- 1 -

Ans. 1:

I II III IV 1 16 52 34 22 2 26 56 8 52 3 76 38 36 30 4 38 52 48 20

Step 1: Subtract the smallest element of each row from every element of the corresponding row

I II III IV 1 0 36 18 6 2 18 48 0 44 3 46 8 6 0 4 18 32 28 0

Step 2: Subtract the smallest element of each column from every element in that column

I II III IV

1 0 28 18 6

2 18 40 0 44

3 46 0 6 0

4 18 24 28 0

Step 3: Drew minimum number of horizontal and vertical lines to cover all the zeros

I II III IV

1 0 28 18 6

2 18 40 0 44

3 46 0 6 0

4 18 24 28 0

The optimal assignment is

1 ─ I = 16

2 ─ III = 8

3 ─ II = 38

4 ─ IV = 20

82 hours

Minimum time taken = 82 hours

Ans. 2:

(a) Consider the following assignment problem: Division N E W S A 14 20 11 19

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B 12 10 15 9 Marketing Executives C 16 19 18 15 D 17 13 15 14

Step 1 Select the minimum element of first row and subtract it from all the elements of the row. On repeating the step with all the rows of the above matrix, we get the following

Division

N E W S A 3 9 0 8 B 3 1 6 0 Marketing Executives C 1 4 3 0 D 4 0 2 1

Step 2 Select the minimum element of first column and subtract it from all the elements of the column. On repeating this step with all the columns of the above matrix; we get the following

Division N E W S A 2 9 0 8 B 2 1 6 0 Marketing Executives C 0 4 3 0 D 3 0 2 1

Step 3 On drawing the minimum number of lines in the above matrix, so as to cover at the zeros, we get the following matrix. Division

N E W S A 2 9 0 8 B 2 1 6 0 Marketing Executives C 0 4 3 0 D 3 0 2 1

Since the minimum number of lines drawn under the step is equal to number of marketing executives or number of divisions, therefore we go over to the final step for determining the required optimal solution. Step 4 For determining the optimal solution scan each row in turn for a single uncovered zero in it, encircle it and pass a line in its column.

Division

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N E W S A 2 9 0 8 B 2 1 6 0 Marketing Executives C 0 4 3 0 D 3 0 2 1

The optimal assignment obtained in this case is as under: Marketing Division Cost Executive Rs. A W 11 B S 09 C N 16 D E 13 Total minimum cost 49

Ans. 5: Using the information that the factory works effectively 7 hours (=420 minutes) a day and the time required by each operator for producing each of the products, we obtain the following production and profit matrices:

Production Matrix (units) Profit Matrix (in Rs.)

Product Product Operator

A B C D

Operator

A B C D

P 70 42 30 35 P 210 84 120 35

Q 60 84 140 105 Q 180 168 560 105

R 70 60 42 42 R 210 120 168 42

S 21 42 28 28 S 63 84 112 28

In order to apply the assignment algorithm for minimizing losses, let us first convert this profit matrix to a loss matrix by subtracting all the elements of the given matrix from its highest element which is equal to Rs.560. The matrix so obtained is given below:

Product Operator

A B C D

P 350 476 440 525

Q 380 392 0 455

R 350 440 392 518

S 497 476 448 532

Now apply the assignment algorithm to the above loss matrix. Subtracting the minimum element of each row from all elements of that row, we get the following matrix:

Operator Product

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A B C D P 0 126 90 175 Q 380 392 0 455 R 0 90 42 168 S 49 28 0 84

Now subtract the minimum element of each column from the elements of that column to get the following matrix:

Product Operator

A B C D

P 0 98 90 91

Q 380 364 0 371

R 0 62 42 84

S 49 0 0 0

Draw the minimum number of lines to cover all zeros. The minimum number of lines to cover all zeros is three which is less than the order of the square matrix (i.e.4) thus the above matrix will not give the optimal solution. Subtract the minimum uncovered element (=62) from all uncovered elements and add it to the elements lying on the intersection of two lines, we get the following matrix:

Product Operator

A B C D

P 0 36 90 29

Q 380 302 0 309

R 0 0 42 22

S 111 0 62 0

The minimum number of lines which cover all zeros is 4 which is equal to the order of the matrix, hence, the above matrix will give the optimal solution. Specific assignments in this case are as below:

Product Operator

A B C D

P 0 36 90 29

Q 380 302 0 309

R 0 0 42 22

S 111 0 62 0

Operator Product Profit (Rs.)

P A 210

Q C 560

R B 120

S D 28

Total Profit (Rs.) 918

Ans. 8:

(i)

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4 12 16 8

20 28 32 24

36 44 48 40

52 60 64 56

Subtracting minimum element – each row.

0 8 12 4

0 8 12 4

0 8 12 4

0 8 12 4

Subtracting minimum element – each column,

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

Minimum no. of lines to cover all zeros = 4 = order of matrix. Hence optional assignment is possible.

Minimum cost = 4 + 28 + 48 + 56 = 136.

= AR1 + BR2 + CR3 + DR4

Since all are zeros, there are 24 solutions to this assignment problem.

Viz. A B C D R1 R2 R3 R4

R2 R3 R4 R1

R3 R4 R1 R2

R4 R1 R2 R3

R1 R3 R4 R2 etc.

A can be assigned in 4 ways, B in 3 ways for each of A’s 4 ways.

(ii) SP – VC = 100 Rs.

A B C D

R1 96 88 84 92

R2 80 72 68 76

R3 64 56 52 60

R4 48 40 36 44

Subtracting the highest term

0 8 12 4

16 24 28 20

32 40 44 36

48 56 60 52

Subtracting minimum term of each row.

0 8 12 4

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0 8 12 4

0 8 12 4

0 8 12 4

Which is the same as the earlier matrix

Maximum contribution = Rs. (96 + 72 + 52 + 44) = Rs. 264. Alternative Solution:

Maximisation of contribution is same as minimizing cost. Hence, same assignments as in (i) will be the optional solution.

Maximum Contribution Rs. (400 – 136) = Rs. 264

(iii) (a) The relative cost of assigning person i to region r does not change by addition or subtraction of a constant from either a row, or column or all elements of the matrix.

(b) Minimising cost is the same as maximizing contribution. Hence, the assignment solution will be the same, applying point (i) above.

(c) Many zero’s represent many feasible least cost assignment. Here, all zeros mean maximum permutation of a 4 4 matrix, viz. 4 3 2 1 = 24 solutions are possible.

Ans. 9: Reducing minimum from each column element (figure in ’000s)

Step 1 Step 2

R1 R2 R3 R4 R1 R2 R3 R4

C1 1 1 C1 0 0

C2 0 0 C2 0 0

C3 0 0 C3 0 0

C4 2 1 C4 1 0

Number of lines to connect all zeros nos. is 4 which is optional.

Alternatively you may also reduce the minimum from each row.

Step 1 Step 2

R1 R2 R3 R4 R1 R2 R3 R4

C1 0 1 C1 0 1

C2 0 0 C2 0 0

C3 1 0 C3 0 0

C4 0 1 C4 0 0

Number of lines to connect all zeros nos. is 4 which is optional.

All diagonal elements are zeros and are chosen. The minimum cost is Rs.15,000 C1 – R1 4,000; C2 – R2 4,000; C3 – R3 2,000; C4 – R4 5,000; (Total) = 15,000.

Ans.10:

Let us first formulate the preference ranking assignment problem.

MANAGERS

Room No. M1 M2 M3 M4 M5

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301 – 4 2 – 1 302 1 1 5 1 2 303 2 – 1 4 – 304 3 2 3 3 3 305 – 3 4 2 –

We have to find an assignment so that total preference ranking is minimum. In a cell (-) indicates that no assignment is to be made in that particular cell. Let us assign a very large ranking value M to all such cells.

Step 1 : From each row, subtract the minimum element of that row, from all the elements of that row to get the following matrix.

MANAGERS

Room No 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

Draw the minimum number of lines in the above table to cover all zeros. In this case the number of such lines is five, so the above matrix will give the optimal solution. The assignment is made as below:

MANAGERS

Rooms No. 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 l 305 M 1 2 0 M

Thus, the assignment is M1 → 302, M2 → 304, M3 → 303, M4 → 305, M5 → 301 and the total minimum ranking = 1 + 2 + 1 + 2 + 1 = 7

Ans. 11:

Dummy machine (M5) is inserted to make it a balanced cost matrix and assume i ts installation cost to be zero. Cost of install at cell M3 (J) and M2 (L) is very high marked as é.

J K L M N

M1

M2

M3

M4

M5 (Dummy)

18

24

é

28

0

22

18

22

16

0

30

é

28

24

0

20

20

22

14

0

22

18

14

16

0

Step 1

Subtract the minimum element of each row from each element of that row

J K L M N

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M1

M2

M3

0

6

é

4

0

8

12

é

14

2

2

8

4

0

0

M4 14 2 10 0 2

M5 (Dummy) 0 0 0 0 0

Step 2

Subtract the minimum element of each column from each element of that column

J K L M N

M1

M2

M3

M4

M5 (Dummy)

0

6

é

14

0

4

0

8

2

0

12

é

14

10

0

2

2

8

0

0

4

0

0

2

0

Step 3

Draw lines to connect the zeros as under:

J K L M N

M1 0 4 12 2 4 M2 6 0 é 2 0

M3 é 8 14 8 0 M4 14 2 10 0 2

M5 (Dummy) 0 0 0 0 0

There are five lines which are equal to the order of the matrix. Hence the solution is optimal. We may proceed to make the assignment as under:

J K L M N

M1

M2

M3

M4

M5 (Dummy)

0

6

e

14

0

4

0

8

2

0

12

e

14

10

0

2

2

8

0

0

4

0

0

2

0

The following is the assignment which keeps the total cost at minimum:

Machines Location Costs Rs.

M1 J 18

M2 K 18

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M3 N 14

M4 M 14

M5 (Dummy) L 0

Total 64

Ans. 12: Since the Executive Director of the 5 star hotel is interested in maximizing the revenue of the hotel, therefore, the objective of the given problem is to identify the preferences of marriage parties about halls so that hotel management could maximize its profit. To solve this problem first convert it to a minimization problem by subtracting all the elements of the given matrix from its highest element which is equal to Rs. 10,000. The matrix so obtained which is known as loss matrix is given below:

Loss matrix/Hall

Marriage party 1 2 3 4 A 0 1000 M M

B 2000 0 2000 5000 C 3000 0 4000 2000

D 0 2000 M M

Now apply the assignment algorithm to the above loss matrix. Subtracting the minimum element of each column from all elements of that column, we get the following matrix.

Loss matrix/Hall

Marriage party 1 2 3 4 A 0 1000 M M

B 2000 0 0 3000 C 3000 0 2000 0

D 0 2000 M M

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. Subtracting the minimum uncovered element (= 1000) from all uncovered elements and add it to the elements lying on the intersection of two lines, we get the following matrix

Marriage party 1 2 3 4

A 0 0 M M

B 3000 0 0 3000 C 4000 0 2000 0

D 0 1000 M M

Since the minimum number of lines to cover all zeros is 4 which is equal to the order of the matrix, the above matrix will give the optimal solution which is given below:

Marriage party 1 2 3 4 A 0 0 M M B 3000 0 0 3000 C 4000 0 2000 0 D 0 1000 M M

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and the optimal schedule is :

Revenue (Rs.) Marriage party A → Hall 2 9,000 B → Hall 3 8,000 C → Hall 4 8,000 D → Hall 1 10,000 Total 35,000

Ans. 14: The following matrix gives the cost incurred if the typist (i = A, B, C, D, E) executes the job (j = P, Q, R, S, T).

Job

Typist 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 elements in turn, the above matrix reduces to

Job

Typist 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 subtract the minimum element of each from all its elements in turn, and draw minimum number of lines horizontal or vertical so as to cover all zeros . All zeros can be covered by four lines as given below:

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 there are only 4 lines (<5) to cover all zeros, optimal assignments cannot be made. The minimum uncovered element is 2.

We subtract the value 2 from all uncovered elements. Add this value to al junction values and leave the other elements undisturbed. The revised matrix to obtained is given below:

2 2 2 4 0

4 2 0 8 0

0 1 2 1 2

0 2 0 2 0

2 0 2 0 2

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Since the minimum no. of lines required to cover al the zeros is only 4(<5), optimal assignment cannot be made at this stage also.

The minimum uncovered element is 1, repeating the usual process again, we get the following matrix.

2 1 2 8 0

4 1 0 7 0

0 0 2 0 2

0 1 0 1 0

3 0 3 0 3

Since the minimum number of lines to cover all zeros is equal to 5, is this matrix will give optimal solution? The optimal assignment is made in the matrix below:

Typist 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 0 0 3

Cost ( Rs.)

Thus typist A is given job 75 T :

Thus typist B is given job 66 R :

Thus typist C is given job 66 Q :

Thus typist D is given job 80 P :

Thus typist E is given job S 112

Total Rs.399

Note: In case the above solution is not unique. Alternate solution also exists.

Ans. 17:

(a) Sum of the proportion = (8 + 7 + 5 + 4) = 24

Assuming Rs. 1,000 as one unit, the effective matrix is as follows:

Effective Matrix Managers

East West North South Z (8/24) 240 = 80 (8/24) 192 = 64 (8/24) 144 = 48 (8/24) 20 = 40

N (7/24) 240 = 70 (7/24) 192 = 56 (7/24) 144 = 42 (7/24) 120 = 35

O (5/24) 240 = 50 (5/24) 192 = 40 (5/24) 144 = 30 (5/24) 120 = 25

P (4/24) 240 = 40 (4/24) 192 = 32 (4/24) 144 = 24 (4/24) 120 = 20

Convert the maximization problem to minimization problem

The resultant loss matrix is as follows:

Loss Matrix

Managers East West North South

M 0 16 32 40

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N 10 24 38 45

O 30 40 50 55 P 40 48 56 60

Row operation

Managers East West North South

M 0 16 32 40

N 0 14 28 35 O 0 10 20 25

P 0 8 16 20

Column operation

Managers East West North South

M 0 8 16 20 N 0 6 12 15

O 0 2 4 5

P 0 0 0 0

Managers

East

West

North

South

M 0 6 14 18

N 0 4 10 13

O 0 0 2 3 P 2 0 0 0

Managers

East

West

North

South

M 0 2 10 14

N 0 0 6 9 O 4 0 2 3 P 6 0 0 0

Managers

East

West

North

South

M 0 2 8 12 N 0 0 4 7 O 4 0 0 1

P 8 2 0 0

Assignment Sales

Rs.

M – East

N – West

O – North P – South

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1,86,000

Ans. 20

The initial matrix relating to nurse-patient combination is as under:

Nurse Patients W X Y

K 10 10 30 L 30 10 20 M 20 30 20

Deducting the lowest element of each row from the other elements of the same row, we get the following matrix:

0 0 20

20 0 10

0 10 0

We deduct the lowest element of each column from the other elements of the same column. Since there is zero in each column, the same matrix will be returned.

Draw lines to connect zeros as under:

0 0 20

20 0 10

0 10 0 There are three lines as required by the order of matrix of three.

Hence the solution is optimal.

Allocation of patients to nurses as under to minimize the cost

0 0

20 0

0 10 K W 10 400 400

L X 10 400 400

M Y 20 800 800

Total minimum cost 1600

(iii) With the introduction of a new patient and a new nurse, the original matrix of nurse-patient combinations will stand revised as under:

Nurse Patients

W X Y Z

K 10 10 30 40

L 30 10 20 40

M 20 30 20 40

N 50 50 50 50

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9

Deducting the lowest element of each row from the other element of the same row, we get the following matrix:

0 0 20 30

20 0 10 30

0 10 0 20

0 0 0 0

Deduct the lowest element of each column from the other elements of the same column. Since there is zero in each column, the same matrix will be returned.

Draw lines to connect zeros as under:

0 0 20 30

20 0 10 30

0 10 0 20

0 0 0 0

There are four lines as required by the order of matrix of four

Hence the solution is optimal.

Proceed to allocate the patients to nurses as under to minimize the cost.

0 0 20

20 0 10

0 10 0

0 0 0

K W 10 400 400

L X 10 400 400 M Y 20 800 800

N Z 50 2000 2000 Total minimum cost 3600

(iv) The cost of new nurse per hour is Rs. 50 in respect of any patient and the cost of the existing nurses for attending to the new patient is Rs. 40 per hour. Both these rates are greater than the values of other elements of existing nurse-

patient combination matrix. Thus the new nurse row and new patient column will have a higher value than the element of the existing matrix. Hence the new nurse can be allocated to the new patient without having to redo the assignment exercise. Hence we need not to a fresh assignment. N will be assigned to patient Z at 50/ hr is Rs. 2000/ week.

This will be the extra minimum cost to the hospital i.e. 2000 + 1600 = 3600.