Linear Programming MCQ Quiz in मल्याळम - Objective Question with Answer for Linear Programming - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Explanation:

Simplex method:

• The simplex method is the most popular method used for the solution of Linear Programming Problems (LPP).
• The Simplex method is a search procedure that shifts through the set of basic feasible solutions, one at a time until the optimal basic feasible solution is identified.
• It can be used for two or more variables as well (always advisable for more than two variables to avoid lengthy graphical procedure).
• The simplex method is not used to examine all the feasible solutions.
• It deals only with a small and unique set of feasible solutions, the set of vertex points (i.e. extreme points/corner points) of the convex feasible space that contain the optimal solution.
• All the resource values or constraints should be non-negative.
• All the inequalities of the constraint should be converted to equalities with the help of slack or surplus variables.

Explanation:

Simplex Method:

• Simplex Method is a standard technique of solving linear programming problems for an optimized solution, typically involving a function and several constraints expressed as inequalities.
• The inequalities define a polygonal region and the solution is typically at one of the verticles.

Some Special Conditions of the Simplex Method:

1. Unbounded Solution: In the simplex method, if in the pivot column all the entries are negative or zero when choosing to leave the variable then the solution is unbounded.

2. Infeasible Solution: In the simplex method, if artificial variables are present in the basis, then the solution obtained is infeasible.

3. Degenerated Solution: In the simplex method, if some of the values in the constant column are zero, then the solution becomes degenerated.

4. Multiple Solution: In the simplex method, if the non-basic variable in the final simplex table showing the optimal solution to a problem, has a net zero contribution, then the condition of multiple optimal solutions arises.

Explanation:

After equating each constraint to zero (0) we get the equations of lines on cartesian coordinate.

On comparing the inequality with (0,0) and shading the common area we get the feasible region as follows,

Hence we can easily conclude that point (60,0) and (0,100) are outside the feasible region.

Additional Information

Points outside the feasible region do not contribute to the objective function.

Concept:

Convert the inequality constraints into equations and find the common points of the bounded region.

Calculations:

Given, LPP:

Max. Z = -0.1 x₁ + 0.5 x₂

2x₁ + 5x₂ ≤ 80

x₁ + x₂ ≤ 20

x₁, x₂ ≥ 0

Convert the inequality constraints into equations, we have

2x₁ + 5x₂ = 80

x₁ + x₂ = 20

2x₁ + 5x₂ = 80 passes through the point (0, 16) and (40, 0)

x₁ + x₂ = 20 passes through the point (0, 20) and (20, 0).

Now, the co ordinates of the point A = (0, 16), B = (20, 0) and C = \(\left(\dfrac{20}{3}, \dfrac{40}{3}\right)\)

Here, maximum value occurs at (0, 16)

For given

Max. Z = -0.1 x₁ + 0.5 x₂

2x₁ + 5x₂ ≤ 80

x₁ + x₂ ≤ 20

x₁, x₂ ≥ 0

to get the optimum solution, the values of x₁, x₂ are (0, 16)

Concept:

Draw the constraints to find the feasible region:

• To draw the inequalities, first, draw the equation form of the inequalities.
• 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 (0,0) else the opposite side of (0,0).

Calculation:

Given: 

The equations are plotted in the graph

As it is clear from graph that there is no point (x₁, x₂) which can lie in both regions, so there is no solution or infeasible solution.

Concept:

In an assignment problem of size n × n, number of decision variables is n².

Calculation:

Given:

n = 6

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).​

Calculation:

Given: The shaded infeasible region is given to us.

• 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.

Explanation:

The transportation problem is about transporting a single item from a set of 'm' supply points to 'n' demand points at minimum cost. The supply and demand quantities as well as the unit cost of transportation are known.

There are two types of transportation problem:

1. Balanced

2. Unbalanced

Balanced: When the total demand is equal to the total supply.

Unbalanced: When the total demand and total supply is not equal.

For solving any transportation problem, the cost matrix should be made balanced (if unbalanced) by assigning unit transportation costs as zero.

Given:

Source supply: 80 + 30 + 40 = 150 units

Destination demand: 30 + 20 + 25 + 55 = 130 units

Since demand at destination is less than the supply by 20 units, a dummy destination of capacity 20 units is needed with unit transportation cost zero.

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.

Concept:

In a transportation problem, the number of allocations required without degeneracy can be determined using a specific formula.

A transportation problem with m" id="MathJax-Element-23-Frame" role="presentation" style="position: relative;" tabindex="0">mm $m$ supply centers and n" id="MathJax-Element-24-Frame" role="presentation" style="position: relative;" tabindex="0">nn $n$ demand centers has a non-degenerate solution if the number of allocations (basic variables) is m+n−1" id="MathJax-Element-25-Frame" role="presentation" style="position: relative;" tabindex="0">m+n−1m+n−1 $m + n - 1$ .

Calculation:

Given:

Number of supply centers, \( m = 4 \)

Number of demand centers, \( n = 5 \)

The formula for non-degenerate allocations is, \( m + n - 1 \)

Applying the formula:

m+n−1=4+5−1=8m+n−1=4+5−1=8" id="MathJax-Element-26-Frame" role="presentation" style="text-align: center; position: relative;" tabindex="0">m+n−1=4+5−1=8

$$m + n - 1 = 4 + 5 - 1 = 8$$

Conclusion:

The number of allocations without degeneracy during an iteration is: 8

The correct answer is:

\( 8 \)