Sr. No. of Question Paper: 1509 F-7 Your Roll No..............
Unique Paper Code : 2362301
Name of the Paper : Introduction of Operational Research and Linear Programming
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Name of the Course : B.Tech Computer Science (Erstwhile FYUP) Allied Course
Semester : III
Duration : 3 Hours Maximum Marks: 75
Instructions for Candidates
- Write your Roll No. on the top immediately on the receipt of this question paper.
- Answer fifteen questions in all.
- All questions carry equal marks.
- Simple calculators are allowed.
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- Explain the importance of operational research in decision making. (5)
- Find all possible basic solutions to the following set of linear equations.
2x1 + 3x2 + x3 = 6
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x1 + 3x2 + 5x3 = 5
Also check whether any of the above solutions is degenerate or not. (5)
- Define the basis and dimension of a vector space. Test whether the set of vectors a1 = [1, 1, 0], a2 = [3, 0, 1], a3 = [5, 2, 1] form a basis of R³? (5)
- Define a convex set. Examine the convexity of the following set :
S = {(x1, x2): x1 + x2 = 1, ?x1, x2 ? R} (5)
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- A firm produces three products A, B and C. It uses two type of raw material I and II of which 5,000 and 7,500 units available. The raw material requirements per unit of the products are given below :
Raw Material Requirement per unit of product A B C I 3 4 5 II 5 3 5 The labour time for each unit of product A is twice that of product B and three times that of product C. The entire labour force of the firm can produce the equivalent of 3,000 units. The minimum demand of the three products is 600, 650 and 500 units respectively. Formulate the problem as a linear programming problem (LPP) that will maximize the profit. (5)
- Consider the following LPP :
Maximize Z = 5x1 + 3x2
Subject to: 3x1 + 5x2 <= 15
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5x1 + 2x2 <= 10
x1, x2 >= 0
- Determine all basic solutions of the problem and classify them as feasible and infeasible. (3)
- Show how the infeasible basic solutions are represented on the graphical solution space. (2)
- Use graphical method to solve the following LPP :
Minimise Z = 3x1 + 4x2
Subject to 3x1 + 4x2 >= 240
2x1 + x2 >= 100
5x1 + 3x2 >= 120
x1, x2 >= 0 (5)
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- Use Big M method to solve the following LPP
Maximize Z = 10x1 + 20x2
Subject to 2x1 + 4x2 >= 16
x1 + 5x2 >= 15
x1, x2 >= 0 (5)
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- How do you identify following in the optimal simplex table ?
- Alternate solution (2)
- Unbounded solution (2)
- Infeasible solution (1)
- Solve the following LPP by dual simplex method
Minimize Z = 3x1 - 2x2 + x3
Subject to 3x1 + x2 + x3 >= 3
-3x1 + 3x2 + x3 >= 6
x1 + x2 + x3 <= 6
x1, x2, x3 >= 0 (5)
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- Consider the following LPP
Maximize Z = 3x1 + 4x2 + x3 + 7x4
Subject to 8x1 + 3x2 + 4x3 + x4 <= 7
2x1 + 6x2 + x3 + 5x4 <= 3
x1 + 4x2 + 5x3 + 2x4 <= 8
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x1, x2, x3, x4 >= 0
Its associated optimal simplex table is given as:
Basic x1 x2 x3 x4 x5 x6 x7 Solution Z 0 169/38 ½ 0 1/38 53/38 0 83/19 x1 1 9/38 ½ 0 5/38 5/38 0 16/19 x4 0 21/19 0 1 -1/19 -1/19 0 5/19 x6 0 159/38 9/2 0 -1/38 -1/38 1 126/19 Obtain the variations in cost coefficients which are permitted without changing the optimal solutions. (5)
- Obtain the dual for the following primal problem :
Minimize Z = 5x1 + 6x2 + x3
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Subject to x1 + 2x2 + x3 = 15
-x1 + 5x2 <= 18
4x1 + 7x2 <= 20
x1, x2 >= 0, x3 unrestricted (5)
- Consider the following LPP :
Maximize Z = 4x1 + 14x2
Subject to 2x1 + 7x2 + x3 = 21
7x1 + 2x2 + x4 = 21
x1, x2, x3, x4 >= 0
Check the optimality and feasibility of the following basic solution :
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Basic variables = (x2, x1), Inverse of the basis matrix = (5)
- Find any three alternate optimal solution (if they exist) for the following LPP:
Maximize Z = 2x1 + 4x2
Subject to x1 + 2x2 <= 5
x1 + x2 <= 4
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x1, x2 >= 0 (5)
- Consider the LPP:
Maximize Z = 2x1 + 4x2 + 4x3 - 3x4
Subject to x1 + x2 + x3 = 4
x1 + 4x2 + x4 = 8
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x1, x2, x3, x4>= 0
By using x3 and x4 as the starting variables, the optimal table is given by
-
Basic x1 x2 x3 x4 Solution Z 2 0 0 3 16 x3 3/4 0 1 -1/4 2 x2 1/4 1 0 1/4 2 Write the associated dual problem, and determine its optimal solution in two ways. (5)
- Solve the following LPP by Two Phase method:
Maximize Z = 3x1 + 2x2
Subject to 2x1 + x2 <= 2
3x1 + 4x2 >= 12
x1, x2 >= 0 (5)
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