linear programming
TRANSCRIPT
GROUP MEMBERS
• YASHFA SOHAIL
• AMNA ABID
• SAMRA NAZIR
• SOBIA ANWAR
• ANAM MANZOOR
Quantitative Techniques
Topic of project:
LINEAR PROGRAMMING
Introduction
Linear programming is a mathematical technique which helps
the manager in making the use of firm’s economic resources
which are in limited supply such as money, material , labor and
space etc.
• These limited resources has to be allocated so as to maximize
profits or to minimize cost.
Objective functionThe objective function is a presentation of the goal to be
achieved, either a profit is to be maximized or a cost is to be
minimized.
The linear model consists of the following
components:
• A set of decision variables.
• An objective function.
• A set of constraints.
Types of constraints
• Operational constraints
• Non negative constraints.
Operational constraints
The operational constraints indicates that the
total amount of each type of economic resources
used has to be consistent with the available
amount of each resource.
Non negative constraints
These constraints indicate that “x” and “y”
cannot less than ZERO. Negative values doesn’t
indicate.
The Importance of Linear Programming
Many real world problems can be approximated by linear
models.
There are well-known successful applications in:
Manufacturing
Marketing
Finance (investment)
Advertising
Agriculture
Question
Maximize: f=15x+12y
Subject to: 3x+2y<24
1/2x+1y<8
x>0,y>0
Consider: 3x+2y=24
X-intercept
3x+2y=24
3x+2(0)=24
3x=24
x=24/3
x=8
Y-intercept
3x+2y=24
3(0)+2y=24
2y=24
y=24/2
y=3
Test point
3x+2y<24
3(0)+2(0)<24
0<24= true
Consider: 1/2x+1y=8
X-intercept
1/2x+1y=8
1/2x+1(0)=8
1/2x=8
X=8*2
X=16
Y-intercept
1/2x+1y=8
½(0)+1y=8
Y=8
Test point
1/2x+1y<8
½(0)+1(0)<8
0<8 true
Point of intersection
3x+2y=24 --- (1)
1/2x+1y=8 ---- (2)
To multiply the equation (2) by 2
2(1/2x+1y)=8*2
X+2y=16 --- (3)
To subtract the equation of 1 & 3
3x+2y=24
x+2y=16
_________
2x=8
x=8/2
x=4
Cont’d
To find “y”
X+2y=16
4+2y=16
2y=16-4
2y=12
Y=12/2
Y=6
F=15x+12y
Corner points:
(8,0) 15(8)+12(0)=120
(0,0) 15(0)+12(0)=0
(0,8) 15(0)+12(8)=96
(4,6) 15(4)+12(6)=132
Interpretation
By looking at the above table we observe that
our objective function is maximized at the point
(4,6)
And the maximum profit is 132. so the optimal
situation is x=4, y=6.
