Graphical Solution Method
When there are two decision variables (let they are x1 and x2) in an LP model the feasible
solution space (FSS) of the model can be shown in a plane (2-dimensional space). One axis is
x1 axis and the other is x2 axis.
Example 1:
Max z = 3x1 + x2
s.t.
x1 <= 4
x1 + x2 <= 7
x1>=0
x2>=0
Our empty plane is:
Then we consider our constraints one by one. Let us consider first constraint: x1<=4.
We should determine the part of our plane that satisfies this constraint. First draw the x1=4
line. This line is drawn in red below.
The red line is the x1=4 line. But, according to our first constraint (x1<=4) the points having
smaller x1 values than 4 are also valid. So we should select the side of the line x1=4 that
satisfies the constraint x1<=4. The invalid side is scrabbled with red dashes. The valid side is
the clean side.
Now, let’s consider our second constraint: x1+x2<=7. First we should draw the line:
x1 + x2 = 7. The line is drawn in blue and the side of the line which is invalid according to the
constrain is scrabbled. The remaining clean part of our plane is the solutions satisfying the
first and second constraints simultaneously.
Now, let’s consider our third constraint: x1>=0. The invalid part is scrabbled below.
Now, the last constraint (x2>=0) is considered and the invalid side is scrabbled below.
The final clean part, the region OABC, is the feasible solution space (FSS) of the model. The
FSS is shown more clearly in the following figure.
FSS
The last work is searching the optimal solution in this FSS region. Our objective function is:
Max z = 3x1 + x2. Let’s draw the line 3x1 + x2 = 3 (i.e., z = 3). The intersection points of the
FSS and 3x1 + x2 = 3 line are the feasible solution with objective function value 3.
Let’s draw another line with a higher z value, for example the line 3x1 + x2 = 6 (i.e., z = 6).
The intersection points of the FSS and 3x1 + x2 = 6 line are the feasible solution with
objective function value 6. When we move the objective function in the direction shown
below the objective function value increases. So, in order to find the feasible solution with the
maximum z value (the optimal solution), we should move the objective function line in this
direction as much as possible.
increasing
direction
Corner point B is the last feasible solution that touches the objective function line when we
move the line in the increasing direction. So, after that point we don’t have any other feasible
solution in the increasing direction of the objective function. Then, corner point B is the
OPTIMAL solution of the model.
increasing
direction
We should determine the coordinates of the corner point B. Point B is the intersection point of
the constraint lines x1=4 and x1 + x2 = 7. We can use these equations and find the
coordinates of point B. We have two unknowns and two equations.
x1 = 4
x1 + x2 = 7
Solution is: x1=4, x2=3.
At this point B(4,3) the optimal objective function value is:
z = 3x1 + x2 = 3*4 + 3 = 15.
Example 2: Search the optimal solution of the following LP model by the graphical solution
method.
Min z = x1 + x2
s.t.
-x1 + x2 <= 2
x1 + x2 >= 4
x2 >= 2
x1, x2 >= 0
The FSS of the model is depicted below. Notice that the right hand side of the FSS is open
ended.
A
FSS
B
The z = x1 + x2 = 7 line and the decreasing direction of the objective function is shown
below.
decreasing
direction
A
B
In order to find the feasible solution(s) with the minimum z value (the optimal solution), we
should move the objective function line in this direction as much as possible. When we move
the objective function in its decreasing direction just before it leaves the FSS it coincides with
a surface of the FSS (the line part between points A and B). At this moment the value of z is 4
(i.e, z = x1 + x2 = 4). So, in this example we have alternative optimal solutions. All the
feasible solutions on this surface (including corner points A and B) are alternate optimal
solutions.
A
B
Example 3: Search the optimal solution of the following LP model by the graphical solution
method.
Max z = x1 + x2
s.t.
-x1 + x2 <= 2
x1 + x2 >= 4
x2 >= 2
x1, x2 >= 0
This example is almost same with the Example 2. The only difference is that in this example
we want to maximize the objective function. So, the FSS and slope of the objective function is
same. Now, we should consider its increasing direction. It is shown below.
increasing
direction
A
B
In order to find the feasible solution(s) with the maximum z value (the optimal solution), we
should move the objective function line in this direction as much as possible. Since, the FSS
is open ended in this direction always we can find a better feasible solution then our current
solution. So, we cannot determine any specific solution as our optimal solution. In this case it
is said that the model has unbounded solution.
Example 4: Search the optimal solution of the following LP model by the graphical solution
method.
Max z = x1 + x2
s.t.
-x1 + x2 <= 2
x1 + x2 <= 1
x2 >= 4
x1, x2 >= 0
When you try to determine the FSS you will see that the entire plane will be cancelled and no
FSS will be remained. In this example the FSS is an empty set. The answer is infeasible
solution. (Hint: Last three constraints cannot be satisfied at the same time)