PESIT Bangalore South Campus

USN
1 P E
I
S
PESIT Bangalore South Campus
Hosur road, 1km before Electronic City, Bengaluru -100
Department of Information Science and Engineering
INTERNAL ASSESSMENT TEST I
Date
: 23-02-2016
Subject & Code : Operation Research, 10IS661
Name of faculty : KARTHIK S
Max Marks: 50
Section
: VI A&B
Time
: 12:00 – 1:30
Note: Answer any 5 full questions
1
What is operation research? Explain the phases of operation research study
Ans: Phases of OR are
1 Observe the problem environment
2 Analyze and define the problem
3 Develop a model
4 Selection of data input
5 solution and testing
6 Implementation of the solution
2
Explain the steps involved in graphical method to solve LPP. Solve the following problem graphically
Max Z = 3x1 + 5x2
Subjected to the conditions x1<=4
2x2<=12
3x1 + 2x2 <=18
x1, x2>=0
Ans: Max Z = 36
B.E 6th Semester
USN
1 P E
PESIT Bangalore South Campus
Hosur road, 1km before Electronic City, Bengaluru -100
Department of Information Science and Engineering
3
Solve the LPP
Max z = 3x1 + 2x2 + 5x3
Subjected to the conditions x1+2x2+x3 <=430
3x1 + 2x3 <=460
x1 + 4x2<=420
x1,x2 and x3 are all non negative variables
Ans:
4
Using Big-M method solve
B.E 6th Semester
I
S
USN
1 P E
PESIT Bangalore South Campus
Hosur road, 1km before Electronic City, Bengaluru -100
Department of Information Science and Engineering
Min Z = 4x1 + 2x2
Subjected to the conditions 3x1 + x2>=27
x1 + x2 >= 21
x1 and x2 are non negative variables
Ans:
5
Solve the following LPP
Max Z = 3x1 + 5x2
B.E 6th Semester
I
S
USN
1 P E
I
S
PESIT Bangalore South Campus
Hosur road, 1km before Electronic City, Bengaluru -100
Department of Information Science and Engineering
Subjected to the conditions x1<=4
2x2<=12
3x1 + 2x2 <=18
x2>=0
Ans: Since x1 is unbounded we can write it as x1 = x1 + - x1- where x1+ and x1- are non negative integers
Solving this using simplex method will result in Max Z=36
6a
A firm manufactures 3 types of products P1, P2 and P3. The profits are Rs.30, Rs.40 and Rs. 20 respectively. The firm has 2 machines
M1 and M2. The processing time for each product in these machines and maximum availability is shown in the table. The products should
undergo processing in both the machines. The firm should manufacture minimum of 100 units of P1, 200 units of P2 and 150 units of P3.
Formulate this problem as a linear problem.
Machine Minutes required
Max
availability
P1
P2
P3
M1
4
3
5
2000
M2
2
2
4
2500
Ans: Max Z = 30x1+40x2+20x3
STC 4x1+3x2+5x3<=2000
2x1+2x2+4x3<=2500
x1>=100, x2>=200 and x3>=150
6b
Explain i. Feasible solution
ii. Feasible region
iii. Optimal solution
Ans:
Feasible solution is a solution which satisfies all the constraints of the LPP
Feasible region, is the set of all possible points (sets of values of the choice variables) of an optimization problem that satisfy the
problem's constraints,
B.E 6th Semester
USN
1 P E
I
S
PESIT Bangalore South Campus
Hosur road, 1km before Electronic City, Bengaluru -100
Department of Information Science and Engineering
An optimal solution to a linear program is a feasible solution with the largest objective function value (for a maximization problem).
B.E 6th Semester