

A4 10 9 28 23

Now in order to obtain the minimum lay-over time we select the corresponding lower value out of

the two tables. For eg: In combining A1-B1 we select 20, which is lower than 24.

Flight B1 B2 B3 B4

A1 20* 19* 6 9*

A2 22 21* 16* 9

A3 16 17 22 17*

A4 10 9* 16 21

*when crew is based at Delhi.

Now we use the Hungarian method to solve the matrix,

Reduced Cost Table,

Flight B1 B2 B3 B4

A1 14 13 (0) 3

A2 13 12 7 (0)

A3 (0) 1 6 1

A4 (1) (0) 7 12

Here there is no need to perform column reduction, as zero is obtained in each column.

The Optimal Solution

Flight No. Flight No. Lay-over Time Crew Based At

A1 B1 6 Hyderabad

A2 B2 9 Hyderabad

A3 B3 16 Hyderabad

A4 B4 9 Delhi

Total 40

16. A company has just developed a new item for which it proposes to undertake a

national television promotional campaign. It has decided to schedule a series of one-

minute commercials during peak audience viewing hours of 1 to 5 PM. To reach the

widest possible audience, the company want to schedule one commercials on each of

the networks and to have only one commercial appear during each of the four one

hour time blocks. The exposure ratings for each hour, which represent the number of

viewers per Rs.10,000 spent, are given below.

Viewing Hours

Network A B C D

1-2PM 27.1 18.1 11.3 9.5

2-3PM 18.9 15.5 17.1 10.6

3-4PM 19.2 18.5 9.9 7.7

4-5PM 11.5 21.4 16.8 12.8

(a) Which network should be scheduled each hour to provide the maximum audience exposure?

(b) How would the schedule change if it is decided not to use network A between 1 and 3 PM?

(a) to solve this problem we multiply each exposure rating by 10 to simplify the calculation. Further

being a minimisation problem each value is subtracted from the largest value to get the opportunity

loss matrix.

Opportunity Loss Matrix

Viewing Hours

Network A B C D

1-2PM 0 90 158 176

2-3PM 82 116 100 165

3-4PM 79 86 172 194

4-5PM 156 57 103 143

Now we use the Hungarian method to solve the above matrix

Row-wise reduction,

Viewing Hours

Network A B C D

1-2PM 0 90 158 176

2-3PM 0 34 18 83

3-4PM 0 7 93 115

4-5PM 99 0 46 86

Column –wise reduction,

Viewing Hours

Network A B C D

1-2PM 0 90 140 93

2-3PM 0 34 0 0

3-4PM 0 7 75 32

4-5PM 99 0 28 3

Since the number of lines covering all zeros is 3, we improve the solution,

Viewing Hours

Network A B C D

1-2PM 0 90 137 90

2-3PM 3 37 0 0

3-4PM 0 7 72 29

4-5PM 99 0 25 0

Further reduction,

Viewing Hours

Network A B C D

1-2PM 0 83 130 83

2-3PM 10 37 0 0

3-4PM 0 0 65 22

4-5PM 106 0 25 0

The Optimal solution,

Viewing Hours

Network

1-2PM A

2-3PM C

3-4PM B

4-5PM D

(b) For the given restriction, the prohibited time-slots are replaced M , row- wise reduction are

shown in the below table,

Viewing Hours

Network A B C D

1-2PM M 0 68 86

2-3PM M 16 0 65

3-4PM 0 7 93 115

4-5PM 99 0 46 86

Column –wise Reduction,

Viewing Hours

Network A B C D

1-2PM M 0 68 86

2-3PM M 16 0 65

3-4PM 0 7 93 115

4-5PM 99 0 46 86

Further Reduction

Viewing Hours

Network A B C D

1-2PM M 0 47 0

2-3PM M 37 0 0

3-4PM 0 28 93 50

4-5PM 78 0 25 0

Ans:

Viewing Hours Network Network

1-2PM B D

2-3PM C C

3-4PM A A

4-5PM D B

17. An airline that operates seven days a week has a time table shown below. Crew must

have a minimum lay-over of 6 hours between flights.Obtain the pairing of flights that

minimises layover time away from home. For any given pairing the crew will be based

at the city thath results in the smaller layover time.

Delhi-Calcutta Calcutta-Delhi

Flight No. Departure Arrival Flight No. Departure Arrival

1 7AM 9AM 101 9AM 11AM

2 9AM 11AM 102 10AM 12AM

3 1.30PM 3.30PM 103 3.30PM 5.30PM

4 7.30PM 9.30PM 104 8PM 10PM

Layover time converted when crew is located in Delhi

Flights 101 102 103 104

1 48 50 13 22

2 44 46 57 18

3 35 37 48 57

4 23 25 36 45

*Layover time can be converted into points by assigning 3- minutes=1 point

Similarly we assign points, when crew is based in Calcutta

Flights 101 102 103 104

1 40 38 27 18

2 44 42 31 22

3 35 51 40 31

4 17 15 36 43

Now we can select the minimum layover time from both the above tables

After solving the matrix by Hungarian method we get the Following Optimal Solution:

Flights 101 102 103 104

1 (0)

2 (0)

3 (0)

4 (0)

18. An office has four workers, and four tasks have to be performed. Workers differ in efficiency and tasks differ in their intrinsic difficulty. Time each worker would take to complete each task is given in the effectiveness matrix. How the tasks should be allocated to each worker so as To minimise the total man-hour?

Tasks

Workers

1 2 3 4

18

23

12

7

A 5 23 14

B 10 25 1

C 35 16 15

D 16 23 21

Solution: I - A, II - C, III - B, IV – D.

19.You are the given information about the cost performing different jobs by different persons.

The job person marking X indicates that the individual involved cannot perform the particular job.

Using this information, state(i)the optimal assignment of the job (ii) the cost of such assignment.

person Job

J1 j2 j3 j4 j5

P1 27 18 X 20 21

P2 31 24 21 12 17

P3 20 17 20 X 16

P4 22 28 20 16 27

Balancing the problem and assigning a high cost to pairing p1-j3 and p3-j4,we have the cost table

given in table below

person Job

J1 j2 j3 j4 j5

P1 27 18 M 20 21

P2 31 24 21 12 17

P3 20 17 20 M 16

P4 22 28 20 16 27

P5(dummy) 0 0 0 0 0

Now we can drive the reduced cost table. The cell with prohibited assignment continue to be shown

with cost element M, since M is defined to be extremely large so that subtraction and addition of a

values does not practically affect it. To test optimality lines are drawn to cover all zeros. Since the

number of lines covering all zeros is less than n, we select the lowest uncovered cell, which equals

4.with this value, we can obtain the revised reduced cost table as below.

Person job

J1 j2 j3 j4 j5

P1 9 0 M 2 3

P2 19 12 9 0 5

P3 4 1 4 M 0

P4 6 12 4 0 11

P5 0 0 0 0 0

Reduced cost table

Person job

J1 j2 j3 j4 j5

P1 9 (0) M 6 3

P2 15 8 5 (0) 1

P3 4 1 4 M (0)

P4 2 8 (0) 0 7

P5 0 0 0 4 0

The number lines covering zeros is equal to 5(=n). The assignment is p1-j2, p2-j4, p3-j5, p4-j3, while

job j1 would remained unassigned. This assignment pattern would cost 18+12+16+20=66 aggregate.

20 A company plans to assigns 5 salesmen to 5 district in which it operates. Estimate of sales

revenue in thousands of rupees for each salesman in different district are given in the following

table. In your option, what should be the placement of salesmen if the objective is to maximise

the expected sales revenue?

Expected sales area

Salesman District

D1 D2 D3 D4 D5

S1 40 46 48 36 48

S2 48 32 36 29 44

S3 49 35 41 38 45

S4 30 46 49 44 44

S5 37 41 48 43 47

Since it is maximisation problem, we would first subtract each of entries in the largest one,which

equals to 49 here. The resultant data given below.

Opportunity Loss Matrix

Salesman District

D1 D2 D3 D4 D5

S1 9 3 1 13 1

S2 1 17 13 20 5

S3 0 14 8 11 4

S4 19 3 0 5 5

S5 12 8 1 6 2

Now we shall proceed as usual

STEP 1: subtract minimum value in each row from every value in the row.

Reduced Cost Table 1

Salesman District

D1 D2 D3 D4 D5

S1 8 2 0 12 0

S2 0 16 12 19 4

S3 0 14 8 11 4

S4 19 3 0 5 5

S5 11 7 0 5 1

STEP 2,3 : subtract minimum value in each column from each value of column.

Reduced Cost Table 2

Salesman District

D1 D2 D3 D4 D5

S1 8 0 0 7 0

S2 0 14 12 14 4

S3 0 12 8 6 4

S4 19 1 0 0 5

S5 11 5 0 0 1

Since the number of lines covering all zeros is fewer than n, we select the least uncovered cell value

which equals 4. With this, we can modify the table as below.

STEP 4,5,6: find improved solution. Test for optimality and make assignment.

Salesman District

D1 D2 D3 D4 D5

S1 12 (0) 0 7 0

S2 (0) 10 5 10 0

S3 0 8 4 2 0

S4 23 1 (0) 0 5

S5 15 5 0 (0) 1

There are more than one optimal assignment possible because the existence of multiple zeros in

different rows and columns. The assignments possible are:

S1-D1, S2-D5, S3-D1, S4-D3, S5-D4 ; or

S1-D2, S2-D1, S3-D5, S4-D3, S5-D4; or

S1-D2, S2-D5, S3-D1, S4-D4, S5-D3; or

S1-D2, S2-D2, S3-D5, S4-D4, S5-D3.

Each of these assignment pattern would lead to expected aggregated sales equal to 231 thousand

rupees.

21. A solicitor firm employs typist on hourly piece- rate basis for their daily work. There are 5

typist and their charges and speed are different. According to an early understanding, only one job

is given to one typist the typist is paid for a full hour even when he works for a fraction of hour.

Find the least cost allocation for the following data:

Typist Rate/hour(Rs) Number of pages Typed/hour

Job No. Of pages

A 5 12 P 199

B 6 14 Q 175

C 3 8 R 145

D 4 10 S 298

E 4 11 T 178

Using the given information we3 first obtain the cost matrix, when the different jobs are performed

by different typist.

If typist A given job P, we should require 199/12=16X7/12 hours, hence be paid for 17 hours @ Rs. 5

per hour. The result in a cost of Rs.85 for the combination.

Total Cost Matrix

Typist Job

P Q R S T

A 85 75 65 125 75

B 90 78 66 132 78

C 75 66 57 114 69

D 80 72 60 120 72

E 76 64 56 112 68

Subtracting the minimum element of each row from all its element, we obtain RCT-1, as below.

Typist Job

P Q R S T

A 20 10 0 60 10

B 24 12 0 66 12

C 18 9 0 57 12

D 20 12 0 60 12

E 20 8 0 56 12

Now subtracting the minimum element of each column from all elements, we get RCT-2, as below

Typist Job

P Q R S T

A 2 2 0 4 0

B 6 4 0 10 2

C 0 1 0 1 2

D 2 4 0 4 2

E 2 0 0 0 2

Here the minimum number of lines cover all zeros is 4, which is smaller than the order,5, of given

matrix .Accordingly the revised table is prepared by considering the least uncovered value , equal to

1, and adjusting it with uncovered cell values and those lying at the intersection of lines. table

containsRCT-3.

Reduced Cost Table 3

Typist Job

P Q R S T

A 2 1 0 3 0

B 6 3 0 9 2

C 0 0 0 0 2

D 2 3 0 3 2

E 3 0 1 0 3

4 lines can cover all the zeros, Accordingly RCT-4 is drawn in table below.

Typist Job

P Q R S T

A 2 1 2 3 (0)

B 4 1 (0) 7 0

C 0 (0) 2 0 2

D (0) 1 0 1 0

E 3 0 3 (0) 3

In this case the minimum number of lines cover all zeros equal 5, which matches with order of the

matrix. Accordingly assignment have been made as below.

Typist Job Cost

A T 75

B R 66

C Q 66

D P 80

E S 112

Total 399

The optimal solution, however is not unique.

22. Well done company has taken third floor of a multi-storeyed building for a rent with a

view to locate one of their zonal offices. There are 5 main rooms in this to be assigned to be five

managers. Each room has its own advantages and disadvantages. Some have windows, some are

closer to wash rooms or to the canteen or secretarial pool. The rooms are of all different sizes and

shapes. Each of the five managers were asked to rank their room preferences amongst the room

301,301,303,304 and 305. Their preferences are recorded as below.

M1 M2 M3 M4 M5

302 302 303 302 301

303 304 301 305 302

304 305 304 304 304

301 305 303

302

Most of the manager did not list all the five rooms since they were not satisfied with some of these

rooms and they have left these from the list. Assuming that their preferences can be quantified by

numbers, find out as to which manager should be assigned to which room so that their total

preference ranking is minimum.

In the first step, we formulate the assignment problem using preference ranks as shown in table

below. Notice that the rooms not ranked by manager are represented by M-as prohibited

assignment.

Room Manager

M1 M2 M3 M4 M5

301 M 4 2 M 1

302 1 1 5 1 2

303 2 M 1 4 M

304 3 2 3 3 3

305 M 3 4 2 M

Since greater preferences are shown by lower numbers, the optimal solution calls for minimising the

total preference ranking. To solve we obtain RCT-1 by subtracting the least value from each value of

the row for each of these rows. As table below.

Room Manager

M1 M2 M3 M4 M5

301 M 3 1 M 0

302 0 0 4 0 1

303 1 M 0 3 M

304 1 0 1 1 1

305 M 1 2 0 M

Since each of columns as well as rows has a zero, lines are drawn to cover all zeros. Further, the

number of being equal to 5, the order of given matrix, assignment can be made, as shown below.

Room Manager

M1 M2 M3 M4 M5

301 M 3 1 M 0

302 0 0 4 0 1

303 1 M 0 3 M

304 1 0 1 1 1

305 M 1 2 0 M

The optimal assignment pattern is:

Room Manager Ranking

301 M5 1

302 M1 1

303 M3 1

304 M2 2

305 M4 2

23.A company has four sales representative who are to assigned to four different sales territories.

The monthly sales increase estimated for each sales representative for different sales territories(

in lakhs or rupees) are shown in table below.

Sales representative

Sales territories

I II III IV

A 200 150 170 220

B 160 120 150 140

C 190 195 190 200

D 180 175 160 190

Suggest the optimal assignment and the total maximum sales increase per month.

If for certain reasons, sales representative B can not be assigned to sales territory III, will the optimal

assignment schedule be different? If so, find that schedule and the effect of total sales.

The given problem being maximization type, need to be converted first into minimization type as

below; wherein the various elements are obtained by subtracting each of element in the given table

from 220.

Opportunity Loss Matrix

I II III IV

A 20 70 50 0

B 60 100 70 80

C 30 25 30 20

D 40 45 60 30

Subtracting minimum value in each row from each row element. RCT-1.

Reduced Cost Table-1

I II III IV

A 20 70 50 0

B 0 40 10 20

C 10 5 10 0

D 10 15 30 0

Subtracting minimum value in each column from each column element. RCT-2.

I II III IV

A 20 65 40 0

B 0 35 0 20

C 10 0 0 0

D 10 10 20 0

Since the minimum number lines to cover all zeros(3) is smaller than matrix order(4), rct-3 IS derived

as revised table.

Reduced Cost Table-3

I II III IV

A 10 55 30 0

B 0 35 0 30

C 10 0 0 10

D 0 0 10 0

With the number of lines to cover all zeros being equal to order of given matrix, assignment can be

made as shown in table below. However the problem has an alternative optimal solution as well.

Both of these are given below:

Alternative 1

Salesman Territory Sales

A IV 220

B I 160

C III 190

D II 175

Total 745

Alternative 2

Salesman Territory Sales

A IV 220

B III 150

C II 195

D I 180

Total 745

If salesman B cannot be assigned to territory III( Alternative 2), then alternative 1 above may be

adopted without adverse on sale.

24. Four sales manager(A,B,C and D) are to be assigned to four sales zones (P,Q,R and S). The

zones differ in their sales potential and sales manager differ in their capabilities. It is estimated

that a bright sales manager operating in each sale zone would bring in the following monthly sales.

Zone P= Rs.150 lakhs

Zone Q=Rs.120 lakhs

Zone R=Rs.100 lakhs

Zone S=Rs.80 lakhs

Since the sales manager differ in their capabilities, their effectiveness on a 10 point scale( on

which a bright sales manager is rated as 10) has been assessed as follows:

Sales manager A: 7 points

Sales manager B: 5 points

Sales manager C: 9 points

Sales manager D: 4 points

Suggest the optimal allocation of sales manager to sales zones to maximize monthly business.

SOLUTION:

The 1st step to determine proportionate sale( in Rs. Lakhs) in the four zones considering

effectiveness of different sales manager. The following table gives such calculation.

Sales manager

Sales zone

P Q R S

A 7/10x150=105 7/10x120=84 7/10X100=70 7/10X80=56

B 5/10X150=75 5/10X150=60 5/10X100=50 5/10x80=40

C 9/10x150=135 9/10x120=108 9/10X100=90 9/10X80=72

D 4/10X150=60 4/10X120=48 4/10X100=40 4/10X80=32

: the optimal solution is-

A to R : Rs. 70 lakhs

B to Q : Rs. 60 lakhs

C to S : Rs.72 lakhs

D to A : Rs.60 lakhs

Rs. 262 lakhs

26. An office has four workers, and four tasks have to be performed. Workers differ in efficiency and tasks differ in their intrinsic difficulty. Time each worker would take to complete each task is given in the effectiveness matrix. How the tasks should be allocated to each worker so as To minimise the total man-hour?

Tasks

Workers

1 2 3 4

A 5 23 14 18

B 10 25 1 23

C 35 16 15 12

D 16 23 21 7

Solution: I - A, II - C, III - B, IV - D

27. A taxi hire company has one taxi at each of five depots a,b,c,d and e. A customer requires a taxi in each town, namely A,B,C,D and E. Distances (in kms) between depots (origins) and towns (Destinations) are given in the following distance matrix:

a b c d e

A 140 110 155 170 180

B 115 100 110 140 155

C 120 90 135 150 165

D 30 30 60 60 90

E 35 15 50 60 85

Solution: A -e, B - c, C - b, D - a, E - d Minimum distance (km) = 180+110+90+30+60=470 km.

28.

Jobs

Workers

1 2 3 4

A 1 19 10 4

B 6 21 7 19

C 31 12 11 1

D 12 19 17 3

Solution- A - I, B - III, C - II, D – IV

29.

Jobs

Workers

1 2 3 4

A 15 17 24 16

B 10 15 12 13

C 17 19 18 16

D 13 20 16 14

Solution- A - II, B -III, C - IV, D - I Minimum cost = 17+12+16+13=58 30.

Jobs

Workers

1 2 3 4

A 8 26 17 11

B 14 29 5 27

C 40 21 20 17

D 19 19 24 10

Solution- I -A, II - C, III - B, IV – D 31.

Find the proper assignment of the assignment problem whose cost matrix is given as under.

1 2 3 4 5

A 10 6 4 8 3

B 2 11 7 7 3

C 5 10 11 4 8

D 6 5 3 2 5

E 11 7 10 11 7

Solution- A -V, B -I, C- IV, D -III, E – II. 32. Solve the following assignment problem.

1 2 3 4

A 8 9 10 11

B 10 11 12 13

C 13 14 15 13

D 9 11 14 10

Solution- A-2, B-3, C-4, D-1 or A-3, B-2, C-4, D-1