MCQExams
0:0:1
CBSE
JEE
NTSE
NEET
Practice
Homework
×
CBSE Questions for Class 12 Commerce Maths Linear Programming Quiz 3 - MCQExams.com
CBSE
Class 12 Commerce Maths
Linear Programming
Quiz 3
Choose the most correct of the following statements relating to primal-dual linear programming problems:
Report Question
0%
Shadow prices of resources in the primal are optimal values of the dual variables.
0%
The optimal values of the objective functions of primal and dual are the same.
0%
If the primal problem has unbounded solution, the dual problem would have infeasibility.
0%
All of the above.
Explanation
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.
To write the dual; it should be ensured that
I. All the primal variables are non-negative.
II. All the bi values are non-negative.
III. All the constraints are $$≤$$ type if it is maximization problem and $$≥$$ type if it is a minimization problem.
Report Question
0%
I and II
0%
II and III
0%
I and III
0%
I, II and III
Explanation
To write the dual, then all the primal variables must be non-negative.
All the constraints are $$\leq$$ type if it ia maximization problem and $$\geq$$ type if it is a minimization problem.
Mark the wrong statement:
Report Question
0%
The primal and dual have equal number of variables.
0%
The shadow price indicates the change in the value of the objective function, per unit increase in the value of the RHS.
0%
The shadow price of a non-binding constraint is always equal to zero.
0%
The information about shadow price of a constraint is important since it may be possible to purchase or, otherwise, acquire additional units of the concerned resource.
Explanation
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 doesn't have equal number of variables.
Which of the following statements about an LP problem and its dual is false?
Report Question
0%
If the primal and the dual both have optimal solutions, the objective function values for both problems are equal at the optimum
0%
If one of the variables in the primal has unrestricted sign, the corresponding constraint in the dual is satisfied with equality
0%
If the primal has an optimal solution, so has the dual
0%
The dual problem might have an optimal solution, even though the primal has no (bounded) optimum
Explanation
if one of the problems(primal, dual) is infeasible then the other problem is infeasible. Hence, the option D is the false statement.
LP theory states that the optimal solution to any problem will lie at
Report Question
0%
the origin.
0%
a corner point of the feasible region.
0%
the highest point of the feasible region.
0%
the lowest point in the feasible region.
0%
none of the above
Explanation
In linear programming, the optimal solution will occur at one or more corner points or on a line segment between two corner points. The corner points occurs only at the vertex of a feasible solution.
Unboundedness is usually a sign that the LP problem.
Report Question
0%
has finite multiple solutions.
0%
is degenerate.
0%
contains too many redundant constraints.
0%
has been formulated improperly.
0%
none of the above.
Explanation
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
An objective function in a linear program can be which of the following?
Report Question
0%
A maximization function
0%
A nonlinear maximization function
0%
A quadratic maximization function
0%
An uncertain quantity
0%
A divisible additive function
Explanation
linear programming problem may be defined as the problem of maximizing or minimizing a linear function subject to linear constraints.
The objective function in a linear program is a maximization function.
A point that satisfies all of a problem's constraints simultaneously is $$a(n)$$
Report Question
0%
maximum profit point.
0%
corner point.
0%
intersection of the profit line and a constraint.
0%
intersection of two or more constraints.
0%
None of the above
Explanation
The set of all the points that satisfy all of the problem's constraints is called feasible region. The points which belong to this region are feasible points.
In the above figure, the shaded blue region is the feasible region. The points in this region are feasible points which satisfy all the constraints.
Feasible region's optimal solution for a linear objective function always includes
Report Question
0%
downward point
0%
upward point
0%
corner point
0%
front point
Explanation
The vertex of the feasible solution is the corner point.
In the above figure, the blue shaded region is the feasible region and the points which highlighted in red color are the vertices of the feasible region which are called corner points.
Therefore, feasible region's optimal solution for linear objective function always includes corner point.
While plotting constraints on a graph paper, terminal points on both the axes are connected by a straight line because:
Report Question
0%
the resources are limited in supply
0%
the objective function as a linear function
0%
the constraints are linear equations or inequalities
0%
all of the above
Explanation
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.
In linear programming context, sensitivity analysis is a technique to
Report Question
0%
Allocate resources optimally.
0%
Minimize cost of operations.
0%
Spell out relation between primal and dual.
0%
Determine how optimal solution to LPP changes in response to problem inputs.
Explanation
A sensitivity analysis is performed to determine the sensitivity of the solution to changes in parameters.
Option D is correct.
Which of the following is an essential condition in a situation for linear programming to be useful?
Report Question
0%
Linear constraints
0%
Bottlenecks in the objective function
0%
Non-homogeneity
0%
Uncertainty
0%
None of the above
Explanation
For linear programming, the constraints must be linear.
Apply linear programming to this problem. A firm wants to determine how many units of each of two products (products D and E) they should produce to make the most money. The profit in the manufacture of a unit of product D is $100 and the profit in the manufacture of a unit of product E is $The firm is limited by its total available labor hours and total available machine hours. The total labor hours per week are 4,Product D takes 5 hours per unit of labor and product E takes 7 hours per unit. The total machine hours are 5,000 per week. Product D takes 9 hours per unit of machine time and product E takes 3 hours per unit. Which of the following is one of the constraints for this linear program?
Report Question
0%
$$5 D + 7 E≤ 5,000$$
0%
$$9 D + 3 E ≥4,000$$
0%
$$5 D + 7 E = 4,000$$
0%
$$5 D + 9 E ≤5,000$$
0%
$$9 D + 3 E ≤5,000$$
Explanation
Given, product D takes 5 hours per unit of labour, and
product E takes 7 hours per unit of labour.
Therefore, to produce D units of product D takes $$5D$$ hours and
to produce E units of product E takes $$7E$$ hours
Given, total labour hours per week are $$4000$$ hours.
Hence, $$5D+7E\leq 4000$$
Given, product D takes 9 hours per unit of machine time, and
product E takes 3 hours per unit of machine time.
Therefore, to produce D units of product D takes $$9D$$ hours and
to produce E units of product E takes $$3E$$ hours
Given, total machine hours per week are $$5000$$ hours.
Hence, $$9D+3E\leq 5000$$
A constraint in an LP model becomes redundant because:
Report Question
0%
two iso-profit line may be parallel to each other
0%
the solution is unbounded
0%
this constraint is not satisfied by the solution values
0%
none of the above
Explanation
A constraint in an LP model becomes redundant when the feasible region doesn't change by the removing the constraint.
For example, $$x+2y\leq 20 $$ and $$2x+4y\leq 40$$ are the constraints.
$$2x+4y\leq 40 \implies 2*(x+2y)\leq 2*20$$
$$\implies x+2y\leq 20$$ which is same as the first constraint.
Therefore
$$2x+4y\leq 40$$ can be removed. By removing this constraint feasible region doesn't change.
If two constraints do not intersect in the positive quadrant of the graph, then
Report Question
0%
The problem is infeasible
0%
The solution is unbounded
0%
One of the constraints is redundant
0%
None of the above
Explanation
Any linear programming problem must have the following properties:-
1. The relationship between variables and constraints must be linear.
2. The constraints must be non-negative.
3.. objective function must be linear.
Non-negativity conditions are used because the variables cannot take negative values. i.e., it is not possible to have negative resources(land, capital, labour cannot be negative).
Because of the non-negativity condition, the feasible region exists only in I quadrant.
In profit objective function, all lines representing same level of profit are classified as
Report Question
0%
iso-objective lines
0%
iso-function lines
0%
iso-profit lines
0%
iso-cost lines
Explanation
An iso-profit line is obtained by equating the objective function with the constant number $$a$$ which results in a linear equation. The graph of the linear equation is a straight line. This straight line represents the $$\text{iso profit line}$$. Each and every point on this iso-profit line has the same objective value $$a$$.
A line parallel to an iso-profit line is also an iso-profit line, but with a different objective function value.
The number of constraints allowed in a linear program is which of the following?
Report Question
0%
Less than 5
0%
Less than 72
0%
Less than 512
0%
Less than 1,024
0%
Unlimited
Explanation
there is no limit on constraints allowed in linear programming.
so the number of constraints is unlimited.
A feasible solution to an LP problem
Report Question
0%
must satisfy all of the problems constraints simultaneously
0%
need not satisfy all of the constraints, only some of them
0%
must be a corner point of the feasible region
0%
must optimize the value of the objective function
Explanation
A feasible solution to an LP problem belongs to the feasible region. Feasible region is the set of all the points that satisfy the problem's constraints including inequalities, equalities and integer constraints.
In the above figure, blue region is the feasible region. All the points in this region are feasible solutions or feasible points which satisfy the given constraints($$4x+3y\leq 480 $$ and $$2x+3y\leq 360$$)
In North west corner rule if the demand in the column is satisfied one must move to the
Report Question
0%
left cell in the next column
0%
right cell in the next row
0%
right cell in the next column
0%
left cell in the next row
Explanation
The North West Corner method is used to compute the initial feasible solution. This method starts from the north west (i.e., upper left) cell.
If the supply or demand in the column is satisfied then we should move to the right cell in the next column.
For example, in the above figure, 20 units are assigned to the first cell that satisfies the demand of $$D$$ while the supply is in surplus. Now move to the $$\text{right cell in the next column}$$ and assign 30 units which are available with source $$C$$.
The __________ is the method available for solving an L.P.P
Report Question
0%
graphical method
0%
least cost method
0%
MODI method
0%
Hungarian method
Explanation
There are different methods to solve an linear programming problem. Such as Graphical method, Simplex method, Ellipsoid method, Interior point methods.
In North west corner rule, if the supply in the row is satisfied one must move
Report Question
0%
down in the next row
0%
up in the next row
0%
right cell in the next column
0%
left cell in the next row
Explanation
The North West Corner method is used to compute the initial feasible solution. This method starts from the north-west (i.e., upper left) cell.
If the supply or demand in the column is satisfied then we should move to the right cell in the next column.
For example, in the above figure, 20 units are assigned to the first cell that satisfies the demand of $$D$$ while the supply is in surplus.
Now move to the right cell in the next column and assign 30 units which are available with source $$A$$.
Now $$\text{move down in the next row}$$ and assign 40 units which are available with source $$B$$.
In North west corner rule the allocation is done in
Report Question
0%
upper left corner
0%
upper right corner
0%
middle cell in the transportation table
0%
cell with the lowest cost.
Explanation
The North West Corner method is used to compute the initial feasible solution. This method starts from the
north west
(i.e., upper left) cell.
For example, in the above figure, 20 units are assigned to the first cell (i.e., upper left) that satisfies the demand of $$D$$.
In Graphical solution the feasible solution is any solution to a LPP which satisfies
Report Question
0%
only objective function.
0%
non-negativity restriction.
0%
only constraint.
0%
all the three
Explanation
The feasible region is the set of all the points that satisfy all the given constraints . The variables of the linear programs must always take the non-negative values (i.e., $$x\geq 0$$ and $$y\geq 0$$). These are used because $$x$$ and $$y$$ are usually the number of items produced and we cannot produce the negative number of items. The least possible number of items could be zero.
Therefore, the feasible solution should satisfy the non-negativity restriction.
Which of the following is not true about feasibility?
Report Question
0%
It cannot be determined in a graphical solution of an LPP
0%
It is independent of the objective function
0%
It implies that there must be a convex region satisfying all the constraints
0%
Extreme points of the convex region gives the optimum solution.
Explanation
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.
In Graphical solution the redundant constraint is
Report Question
0%
which forms the boundary of feasible region.
0%
which do not optimizes the objective function.
0%
which does not form boundary of feasible region
0%
which optimizes the objective function.
Explanation
A constraint in an LP model becomes redundant when the feasible region doesn't change by the removing the constraint.
For example, $$2x+y\geq 10$$ and $$6x+3y\geq 30$$ are constraints.
$$6x+3y\geq 30 \implies 3\times (2x+y)\geq 3\times 10$$
$$\implies 2x+y\geq 10$$ which is same as the first constraint.
Therefore, $$ 6x+3y \geq 30$$ can be removed. By removing this constraint the feasible region doesn't change.
If the boundary of the feasible region is removed then feasible solution set changes. Hence, redundant constraint cannot be the boundary of the feasible region.
In a graphical solution, the feasible region is:
Report Question
0%
where all the constraints are satisfied simultaneously.
0%
any one constraint is satisfied.
0%
only the first constraint is satisfied.
0%
any one of the above condition.
Explanation
In graphical solution, the feasible region is the set of all possible points that satisfy the problem's constraints including inequalities, equalities and integer constraints.
In the above figure, the blue shaded region is the feasible region. The feasible region is the intersection of the given constraints ($$4x+3y\leq 480$$ and $$2x+3y\leq360$$). I.e., this region contains all the points which satisfy the given constraints.
One disadvantage of using North-West Corner rule to find initial solution to the transportation problem is that
Report Question
0%
It is complicated to use
0%
It does not take into account cost of transportation
0%
It leads to a degenerate initial solution
0%
All of the above
Explanation
The north-west corner rule is a method adopted to compute the initial feasible solution of the transportation problem. The name north-west corner is given to this method because the basic variable is selected from entrance left corner.
In the case of transportation problem the north-west corner rule does not take into account the cost of transportation
One disadvantage of using north west corner rule to find initial solution to the transportation problem is that
Report Question
0%
it is difficult to use.
0%
it does not take into account cost of transportation.
0%
it leads to a degenerate initial solution.
0%
.transportation cost is maximum.
Explanation
The $$\text{North West Corner Rule}$$
is used to compute the initial feasible solution for the transportation problem. This method doesn't take the $$\text{shipping cost}$$
shipping cost
into consideration. As a result, to obtain the initial solution requires several iterations before an optimum solution is obtained.
Which of the following is not a corner point $$(x,y)$$ in the formulation of the given LPP?
Report Question
0%
$$(100,100)$$
0%
$$(200,170)$$
0%
$$(200,100)$$
0%
$$(150,100)$$
Explanation
let the number of normal calculators produced in a day be $$x$$ and
the number of scientific calculators produced in a day be $$y$$
the minimum of total calculators to be produced per day is $$200\implies x+y\geq 200$$
Given, the minimum number of
normal calculators to be produced per day is $$100 \implies x\geq 100$$ and
the minimum number of scientific
calculators to be produced per day is $$80 \implies y\geq 80$$
Also g
iven, the maximum number of
normal calculators can be produced per day is $$200 \implies x\leq 200$$ and
the maximum number of scientific
calculators can be produced per day is $$170 \implies y\leq 170$$
A normal calculator incurred a loss of $$Rs.2$$
For $$x$$ normal calculators, the loss is $$Rs.2x$$
A scientific calculator gained a profit of $$Rs.5$$
For $$xy$$ scientific calculators, the gain is $$Rs.5y$$
Therefore, profit of the manufacturer $$P=5y-2x$$
Corner points are vertices of the feasible region.
first draw the graph for the equations
$$x+y = 200$$
$$x=100$$
$$y=80$$
$$x=200$$
$$y=170$$
for
$$x+y = 200$$
substituting $$y=0 \implies x=200$$
substituting $$x=0 \implies y=200$$
From the figure,
option D I.e., $$(150,100)$$ is not the cornet point.
A farmer has $$10$$ acres of land to plant wheat and rye. He has to plant atleast $$7$$ acres. Each acre of wheat costs $$\$200$$ and each acre of rye costs $$\$100$$ to plant. He has only $$\$1200$$ to spend. Moreover, the farmer has to get the planting done in $$12$$ hours and it takes $$1$$ hour to plant an acre of wheat and $$2$$ hours to plant an acre of rye. An acre of wheat yields a profit of $$\$500$$ and an acre of rye yields a profit of $$\$300$$.
$$($$Take $$x$$ and $$y$$ as the acres of wheat and rye planted respectively$$)$$.
What is the maximum profit that the farmer can make?
Report Question
0%
$$\$2500$$
0%
$$\$2800$$
0%
$$\$3100$$
0%
$$\$3200$$
Explanation
let $$x$$ be the acres of wheat planted and
$$y$$ be the acres of rye planted
Given that there are a total of $$10$$ acres of land to plant.
Atleast $$7$$ acres is to be planted i.e., $$x+y\geq 7$$
Given that the cost to plant one acre of wheat is $$\$200$$
Therefore, the cost for $$x$$ acres of wheat is $$200x$$
Given that the cost to plant one acre of rye is $$\$100$$
Therefore, the cost for $$y$$ acres of rye is $$100y$$
Given that, an amount for planting wheat and rye is $$\$1200$$
Therefore the total cost to plant wheat and rye is $$200x+100y\leq 1200\implies 2x+y\leq 12$$
Given that, the time taken to plant one acre of wheat is $$1$$ hr
Therefore, the time taken to plant $$x$$ acres of wheat is $$x$$ hrs
Given that, the time taken to plant one acre of rye is $$2$$ hrs
Therefore, the time taken to plant $$y$$ acres of rye is $$2y$$ hrs
Given that, the total time for planting is $$12$$ hrs
Therefore, the total time to plant wheat and rye is $$x+2y\leq 12$$
Given that, one acre of wheat yields a profit of $$\$500$$
Therefore, the profit from $$x$$ acres of wheat is $$500x$$
Given that, one acre of rye yields a profit of $$\$300$$
Therefore, the profit from $$y$$ acres of wheat is $$300y$$
therefore the total profit from the wheat and rye is $$P=500x+300y$$
In the above figure, the blue shaded region is the feasible region with three corner points.
$$(4,4), (2,5), (5,2)$$
Now substituting the corner points the profit equation,
substituting
$$(4,4) \implies P=500x+300y=500(4)+300(4)=3200$$
substituting
$$(2,5) \implies P=500x+300y=500(2)+300(5)=2500$$
substituting
$$(5,2) \implies P=500x+300y=500(5)+300(2)=3100$$
$$\$3200 $$ is the maximum profit.
0:0:1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
0
Answered
0
Not Answered
0
Not Visited
Correct : 0
Incorrect : 0
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Practice Class 12 Commerce Maths Quiz Questions and Answers
<
>
Support mcqexams.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page
