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
© The Institute of Chartered Accountants of India
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.”
© The Institute of Chartered Accountants of India
© The Institute of Chartered Accountants of India
12.4
Advanced Management Accounting
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 stepsStep-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 Institute of Chartered Accountants of India
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?
© The Institute of Chartered Accountants of India
12.6
(ii)
Advanced Management Accounting
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.
© The Institute of Chartered Accountants of India
© The Institute of Chartered Accountants of India
12.8
Advanced Management Accounting
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 followsComputer Time (in minutes)
Required by Software Professionals
Software Programs
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 belowA
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-
© The Institute of Chartered Accountants of India
The Assignment Problem
A
B
C
1
30
5
0
2
0
10
60
3
0
0
20
12.9
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 belowProgram
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 jobsMechanist
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.
© The Institute of Chartered Accountants of India
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 problem
reduces toMechanist
Job1
Job 2
Job 3
Job 4
Job 5
A
8
1
1
0
6
B
7
5
6
0
5
C
5
3
4
0
2
D
1
3
6
0
2
E
3
4
3
0
4
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.
Mechanist
Job1
Job 2
Job 3
Job 4
Job 5
A
7
0
0
0
4
B
6
4
5
0
3
C
4
2
3
0
0
D
0
2
5
0
0
E
2
3
2
0
2
Since the 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 matrixMechanist
Job1
Job 2
Job 3
Job 4
Job 5
A
7
0
0
2
6
B
4
2
3
0
3
C
2
0
1
0
0
D
0
2
5
2
2
E
0
1
0
0
2
© The Institute of Chartered Accountants of India
The Assignment Problem
12.11
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 belowMechanist
Job1
Job 2
Job 3
Job 4
Job 5
A
7
0
0
2
6
B
4
2
3
0
3
C
2
0
1
0
0
D
0
2
5
2
2
E
0
1
0
0
2
The optimal assignment is given belowMechanist
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 belowContractors
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
© The Institute of Chartered Accountants of India
12.12
Advanced Management Accounting
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 toContractors
Job A
Job B
Job C
Job D
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.
Contractors
Job A
Job B
Job C
Job D
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 belowContractors
Job A
Job B
Job C
Job D
1
30
10
0
10
2
0
10
40
0
3
0
30
20
0
4
0
0
0
10
© The Institute of Chartered Accountants of India
The Assignment Problem
12.13
The optimal assignment isContractor
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
© The Institute of Chartered Accountants of India
12.14
Advanced Management Accounting
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-
© The Institute of Chartered Accountants of India
The Assignment Problem
12.15
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.
© The Institute of Chartered Accountants of India
12.16
Advanced Management Accounting
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 M 1, 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.
© The Institute of Chartered Accountants of India
The Assignment Problem
12.17
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 rowJ
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 columnJ
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-
© The Institute of Chartered Accountants of India
12.18
Advanced Management Accounting
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 underJ
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 minimumMachines
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:
© The Institute of Chartered Accountants of India
The Assignment Problem
"
12.19
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-
© The Institute of Chartered Accountants of India
12.20
Advanced Management Accounting
Typist
Job P
Job Q
Job R
Job S
Job 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 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 belowTypist
Job P
Job Q
Job R
Job S
Job T
A
2
2
0
4
0
B
6
4
0
10
2
C
0
1
0
1
2
D
2
4
0
4
2
E
2
0
0
0
2
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 belowTypist
Job P
Job Q
Job R
Job S
Job T
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 Institute of Chartered Accountants of India
The Assignment Problem
12.21
The minimum uncovered element is 2. Repeating the usual process again, we get the
following matrixTypist
Job P
Job Q
Job R
Job S
Job T
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 belowTypist
Job P
Job Q
Job R
Job S
Job T
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.
© The Institute of Chartered Accountants of India
12.22
Advanced Management Accounting
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 tableS1
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
matrixBrands
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
© The Institute of Chartered Accountants of India
The Assignment Problem
12.23
Now subtract the minimum elements of each column from all elements of that columnManufacturing Setups
Brands
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 matrixManufacturing Setups
Brands
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 belowBrand
Setup
Cost
B1
S1
4
B2
S5
5
B3
S4
6
B4
S3
7
B5
S2
5
Total
27
© The Institute of Chartered Accountants of India
12.24
Advanced Management Accounting
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
69
75
-
-
0
Bhaskar
-
76
61
80
0
Chandru
70
80
65
72
0
-
-
-
71
0
80
76
66
70
0
Anand
Dorai
Easwar
Step 1:
As there is a zero in each row, we go straight to the column reductionBack Stroke
Breast Stroke
Free Style
Butterfly
Dummy
Anand
0
0
-
-
0
Bhaskar
-
1
0
10
0
Chandru
1
5
4
2
0
Dorai
-
-
-
1
0
11
1
5
0
0
Easwar
© The Institute of Chartered Accountants of India
The Assignment Problem
12.25
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 beBack Stroke
Breast Stroke
Free Style
Butterfly
Dummy
Anand
0
0
-
-
1
Bhaskar
-
1
0
11
1
Chandru
0
4
3
2
0
Dorai
-
-
-
1
0
10
0
4
0
0
Easwar
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
belowSwimmer
Swimming Style
Anand
Breast Stroke
75
Bhaskar
Free Style
61
Chandru
Back Stroke
70
Dorai
Dummy (not participate)
Easwar
Butterfly
Total Minimum Time in the relay
Time (Seconds)
70
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-
© The Institute of Chartered Accountants of India
12.26
Advanced Management Accounting
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 tableTown
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 followingTown
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
© The Institute of Chartered Accountants of India
The Assignment Problem
12.27
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 matrixTown
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 getTown
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-
© The Institute of Chartered Accountants of India
12.28
Advanced Management Accounting
Depots
Town
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
belowThis optimal assignment isCustomer 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
© The Institute of Chartered Accountants of India
The Assignment Problem
North
1,44,000
South
1,20,000
12.29
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 followsEffective Matrix
Manager
M
N
O
P
Zones
East
West
North
South
80
64
48
40
[(8/24) × 240]
[(8/24) × 192]
[(8/24) × 144]
[(8/24) × 120]
70
56
42
35
[(7/24) × 240]
[(7/24) × 192]
[(7/24) × 144]
[(7/24) × 120]
50
40
30
25
[(5/24) × 240]
[(5/24) × 192]
[(5/24) × 144]
[(5/24) × 120]
40
32
24
20
[(4/24) × 240]
[(4/24) × 192]
[(4/24) × 144]
[(4/24) × 120]
Convert the maximization problem to minimization problem. The resultant loss matrix is as
follows-
© The Institute of Chartered Accountants of India
12.30
Advanced Management Accounting
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
Manager
East
West
North
South
M
0
16
32
40
N
0
14
28
35
O
0
10
20
25
P
0
8
16
20
Manager
East
West
North
South
M
0
8
16
20
N
0
6
12
15
O
0
2
4
5
P
0
0
0
0
Row Operation
Column Operation
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.
Manager
East
West
North
South
M
0
6
14
18
N
0
4
10
13
O
0
0
2
3
P
2
0
0
0
© The Institute of Chartered Accountants of India
The Assignment Problem
12.31
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.
Manager
East
West
North
South
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.
Manager
East
West
North
South
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)-
© The Institute of Chartered Accountants of India
12.32
Advanced Management Accounting
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.
Operator
Weekly Output in Lathe
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.
© The Institute of Chartered Accountants of India
The Assignment Problem
Operator
12.33
Weekly Output in Lathe
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 matrixOperator
Weekly Output in Lathe
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 tableOperator
Weekly Output in Lathe
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 Institute of Chartered Accountants of India
12.34
Advanced Management Accounting
The minimum number of lines to cover all zeros is 4 which is less than 5. Repeating the above
step once again, we thus getWeekly Output in Lathe
Operator
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 belowOperator
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-
© The Institute of Chartered Accountants of India
The Assignment Problem
12.35
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 belowLoss 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 columnLoss Matrix/Hall
Marriage 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
© The Institute of Chartered Accountants of India
12.36
Advanced Management Accounting
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 matrixLoss Matrix/Hall
Marriage Party
1
2
3
4
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 belowLoss Matrix/Hall
Marriage Party
1
2
3
4
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 isMarriage Party
Hall
A
B
C
D
2
3
4
1
Total
Revenue (`)
22,500
20,000
20,000
25,000
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:
© The Institute of Chartered Accountants of India
The Assignment Problem
A
B
C
D
R1
4
12
16
8
R2
20
28
32
24
R3
36
44
48
40
R4
52
60
64
56
12.37
(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 rowA
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 columnA
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.
© The Institute of Chartered Accountants of India
12.38
Advanced Management Accounting
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 belowA
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 matrixA
B
C
D
R1
0
8
12
4
R2
16
24
28
20
R3
32
40
44
36
R4
48
56
60
52
© The Institute of Chartered Accountants of India
The Assignment Problem
12.39
Subtracting minimum term of each rowA
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:
© The Institute of Chartered Accountants of India
12.40
Advanced Management Accounting
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 followsI
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.
© The Institute of Chartered Accountants of India
The Assignment Problem
12.41
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
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
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
Step III:
(i)
Matrix after Step II -
(ii)
Based on the Matrix after Step II most difficult task for operator I, II and V are as followsOperator I
=
T2 (9 hours)
Operator II
=
T2 (6 hours)
Operator V
=
T2 (6 hours)
© The Institute of Chartered Accountants of India
12.42
Advanced Management Accounting
(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 beI
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-
(i)
J1
J2
J3
J4
O1
12
10
10
8
O2
14
12
15
11
O3
6
10
16
4
O4
8
10
9
7
How should the manager assign the jobs so that the total time needed for all four jobs
is minimum?
© The Institute of Chartered Accountants of India
The Assignment Problem
(ii)
12.43
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 matrixJ1
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 columnJ1
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 isOperator
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).
© The Institute of Chartered Accountants of India
12.44
Advanced Management Accounting
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 obtainedJ1
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 matrixJ1
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 matrixJ1
J2
J3
J4
O1
3
0
0
0
O2
1
M
1
0
O3
0
3
9
0
O4
0
1
0
0
© The Institute of Chartered Accountants of India
The Assignment Problem
12.45
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 belowOperator
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?
© The Institute of Chartered Accountants of India
12.46
Advanced Management Accounting
"
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 toBatsman
P
Q
R
S
T
Batting Position
I
II
III
IV
V
20
18
10
40
2
20
30
12
41
0
25
44
20
40
1
35
35
0
42
5
10
33
10
35
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 matrixBatsman
P
Q
R
S
T
Batting Position
I
II
III
IV
V
10
0
10
5
2
10
12
12
6
0
15
26
20
5
1
25
17
0
7
5
0
15
10
0
7
Now perform column operation for all columns by subtracting the minimum element of each
column from all elements of that columnBatsman
Batting Position
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 Institute of Chartered Accountants of India
The Assignment Problem
12.47
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 belowBatting Position
Batsman
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 belowBatsman
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 matrixBatsman
Batting Position
I
II
III
IV
V
Dummy
P
20
20
25
35
10
0
Q
18
30
44
35
33
0
© The Institute of Chartered Accountants of India
12.48
Advanced Management Accounting
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 matrixBatsman
Batting Position
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 belowBatsman
Batting Position
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
© The Institute of Chartered Accountants of India
The Assignment Problem
12.49
Batting Position
Batsman
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
Batting Position
Batsman
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 belowBatsman
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.
© The Institute of Chartered Accountants of India