While obtaining the optimal solution to the LP problem by the graphical method, the statement of the following theorems of linear programming is used.
GRAPHICAL SOLUTION PROCEDURE OF
LP PROBLEMS
While obtaining the optimal solution to the LP
problem by the graphical method, the statement of the following theorems of
linear programming is used.
1. The
collection of all feasible solutions to an LP problem constitutes a convex set
whose extreme points correspond to the basic feasible solutions.
2. There are
a finite number of basic feasible solutions within the feasible solution space.
3. If the
convex set of the feasible solutions of the system Ax=b, x>=0, is a convex
polyhedron, then at least one of the extreme points gives an optimal solution.
4. If the
optimal solution occurs at more than one extreme point, then the value of the
objective function will be the same for all convex combinations of these
extreme points.
METHOD-1: Extreme Point Enumeration Approach
This
solution method for an LP problem is divided into five steps.
Step 1
State the given problem in the
mathematical form.
Step 2
Graph the constraints, by temporarily ignoring the
inequality sign and decide about the area of feasible solutions according to
the inequality sign of the constraints. Indicate the area
of feasible solutions by a shaded area, which forms a convex polyhedron.
Step 3
Determine
the coordinates of the extreme points of the feasible solution space.
Step 4
Evaluate
the value of the objective function at each extreme point.
Step 5
Determine
the extreme point to obtain the optimum (best) value of the objective function.
Example-1
Use the
graphical method to solve the following LPP problem:
Maximize z = 15X1 + 10X2
Subject to constraints
4X1+6X2 <=360
3X1+0X2<=180
0X1+5X2 <=200
X1, X2>=0
Solution
Step 1
State the problem in mathematical
form.
The
problem we have considered here is already in mathematical form [given in terms
of mathematical symbols and notations]
Step 2
Plot the constraints on the graph paper and find
the feasible region.
We shall treat x1 as the horizontal axis and x2 as
the vertical axis. Each constraint will be plotted on the graph by treating it
as a linear equation and then appropriate inequality conditions will be used to
mark the area of the feasible solutions.
Consider
the first constraint 4X1+6X2<=360 .Treated it as the equation,
This equation indicates that when it is plotted,
the graph cuts X1 axis [intercept] at 90 and X2 axis [x2 intercept] at 60.
These two points are then connected by a straight line as shown in figure. But the inequality and
non-negativity condition can only be satisfied by the shaded area (feasible
region) as shown in figure below:
Similarly
the constraints 3X1<=180 and 5X2<=200 are also plotted on the same graph,
which is shown above and the required region is indicated by the shaded area,
which is common to all the 3 constraint as shown in figure below:
Since all constraints have been graphed, the area
which is bounded by all the constraints lines including all the boundary points
is called the feasible region or solution space. The feasible region is shown
in fig by the area OABCD. Step 3 Determine the coordinates of
extreme points Since the
optimal value of the objective function occurs at one of the extreme points of
the feasible region, it is necessary to determine their coordinates. The
coordinates of extreme points of the feasible region are: O = (0, 0), A= (60, 0) B= (60, 20), C= (30, 40), D= (0, 40) Step 4 Evaluate the value of the
objective at extreme points Now, the next step is to evaluate the objective
function value at each extreme point of the feasible region, which is shown
below:
Step 5 Determining the optimal value of the objective
function Since our desire it to maximize the value of the objective function,
Z, therefore from step 4 we can conclude that maximum value of Z=1100 is
achieved at the point B (60, 20). Hence the optimal solution to the given LP
problem is: X1=60, X2=20 and Value of the objective function,
Z=1100 Example-2: Use the
graphical method to solve the following LP
problem Minimize Z=20X1+10X2 Subject to constraints,
X1+2X2 ≤
40 3X1+X2 ≥
30 4X1+3X2 ≥
60 And X1,
X2 ≥ 0 Solution Step 1 State the problem in mathematical
form. The
problem we have considered here is already in mathematical form [given in terms
of mathematical symbols and notations] Step 2 Plot the constraints on the graph paper and find
the feasible region. We shall
treat x1 as the horizontal axis and x2 as the vertical axis. Each constraint
will be plotted on the graph by treating it as a linear equation and then
appropriate inequality conditions will be used to mark the area of the feasible
solutions. The constraints are written in
the following form; X1/40 + X2/(40/2) ≤ 1 or | X1/40 +
X2/20 ≤ 1----- | (1) |
X1/(30/3) + X2/3 ≥ 1 or | X1/10 +
X2/30 ≤ 1---- | (2) |
X1/(60/4) + X2/(60/3) ≥ 1 or X1/15 + X2/20 ≥
1-------------- | (3) |
Graph each constraint by first treating it as a
linear equation. Then use the inequality condition of each constraint to make
the feasible region as shown in following figure.
Step 3 Determine the coordinates of
extreme points Since the
optimal value of the objective function occurs at one of the extreme points of
the feasible region, it is necessary to determine their coordinates. The
coordinates of extreme points of the feasible region are: A= (15, 0), B= (40, 0),C= (4, 18)
and D= (6, 12). Step 4 Evaluate
the value of the objective at extreme points Now, the next step is to evaluate
the objective function value at each extreme point of the feasible region,
which is shown below:
Therefore,
the minimum value of the objective function Z=240 occurs at the extreme point D
(6, 12).Hence, the optimal solution to the given LP problem
is: X1 = 6, X2 = 12 And the value of the objective function is Z = 240.
Tags : Operations Management - Introduction to Operations Research
Last 30 days 4826 views