Computer-Networks

Question 1

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.
       Computer-Networks       Data-Link-Layer       GATE 2004-IT
Question 2

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
       Computer-Networks       General       GATE 2004-IT
Question 2 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 3

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
       Computer-Networks       General       GATE 2004-IT
Question 3 Explanation: 
Note: Out of syllabus.
Question 4

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
       Computer-Networks       Network-Security       GATE 2004-IT
Question 4 Explanation: 
Sender can encrypts using the receiver public key and receiver decrypts it using his own private key.
Question 5

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
       Computer-Networks       Network-Layer       GATE 2004-IT
Question 5 Explanation: 
Maximum no. of hosts = 2(no. of bits in HID) – 2
= 26– 2
= 64 – 2
= 62
Question 6

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
       Computer-Networks       Network-Layer       GATE 2004-IT
Question 6 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 7

In TCP, a unique sequence number is assigned to each

A
byte
B
word
C
segment
D
message
       Computer-Networks       Network-Layer       GATE 2004-IT
Question 7 Explanation: 
In TCP, a unique sequence number is assigned to each byte.
Question 8

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
       Computer-Networks       Network-Layer       GATE 2004-IT
Question 8 Explanation: 
Note: Out of syllabus.
Question 9

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
       Computer-Networks       Application-Layer       GATE 2004-IT
Question 9 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 10

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
       Computer-Networks       General       GATE 2004-IT
Question 10 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 11

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
       Computer-Networks       Data-Link-Layer       GATE 2004-IT
Question 11 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 12

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
       Computer-Networks       ARQ-Protocol       GATE 2004-IT
Question 12 Explanation: 
For such a scheme to work properly, we will need a total of M+N distinct sequence numbers.
Question 13

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
       Computer-Networks       Network-Layer       GATE 2004-IT
Question 13 Explanation: 
Note: Out of syllabus.
Question 14

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
       Computer-Networks       Network-Layer       GATE 2004-IT
Question 14 Explanation: 
Question 15

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
       Computer-Networks       Data-Link-Layer       GATE 2004-IT
Question 15 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 16

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
       Computer-Networks       Network-Layer       GATE 2004-IT
Question 16 Explanation: 
Destination port number is not present at IP header.
Question 17

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
       Computer-Networks       Network-Layer       GATE 2004-IT
Question 17 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 18

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
       Computer-Networks       Network-Layer       GATE 2004-IT
Question 18 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 19

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
       Computer-Networks       Application-Layer-Protocol       GATE 2020       Video-Explanation
Question 19 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 20

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
       Computer-Networks       Routers       GATE 2020       Video-Explanation
Question 20 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 21

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
       Computer-Networks       IP-Address       GATE 2020       Video-Explanation
Question 21 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 22

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
       Computer-Networks       TCP-Congestion-Window       GATE 2020       Video-Explanation
Question 22 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 23
Consider an enterprise network with two Ethernet segments, a web server and a firewall, connected via three routers as shown below.

What is the number of subnets inside the enterprise network?
A
3
B
12
C
6
D
8
       Computer-Networks       Subnetting       GATE 2022       Video-Explanation
Question 23 Explanation: 
It can be seen as Non equal subneting.
Where router 2 (via Firewall) has the half of the addresses.
Router 1 (via web server) has other half which is further divided into two subnets which is web server and router 3. So total of 3 subnets possible.
Question 24
Consider the resolution of the domain name www.gate.org.in by a DNS resolver. Assume that no resource records are cached anywhere across the DNS servers and that iterative query mechanism is used in the resolution. The number of DNS query-response pairs involved in completely resolving the domain name Is_____________.
A
3
       Computer-Networks       DNS       GATE 2022       Video-Explanation
Question 24 Explanation: 
In the iterative query the DNS resolver go to the following three servers
1.root server, 2. TLD DNS server, 3. authoritative server.
So there will be three pairs of request and response here.
Question 25
Consider routing table of an organization’s router shown below:

Which of the following prefixes in CIDR notation can be collectively used to correctly aggregate all of the subnets in the routing table?
A
12.20.164.0/20
B
12.20.164.0/22
C
12.20.164.0/21
D
12.20.168.0/22
       Computer-Networks       Subnetting       GATE 2022       Video-Explanation
Question 25 Explanation: 
Do bit wise and with the given subnet and subnets mask then compare to given choices IP address range.
Question 26
Consider a network with three routers P, Q, R shown in the figure below. All the links have cost of unity.

The routers exchange distance vector routing information and have converged on the routing tables, after which the link Q−R fails. Assume that P and Q send out routing updates at random times, each at the same average rate. The probability of a routing loop formation (rounded off to one decimal place) between P and Q, leading to count-to-infinity problem, is___________.
A
1
       Computer-Networks       Routing       GATE 2022       Video-Explanation
Question 26 Explanation: 
If asked for R, Probablity = 1
If asked for R, from among P, Q, R
Probability = 1 /3
Question 27
Consider a 100 Mbps link between an earth station (sender) and a satellite (receiver) at an altitude of 2100 km. The signal propagates at a speed of 3×10 8 m/s. The time taken (in milliseconds, rounded off to two decimal places) for the receiver to completely receive a packet of 1000 bytes transmitted by the sender is_________
A
7.08 ms
       Computer-Networks       Transmission-and-Propagation-Delay       GATE 2022       Video-Explanation
Question 27 Explanation: 
The time required for the receiver to receive the packet is transmission time + propagation time i.e. Tt +Tp
Tt = L / B => 1000 x 8 bits / 10^8 bps = 0.08 ms
Tp = D / V => 2100 x 1000 m / 3 x10^8 ms = 7 ms
Therefore, Total time = 7.08 ms
Question 28
Consider the data transfer using TCP over a 1 Gbps link. Assuming that the maximum segment lifetime (MSL) is set to 60 seconds, the minimum number of bits required for the sequence number field of the TCP header, to prevent the sequence number space from wrapping around during the MSL is____________.
A
33
       Computer-Networks       Bandwidth       GATE 2022       Video-Explanation
Question 28 Explanation: 
Bandwidth = 1Gbps =2^30 / 8 bps
1 sec = 20^30 / 8 bytes
=> 1 sec = 20^30 / 8 sequence number
=> 60 sec =20^30 x 60 / 8 bytes
Number of sequence bits required = log (20^30 x 60 / 8) => 33
Question 29

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
       Computer-Networks       IPv4       GATE 2003
Question 29 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 30

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
       Computer-Networks       Transport Layer Protocol       GATE 2003
Question 30 Explanation: 
End to end connectivity is the required functionality provided by Transport protocol.
Question 31

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
       Computer-Networks       IP-Address       GATE 2003
Question 31 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 32

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
       Computer-Networks       Ethernet       GATE 2003
Question 32 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 33

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
       Computer-Networks       Sliding-Window-Protocol       GATE 2003
Question 33 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 34

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
       Computer-Networks       Match-the-Following       GATE 2004
Question 34 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 35

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
       Computer-Networks       Bridge-and-Router       GATE 2004
Question 35 Explanation: 
A bridge use MAC addresses (DLL layer) and router uses IP addresses (network layer).
Question 36

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
       Computer-Networks       Serial-Communication       GATE 2004
Question 36 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 37

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
       Computer-Networks       Ethernet       GATE 2004
Question 37 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 38

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
       Computer-Networks       IP-Address       GATE 2004
Question 38 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 39

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
       Computer-Networks       IPv4-and-TCP       GATE 2004
Question 39 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 40

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
       Computer-Networks       IPv4-and-TCP       GATE 2004
Question 40 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 41

Packets of the same session may be routed through different paths in:

A
TCP, but not UDP
B
TCP and UDP
C
UDP, but not TCP
D
Neither TCP nor UDP
       Computer-Networks       Routing       GATE 2005
Question 41 Explanation: 
Both TCP and UDP uses IP, which is a datagram service.
Question 42

The address resolution protocol (ARP) is used for:

A
Finding the IP address from the DNS
B
Finding the IP address of the default gateway
C
Finding the IP address that corresponds to a MAC address
D
Finding the MAC address that corresponds to an IP address
       Computer-Networks       Network-Addressing       GATE 2005
Question 42 Explanation: 
Address Resolution Protocol (ARP) is a request and reply protocol used to find MAC address from IP address.
Question 43

The maximum window size for data transmission using the selective reject protocol with n-bit frame sequence numbers is:

A
2n
B
2n-1
C
2n – 1
D
2n-2
       Computer-Networks       Sliding-Window-Protocol       GATE 2005
Question 43 Explanation: 
In Selective Reject (or Selective Repeat), maximum size of window must be half of the maximum sequence number = 2n/ 2 = 2n-1
For Go-back N, the maximum window size can be 2n – 1.
Question 44

In a network of LANs connected by bridges, packets are sent from one LAN to another through intermediate bridges. Since more than one path may exist between two LANs, packets may have to be routed through multiple bridges. Why is the spanning tree algorithm used for bridge-routing?

A
For shortest path routing between LANs
B
For avoiding loops in the routing paths
C
For fault tolerance
D
For minimizing collisions
       Computer-Networks       Bridges       GATE 2005
Question 44 Explanation: 
In a spanning tree, there is a unique path from a source to the destination, which avoids loops, since it is a tree, and contains all the nodes.
Question 45

An organization has a class B network and wishes to form subnets for 64 departments. The subnet mask would be:

A
255.255.0.0
B
255.255.64.0
C
255.255.128.0
D
255.255.252.0
       Computer-Networks       IP-Address       GATE 2005
Question 45 Explanation: 
Organization have 64 departments, and to assign 64 subnet we need 6 bits for subnet. In Class B network first two octet are reserved for NID, so we take first 6 bit of third octet for subnets and subnet mask would be 255.255.11111100.00000000 = 255.255.252.0
Question 46

In a packet switching network, packets are routed from source to destination along a single path having two intermediate nodes. If the message size is 24 bytes and each packet contains a header of 3 bytes, then the optimum packet size is:

A
4
B
6
C
7
D
9
       Computer-Networks       Network-Switching       GATE 2005
Question 46 Explanation: 

Packet size have minimum header overhead.
Question 47

Suppose the round trip propagation delay for a 10 Mbps Ethernet having 48-bit jamming signal is 46.4 μs. The minimum frame size is:

A
94
B
416
C
464
D
512
       Computer-Networks       Ethernet       GATE 2005
Question 47 Explanation: 
Given RTT = 46.4 μs, B.w. = 10 Mbps
Round trip propagation delay is RTT = 2*Tp
Minimum frame size of Ethernet can be found by using formula Tt = 2*Tp
Let L is minimum frame size. Then L / 10Mbps = 46.4 μs
L = 464 Kbits
It has nothing to do with jamming signal.
Question 48

Let E1 and E2 be two entities in an E/R diagram with simple single-valued attributes. R1 and R2 are two relationships between E1 and E2, where R1 is one-to-many and R2 is many-to-many. R1 and R2 do not have any attributes of their own. What is the minimum number of tables required to represent this situation in the relational model?

A
2
B
3
C
4
D
5
       Computer-Networks       ER-Model       GATE 2005
Question 48 Explanation: 
R1 and R2 two relationships between E1 and E2.
R1 is one to many.
R2 is many to many.
→ E1 and E2 have separate table because they need to store multiple values.
→ R2 also have separate table by considering Primary keys E1 and E2 as foreign keys.
→ R1 is converted to many side table i.e., E2 as Primary key and E1 as Foreign key.
So, totally we need 3 tables to store the value.
Question 49

For which one of the following reasons does Internet Protocol (IP) use the time-to- live (TTL) field in the IP datagram header?

A
Ensure packets reach destination within that time
B
Discard packets that reach later than that time
C
Prevent packets from looping indefinitely
D
Limit the time for which a packet gets queued in intermediate routers
       Computer-Networks       IP-Header       GATE 2006
Question 49 Explanation: 
It prevent infinite looping over network . Because each router decrease its value by one and when this (TTL) value become zero , then it is discarded by router silently.
Question 50

Station A uses 32 byte packets to transmit messages to Station B using a sliding window protocol. The round trip delay between A and B is 80 milliseconds and the bottleneck bandwidth on the path between A and B is 128 kbps. What is the optimal window size that A should use?

A
20
B
40
C
160
D
320
       Computer-Networks       Sliding-Window-Protocol       GATE 2006
Question 50 Explanation: 
Tt = L / B = 32 bytes/ 128 kbps = 32*8/128 ms = 2 ms
Round trip delay = 2 * Tp = 80 ms (given)
Optimal window size is = (Tt + 2*Tp) / Tt = 82 / 2 = 41
Option is not given, closest option is 40.
There are 50 questions to complete.

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