Examples 3-Operation Research-Handouts, Lecture notes of Operational Research

Operations Research (OR) refers to the science of decision making. This course elaborate like linear, nonlinear and discrete optimization. This lecture handout was provided by Sir Avikshit Gupte. It includes: Example, Allocate, Depending, Efficiency, Individual, Minimize, Manager, Depending, Different, Solution

Typology: Lecture notes

2011/2012

Uploaded on 08/06/2012

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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
Job
Worker
A B C D
1
10
20
18
14
2
15
25
9
25
3
30
19
17
12
4
19
24
20
10
Solution: The following steps are followed to find an optimal 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
Job
Worker
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
Job
Worker
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.
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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

Job

Worker A B C D 1 10 20 18 14 2 15 25 9 25 3 30 19 17 12 4 19 24 20 10

Solution: The following steps are followed to find an optimal 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

Job

Worker 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

Job

Worker 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

Job

Worker A B C D 1 0 3 8 4

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

Job

Worker 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: