The linear programming problems (LPP) discussed in the previous section possessed unique solutions. This was because the optimal value occurred at one of the extreme points (corner points). But situations may arise, when the optimal solution obtained is not unique.
This case may arise when the line representing the objective function is parallel to one of the lines bounding the feasible region. The presence of multiple solutions is illustrated through the following graphical method example.
Maximize z = x1 + 2x2
subject to
x1 ≤ 80
x2 ≤ 60
5x1 + 6x2 ≤ 600
x1 + 2x2 ≤ 160
x1, x2 ≥ 0.
In the above figure, there is no unique outer most corner cut by the objective function line. All points from P to Q lying on line PQ represent optimal solutions and all these will give the same optimal value (maximum profit) of Rs. 160. This is indicated by the fact that both the points P with co-ordinates (40, 60) and Q with co-ordinates (60, 50) are on the line x1 + 2x2 = 160. Thus, every point on the line PQ maximizes the value of the objective function and the problem has multiple solutions.
In some cases, there is no feasible solution area, i.e., there are no points that satisfy all constraints of the problem. An infeasible LP problem with two decision variables can be identified through its graph. For example, let us consider the following linear programming problem (LPP).
Minimize z = 200x1 + 300x2
subject to
2x1 + 3x2 ≥
1200
x1 + x2 ≤ 400
2x1 + 1.5x2 ≥
900
x1, x2 ≥ 0
The region located on the right of PQR includes all solutions, which satisfy the first and the third constraints. The region located on the left of ST includes all solutions, which satisfy the second constraint. Thus, the problem is infeasible because there is no set of points that satisfy all the three constraints.
It is a solution whose objective function is infinite. If the feasible region is unbounded then one or more decision variables will increase indefinitely without violating feasibility, and the value of the objective function can be made arbitrarily large. Consider the following model:
Minimize z = 40x1 + 60x2
subject to
2x1 + x2 ≥
70
x1 + x2 ≥ 40
x1 + 3x2 ≥ 90
x1, x2 ≥ 0
The point (x1, x2) must be somewhere in the solution space as shown in the figure by shaded portion.
The three extreme points (corner points) in the finite plane are:
P = (90, 0); Q = (24, 22) and R = (0, 70)
The values of the objective function at these extreme points are:
Z(P) = 3600, Z(Q) = 2280 and Z(R) = 4200
In this case, no maximum of the objective function exists because the region has no boundary for increasing values of x1 and x2. Thus, it is not possible to maximize the objective function in this case and the solution is unbounded.
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