C 12 The Assignment Problem - ICAI Knowledge … 12 The Assignment Problem Basic Concepts Assignment...

CHAPTER 12

The Assignment Problem

Basic Concepts Assignment Algorithm “The Assignment Problem is another special case of LPP. It occurs when m jobs are to be assigned to n facilities on a one-to-one basis with a view to optimising the resource required.” Steps for Solving the Assignment Problem “Assignment problem can be solved by applying the following steps: Step-1: Subtract the minimum element of each row from all the elements in that row. From each column of the matrix so obtained, subtract its minimum element. The resulting matrix is the starting matrix for the following procedure. Step-2: Draw the minimum number of horizontal and vertical lines that cover all the zeros. If this number of lines is n, order of the matrix, optimal assignment can be made by skipping steps 3 and 4 and proceeding with step 5. If, however, this number is less than n, go to the next step. Step-3: Here, we try to increase the number of zeros in the matrix. We select the smallest element out of these which do not lie on any line. Subtract this element from all such (uncovered) elements and add it to the elements which are placed at the intersections of the horizontal and vertical lines. Do not alter the elements through which only one line passes. Step-4: Repeat steps 1, 2 and 3 until we get the minimum number of lines equal to n. Step-5(A): Starting with first row, examine all rows of matrix in step 2 or 4 in turn until a row containing exactly one zero is found. Surround this zero by ( ), indication of an assignment there. Draw a vertical line through the column containing this zero. This eliminates any confusion of making any further assignments in that column. Process all the rows in this way. Step-5(B): Apply the same treatment to columns also. Starting with the first column, examine all columns until a column containing exactly one zero is found. Mark ( ) around this zero and draw a horizontal line through the row containing this marked zero. Repeat steps 5A and B, until one of the following situations arises: - No unmarked ( ) or uncovered (by a line) zero is left, - There may be more than one unmarked zero in one column or row. In this case, put

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12.2 Advanced Management Accounting

around one of the unmarked zero arbitrarily and pass 2 lines in the cells of the remaining zeros in its row and column. Repeat the process until no unmarked zero is left in the matrix.”

Unbalanced Assignment Problems “Like the unbalanced transportation problems there could arise unbalanced assignment problems too. They are to be handled exactly in the same manner i.e., by introducing dummy jobs or dummy men, etc.”



Question-1 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?

Solution: 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. Question-2 Prescribe the steps to be followed to solve an assignment problem.

Solution: The assignment problem can be solved by applying the following steps- Step-1: Subtract the minimum element after row operation of each row from all the elements in that row. From each column of the matrix so obtained, subtract its minimum element. The resulting matrix is the starting matrix for the following procedure. Step-2: Draw the minimum number of horizontal and vertical lines that cover all the zeros. If this number of lines is n, order of the matrix, optimal assignment can be made by skipping steps 3 and 4 and proceeding with step 5. If, however, this number is less than n, go to the next step Step-3: Here, we try to increase the number of zeros in the matrix. We select the smallest element out of these which do not lie on any line. Subtract this element from all such (uncovered) elements and add it to the elements which are placed at the intersections of the horizontal and vertical lines. Do not alter the elements through which only one line passes. Step-4: Repeat steps 1, 2 and 3 until we get the minimum number of lines equal to n.


The Assignment Problem 12.5

Step-5(A): Starting with first row, examine all rows of matrix in step 2 or 4 in turn until a row containing exactly one zero is found. Surround this zero by ( ), indication of an assignment there. Draw a vertical line through the column containing this zero. This eliminates any confusion of making any further assignments in that column. Process all the rows in this way. Step5 (B): Apply the same treatment to columns also. Starting with the first column, examine all columns until a column containing exactly one zero is found. Mark ( ) around this zero and draw a horizontal line through the row containing this marked zero. Repeat steps 5A and B, until one of the following situations arises: - No unmarked ( ) or uncovered (by a line) zero is left, - There may be more than one unmarked zero in one column or row. In this case, put

around one of the unmarked zero arbitrarily and pass 2 lines in the cells of the remaining zeros in its row and column. Repeat the process until no unmarked zero is left in the matrix.

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?

Solution: 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. Question-4 Answer the following independent situations relating to an assignment problem with a minimization objective: (i) Just after row and column minimum operations, we find that a particular row has 2

zeroes. Does this imply that the 2 corresponding numbers in the original matrix before any operation were equal? Why?



(ii) Under the usual notation, where a32 means the element at the intersection of the 3rd row and 2nd column, we have, in a assignment problem, and ~2 figuring in the optimal solution. What can you conclude about the remaining assignments? Why?

Solution: (i) 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.

(ii) 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.



Assignment – Minimization Question-1

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. The minimum time elements in row 1, 2 and 3 are 70, 50 and 110 respectively. Subtract these elements from all elements in their respective row. The reduced matrix is shown below-

A B C

1 30 15 0

2 0 20 60

3 0 10 20

The minimum time elements in columns A, B and C are 0, 10, and 0 respectively. Subtract these elements from all the elements in their respective columns to get the reduced time matrix as shown below-



A B C

1 30 5 0

2 0 10 60

3 0 0 20

The minimum number of horizontal and vertical lines to cover all zeros is 3, which is equal to the order of the matrix. The Pattern of assignments among software professionals and programs with their respective time (in minutes) is given below-

Program Software Professionals Time (in Minutes)

1 C 70

2 A 50

3 B 120

Total 240 Question-2 A Production supervisor is considering, how he should assign five jobs that are to be performed, to five mechanists working under him. He wants to assign the jobs to the mechanists in such a manner that the aggregate cost to perform the jobs is the least. He has following information about the wages paid to the mechanists for performing these jobs-

Mechanist Job1 Job 2 Job 3 Job 4 Job 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

Assign the jobs to the mechanists so that the aggregate cost is the least.



Solution: The given problem is a balanced minimization problem. Subtracting minimum element of each row from all the elements of that row, the given problem reduces to-


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

Subtract the minimum element of each column from all the elements of that column. Draw the minimum number of lines horizontal or vertical so as to cover all zeros.


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 minimum number of lines covering all zeros is equal to 4 which is less than the number of columns / rows (=5), the above table will not provide optimal solution. Subtract the minimum uncovered element (=2) from all uncovered elements and add to the elements lying on the intersection of two lines, we get the following matrix-


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



Since the minimum number of horizontal and vertical lines to cover all zeros is equal to five which is equal to the order of the matrix, the above table will give the optimal solution. The optimal assignment is made below-


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

The optimal assignment is given below-

Mechanist Job Wages

A 2 3

B 4 2

C 5 4

D 1 3

E 3 9

Total 21

The total least cost associated with the optimal mechanist-job assignment equals to 21. Question-3

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 Job B Job C Job 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. Subtracting the minimum element of each row from all its elements in turn, the given problem reduces to-


1 40 20 0 10

2 10 20 40 0

3 10 40 20 0

4 10 10 0 10

Now subtract the minimum element of each column from all its elements in turn. Draw the minimum number of lines horizontal or vertical so as to cover all zeros.


1 30 10 0 10

2 0 10 40 0

3 0 30 20 0

4 0 0 0 10

Since the minimum number of lines to cover all zeros is equal to 4(order of the matrix), this matrix will give optimal solution. The optimal assignment is made in the matrix below-


1 30 10 0 10

2 0 10 40 0

3 0 30 20 0

4 0 0 0 10



The optimal assignment is-

Contractor Job Cost (‘000 `)

1 C 80

2 A 80

3 D 100

4 B 90

Hence, total minimum cost of the project will be `3,50,000. Question-4 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. How should the task be allocated one to one man so that the total man-hours are minimised?

I II III IV

1 16 52 34 22

2 26 56 8 52

3 76 38 36 30

4 38 52 48 20

Solution:

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

Since, No. of lines are equal to order of matrix, hence, solution is optimal.

1 I 16 hrs.

2 III 8 hrs.

3 II 38 hrs.

4 IV 20 hrs.

Total 82 hrs. Minimum time taken is 82 hrs. Question-5

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-



Bids for Routes (` )

Company / 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: Step 1: Reducing minimum from each column element (figure in ’000s)-

R1 R2 R3 R4

C1 1 1 − −

C2 − 0 − 0

C3 0 − 0 −

C4 − − 2 1

Step 2: Reducing minimum from each row element (figure in ’000s)-

R1 R2 R3 R4

C1 0 0 − −

C2 − 0 − 0

C3 0 − 0 −

C4 − − 1 0

Number of lines to connect all zeros nos. is 4 which is optimal. All diagonal elements are zeros and are chosen.



Company Route (`)

C1 R1 4,000

C2 R2 4,000

C3 R3 2,000

C4 R4 5,000

Total 15,000

The minimum cost is ` 15,000 Question-6

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



J K L M N

M1 18 22 30 20 22

M2 24 18 M 20 18

M3 M 22 28 22 14

M4 28 16 24 14 16

M5 (Dummy) 0 0 0 0 0

Step 1:

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

J K L M N

M1 0 4 12 2 4

M2 6 0 M 2 0

M3 M 8 14 8 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 0 4 12 2 4

M2 6 0 M 2 0

M3 M 8 14 8 0

M4 14 2 10 0 2

M5 (Dummy) 0 0 0 0 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 M 2 0

M3 M 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 0 4 12 2 4

M2 6 0 M 2 0

M3 M 8 14 8 0

M4 14 2 10 0 2

M5 (Dummy) 0 0 0 0 0

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

Machines Location Costs (`)

M1 J 18

M2 K 18

M3 N 14

M4 M 14

M5 (Dummy) L 0

Total 64 Question-7

A private firm employs typists on hourly piece rate basis for their daily work. Five typists are working in that firm and their charges and speeds are different. On the basis of some earlier understanding, only one job is given to one typist and the typist is paid for full hours even when he or she works for a fraction of an hour. Find the least cost allocation for the following data:



Typist Rate per hour (`) No. of pages typed per hour

A 5 12

B 6 14

C 3 8

D 4 10

E 4 11

Job No. of pages

P 199

Q 175

R 145

S 298

T 178

Solution: 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).

Typist Job P Job Q Job R Job S Job 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-




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 column 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-


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

Since there are only 4 lines (<5) to cover all zeros, optimal assignment cannot be made. The minimum uncovered element is 1. We subtract the value 1 from all uncovered elements, add this value to all intersections of two lines values and leave the other elements undisturbed. The revised matrix so obtained is given below-


A 3 2 1 4 0

B 6 3 0 9 1

C 0 0 0 0 1

D 2 3 0 3 1

E 3 0 1 0 2

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



The minimum uncovered element is 2. Repeating the usual process again, we get the following matrix-


A 1 0 1 2 0

B 4 1 0 7 1

C 0 0 2 0 3

D 0 1 0 1 1

E 3 0 3 0 4

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


A 1 0 1 2 0

B 4 1 0 7 1

C 0 0 2 0 3

D 0 1 0 1 1

E 3 0 3 0 4

Typist Job Cost (`)

A T 75

B R 66

C Q 66

D P 80

E S 112

Total 399

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



Question-8

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-

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

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.

Solution: This is an assignment problem whose objective is to assign on manufacturing set up to one brand so that the total cost of production is minimum. To determine the appropriate assignment, let us apply the assignment algorithm. Subtract the minimum element of each row from all elements of that row to get the following matrix-

Brands Manufacturing Setups

S1 S2 S3 S4 S5

B1 0 2 3 1 7

B2 4 0 3 6 2

B3 4 1 0 2 5

B4 2 5 0 4 3

B5 2 0 4 3 6



Now subtract the minimum elements of each column from all elements of that column-


S1 S2 S3 S4 S5

B1 0 2 3 0 5

B2 4 0 3 5 0

B3 4 1 0 1 3

B4 2 5 0 3 1

B5 2 0 4 2 4

The minimum number of lines drawn to cover all zeros is equal to 4 which is one less than the order of the matrix (= 5), the above table will not yield the optimal assignment. For obtaining the optimal assignment, we increase the number of zeroes 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-


S1 S2 S3 S4 S5

B1 0 3 4 0 5

B2 4 1 4 5 0

B3 3 1 0 0 2

B4 1 5 0 2 0

B5 1 0 4 1 3

Since the minimum number of lines required to cover all zeros is five, the above table will give the optimal solution. The required assignment is made as below-

Brand Setup Cost B1 S1 4 B2 S5 5 B3 S4 6 B4 S3 7 B5 S2 5 Total 27



Question-9

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 storke 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. Which swimmers should be assigned to which swimming style? Who will be in the relay?

Solution: Let us first create the assignment matrix with time expressed in seconds. Also it is an unbalanced assignment problem hence a dummy style is added to balance it.

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 1:

As there is a zero in each row, we go straight to the column reduction-


Anand 0 0 - - 0

Bhaskar - 1 0 10 0

Chandru 1 5 4 2 0

Dorai - - - 1 0

Easwar 11 1 5 0 0



Step 2:

The minimum number of lines to cover all zeros is 4, which is less than 5, the order of the square matrix. Hence, the above matrix will not provide the optimal solution. Thus we try to increase the number of zeros. Select the minimum uncovered element by these lines (which is 1). Subtract it from all the uncovered elements and add it to the elements lying on the intersection of lines, as drawn above. The revised matrix will be-


Anand 0 0 - - 1

Bhaskar - 1 0 11 1

Chandru 0 4 3 2 0

Dorai - - - 1 0

Easwar 10 0 4 0 0

As the minimum number of lines to cover all zeros is 5, which is equal to the order of the square matrix, the above matrix will provide the optimal solution. The assignment is made below-

Swimmer Swimming Style Time (Seconds)

Anand Breast Stroke 75

Bhaskar Free Style 61

Chandru Back Stroke 70

Dorai Dummy (not participate) -

Easwar Butterfly 70

Total Minimum Time in the relay 276

Dorai will be out of the relay. Question-10

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

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?

Solution: The given problem is a balance minimization assignment problem. Let us apply the assignment algorithm to find the optimal assignment. Subtracting the smallest element of each row from all the elements of that row, we get the following table-

Town Depots

A B C D E V 0 2 7 12 5 W 0 3 11 14 4 X 0 4 12 12 4 Y 0 0 3 5 1 Z 0 0 5 15 0

Subtracting the smallest element of each column from all the elements of that column, wet get the following-

Town Depots

A B C D E

V 0 2 4 7 5

W 0 3 8 9 4

X 0 4 9 7 4

Y 0 0 0 0 1

Z 0 0 2 10 0



Draw the minimum number of lines to cover all zeros. 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. 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 matrix-

Town Depots

A B C D E

V 0 0 2 5 3

W 0 1 6 7 2

X 0 2 7 5 2

Y 2 0 0 0 1

Z 2 0 2 10 0

Against, the minimum number of lines of cover all zeros is 4, which is less than the order of the matrix. Subtract the uncovered element (=2) from all the uncovered element and add it to the elements lying on the intersection of two lines, we get-

Town Depots

A B C D E

V 0 0 0 3 1

W 0 1 4 5 0

X 0 2 5 3 0

Y 4 2 0 0 1

Z 4 2 2 10 0

Since the minimum number of lines to cover all zeros is 4 which is less than the order of the matrix, hence, the above matrix will not give the optimal solution. Subtracting the uncovered element (=1) from all the uncovered elements and adding it to the elements lying on the intersection of two lines, we get-



Town Depots

A B C D E

V 1 0 0 3 2

W 0 0 3 4 0

X 0 1 4 2 0

Y 5 2 0 0 2

Z 4 1 1 9 0

Since the minimum number of lines of cover all zeros is 5 which is equal to the order of the matrix, the above table will give the optimal assignment. The optimal assignment is made below- This optimal assignment is-

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

Hence the minimum total travelled distance equals to 45 kms. Assignment - Maximization 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 (`)

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: Sum of the proportion is 24 (8 + 7 + 5 + 4). Assuming `1,000 as one unit, the effective matrix is as follows- Effective Matrix

Manager Zones

East West North South

M 80 [(8/24) × 240]

64 [(8/24) × 192]

48 [(8/24) × 144]

40 [(8/24) × 120]

N 70 [(7/24) × 240]

56 [(7/24) × 192]

42 [(7/24) × 144]

35 [(7/24) × 120]

O 50 [(5/24) × 240]

40 [(5/24) × 192]

30 [(5/24) × 144]

25 [(5/24) × 120]

P 40 [(4/24) × 240]

32 [(4/24) × 192]

24 [(4/24) × 144]

20 [(4/24) × 120]

Convert the maximization problem to minimization problem. The resultant loss matrix is as follows-



Loss 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

Row Operation


M 0 16 32 40

N 0 14 28 35

O 0 10 20 25

P 0 8 16 20

Column Operation


M 0 8 16 20

N 0 6 12 15

O 0 2 4 5

P 0 0 0 0

No. of lines are 2 which is less than the order of matrix, hence, this is not an optimal solution. Lowest uncovered element 2 shall be deducted from all uncovered cells value and added to the value at intersections.


M 0 6 14 18

N 0 4 10 13

O 0 0 2 3

P 2 0 0 0



Again no. of lines covering zeros are not equal to the order of matrix, therefore, lowest uncovered element 2 shall be deducted from all uncovered cells value and added with value at intersections.


M 0 6 12 16

N 0 4 8 11

O 0 0 0 1

P 4 2 0 0

Once again no. of lines covering zeros are not equal to the order of matrix, therefore, lowest uncovered element 4 shall be deducted from all uncovered cells value and added with value at intersections.


M 0 2 8 12

N 0 0 4 7

O 4 0 0 1

P 8 2 0 0

Now the number of lines covering zeros are equal to the order of matrix, hence, this is the optimal solution.

Assignment Sales (`)

M - East 80,000

N - West 56,000

O - North 30,000

P - South 20,000

Total 1,86,000 Question-12 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 ` 25. Find the maximum profit per week.

Solution: The given assignment problem is a maximization problem. Let us 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.


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

The assignment procedure is now applies to above problem. Subtract the minimum element of each row from all the elements of that row, and repeat this step with all the rows of the table.

Operator Weekly output in lathe

L1 L2 L3 L4 L5

P 16 14 9 4 0 Q 21 17 11 6 0 R 16 11 4 0 5 S 21 18 11 5 0 T 17 13 10 5 0

Repeat the above step with the columns of the table also. Subtract the minimum element of each column from all the elements of that column.




L1 L2 L3 L4 L5

P 0 3 5 4 0

Q 5 6 7 6 0

R 0 0 0 0 5

S 5 7 7 5 0

T 1 2 6 5 0

Since the minimum number of lines to cover all zeros is 3 which is less than 5 (order of the square matrix), the above matrix will not give the optimal solution. Hence we try to increase the number of zeros. Subtract the least uncovered element which is 2 from all uncovered elements and add it to all the elements lying on the intersection of two lines. We get the following matrix-


L1 L2 L3 L4 L5

P 0 1 3 2 0

Q 5 4 5 4 0

R 2 0 0 0 7

S 5 5 5 3 0

T 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-


L1 L2 L3 L4 L5

P 0 2 3 2 1

Q 4 4 4 3 0

R 2 1 0 0 8

S 4 5 4 2 0

T 0 0 3 2 0



The minimum number of lines to cover all zeros is 4 which is less than 5. Repeating the above step once again, we thus get-


L1 L2 L3 L4 L5

P 0 0 1 0 1

Q 4 2 2 1 0

R 4 1 0 0 10

S 4 3 2 0 0

T 2 0 3 2 2

The minimum number of lines to cover all zeros is 5. Hence the above matrix will give the optimal solution. The optimal assignment is made as below-

Operator Lathe Machine Weekly Output

P L1 20

Q L5 40

R L3 35

S L4 37

T L2 28

Total 160

The maximum profit per week is `4,000 (` 25 × 160). Question-13 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 ` 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-



Revenue/Hall 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. Decide on an allocation that will maximize the revenue to your hotel.

Solution:

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. The matrix so obtained which is known as loss matrix is given below-

Loss Matrix/Hall Marriage Party 1 2 3 4

A 0 2,500 X X

B 5,000 0 5,000 12,500

C 7,500 0 10,000 5,000

D 0 5,000 X X

Now we can apply the assignment algorithm to find optimal solution. Subtracting the minimum element of each column from all elements of that column-


A 0 2,500 X X

B 5,000 0 0 7,500

C 7,500 0 5,000 0

D 0 5,000 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. Subtracting the minimum uncovered element (2,500) from all uncovered elements and add it to the elements lying on the intersection of two lines, we get the following matrix-


A 0 0 X X

B 7,500 0 0 7,500

C 10,000 0 5,000 0

D 0 2,500 X X

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-


A 0 0 X X

B 7,500 0 0 7,500

C 10,000 0 5,000 0

D 0 2,500 X X

Optimal Schedule is- Marriage Party Hall Revenue (`)

A 2 22,500 B 3 20,000 C 4 20,000 D 1 25,000

Total 87,500

Assignment - Miscellaneous Question-14

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

(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:

(i)

Subtracting minimum element of each row-

A B C D

R1 0 8 12 4

R2 0 8 12 4

R3 0 8 12 4

R4 0 8 12 4

Subtracting minimum element of each column-

A B C D

R1 0 0 0 0

R2 0 0 0 0

R3 0 0 0 0

R4 0 0 0 0

Minimum no. of lines to cover all zeros are 4 (equal to order of matrix). Hence optional assignment is possible.



Minimum Cost = AR1 + BR2 + CR3 + DR4 = 4 + 28 + 48 + 56 = 136 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

A can be assigned in 4 ways, B in 3 ways for each of A’s 4 ways. (ii) Given - Sales Price `200, Variable Cost Excluding Selling Cost `100 The contribution matrix is given below-

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 all the cells value with the highest cell value i.e. 96 to make it loss matrix-

A B C D

R1 0 8 12 4

R2 16 24 28 20

R3 32 40 44 36

R4 48 56 60 52



Subtracting minimum term of each row-

A B C D

R1 0 8 12 4

R2 0 8 12 4

R3 0 8 12 4

R4 0 8 12 4

This is the same as the earlier matrix. Maximum Contribution = AR1 + BR2 + CR3 + DR4 = `96 + `72 + `52 + `44 = `264 (iii) (a) The relative cost of assigning person i to region r do 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. 24 solutions (4 × 3 × 2 × 1) are possible. Question-15

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 preformed 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:



Operators

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

(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?

(v) Can the manager go ahead with his assignment to correctly arrive at the optimal assignment, or should he start afresh after introducing the dummy task at the beginning?

Solution:

Matrix after Step V is as follows-

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 Junction values at T5 (Dummy) is 3, it implies 3 was the minimum uncovered element. Now we do the reverse steps.



Previous step was i.e. Step IV:

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

T5 (Dummy) 0 0 0 0 0

Step III:

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

T5 (Dummy) 0 0 0 0 0

(i)

Matrix after Step II -

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

(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)



(iii) Based of the Matrix after Step II the most efficient operator is Operator IV. (iv) If the Manager’s error is not known, then assignment would be-

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, other I or II can be denied the job.

(v)

Yes, the Manager can go ahead with the optimal assignment. Row minimum is not affected by when the dummy was introduced. Column minimum was affected. But in the process, more zeros were generated to provide better solution.

Question-16

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

(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: 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. Subtract the minimum element of each row from all elements of that row to get the following matrix-

J1 J2 J3 J4

O1 4 2 2 0

O2 3 1 4 0

O3 2 6 12 0

O4 1 3 2 0

Now subtract the minimum element of each column from all elements of that column-

J1 J2 J3 J4

O1 3 1 0 0

O2 2 0 2 0

O3 2 5 10 0

O4 0 2 0 0

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.

Thus the optimal assignment in this part of the question is-

Operator Job

O1 J3

O2 J2

O3 J4

O4 J1

The total time taken by four operators to perform the jobs is 34 (10+ 12 + 4 + 8).



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. Now apply the assignment algorithm to the following matrix so obtained-

J1 J2 J3 J4

O1 12 10 10 8 O2 14 M 15 11 O3 6 10 16 4 O4 8 10 9 7

Perform row and column operations to the above matrix as mentioned in part (i) of the problem. We thus have the following matrix-

J1 J2 J3 J4

O1 4 2 2 0 O2 3 M 4 0 O3 2 6 12 0 O4 1 3 2 0

J1 J2 J3 J4

O1 3 0 0 0

O2 2 M 2 0

O3 1 4 10 0

O4 0 1 0 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-

J1 J2 J3 J4

O1 3 0 0 0

O2 1 M 1 0

O3 0 3 9 0

O4 0 1 0 0



Since the minimum number of lines required to cover all zeros is four, so the above matrix will give the optimal solution. The required assignment is made as below-

Operator Job

O1 J2

O2 J4

O3 J1

O4 J3

The minimum time required is 36 (10 + 11 + 6 + 9). Additional total time required will be 2 (36 – 34) units of time.

Question-17

The Captain of a cricket team has to allot five middle batting positions to five batsmen. The average runs scored by each batsman at these positions are as follows:

Batsman Batting Position

I II III IV V

P 40 40 35 25 50

Q 42 30 16 25 27

R 50 48 40 60 50

S 20 19 20 18 25

T 58 60 59 55 53

(i) Find the assignment of batsmen to positions, which would give the maximum number of runs.

(ii) If another batsman ‘U’ with the following average runs in batting positions as given below :

Batting position: I II III IV V

Average runs: 45 52 38 50 49

Is added to the team, should he be included to play in the team? If so, who will be replaced by him?



Solution: (i) The given problem is a maximization assignment problem. Let us first convert it into a minimisation assignment problem by subtracting all the elements of the given matrix from the largest element of the matrix i.e. 60. The matrix then reduces to-


I II III IV V P 20 20 25 35 10 Q 18 30 44 35 33 R 10 12 20 0 10 S 40 41 40 42 35 T 2 0 1 5 7

We now carry out row operation of all rows by subtracting from all elements of a row, the smallest element of that row to get the following matrix-


I II III IV V P 10 10 15 25 0 Q 0 12 26 17 15 R 10 12 20 0 10 S 5 6 5 7 0 T 2 0 1 5 7

Now perform column operation for all columns by subtracting the minimum element of each column from all elements of that column-


I II III IV V

P 10 10 14 25 0

Q 0 12 25 17 15

R 10 12 19 0 10

S 5 6 4 7 0

T 2 0 0 5 7



The minimum number of horizontal and vertical lines to cover all zeros is 4 which is less than the order of the matrix (= 5), the above table will not give the optimal solution. We subtract the minimum uncovered element ‘4’ from all uncovered elements and add this element to the elements lying on the intersection of two lines. The new matrix is given below-


I II III IV V

P 10 6 10 25 0

Q 0 8 21 17 15

R 10 8 15 0 10

S 5 2 0 7 0

T 6 0 0 9 11

Now the minimum number of lines to cover all zeros is 5 which is equal to the order of the matrix. Thus, the above matrix will give the optimal solution. The assignments are made as below-

Batsman Batting Position Runs

P V 50

Q I 42

R IV 60

S III 20

T II 60

Total 232

(ii)

We now include another batsman U in the minimization matrix by subtracting average runs of batsman ‘U’ from 60. We also introduce a dummy batting position to balance the given assignment problem. We thus, thus, get the following matrix-


I II III IV V Dummy

P 20 20 25 35 10 0

Q 18 30 44 35 33 0



R 10 12 20 0 10 0

S 40 41 40 42 35 0

T 2 0 1 5 7 0

U 15 8 22 10 11 0

Now perform the column operation for all columns by subtracting the smallest element of each column from all elements of that column to get the following matrix-


I II III IV V Dummy

P 18 20 24 35 3 0

Q 16 30 43 35 26 0

R 8 12 19 0 3 0

S 38 41 39 42 28 0

T 0 0 0 5 0 0

U 13 8 21 10 4 0

The minimum number of lines to cover all zeros is only 3 and order of matrix is 6. We subtract the minimum uncovered element ‘3’ from all uncovered elements and add this element to the elements lying on the intersection of two lines. The new matrix is given below-


I II III IV V Dummy

P 15 17 21 32 0 0

Q 13 27 40 32 23 0

R 8 12 19 0 3 3

S 35 38 36 39 25 0

T 0 0 0 5 0 3

U 10 5 18 7 1 0




I II III IV V Dummy

P 10 12 16 27 0 0

Q 8 22 35 27 23 0

R 8 12 19 0 8 8

S 30 33 31 34 25 0

T 0 0 0 5 5 8

U 5 0 13 2 1 0


I II III IV V Dummy

P 2 4 8 27 0 0

Q 0 14 27 27 23 0

R 0 4 11 0 8 8

S 22 25 23 34 25 0

T 0 0 0 13 13 16

U 5 0 13 10 9 8

Now the minimum number of lines to cover all zeros is 6 which is equal to the order of matrix. Hence the assignments are made as below-

Batsman Batting Position Runs

P V 50 Q I 42 R IV 60 S Dummy - T III 59 U II 52 Total 263

Hence, batsman U will be included in the team at position II and he will replace batsman S. Total runs will be 263.


C 12 The Assignment Problem - ICAI Knowledge … 12 The Assignment Problem Basic Concepts Assignment...

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