U.G.C. NET Exam. 4 December, 2019 Paper II (COMPUTER SCIENCE & APPLICATIONS)

Total Questions: 100

1. A basic feasible solution of an m × n transpiration problem is said to be nondegenerate. if basic feasible solution contains exactly ______ number of individual allocation is ______ positions.

Correct Answer: (b) m + n – 1, independent
Solution:

Transportation problem is crucial part of linear programming problem which can be connected for required sources of supply to responding destination of demand, with the end goal that the aggregate transpiration cost to be limited
The essential phase of any transportation problem as initial basic feasible solution. Initial basic feasible solution must be feasible i.e. it must satisfy all the supply and demand constraints. The number of positive allocations must be equal to m + n – 1 where m is number of rows and n is number of columns.
Non degenerate basic feasible solution : A basic feasible solution is non-degenerate it has exactly m + n – 1 positive allocations in individual positions. It the allocations are less than required number, then it is known as degenerate basic feasible solution. This solution is not easy to modify, because it is impossible to draw a closed loop for each occupied cell.

2. Consider the following Linear programming problem (LPP) :

Correct Answer: (c)
Solution:

3. Answer the following question

Correct Answer: (b) B
Solution:

4. Answer the following question

Correct Answer: (a) A(GLB – 3,LUB – 36); B(GLB – 1,LUB – 20)
Solution:

5. Let P be the set of all people. Let R be a binary relation on P such that (a, b) is in R if a is a brother of b. Is R symmetric transitive, an equivalence relation, a partial order relation?

Correct Answer: (c) NO,YES,NO,NO
Solution:

R is a relation on P such that (a, b) is in R if a is a brother of b.
Case 1 : check symmetric relation (a, b) – a is the brother of b, (b, a) – b may or may not be brother of a, so given relation is not symmetric.
Case 2 : Check transivity If a is the brother of b, b is the brother of c, then a will also be brother of c. It means if (a, b) and (b, c) then (a, c). So given relation is transitive.
Case 3 : Equivalence. As relation is not symmetric, So it cannot be equivalence relation.
Case 4 : Partial order relation for this, we have to check anti symmetric property. i.e. if (a, b) is in relation than (b, a) should not be in relation unless a = b, Here it is not partial order relation.

6. The weight of minimum spanning tree in graph G, calculated using Kruskal's algorithm is :

Correct Answer: (b) 15
Solution:

Minimum spanning Tree : → A minimum spanning tree (MST) or minimum weight spanning tree is a subset of the edges of a connected, edge-weighted undirected graph that connects all the vertices together, without any cycles and with the minimum possible total edge weight. So a minimum spanning tree or minimum weight spanning tree is a subset of the edges (V - 1) of a connected, edge weighted undirected graph G (V, E) that connects all the vertices together, without any cycles and with the minimum possible total edge weight.
Edge set for the given graph = {2, 3, 4, 5, 6, 7, 8} for 5
vertices, we need 4 edges in MST. So edge set for MST
= {2, 3, 4, 6}
Minimum spanning tree

7. A tree has 2n vertices of degree 1, 3n vertices of degree 2, n vertices of degree 3. Determine the number of vertices and edges in tree.

Correct Answer: (a) 12,11
Solution:

Degree of 2n vertices = 1
Degree of 3n vertices = 2
Degree of n vertices = 3
Formula –
Handshaking theorem : sum of the degree of vertices =
2 × number of edges (E) In a tree, number of edges (E)
= number of vertices – 1
Calculation – using 1st formula
2n ×1 + 3n × 2 + n × 3 = 2E
11n = 2E
E = 2n + 3n + n – 1
E = 6n – 1
Comparing above two equation for E.
11n = 12n – 2
n = 2
number of vertices = 6n= 12
number of edges = 12 – 1 = 11

8. How many reflexive relations are there on a set with 4 elements?

Correct Answer: (b)
Solution:

The number of reflexive relation on an n-

9. A non-pipelined system takes 30ns to process a task. The same task can be processed in a foursegment pipeline with a clock cycle of 10ns. Determine the speed up of the pipeline for 100 tasks.

Correct Answer: (d) 2.91
Solution:

For a non-pipelines system.

10. A computer a memory unit of 512 K works of 32 bits each. A binary instruction code is stored in one word of the memory. The instruction has four parts: an addressing mode field to specify one of the two-address sing mode (direct and indirect), an operation code, a register code part to specify one of the 256 registers and an address part. How many bits are there in addressing mode part, opcode part, register code part and the address part?

Correct Answer: (c) 1, 4, 8, 19
Solution:

Memory unit = 512 k words
1 word = 32 bits = 4B
Binary instruction code stored in 1 word of memory
Instruction divided as follows.