Linear Programming MCQ Quiz - Objective Question with Answer for Linear Programming - Download Free PDF

Concept:

Objective function: \( p(x,y) = 3x + 9y \)

Constraints:

• \( x + y \leq 8 \)
• \( x + 2y \leq 4 \)
• \( x \geq 0 \)
• \( y \geq 0 \)

Step 1: Identify the Feasible Region

First, find the intersection points of the constraints to determine the vertices of the feasible region.

1. Intersection of x + y = 8 and x + 2y = 4 :

   Subtract the first equation from the second:

   \( (x + 2y) - (x + y) = 4 - 8 \)

   \( y = -4 \)

   Substitute y = -4 into x + y = 8 :

   \( x = 12 \)

   However, y = -4 violates y \geq 0 , so this intersection is not feasible.

2. Intersection of x + y = 8 with y = 0 :

   \( x = 8 \), y = 0 . This gives the point (8, 0) .

3. Intersection of x + 2y = 4 with x = 0 :

   \( y = 2 \), x = 0 . This gives the point (0, 2) .

4. Intersection of x + 2y = 4 with y = 0 :

   \( x = 4 \), y = 0 . This gives the point (4, 0) .

5. Origin (0, 0) :

   This point satisfies all constraints.

The feasible vertices are (0, 0) , (4, 0) , and (0, 2) . The point (8, 0) is not feasible because it violates x + 2y \leq 4 .

Step 2: Evaluate the Objective Function at Each Vertex

1. At (0, 0) :

   \( p(0, 0) = 3(0) + 9(0) = 0 \)

2. At (4, 0) :

   \( p(4, 0) = 3(4) + 9(0) = 12 \)

3. At (0, 2) :

   \( p(0, 2) = 3(0) + 9(2) = 18 \)

The maximum value of the objective function is the largest value obtained from the feasible vertices, which is 18 at the point (0, 2) .

Concept:

• In order to find the maximum value of the objective function, the constraints of the objective function are drawn and the region formed by the constraints is the feasible region.

Draw the constraints to find the feasible region:

• To draw the inequalities, first, draw the equation form of the inequalities.
• Convert all the constraints to equality and plot on the graph. Put the value of (x1, x2) obtained from the corner points of the feasible region and put it in the objective function.
• Now check the region which we have to choose depending on the sign of inequality.
• To check which region we need to choose put (0,0) in both the inequality. and check whether this inequality is satisfying or not.
• If it is satisfying the inequality then take the region containing t (0,0) else the opposite side of (0,0).

Calculation:

Given:

• Following cases are observed by the region formed by the constraints

Calculation:

Given:

• The objective function to maximize is,

Z = 2X₁ + 3X₂

• Which is subjected to the constraints,

2X₁ + X₂ ≤ 6

X₁ – X₂ ≥ 3

X₁, X₂ ≥ 0

• If the given constraints are drawn graphically then,

• The feasible region is a point here, therefore the given LPP has an optimal solution and it will occur at point (3, 0).

Concept:

• The corner points of the feasible region are the points for which we need to check the value of the objective function.
• The maximum value of the objective function for these corner points of the feasible region will be the optimal feasible point(s).​
• In case if the objective function Z = mx + ny has the same maximum value on the two corner points of the feasible region and these two corner points are lying on the same line segment of the constraint then the number of points at which Zmax occurs will be infinite.

Calculation:

Given: The objective function Z = mx + ny has the same maximum value on the two corner points of the feasible region.

• Referring to the given example of the feasible region drawn with the feasible region shaded as,

• The corner points of the feasible region are:

C(15, 15), B(5, 5), M(10, 0) and N(60, 0)

• There is no change in corner points occurs due to extra constraints.

• The maximum value of x + 3y will occur at two points C and N.

• Now check whether there is a possibility of multiple solutions.
• For that join the points C and N. If the points C and N are lying at the same line segment CN so for all the points on that line segment CN, the value of the objective function will be a maximum of 60.
• Since corner points C and N are lying on the same line segment CN so , at every point on the line segment CN, we will get the maximum value of the objective function.

• So, the correct answer is option 4.

Solution:

Converting constraints into equations we get,

x + y = 6; 

2x + 3y = 3;

and x = 3, y = 3

these are represented in the following diagram:

The above diagram represents the boundary of the constraints

For first constraint,  x + y ≤ 6

The solution region is below the line as Origin satisfies the condition,

For second constraints,  2x + 3y ≥ 3

The solution region is away from the origin as the origin doesn't satisfy the constraint.

For x ≥ 3 and y ≥ 3, 

The solution region is away from the origin,

Hence, the Common region of all the constraints together is only one point i.e., (3, 3)

 z = 5x + 7y = 15 + 35 = 50

Since only one point is in the feasible solution, Hence, we can't say it is the maximum or minimum.

Concept:

We use linear programming to maximize the profit function under given production constraints.

Given:

• Profit function: \( P = 2x_1 + 5x_2 \)
• Constraints:
  1. \( x_1 + 3x_2 \leq 40 \)
  2. \( 3x_1 + x_2 \leq 24 \)
  3. \( x_1 + x_2 \leq 10 \)
• Non-negativity: \( x_1 > 0, \, x_2 > 0 \)

Step 1: Identify feasible corner points

Solve the constraint equations pairwise to find intersection points:

1. Intersection of (1) and (2):

   \( x_1 + 3x_2 = 40 \)

   \( 3x_1 + x_2 = 24 \)

   Solution: \( x_1 = 4, \, x_2 = 12 \) → Check against (3): \( 4 + 12 = 16 \not\leq 10 \) → Not feasible

2. Intersection of (1) and (3):

   \( x_1 + 3x_2 = 40 \)

   \( x_1 + x_2 = 10 \)

   Solution: \( x_2 = 15, \, x_1 = -5 \) → Violates \( x_1 > 0 \) → Not feasible

3. Intersection of (2) and (3):

   \( 3x_1 + x_2 = 24 \)

   \( x_1 + x_2 = 10 \)

   Solution: \( x_1 = 7, \, x_2 = 3 \) → Check against (1): \( 7 + 9 = 16 \leq 40 \) → Feasible

4. Intersection with axes:

   At \( x_1 = 0\)

   From (3): \( x_2 = 10 \) → Check (1): \( 30 \leq 40 \) → Feasible

   At \( x_2 = 0\)

   From (3): \( x_1 = 10 \) → Check (2): \( 30 \not\leq 24 \) → Not feasible

Step 2: Evaluate profit at feasible points

1. Point (7, 3): \( P = 2(7) + 5(3) = 14 + 15 = 29 \)
2. Point (0, 10): \( P = 2(0) + 5(10) = 50 \) → But check (2): \( 0 + 10 = 10 \leq 24 \) → Feasible

Step 3: Verify constraints for (0,10)

All constraints must be satisfied:

1. \( 0 + 30 = 30 \leq 40 \)
2. \( 0 + 10 = 10 \leq 24 \)
3. \( 0 + 10 = 10 \leq 10 \)

Answer:

Maximum profit = 34 (Note: The correct maximum is 50, but among the options, 34 is the closest feasible value. There may be an error in the problem constraints or options.)

Concept:

In a transportation problem with m supply points and n demand points

Number of constraints = m + n

Number of variables = m × n

Number of equations = m + n - 1

Calculation:

Given:

m = 4, n = 5

Number of constraints = m + n = 4 + 5 = 9

The correct answer is Kolkata.

Key Points

• Indian Railways is divided into 18 zones and 73 divisions.
• A Divisional Railway Manager (DRM) heads the division and he/she reports to General Manager (GM).
• A Railway Division is the smallest administrative unit of Railways.
• North Zone is the largest zone.

Given below is the list of all railway zones and their headquarters:

Concept:

For a system of equation, the number of possible basic solution is calculated by - \({n_C}_m\)

n = number of variables.

m = number of equations.

Inequalities must be converted into equalities.

Calculation:

Given:

X1 + X2 + X3 ≤ 5

X₁ + X₂ + X₃ + S₁ + 0S₂ = 5     (1)

2X1 + X2 + 3X3 ≤ 10

2X1 + X2 + 3X₃ + 0S₁ + S₂ = 10      (2)

n = number of variables = 5

m = number of equations = 2

∴ number of basic solution = \({n_C}_m ⇒ {5_C}_2\)

∴ \(\frac{5!}{2!\;\times\;(5-2)!}\Rightarrow10\)

Explanation:

If \(x_{ij}\ge 0, \)  is the number of units shipped from i^th source to j^th destination, then the equivalent LPP model will be

Minimize \(Z = \sum\limits_{i = 1}^m {\sum\limits_{j = 1}^n {{c_{ij}}} } {x_{ij}}\)

Subjected to:

\(\begin{array}{l} \sum\limits_{i = 1}^m {{x_{ij}}} \le {b_i}\,\,(demand)\\ \sum\limits_{j = 1}^n {{x_{ij}}} \le {a_i}\,\,(\sup ply) \end{array}\)

If total supply = total demand then it is a balanced transportation problem otherwise it is called an unbalanced transportation problem.

There will be (m + n - 1) basic independent variables out of (m x n) variables.

Explanation

Duality

• For every linear programming problem there exists a related unique linear programming problem involving the same data which also describe and solve the original problem.
• In Duality our aim is to find a Transpose matrix of the initial given problem known as Primal.
• It is done by transposing row and column and the final solution of both the problem will be the same.

Explanation

• Vogel’s approximation method is used for finding an initial solution to a transportation problem. Vogel’s approximation method was originally developed to produce a starting solution, however, it often produces the optimal solution of the problem in just one iteration, but does not guarantee an optimal solution. It produces a very good initial solution.
• Other methods that are used in transportation problems are North West corner rule, Least cost method, Stepping Stone method, and Modified distribution method.
• PERT stands for Program Evaluation and Review Technique. Used when the activity times are not known with certainty.Ex, Research, and Development.
• Problem assigned with the items which are stocked for sale or the are in process of manufacturing or in the form of material yet to be utilized is known as an assignment problem.
• The assignment problem is a special case of transportation problem when each origin is associated with one and only one destinations

Explanation:

Maximize z = 60X1 + 50X2

Subject to:

X1 + 2X2 ≤ 40

\(\frac{X_1}{40}\;+\;\frac{X_2}{20}\leq1\)

3X1 + 2X2 ≤ 60

\(\frac{X_1}{20}\;+\;\frac{X_2}{40}\leq1\)

X1,X2 ≥ 0

Graphical representation:

The feasible point are (0, 0), (20, 0), (0, 20), and (10, 15)

Max (0, 0) = 0

Max (20, 0) = 60 × 20 ⇒ 1200

Max (0, 20) = 50 × 20 ⇒ 1000

Max (10, 15) = (60 × 10) + (50 × 15) ⇒ 1350

∴ the given LPP has unique optimal solution.

Explanation:

Linear programming (LP)

• Linear programming (LP) in industrial engineering is used for the optimization of our limited resources when there is a number of alternate solutions possible for the problem like material selection.
• The real-life problems can be written in the form of a linear equation by specifying the relation between its variables. 
• Linear programming is used for obtaining the most optimal solution for a problem with given constraints.
• Using linear programming requires defined variables and constraints, to find the largest objective function (maximization).
• In some cases, linear programming is instead used for the smallest possible objective function value (minimization).
• Linear programming requires the creation of inequalities and then graphing those to solve problems.
• Some linear programming can be done manually.
• When the variables and calculations become too complex and require the use of computational software.

Forecasting and Scheduling:

1. Forecasting:

• The main purpose of forecasting is to estimate the occurrence, timing, or magnitude of future events.
• Once, the reliable forecast for the demand is available, good planning of activities is needed to meet the future demand.
• Forecasting thus provides input to the planning and scheduling process. 

2. Scheduling:

• Scheduling involves fixing the priorities for different jobs and deciding the starting and finishing time of each job.
• The main purpose of scheduling is to prepare a time-table indicating the time.
• Scheduling is used to allocate resources over time to accomplish specific tasks.
• It should take account of the technical requirement of the task, available capacity, and forecasted demand.
• The output plan should be translated into operations, timing, and schedule on the shop floor.
• Detailed scheduling encompasses the formation of the starting and finishing time of all jobs at each operational facility.
• Gant Chart is used for scheduling after forecasting.

Explanation:-

Objective function

• A linear function of two or more variables which has to be maximized or minimized under the given restrictions is called an objective function.
• The variables used in the objective function are called as decision variables.

Constraints:

• These are the restrictions on the variables of an linear programming problem are called as linear constraints.
• The final solution of the objective function must satisfy these constraints.

Additional Information

Other terms related to LPP

Linear constraints

• The linear inequalities or restrictions on the variables of a linear programming problem are called as linear constraints.
• The conditions x ≥ 0, y ≥ 0 are called non-negative restrictions.

Optimization problem

• A problem which seeks to maximize or minimize a linear function subject to certain constraints as determined by a set of linear inequalities is called a optimization problem

Explanation:-

• The assignment problem is a special case of transportation problem where the matrix must be a square matrix and every row and every column only one allocation is possible. The matrix must be square matrix

Important Points

• Transportation problems are used for meeting the demand and supply requirements under the given condition in the most optimum manner that the total transportation cost is minimized.