02 graphical solution of two variable lpps
TRANSCRIPT
Problem 6 Problem Set 2.3A Page 26(Modified)
Electra produces two types of electric motors,
each on a separate assembly line. The respective
daily capacities of the two lines are 150 and 200
motors. Type I motor uses 2 units of a certain
electronic component, and type II motor uses
only 1 unit. The supplier of the component can
provide 400 pieces a day.The profits per motor
of types I and II are $8 and $5 respectively.
Formulate the problem as a LPP and find the
optimal daily production.
Let the company produce x1 type I motors
and x2 type II motors per day.
1 28 5z x x
Subject to the constraints 1
2
1 2
1 2
150
200
2 400
, 0
x
x
x x
x x
The objective is to find x1 and x2 so as to
Maximize the profit
We shall use graphical method to solve the
above problem.
Step 1 Determination of the Feasible solution space
The non-negativity restrictions tell that the
solution space is in the first quadrant.
Then we replace each inequality constraint
by an equality and then graph the resulting
line (noting that two points will determine a
line uniquely).
Next we note that each constraint line
divides the plane into two half-spaces and
that on one half-space the constraint will be
≤ and on the other it will be ≥. To determine
the “correct” side we choose a reference
point and see on which side it lies.
(Normally (0,0) is chosen. If the constraint
line passes through (0,0), then we choose
some other point.) Doing like this we would
get the feasible solution space.
Step 2 Determination of the optimal solution
The determination of the optimal solution
requires the direction in which the objective
function will increase (decrease) in the case of
a maximization (minimization) problem. We
find this by assigning two increasing
(decreasing) values for z and then drawing the
graphs of the objective function for these two
values. The optimum solution occurs at a point
beyond which any further increase (decrease)
of z will make us leave the feasible space.
Graphical solution
z=1800
z=1700
z=1200
z=400
z=1000
(100,200)
(150,100)
(150,0)
(0,200)
x1
x2
Maximize z=8x1+5x2
Subject to the constraints
2x1+x2 400
x1 150
x2 200
x1,x2 0
Optimum
=1800 at
Feed Mix Problem
Minimize z = 2x1 + 3x2
Subject to x1 + 3 x2 15
2 x1 + 2 x2 20
3 x1 + 2 x2 24
x1, x2 0
Optimum = 22.5 at (7.5, 2.5)
(7.5,2.5)
(4,6)
(0,12)
(15,0) Minimum at
z = 42
z = 39
z= 36
z = 30 z = 26
z = 22.5
Graphical Solution of Feed Mix Problem
Maximize z = 2x1 + x2
Subject to x1 + x2 ≤ 40
4 x1 + x2 ≤ 100
x1, x2 ≥ 0
Optimum = 60 at (20, 20)
(20,20)
(0,40)
(25,0)
z maximum at
Maximize z = 2x1 + x2
Subject to 2
1 2
1 2
1 2
1 2
10
2 5 60
18
3 44
, 0
x
x x
x x
x x
x x
[1]
[2]
(5,10)
[3]
(10,8)
[4]
(13,5)
(14.6,0)
(0,10)
z is maximum at (13, 5)
Max z = 31
z = 4
z = 10
z = 20
z = 28
z = 31
Exceptional Cases
Usually a LPP will have a unique optimal
solution. But there are problems where
there may be no solution, may have
alternative optimum solutions and
“unbounded” solutions. We graphically
explain these cases in the following slides.
We note that the (unique) optimum
solution occurs at one of the “corners” of
the set of all feasible points.
Alternative optimal solutions
1 210 5z x x
Subject to the constraints 1
2
1 2
1 2
150
200
2 400
, 0
x
x
x x
x x
Consider the LPP
Maximize
Graphical solution
z=2000
z=1500
z=400
z=1000
(100,200)
(150,100)
(150,0)
(0,200)
x1
x2
Maximize z=10x1+5x2
Subject to the constraints
2x1+x2 400
x1 150
x2 200
x1,x2 0
z maximum
=2000 at
z=600
Thus we see that the objective function z
is maximum at the corner (150,200) and
also has an alternative optimum solution at
the corner (100,200). It may also be noted
that z is maximum at each point of the line
segment joining them. Thus the problem
has an infinite number of (finite) optimum
solutions. This happens when the objective
function is “parallel” to one of the
constraint equations.
Maximize z = 5x1 + 7 x2
Subject to
2 x1 - x2 ≤ -1
- x1 + 2 x2 ≤ -1
x1, x2 ≥ 0
No feasible solution
Maximize z = x1 + x2
Subject to
- x1 + 3 x2 ≥ 30
- 3 x1 + x2 ≤ 30
x1, x2 ≥ 0
unbounded solution z=20
z=30
z=50
z=70
Let x = (x1, x2) and y = (y1, y2) be two pints in
the x1 x2 plane. Any point t = (t1, t2) on the line
segment joining them is of the form
t1 = (1 - ) x1 + y1
t2 = (1 - ) x2 + y2
for some : 0 1
= 0 corresponds the „left‟ endpoint x
= 1 corresponds the „right‟ endpoint y
More generally, let x=(x1, x2, …, xn) and
y=(y1, y2, …, yn) be two points in the n-
dimensional space Vn. The line segment
joining x and y is the set of all points of
the form t = (1 - ) x + y, for all :
0 1 in the sense that the ith
coordinate of t, namely ti is given by
ti = (1 - ) xi + yi for all i = 1, 2, …n.
Convex sets
A subset S in the n-space Vn is called
convex if x, y belong to S implies the line
segment joining them also lies in S.
convex subset NOT convex
In Vn, the “half spaces”
x=(x1, x2, …, xn) | a1x1+a2x2 + …+ anxn ≤ b
x=(x1, x2, …, xn) | a1x1+a2x2 + …+ anxn ≥ b
x=(x1, x2, …, xn) | a1x1+a2x2 + …+ anxn = b
are all convex.
and the “hyperplane”
Theorem: The intersection of any number
of convex sets is a convex set.
Corollary: The set of all feasible solutions
of an LPP is a convex subset.
Extreme Points (Vertices=corners)
Let S be a convex set. A point t in S is
called an extreme point of S if it is not
strictly between two distinct points of S.
In other words, whenever x, y are two
points in S, we cannot find a : 0 < < 1
such that t = (1 - ) x + y
Extreme point Every point on the
boundary is an Extreme
point
Extreme point
Existence of extreme points
Theorem: Let S be a nonempty, closed,
convex set that is either bounded from
above or bounded from below. Then S
has at least one extreme point.
Extreme Points and LPP
Suppose the set SF of all feasible solutions
of a LPP is nonempty and bounded. (We
already “know” it is closed and convex.)
Then the optimum solution of the LPP
occurs at a corner (=extreme point) of SF.
Proof: You may see the book by JC Pant.