Some Special Cases of Graphical Solution Methods of Lp Problems
Some Special Cases of Graphical Solution Methods of
Lp Problems
Case-1: No Feasible Solution space
Consider the following LPP
Max Z= 4X1 + 4 X2
Subject to
2X1 + 2X2 >= 12
3X1 + 3X2
<= 9
X1, X2
>= 0
The
solution is arrived through the graphical method; various steps are given
below.
We plot the first constraint in the graphical plane
Consider 2X1 + 2X2 = 12 If X1 = 0, then X2 = 6 Therefore, one point on the line
is (0, 6) If X2 =0, then X1 = 6 Therefore another point on the
line is (6, 0) We will
fix with the help of these two points in the graphical plane, which is shown
here. Then, based on the inequality sign, the required region is marked here. In similar way, we fixed the
second constraint 3X1 + 3X2 = 9 If X1 = 0, then X2 = 3
Therefore, one point on the line is (0, 3) If X2 =0, then X1 = 3 Therefore another point on the line is (3, 0) We will fix straight line with the help of these
two points in the graphical plane, which is shown here. Then, based on the
inequality sign, the required region is marked here. Then the
combined graphical solution space is arrived and shown here in the diagram.
You can
note that there is no common solution area between these two constraints. Hence this kind of problem is
categorized as ‘No Feasible solution’. Case-2: Unbounded
Solution space Consider the following LPP Max Z= 4X1 + 4 X2 Subject to 2X1 + 2X2 >= 12 3X2 <= 9 X1, X2 >= 0 The
solution is arrived through the graphical method; various steps are given
below. We plot the first constraint in
the graphical plane
Consider 2X1 + 2X2 = 12 If X1 = 0, then X2 = 6 Therefore, one point on the line is (0, 6) If X2 =0, then X1 = 6 Therefore another point on the line is (6, 0) We will fix with the help of these two points in the
graphical plane, which is shown here. Then, based on the inequality sign, the
required region is marked here. In similar way, we fixed the second constraint; | 3X2 | = 9 |
Since, | X1 = 0, | |
then | X2 | = 3 |
Therefore, the straight line is parallel to X1 axis
and passes through X2=3 Now, we fixed the straight line with the help of
this point in the graphical plane, which is shown here. Then, based on the
inequality sign, the required region is marked here. Then the combined graphical solution space is
arrived and shown below in the diagram.
You can note that the solution area between these
two constraints is not bounded one; that is there is no finite solution space
and this has implied that the solution can be infinitely increased with the
limited resources. This is not obviously possible. Hence this kind of problem
is categorized as ‘unbounded feasible solution’. Case-3: Unbounded Solution
space-possibility of optimum solution Consider
the following LPP Min Z= 4X1 + 4 X2 Subject to 2X1 +
0X2<= 12 0X1+ 3X2
<= 9 X1, X2
>= 0 The
solution is arrived through the graphical method; various steps are given
below.
Consider the first constraint-2X1
+ <= 12 From this constraint, we consider
the following straight line equation, | 2X1 = 12 |
Since, | X2 = 0, |
then | X1 = 6 |
Therefore, the straight line is
parallel to X2 axis and passes through X1=6 Then the
required region for the constraint, 2X1 + <= 12 is deducted by substituting
an arbitrary point – say origin (0, 0) and check whether it is satisfied by the
constraint. It is shown below in the diagram. In a
similar way, the second constraint is considered; 0X1+ 3X2 <= 9 From this
constraint, we consider the following straight line equation, | 3X2 = 9 |
Since, | X1 | = 0, |
then | X2 | = 3 |
Thus, you
can sense that the straight line is parallel to X1 axis and passes through X2=3
Then the required region for the constraint, 3X2
<= 9 is deducted by substituting an arbitrary point – say origin (0, 0) and
check whether it is satisfied by the constraint. It is shown above in the diagram. In the
diagram in the right hand side, the combined graphical solution space is shown.
You can
notice that there exists an optimum solution, even though the solution space is
not bounded.
At X1=0 & X2= 2, there is
minimum value obtained for the problem. Case-4: Multiple
Optimum Solutions -possibility of more than one optimum solution Consider the following LPP. Max Z = X1 + 2X2 Subject to X1 + 2X2
≤ 10 X1 + X2 ≥
1 X2 ≤ 4 X1 &
X2 ≥0 The
solution is arrived through the graphical method; various steps are given
below. Consider
the first constraint X1 + 2X2 ≤ 10 We plot
the first constraint in the graphical plane;
Consider the straight line equation, | X1 + 2X2 = 10 |
If | X1 | = 0, |
then | X2 | = 5 |
Therefore, one point on the line is (0, 5) Therefore another point on the line is (10, 0) We will
fix with the help of these two points in the graphical plane, which is shown
here. Then, based on the inequality sign, the required region is marked here.
Consider the second constraint,
X1 + X2 ≥ 1 Now the
respective straight line equation is considered for fixing the line. | X1 + X2 = 1 |
If | X1 | = 0, |
then | X2 | = 1 |
Therefore, one point on the line
is (0, 1) Therefore another point on the
line is (1, 0). We will
fix with the help of these two points in the graphical plane, which is shown
here. To fix the required region, we take arbitrary point, (0, 0) and
substituting the values, in the second constraint, we find that it is not
satisfying the inequality. Therefore, the region, above the straight line forms
the required region.
In a similar way, the third constraint is also
placed in the graphical plane and the combined picture of all the straight
lines is also placed below with the solution space. You can notice that the present problem, at two
points, there is a possibility of getting maximum value for the objective
function. At X1= 2
& X2 = 4, the objective function value, Z = 10 And X1 = 10 & X2=0,
again the objective function value, Z = 10 These
kinds of problems are called as linear programming problem with multiple optimum
solutions.
Tags : Operations Management - Introduction to Operations Research
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