Linear Programming Test

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Linear Programming Test - 2

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Question 1
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In a LPP, the objective function is always

A
cubic

B
quadratic

C
Linear

D
constant

Solution

In a LPP, the objective function is always linear.
Question 2
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Maximize Z = –x + 2y, subject to the constraints: x ≥3, x + y ≥5, x + 2y ≥6, y ≥0.

A
Maximum Z = 12 at (2, 6)

B
Z has no maximum value

C
Maximum Z = 10 at (2, 6)

D
Maximum Z = 14 at (2, 6)

Solution

Objective function is Z = - x + 2 y ……………………(1).
The given constraints are : x ≥3, x + y ≥5, x + 2y ≥6, y ≥0.
Corner points Z = - x + 2y

Here , the open half plane has points in common with the feasible region .
Therefore , Z has no maximum value.
Question 3
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Two godowns A and B have grain capacity of 100 quintals and 50 quintals respectively. They supply to 3 ration shops, D, E and F whose requirements are 60, 50 and 40 quintals respectively. The cost of transportation per quintal from the godowns to the shops are given in the following table:

How should the supplies be transported in order that the transportation cost is minimum? What is the minimum cost?

A
From A : 12,52, 40 units; From B: 50,0,0 units to D, E and F respectively and minimum cost = Rs 530

B
From A : 10,52, 42 units; From B: 50,0,0 units to D, E and F respectively and minimum cost = Rs 550

C
From A : 10,50, 40 units; From B: 50,0,0 units to D, E and F respectively and minimum cost = Rs 510

D
From A : 10,53, 44 units; From B: 50,0,0 units to D, E and F respectively and minimum cost = Rs 570

Solution

Let the number of units of grain transported from godown A to D = x And the number of units of grain transported from godown A to E = y Therefore , the number of units of grain transported from godown A to F = 100 –(x+y) Therefore , the number of units of grain transported from godown B to D = 60 –x The number of units of grain transported from godown B to E = 50 –y The number of units of grain transported from godown B to F = x + y –60 . Here , the objective function is :Minimise Z = 2.5x + 1.5y + 410 . , subject to constraints : 60 - x ≥0,50 - y ≥0 ,100 –(x + y) ≥0 , (x + y) - 60 ≥0 , x,y ≥0.

Here Z = 510 is minimum.i.e. From A : 10,50, 40 units; From B: 50,0,0 units to D, E and F respectively and minimum cost = Rs 510 .
Question 4
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Corner points of the feasible region for an LPP are (0, 2), (3, 0), (6, 0), (6, 8) and (0, 5). Let F = 4x + 6y be the objective function. Maximum of F –Minimum of F =

A
60

B
48

C
42

D
18

Solution

Here the objective function is given by : F = 4x +6y .

Maximum of F –Minimum of F = 72 –12 = 30 .
Question 5
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A linear programming problem is one that is concerned with

A
finding the upper limits of a linear function of several variables

B
finding the lower limit of a linear function of several variables

C
finding the limiting values of a linear function of several variables

D
finding the optimal value (maximum or minimum) of a linear function of several variables

Solution

A linear programming problem is one that is concerned with finding the optimal value (maximum or minimum) of a linear function of several variables .
Question 6
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Which of the following types of problems cannot be solved by linear programming methods

A
Diet problems

B
Transportation problems

C
Manufacturing problems

D
Traffic signal control

Solution

Traffic signal control types of problems cannot be solved by linear programming methods .
Question 7
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The optimal value of the objective function Z = ax + by may or may not exist, if the feasible region for a LPP is

A
a circle

B
a polygon

C
Bounded

D
Unbounded

Solution

The optimal value of the objective function Z = ax + by may or may not exist, if the feasible region for a LPP is unbounded.
Question 8
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One kind of cake requires 200g of flour and 25g of fat, and another kind of cake requires 100g of flour and 50g of fat. Find the maximum number of cakes which can be made from 5kg of flour and 1 kg of fat assuming that there is no shortage of the other ingredients used in making the cakes.

A
Maximum number of cakes = 30 , 20 of kind one and 10 cakes of another kind

B
Maximum number of cakes = 32 , 20 of kind one and 12 cakes of another kind

C
Maximum number of cakes = 34 , 27 of kind one and 7 cakes of another kind

D
Maximum number of cakes = 33 , 22 of kind one and 11 cakes of another kind

Solution

Let number of cakes of first type = x
And number of cakes of second type = y
Therefore , the above L.P.P. is given as :
Minimise , Z = x +y , subject to the constraints : 200x +100y ≤5000 and. 25x +50y ≤1000, i.e. 2x + y ≤50 and x +2y ≤40 x, y ≥0.

i.e Maximum number of cakes = 30 , 20 of kind one and 10 cakes of another kind .
Question 9
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An oil company has two depots A and B with capacities of 7000 L and 4000 L respectively. The company is to supply oil to three petrol pumps, D, E and F whose requirements are 4500L, 3000L and 3500L respectively. The distances (in km) between the depots and the petrol pumps is given in the following table:
Assuming that the transportation cost of 10 litres of oil is Re 1 per km, how should the delivery be scheduled in order that the transportation cost is minimum?What is the minimum cost?

A
From A: 550, 3000 and 3500 litres; From B: 4000, 0, 0 litres to D, E and F respectively; Minimum cost = Rs 4800

B
From A: 540, 3000 and 3500 litres; From B: 4000, 0, 0 litres to D, E and F respectively; Minimum cost = Rs 4700

C
From A: 500, 3000 and 3500 litres; From B: 4000, 0, 0 litres to D, E and F respectively; Minimum cost = Rs 4400

D
From A: 520, 3000 and 3500 litres; From B: 4000, 0, 0 litres to D, E and F respectively; Minimum cost = Rs 4600

Solution

Here objective function is : Z = 3x + y + 39500 , subject to constraints : : x + y ≤7000, x ≤4500, x + y ≥3500, , y ≤3000x , x,y ≥0

Here Z = 4400 is minimum.i.e. . From A: 500, 3000 and 3500 litres; From B: 4000, 0, 0 litres to D, E and F respectively; Minimum cost = Rs 4400 .
Question 10
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Corner points of the feasible region determined by the system of linear constraints are (0, 3), (1, 1) and (3, 0). Let Z = px+qy, where p, q >0. Condition on p and q so that the minimum of Z occurs at (3, 0) and (1, 1) is

A
p = 2q

B
p = q

C
p = 3q

D
p = q/2

Solution

We have Z = px + qy , At (3, 0) Z = 3p ……………………………….(1) At (1 , 1) Z = p + q …………………………(2) Therefore , from (1) and (2) : We have : p = q/2 .