# Operations Research

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Operations Research (MTH601)
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Subject to restrictions,
Row restrictions
x11 + x12 + x13 + x14 = 1
for job 1
x21 + x22 + x23 + x24 = 1
for job 2
x31 + x32 + x33 + x34 = 1
for job 3
x41 + x42 + x43 + x44 = 1
for job 4
Column restrictions
x11 + x21 + x31 + x41 = 1
for person 1
x12 + x22 + x32 + x42 = 1
for person 2
x13 + x23 + x33 + x43 = 1
for person 3
x14 + x24 + x34 + x44 = 1
for person 4
xij = 0 or 1
and
In general,
xi1 + xi2 + ... + xin = 1, for i = 1, 2, ..., n
x1j + x2j + ... + xnj = 1, for j = 1, 2, ..., n
and
When compared with a transportation problem, we see that ai = 1 and bj = 1 for all rows and columns, xij = 0 or
1.
We shall not attempt simplex algorithem or the transportation algorithm to get a
solution to an assignment problem. Certain systematic procedure has been devised so as to
obtain the solution to the problem with ease.
SOLUTION OF AN ASSIGNMENT PROBLEM
The solution to an assignment problem is based on the following theorem.
Theorem:
If in an assignment problem we add a constant to every element of a row or column in the
effectiveness matrix then an assignment that minimizes the total effectiveness in one matrix also minimizes the
total effectiveness in the other matrix.
Example:
A works manager has to allocate four different jobs to four workmen. Depending on the
efficiency and the capacity of the individual the times taken by each differ as shown in the table 2. How should
the tasks be assigned one jot to a worker so as to minimize the total man-hours?
Table 2
Worker
Job
A
B
C
D
1
10
20
18
14
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2
15
25
9
25
3
30
19
17
12
4
19
24
20
10
The following steps are followed to find an optimal solution.
Solution:
STEP 1:Consider each row. Select the minimum element in each row. Subtract this smallest element form all
the elements in that row. This results in the table 3.
Table 3
Worker
Job
A
B
C
D
1
0
10
8
4
2
6
16
0
16
3
18
7
5
0
4
9
14
10
0
STEP 2: We subtract the minimum element in each column from all the elements in its column. Thus we obtain
table 4
Table 4
Worker
Job
A
B
C
D
1
0
3
8
4
2
6
9
0
16
3
18
0
5
0
4
9
7
10
0
STEP 3:In this way we make sure that in the matrix each row and each column has atleast one zero element.
Having obtained atleast one zero in each row and each column, we assign starting from first row.
In the first row, we have a zero in (1, A). Hence we assign job 1 to the worker A. This assignment is indicated
by a square . All other zeros in the column are crossed (X) to show that the other jobs cannot be assigned to
worker A as he has already been assigned. In the above problem we do not have other zeros in the first column
A.
Proceed to the second row. We have a zero in (2, C). Hence we assign the job 2 to worker C, indicating by a
square . Any other zero in this column is crossed (X).
Proceed to the third row. Here we have two zeros corresponding to (3, B) and (3, D). Since there is a tie for the
job 3, go to the next row deferring the decision for the present. Proceeding to the fourth row, we have only one
zero in (4, D). Hence we assign job 4 to worker D. Now the column D has a zero in the third row. Hence cross
(3, D). All the assignments made in this way are as shown in table 5.
Table 5
Worker
Job
A
B
C
D
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1
0
3
8
4
2
6
9
0
14
3
18
0
5
0
4
9
7
10
0
STEP 4: Now having assigned certain jobs to certain workers we proceed to the column 1. Since there is an
assignment in this column, we proceed to the second column. There is only one zero in the cell (3, B); we assign
the jobs 3 to worker B. Thus all the four jobs have been assigned to four workers. Thus we obtain the solution to
the problem as shown in the table 6.
Table 6
Worker
Job
A
B
C
D
1
0
3
8
4
2
6
9
0
16
3
18
0
5
0
4
9
7
10
0
The assignments are
Job to Worker
1
A
2
C
3
B
4
D
We summarise the above procedure as a set of following rules:
a.
Subtract the minimum element in each row from all the elements in its row to make sure that at least
we get one zero in that row.
b.
Subtract the minimum element in each column from all the elements in its
column
in
the
above
reduced matrix, to make sure that we get at least one zero in each column.
c.
Having obtained at least one zero in each row and atleast one zero in each column, examine rows
successively until a row with exactly one unmarked zero is found and mark ( ) this zero, indicating
that assignment in made there. Mark (X) all other zeros in the same column, to show that they cannot
be used to make other assignments. Proceed in this way until all rows have been examined. If there is a
tie among zeros defer the decision.
Next consider columns, for single unmarked zero, mark them ( ) and
d.
mark (X) any other unmarked
zero in their rows.
e.
Repeat (c) and (d) successively until one of the two occurs.
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1) There are no zeros left unmarked.
2) The remaining unmarked zeros lie atleast two in each row and column. i.e., they occupy
corners of a square.
If the outcome is (1), we have a maximal assignment. In the outcome (2) we use arbitrary assignments. This
process may yield multiple solutions.
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MULTIPLE SOLUTIONS
Example 1 Given the following matrix, find the optimal assignment.
1
2
3
4
5
5
0
3
2
6
1
0
0
5
4
7
2
0
3
0
4
0
3
0
1
0
3
0
4
6
5
0
0
0
5
Solution: Note that all the rows and columns have at least one zero.
Row 1 has a single zero in column 2. So make an assignment, delete (mark X) the second zero in
0 denotes assignment
column 2. This is shown in table 7.
Table 7
1
2
3
4
5
1
0
5
3
2
6
2
0
0
5
4
7
3
0x
3
0
4
0
4
0x
1
0
3
0
5
0
6
5
0x
0x
Row 2 has a single zero in the first column. So make an assignment and delete the remaining zeros in column 1
as shown in table 8.
Table 8
1
2
3
4
5
1
0
5
3
2
6
2
0
0x
5
4
7
3
0x
3
0
4
0
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4
0x
1
0
3
0
5
6
5
0
0
0
Row 3, 4 and 5 have more than a single zero. So we skip these rows and examine the columns. Columns 3 has
three zeros and so omit it. Column 4 has a single zero in row 5. So we make an assignment, deleting the
remaining zeros in row 5. The result is as shown in table 9.
Table 9
1
2
3
4
5
1
0
5
3
2
6
2
0
0x
5
4
7
0x
3
0
4
0
3
0x
1
0
3
0
4
0
5
6
5
0x
0x
Now we have two zeros in rows 3 and 4 in columns 3 and they occupy the corners of a square. An arbitrary
assignment has to be made. If we make an assignment in (3, 3) and delete the remaining zero in row 3 and in
column 3, this leaves one zero in the position (4, 5) and an assignment is made there. Thus we have a solution to
the problem as in table 10.
Table 10
1
2
3
4
5
1
0
3
2
6
5
0
0x
5
4
7
2
3
0
0x
3
4
0x
0
0
1
0x
3
4
0
5
0x
6
5
0
One more assignment (as a solution) is possible in this problem. (i.e.) we could have made an assignment at (3,
5) deleting other zero in the row 3 and zero in column 5 and making the last assignment at (4, 3). This is shown
below in table 11.
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Table 11
1
2
3
4
5
1
0
3
2
6
5
2
0
0x
5
4
7
3
0
0x
3
0x
4
4
0
3
0x
0x
1
5
0
0x
6
5
0x
All the above markings can be done in a single matrix itself. We need not write the matrix
Remark:
repeatedly. This is done only to clarify the presentation.
HUNGARIAN ALGORITHM
In the two examples above, the first one gave a solution leaving no zeros. It was a case of no ambiguity
and in the second, we had more zeros and the tie was broken arbitrarily. Sometimes if we proceed in the steps
explained above, we get a maximal assignment, which does not contain an assignment in every row or column.
We are faced with a question of how to solve the problem. In such a case, the effectiveness matrix has to be
modified, so that after a finite number of iterations an optimal assignment will be in sight. The following is the
algorithm to solve a problem of this kind and this is known as Hungarian algorithm. The systematic procedure is
explained in different steps and a problem is solved as an illustration.
STEP 1: Starting with a maximal assignment mark () all rows for which assignments have not been made.
STEP 2:Mark () columns not already marked which have zeros in the marked-rows.
STEP 3:Mark () rows not already marked which have assignments in the marked columns.
STEP 4:Repeat steps 2 and 3 until the chain of markings ends.
STEP5: Draw lines through all unmarked rows and though all marked columns. (Check: If the above steps have
been carried out correctly, there should be as many lines as there were assignments in the maximal assignment
and we have atleast one line passing through every zero.) This is a method of drawing minimum number of lines
that will pass through all zeros. Thus all the zeros are covered.
STEP 6:Now examine the elements that do not have a line through them. Choose the smallest element and
subtract it form all the elements the intersection or junction of two lines. Leave the remaining elements of the
matrix unchanged.
STEP 7:Now proceed to make an assignment. If a solution is obtained with an assignment for every row, then
this will be the optimal solution. Otherwise proceed to draw minimum number of lines to cover all zeros as
explained in steps 1 to 5 and repeat iterations if needed.
Consider the following example.
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Example 1
Solve the following assignment problem to minimize the total cost represented as elements in the
matrix (cost in thousand rupees).
Contractor
1
2
3
4
Building
A
48
48
50
44
B
56
60
60
68
C
96
94
90
85
D
42
44
54
46
Solution:
STEP 1:Choose the least element in each row of the matrix and subtract the same from all the elements in each
row so that each row contains atleast one zero. Thus we have table 12.
Table 12
Contractor
1
2
3
4
Building
A
4
4
6
0
B
0
4
4
12
C
11
9
5
0
D
0
2
12
4
STEP 2:Choose the least element in each column and subtract the same from all the elements in that column to
ensure that there is atleast one zero in each column. Thus we have table 13.
Table 13
Contractor
1
2
3
4
Building
A
4
2
2
0
B
2
0x
12
0
C
11
7
1
0x
D
0x
8
4
0
STEP 3:We make the assignment in each row and column as explained previously. This results in table 1
Table 14
Contractor
1
2
3
4
Building
A
4
2
2
0
B
2
0x
12
0
C
11
7
1
0x
D
0x
8
4
0
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Here we have only three assignments. But we must have four assignments. With this maximal assignment
we have to draw the minimum number of lines to cover all the zeros. This is carried out as explained in steps 4
to 9. Refer table 15.
Table 15
Contractor
1
2
3
4
Building
A
4
2
2
0
B
2
0x
12
0
C
11
7
1
0x
D
0x
8
4
0
STEP 4:Mark () the unassigned row (row C).
STEP 5:Against the marked row C, look for any 0 element and mark that column (column 4).
STEP 6:Against the marked column 4, look for any assignment and mark that row (row A).
STEP 7:Repeat steps 6 and 7 until the chain of markings ends.
STEP 8:Draw lines through all unmarked rows (row B and row D) and through all marked columns (column 4).
(Check: There should be only three lines to cover all the zeros.)
STEP 9:Select the minimum from the elements that do not have a line through them. In this example we have 1
as the minimum element, subtract the same from all the elements that do not have a line through them and add
this smallest element at the intersection of two lines. Thus we have the matrix shown in table 16.
Table 16
Contractor
1
2
3
4
Building
A
3
1
1
0
B
2
0x
13
0
C
10
6
0x
0
D
0x
8
5
0
STEP 10: Go ahead with the assignment with the usual procedure. This is carried out in the table 16. Thus we
have four assignments.
Building A is alloted to contractor 4
Building B is alloted to contractor 1
Building C is alloted to contractor 3
Building D is alloted to contractor 2
Total cost is 44 + 56 + 90 + 44 = Rs. 234 thousands.
An airline that operates seven days a week has a timetable given below. Crews must have a
Example:
minimum layover time of 6 hours between flights. Obtain the pairing of flights that minimizes layover time
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away from home. For any given pairing the crew will be based at the city that results in the smaller layover.
For each pair also mention the city where the crew should be placed.
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Flight No.
Karachi to
1
7.00 a.m.
9.00 a.m.
2
9.00 a.m.
11.00 a.m.
3
1.30 p.m.
3.30 p.m.
4
7.30 p.m.
9.30 p.m.
Flight No.
101
9.00 a.m.
11.00 a.m.
102
10.00 a.m.
12.00 Noon
103
3.30 p.m.
5.30 p.m.
104
8.00 p.m.
10.00 p.m.
Solution:
STEP 1: We prepare a matrix in which all the flights reaching Islamabad are paired with flights with No's 101,
102, 103 and 104. The elements in the matrix represent the time taken (hrs) by Karachi based crew in
Flight No.
101
102
103
104
1
24
25
6.5
11
2
22
23
28.5
9
3
17.5
18.5
24
28.5
4
11.5
12.5
18
22.5
Explanation: The crew in the first flight No. 1 leaving Karachi arrives at Islamabad at 9.00 a.m. If the crew is
to return by flight No. 101, it is not possible on the same day and it has to stay for 24 hours (layover time).
Similarly to return by flight No. 102, 103 and 104, the same crew takes 25 hours, 6.5 hours, 11 hours
respectively. In the same way we calculate the time spent by the crew in flights 2, 3 and 4 to pair with flights
with No. 101, 102, 103 and 104. This is shown in Table 17.
STEP 2:We prepare a matrix in which all the flights reaching Karachi are paired with flights with numbers
1,2,3 and 4. The elements in the matrix indicate the time spent by Islamabad based crew in Karachi. This is
shown in table 18.
Table 18
Flight No.
101
102
103
104
1
20
19
13.5
9
2
22
21
15.5
11
3
26.5
25.5
20
15.5
4
8.5
7.5
26
21.5
STEP 3:Comparing the corresponding elements of the two matrices (tables 17 and 18), we choose the minimum
element and indicate the base at the top of the element (Karachi base-D, Islamabad base-C,). Thus we prepare
the table 19. If the crew is placed in either city it is denoted by *.
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Table 19
Flight No.
101
102
103
104
C
C
C
9C
1
20
19
6.5
21C  15.5C
9D
2
22*
17.5D  18.5D  20C  15.5C
3
8.5C   7.5C
18D  21.5C
4
STEP 4:We proceed with the usual steps of assignment problem. Thus we have the following tables 20 and 21.
Table 20
Flight No.
101
102
103
104
1
13.5
12.5
0
2.5
2
13
12
6.5
0
3
2
3
5
0
4
1
0
10.5
14
Table 21
Flight No.
101
102
103
104
1
12.5
12.5
0
25
2
12
12
6.5
0
3
1
3
5
0
4
0
0
10.5
14
We have the following assignments shown in table 22. The minimum number of lines (3), are drawn to cover all
zero following Hungarian algorithm.
Table 22
Flight No.
101
102
103
104
1
12.5
12.5
0
26
2
11
11
5.5
0
3
0
2
3.5
0
4
0
0
10.5
15
STEP 5
Select the smallest element from those, which have no lines through them. In this case this is 1. Hence subtract
the same from all the elements that do not have a line through them and add the same at the intersection of two
lines. Thus we have the table 23.
Table 23
Flight No.
101
102
103
104
1
12.5
12.5
0
26
2
11
11
5.5
0
3
0
2
3.5
0
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4
0
0
10.5
15
We assign rowwise and then columnwise. We have the table 24
Table 24
Flight No.
101
102
103
104
1
12.5
12.5
26
0
2
11
11
5.5
0
3
2
3.5
0x
0
4
0x
10.5
15
0
The result can be given as,
Flight no. 1 is paired with 103 with base at Karachi.
Flight no. 2 is paired with 104 with base at Karachi.
Flight no. 3 is paired with 101 with base at Karachi.
Flight no. 4 is paired with 102 with base at Islamabad.
MAXIMIZATION IN ASSIGNMENT MODEL
The problem of maximization is carried out similar to the case of minization making a slight modification. The
required modification is to multiply all elements in the matrix by -1, based on the concept that minizing the
negative of a function is equivalent to maximize the original function. This case is illustrated in the following
example.
Example:
Six salesmen are to be allocated to six sales regions. The earning of each salesman at each
region is given below. How can you find an allocation, so that the earnings will be maximum?
Region
Salesman
1
2
3
4
5
6
15
35
0
25
10
45
A
40
5
45
20
15
20
B
25
60
10
65
25
10
C
25
20
35
10
25
60
D
30
70
40
5
40
50
E
10
25
30
40
50
15
F
(Figure are in Rs. 1000)
In this problem the objective is to maximize the earnings of six salesmen sent to different
Solution:
regions. The first step is to multiply all the elements by (-1) and apply the method for minimization. So we have
the starting table as shown in table 25.
Table 25
Region
Salesman
1
2
3
4
5
6
-15
-35
0
-25
-10
-45
A
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-40
-5
-45
-20
-15
-20
B
-25
-60
-10
-65
-25
-10
C
-25
-20
-35
-10
-25
-60
D
-30
-70
-40
-5
-40
-50
E
-10
-25
-30
-40
-50
-15
F
The second step is to follow the procedure of subtracting the least element from each row and then in each
column to ensure that there is atleast one zero in each row and in each column. This is presented in tables 26 and
27.
Table 26
Region
Salesman
1
2
3
4
5
6
30
10
45
20
35
0
A
5
40
5
25
30
25
B
40
5
55
0
40
55
C
35
40
25
50
35
0
D
40
0
30
65
30
20
E
40
25
20
10
0
35
F
Table 27
Region
Salesman
1
2
3
4
5
6
25
10
45
20
35
0
A
0
40
0
25
30
25
B
35
5
55
0
40
55
C
30
40
25
50
35
0
D
35
0
30
65
30
20
E
35
25
20
10
0
35
F
The third step is to assign salesmen to regions and test for optimality with the usual method as in table 28.
Table 28
Region
Salesman
1
2
3
4
5
6
25
10
45
20
35
A
0
40
0x
25
30
25
B
0
35
5
55
40
55
C
0
30
40
25
50
35
0x
D
35
30
65
30
20
E
0
35
25
20
10
35
F
0
Since we have six assignments, we have an optimal solution as given below in tables 29 to 31.
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Salesman A is assigned to region 1,
B to region 3,
C to region 4,
D to region 6,
E to region 2 and
F to region 5
Total earnings are 15 + 45 + 65 + 60 + 70 + 50 = Rs. 3.5 (in thousands)
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First iteration
Table 29
Salesman
Region
1
2
3
4
5
6
15
0x
35
10
25
0x
A
40
0x
25
30
35
B
0
35
5
55
40
65
C
0
20
30
15
40
25
D
0
35
30
65
30
30
E
0
35
25
20
10
45
F
0
Table 30
Second iteration
Salesman
Region
1
2
3
4
5
6
5
0x
25
0x
15
0x
A
50
0x
25
30
45
B
0
35
15
55
40
75
C
0
10
30
5
30
15
D
0
25
30
55
20
30
E
0
35
35
20
10
55
F
0
Table 31
Third iteration
Salesman
Region
1
2
3
4
5
6
0x
20
5
10
0x
A
0
0x
55
30
30
45
B
0
30
15
50
35
75
C
0
5
30
0x
30
10
D
0
20
15
55
15
30
E
0
35
40
20
15
60
F
0
IMPOSSIBLE ASSIGNMENT
Sometimes in an assignment model we are not able to assign some jobs to some persons. For example
if machines are to be allocated to locations and if a machine cannot be accommodated in a particular location,
then it i an impossible assignment. To solve the problem in such situations we attach a very highly prohibited
(say ) cost to the cell in the matrix so that there is absolutely no chance to get the assignment with infinite cost
in the optimal solution.
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Example
The processing times in hours for the jobs when allocated to the different machines are indicated
below. When a job is not possible to be made in a particular manner, it is indicated as '-'
Machine
Jobs
I
II
III
IV
V
3
-
8
-
8
A
4
7
15
18
8
B
8
12
-
-
12
C
5
5
8
3
6
D
10
12
15
10
-
E
Allocate the machines for the jobs so that the total processing time is minimum.
Solution:
We have the impossible assignments marked as -. We introduce deliberately a high prohibitive time
(say ) in those places and proceed with the usual steps of solution procedure for assignment problem. Thus we
have the
table 32.
Table 32
Machine
Jobs
I
II
III
IV
V
8
8
3
A
4
7
15
18
8
B
12
8
12
C
5
5
8
3
6
D
10
12
15
10
E
Reducing the matrix in rows and columns so as to have atleast one zero in each row and in each
column, we have the following tables 33 and 34
Table 33
Machine
Jobs
I
II
III
IV
V
8
5
0
A
0
3
11
14
4
B
4
0
4
C
2
2
5
0
3
D
0
2
5
0
E
Table 34
Machine
Jobs
I
II
III
IV
V
0
2
0
A
0
1
6
14
1
B
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1
0
2
C
2
0
0
0
0
D
0
0
0
0
E
Next we proceed to assign jobs to machines as in table 35.
Table 35
Machine
Jobs
I
II
III
IV
V
2
0x
A
0
1
6
14
1
B
0
1
0x
2
C
2
0x
0x
0x
D
0
0x
0x
0x
E
0
From 35 we see that there are only four assignments but we should have five assignments.
We now proceed with drawing minimum number of lines to cover all zeros as in table 36.
Table 36
Machine
Jobs
I
II
III
IV
V
0
2
0x
A
1
6
14
1
B
0
1
0x
2
C
2
0x
0x
0x
D
0
0x
0x
0x
E
0
Now subtract the least element from the elements that do not have a line through them. In this case this element
is 1. We subtract it from all elements, which do not have a line through them and add the same at the
intersection of two lines. Proceeding in this way we have the table 37.
Table 37
Machine
Jobs
I
II
III
IV
V
2
1
A
0
0x
5
13
0x
B
0
0
0x
1
C
3
0x
0x
D
0
0
1
0x
0x
E
0
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(Note: Since we have number of zeros occupying corner of a square, we have multiple solutions).
Job A is assigned to machine III, B to machine I, C to machine V, D to machine II, E to machine IV.
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REVIEW QUESTIONS
1.
Three jobs A, B, C are to be assigned to three machines X, Y, Z. The processing costs (in Rs.) are as
given in the matrix shown below. Find the allocation, which will minimize the overall processing cost.
Machine
jobs
A
B
C
19
28
31
X
11
17
16
Y
12
15
13
Z
2.
A college department chairman has the problem of providing teachers for all courses offered by his
department at the highest possible level of educational 'quality'. He has one professor, two associate
professors, and one teaching assistant (TA) available. Four courses must be offered and, after
appropriate introspection and evaluation he has arrived at the following relative ratings (100 = basic
rating) regarding the ability of each instructor to teach the four courses, respectively.
Courses
1
2
3
4
Prof. 1
60
40
60
70
Prof. 2
20
60
50
70
Prof. 3
20
30
40
60
T.A.
30
10
30
40
How should he assign his staff to the courses to maximize educational quality in his department?
3.
(a)  Explain the Hungarian method of solving an assignment problem for minimization.
(b)  Solve the following assignment problem for minimization with cost (in rupees) matrix as:
Machine
Jobs
A
B
C
D
E
4
10
3
4
8
1
7
2
6
7
7
2
10
5
8
11
4
3
3
6
5
3
2
4
10
7
3
5
7
5
4.
State the linear programming formulation of an assignment problem.
Four Jobs can be processed on four different machines, one job on one
machine. Resulting profits vary with assignments. They are given
below.
Machine
Jobs
A
B
C
D
42  35  28
21
I
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30  25  20
15
II
30
2
20
15
III
24  20  16
12
IV
Find the optimum assignment of jobs to machines and the corresponding profit.
5.
Five men are available to do five different jobs. From past records the time in hours that each man
takes for each job is known and is given below.
Jobs
Men
I
II
III
IV
V
3
10
3
8
2
A
7
9
8
7
2
B
5
7
6
4
2
C
5
3
8
4
2
D
6
4
10
6
2
E
Find the assignment of men to jobs that will minimize the total time taken.
6.
Pearl Corporation has four plants each of which can manufacture any one of four products. Production
costs differ from one plant to another as do sales revenue. Given the revenue and cost data below,
obtain which product each plant should produce to minimize the profit.
Sales Revenue
Production Cost
Plant
Product
Product
1
2
3
4
1
2
3
4
50
68
49
62
49
60
45
61
A
60
70
51
74
55
63
45
69
B
55
67
53
70
52
62
49
58
C
58
65
54
69
55
64
48
66
D
7.
Five different machines can process any of the five required jobs with different profits resulting from
each assignment.
Machine
Job
A
B
C
D
E
1
130
137
140
128
140
2
140
124
127
121
136
3
140
132
133
-
135
4
-
138
140
136
136
5
129
-
141
134
139
Find the maximum profit possible through optimum assignments.
8.
A construction company has five bulldozers at different locations and one bulldozer is required at three
different constructions sites. If the transportation cost are as shown, determine the optimum shipping
schedule.
Shipping cost ('000 Rs)
Location
Construction site
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A
B
C
1
20
30
40
2
70
60
40
3
30
50
80
4
40
60
50
5
40
60
30
9.
A manager has the problem of assigning four new machines to three production facilities. The
respective profits derived are as shown. If only one machine is assigned to a production facility
determine the optimal assignment.
Profits ('000 Rs)
Machine
Production facility
1
2
3
A
10
10
14
B
10
11
13
C
12
10
10
D
13
12
11
10.
Solve the following unbalanced assignment problem of minimizing total time for doing all the jobs.
Jobs
Operator
1
2
3
4
5
1
6
2
5
2
6
2
2
5
8
7
7
3
7
8
6
9
8
4
6
2
3
4
5
5
9
3
8
9
7
6
4
7
4
6
8
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Segment VII: Queuing Theory
Lectures 38 - 39
225