Computer-Networks

Question 1
A TCP server application is programmed to listen on port number P on host S. A TCP client is connected to the TCP server over the network. Consider that while the TCP connection was active, the server machine S crashed and rebooted. Assume that the client does not use the TCP keepalive timer. Which of the following behaviours is/are possible?
A
If the client was waiting to receive a packet, it may wait indefinitely.
B
If the client sends a packet after the server reboot, it will receive a RST segment.
C
The TCP server application on S can listen on P after reboot.
D
If the client sends a packet after the server reboot, it will receive a FIN segment.
Question 1 Explanation: 
  1. True
    Since broken connections can only be detected by sending data, the receiving side will wait forever. This scenario is called a “half-open connection” because one side realizes the connection was lost but the other side believes it is still active.
  2. True
    The situation resolves itself when client tries to send data to server over the dead connection, and server replies with an RST packet (not FIN).
  3. True
    Yes, a TCP Server can listen to the same port number even after reboot.  For example, the SMTP service application usually listens on TCP port 25 for incoming requests. So, even after reboot the port 25 is assigned to SMTP.
  4. False
    The situation resolves itself when client tries to send data to server over the dead connection, and server replies with an RST packet (not FIN), causing client to finally to close the connection forcibly.
    FIN is used to close TCP connections gracefully in each direction (normal close of connection), while TCP RST is used in a scenario where TCP connections cannot recover from errors and the connection needs to reset forcibly.
Question 2
Consider the sliding window flow-control protocol operating between a sender and a receiver over a full-duplex error-free link. Assume the following:
  • The time taken for processing the data frame by the receiver is negligible.
  • The time taken for processing the acknowledgement frame by the sender is negligible.
  • The sender has an infinite number of frames available for transmission.
  • The size of the data frame is 2,000 bits and the size of the acknowledgment frame is 10 bits.
  • The link data rate in each direction is 1 Mbps (=106bits per second).
  • One way propagation delay of the link is 100 milliseconds.
The minimum value of the sender’s window size in terms of the number of frames, (rounded to the nearest integer) needed to achieve a link utilization of 50% is ______.
A
51
Question 2 Explanation: 

Tt(packet) = L / B.W => 2000 bits / 10^6 bps = 2  x 10^-3 sec = 2 millisec


Tt(Ack) = L / B.W. => 10 bits / 10^6 bps = 10^-5 sec = 10^-2 millisec = 0.01 millisec


Tp = 100 millisec


Total time = Tt(packet) + 2 x Tp + Tt(Ack)

=> 2 + 2 x 100 + 0.01 = 202.01 millisec


Efficiency = 50 % = ½


Efficiency = Useful Time /  Total time


½ = n x Tt / Total time 

 

=> 2 x n x Tt =  Total time

=>2 x n x 2 = 202.01  

=> n = 202.01 / 4 => 50.50

 

For minimum, we have to take ceil, Hence size of window = 51

Question 3
Consider two hosts P and Q connected through a router R. The maximum transfer unit (MTU) value of the link between P and R is 1500 bytes, and between R and Q is 820 bytes. A TCP segment of size 1400 bytes was transferred from P to Q through R, with IP identification value as 0x1234. Assume that the IP header size is 20 bytes. Further, the packet is allowed to be fragmented, i.e., Don’t Fragment (DF) flag in the IP header is not set by P. Which of the following statements is/are correct?
A
If the second fragment is lost, R will resend the fragment with the IP identification value 0x1234.
B
If the second fragment is lost, P is required to resend the whole TCP segment.
C
Two fragments are created at R and the IP datagram size carrying the second fragment is 620 bytes.
D
TCP destination port can be determined by analysing only the second fragment.
Question 3 Explanation: 
Question 4
Consider the following two statements.
S1: Destination MAC address of an ARP reply is a broadcast address.
S2: Destination MAC address of an ARP request is a broadcast address.
Which of the following choices is correct?
A
Both S1and S2are true.
B
S1is true and S2is false.
C
S1is false and S2is true.
D
Both S1and S2are false.
Question 4 Explanation: 
ARP request is Broadcasting. ARP reply is unicasting.
Question 5

What is the distance of the following code 000000, 010101, 000111, 011001, 111111?

A
2
B
3
C
4
D
1
Question 5 Explanation: 
Distance = Minimum hamming distance = 2
010101 ⊕ 011001 = 001100
Question 6

Assume that you have made a request for a web page through your web browser to a web server. Initially the browser cache is empty. Further, the browser is configured to send HTTP requests in non-persistent mode. The web page contains text and five very small images. The minimum number of TCP connections required to display the web page completely in your browser is ______.

A
6
Question 6 Explanation: 
In non-persistent HTTP connection for every object, there is a TCP connection established. Therefore, 1 TCP connection for text and 5 TCP connections for images required.
Hence, 1 Text + 5 Image = 6 Objects
Question 7

Consider the following statements about the functionality of an IP based router.

    I. A router does not modify the IP packets during forwarding.
    II. It is not necessary for a router to implement any routing protocol.
    III. A router should reassemble IP fragments if the MTU of the outgoing link is larger than the size of the incoming IP packet.

Which of the above statements is/are TRUE?

A
I and II only
B
II only
C
I only
D
II and III only
Question 7 Explanation: 
I: The packet contains Header and data. The router modifies the header details like TTL.
II: Is True.
III: Reassemble is not necessary at the router.
Question 8

An organization requires a range of IP addresses to assign one to each of its 1500 computers. The organization has approached an Internet Service Provider (ISP) for this task. The ISP uses CIDR and serves the requests from the available IP address space 202.61.0.0/17. The ISP wants to assign an address space to the organization which will minimize the number of routing entries in the ISP’s router using route aggregation. Which of the following address spaces are potential candidates from which the ISP can allot any one to the organization?

I. 202.61.84.0/21
II. 202.61.104.0/21 
III. 202.61.64.0/21
IV. 202.61.144.0/21 
A
I and II only
B
III and IV only
C
II and III only
D
I and IV only
Question 8 Explanation: 
Given CIDR IP is 202.61.0.0/17 and for HID 32 - 17 = 15 bits can be used.
And to Assign an IP address for 1500 computer, we require 11 bit from HID part.
So NID + SID = 17 + 4 = 21 bits and HID = 11 bits
NID HID
202.61.0 0000 000.00000000
So, from the given option, possible IP Address is
I. 84 -> 0 1010 100 (Because in HID bit 1 is not possible)
II. 104 -> 0 1101 000
III. 64 -> 0 1000 000
IV. 144 -> 1 0010 000 (Because in NID bit 1 is not possible )
Question 9

Consider a TCP connection between a client and a server with the following specifications: the round trip time is 6 ms, the size of the receiver advertised window is 50 KB, slow start threshold at the client is 32 KB, and the maximum segment size is 2 KB. The connection is established at time t=0. Assume that there are no timeouts and errors during transmission. Then the size of the congestion window (in KB) at time t+60 ms after all acknowledgements are processed is ______.

A
44
Question 9 Explanation: 
Threshold = 32 Kb, MSS = 2KB, RTT = 6ms
Here, t + 60 is nothing but at the 10 RTT (60/6 = 10), but here it’s asking after all acknowledgement are processed it means after the 10th RTT, i.e., at the 11RTT.
1st transmission: 2 KB
2nd transmission: 4 KB
3rd transmission: 8 KB
4th transmission: 16 KB
5th transmission: 32 KB (Threshold reached)
6th transmission: 34 KB
7th transmission: 36 KB
8th transmission: 38 KB
9th transmission: 40 KB
10th transmission: 42 KB
At the completion of 10th transmission RTT = 10*6 = 60 ms
For the 11th transmission, The congestion window size is 44 KB.
Question 10

Which of the following assertions is FALSE about the Internet Protocol (IP)?

A
It is possible for a computer to have multiple IP addresses
B
IP packets from the same source to the same destination can take different routes in the network
C
IP ensures that a packet is discarded if it is unable to reach its destination within a given number of hops
D
The packet source cannot set the route of an outgoing packets; the route is determined only by the routing tables in the routers on the way
Question 10 Explanation: 
Because in strict source routing or loose source routing path is set by the source not by router and main task of router is to check outgoing path with the help of forwarding table inside it.
Question 11

Which of the following functionalities must be implemented by a transport protocol over and above the network protocol?

A
Recovery from packet losses
B
Detection of duplicate packets
C
Packet delivery in the correct order
D
End to end connectivity
Question 11 Explanation: 
End to end connectivity is the required functionality provided by Transport protocol.
Question 12

The subnet mask for a particular network is 255.255.31.0. Which of the following pairs of IP addresses could belong to this network?

A
172.57.88.62 and 172.56.87.233
B
10.35.28.2 and 10.35.29.4
C
191.203.31.87 and 191.234.31.88
D
128.8.129.43 and 128.8.161.55
Question 12 Explanation: 
To find whether hosts belong to same network or not , we have to find their net id, if net id is same then hosts belong to same network and net id can be find by ANDing subnet mask and IP address.
128.8.129.43 (Bitwise AND) 255.255.31.0 = 128.8.1.0
128.8.161.55 (Bitwise AND) 255.255.31.0 = 128.8.1.0
Question 13

A 2 km long broadcast LAN has 107 bps bandwidth and uses CSMA/CD. The signal travels along the wire at 2×108 m/s. What is the minimum packet size that can be used on this network?

A
50 bytes
B
100 bytes
C
200 bytes
D
None of the above
Question 13 Explanation: 
Minimum packet size for a CSMA/CD LAN is the frame which cover whole RTT(round trip time). i.e. Tt = 2Tp
d= 2 km = 2 x 103 m, v = 2 x 108 m/s, B= 107
Tp = d / v = 2 x 103 /(2 x 108 ) seconds = 10-5 seconds
Let L bits be minimum size of frame, then Tt = t L / B = L / 107 seconds
Now, Tt = 2Tp
L/107 = 2 x 10-5 = 200 bits = (200 / 8) bytes = 25 bytes
Question 14

Host A is sending data to host B over a full duplex link. A and B are using the sliding window protocol for flow control. The send and receive window sizes are 5 packets each. Data packets (sent only from A to B) are all 1000 bytes long and the transmission time for such a packet is 50 µs. Acknowledgement packets (sent only from B to A) are very small and require negligible transmission time. The propagation delay over the link is 200 µs. What is the maximum achievable throughput in this communication?

A
7.69 × 106 bps
B
11.11 × 106 bps
C
12.33 × 106 bps
D
15.00 × 106 bps
Question 14 Explanation: 
Given, Tt = 50 μs, Tp = 200 μs, L = 1000 bytes, N = 5,
Transmission rate , Tt = L / B.W
Therefore, B.W. = L / Tt = 1000 bytes/ 50 μs = 8000 bits / 50 μs=160 Mbps
Efficiency = N / 1 + 2a, where a = Tp / Tt
Efficiency = 5 * 50 / (50+400) = 250/450 = 5/9
Maximum achievable throughput = Efficiency * B.W = (5/9)*160 Mbps = 88.88 Mbps = = 11.11 x 106 bytes per second
*Actual option should be in bytes per second.
Question 15

Choose the best matching between Group 1 and Group 2.

   Group-1   	  	  Group-2   
 P. Data link          1. Ensures reliable transport of data
                          over a physical point-to-point link
 Q. Network layer      2. Encoder/decodes data for physical
                          transmission
 R. Transport layer    3. Allows end-to-end communication
                          between two processes
                       4. Routes data from one network
                          node to the next
A
P - 1, Q - 4, R - 3
B
P - 2, Q - 4, R - 1
C
P - 2, Q - 3, R - 1
D
P - 1, Q - 3, R - 2
Question 15 Explanation: 
Data Link Layer :: Second layer of the OSI Model, Responsible for Hop to Hop connection or point to point connection.
Transport Layer :: Fourth layer of the OSI Model, Responsible for Service point addressing/Socket to socket connection or end to end connection with full reliability.
Network Layer :: Third layer of the OSI Model, Responsible for Host to Host.
Question 16

Which of the following is NOT true with respect to a transparent bridge and a router?

A
Both bridge and router selectively forward data packets
B
A bridge uses IP addresses while a router uses MAC addresses
C
A bridge builds up its routing table by inspecting incoming packets
D
A router can connect between a LAN and a WAN
Question 16 Explanation: 
A bridge use MAC addresses (DLL layer) and router uses IP addresses (network layer).
Question 17

How many 8-bit characters can be transmitted per second over a 9600 baud serial communication link using asynchronous mode of transmission with one start bit, eight data bits, two stop bits, and one parity bit?

A
600
B
800
C
876
D
1200
Question 17 Explanation: 
In Serial port communication baud rate = bit rate.
So bit rate is 9600 bps.
To send one char we need to send (1 + 8 + 2 +1) = 12
So total char send = 9600 / 12 = 800
Question 18

A and B are the only two stations on an Ethernet. Each has a steady queue of frames to send. Both A and B attempt to transmit a frame, collide, and A wins the first backoff race. At the end of this successful transmission by A, both A and B attempt to transmit and collide. The probability that A wins the second backoff race is:

A
0.5
B
0.625
C
0.75
D
1.0
Question 18 Explanation: 
A has 5 chances to win out of 8 combinations.
The probability that A wins the second back-off race = 5/8 = 0.625
More explanation in the video.
Question 19

The routing table of a router is shown below:

 Destination     Sub net mask 	     Interface
 128.75.43.0 	 255.255.255.0 	        Eth0
 128.75.43.0 	 255.255.255.128 	Eth1
 192.12.17.5 	 255.255.255.255 	Eth3
 Default 	  	                Eth2 

On which interfaces will the router forward packets addressed to destinations 128.75.43.16 and 192.12.17.10 respectively?

A
Eth1 and Eth2
B
Eth0 and Eth2
C
Eth0 and Eth3
D
Eth1 and Eth3
Question 19 Explanation: 
Router decides route for packet by ANDing subnet mask and IP address.
If results of ANDing subnet masks and IP address are same then subnet mask with higher number of 1s is preferred.
IP address 128.75.43.16 is AND with 255.255.255.0 results 128.75.43.0 Net ID which is similar to destination of this mask, but ANDing 128.75.43.16 with 255.255.255.128 also results same destination. So, here, mask with higher number of one is considered and router will forward packet to Eth1.
ANDing 192.12.17.10 with three subnet mask in table does not result in destination Net ID so router will forward this packet to default network via Eth2.
Question 20

Consider three IP networks A, B and C. Host HA in network A sends messages each containing 180 bytes of application data to a host HC in network C. The TCP layer prefixes a 20 byte header to the message. This passes through an intermediate net­work B. The maximum packet size, including 20 byte IP header, in each network is:

     A : 1000 bytes 
     B : 100 bytes 
     C : 1000 bytes

The network A and B are connected through a 1 Mbps link, while B and C are connected by a 512 Kbps link (bps = bits per second).

Assuming that the packets are correctly delivered, how many bytes, including headers, are delivered to the IP layer at the destination for one application message, in the best case? Consider only data packets.

A
200
B
220
C
240
D
260
Question 20 Explanation: 
Application data is 180 bytes. TCP layer add 20 bytes to it and passes to IP layer, data for IP layer becomes 200 byte. HA send packet by adding 20 byte of IP header. So total size of IP packet is 220 bytes. Since, maximum packet size for packet in network A is 1000 bytes, there will be no fragmentation at network A. IP Layer at Network B removes IP header and receive 200 bytes of data. Network B has Maximum packet size 100 bytes including 20 byte IP header, network B divide data in 80 bytes fragments and add 20 byte of IP header to it.
Data will be divided in three packets as:
First packet: 80 bytes + 20 byte of header
Second packet: 80 bytes + 20 byte of header
Third packet: 40 bytes + 20 byte of header
Note: Defragmentation (grouping of fragments) is done only at destination.
HC will receive total 260 bytes including header.
Question 21

Consider three IP networks A, B and C. Host HA in network A sends messages each containing 180 bytes of application data to a host HC in network C. The TCP layer prefixes a 20 byte header to the message. This passes through an intermediate net­work B. The maximum packet size, including 20 byte IP header, in each network is:

     A : 1000 bytes 
     B : 100 bytes 
     C : 1000 bytes

The network A and B are connected through a 1 Mbps link, while B and C are connected by a 512 Kbps link (bps = bits per second).

What is the rate at which application data is transferred to host HC? Ignore errors, acknowledgements, and other overheads.

A
325.5 Kbps
B
354.5 Kbps
C
409.6 Kbps
D
512.0 Kbps
Question 21 Explanation: 
HC will receive 260 bytes in which only 180 bytes are of application data.
Application data is transferred at rate of (180/260) x 512 Kbps = 354.46 Kbps
Question 22
Consider the three-way handshake mechanism followed during TCP connection established between hosts P and Q. Let X and Y be two random 32-bit starting sequence numbers chosen by P and Q respectively. Suppose P sends a TCP connection request message to Q with a TCP segment having SYN bit = 1, SEQ number = X, and ACK bit = 0. Suppose Q accepts the connection request. Which one of the following choices represents the information present in the TCP segment header that is sent by Q to P?
A
SYN bit = 1, SEQ number = X+1, ACK bit = 0, ACK number = Y, FIN bit = 0
B
SYN bit = 0, SEQ number = X+1, ACK bit = 0, ACK number = Y, FIN bit = 1
C
SYN bit = 1, SEQ number = Y, ACK bit = 1, ACK number = X+1, FIN bit = 0
D
SYN bit = 1, SEQ number = Y, ACK bit = 1, ACK number = X, FIN bit = 0
Question 22 Explanation: 

 

Q will send the SYN bit = 1 to the connection establishment.

Q Seq number will be Y different from X

ACK bit = 1 because sending the ACK

ACK number = X+1 (Next seq number id)

FIN bit = 0 (Because establishing the connection) 

Question 23
Consider the cyclic redundancy check (CRC) based error detecting scheme having the generator polynomial X3+X+1. Suppose the message m4m3m2m1m0=11000 is to be transmitted. Check bits c2c1c0are appended at the end of the message by the transmitter using the above CRC scheme. The transmitted bit string is denoted by m4m3m2m1m0c2c1c0. The value of the checkbit sequence c2c1c0is
A
111
B
100
C
101
D
110
Question 23 Explanation: 
Question 24
Consider a computer network using the distance vector routing algorithm in its network layer. The partial topology of the network is as shown below.   The objective is to find the shortest-cost path from the router R to routers P and Q. Assume that R does not initially know the shortest routes to P and Q. Assume that R has three neighbouring routers denoted as X, Y and Z. During one iteration, R measures its distance to its neighbours X, Y and Z as 3, 2 and 5, respectively. Router R gets routing vectors from its neighbours that indicate that the distance to router P from routers X, Y and Z are 7, 6 and 5, respectively. The routing vector also indicates that the distanceto router Q from routers X, Y and Z are 4, 6 and 8, respectively. Which of the following statement(s) is/are correct with respect to the new routing table of R, after updation during this iteration
A
The next hop router for a packet from R to P is Y.
B
The distance from R to Q will be stored as 7.
C
The next hop router for a packet from R to Q is Z.
D
The distance from R to P will be stored as 10.
Question 24 Explanation: 

Given R gets the distance vector (3,2,5)

After the one iteration distance vector from X to P, Y to P,  and Z to P is (7, 6, 5) respectively

The distance vector from R to P via X Y Z is (3+7, 2+6, 5+5) =(10, 8, 10)

So Take minimum distance from R to P which is 8 via Y

After the iteration distance vector from X to Q, Y to Q, Z to Q is ( 4, 6, 8) respectively

The distance vector from R to Q via X Y Z is (3+4, 2+6, 5+8) = (7, 8 13)

So Take minimum distance from R to Q  which is 7 via X.



Question 25
Consider a network using the pure ALOHA medium access control protocol, where each frame is of length 1,000 bits. The channel transmission rate is 1 Mbps (= 106bits per second). The aggregate number of transmissions across all the nodes (including new frame transmissions and retransmitted frames due to collisions) is modelled as a Poisson process with a rate of 1,000 frames per second. Throughput is defined as the average number of frames successfully transmitted per second. The throughput of the network (rounded to the nearest integer) is _________. 
A
135
Question 25 Explanation: 

1 frames takes = Tt = L / B.w. => 1000 / 10^6 = 1 millisec

1000 frame Tt = 1000 x 1 millisec = 1 sec

In 1 sec, 1000 frames sends, which is 1 millisec per frame.

So, G = 1

 

Efficiency of Pure Aloha (η) = G x e-2G

where G = Number of requests per time slot willing to transmit.

e = Mathematical constant approximately equal to 2.718

So, η = 1 x 2.718(-2 x 1) = 0.1353

Therefore, In 1 sec1000 frames = 0.1353 x 1000 = 135.3(closest integer) =>135

 

Throughput =>  135

Question 26
Consider the string abbccddeee. Each letter in the string must be assigned a binary code satisfying the following properties:
  1. For any two letters, the code assigned to one letter must not be a prefix of the code assigned to the other letter.
  2. For any two letters of the same frequency, the letter which occurs earlier in the dictionary order is assigned a code whose length is at most the length of the code assigned to the other letter.
Among the set of all binary code assignments which satisfy the above two properties, what is the minimum length of the encoded string?
A
21
B
30
C
23
D
25
Question 26 Explanation: 

Input String : abbccddeee

The character frequencies are

Character

a

b

c

d

e

Frequency

1

2

2

2

3

Binary Code

?

?

?

?

?


Question 27

Which one of the following statements is FALSE?

A
Packet switching leads to better utilization of bandwidth resources than circuit switching.
B
Packet switching results in less variation in delay than circuit switching.
C
Packet switching requires more per packet processing than circuit switching.
D
Packet switching can lead to reordering unlike in circuit switching.
Question 28

Which one of the following statements is FALSE?

A
TCP guarantees a minimum communication rate
B
TCP ensures in-order delivery
C
TCP reacts to congestion by reducing sender window size
D
TCP employs retransmission to compensate for packet loss
Question 28 Explanation: 
Option B:
Sequence numbers can allow receivers to discard duplicate packets and properly sequence reordered packets.
Option C:
If the congestion is deleted, the transmitter decreases the transmission rate by a multiplicative factor.
Option D:
Acknowledgement allows the sender to determine when to retransmit lost packets.
Question 29

Which one of the following statements is FALSE?

A
HTTP runs over TCP
B
HTTP describes the structure of web pages
C
HTTP allows information to be stored in a URL
D
HTTP can be used to test the validity of a hypertext link
Question 29 Explanation: 
Note: Out of syllabus.
Question 30

A sender is employing public key cryptography to send a secret message to a receiver. Which one of the following statements is TRUE?

A
Sender encrypts using receiver’s public key
B
Sender encrypts using his own public key
C
Receiver decrypts using sender’s public key
D
Receiver decrypts using his own public key
Question 30 Explanation: 
Sender can encrypts using the receiver public key and receiver decrypts it using his own private key.
Question 31

A subnet has been assigned a subnet mask of 255.255.255.192. What is the maximum number of hosts that can belong to this subnet?

A
14
B
30
C
62
D
126
Question 31 Explanation: 
Maximum no. of hosts = 2(no. of bits in HID) - 2
= 26- 2
= 64 - 2
= 62
Question 32

A host is connected to a Department network which is part of a University network. The University network, in turn, is part of the Internet. The largest network in which the Ethernet address of the host is unique is:

A
the subnet to which the host belongs
B
the Department network
C
the University network
D
the Internet
Question 32 Explanation: 
The answer is option (D), in a specified LAN technology - Ethernet is mentioned here. So, MAC addresses will be specifically taken as physical address which is unique in the entire world.
Question 33

In TCP, a unique sequence number is assigned to each

A
byte
B
word
C
segment
D
message
Question 33 Explanation: 
In TCP, a unique sequence number is assigned to each byte.
Question 34

Which of the following objects can be used in expressions and scriplets in JSP (Java Server Pages) without explicitly declaring them?

A
session and request only
B
request and response only
C
response and session only
D
session, request and response
Question 34 Explanation: 
Note: Out of syllabus.
Question 35

Consider the following statements:
I. telnet, ftp and http are application layer protocols.
II.l EJB (Enterprise Java Beans) components can be deployed in a J2EE (Java2 Enterprise Edition) application server.
III. If two languages conform to the Common Language Specification (CLS) of the Microsoft.NET framework, then a class defined in any one of them may be inherited in the other.
Which statements are true?

A
l and II only
B
II and III only
C
l and III only
D
I, II and III
Question 35 Explanation: 
If two languages conform to the common language specification (CLS) of the Microsoft.NET framework.
Then there are certain compliance rules which may be used for inheritance. So other statement (I) and (II) are True.
Question 36

A serial transmission T1 uses 8 information bits, 2 start bits, 1 stop bit and 1 parity bit for each character. A synchronous transmission T2 uses 3 eight bit sync characters followed by 30 eight bit information characters. If the bit rate is 1200 bits/second in both cases, what are the transfer rates of T1 and T2?

A
100 characters/sec, 153 characters/sec
B
80 characters/sec, 136 characters/sec
C
100 characters/sec, 136 characters/sec
D
80 characters/sec, 153 characters/sec
Question 36 Explanation: 
T1: 1 char = (8 + 2 + 1 + 1) = 12 bits
Transfer rate = 1200/12 = 100 char/sec
T2: Transfer character in bits = 24 + 240 = 264 bits
In 264 = 30 characters
Then in 1200 = ? 264/30 = 1200/x
x = 136.3 char/sec
So, correct option is (C).
Question 37

In a data link protocol, the frame delimiter flag is given by 0111. Assuming that bit stuffing is employed, the transmitter sends the data sequence 01110110 as

A
01101011
B
011010110
C
011101100
D
0110101100
Question 37 Explanation: 
In the data link layer, bits stuffing is employed then bit stuffing is done using the flag delimiter. If there is a flag of n bits then we will compare the data sequence with the flag and for every n-1 bits matched found, a bit 0 is stuffed in the data sequence.
Thus using the above logic,
Delimiter flag: 0111
Data sequence: 01110110
So, for a flag of 4 bits we will compare data sequence with a pattern of 3 bits, i.e., 011.
0 1 1 0 1 0 1 1 0 0
In the above pattern the underlined bits are found matched. Hence, 0 in italics is stuffed. Thus resulting in the data sequence as 0110101100 which is option (D).
Question 38

In a sliding window ARQ scheme, the transmitter's window size is N and the receiver's window size is M. The minimum number of distinct sequence numbers required to ensure correct operation of the ARQ scheme is

A
min (M,N)
B
max (M,N)
C
M + N
D
MN
Question 38 Explanation: 
For such a scheme to work properly, we will need a total of M+N distinct sequence numbers.
Question 39

Consider a 10 Mbps token ring LAN with a ring latency of 400 µs. A host that needs to transmit seizes the token. Then it sends a frame of 1000 bytes, removes the frame after it has circulated all around the ring, and finally releases the token. This process is repeated for every frame. Assuming that only a single host wishes to transmit, the effective data rate is

A
1Mbps
B
2Mbps
C
5Mbps
D
6Mbps
Question 39 Explanation: 
Note: Out of syllabus.
Question 40

A 20 Kbps satellite link has a propagation delay of 400 ms. The transmitter employs the "go back n ARQ" scheme with n set to 10. Assuming that each frame is 100 bytes long, what is the maximum data rate possible?

A
5 Kbps
B
10 Kbps
C
15 Kbps
D
20 Kbps
Question 40 Explanation: 
Question 41

Consider a simplified time slotted MAC protocol, where each host always has data to send and transmits with probability p = 0.2 in every slot. There is no backoff and one frame can be transmitted in one slot. If more than one host transmits in the same slot, then the transmissions are unsuccessful due to collision. What is the maximum number of hosts which this protocol can support, if each host has to be provided a minimum through put of 0.16 frames per time slot?

A
1
B
2
C
3
D
4
Question 41 Explanation: 
Let there be N such hosts. Then when one host is transmitting then others must be silent for successful transmission. So throughput per host,
0.16 = 0.2 × 0.8N-1
⇒ 0.8 = 0.8N-1
⇒ N = 2
Question 42

In the TCP/IP protocol suite, which one of the following is NOT part of the IP header?

A
Fragment Offset
B
Source IP address
C
Destination IP address
D
Destination port number
Question 42 Explanation: 
Destination port number is not present at IP header.
Question 43

A TCP message consisting of 2100 bytes is passed to IP for delivery across two networks. The first network can carry a maximum payload of 1200 bytes per frame and the second network can carry a maximum payload of 400 bytes per frame, excluding network overhead. Assume that IP overhead per packet is 20 bytes. What is the total IP overhead in the second network for this transmission?

A
40 bytes
B
80 bytes
C
120 bytes
D
160 bytes
Question 43 Explanation: 
At Router-1:
2120B reach R1's network layer. It removes original IP header, fragments data part at IP and then appends IP header to all fragments and forwards . So, it divides 2100 Bytes into two fragments of size 1200 and 900. And both fragments are sent to R2.
At Router-2:
Both fragments that reach R2 exceed MTU at R2. So, both are fragmented. First packet of 1200B is fragmented into 3 packets of 400 Bytes each. And second packet of 900B is fragmented into 3 fragments of 400, 400 and 100 Bytes respectively.
So, totally 6 packets reach destinations.
So, total IP overhead = 6 × 20 = 120 Bytes
Question 44

Suppose that the maximum transmit window size for a TCP connection is 12000 bytes. Each packet consists of 2000 bytes. At some point of time, the connection is in slow-start phase with a current transmit window of 4000 bytes. Subsequently, the transmitter receives two acknowledgements. Assume that no packets are lost and there are no time-outs. What is the maximum possible value of the current transmit window?

A
4000 bytes
B
8000 bytes
C
10000 bytes
D
12000 bytes
Question 44 Explanation: 
Since maximum transmit window size = 12000 B
and packet size =2000 B (or MSS)
Receiver window size = 6 MSS and
Current sender window size = 2 MSS
Slow start threshold = receiver window/2 = 3 MSS
Now current sender window size = 2 MSS <3 MSS,
which implies transmission is in slow start phase.
After receiving first Ack: Current sender window should increase exponentially to 4 MSS but since threshold = 3 MSS, current sender window size goes to threshold which is 3 MSS, then after receiving second Ack: Since now it is in congestion avoidance phase, sender window size increases linearly which makes current sender window
= 4 MSS
= 4 × 2000 B
= 8000 B
Question 45

Traceroute reports a possible route that is taken by packets moving from some host A to some other host B. Which of the following options represents the technique used by traceroute to identify these hosts

A
By progressively querying routers about the next router on the path to B using ICMP packets, starting with the first router
B
By requiring each router to append the address to the ICMP packet as it is forwarded to B. The list of all routers en-route to B is returned by B in an ICMP reply packet
C
By ensuring that an ICMP reply packet is returned to A by each router en-route to B, in the ascending order of their hop distance from A
D
By locally computing the shortest path from A to B
Question 45 Explanation: 
Traceroute works by sending packets with gradually increasing TTL value, starting with TTL value of 1. The first router receives the packet, decrements the TTL value and drops the packet because it then has TTL value zero. The router sends an ICMP time exceeded message back to the source. The next set of packets are given a TTL value of 2.
So the first router forwards the packets, but the second router drops them and replies with ICMP time exceeded. Proceeding in this way, traceroute uses the returned ICMP time exceeded messages to build a list of routers that packets traverse, until the destination is reached and returns an ICMP echo reply message.
Question 46

Which of the following statements is TRUE about CSMA/CD

A
IEEE 802.11 wireless LAN runs CSMA/CD protocol
B
Ethernet is not based on CSMA/CD protocol
C
CSMA/CD is not suitable for a high propagation delay network like satellite network
D
There is no contention in a CSMA/CD network
Question 46 Explanation: 
For CSMA/CD requires that sender is to be transmitting atleast till the first bit reaches the receiver. So the collision will be eliminated in case if it is present.
For networks with high propagation delay this time becomes too long hence the minimum packet size required becomes too big to be feasible.
Question 47

Which of the following statements is FALSE regarding a bridge?

A
Bridge is a layer 2 device
B
Bridge reduces collision domain
C
Bridge is used to connect two or more LAN segments
D
Bridge reduces broadcast domain
Question 47 Explanation: 
Bridge devices works at the data link layer of the open system interconnected (OSI) model, connecting two different networks together and providing communication between them. So, option A, C are true.
The bridge acts as a interface between two networks and speed the traffic between them and there by reduces the collision domain.
So, option B also True.
Question 48

Count to infinity is a problem associated with

A
link state routing protocol.
B
distance vector routing protocol.
C
DNS while resolving host name.
D
TCP for congestion control.
Question 48 Explanation: 
Distance vector routing protocol uses the Bellman-Ford algorithm and Ford-Fulkerson algorithm.
The Bellman-Ford algorithm does not prevent routing loops from happening and suffers from the count-to-infinity problem.
Question 49

A network with CSMA/CD protocol in the MAC layer is running at 1 Gbps over a 1 km cable with no repeaters. The signal speed in the cable is 2 × 108 m/sec. The minimum frame size for this network should be

A
10000 bits
B
10000 bytes
C
5000 bits
D
5000 bytes
Question 49 Explanation: 
For CSMA/CD protocol we know that minimum frame size required is,
L ≤ 2×Tp×B
L ≤ 2×(d/v)×B
d = 1Km = 1000m
v = 2×103 m/s
B = 109 bps
By solving the above equation we will set the value of L as,
10000 bits.
Question 50

A channel has a bit rate of 4 kbps and one-way propagation delay of 20 ms. The channel uses stop and wait protocol. The transmission time of the acknowledgement frame is negligible. To get a channel efficiency of at least 50%, the minimum frame size should be

A
80 bytes
B
80 bits
C
160 bytes
D
160 bits
Question 50 Explanation: 
There are 50 questions to complete.

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