assignment chapter - q & a compilation by niraj thapa

1

THE ASSIGNMENT PROBLEM

6

4

8

5 5

7

5

8

4

0

1

2

3

4

5

6

7

8

9

Nov-11 May-12 Nov-12 May-13 Nov-13 May-14 Nov-14 May-15 Nov-15 May-16

Examwise Marks Disrtibution-Assignment

MA

RK

S

YEAR OF EXAMINATION

Let us start the topic with some questions relating to the topic.

Issue 1: If in a printing press there is one machine and one operator is there to operate. How would the

manager employ the worker? Obviously, the only operator shall operate the machine.

Issue 2: If there are two machines in the press and two operators are engaged at different rates to operate

them. Which operator should operate which machine for minimising the total cost?

Issue 3: If there are n machines available and n persons are engaged at different rates to operate them.

Which operator should be assigned to which machine to ensure maximum efficiency?

The answer to Issue No.2 & 3 is based on the Principle of Assignment Technique (Hungarian Method).

Assignment is optimal if it is in the interest of the press (i.e. the press should be able to maximise its total profit

or minimise the total operating cost)

The problem of assignment arises because available resources such as men, machines, etc. have varying

degrees of efficiency for performing different activities

An assignment is an act to allot the given number of jobs to operators. Assignment problem is one of the

special cases of transportation problems. The objective of the assignment problem is to minimize the cost

(i.e., maximising revenue/profits) or time of completing a number of jobs by a number of persons.

An important characteristic of the assignment problem is the number of sources is equal to the number of

destinations .It is explained in the following way.

Only one job is assigned to person

Each person is assigned with exactly one job

The assignment problem involves following steps to get optimal solution.

I. Initial Solution

II. Optimality Test

III. Assigning Jobs

EXPLAINED

I. Initial Solution

The problem will specify jobs, operators and their associated costs. The no. of jobs should be equal

to the no. of operators. It should be in matrix form.

Start with a balanced minimisation* matrix.

a) Row Reduction (The minimum cost of each row should be deducted from all cell cost of that row)

b) Column Reduction (Deduct minimum cost of each column from all cost cell of that column)

(Perform column operation in matrix obtained after Row Reduction)

II. Optimality Test

The matrix obtained after initial solution is to be tested for optimality.

In this step the objective is to cover maximum zeros by drawing minimum no. of straight lines

(straight lines means not diagonal lines)

Then check whether the no. of straight lines drawn equals no. of row (column).

If the answer is affirmative# then the solution is optimal.

INTRODUCTION

STEPS INVOLVED

1

III. Assigning Jobs

Start from the first row and see if there is single zero$ row, if yes then allot job over that zero by

marking square sign over this zero and cross (X) all zeros against the column where is allotted.

Repeat the same process for remaining rows and proceed in a similar way for all columns. Finally write the combination of operators/workers and job along with cost and take the total cost. The total cost will be minimum.

1) *Treatment of Unbalanced Maximisation Matrix

A matrix is said to be balanced if No. of rows = No. of columns, otherwise it will be an unbalanced matrix.

An unbalanced matrix is to be balanced by introducing a dummy row or column (whichever is necessary)

so as to proceed further.

A minimisation matrix has the objective of minimising cost or time. A matrix with the objective of

maximising profit/revenue is a maximisation matrix. A maximisation matrix is to be converted in to the

minimisation matrix by the following steps:

a) Select the largest cell cost from the entire matrix

b) Deduct all cell costs from the cell element selected in (a) above and proceed further.

In case of unbalanced matrix with maximisation objective, we do have the options either to

a) Minimise the matrix and then balance the matrix OR

b) Balance the matrix and minimise the matrix

(There will be no change in optimal solution)

2) # What if no. of straight lines drawn ≠ no. of rows (columns)

In such cases the solution is to be improved before assigning jobs and should be tested for optimality. How to improve the matrix? Pick the minimum uncovered cell cost. (Uncovered element means such element not touched by st.

line) Deduct the minimum cell cost from uncovered cells Add the said minimum cell cost to intersecting cells (Intersecting cell means two lines have met each

other in that cell) In cells where there is single line (i.e. no intersection) no treatment is required. Copy the elements as it

is in the upcoming matrix. 3) $ What if there are more than one zeros in a row/column? In such cases we shall move forward to next row and subsequent rows thereon and make allotment

where there is one zero. If all rows are finished without any single zero and again we move forward

column wise and proceed further.

4) What if few zeros are left uncovered either by square sign or cross sign (X)

This case arises when tie appears. Select any one zero arbitrarily and assign it and mark two cross sign

(X) against this zero in Row & Column. Now assign one remaining zero.

5) If a constant is added/multiplied/divided/ subtracted to every element of the matrix in an assignment

problem then an assignment which minimises the total cost for the new matrix will also minimize the

total cost matrix. (i.e. there will be no impact in final solution)

ISSUES IN ABOVE STEPS

2

STEPS IN ASSGNMENT PROBLEM AT A GLANCE

Start

Write the problem

in matrix form

Is it a

balanced

problem?

Add Dummy

Row/Column

Is it a

maximisation

Problem?

Convert it into a

minimisation problem

Obtain reduced cost matrix by

Row & Column Operation

Make assignments on one-to-one match basis considering zeros in Rows/columns

Stop

NO

YES

NO

YES

3

Question -1

Prescribe the steps to be followed to solve an assignment problem.

Answer:

The steps involved in assignment problem can be solved by following steps:

Step-1:

Take a balanced minimization matrix and perform Row Reduction Operation by subtracting the minimum cost of each row from all cell cost of that row and conduct Column Reduction Operation by deducting minimum cost of each column from all cost cell of that column.

Step-2: The matrix obtained after initial solution is to be tested for optimality.

In this step the objective is to cover maximum zeros by drawing minimum no. of straight lines (straight

lines mean not diagonal lines)

Then check whether the no. of straight lines drawn equals no. of row (column).

If the answer is affirmative then the solution is optimal

Step-3:

Assigning Jobs Start from the first row and see if there is single zero row, if yes then allot job over that zero by marking square sign over this zero and cross (X) all zeros against the column where is allotted.

Repeat the same process for remaining rows and proceed in a similar way for all columns. Finally write the combination of operators/workers and job along with cost and take the total cost. The total cost will be minimum.

(Read Introduction portion for detailed study)

Question -2

Explain following statement “Assignment is special case of transportation problem; it can also be solved by transportation methods” Answer: The assignment problem is special case of transportation problem; it can also be solved by transportation method. But the solution obtained by applying this method would be severely degenerate. This is because the optimality test in the transportation method requires that there must be m+n-1 allocations/assignments. But due to the special structure of assignment problem of order n × n, any solution cannot have more than n assignments. Thus, the assignment problem is naturally degenerate. In order to remove degeneracy, (n-1)* number of dummy allocations will be required in order to proceed with the transportation method. Thus, the problem of degeneracy at each solution makes the transportation method computationally inefficient for solving an assignment problem. (*) m+n-1-n n+n-1 – n 2n-1 - n n-1 Question -3

In an assignment problem to assign jobs to men to minimize the time taken, suppose that one man does

not know how to do a particular job, how will you eliminate this allocation from the solution?

Answer: The objective of assignment problem is to minimize time the total time take to perform a particular task or to minimize the overall cost so, in an assignment minimization problem, if one task cannot be assigned to one person, introduce a prohibitively large cost for that allocation, say M, where M has a high the value. Then, while doing the row minimum and column minimum operations, automatically this allocation will get eliminated.

THEORY QUESTIONS [EXAM-SM-PM-RTP-OTHERS]

4

Question -4

Just after row and column minimum operations, we find that a particular row has 2 zeros. Does this imply that the 2 corresponding numbers in the original matrix before any operation were equal? Why? Answer: Under the Hungarian Assignment Method, the prerequisite to assign any job is that each row and column must have a zero value in its corresponding cells. If any row or column does not have any zero value then to obtain zero value, each cell values in the row or column is subtracted by the corresponding minimum cell value of respective rows or columns by performing row or column operation. This means if any row or column have two or more cells having same minimum value then these row or column will have more than one zero. However, having two zeros does not necessarily imply two equal values in the original assignment matrix just before row and column operations. Two zeroes in a same row can also be possible by two different operations i.e. one zero from row operation and one zero from column operation. Question -5

Under the usual notation, where a32 means the element at the intersection of the 3rd row and 2nd column, we have, in a 4 × 4 assignment. What can you conclude about the remaining assignments? Why? Answer: The order of matrix in the assignment problem is 4 × 4. The total assignment (allocations) will be four. In the assignment problem when any allocation is made in any cell then the corresponding row and column become unavailable for further allocation. Hence, these corresponding row and column are crossed mark to show unavailability. In the given assignment matrix two allocations have been made in a24 (2nd row and 4th column) and a32 (3rd row and 2nd column). This implies that 2nd and 3rd row and 4th and 2nd column are unavailable for further allocation. Therefore, the other allocations are at either at a11 and a43 or at a13 and a41. Question -6

Explain the following terms:

a) Balanced Problem b) Unbalanced Prob. c) Dummy d) Infeasible Assignment e) Maximisation Prob.

Answer:

Balanced Problem

An assignment problem is said to be balanced if the no. of rows = no. of columns (i.e. No. of jobs = no. of workers)

Unbalanced Problem

If in an assignment problem the no. of rows is not equal to no. of columns the problem is said to be unbalanced problem. Here, no. of facilities is not equal to the no. of jobs. Such matrix is to be balanced by inserting requisite no. of dummy row(s)/column(s).

Dummy A dummy is an imaginary job/facility with all cell element being Zero which is introduced to make an unbalanced problem balanced. In case final allotment is in dummy row/column, then it and indication that the particular operator has not been assigned any task.

Infeasible Assignment

Sometimes, it happens that a particular person is unable to perform a specific job/task or a specific job cannot be performed in a particular machine. In such cells a very high cost is assigned so as to avoid the infeasibility and continue the solution.

Maximisation Problem

In a maximisation problem the objective is to maximise the sales/revenue/profit. This is to be converted into minimisation problem. A maximisation matrix is to be converted in to the minimisation matrix by the following steps: a) Select the largest cell cost from the entire matrix b) Deduct all cell costs from the cell element selected in (a) above.

5

liahf;auofahofhuaofuafaw

Question -1

An Accounts Officer has 4 subordinates and 4 tasks. The subordinates differ in efficiency. The tasks also differ in their intrinsic difficulty. His estimates of the time each would take to perform each task is given in the matrix below. How should the tasks be allocated one to one man, so that the total man hours are minimized?

Solution:

The given problem is a balanced minimisation problem.

Step I: Initial Solution

a) Row Reduction

0 18 9 3

9 2 0 22

23 4 3 0

9 16 14 0

b) Column Reduction

0 14 9 3

9 20 0 22

23 0 3 0

9 12 14 0

Step II: Optimality Test

We draw minimum no. of straight lines to cover maximum zeros.

0 14 9 3 9 20 0 22

23 0 3 0

9 12 14 0

Since the no. of lines = No. of row/column=4, optimal solution is possible.

Step III: Assignment Final Solution

0 14 9 3

9 20 0 22

23 0 3 0

9 12 14 0

Sub

ord

inat

es Tasks

I II III IV

1 8 26 17 11

2 13 28 4 26

3 38 19 18 15

4 19 26 24 10

Subordinates Tasks Time

1 I 8

2 III 4

3 II 19

4 IV 10

Total Time 41

PRACTICAL QUESTIONS [EXAM-SM-PM-RTP-OTHERS]

6

Question -2

A manager has 5 jobs to be done. The following matrix shows the time taken by the j-th job (j = 1,2...5) on the i-th machine (i = I,II,III…V). Assign 5 jobs to the 5 machines so that the total time taken is minimized.

Jobs

Machines 1 2 3 4 5

I 9 3 4 2 10

II 12 10 8 11 9

III 11 2 9 0 8

IV 8 0 10 2 1

V 7 5 6 2 9

Solution:

The given problem is a balanced minimisation problem.


a) Row Reduction

7 1 2 0 8

4 2 0 3 1

11 2 9 0 8

8 0 10 2 1

5 3 4 0 7 b) Column Reduction

3 1 2 0 7

0 2 0 3 0

7 2 9 0 7

4 0 10 2 0

1 3 4 0 6



3 1 2 0 7

0 2 0 3 0

7 2 9 0 7

4 0 10 2 0

1 3 4 0 6

Since, the no. of lines ≠ No. of row/column, optimal solution is not possible. We further improve the

matrix.

.

Pick the minimum uncovered cell cost. Deduct the minimum cell cost from

uncovered cells Add the said minimum cell cost to

intersecting cells Copy the remaining elements in the

coming matrix as it is.

7

Here, the minimum uncovered cell element is 1, we shall proceed further with the steps stated above.

Finally we get the following matrix.

2 0 1 0 6

0 2 0 4 0

6 1 8 0 6

4 0 10 3 0

0 2 3 0 5

We again draw minimum no. of straight lines to cover maximum zeros.

2 0 1 0 6

0 2 0 4 0

6 1 8 0 6

4 0 10 3 0

0 2 3 0 5

Since, the no. of lines = No. of row/column = 5, optimal solution is possible.

Step III: Assignment

2 0 1 0 6

0 2 0 4 0

6 1 8 0 6

4 0 10 3 0

0 2 3 0 5

Final Answer

Machines Jobs Time

I 2 3

II 3 8

III 4 0

IV 5 1

V 1 7

Total Time 19

8

Question -3

5 salesmen are to be assigned to 5 districts. Estimates of sales revenue in thousands of rupees for each salesman are given below. Find the assignment pattern that maximizes the sales revenue.

A B C D E

1 32 38 40 28 40

2 40 24 28 21 36

3 41 27 33 30 37

4 22 38 41 36 36

5 29 33 40 35 39

Solution:

The given problem is a balanced maximisation problem, so at first we need to convert the matrix into

cost/loss matrix.

The cost matrix will be as:

9 3 1 13 1

1 17 13 20 5

0 14 8 11 4

19 3 0 5 5

12 8 1 6 2


a) Row Reduction

8 2 0 12 0

0 16 12 19 4

0 14 8 11 4

19 3 0 5 5

11 7 0 5 1

b) Column Reduction

8 0 0 7 0

0 14 12 14 4

0 12 8 6 4

19 1 0 0 5

11 5 0 0 1



8 0 0 7 0 0 14 12 14 4

0 12 8 6 4

19 1 0 0 5

11 5 0 0 1

A maximization matrix is to be converted in to the

minimization matrix:


b) Deduct all cell costs from the cell element

selected in (a) above and proceed further.

9

Since, the no. of lines drawn ≠ No. of row/column, optimal solution is not possible. We further improve

the matrix. The minimum uncovered element is 4 that is subtracted from all elements and added to all

elements at intersections. This yields the following matrix in which 5 lines are needed to cover all zeros.

We again draw minimum no. of straight lines to cover maximum zeros.



Here by following the steps we first allot at cell12 [12 means 1st Row 2nd Column] now tie appears as there

are more than one zeros in remaining rows/columns.

So, we arbitrarily allot assignment in Cell21

[Note: You may allot in any cells containing Zero other than elements at Column 2 & Row 1]

Again tie appears in Cell43, Cell44, and Cell53 & Cell54

We make arbitrary allotment in Cell54 and proceed further.

Final Answer

Salesman District Sales (Rs. In ‘000)

1 B 38

2 A 40

3 E 37

4 C 41

5 D 35

Total Sales 191

12 0 0 7 0

0 10 8 10 0

0 8 4 2 0

23 1 0 0 5

15 5 0 0 1

12 0 0 7 0

0 10 8 10 0

0 8 4 2 0

23 1 0 0 5

15 5 0 0 1

12 0 0 7 0

0 10 8 10 0

0 8 4 2 0

23 1 0 0 5

15 5 0 0 1





10

Question -4

To stimulate interest and provide an atmosphere for intellectual discussion, a finance faculty in a management school decides to hold special seminars on four contemporary topics– leasing, portfolio management, private mutual funds, swaps and options. Such seminars should be held once a week in the afternoons. However, scheduling these seminars (one for each topic, and not more than one seminar per afternoon) has to be done carefully so that the number of students unable to attend is kept to a minimum. A careful study indicates that the number of students who cannot attend a particular seminar on a specific day is as follows:

Leasing Portfolio

Management Private

Mutual Funds Swaps & Options

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:

The given problem is an unbalanced minimisation problem.

The matrix is unbalanced since No. of rows (5) ≠ No. of columns (4). We add a dummy column with all its

elements 0 to balance the matrix. The above matrix will be as follows:

Leasing Portfolio

Management Private

Mutual Funds Swaps & Options 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


a) Row Reduction is not required since all rows have at least one zero element.

[Even if we proceed for row operation there will be no change in the matrix]

b) Column Reduction

40 20 50 0 0

30 10 30 10 0

50 0 20 0 0

20 10 10 10 0

0 0 0 10 0

11



40 20 50 0 0

30 10 30 10 0

50 0 20 0 0

20 10 10 10 0

0 0 0 10 0

Since the no. of lines ≠ No. of row/column, optimal solution is not possible, we further improve the

matrix. The matrix after improvement will be as follows:

40 20 50 0 10 20 0 20 0 0

50 0 20 0 10

10 0 0 0 0

0 0 0 10 10


40 20 50 0 10

20 0 20 0 0

50 0 20 0 10

10 0 0 0 0

0 0 0 10 10

Final Answer

Days Seminars No. of Students

Monday Swaps & Options 20

Tuesday Dummy 0

Wednesday Portfolio Management 20

Thursday Private Mutual Funds 20

Friday Leasing 10

Total 70 The total number of students who will be missing at least one seminar = 70





12

Question -5

A solicitor's firm employs typists on hourly piece-rate basis for their daily work. There are five typists for service and their charges and speeds are different. According to an earlier understanding only one job is given to one typist and the typist is paid for full hours even if he works for a fraction of an hour. Find the least cost allocation for the following data:

Solution: First of all we shall find the no. of hours required by each typist to complete the particular jobs. The matrix will be as follows:

Now we shall change the above matrix into cost matrix by multiplying the cell elements by hourly rate.

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

The matrix is a balanced minimisation matrix we shall proceed further.


a) Row Reduction

20 10 0 60 10

24 12 0 66 12

18 9 0 57 12

20 12 0 60 12

20 8 0 56 12

Job No. of Pages

P 199

Q 175

R 145

S 298

T 178

Typist Rate/hour

(Rs) No. of pages typed/hour

A 5 12

B 6 14

C 3 8

D 4 10

E 4 11

P Q R S T

A 17 15 13 25 15

B 15 13 11 22 13

C 25 22 19 38 23

D 20 18 15 30 18

E 19 16 14 28 17

P Q R S T

A 199/12 175/12 145/12 298/12 178/12

B 199/14 175/14 145/14 298/14 178/14

C 199/8 175/8 145/8 298/8 178/8

D 199/10 175/10 145/10 298/10 178/10

E 199/11 175/11 145/11 298/11 178/11

13

b) Column Reduction

2 2 0 4 0

6 4 0 10 2

0 1 0 1 2

2 4 0 4 2

2 0 0 0 2



2 2 0 4 0 6 4 0 10 2

0 1 0 1 2

2 4 0 4 2

2 0 0 0 2

Since the no. of lines ≠ No. of row/column, optimal solution is not possible. We further improve the

matrix for obtaining optimal solution.

2 1 0 3 0

6 3 0 9 2

0 0 0 0 2

2 3 0 3 2

3 0 1 0 3

Again, the minimum number of lines required to cover all the zeros is only 4 (< 5), optimal assignment

cannot be made at this stage also. We again improve the matrix.

1 0 0 2 0

5 2 0 8 2

0 0 1 0 3

1 2 0 2 2

3 0 2 0 4

Again, the minimum number of lines required to cover all the zeros is only 4 (< 5), optimal assignment

cannot be made at this stage also. We again improve the matrix.

1 0 1 2 0

4 1 0 7 1

0 0 2 0 3

0 1 0 1 1

3 0 3 0 4 Since the no. of lines = No. of row/column=5, optimal solution is possible.

14


1 0 1 2 0

4 1 0 7 1

0 0 2 0 3

0 1 0 1 1

3 0 3 0 4

Final Answer

Typist Job Amount

A T 75

B R 66

C S 114

D P 80

E Q 64

Total 399

[Alternative solution also exists since there appears tie.]

15

Question -6

WELLDONE Company has taken the third floor of a multi-storied building for rent with a view to locate one of their zonal offices. There are five main rooms in this floor to be assigned to five managers. Each room has its own advantages and disadvantages. Some have windows, some are closer to the washrooms or to the canteen or secretarial pool. The rooms are of all different sizes and shapes. Each of the five managers was asked to rank their room preferences amongst the rooms 301, 302, 303,304 and 305. Their preferences were recorded in a table as indicated below:

MANAGER

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

and they have left off 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 a

minimum.

Solution:

Let us frame the matrix showing the ranks of the rooms.

Room No. MANAGER

M1 M2 M3 M4 M5

301 - 4 2 - 1

302 1 1 5 1 2

303 2 - 1 4 -

304 3 2 3 3 3

305 - 3 4 2 -

In a cell (-) indicates that no assignment is to be made in that particular cell. We shall ignore (-) cells for

the purpose of solving the problem. The given problem is a balanced minimisation problem, we proceed

further.


a) Row Reduction

- 3 1 - 0

0 0 4 0 1

1 - 0 3 -

1 0 1 1 1

- 1 2 0 -

b) Column Reduction: Not required

16



- 3 1 - 0

0 0 4 0 1

1 - 0 3 -

1 0 1 1 1

- 1 2 0 -



Final Answer

- 3 1 - 0

0 0 4 0 1

1 - 0 3 -

1 0 1 1 1

- 1 2 0 -

Room No. Manager Rank

301 M5 1

302 M1 1

303 M3 1

304 M2 2

305 M4 2

Total Min. Ranking 7

17

Question -7

An organisation producing 4 different products viz. A, B, C and D having 4 operators viz. P,Q,R and S, who are capable of producing any of the four products, work effectively 7 hours a day. The time (in minutes) required for each operator for producing each of the product are given in the cells of the following matrix along with profit (Rs. per unit).

Product

Operator A B C D

P 6 10 14 12

Q 7 5 3 4

R 6 7 10 10

S 20 10 15 15

Profit/Unit 3 2 4 1 Find out the assignment of operators to products which will maximize the profit.

Solution:

Total time available to each operator = 7 hours = (7x60) minutes = 420 minutes.

At first we calculate the no. of units produced by each operator and then make a profit matrix and finally

convert the matrix into loss/cost matrix.

The above formulate matrix is a balanced maximisation matrix, we will convert the matrix into loss

matrix which will be as:

350 476 440 525

380 392 0 455

350 440 392 518

497 476 448 532


a) Row Reduction

0 126 90 175

380 392 0 455

0 90 42 168

49 28 0 84

Operator

Product (Units)

A B C D

P 70 42 30 35

Q 60 84 140 105

R 70 60 42 42

S 21 42 28 28

Operator

Profit Matrix (In Rs.)

A B C D

P 210 84 120 35

Q 180 168 560 105

R 210 120 168 42

S 63 84 112 28

18

b) Column Reduction

0 98 90 91

380 364 0 371

0 62 42 84

49 0 0 0



0 98 90 91

380 364 0 371 0 62 42 84

49 0 0 0

Since the no. of lines ≠ No. of row/column, optimal solution is not possible. We further improve the

matrix. The improved matrix will be:

0 36 90 29

380 302 0 309

0 0 42 22

111 0 62 0


Step III: Assignment Final Answer

Operator Products Profit

P A 210

Q C 560

R B 120

S D 28

Total 918

0 36 90 29

380 302 0 309

0 0 42 22

111 0 62 0

19

Question -8

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

products. The processing time varies from operator to operator. The firm records 8 hours a day and allow

30 minutes for lunch. The processing time in minutes and the profit for each of the products are given

below:

Products

Operators 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/Unit 8 6 5 4 Find the optimal assignment of products to operators.

Solution:

Productive time per worker = [8hrs. x 60]-30 = 450 mins.

Calculating the no. of units of produced by each operators.

A B C D

1 30 50 45 75

2 45 75 50 75

3 18 30 30 50

4 30 50 45 45

Since we are given the profit per unit of each product, the profit matrix is computed as given below:

Profit Matrix

A B C D

1 240 300 225 300

2 360 450 250 300

3 144 180 150 200

4 240 300 225 180

Converting the contribution matrix into cost matrix.

210 150 225 150

90 0 200 150

306 270 300 250

210 150 225 270

20


a) Row Reduction

60 0 75 0

90 0 200 150

56 20 50 0

60 0 75 120

b) Column Reduction

4 0 25 0

34 0 150 150

0 20 0 0

4 0 25 120



4 0 25 0

34 0

150 150

0 20 0 0

4 0 25 120

The minimum number of lines required to cover all the zeros is only 3(< 4), optimal assignment cannot

be made at this stage also. We improve the matrix.

0 0 21 0

30 0 146 150

0 24 0 4

0 0 21 120



Operators Products Profit 1 D 300 2 B 450 3 C 150 4 A 240

Total 1140

0 0 21 0

30 0 146 150

0 24 0 4

0 0 21 120

21

Question -9

Five lathes are to be allotted to five operators (one for each). The following table gives weekly output figures (in pieces)-

Operator Weekly Output in Lathe

L1 L2 L3 L4 L5

P 20 22 27 32 36

Q 19 23 29 34 40

R 23 28 35 39 34

S 21 24 31 37 42

T 24 28 31 36 41 Profit per piece is Rs. 25. Required: Find the maximum profit per week. Solution: The given assignment problem is a maximization problem. We convert it into an opportunity loss matrix by subtracting all the elements of the given table from the highest element of the table that is 42. The cost matrix will be as follows:


a) Row Reduction

16 14 9 4 0

21 17 11 6 0

16 11 4 0 5

21 18 11 5 0

17 13 10 5 0

b) Column Reduction

0 3 5 4 0

5 6 7 6 0

0 0 0 0 5

5 7 7 5 0

1 2 6 5 0

Operator L1 L2 L3 L4 L5

P 22 20 15 10 6

Q 23 19 13 8 2

R 19 14 7 3 8

S 21 18 11 5 0

T 18 14 11 6 1

22



0 3 5 4 0

5 6 7 6 0

0 0 0 0 5

5 7 7 5 0 1

2 6 5 0

Since the no. of lines (3) ≠ No. of row/column (5), optimal solution is not possible at this stage. We

further improve the matrix.

0 1 3 2 0

5 4 5 4 0

2 0 0 0 7

5 5

5 3 0 1

0 4 3 0

Again, the minimum number of lines drawn to cover all zeros is 4. Repeating the above steps once again,

we get the following table-

0 1 1 0 0

5 4 3 2 0

4 2 0 0 9

5 5

3 1 0

1 0 2 1 0

The minimum number of lines to cover all zeros is 4 which is less than 5. We further improve the

matrix.

0 2 1 0 1

4 4 2

1 0 4

3 0 0

10

4 5

2 0 0

0 0 1 0 0 Since the no. of lines = No. of row/column=4, optimal solution is possible.

23


0 2 1 0 1

4 4 2 1 0

4 3 0 0 10

4 5 2 0 0

0 0 1 0 0

The maximum profit per week is Rs.4,000 (25 × 160).

Operator Lathe

Machine Weekly Output

P L1 20 Q L5 40 R L3 35 S L4 37 T L2 28

Total 160

24

Question -10

A manufacturing company has four zones A, B, C, D and four sales engineers P, Q, R, S respectively for

assignment. Since the zones are not equally rich in sales potential, therefore it is estimated that a

particular engineer operating in a particular zone will bring the following sales:

Zone A: 4,20,000

Zone B: 3,36,000

Zone C: 2,94,000

Zone D: 4,62,000

The engineers are having different sales ability. Working under the same conditions, their yearly sales are

proportional to 14, 9, 11 and 8 respectively. The criteria of maximum expected total sales is to be met by

assigning the best engineer to the richest zone, the next best to the second richest zone and so on.

Find the optimum assignment and the maximum sales.

Solution:

Calculating Sales Matrix

Sales Man Ratio A B C D

P 14 140000 112000 98000 154000

Q 9 90000 72000 63000 99000

R 11 110000 88000 77000 121000

S 8 80000 64000 56000 88000

Total 42 420000 336000 294000 462000

Dividing all the elements of the matrix by 1000 we get the following matrix.

Converting the revenue matrix into loss matrix we get the following matrix.

Sales Man A B C D

P 140 112 98 154

Q 90 72 63 99

R 110 88 77 121

S 80 64 56 88

Sales Man Ratio A B C D

P 14 140 112 98 154

Q 9 90 72 63 99

R 11 110 88 77 121

S 8 80 64 56 88

Total 42 420 336 294 462

If a constant is added/multiplied/divided/

subtracted to every element of the matrix in an

assignment problem then an assignment which

minimises the total cost for the new matrix will also minimize the total cost matrix. (i.e. there

will be no impact in final solution)

25


a) Row Reduction

14 42 56 0

9 27 36 0

11 33 44 0

8 24 32 0

b) Column Reduction

6 18 24 0

1 3 4 0

3 9 12 0

0 0 0 0



6 18 24 0

1 3 4 0

3 9 12 0

0 0 0 0

The minimum number of lines required to cover all the zeros is only 2 (< 4), optimal assignment cannot

be made at this stage. We improve the matrix and draw minimum no. of straight lines to cover max. zeros

Again, the min. number of lines required to cover all the zeros is only 3 (< 4), optimal assignment cannot be made at this stage also. We improve the matrix and draw min. no. of straight lines to cover max. Zeros.

5 15 21 0

0 0 1 0

2 6 9 0

2 0 0 3

Again, the min. number of lines required to cover all the zeros is only 3 (< 4), optimal assignment cannot be made at this stage also. We improve the matrix and draw min. no. of straight lines to cover max. zeros.


5 17 23 0 0 2 3 0

2 8 11 0

0 0 0 1

3 13 19 0

0 0 1 2

0 4 7 0

2 0 0 5

26


Final Answer

Question -11

A company has four zones open and four marketing managers available for assignment. The zones are not equal in sales potentials. It is estimated that a typical marketing manager operating in each zone would bring in the following Annual sales: Zones Rs. East………………………………………………. 2,40,000 West……………………………………………… 1,92,000 North……………………………………………... 1,44,000 South…………………………………………….. 1,20,000 The four marketing manages are also different in ability. It is estimated that working under the same conditions, their yearly sales would be proportionately as under: Manager M………………………………………. 8 Manager N………………………………………. 7 Manager O………………………………………. 5 Manager P………………………………………. 4 Required If the criterion is maximum expected total sales, find the optimum assignment and the maximum sales. Solution: Calculating Sales Matrix

Sales Man

Ratio East West North South

M 8 80000 64000 48000 40000

N 7 70000 56000 42000 35000

O 5 50000 40000 30000 25000

P 4 40000 32000 24000 20000

Total 24 240000 192000 144000 120000

3 13 19 0

0 0 1 2

0 4 7 0

2 0 0 5

Sales Man Zone Sales Amt.

P D 154000

Q B 72000

R A 110000

S C 56000

Total 392000

27

Dividing all the elements of the matrix by 1000 we get the following matrix.

Sales Man

Ratio East West North South

M 8 80 64 48 40

N 7 70 56 42 35

O 5 50 40 30 25

P 4 40 32 24 20

Total 24 240 192 144 120

Converting the revenue matrix into loss matrix we get the following matrix.

Manager East West North South

M 0 16 32 40

N 10 24 38 45

O 30 40 50 55

P 40 48 56 60 Step I: Initial Solution

a) Row Reduction

0 16 32 40

0 14 28 35

0 10 20 25

0 8 16 20 b) Column Reduction

0 8 16 20

0 6 12 15

0 2 4 5

0 0 0 0 Step II: Optimality Test


0 8 16 20

0 6 12 15

0 2 4 5

0 0 0 0

Since the no. of lines (2) ≠ No. of row/column (4), optimal solution is not possible. We further improve

the matrix.

If a constant is added/multiplied/divided/

subtracted to every element of the matrix in an

assignment problem then an assignment which

minimises the total cost for the new matrix will

also minimize the total cost matrix. (i.e. there

will be no impact in final solution)

28

0 6 14 18

0 4 10 13

0 0 2 3

2 0 0 0 Again no. of lines covering zeros are not equal to the order of matrix, we further improve the matrix.

Again no. of lines covering zeros are not equal to the order of matrix, we further improve the matrix



0 2 10 14

0 0

6 9

4 0 2 3

6 0 0 0

0 2 8 12

0 0

4 7

4 0 0

1

8 2 0 0

0 2 8 12

0 0 4 7

4 0 0 1

8 2 0 0

Sales Man Zones Sales

M East 80,000

N West 56,000

O North 30,000

P South 20,000

Total 1,86,000

29

Question -12

XYZ airline operating 7 days a week has given the following timetable. Crews must have a minimum layover

of 5 hours between flights. Obtain the pairing flight that minimizes layover time away from home. For any

given pairing the crew will be based at the city that results in the smaller layover.

Solution:

To begin with, let us first assume that the crew is based at Chennai. The flight A1, which starts from Chennai

at 6 AM, reaches Mumbai at 8 AM. The schedule time for the flight at Mumbai is 8 AM. Since the minimum

layover time for crew is 5 hours, this flight can depart only on the next day i.e. the layover time will be 24

hours. Similarly, layover times for other flights are also calculated and given in the following table.

The layover times for various flight connections when crew is assumed to be based at Mumbai are similarly

calculated in the following table.

Flight No. Crew based at Mumbai

B1 B2 B3 B4

A1 20 19 14 9

A2 22 21 16 11

A3 28 27 22 17

A4 10 9 28 23

Chennai-Mumbai Mumbai - Chennai

Flight No. Depart. Arrive Flight No. Depart. Arrive

A1 6 AM 8 AM B1 8 AM 10 AM

A2 8 AM 10 AM B2 9 AM 11 AM

A3 2 PM 4 PM B3 2 PM 4 PM

A4 8 PM 10 PM B4 7 PM 9 PM

Flight No. Crew based at Chennai

B1 B2 B3 B4

A1 24 25 6 11

A2 22 23 28 9

A3 16 17 22 27

A4 10 11 16 21

30

Now since the crew can be based at either of the places, minimum layover times can be obtained for different flight numbers by selecting the corresponding lower value out of the above two tables. The resulting table is as given below

A * with an entry in the above table indicates that it corresponds to layover time when the crew is based at Mumbai. We will now apply the assignment algorithm to find the optimal solution. Subtracting the minimum element of each row from all the elements of that row, we get the following matrix.


a) Row Reduction

Flight No.

Flight No. B1 B2 B3 B4

A1 14 13 0 3

A2 13 12 7 0

A3 0 1 6 1

A4 1 0 7 12

b) Column Reduction: Not required



Flight No.


A1 14 13 0 3

A2 13 12 7 0

A3 0 1 6 1

A4 1 0 7 12



Flight No.


A1 14 13 0 3

A2 13 12 7 0

A3 0 1 6 1

A4 1 0 7 12

Flight No.


A1 20* 19* 6 9*

A2 22 21* 16* 9

A3 16 17 22 17*

A4 10 9* 16 21

From Flight To Flight

Lay Over Time

A1 B3 6

A2 B4 9

A3 B1 16

A4 B2 9

Total 40

31

Question -13

An Electronic Data Processing (EDP) centre has three expert Software professionals. The Centre wants

three application software programs to be developed. The head of EDP Centre estimates the computer

time in minutes required by the experts for development of Application Software Programs as follows-

Software Programs

Computer Time (in minutes) Required by Software Professionals

A B C

1 100 85 70

2 50 70 110

3 110 120 130 Assign the software professionals to the application software programs to ensure minimum usage of computer time. Solution: The given problem is a balanced minimization assignment problem.


a) Row Reduction

30 15 0

0 20 60

0 10 20

b) Column Reduction

30 5 0

0 10 60

0 0 20



30 5 0

0 10 60

0 0 20



30 5 0

0 10 60

0 0 20

Software Programs Experts Time

1 C 70

2 A 50

3 B 120

Total 240

32

Question -14

A project consists of four (4) major jobs, for which four (4) contractors have submitted tenders. The tender amounts, in thousands of rupees, are given below-

Contractors Job

A B C D

1 120 100 80 90

2 80 90 110 70

3 110 140 120 100

4 90 90 80 90

Find the assignment, which minimizes the total cost of the project. Each contractor has to be assigned one job. Solution: The given problem is a balanced minimization problem. Step I: Initial Solution

a) Row Reduction

40 20 0 10

10 20 40 0

10 40 20 0

10 10 0 10

b) Column Reduction

30 10 0 10

0 10 40 0

0 30 20 0

0 0 0 10



30 10 0 10

0 10 40 0

0 30 20 0

0 0 0 10



30 10 0 10

0 10 40 0

0 30 20 0

0 0 0 10

(Note: Alternative solution also exists)

Contractors Job Cost (Rs.'000)

1 C 80

2 D 70

3 A 110

4 B 90

Total 350

33

Question -15

A Marketing Manager has 4 subordinates and 4 tasks. The subordinates differ in efficiency. The tasks also differ in their intrinsic difficulty. His estimates of the time each subordinate would take to perform each task is given in the matrix below.

I II III IV

1 16 52 34 22

2 26 56 8 52

3 76 38 36 30

4 38 52 48 20 How should the task be allocated one to one man so that the total man-hours are minimised? Solution: The given problem is a balanced minimization problem. Step I: Initial Solution

a) Row Reduction

0 36 18 6

18 48 0 44

46 8 6 0

18 32 28 0 b) Column Reduction

0 28 18 6

18 40 0 44

46 0 6 0

18 24 28 0 Step II: Optimality Test


0 28 18 6

18 40 0 44

46 0 6 0

18 24 28 0 Since the no. of lines = No. of row/column=4, optimal solution is possible.


0 28 18 6

18 40 0 44

46 0 6 0

18 24 28 0

Operator Task Time (Hrs) 1 I 16 2 III 8 3 II 38 4 IV 20

Total 82

34

Question -16 Five swimmers are eligible to compete in a relay team which is to consist of four swimmers swimming four different swimming styles ; back stroke breast stroke, free style and butterfly. The time taken for the five swimmers – Anand, Bhaskar, Chandru, Dorai and Easwar- to cover a distance of 100 meters in various swimming styles are given below in minutes : seconds. Anand swims the back stroke in 1:09, the breast stroke in 1:15 and has never competed in the free style or butterfly. Bhaskar is a free style specialist averaging 1:01 for the 100 metres but can also swim the breast stroke in 1:16 and butterfly in 1:20. Chandru swims all styles – back stroke 1:10, butterfly 1:12, free style 1:05 and breast stroke 1:20. Dorai swims only the butterfly 1:11 while Easwar swims the back stroke 1:20, the breast stroke 1:16, the free style 1:06 and the butterfly 1:10. Required Which swimmers should be assigned to which swimming style? Who will be in the relay? Solution: First of all we shall make assignment matrix with time expressed in seconds. The matrix table will be:

Back

Stroke Breast Stroke

Free Style Butterfly

Anand 69 75 - -

Bhaskar - 76 61 80

Chandru 70 80 65 72

Dorai - - - 71

Easwar 80 76 66 70 Since the matrix is unbalanced, we make the matrix balanced by inserting a dummy column and proceed further.

Back

Stroke Breast Stroke

Free Style Butterfly Dummy

Anand 69 75 - - 0

Bhaskar - 76 61 80 0

Chandru 70 80 65 72 0

Dorai - - - 71 0

Easwar 80 76 66 70 0 Step I: Initial Solution

a) Row Reduction: Not Required [Since each row has at least one element as Zero] b) Column Reduction

0 0 - - 0

- 1 0 10 0

1 5 4 2 0

- - - 1 0

11 1 5 0 0

35



0 0 - - 0

- 1 0 10 0

1 5 4 2 0

- - - 1 0

11 1 5 0 0



0 0 - - 1

- 0 0 10 0 0 4 4 2 0

- - - 1 0

10 0 5 0 0



Question -17

ABC Company is engaged in manufacturing 5 brands of packet snacks. It is having five manufacturing

setups, each capable of manufacturing any of its brands, one at a time. The cost to make a brand on these

setups vary according to following table-

Required: Assuming five setups are S1, S2, S3, S4, and S5 and five brands are B1, B2, B3, B4, and B5, Find the optimum assignment of the products on these setups resulting in the minimum cost.

Swimmers Style Time

Anand Back Stroke 75

Bhaskar Free Style 61

Chandru Back Stroke 70

Dorai Dummy 0

Easwar Butterfly 70

Total Time (Seconds) 276

0 0 - - 1

- 0 0 10 0

0 4 4 2 0

- - - 1 0

10 0 5 0 0

S1 S2 S3 S4 S5

B1 4 6 7 5 11

B2 7 3 6 9 5

B3 8 5 4 6 9

B4 9 12 7 11 10

B5 7 5 9 8 11





36

Solution:


a) Row Reduction

0 2 3 1 7

4 0 3 6 2

4 1 0 2 5

2 5 0 4 3

2 0 4 3 6

b) Column Reduction

0 2 3 0 5

4 0 3 5 0

4 1 0 1 3

2 5 0 3 1

2 0 4 2 4



0 2 3 0 5

4 0 3 5 0

4 1 0 1 3

2 5 0 3 1

2 0 4 2 4



0 3 4 0 6

3 0 3 4 0

3 1 0 0 3

1 5 0 2 1

1 0 4 1 4



Brands Setup Costs

B1 S1 4

B2 S5 5

B3 S4 6

B4 S3 7

B5 S2 5

Total 27

0 3 4 0

6

3 0

3 4 0

3 1 0 0 0

1 5 0 2 1

1 0 4 1 4





37

Question -18

A factory is going to modify of a plant layout to install four new machines M1, M2, M3 and M4. There are 5 vacant places J, K, L, M and N available. Because of limited space machine M2 cannot be placed at L and M3 cannot be placed at J. The cost of locating machine to place in Rupees is shown below:

J K L M N

M1 18 22 30 20 22

M2 24 18 -- 20 18

M3 -- 22 28 22 14

M4 28 16 24 14 16 Required Determine the optimal assignment schedule in such a manner that the total costs are kept at a minimum.

Solution:

The matrix is unbalanced, we make the matrix balanced by inserting a dummy row and proceed

further.

J K L M N

M1 18 22 30 20 22

M2 24 18 -- 20 18

M3 -- 22 28 22 14

M4 28 16 24 14 16

M5 (Dummy) 0 0 0 0 0


a) Row Reduction

0 4 12 2 4

6 0 -- 2 0

-- 8 14 8 0

14 2 10 0 2

0 0 0 0 0

b) Column Reduction: Not Required

38





Question -19

A car hiring company has one car at each of the five depots A, B C, D and E. A customer in each of the five

towns V, W, X, Y and Z requires a car. The distance in kms, between depots (origin) and the town

(destination) are given in the following table-

Town

Depots

A B C D E

V 3 5 10 15 8

W 4 7 15 18 8

X 8 12 20 20 12

Y 5 5 8 10 6

Z 10 10 15 25 10

Required: Find out as to which car should be assigned to which customer so that the total distance travelled is a minimum. How much is the total travelled distance?

0 4 12 2 4

6 0 -- 2 0

-- 8 14 8 0 14 2 10 0 2

0 0 0 0 0

Machine Places Cost

M1 J 18

M2 K 18

M3 N 14

M4 M 14

M5 (Dummy) L 0

Total Cost 64

0 4 12 2 4

6 0 -- 2 0

-- 8 14 8 0

14 2 10 0 2

0 0 0 0 0

39

Solution:

The given matrix is a balanced matrix, so we proceed step wise without any improvement to the matrix.


a) Row Reduction

0 2 7 12 5

0 3 11 14 4

0 4 12 12 4

0 0 3 5 1

0 0 5 15 0

b) Column Reduction

0 2 4 7 5

0 3 8 9 4

0 4 9 7 4

0 0 0 0 1

0 0 2 10 0



0 2 4 7 5

0 3 8 9 4

0 4 9 7 4

0 0 0 0 1

0 0 2 10 0

Since the number of lines (=3) is not equal to the order of the matrix (which is 5), the above matrix will not give the optimal solution. We subtract the minimum uncovered element (=2) from all uncovered elements and add it to the elements lying on the intersection of two lines. We get the following improved matrix-

0 0 2 5 3

0 1 6 7 2

0 2 7 5 2

2 0 0 0 1

2 0 2 10 0

Again, the minimum number of lines required to cover all the zeros is less than the rows/column, optimal assignment cannot be made at this stage also. We again improve the matrix.

0 0 0 3 3 0 1 4 5 2

0 2 5 3 2 4 2 0 0 3 2 0 0 8 0

40

Again, the minimum number of lines required to cover all the zeros is less than the rows/column, optimal assignment cannot be made at this stage also. We again improve the matrix.

1 0 0 3 3

0 0 3 4 1 0 1 4 2

1

5 2 0 0 3

3 0 0 8 0



Question -20 Imagine yourself to be the Executive Director of a 5-Star Hotel which has four banquet halls that can be used for all functions including weddings. The halls were all about the same size and the facilities in each hall differed. During a heavy marriage season, 4 parties approached you to reserve a hall for the marriage to be celebrated on the same day. These marriage parties were told that the first choice among these 4 halls would cost Rs.25,000 for the day. They were also required to indicate the second, third and fourth preferences and the price that they would be willing to pay. Marriage party A & D indicated that they won’t be interested in Halls 3 & 4. Other particulars are given in the following table-

Marriage Party Hall 1 Hall 2 Hall 3 Hall 4

A 25,000 22,500 X X

B 20,000 25,000 20,000 12,500

C 17,500 25,000 15,000 20,000

D 25,000 20,000 X X Where X indicates that the party does not want that hall. Required Decide on an allocation that will maximize the revenue to your hotel.

Customer at Town

Car at Depot

Distance (Km.)

V C 10

W B 7

X A 8

Y D 10

Z E 10

Total 45

1 0 0 3 3

0 0 3 4 1

0 1 4 2 1

5 2 0 0 3

3 0 0 8 0

41

Solution:

We are given a preference matrix stating the amount that the parties are willing to pay for each hall. The

objective of the problem is to maximise the revenue of the hotel. Hence, we are given a revenue matrix.

The cost matrix will be as:

Marriage Party Hall 1 Hall 2 Hall 3 Hall 4

A 0 2500 X X

B 5000 0 5000 12500

C 7500 0 10000 5000

D 0 5000 X X


a) Row Reduction: Not required [Each row contains at least ONE Zero

b) Column Reduction

0 2500 X X

5000 0 0 7500

7500 0 5000 0

0 5000 X X



0 2500 X X

5000 0 0 7500

7500 0 5000 0

0 5000 X X

The minimum number of lines to cover all zeros is 3 which is less than the order of the square matrix

(i.e.4), the above matrix will not give the optimal solution, and further improvement is required.

The improved matrix will be as:

0 0 X X

7500 0 0 7500

10000 0 5000 0

0 2500 X X

A maximization matrix is to be converted in to the

minimization matrix:


b) Deduct all cell costs from the cell element

selected in (a) above and proceed further.

Pick the minimum uncovered cell cost. In this case its 2500

Deduct the minimum cell cost from uncovered cells Uncovered cell costs are Cell12 & Cell42

with values 2500 & 5000 resp. Add the said minimum cell cost to intersecting cells

Intersecting Cells are Cell21 & Cell31

2500 is to be added to both cells. Copy the remaining elements in the coming matrix

as it is. Copy the remaining values as it is in the improved matrix.

42


0 0 X X

7500 0 0 7500

10000 0 5000 0

0 2500 X X



0 0 X X

7500 0 0 7500

10000 0 5000 0

0 2500 X X

Marriage Party Hall Revenue

A 2 22,500

B 3 20,000

C 4 20,000

D 1 25,000

Total Revenue 87,500

43

Question -21 [Important] A 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. Cost (in ₹000) of travelling from one city to another is given below:

Required

Find out the ‘Least Cost Route’.

Solution:

Based on the requirement of the question the given matrix is a minimisation matrix. While solving the question remember the additional requirement of the question i.e. He wishes to start from a particular city, visit each city once and then return to his starting point


a) Row Reduction

- 3 12 18 0

12 - 3 18 0

12 9 - 0 9

24 0 6 - 3

0 6 3 21 -

b) Column Reduction

- 3 9 18 0

12 - 0 18 0

12 9 - 0 9

24 0 3 - 3

0 6 0 21 -



- 3 9 18 0

12 - 0 18 0

12 9 - 0 9

24 0 3 - 3

0 6 0 21 -

P Q R S T

P - 5 14 20 2

Q 17 - 8 23 5

R 23 20 - 11 20

S 35 11 17 - 14

T 2 8 5 23 -

44



P Q R S T

P - 3 9 18 0

Q 12 - 0 18 0 R 12 9 - 0 9

S 24 0 3 - 3

T 0 6 0 21 -

At this stage no doubt we arrive at minimum cost but the additional condition does not get fulfilled.

The above solution is optimum solution with two routes:

I. Starts from P & reaches destination T & finally comes back to P and

II. Starts from Q & reaches R & then reaches to S & Comes back to Q

To solve this problem we have to bring next minimum element in the matrix i.e.3. Now the possible new

assignments are:

P Q R S T P - 3 9 18 0 Q 12 - 0 18 0 R 12 9 - 0 9

S 24 0 3 - 3

T 0 6 0 21 -

Note:

Next lowest cost is 3 and Cell (P,Q), Cell(S,R) & Cell(S,T) has same cell cost 3. We may allot in any cells

provided that the additional condition gets fulfilled.

On allotting to Cell(S,R) the additional condition does not get fulfilled.

Allotment may be made at Cell(S,T), the overall cost will be same when assignment is made in Cell(P,Q) or

in Cell(S,T).

Final Answer

From To Cost

P Q 5000

Q R 8000

R S 11000

S T 14000

T P 2000

Total Cost 40000

45

Question-22

A 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. Cost (in ` '000) of travelling from one city to another is given below-

Required

Find out the least cost route.

Solution:

Based on the requirement of the question the given matrix is a minimisation matrix. While solving the question remember the additional requirement of the question i.e. He wishes to start from a particular city, visit each city once and then return to his starting point


a) Row Reduction

Cities P Q R S T

P - 2 8 0 2

Q 2 - 6 0 2

R 4 2 - 4 0

S 0 0 8 - 8

T 0 0 2 6 -

b) Column Reduction

Cities P Q R S T

P - 2 6 0 2

Q 2 - 4 0 2

R 4 2 - 4 0

S 0 0 6 - 8

T 0 0 0 6 -



Cities P Q R S T

P - 2 6 0 2

Q 2 - 4 0 2

R 4 2 - 4 0

S 0 0 6 - 8

T 0 0 0 6 -

P Q R S T

P - 6 12 4 6

Q 6 - 10 4 6

R 12 10 - 12 8

S 4 4 12 - 12

T 6 6 8 12 -

From

To

46




Cities P Q R S T

P - 0 4 0 2

Q 0 - 2 0 2

R 2 0 - 4 0

S 0 0 6 - 10

T 0 0 0 8 -

Again we draw minimum no. of straight lines to cover maximum zeros.



Cities P Q R S T

P - 0 4 0 2

Q 0 - 2 0 2

R 2 0 - 4 0

S 0 0 6 - 10

T 0 0 0 8 -

After making allotment in Cell53 and Cell35 there appears tie. We allot in Cell41.

The next lowest cost cell is 2 so we make allotment in cells where cost is 2.

Possible cells are: Cell 31 Cell 15 Cell 25 Cell 23

We make allotment in all cells with cell cost 2.

The routes and their associated costs are as follows:

From To Cost

P S 4000

S Q 4000

Q P 6000

R T 8000

T R 8000

Total Cost 30000

Cities P Q R S T

P - 0 4 0 2

Q 0 - 2 0 2

R 2 0 - 4 0

S 0 0 6 - 10

T 0 0 0 8 -





The least cost comes to 30000 but this solution does not

fulfil the additional condition of the question ie.

Travelling pattern.

So we shall proceed further to find the alternative route.

The next lowest cost cell is 2 so we make allotment in cells

where cost is 2.

Possible cells are: Cell 31 Cell 15 Cell 25 Cell 23

47

We make allotment in all cells with cell cost 2.

Case I: Allotment made in Cell 15 Case II Allotment made in Cell 31

Cost Sheet: Cost Sheet:

Case III: Allotment made in Cell 25 Case IV: Allotment made in Cell 23

Cost Sheet: Cost Sheet:

Hence, case I fulfils both the conditions, the travelling route is selected as per case I.

Cities P Q R S T

P - 0 4 0 2

Q 0 - 2 0 2

R 2 0 - 4 0

S 0 0 6 - 10

T 0 0 0 8 -

Cities P Q R S T

P - 0 4 0 2

Q 0 - 2 0 2

R 2 0 - 4 0

S 0 0 6 - 10

T 0 0 0 8 -

From To Cost

P S 4000

S Q 4000

Q R 10000

R T 8000

T P 6000

Total 32000

From To Cost

P R 12000

R T 8000

T S 12000

S Q 4000

Q P 6000

Total 42000

Cities P Q R S T

P - 0 4 0 2

Q 0 - 2 0 2

R 2 0 - 4 0

S 0 0 6 - 10

T 0 0 0 8 -

Cities P Q R S T

P - 0 4 0 2

Q 0 - 2 0 2

R 2 0 - 4 0

S 0 0 6 - 10

T 0 0 0 8 -

From To Cost

P S 4000

S R 12000

R T 8000

T Q 6000

Q P 6000

Total 36000

From To Cost

P S 4000

S T 12000

T R 8000

R Q 10000

Q P 6000

Total 40000

48

Question -23

The cost matrix giving selling costs per unit of a product by salesman A, B, C and D in regions R1, R2, R3

and R4 is given below:

A B C D

R1 4 12 16 8

R2 20 28 32 24

R3 36 44 48 40

R4 52 60 64 56

Required:

(i) Assign one salesman to one region to minimise the selling cost.

(ii) If the selling price of the product is ₹200 per unit and variable cost excluding the selling cost given in the table is ₹100 per unit, find the assignment that would maximise the contribution. (iii) What other conclusion can you make from the above? Solution: Answer to Point (i)


a) Row Reduction b) Column Reduction



0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0 Since the no. of lines = No. of row/column=4, optimal solution is possible.


0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 8 12 4

0 8 12 4

0 8 12 4

0 8 12 4

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

Region Sales

Person Selling

Cost

R1 A 4

R2 B 28

R3 C 48

R4 D 56

Total 136

49

Answer to Point (ii)

Particulars Amount

Sales 200

Less: Variable cost before Selling Cost 100

Contribution before Selling Cost 100

Forming contribution matrix: After Including Selling Costs.

96 88 84 92

80 72 68 76

64 56 52 60

48 40 36 44

Since, the matrix is a maximisation matrix we will convert the matrix into a loss matrix.

0 8 12 4

16 24 28 20

32 40 44 36

48 56 60 52


a) Row Reduction

0 8 12 4

0 8 12 4

0 8 12 4

0 8 12 4

This matrix is the same matrix that we have derived in point no. (1) In Step I

So, final answer shall be:

Answer to Point (iii)

i) If a constant is added/multiplied/divided/ subtracted to every element of the matrix in an

assignment problem then an assignment which minimises the total cost for the new matrix will also

minimize the total cost matrix. (i.e. there will be no impact in final solution)

ii) Minimising cost is the same as maximizing contribution.

iii) Many zero’s represent many feasible least cost assignment. Here, all zeros mean maximum

permutation of a 4 X 4 matrix, viz. 24 solutions (4 X 3 X 2 X1) are possible.

Region Sales Person Contribution

R1 A 96

R2 B 72

R3 C 52

R4 D 44

Total 264

50

Question - 24

R3C2 denotes the element at the intersection of the third row and 2nd column. Under this notation, R1C1, R2C1, R3C1, R3C2, R3C3, R4C3, R4C4, were the only zero elements in a 4x4 minimization assignment problem after the row minimum and column minimum operations. (i) In the next step to draw lines to cover zeroes, a student drew 4 horizontal lines covering rows R1, R2,

R3, and R4. Will he arrive at the optimal assignment at the next step? Why? Explain the concept.

(ii) Independent of (i), if you are given the additional information that R2C2 element is lesser than the Row 1 and Row 2 non-zero values, how will you arrive at the optimal solution?

Solution: Answer to point no. (i) Framing a 4 X 4 matrix based on the information given. The matrix obtained is a matrix after Row & Column Operation. This means Step I (as per our steps) has been completed and Step II is to be started.

C1 C2 C3 C4

R1 0

R2 0

R3 0 0 0

R4 0 0 Now, let’s continue to Step II: The step performed by the student is as follows: Actually to be done: Correct Solution

C1 C2 C3 C4

R1 0

R2 0

R3 0 0 0

R4 0 0

Assignment is made in the cell which has zero (0) element and other zero elements in the corresponding row and column is crossed i.e. no further assignment is possible in that row and column in which first assignment has already been made. Analysis of the operation performed by the student- i) The first (assuming) assignment has been made in cell R1C1, leaving no further scope for assignment

in R1 and C1. However, in the above matrix a further assignment has also been made in R2C1, which could have not been done. It makes the above solution invalid and it will not arrive at the optimal assignment.

ii) However, assignment in R3C2 or R3C3 and R4C3 or R4C4 is possible provided no other assignment is made in the corresponding row and column.

C1 C2 C3 C4

R1 0

R2 0

R3 0 0 0

R4 0 0

51

Answer to point no. (ii) Let us assume that the non-zero values in Row 1 and Row 2 except value at R2C2 is “x" and no assignments are possible in cells R4C1, R4C2 and R3C4. The following assignment matrix could be as below:




We draw minimum no. of straight lines to cover

maximum zeros.



The optimal assignment will be at R1C1, R2C2, R3C3 and R4C4.

C1 C2 C3 C4

R1 0 N N N

R2 0 (N-1) N N

R3 0 0 0

R4 0 0

C1 C2 C3 C4

R1 0 1 1 1

R2 0 0 1 1

R3 N-1 0 0

R4 0 0

C1 C2 C3 C4

R1 0 1 1 1

R2 0 0 1 1

R3 N-1 0 0

R4 0 0

C1 C2 C3 C4

R1 0 1 1 1

R2 0 0 1 1

R3 N-1 0 0

R4 0 0

Pick the minimum uncovered cell cost. Here it is (N-1)

Deduct the minimum cell cost from uncovered cells Uncovered Cells are those cells which are not touched by even a single line.

Add the said minimum cell cost to intersecting cells Intersecting Cells are R3C1 & R4C1

Copy the remaining elements in the coming matrix as it is. Cells touched by single line are to be incorporated in same form.

52

Question - 25

A BPO company is taking bids for 4 routes in the city to ply pick-up and drop cabs. Four companies have

made bids as detailed below-

Co./Routes R1 R2 R3 R4

C1 4,000 5,000 − −

C2 − 4,000 − 4,000

C3 3,000 − 2,000 −

C4 − − 4,000 5,000

Each bidder can be assigned only one route. Determine the minimum cost that the BPO should incur.

Solution: For simplicity in calculation we divide all the figures by 1000 and proceed further



1 1 − −

− 0 − 0

0 − 0 −

− − 2 1 Step II: Optimality Test


0 0 − −

− 0 − 0

0 − 0 −

− − 1 0



0 0 − −

− 0 − 0

0 − 0 −

− − 1 0

Company Route Cost

C1 R1 4,000

C2 R2 4,000

C3 R3 2,000

C4 R4 5,000

Total 15,000

0 0 − −

− 0 − 0

0 − 0 −

− − 1 0

53

Question - 25

A manager was asked to assign tasks to operators (one task per operator only) so as to minimize the time

taken. He was given the matrix showing the hours taken by the operators for the tasks.

First, he performed the row minimum operation. Secondly, he did the column minimum operation. Then,

he realized that there were 4 tasks and 5 operators. At the third step he introduced the dummy row and

continued with his fourth step of drawing lines to cover zeros. He drew 2 vertical lines (under operator

III and operator IV) and two horizontal lines (aside task T4 and dummy task T5). At step 5, he performed

the necessary operation with the uncovered element, since the number of lines was less than the order of

the matrix. After this, his matrix appeared as follows:

Tasks I II III IV V

T1 4 2 5 0 0

T2 6 3 3 0 3

T3 4 0 0 0 1

T4 0 0 5 3 0

T5 (Dummy) 0 0 3 3 0

Required

(i) What was the matrix after step II? Based on such matrix, ascertain (ii) and (iii) given below. (ii) What was the most difficult task for operators I, II and V? (iii) Who was the most efficient operators? (iv) If you are not told anything about the manager’s errors, which operator would be denied any task? Why? Solution: Answer to Point (i) Let us visualise what the manager did while solving this problem:

Tasks I II III IV V

T1 4 2 5 0 0

T2 6 3 3 0 3

T3 4 0 0 0 1

T4 0 0 5 3 0

T5 (Dummy) 0 0 3 3 0

At the end he improved the matrix with lowest uncovered cell element. Going in backward way: (i) The lowest uncovered cell is added to intersection point. (ii) Cell elements of Dummy row/column are always Zero (iii) Whenever any element is added to Zero the resulting figure will be the same element. (iv)Similarly, Cell 53 & Cell 54 are cell elements of Dummy row and after improvement they became 3. This proves that the minimum uncovered cell cost is 3





54

Now the previous matrix is as follows (This is obtained by Deducting 3 from intersecting points and adding 3 to uncovered cell costs and leaving remaining elements as it is)

Tasks I II III IV V

T1 7 5 5 0 3

T2 9 6 3 0 6

T3 7 3 0 0 4

T4 0 0 2 0 0

Answer to Point (ii) Based on the Matrix after Step II most difficult task for operator I, II and V are as follows- Operator I = T2 (9 hours) Operator II = T2 (6 hours) Operator V = T2 (6 hours) Answer to Point (iii) Based on the Matrix after Step II the most efficient operator is Operator IV. Answer to Point (iv) If the Manager’s error is not known, then assignment would be-

Tasks I II III IV V

T1 4 2 5 0 0

T2 6 3 3 0 3

T3 4 0 0 0 1

T4 0 0 5 3 0

T5 (Dummy) 0 0 3 3 0

We continue the assignment; T1 – V, T2 – IV, T3 – III are fixed. Between T4 and T5, I or II can be allotted. So, operator I or II can be denied the job.

55

Question - 26 Four operators O1, O2, O3 and O4 are available to a manager who has to get four jobs J1, J2, J3 and J4 done by assigning one job to each operator. Given the times needed by different operators for different jobs in the matrix below-

J1 J2 J3 J4

O1 12 10 10 8

O2 14 12 15 11

O3 6 10 16 4

O4 8 10 9 7

Required (i) How should the manager assign the jobs so that the total time needed for all four jobs is minimum? (ii) If job J2 is not to be assigned to operator O2 what should be the assignment and how much additional total time will be required? Solution: Answer to point (i) This is an assignment problem whose objective is to assign one job to one operator, so that total time needed for all four jobs is minimum. To determine appropriate assignment of jobs and operators, let us apply the assignment algorithm.


a) Row Reduction

4 2 2 0

3 1 4 0

2 6 12 0

1 3 2 0

b) Column Reduction



3 1 0 0

2 0 2 0

2 5 10 0

0 2 0 0

3 1 0 0

2 0 2 0

2 5 10 0

0 2 0 0

56

The minimum number of lines drawn to cover all zeros is equal to 4. Since the number of lines drawn viz., 4 is equal to the number of jobs or the number of operators, so we proceed for making the optimal assignment.


Answer to point (ii) If job J2 is not to be assigned to operator O2 then this objective can be achieved by replacing the time for cell (O2, J2) by a very large time estimate say M (Alternatively we may leave that cell blank and proceed further). Now apply the assignment algorithm to the following matrix so obtained-

12 10 10 8

14 M 15 11

6 10 16 4

8 10 9 7


4 2 2 0

3 M 4 0

2 6 12 0

1 3 2 0


Since the minimum number of lines drawn in the above matrix to cover all the zeroes is 3 which is less than the number of operators or jobs, therefore the above table will not yield the optimal assignment. For obtaining the optimal assignment we increase the number of zeros by subtracting the minimum uncovered element from all uncovered elements and adding it to elements lying at the intersection of two lines, we get the following matrix-

3 1 0 0

2 0 2 0

2 5 10 0

0 2 0 0

Operator Job Time

O1 J3 10

O2 J2 12

O3 J4 4

O4 J1 8

Total Time 34

3 0 0 0

2 M 2 0

1 4 10 0

0 1 0 0

3 0 0 0

2 M 2 0

1 4 10 0

0 1 0 0

57

3 0 0 0

1 M 1 0 0 3 9 0

0 1 0 0



Additional total time required will be 2 (36 – 34) units of time.

Operator Job Time

O1 J2 10

O2 J4 11

O3 J1 6

O4 J3 9

Time 36

3 0 0 0

1 M 1 0

0 3 9 0

0 1 0 0

assignment chapter - q & a compilation by niraj thapa

Education