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Multiple or Alternative optimal Solutions
In some of the linear programming problems we face a situation that the final basic solution to the problem need not be only one, but there may be alternative or infinite basic solutions, i.e., with different product mixes, we have the same value of the objective function line (namely the profit). This case occurs when the objective function line is parallel to a binding constraint line. Then the objective function takes the same optimal value at more than one basic solution. These are called alternative basic solutions. Any weighted average of the basic optimal solutions should also yield an alternative non-basic feasible solution, which implies that the problem will have multiple or infinite number of solutions without further change in the objective function. This is illustrated in the following example.
Example Maximize Z = 3 x + 2 y Subject to x < 40 y < 60 3 x + 2 y < 180 x, y > 0
Solution:
With graphical approach as in figure it is evident that x = 20, y = 60 and x = 40, y = 30 are both basic feasible solutions and Z* = 180, Since the two straight lines representing the objective function and the third constraint are parallel, we observe that as the line of objective function is moved parallel in the solution space in order to maximize the value, this objective line will coincide with the line representing the third constraint. This indicates that not only the two points (20, 60) and (40, 30) are the only basic solutions to the linear programming problem, but all points in the entire line segment between these two extreme points are basic optimal solution. Indeed we have infinite points and hence multiple non-basic feasible solution to the linear programming problem.
The same problem is now approached through simplex method to see how the simplex method provides a clue for the existence of other optimal solutions. We introduce slack variables to convert the problem into a standard form, which is presented below:
Z - 3 x - 2 y = 0 (0) x + S 1 = 40 (1) y + S 2 = 60 (2) 3 x + 2 y + S 3 = 180 (3)
The above equations are conveniently set down in the initial table and further iterations are carried out as shown in the following tables.
Initial Table
Eqn. No.
Basic Variable
Coefficient of RHS Ratio Z x y S 1 S 2 S 3 0 - 1 -3 -2 0 0 0 0 40 1 S 1 0 1 0 1 0 0 40 40 2 S 2 0 0 1 0 1 0 60 ∞ 3 S 3 0 3 2 0 0 1 180 60
First Iteration: x enters and S 2 leaves the basis.
Eqn. No.
Basic Variable
Coefficient of RHS Z x y S 1 S 2 S 3 0 1 0 0 0 0 -2 1 180 1 x 0 1 0 0 -2/3 1/3 20 2 S 1 0 0 0 3/2 1 -1/2 30 3 y 0 0 1 0 1 0 60
Hence x = 20, y = 60, S 1 = 20, S 2 = S 3 = 0 is another optimal basic feasible solution. Still another iteration can also be performed, as the current objective row has a zero coefficient for a non-basic variable. So we conclude that we have only two optimal basic feasible solutions for the problem. However, any weighted average of optimal solutions must also be optimal non-basic feasible solutions. Indeed there are infinite numbers of such solutions, corresponding to the points on the line segment between the two optimal extreme points.
The same idea can also be checked by counting the number of zero coefficients in the objective row in the optimal table. In all linear programming problems employing simplex method, there will be as many zero coefficients in the objective row as there are basic variables. But if we find that the optimal table contains more zero coefficients in the objective row than the number of variables in the basis, this is a clear indication that there will be yet another optimal basic feasible solution.
Non-existing feasible solution
This happens when there is no point in the solution space satisfying all the constraints. In this case the constraints may be contradictory or there may be inconsistencies among the constraints. Thus the feasible solution space is empty and the problem has no feasible graphical approach and then by the simplex method.
Example (no feasible solution)
Maximize Z = 3 x = 4 y Subject to 2 x + y < 4 x + 2 y > 12 x , y > 0 Solution: Introduce slack variables and artificial variable (for the constraint of the type > ).
We have the standard form as
Z - 3 x - 4 y = 0 2 x + y + S 1 = 4 x + 2 y - S 2 + A = 12
Since artificial variable is introduced in the last constraint, a penalty of + MA is added to the L.H.S of the objective row. Hence the objective function equation becomes, Z - 3 x - 4 y + MA = 0
Again the objective function should not contain the coefficients of basic variable. Thus we multiply the last constraint with ( -M ) and add to the above equation. Thus we have,
Z - 3 x - Mx - 4 y - 2 My + MS 2 = -12 M as the objective function equation.
Starting Table Eq. No.
Basic Variable
Coefficient of RHS Ratio Z x y S 1 S 2 A 0 - 1 - M -3 -2 M -4 0 M 0 -12 M 1 S 1 0 2 1 1 0 0 4 4 2 A 0 1 2 0 -1 1 12 6
First Iteration: y enters and S 1 leaves the basis.
Eq. No.
Basic Variable
Coefficient of RHS Z x y S 1 S 2 A 0 - 1 - 3 M+
0 2 M +4 M 0 -4 M+ 16
1 y 0 2 1 1 0 0 4 2 A 0 -3 0 -2 -1 1 4
The last iteration reveals that we cannot further proceed with maximization as there is no negative coefficient in the objective function row, but the final solution has the artificial variable in the basis with value at a positive level (equal to 4). This is the indication that the second constraint is violated and hence the problem has no feasible optimal solution.
Therefore a linear programming problem has no feasible optimal solution if an artificial variable appears in the basis in the optimal table.
Unrestricted Variables
Eq. No.
Basic Variable
Coefficient of RHS Ratio Z m n y S 1 S 2 S 3 0 - 1 -2 2 -5 0 0 0 0 1 S 1 0 1 -1 0 1 0 0 4 ∞ 2 S 2 0 0 0 1 0 1 0 3 3 3 S 3 0 1 -1 1 0 0 1 6 6
First Iteration: y enters and S 2 leaves the basis.
Eq. No.
Basic Variable
Coefficient of RHS Ratio Z m n y S 1 S 2 S 3 0 - 1 -2 2 0 0 5 0 15 1 S 1 0 1 -1 0 1 0 0 4 4 2 y 0 0 0 1 0 1 0 3 ∞ 3 S 3 0 1 -1 0 0 -1 1 3 3
Second Iteration: m enters and S 3 leaves the basis.
Eq. No.
Basic Variable
Coefficient of RHS Z m n y S 1 S 2 S 3 0 - 1 0 0 0 0 3 2 21* 1 S 1 0 0 0 0 0 1 -1 1 2 y 0 0 0 1 0 1 0 3 3 m 0 1 -1 0 0 -1 1 3
Therefore the solution to the original problem will be
Z* = 21
x = m - n = 3 y = 3 S 1 = 1 and S 2 = S 3 = 0
If the original problem has more than one variable which is unrestricted in sign, then the procedure systematized by replacing each such unrestricted variable x i by x j = m j - n where m j > 0 and n > 0, as before, but n is the same variable for all relevant j.
REVIEW QUESTIONS
Solve the Following Problems by the Simplex Method
- Minimize Z = 2 x + y Subject to x + 4 y < 1 x + 2 y > 4 x , y > 0
- Maximize Z = 4 x 1 + 2 x 2 + x 3 Subject to x 1 + x 2 < 1 x 1 + x 3 < 1 x 1 , x 2 , x 3 > 0
- Maximize Z = 4x 1 + 3x Subject to 4 x 1 + 2 x 2 < 10 6 x 1 + 8 x 2 < 24 x 1 > 0, x 2 > 1.
- Maximize Z = 3 x 1 + x 2 Subject to 2 x 1 + x 2 > 4 x 2 > 2
- Maximize Z = 2 x 1 + 3 x 2 + 5 x 3 Subject to 3 x 1 + 10 x 2 + 5 x 3 15 33 x 1 - 10 x 2 + 9 x 3 < 33 x 1 + 2 x 2 + x 3 > 4 x 1 , x 2 , x 3 > 0 6. Use the simplex method to demonstrate that the following problem has an unbounded optimal solution.
Maximize Z = 4 x 1 + x 2 + 3 x 3 + 5 x 4
Subject to -4 x 1 + 6 x 2 + 5 x 3 - 4 x 4 < 20 3 x 1 - 2 x 2 + 4 x 3 + x 4 < 10 8 x 1 - 3 x 2 + 3 x 3 + 2 x 4 < 20 x 1 , x 2 , x 3 , x 4 > 0
- How do you identify multiple solutions in a linear programming problem using simplex procedure?
b) Maximize Z = 3 x 1 + 2 x 2 Subject to x 1 < 4 x 2 < 6 3 x 1 + 2 x 2 < 18 x 1 , x 2 > 0
- Maximize Z = 5 u + 6 v
Subject to 3 u + v < 1 3 u + 4 v < 0
and u and v are unrestricted.
