Linear Programming Test

Result

Linear Programming Test - 23

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

In North west corner rule the allocation is done in

A
upper left corner

B
upper right corner

C
middle cell in the transportation table

D
cell with the lowest cost.

Solution

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 $$D$$.
Question 2
1 / -0

In Graphical solution the feasible solution is any solution to a LPP which satisfies

A
only objective function.

B
non-negativity restriction.

C
only constraint.

D
all the three

Solution

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.
Question 3
1 / -0

Which of the following is not true about feasibility?

A
It cannot be determined in a graphical solution of an LPP

B
It is independent of the objective function

C
It implies that there must be a convex region satisfying all the constraints

D
Extreme points of the convex region gives the optimum solution.

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.
Question 4
1 / -0

In Graphical solution the redundant constraint is

A
which forms the boundary of feasible region.

B
which do not optimizes the objective function.

C
which does not form boundary of feasible region

D
which optimizes the objective function.

Solution

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.
Question 5
1 / -0

In a graphical solution, the feasible region is:

A
where all the constraints are satisfied simultaneously.

B
any one constraint is satisfied.

C
only the first constraint is satisfied.

D
any one of the above condition.

Solution

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.
Question 6
1 / -0

One disadvantage of using North-West Corner rule to find initial solution to the transportation problem is that

A
It is complicated to use

B
It does not take into account cost of transportation

C
It leads to a degenerate initial solution

D
All of the above

Solution

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
Question 7
1 / -0

One disadvantage of using north west corner rule to find initial solution to the transportation problem is that

A
it is difficult to use.

B
it does not take into account cost of transportation.

C
it leads to a degenerate initial solution.

D
.transportation cost is maximum.

Solution

The $$\text{North West Corner Rule}$$ is used to compute the initial feasible solution for the transportation problem. This method doesn't take the $$\text{shipping cost}$$ into consideration. As a result, to obtain the initial solution requires several iterations before an optimum solution is obtained.
Question 8
1 / -0

Directions For Questions

A company manufactures two types of calculators - normal calculator and scientific calculator. Due to demand in the market, the company must produce a minimum of $$100$$ normal and $$80$$ scientific calculators daily. The constraints on the manufacturing limits the production to a maximum of $$200$$ normal and $$170$$ scientific calculators. Also, a minimum of total $$200$$ calculators must be produced in a day. A normal calculator sold incurs a loss of Rs. $$2$$, whereas a scientific calculator gains a profit of Rs. $$5$$.

$$($$Take $$x$$ and $$y$$ as the quantity of normal and scientific calculators produced in a day respectively.$$)$$

...view full instructions

Which of the following is not a corner point $$(x,y)$$ in the formulation of the given LPP?

A
$$(100,100)$$

B
$$(200,170)$$

C
$$(200,100)$$

D
$$(150,100)$$

Solution

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 given, 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.
Question 9
1 / -0

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?

A
$$\$2500$$

B
$$\$2800$$

C
$$\$3100$$

D
$$\$3200$$

Solution

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.
Question 10
1 / -0

Directions For Questions

A company manufactures two types of calculators - normal calculator and scientific calculator. Due to demand in the market, the company must produce a minimum of $$100$$ normal and $$80$$ scientific calculators daily. The constraints on the manufacturing limits the production to a maximum of $$200$$ normal and $$170$$ scientific calculators. Also, a minimum of total $$200$$ calculators must be produced in a day. A normal calculator sold incurs a loss of Rs. $$2$$, whereas a scientific calculator gains a profit of Rs. $$5$$.

$$($$Take $$x$$ and $$y$$ as the quantity of normal and scientific calculators produced in a day respectively.$$)$$

...view full instructions

The maximum profit that can be made in a day is:

A
Rs. $$450$$

B
Rs. $$800$$

C
Rs. $$650$$

D
Rs. $$525$$

Solution

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 given, 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 $$P=5y-2x$$

In the above figure, the blue shaded region is the feasible region with five corner points.$$(100,170), (200,170), (200,80), (120,80), (100,100)$$

$$(100,170)$$ is the point where $$x=100$$ intersects $$y=170$$
$$(200,170)$$ is the point where $$x=200$$ intersects $$y=170$$
$$(200,80)$$ is the point where $$x=200$$ intersects $$y=80$$
$$(120,80)$$ is the point where $$x+y=200$$ intersects $$y=80$$
I.e., substituting $$y=80 \implies x+80 =200 \implies x=200-80 \implies x=120$$
$$(100,100)$$ is the point where $$x+y=200$$ intersects $$x=100$$
I.e., substituting $$x=100 \implies 100+y =200 \implies x=200-100 \implies x=100$$

Now substituting the corner points the profit equation,
substituting $$(100,170) \implies P=5\times 170-2\times 100=650$$

substituting $$(200,170) \implies P=5\times 170-2\times 200=450$$

substituting $$(200,80) \implies P=5\times 80-2\times 200=0$$

substituting $$(120,80) \implies P=5\times 80-2\times 120=160$$

substituting $$(100,100) \implies P=5\times 100-2\times 100=300$$

$$650 $$ is the maximum profit