JUNE 2013 - PAPER III
14. A data file of 1,00,000 characters contains only the characters g-l, with the frequencies as indicated in table :
| g | h | i | j | k | l | |
| Frequency in thousand | 45 | 13 | 12 | 16 | 9 | 5 |
(A) 2,52,000 bits
(B) 2,64,000 bits
(C) 2,46,000 bits
(D) 2,24,000 bits
Ans:- D
Explanation :-
The explanation is quite lengthy. so, download the following ppt and read through it to understand the steps.
Huffman tree.ppt
15. A vertex cover of an undirected graph G(V,E) is a subset V1 such that
(A) Each pair of vertices in V1 is connected by an edge.
(B)If (u,v) E then u V1and v V1
(C) If (u,v) E then u V1 and v V1
(D) All pairs of vertices in V1 are not connected by an edge
Ans:- C
Explanation :-
A vertex cover in a graph G is a set of vertices C such that each edge of G contains at least one vertex in C.
An edge will be formed by two vertices. For each edge of G, at least one vertex should be in C, where C stands for the vertex cover. So, in the question above, vertex cover is represented by the subset V1. V1 is a subset of vertices V. Let us consider the edge (u,v). According to the vertex cover definition, any one vertex, either u or v should belong to V1. So, option C is the correct answer.
16. In a fully connected mesh network with n devices, there are ________ physical channels to link all devices.
(A) n(n-1)/2
(B) n(n+1)/2
(C ) 2n
(D ) 2n+1
Ans:- B
17. The baud rate of a signal is 600 baud/second. If each signal unit carries 6 bits, then the bit rate of a signal is
(A) 3600
(B)100
(C )6/600
(D) None of the above
Ans:- A
Explanation :-
Bit rate = baud rate X n
Baud rate = 600 baud/second.
n= number of bits = 6
bit rate = 600 X 6 = 3600
So, the correct answer is A.
18. Match the following :
a. Data link layer i. Flow control
b. Network layer ii. Node to node delivery
c. Transport layer iii. Mail services
d. Application layer iv. Routing
Codes :
a b c d
(A) ii i iv iii
(B) ii iv i iii
(C) ii i iii iv
(D) ii iv iii i
Ans:- B
Explanation:-
Refer to my post on “Computer networks”. I have solved many questions of the same kind. The most obvious approach is to associate the easy ones first. So, it will be Network layer to routing, Application layer to mail services. b-iv, d – iii, There is only one option which satisfies it and it is B. So, data link layer will be Node to node delivery and transport layer will be flow control. So, the correct answer is B.
19. An image is 1024 ∗ 800 pixels with 3 bytes/pixel. Assume the image is uncompressed. How long does it take to transmit it over a 10-Mbps Ethernet ?
(A) 196.6 seconds
(B) 19.66 seconds
(C) 1.966 seconds
(D) 0.1966 seconds
Ans:- C
Explanation:-
The size of the image is 1024 * 800 pixels = 819200 pixels.
No of bytes / pixel = 3
So, total no of bytes = 819200 X 3 = 2457600 bytes
Therefore, Total no of bits = 2457600 X 8 = 19660800 bits
The data rate of a computer network connection is normally measured in units of bits per second(bps).
One kilobit per second equals 1000 bits per second.
One megabit per second(Mbps) equals 1000 kbps = 1000 X 1000 bps .
10 Mbps = 1000 X 1000 X 10 = 10000000.
Therefore, time taken for transmission = 19660800 / 10000000 = 1.96608 seconds.
Hence, the correct answer is C.
20. The ________ measures the relative strengths of two signals or a signal at two different points.
(A) frequency
(B)attenuation
(C)throughput
(D)decibel
Ans:- D
Explanation:-
Attenuation means loss of energy. When a signal, simple or complex, travels through a medium, it loses some of its energy so that it can overcome the resistance of the medium.
To show that a signal has lost or gained strength, engineers use the concept of decibel. The decibel(db) measures the relative strengths of two signals or a signal at two different points. dB is –ve if a signal is attenuated and +ve if a signal is amplified.
21. Which one of the following media is multidrop ?
(A) Shielded Twisted pair cable
(B) Unshielded Twisted pair cable
(C ) Thick Coaxial cable
(D )Fiber optic cable
Ans:- C
22. What is the baud rate of the standard 10 Mbps Ethernet ?
(A) 10 megabaud
(B) 20 megabaud
(C) 30 megabaud
(D) 40 megabaud
Ans:- B
The data rate of the standard 10 Mbps Ethernet is 10mbps. The baud rate is twice as that. So it is 20 megabaud.
23. At any iteration of simplex method, if Δj (Zj – Cj) corresponding to any non-basic variable Xj is obtained as zero, the solution under the test is
(A) Degenerate solution
(B) Unbounded solution
(C) Alternative solution
(D) Optimal solution
Ans:- C
24. A basic feasible solution to a m-origin, n-destination transportation problem is said to be _________ if the number of positive allocations are less than m + n – 1.
(A) degenerate
(B) non-degenerate
(C) unbounded
(D) unbalanced
Ans:- A
Explanation:-
If a basic feasible solution contains less than m+n-1 non negative allocations, then it is said to be degenerate. So, the correct answer is option A.
Download the above post from the following link.
JUNE 2013 - PAPER III - PART II
I think and of Q16 is A but not B
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