Maths M.C.Q (1 Marks)

2. The quantity in column B is greater.

2. a function to be optimized

Solution:

The objective function of a linear programming problem is either to be maximized or minimized i.e. objective function is to be optimized.

1. Silver WagonR

Solution:

The number of silver Vento car = 37 (from the table)

Twice the number of silver Vento cars = 2 × 37 = 74

Now from table we can see that silver WagonR is only car type having 74 cars

2. open half plane not containing the origin

Solution:

On putting x = 0, y = 0 in the given inequality, we get 0 > 5, which is absurd.

Therefore, the solution set of the given inequality does not include the origin.

Thus, the solution set of the given inequality consists of the open half plane not containing the origin.

5. Understand the managerial problem being faced.

Solution:

The first step in formulating an linear programming problem is to understand the managerial problem being faced i.e., determine the quantities that are needed to solve the problem.

2. Capital budgeting models

Solution:

In linear programming, Capital budgeting models to minimize the total capital cost. 

The solutions from the model are used to decide the best combination of capital resources and best times to start and finish projects and to determine the net capital cost.

4. The dual problem might have an optimal solution, even though the primal has no (bounded) optimum.

Solution:

If one of the problems (primal, dual) is infeasible then the other problem is infeasible.

2. A function to be optimized

Solution:

The objective of Linear Programming Problems (LPP) is to minimize or maximize the function.

1. Single objective function

Solution:

The problem associated with LLP is single objective.

2. Transportation models

Solution:

In linear programming, transportation model are applied to problems related to the study of efficient transportation routes.

For oil companies, how effectively the available resources are transported to different destinations with minimum cost.

1. Half plane not containing the origin

3. If a LPP admits two optimal solution it has an infinite number of optimal solutions

Solution:

Optimal solution of LPP has three types.

1. Unique
2. Infinite
3. Does not exist.

Hence, it has infinite solution if it admits two optimal solution.

3. given by corner points of the feasible region

Solution:

It is known that the optimal value of the objective function is attained at any of the corner point.

Thus, the potimal value of the objective function is attined at the points given by corner points of the feasible region.

4. Any point on the line segment joining the points (0, 2) and (3, 0).

Solution:

3. The constraints are linear equations or inequalities.

Solution:

The graph of the linear equation is a straight line.

If the terminal points are connected by a straight line then the given constraints are linear equations which may include inequalities.

4. All of the above.

Solution:

From the primal - dual relationship, The shadow prices of resources in the primal are optimal values of the dual variables.

If one of the problems has an optimal feasible solution then the other problem also has an optimal feasible solution.

The optimal objective function value is same for both primal and dual problems.

If one problem has an unbounded solution then the other problem is infeasible.

1. Subset of mathematical programming

Solution:

Linear programming is an extremely powerful tool for addressing a wide range of applied optimization problems.

A short list of application areas is resource allocation, production scheduling, warehousing, layout, transportation scheduling, facility location, flight crew scheduling, portfolio optimization, parameter estimation.

So, linear programming is used to subset mathematical programming.

3. At an infinite number of points.

2. The quantity in column B is greater.

Solution:

Hence, maxmimum value of Z = 300 < 325

So, the quantity in column B is greater.

2. The value of the objective function for a maximization problem will likely be less than that for the simplex solution.

Solution:

Solving an integer programming problem by rounding off answers obtained by solving it as a linear programming problem, we find that the value of the objective function for a maximization problem will likely be less than that for the simplex solution.

3. Corner points of the feasible region

Solution:

Any point in the feasible region that gives the optimal value (maximum or minimum) of the objective function is called an optimal solution.

1. The primal and dual have equal number of variables.

Solution:

The number of variables in dual is equal to the number of constraints in the primal and the number of variables in primal is equal to the number of constraints in the dual.

Therefore, the primal and dual doesnt have equal number of variables.

4. Only firstquadrant.

4. Determine how optimal solution to LPP changes in response to problem inputs.

Solution:

A sensitivity analysis is performed to determine the sensitivity of the solution to changes in parameters.

1. {x : |x| = 5}

Solution:

|x| = 5 is not a convex set as any two points from negative and positive x-axis if are joined will not lie in set.

1. The problems is to be re - evaluated.

3. bounded in first quadrant

Solution:

Converting the given inequations into equations, we obtain

y = 6, x + y = 3, x = 0 and y = 0

y = 6 is the line passing through (0, 6) and parallel to the X axis.

The region below the line y = 6 will satisfy the given inequation.

The line x + y = 3 meets the coordinate axis at A(3, 0) and B(0, 3).

Join these points to obtain the line x + y = 3

Clearly, (0, 0) satisfies the inequation x + y ≤ 3.

So, the region in xy-plane that contains the origin represents the solution set of the given equation.

Region represented by x ≥ 0 and y ≥ 0:

Since, every point in the first quadrant satisfies these inequations.

So, the first quadrant is the region represented by the inequations.

These lines are drawn using a suitable scale.

The shaded region represents the feasible region of the given LPP, which is bounded in the first quadrant.

3. Atleast one of the corner points.

Solution:

Let Z = ax + by be the objective function

When Z has optimum value(maximum or minimum), where the variables

x and y are subject to constraints described by linear inequalities, this optimum value must occur at a corner points of the feasible region.

Thus, the function attains its optimum value at one of the corner points.

2. (0, 8)

2. 336 at (6, 4)

4. q = 3p

2. 60 at (4, 2)

2. 600

4. Has been formulated improperly.

Solution:

A linear programming problem is said to have unbounded solution if it has infinite number of solutions.

I.e., the problem has been formulated improperly.

4. None of the above

Solution:

If an iso profit line which is yielding the optimal solution coincide with a constant line; then

→ the solution will b bounded, i.e there will be a definite bounded area where the solution would be optional.

→ Since the area is bounded,the solution is feasible

→ And the constant which coincides is not a redundant

Hence None of above is the answer.

3. Convex Polygon

Solution:

A bounded feasible region will have both a maximum value and a minimum value for the objective function. It is bounded if it can be enclosed in any shape.

A convex polygon is a simple not self-intersecting closed shape in which no line segment between two points on the boundary ever goes outside the polygon.

Hence, the answer is convex polygon.

5. Unlimited

Solution:

There is no limit on constraints allowed in linear programming.

so the number of constraints is unlimited.

2. x₁ = 2, x₂ = 6, Z = 36

Solution:

We need to maximize the function Z = 3x₄ + 5x₂

First, we will convert the given inequations into equations, we obtain the following equations:

3x₁ + 2x₂ = 18, x₁ = 4, x₂ = 6, x₁ = 0 and x₂ = 0

Region represented by 3x₁ + 2x₂ ≤ 18:

The line 3x₁ + 2x₂ = 18 meets the coordinate axes at A(6, 0) and B(0, 9) respectively.

By joining these points we obtain the line 3X1 + 2x2 = 18.

Clearly (0, 0) satisfies the inequation 3x₁ + 2x₂ = 18.

So the region in the plane which contain the origin represents the solution set of the inequation 3x₁ + 2x₂ ≤ 18.

Region represented by x₁ ≤ 4:

The line x₁ = 4 is the line that passes through C(4, 0) and is parallel to the Y axis.

The region to the left of the line x₁ = 4 will satisfy the inequation x₁ ≤ 4.

Region represented by x₂ ≤ 6:

The line x₂ = 6 is the line that passes through D(0, 6) and is parallel to the X axis.

The region below the line x₂ = 6 will satisfy the inequation X₂ ≤ 6.

Region represented by x₁ ≥ 0 and x₂ ≥ 0:

Since, every point in the first quadrant satisfies these inequations.

So, the first quadrant is the region represented by the inequations x₁ ≥ 0 and x₂ ≥ 0.

The feasible region determined by the system of constraints, 3x₁ + 2x₂ ≤ 18, x₁ ≤ 4, x₂ ≤ 6, x₁ ≥ 0 and x₂ ≥ 0 are as follows

Corner points are O(0, 0), D(0, 6), F(2, 6), E(4, 3) and C(4, 0).

The values of the objective function at these points are given in the following table.

We see that the maximum value of the objective function Z is 36 which is at F(2, 6).

1. Feasible solutions

Solution:

In order for a linear programming problem to have a unique solution, the solution must exist at the intersection of two or more constraints.

Then the problem becomes convex and has a single optimum(maximum or minimum) solution.

Therefore the convex set of equations is included in the feasible region.

2. Corner points

Solution:

The fundamental theorem of programming (i.e., Convex Polygon Theorem) states that the optimum value(maximum or minimum) of a linear programming problem over a convex region occur at the corner points.

2. -16

Solution:

the feasible region as show in the figure, has objective function F= 3x - 4y

We have minimum value of F is -16at (0, 4).

1. Have no feasible solution

Solution:

If there is no point in the feasible set, there is no feasible solution of the linear programming model.

In linear programming, lack of points for a solution set is said to have no feasible solution.

4. q = 3p

Solution:

The maximum value of Z is unique.

It is given that the maximum value of Z occurs at two points (3, 4) and (0,5).

Value of Z at (3, 4) = Value of Z at (0,5)

= p(3) + q(4) = p(0) + 7(5)

= 3p + 4q = 5q

= q = 3p

3. at any vertex of feasible region

Solution:

In linear programming problem we substitute the coordinates of vertices of feasible region in the objective function and then we obtain the maximum or minimum value.

Therefore, the value of objective function is maximum under linear constraints at any vertex of feasible region.

1. It cannot be determined in a graphical solution of an LPP. 

Solution:

There are various methods to solve the linear programming problems namely simplex method, ellipsoid method, graphical method, interior points method, etc.

Therefore a linear programming problem can be solved using the graphical method.

Hence, the feasibility of the linear programming problem can be determined by the graphical method.