Linear Programming Formulation Questions and Answers

Q: In graphical solution method of LPP, the redundant constraint is the one :

A redundant constraint is one which does not form the boundary of the feasible region.In the graphical method, the feasible region is the intersection of all the constraint half-planes. A constraint is redundant when the other constraints already imply it: its line lies wholly outside the feasible region, so removing it leaves the region unchanged and it contributes no edge.Constraints whose lines do form the boundary are binding somewhere and are not redundant. Optimising the objective is a property of points (vertices), not of constraints, so the last two options do not describe redundancy.

21.

In graphical solution method of LPP, the redundant constraint is the one :

A.Which forms the boundary of the feasible region
B.Which does not form the boundary of the feasible region
C.Which does not optimize the objective function
D.Which optimizes the objective function

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: B

A redundant constraint is one which does not form the boundary of the feasible region.

In the graphical method, the feasible region is the intersection of all the constraint half-planes. A constraint is redundant when the other constraints already imply it: its line lies wholly outside the feasible region, so removing it leaves the region unchanged and it contributes no edge.

Constraints whose lines do form the boundary are binding somewhere and are not redundant. Optimising the objective is a property of points (vertices), not of constraints, so the last two options do not describe redundancy.

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22.

In a Linear Programming Problem, suppose there are 3 basic variables and 2 non-basic variables, then the possible number of basic solutions are

A.6
B.8
C.10
D.12

Exam Year(s)

UGC NET - 2012,

Answer: C

A linear programming problem with 3 basic variables and 2 non-basic variables has 5 variables in all, so the number of constraints is 3, one for each basic variable. A basic solution is found by choosing 3 of the 5 variables as basic and setting the other two to zero, and the possible number of such choices is

\[ \binom{5}{3} = \frac{5!}{3!\,2!} = 10 \]

The options 6, 8 and 12 do not equal this count. The keyed answer is 10.

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23.

If dual has an unbounded solution, then its corresponding primal has

A.no feasible solution
B.unbounded solution
C.feasible solution
D.none of these

Exam Year(s)

UGC NET - 2012,

Answer: A

If the dual problem has an unbounded solution, then the primal problem has no feasible solution. By the duality theorem, a feasible primal with a bounded objective gives a dual with the same optimum, and if one problem is unbounded, the other cannot have any feasible point at all. A primal with an unbounded solution would need the dual to be infeasible, which is the converse, and a feasible primal would force a bounded dual. The keyed answer is no feasible solution.

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24.

The feasible region represented by the constraints \(x_{1} - x_{2} \le 1\), \(x_{1} + x_{2} \ge 3\), \(x_{1} \ge 0\), \(x_{2} \ge 0\) of the objective function Max \(Z = 3x_{1} + 2x_{2}\) is

A.A polygon
B.Unbounded feasible region
C.A point
D.None of these

Exam Year(s)

UGC NET - 2012,

Answer: B

The constraints are \(x_{1} - x_{2} \le 1\), \(x_{1} + x_{2} \ge 3\) and both variables non-negative. Along the line \(x_{1} - x_{2} = 1\), the point \((x_{2} + 1, x_{2})\) satisfies the second constraint once \(x_{2} \ge 1\) and can be taken as large as one pleases, so the feasible region extends without limit and no constraint closes it. It is therefore an unbounded feasible region, option (B), not a polygon or a single point, which are bounded.

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