Question 1
Which one of the following is not a client server application?
A
Internet chat
B
Web browsing
C
E-mail
D
ping
GATE CS 2010    Misc Topics in Computer Networks    
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Question 1 Explanation: 
Ping is not a client server application. Ping is a computer network administration utility used to test the reachability of a host on an Internet Protocol (IP). In ping, there is no server that provides a service.
Question 2
Match the following:

(P) SMTP     (1) Application layer
(Q) BGP      (2) Transport layer
(R) TCP      (3) Data link layer
(S) PPP      (4) Network layer
             (5) Physical layer 
A
P – 2 Q – 1 R – 3 S – 5
B
P – 1 Q – 4 R – 2 S – 3
C
P – 1 Q – 4 R – 2 S – 5
D
P – 2 Q – 4 R – 1 S – 3
Misc Topics in Computer Networks    GATE-CS-2007    
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Question 2 Explanation: 
Question 3
In the following pairs of OSI protocol layer/sub-layer and its functionality, the INCORRECT pair is
A
Network layer and Routing
B
Data Link Layer and Bit synchronization
C
Transport layer and End-to-end process communication
D
Medium Access Control sub-layer and Channel sharing
Misc Topics in Computer Networks    GATE-CS-2014-(Set-3)    
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Question 3 Explanation: 
1) Yes, Network layer does Rotuing
2) No, Bit synchronization is provided by Physical Layer
3) Yes, Transport layer provides End-to-end process 
   communication
4) Yes, Medium Access Control sub-layer of Data Link Layer provides
   Channel sharing. 
Question 4
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
Misc Topics in Computer Networks    GATE-CS-2004    
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Question 4 Explanation: 
Data link layer is the second layer of the OSI Model. This layer is responsible for data transfer between nodes on the network and providing a point to point local delivery framework. So, P matches with 1. Network layer is the third layer of the OSI Model. This layer is responsible for forwarding of data packets and routing through intermediate routers. So, Q matches with 4. Transport layer is the fourth layer of the OSI Model. This layer is responsible for delivering data from process to process. So, R matches with 3.   Thus, A is the correct option.   Please comment below if you find anything wrong in the above post.
Question 5
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
Misc Topics in Computer Networks    GATE-CS-2004    
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Question 6
Host A sends a UDP datagram containing 8880 bytes of user data to host B over an Ethernet LAN. Ethernet frames may carry data up to 1500 bytes (i.e. MTU = 1500 bytes). Size of UDP header is 8 bytes and size of IP header is 20 bytes. There is no option field in IP header. How may total number of IP fragments will be transmitted and what will be the contents of offset field in the last fragment?
A
6 and 925
B
6 and 7400
C
7 and 1110
D
7 and 8880
Misc Topics in Computer Networks    GATE-CS-2015 (Set 2)    
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Question 6 Explanation: 
UDP data = 8880 bytes
UDP header = 8 bytes
IP Header = 20 bytes

Total Size excluding IP Header = 8888 bytes.

Number of fragments  = ⌈ 8888 / 1480 ⌉ 
                     = 7
Refer the Kurose book slides on IP (Offset is always scaled by 8)

Offset of last segment = (1480 * 6) / 8 = 1110 
Question 7
Since it is a network that uses switch, every packet goes through two links, one from source to switch and other from switch to destination. Since there are 10000 bits and packet size is 5000, two packets are sent. Transmission time for each packet is 5000 / 1077 bits per second links. Each link has a propagation delay of 20 microseconds. The switch begins forwarding a packet 35 microseconds after it receives the same. If 10000 bits of data are to be transmitted between the two hosts using a packet size of 5000 bits, the time elapsed between the transmission of the first bit of data and the reception of the last bit of the data in microseconds is _________.
A
1075
B
1575
C
2220
D
2200
Misc Topics in Computer Networks    GATE-CS-2015 (Set 3)    
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Question 7 Explanation: 
Sender host transmits first packet to switch, the transmission time is 5000/107 which is 500 microseconds. After 500 microseconds, the second packet is transmitted. The first packet reaches destination in 500 + 35 + 20 + 20 + 500 = 1075 microseconds. While the first packet is traveling to destination, the second packet starts its journey after 500 microseconds and rest of the time taken by second packet overlaps with first packet. So overall time is 1075 + 500 = 1575.
Question 8
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
Misc Topics in Computer Networks    GATE-IT-2004    
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Question 9
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
Misc Topics in Computer Networks    GATE-IT-2004    
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Question 9 Explanation: 
HTML describes structure of page not HTTP. HTTP is the set of rules for transferring files (text, graphic images, sound, video, and other multimedia files) on the World Wide Web.
Question 10
A serial transmission Ti 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 Ti 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
Misc Topics in Computer Networks    GATE-IT-2004    
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Question 10 Explanation: 

Serial communication :
Total number of bits transmitted = 8 + 2 + 1 + 1 = 12 bits Bit rate = 1200 / second Transfer Rate = 1200 * (8/12) = 800 bits/sec = 100 bytes/sec = 100 characters/sec
Synchronous transmission :
Total number of bits transmitted = 3 + 30 = 33 bits Transfer Rate = 1200 * (30/33) = 136 characters/sec
 
Thus, option (C) is correct.
 
Please comment below if you find anything wrong in the above post.
Question 11
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
Misc Topics in Computer Networks    GATE-IT-2004    
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Question 11 Explanation: 
  In general sliding window ARQ scheme , the sending process sends a number of frames without worrying about receiving an ACK(acknowledgement) packet from the receiver. The sending window size in general is N and receiver window is 1. This means it can transmit N frames to its peer before requiring an ACK. The receiver keeps track of the sequence number of the next frame it expects to receive and sends that number with ever ACK it sends. But in case of the question the sender window size is N and receiver is M so the receiver will accept M frames instead of 1 frame in general. Thus sending M sequence numbers attached with the acknowledgement. Hence, for such a scheme to work properly we will need a total of M+ N distinct sequence numbers.   This solution is contributed by Namita Singh.
Question 12
Which one of the following protocols is NOT used to resolve one form of address to another one?
A
DNS
B
ARP
C
DHCP
D
RARP
Misc Topics in Computer Networks    GATE-CS-2016 (Set 1)    
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Question 12 Explanation: 
DHCP is used to assign IP dynamically. All others are used to convert one address to other.
Question 13
Identify the correct sequence in which the following packets are transmitted on the network by a host when a browser requests a webpage from a remote server, assuming that the host has just been restarted.
A
HTTP GET request, DNS query, TCP SYN
B
DNS query, HTTP GET request, TCP SYN
C
DNS query, TCP SYN, HTTP GET request
D
TCP SYN, DNS query, HTTP GET request
Misc Topics in Computer Networks    GATE-CS-2016 (Set 2)    
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Question 13 Explanation: 
Step 1 : Whenever the client request for a webpage, the query is made in the form say www.geeksforgeeks.org. As soon as the query is made the server makes the DNS query to identify the Domain Name Space. DNS query is the process to identify the IP address of the DNS such as www.org. The client’s computer will make a DNS query to one of its internet service provider’s DNS server. Step 2 : As soon as DNS server is located a TCP connection is to be established for the further communication. The TCP protocol requests the server to establishing a connection by sending a TCP SYN message. Which is further responded by the server using SYN_ ACK from server to client and then ACK back to server from client (3- way hand shaking protocol). Step 3 : Once the connection has been established the HTTP protocol comes into picture. It requests for the webpage using its GET method and thus, sending an HTTP GET request. Hence, the correct sequence for the transmission of packets is DNS query, TCP SYN, HTTP GET request. This explanation has been contributed by Namita Singh.
Question 14
Consider the following statements about the timeout value used in TCP. i. The timeout value is set to the RTT (Round Trip Time) measured during TCP connection establishment for the entire duration of the connection. ii. Appropriate RTT estimation algorithm is used to set the timeout value of a TCP connection. iii. Timeout value is set to twice the propagation delay from the sender to the receiver. Which of the following choices hold?
A
(i) is false, but (ii) and (iii) are true
B
(i) and (iii) are false, but (ii) is title
C
(i) and (ii) are false, but (iii) is true
D
(i), (ii) and (iii) are false
Transport Layer    Misc Topics in Computer Networks    Gate IT 2007    
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Question 14 Explanation: 
  Time-out timer in TCP: One can’t use static timer used in data link layer (DLL), which is HOP to HOP connection, since nobody knows how many hops are there in the path form sender to receiver as it uses IP service and path may vary time to time. So, dynamic timers are used in TCP. Time-out timer should increase or decrease depending on traffic to avoid unnecessary congestion due to retransmissions. There are three algorithms are for this purpose: 1. Basic algorithm 2. Jacobson’s algorithm 3. Karl’s modification. Solution:
  1. The timeout value is set to the RTT (Round Trip Time) measured during TCP connection establishment for the entire duration of the connection.- FALSE The timeout value can’t be fixed for entire duration as it will turn timer to static timer, we need dynamic timer for timeout.
  1. Appropriate RTT estimation algorithm is used to set the timeout value of a TCP connection.-TRUE Yes, all three algorithm are appropriate RTT estimation algorithm used to set timeout value dynamically.
  1. Timeout value is set to twice the propagation delay from the sender to the receiver.-FALSE This statement is false because, timeout value is set to twice the propagation delay in data link layer where, hop to hop distance is known, not in TCP layer.
This solution is contributed by Sandeep pandey.
Question 15
A firewall is to be configured to allow hosts in a private network to freely open TCP connections and send packets on open connections. However, it will only allow external hosts to send packets on existing open TCP connections or connections that are being opened (by internal hosts) but not allow them to open TCP connections to hosts in the private network. To achieve this the minimum capability of the firewall should be that of
A
A combinational circuit
B
A finite automaton
C
A pushdown automaton with one stack
D
A pushdown automaton with two stacks
Misc Topics in Computer Networks    Network Security    Gate IT 2007    
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Question 15 Explanation: 
A) A combinational circuit => Not possible, because we need memory in Firewall, Combinational ckt has none.
B) A finite automaton => We need infinite memory, there is no upper limit on Number of TCP ckt so Not this.
C) A pushdown automaton with one stack => Stack is infinite. Suppose we have 2 connections , we have pushed details of those on stack we can not access the details of connection which was pushed first, without popping it off. So Big NO.
D) pushdown automaton with two stacks => This is TM. It can do everything our normal computer can do so Yes. Firewall can be created out of TM.
Question 16
How many bytes of data can be sent in 15 seconds over a serial link with baud rate of 9600 in asynchronous mode with odd parity and two stop bits in the frame?
A
10,000 bytes
B
12,000 bytes
C
15,000 bytes
D
27,000 bytes
Data Link Layer    Misc Topics in Computer Networks    Gate IT 2008    
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Question 16 Explanation: 
  Background required - Physical Layer in OSI Stack In serial Communications, information is transferred in or out one bit at a time. The baud rate specifies how fast data is sent over a serial line. It’s usually expressed in units of bits-per-second (bps). Each block (usually a byte) of data transmitted is actually sent in a packet or frame of bits. Frames are created by appending synchronization and parity bits to our data. frame
"9600 baud" means that the serial port is capable of transferring a 
maximum of 9600 bits per second.

1 sec--------> 9600 bits
15 sec------->9600*15 bits

Total Data To send in 1 frame = 1 bit(start) + 8 bits(char size) + 1 bit(Parity) + 2 bits(Stop) 

= 12 bits.

Number of 8-bit characters that can be transmitted per second  = (9600 * 15)/12 = 12000 bytes.
This explanation is contributed by Pranjul Ahuja.
Question 17
Provide the best matching between the entries in the two columns given in the table below:
2008_20
A
I-a, II-d, III-c, IV-b
B
I-b, II-d, III-c, IV-a
C
I-a, II-c, III-d, IV-b
D
I-b, II-c, III-d, IV-a
Misc Topics in Computer Networks    Gate IT 2008    
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Question 17 Explanation: 
DNS - Allows caching of entries at local server.
Question 18
Which protocol will be used to automate the IP configuration mechanism which includes IP address, subnet mask, default gateway, and DNS information?
A
SMTP
B
DHCP
C
ARP
D
TCP/IP
Misc Topics in Computer Networks    GATE 2017 Mock    
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Question 18 Explanation: 
DHCP (Dynamic Host Configuration Protocol) is used to provide IP information to the hosts on the network along with the information regarding IP address, subnet mask, default gateway and DNS information.
Question 19
In Go–back 3 flow control protocol every 6th packet is lost. If we have to send 11 packets. How many transmissions will be needed ?
A
10
B
17
C
12
D
9
Misc Topics in Computer Networks    GATE 2017 Mock    
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Question 19 Explanation: 
In Go back N, if we don’t receive acknowledgement for a packet, whole window of that packet is sent again. As a packet is received window is slided. Here, window size is 3. Initially window will contain 1,2,3 then as acknowledgement of 1 is received window slides so 4 is transmitted. Now,when 4th packet’s acknowledgement is received 7th packet is sent and when 5th packet’s acknowledgement is received 8th packet is sent. Now, as acknowledgement of 6 is not received so the window of 6 i.e. 6,7,8 packets are retransmitted.Now the 6th packet from there is 9, so 9,10 will be retransmitted. These are the serial transmissions of packets: 1 2 3 4 5 6 7 8 6 7 8 9 10 11 9 10 11 . Hence total 17 transmissions are needed. Packets in bold in the above were failed transmissions. Hence their window (underlined) was resent.
Question 20
What will be the total minimum bandwidth of the channel required for 7 channels of 400 kHz bandwidth multiplexed together with each guard band of 20 kHz?
A
2800 khz
B
2600 khz
C
3600 khz
D
2920 khz
Misc Topics in Computer Networks    GATE 2017 Mock    
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Question 20 Explanation: 
(for 6 guard band 20 * 6 = 120) + (for 7 channels 400* 7= 2800)
= 120+ 2800 = 2920 kHz
There are 20 questions to complete.

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