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Past Year Questions: Linear Programming - Mechanical Engineering MCQ


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20 Questions MCQ Test - Past Year Questions: Linear Programming

Past Year Questions: Linear Programming for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Past Year Questions: Linear Programming questions and answers have been prepared according to the Mechanical Engineering exam syllabus.The Past Year Questions: Linear Programming MCQs are made for Mechanical Engineering 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Past Year Questions: Linear Programming below.
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Past Year Questions: Linear Programming - Question 1

In an assembly line for assembling toys, five workers are assigned tasks which take times of 10, 8, 6, 9 and 10 minutes respectively. The balance delay for line is

[1996]

Detailed Solution for Past Year Questions: Linear Programming - Question 1

Assuming cycle time = 10 + 8 + 6 + 9 + 10 = 43

= 14%

Past Year Questions: Linear Programming - Question 2

If at the optimum in a linear programming problem, a dual variable corresponding to a particular primal constraint is zero, then it means that

[1996]

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Past Year Questions: Linear Programming - Question 3

A manufacturer produces two types of products, 1 and 2, at production levels of x1 and x2 respectively. The profit is given is 2x1 + 5x2. The production constraints are:
x1 + 3x2 < 40
3x1 + x2 < 24
x1 + x2 < 10
x1 > 0 x2 > 0
The maximum profit which can meet the constraints is

[2003]

Detailed Solution for Past Year Questions: Linear Programming - Question 3

Feasible Region (0– A – B – C– 0) At point (A) 0, 10) Z = 2 (0) + 5 (10) = 50
At point B (7,3) Z = 2(7) + 5(3) = 2y At point C (8, 0) Z = 2(16) + 5(0) = (16) 2
⇒32
Hence maximum profit is Zmax = 50
No correct option is given.

Past Year Questions: Linear Programming - Question 4

A component can be produced by any of the four processes, I, II, III and IV. Process I has fixed cost of Rs. 20 and variable cost of Rs. 3 per piece.Process II has a fixed cost of Rs. 50 and variable cost of Rs. 1 per piece. Process III has a fixed cost of Rs. 40.00 and variable cost of Rs. 2 per piece. Process IV has fixed cost of Rs. 10 and Variable cost Rs. 4 per piece. If company wishes to produce 100 pieces of the component, from economic point of view it should choose

[2005 : 2 Marks]

Detailed Solution for Past Year Questions: Linear Programming - Question 4

 


Total cost = Fixed Cost (FC) + Number of piece (n) × Variable Cost per piece (VC) TC = FC + (n) x V.C
For I
TCI = 20+ (100) 3 = 320
For II
TCII = 50 + (100) 1 = 150
For III
TCIII = 40 + 100 (2) = 240
ForIV
TCIV = 10 + 100 (4) = 410
So from economical point of view, one should chose process II

Past Year Questions: Linear Programming - Question 5

A company has two factories S1, S2 and two warehouses D1, D2. The supplies from S1 and S2 are 50 and 40 units respectively. Warehouse D1 requires a minimum of 20 units and a maximum of 40 units. Warehouse D2 requires a minimum of 20 units and, over and above, it can take as much as can be supplied. A balanced transportation problem is to be formulated for the above situation. The number of supply points, the number of demand points, and the total supply (or total demand) in the balanced transportation problem respectively are

[2005]

Detailed Solution for Past Year Questions: Linear Programming - Question 5

Total no of supply point 
⇒ m + n – 1
⇒ 2 + 2 – 1
⇒ 3
Total no of Demand point = 4
(x11, x12, x21, x22)
Total supply = Total Demand Þ 90 units

Past Year Questions: Linear Programming - Question 6

If an additional constraint X1 + X2 < 5 is added, the optimal solution is

[2005]

 

Detailed Solution for Past Year Questions: Linear Programming - Question 6

Max z = x1 + x2

As feasibly region remains the same solution remains the same (4/3, 4/3).
Hence, the correct option is (b).

Past Year Questions: Linear Programming - Question 7

Let Y1 and Y2 be the decision variables of the dual and v1 and v2 be the slack variables of the dual of the given linear programming problem.The optimum dual variables are

[2005]

Detailed Solution for Past Year Questions: Linear Programming - Question 7

The optimal dual variables are V1 & V2 .

Past Year Questions: Linear Programming - Question 8

A firm is required to procure three items (P, Q, and R). The prices quoted for these items (in Rs.) by suppliers S1,  S2 and S3 are given in table. The management policy requires that each item has to be supplied by only one supplier and one supplier supply only one item.The minimum total cost (in Rs.) of procurement to the firm is

[2006]

Past Year Questions: Linear Programming - Question 9

Consider the following Linear Programming Problem (LPP):
Maximize z = 3x1 + 2x2,
Subject to x1 < 4 x2 < = 6
3x1 + 2x2 < 18
x1 > 0, x2 > 0

[2009]

Detailed Solution for Past Year Questions: Linear Programming - Question 9

z= 3 × 1 + 2 × 2
x1 ≤ 4
x2 ≤ 6
3x1 + 2x≤ 18
x1,x2 ≥ 0

Past Year Questions: Linear Programming - Question 10

One unit of product P1 requires 3 kg of resource R1 and 1 kg resource R2. One unit of product P2 requires 2 kg of resource R1 and 2 kg of resource R2. The profits per unit by selling product P1 and P2 and Rs. 12000 and Rs 3000 respectively. The manufacturer has 90 kg of resource R1, and 100 kg of resource R2.

The unit worth of resource R2, i.e. dual price of resource R2 in Rs per kg is

[2011]

Detailed Solution for Past Year Questions: Linear Programming - Question 10

Because second constraint is redundant in nature. Therefore, resource R2 has no effect on the feasible solution.

Past Year Questions: Linear Programming - Question 11

The manufacturer can make a maximum profit of  Rs.

[2011]

Detailed Solution for Past Year Questions: Linear Programming - Question 11

z = 2000 P1 + 3000 P2
Subjected to 
3P1 +2P2 ≤ 90
P1 + 2P2 ≤ 100
P1, P2 ≥ 0


At point A (30, 0) Z = 30 × 2000 × 3000 × 0 = 6000
At point B (0, 45) Z = 2000× 0 + 3000 × 45 = 135000
Hence Maximum Profit
[Zmax = 135000]

Past Year Questions: Linear Programming - Question 12

A linear programming problem is shown below:
Maximise 3x + 7y
Subjeot to 3x + 7y < 10
4x + 6y < 8 x, y > 0

[2013]

Detailed Solution for Past Year Questions: Linear Programming - Question 12

z = 3x + 7y Constraints 3x + 7y ≤ 10 4x + 6y < 8; x, y ≥ 0 Corresponding equations 3x + 7y = 10; 4x + 6y =8

A  (0, 4/3) z = 9.23 B (2, 0) z = 6 Thus, exactly one optimal solution.
Hence, the correct option is (b).

Past Year Questions: Linear Programming - Question 13

Maximize z = 15x1 + 20x2
Subject to 12x1 + 4x2 >  36
12x1 + 6x2 > 24 x1,
x2 > 0
The above linear programming problem has

Detailed Solution for Past Year Questions: Linear Programming - Question 13

zmax = 16x1 + 20x2
12x1 + 4x2 ≥36
12x1 + 6x2 ≥24 x1
x2 ≥ 0

Past Year Questions: Linear Programming - Question 14

For the linear programming problem:
Maximize z = 3x1 + 2x2
Subject to –2x1 + 3x2 < 9x1 – 5x2 > – 20x1, x2 > 0
The above problem has

[2016]

Detailed Solution for Past Year Questions: Linear Programming - Question 14

Maximize Z= 3X+ 2X2
Subject to
-2X+ 3X≤ 9
X1 - 5X≥ 20
XP X2 ≥ 0


The LPP is  unbounded.

Past Year Questions: Linear Programming - Question 15

For the standard transportation linear program with m sources and n destinations and total supply equaling total demand, an optimal solution (lowest cost) with the smallest number of non-zero xij values (amounts from source i to destination j) is desired. The best upper bound for this number is

[2008]

 

Detailed Solution for Past Year Questions: Linear Programming - Question 15

In such an L .P.P, m × n variables are th ere an d m + n equations/constraints are there (satisfying the demand-supply requirements). But one constraint is removed as total supply equals total demand.  The best upper bound xij values is (m + n  1).

Past Year Questions: Linear Programming - Question 16

After introducing slack variables s and t, the initial basic feasible solution is represented by the table below (basic variables are s = 6 and t = 6, and the objective function value is 0).

After some simplex iterations, the following table is obtained

From this, one can conclude that

[2008]

Detailed Solution for Past Year Questions: Linear Programming - Question 16

z = 4x + 6y
3x + 2y ≤ 6
2x + 3y ≤ 6
x, y > 0

Feasible region (O–A–B–C–0) Since, Slope of objective function is equal to the slope of constraint Hence LPP has multiple optimal solution
At B (6/5, 6/5)
Z = 12
At C (0,2)
Z = 12

Past Year Questions: Linear Programming - Question 17

The dual for the LP in Q. 21 is

[2008]

Past Year Questions: Linear Programming - Question 18

 Simplex method of solving linear programming problem uses

[2010]

Past Year Questions: Linear Programming - Question 19

A company produces two types of toys : P and Q. Production time of Q is twice that of P and the company has a maximum of 2000 time units per day. The supply of raw material is just sufficient to produce 1500 toys (of any type) per day. Toy type Q requires an electric switch which is available @ 600 pieces per day only.The company makes a profit of Rs. 3 and Rs. 5 on type P and Q respectively. For maximization​of profits, the daily production quantities of P and Q toys should respectively be

[2004]

Detailed Solution for Past Year Questions: Linear Programming - Question 19

Zmax = 3P +5Q
subject to 
P+ 2Q ≤ 2000
P+Q ≤ 1500
Q ≤ 600
P,Q ≥ 0



Feasible solution (O A B C D) Since At point A (1500,0) Z = 3 × 1500 + 5 × 0 = 4500
At Point B (1000, 500) Z = 3 × 1000 + 5 × 500 = 5500
At point C (800, 600) Z = 3 × 800 + 5 × 600 = 5400
At Point O (0, 600) Z = 3 × 0 + 5 × 600 = 3000
Hence Z in maximum at B (1000,500) P = 1000 units Q = 500 units

Past Year Questions: Linear Programming - Question 20

The manufacturer can make a maximum profit of  Rs.

[2011]

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