21531sm Finalnew Vol2 Cp12 CHAPTER 12

8/11/2019 21531sm Finalnew Vol2 Cp12 CHAPTER 12

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CHAPTER12

The Assignment Problem

Basic Concepts

As si gn ment Al go ri thm

The Assignment Problem is another special case of LPP. It occurs when m jobs are to beassigned 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. Fromeach column of the matrix so obtained, subtract its minimum element. The resulting matrixis 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 byskipping 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 smallestelement 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 ofthe horizontal and vertical lines. Do not alter the elements through which only one linepasses.

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 arow containing exactly one zero is found. Surround this zero by ( ), indication of anassignment there. Draw a vertical line through the column containing this zero. Thiseliminates any confusion of making any further assignments in that column. Process allthe 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

The Institute of Chartered Accountants of India


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12.2 Advanced Management Account ing

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

Unbalanced Assignment Problems

Like the unbalanced transportation problems there could arise unbalanced assignmentproblems too. They are to be handled exactly in the same manner i.e., by introducingdummy jobs or dummy men, etc.



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Question-1

In an assignment problem to assign jobs to men to minimize the time taken, suppose that oneman does not know how to do a particular job, how will you eliminate this allocation from thesolution?

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 thisallocation 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 thatrow. From each column of the matrix so obtained, subtract its minimum element. The resultingmatrix 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 thisnumber of lines is n, order of the matrix, optimal assignment can be made by skipping steps 3and 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 outof these which do not lie on any line. Subtract this element from all such (uncovered) elementsand add it to the elements which are placed at the intersections of the horizontal and verticallines. 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.



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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 containingexactly one zero is found. Surround this zero by ( ), indication of an assignment there. Drawa vertical line through the column containing this zero. This eliminates any confusion ofmaking 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, examineall columns until a column containing exactly one zero is found. Mark ( ) around this zero anddraw 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 remainingzeros in its row and column. Repeat the process until no unmarked zero is left in thematrix.

Question-3

In an assignment problem to assign jobs to men to minimize the time taken, suppose that oneman does not know how to do a particular job, how will you eliminate this allocation from thesolution?

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 thisallocation will get eliminated.

Question-4

Answer the following independent situations relating to an assignment problem with aminimization objective:

(i) Just after row and column minimum operations, we find that a particular row has 2zeroes. Does this imply that the 2 corresponding numbers in the original matrix beforeany operation were equal? Why?



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(ii) Under the usual notation, where a 32 means the element at the intersection of the 3rdrow and 2 nd column, we have, in a assignment problem, and ~2 figuring in the optimalsolution. What can you conclude about the remaining assignments? Why?

Solution:

(i) Under the Hungarian Assignment Method, the prerequisite to assign any job is that eachrow and column must have a zero value in its corresponding cells. If any row or columndoes not have any zero value then to obtain zero value, each cell values in the row orcolumn is subtracted by the corresponding minimum cell value of respective rows orcolumns by performing row or column operation. This means if any row or column havetwo or more cells having same minimum value then these row or column will have morethan one zero. However, having two zeros does not necessarily imply two equal values inthe original assignment matrix just before row and column operations. Two zeroes in asame row can also be possible by two different operations i.e. one zero from rowoperation 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 thecorresponding row and column become unavailable for further allocation. Hence, thesecorresponding row and column are crossed mark to show unavailability. In the givenassignment matrix two allocations have been made in a 24 (2nd row and 4 th column) and a 32 (3rd row and 2 nd column). This implies that 2 nd and 3 rd row and 4 th and 2 nd column are

unavailable for further allocation.Therefore, the other allocations are at either at a11 and a43 or at a13 and a41.



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Assignment MinimizationQuestion-1

An Electronic Data Processing (EDP) centre has three expert Software professionals. TheCentre wants three application software programs to be developed. The head of EDP Centreestimates the computer time in minutes required by the experts for development of ApplicationSoftware Programs as follows-

Software Programs

Computer Time (in minutes)

Required by Software Professionals

A B C1 100 85 70

2 50 70 110

3 110 120 130

Assign the software professionals to the application software programs to ensure minimumusage 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 theseelements 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. Subtractthese elements from all the elements in their respective columns to get the reduced timematrix as shown below-



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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 tothe order of the matrix.

The Pattern of assignments among software professionals and programs with their respectivetime (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 beperformed, to five mechanists working under him. He wants to assign the jobs to themechanists in such a manner that the aggregate cost to perform the jobs is the least. He hasfollowing 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 4E 9 10 9 6 10

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



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12.10 Advanced Management Accounting

Solution:

The given problem is a balanced minimization problem.

Subtracting minimum element of each row from all the elements of that row, the given problemreduces 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 theminimum 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 thenumber of columns / rows (=5), the above table will not provide optimal solution. Subtract theminimum uncovered element (=2) from all uncovered elements and add to the elements lyingon 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



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Since the minimum number of horizontal and vertical lines to cover all zeros is equal to fivewhich is equal to the order of the matrix, the above table will give the optimal solution. Theoptimal 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



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Find the assignment, which minimizes the total cost of the project. Each contractor has to beassigned one job.

Solution:

The given problem is a balanced minimization problem. Subtracting the minimum element ofeach row from all its elements in turn, the given problem reduces to-


1 40 20 0 10

2 10 20 40 03 10 40 20 0

4 10 10 0 10

Now subtract the minimum element of each column from all its elements in turn. Draw theminimum 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), thismatrix 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 04 0 0 0 10



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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 subordinatewould take to perform each task is given in the matrix below. How should the task be allocatedone to one man so that the total man-hours are minimised?

I II III IV

1 16 52 34 22

2 26 56 8 523 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



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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. Fourcompanies have made bids as detailed below-



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Bids for Routes ( )

Company / Routes R 1 R 2 R 3 R 4

C 1 4,000 5,000

C 2 4,000 4,000

C 3 3,000 2,000

C 4 4,000 5,000

Each bidder can be assigned only one route. Determine the minimum cost that the BPOshould 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.



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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 inRupees is shown below:

(` )

J K L M N

M 1 18 22 30 20 22M 2 24 18 -- 20 18

M 3 -- 22 28 22 14

M 4 28 16 24 14 16

Required:

Determine the optimal assignment schedule in such a manner that the total costs are kept at aminimum.

Solution:

Dummy machine (M 5) is inserted to make it a balanced cost matrix and assume its installationcost to be zero. Cost of install at cell M 3 (J) and M 2 (L) is very high marked as M.



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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-



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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 areworking in that firm and their charges and speeds are different. On the basis of some earlierunderstanding, only one job is given to one typist and the typist is paid for full hours evenwhen he or she works for a fraction of an hour. Find the least cost allocation for the followingdata:



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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 matrixreduces to-



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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 drawminimum number of lines horizontal or vertical so as to cover all zeros. All zeros can becovered 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 (


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The minimum uncovered element is 2. Repeating the usual process again, we get thefollowing 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 optimalsolution. 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.



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Question-8

ABC company is engaged in manufacturing 5 brands of packet snacks. It is having fivemanufacturing setups, each capable of manufacturing any of its brands, one at a time. Thecost to make a brand on these setups vary according to following table-

S 1 S 2 S 3 S 4 S 5

B 1 4 6 7 5 11

B 2 7 3 6 9 5

B 3 8 5 4 6 9

B 4 9 12 7 11 10

B 5 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, Findthe 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 onebrand 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 followingmatrix-

BrandsManufacturing 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



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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 4The minimum number of lines drawn to cover all zeros is equal to 4 which is one less than theorder of the matrix (= 5), the above table will not yield the optimal assignment. For obtainingthe optimal assignment, we increase the number of zeroes by subtracting the minimumuncovered element from all uncovered elements and adding it to elements lying at theintersection 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 givethe 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



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Question-9

Five swimmers are eligible to compete in a relay team which is to consist of four swimmersswimming 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 covera 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 thefree style or butterfly. Bhaskar is a free style specialist averaging 1:01 for the 100 metres butcan also swim the breast stroke in 1:16 and butterfly in 1:20. Chandru swims all styles backstorke 1:10, butterfly 1:12, free style 1:05 and breast stroke 1:20. Dorai swims only thebutterfly 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 anunbalanced 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



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Step 2:

The minimum number of lines to cover all zeros is 4, which is less than 5, the order of thesquare matrix. Hence, the above matrix will not provide the optimal solution. Thus we try toincrease the number of zeros. Select the minimum uncovered element by these lines (which is1). Subtract it from all the uncovered elements and add it to the elements lying on theintersection 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 thesquare matrix, the above matrix will provide the optimal solution. The assignment is madebelow-

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 ineach 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-



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TownDepots

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 distancetravelled is a minimum. How much is the total travelled distance?

Solution:

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

TownDepots

A B C D E

V 0 2 7 12 5W 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 getthe following-

TownDepots

A B C D EV 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



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Draw the minimum number of lines to cover all zeros. Since the number of lines (=3) is notequal 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 theelements lying on the intersection of two lines, we get the following matrix-

TownDepots

A B C D E

V 0 0 2 5 3

W 0 1 6 7 2

X 0 2 7 5 2Y 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 ofthe matrix. Subtract the uncovered element (=2) from all the uncovered element and add it tothe elements lying on the intersection of two lines, we get-

TownDepots

A B C D E

V 0 0 0 3 1W 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 thematrix, hence, the above matrix will not give the optimal solution. Subtracting the uncoveredelement (=1) from all the uncovered elements and adding it to the elements lying on theintersection of two lines, we get-



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TownDepots

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 thematrix, the above table will give the optimal assignment. The optimal assignment is madebelow-

This optimal assignment is-

Customer at Town Car at Depot Distance (Km.)

V C 10

W B 7

X A 8

Y D 10Z E 10

Total 45

Hence the minimum total travelled distance equals to 45 kms.

Assignment - MaximizationQuestion-11

A company has four zones open and four marketing managers available for assignment. Thezones are not equal in sales potentials. It is estimated that a typical marketing manageroperating in each zone would bring in the following Annual sales:

Zones (`)

East 2,40,000

West 1,92,000



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North 1,44,000

South 1,20,000The four marketing manages are also different in ability. It is estimated that working under thesame 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 themaximum sales.

Solution:

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

Effective Matrix

Manager ZonesEast 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 asfollows-



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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 20N 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 tothe value at intersections.


M 0 6 14 18N 0 4 10 13

O 0 0 2 3

P 2 0 0 0



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


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, lowestuncovered element 4 shall be deducted from all uncovered cells value and added with value atintersections.


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 theoptimal solution.

As si gnment 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 weeklyoutput figures (in pieces)-



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OperatorWeekly Output in Lathe

L1 L 2 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 opportunityloss matrix by subtracting all the elements of the given table from the highest element of thetable 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 ofeach row from all the elements of that row, and repeat this step with all the rows of the table.

OperatorWeekly 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 ofeach column from all the elements of that column.



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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 thesquare matrix), the above matrix will not give the optimal solution. Hence we try to increasethe number of zeros. Subtract the least uncovered element which is 2 from all uncoveredelements and add it to all the elements lying on the intersection of two lines. We get thefollowing 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 stepsonce 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



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Revenue/HallMarriage 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 willmaximize the revenue to your hotel.

Solution:

The objective of the given problem is to identify the preferences of marriage parties abouthalls so that hotel management could maximize its profit.

To solve this problem first convert it to a minimization problem by subtracting all the elementsof the given matrix from its highest element. The matrix so obtained which is known as lossmatrix 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 minimumelement of each column from all elements of that column-

Loss Matrix/Hall

Marri age Party 1 2 3 4

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



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The minimum number of lines to cover all zeros is 3 which is less than the order of the squarematrix (i.e.4), the above matrix will not give the optimal solution. Subtracting the minimumuncovered element (2,500) from all uncovered elements and add it to the elements lying onthe intersection of two lines, we get the following matrix-

Loss Matrix/Hall


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 thematrix, the above matrix will give the optimal solution which is given below-

Loss Matrix/Hall


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

Total 87,500

Assignment - MiscellaneousQuestion-14

The cost matrix giving selling costs per unit of a product by salesman A, B, C and D in regionsR1, R2, R3 and R4 is given below:



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A B C D

R 1 4 12 16 8

R 2 20 28 32 24

R 3 36 44 48 40

R 4 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 theselling cost given in the table is 100 per unit, find the assignment that wouldmaximise 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 4R2 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 optionalassignment is possible.



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Minimum Cost = AR 1 + BR 2 + CR 3 + DR 4

= 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 As 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



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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 = AR 1 + BR 2 + CR 3 + DR 4

= ` 96 + ` 72 + ` 52 + ` 44

= ` 264

(iii)

(a) The relative cost of assigning person i to region r do not change by addition orsubtraction 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 solutionwill be the same, applying point (i) above.

(c) Many zeros 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 tominimize the time taken. He was given the matrix showing the hours taken by the operators forthe tasks.

First, he preformed the row minimum operation. Secondly, he did the column minimumoperation. Then, he realized that there were 4 tasks and 5 operators. At the third step heintroduced 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 (asidetask T4 and dummy task T5). At step 5, he performed the necessary operation with theuncovered element, since the number of lines was less than the order of the matrix. After this,his matrix appeared as follows:



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Operators

Tasks I II III IV V

T 1 4 2 5 0 0

T 2 6 3 3 0 3

T 3 4 0 0 0 1

T 4 0 0 5 3 0

T 5 (Dummy) 0 0 3 3 0

(i) What was the matrix after step II? Based on such matrix, ascertain (ii) and (iii) givenbelow.

(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 managers errors, which operator would be deniedany task? Why?

(v) Can the manager go ahead with his assignment to correctly arrive at the optimalassignment, 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 T 5 (Dummy) is 3, it implies 3 was the minimum uncovered element.

Now we do the reverse steps.



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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 = T 2 (9 hours)

Operator II = T 2 (6 hours)

Operator V = T 2 (6 hours)



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(iii)

Based of the Matrix after Step II the most efficient operator is Operator IV.

(iv)

If the Managers 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; T 1 V, T 2 IV, T 3 III are fixed. Between T 4 and T 5, I or II canbe 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 bywhen the dummy was introduced. Column minimum was affected. But in the process, morezeros 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 differentoperators for different jobs in the matrix below-

J 1 J 2 J 3 J 4

O 1 12 10 10 8

O 2 14 12 15 11

O 3 6 10 16 4

O 4 8 10 9 7

(i) How should the manager assign the jobs so that the total time needed for all four jobsis minimum?



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(ii) If job J2 is not to be assigned to operator O2 what should be the assignment and howmuch additional total time will be required?

Solution:

This is an assignment problem whose objective is to assign one job to one operator, so thattotal time needed for all four jobs is minimum. To determine appropriate assignment of jobsand operators, let us apply the assignment algorithm. Subtract the minimum element of eachrow from all elements of that row to get the following matrix-

J 1 J2 J 3 J 4

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-

J 1 J2 J 3 J 4

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 linesdrawn viz., 4 is equal to the number of jobs or the number of operators, so we proceed formaking the optimal assignment.

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

Operator Job

O1 J 3

O2 J 2

O3 J 4

O4 J 1

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



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If job J 2 is not to be assigned to operator O 2 then this objective can be achieved by replacingthe time for cell (O 2, J 2) by a very large time estimate say M. Now apply the assignmentalgorithm to the following matrix so obtained-

J1 J2 J 3 J 4

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 theproblem. We thus have the following matrix-

J1 J2 J 3 J 4

O1 4 2 2 0

O2 3 M 4 0

O3 2 6 12 0

O4 1 3 2 0

J 1 J2 J 3 J 4

O1 3 0 0 0O2 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 3which is less than the number of operators or jobs, therefore the above table will not yield theoptimal assignment. For obtaining the optimal assignment we increase the number of zeros bysubtracting the minimum uncovered element from all uncovered elements and adding it toelements lying at the intersection of two lines, we get the following matrix-

J 1 J2 J 3 J 4

O1 3 0 0 0

O2 1 M 1 0

O3 0 3 9 0

O4 0 1 0 0



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

Operator Job

O1 J 2

O2 J 4

O3 J 1

O4 J 3

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. Theaverage runs scored by each batsman at these positions are as follows:

BatsmanBatting 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 ofruns.

(ii) If another batsman U with the following average runs in batting positions as givenbelow :

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 bereplaced by him?



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Solution:

(i)

The given problem is a maximization assignment problem. Let us first convert it into aminimisation assignment problem by subtracting all the elements of the given matrix from thelargest element of the matrix i.e. 60. The matrix then reduces to-


I II III IV V

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

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


I II III IV V

P 10 10 15 25 0Q 0 12 26 17 15R 10 12 20 0 10S 5 6 5 7 0T 2 0 1 5 7

Now perform column operation for all columns by subtracting the minimum element of eachcolumn 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



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The minimum number of horizontal and vertical lines to cover all zeros is 4 which is less thanthe order of the matrix (= 5), the above table will not give the optimal solution. We subtract theminimum uncovered element 4 from all uncovered elements and add this element to theelements 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 10S 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 thematrix.

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 ofbatsman 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



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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 eachcolumn 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 theelements 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



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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.

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21531sm Finalnew Vol2 Cp12 CHAPTER 12

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