Computer-Networks

Question 1

Which of the following protocol pairs can be used to send and retrieve e-mails (in that order)?

A
SMTP, MIME
B
SMTP, POP3
C
IMAP, POP3
D
IMAP, SMTP
       Computer-Networks       Application-Layer-Protocol       GATE 2019       Video-Explanation
Question 1 Explanation: 
SMTP & POP3 are the protocols which are responsible for the email communication, SMTP is responsible for outgoing mail & POP3 is responsible for retrieving mail.
POP3: Post Office Protocol (Responsible for retrieve email)
SMTP: Simple Mail Transfer Protocol (Responsible for send Email)
IMAP: Internet Message Access protocol (Responsible for store and view)
MIME: Multi purpose Internet Mail Extensions (For media)
Question 2

The value of 351 mod 5 is ______.

A
3
B
5
C
2
D
1
       Computer-Networks       Network-Security       GATE 2019       Video-Explanation
Question 2 Explanation: 
351 mod 5
⇒ 31 = 3 ⇒ 3 mod 5 = 3
32 ⇒ 9 mod 5 = 4
33 ⇒ 27 mod 5 = 2
34 ⇒ 81 mod 5 = 1
35 ⇒ 243 mod 5 = 3
For every four numbers sequence is repeating.
So, (51 % 4) = 3
⇒ 33 = 27
⇒ 27 mod 5 = 2
Question 3

Consider three machines M, N and P with IP addresses 100.10.5.2, 100.10.5.5, and 100.10.5.6 respectively. The subnet mask is set to 255.255.255.252 for all the three machines. Which one of the following is true?

A
M, N, and P all belong to the same subnet
B
Only M and N belong to the same subnet
C
M, N and P belong to three different subnets
D
Only N and P belong to the same subnet
       Computer-Networks       IP-Address       GATE 2019       Video-Explanation
Question 3 Explanation: 
Take each IP and do bitwise AND with the given Subnet Mask. If we get the same network ID for the given IP’S then it will belong to the same subnet.

Therefore, N and P belong to the same subnet.
Question 4

In an RSA cryptosystem, the value of the public modulus parameter n is 3007. If it is also known that Φ(n) = 2880, where Φ() denotes Euler’s Quotient Function, then the prime factor of n which is greater than 50 is ______.

A
107
B
97
C
45
D
92
       Computer-Networks       Network-Security       GATE 2019       Video-Explanation
Question 4 Explanation: 
It can be solved by Hit and trial method in less time.
n = 3007, fi(n) = 2880 → fi(n) = (p – 1) (q – 1),
where p, q are prime factor of n.
The unit place of n is 7, it is a prime number and factor will be
1.7=7
11*17
21*37
31*47
….
31*97 =>3007
n = 3007 => 31*97
Therefore, 31 & 97 are the two prime numbers, which is satisfying the condition and 97 is greater than 50.
So, 97 is the correct answer.
Other methods:
When ϕ(n) is given when n=pq where p and q are prime numbers, then we have
ϕ(n) = (p−1)(q−1) = pq−(p+q)+1
But pq=n,
therefore, ϕ(n) = n−(p+q)+1 and p+q = n+1−ϕ(n).
Now, p and q are the roots of the equation,
x2 − (p+q)x + pq = (x-p)(x-q)
Substituting for p+q and pq in the above equation
x2 – (n+1-ϕ(n))x + n
Question 5

Consider that 15 machines need to be connected in a LAN using 8-port Ethernet switches. Assume that these switches do not have any separate uplink ports. The minimum number of switches needed is _____.

A
3
B
7
C
1
D
5
       Computer-Networks       Ethernet       GATE 2019       Video-Explanation
Question 5 Explanation: 
In 8 port Ethernet switch one port for the network connection and remaining 7 port for the machine.
Therefore, the total required number of the switches = Ceil (15 /7) = 3
Question 6

Suppose that in an IP-over-Ethernet network, a machine X wishes to find the MAC address of another machine Y in its subnet. Which one of the following techniques can be used for this?

A
X sends an ARP request packet to the local gateway’s IP address which then finds the MAC address of Y and sends to X
B
X sends an ARP request packet with broadcast IP address in its local subnet
C
X sends an ARP request packet to the local gateway’s MAC address which then finds the MAC address of Y and sends to X
D
X sends an ARP request packet with broadcast MAC address in its local subnet
       Computer-Networks       ARP-RARP       GATE 2019       Video-Explanation
Question 6 Explanation: 
Address Resolution Protocol (ARP) is a protocol for mapping an Internet Protocol address (IP address) to a physical machine address (MAC) that is recognized in the local network.
Since both are present in the same subnet thus an ARP request packet can be sent as broadcast MAC address, all will see but the only destination will reply as a unicast reply.
Video Reference :
http://eclassesbyravindra.com/mod/page/view.php?id=147
Question 7

Consider a long-lived TCP session with an end-to-end bandwidth of 1 Gbps (= 109 bits/second). The session starts with a sequence number of 1234. The minimum time (in seconds, rounded to the closest integer) before this sequence number can be used again is _________.

A
33
B
34
C
35
D
36
       Computer-Networks       TCP       GATE 2018       Video-Explanation
Question 7 Explanation: 
In TCP, Sequence number field is 32 bit, which means 232 sequence number per byte are possible. Whatever be the starting sequence number the possible number will be 232bytes
The process of using all the sequence number and repeating a previously used sequence number.
The time taken to wrap around is called wrap around time:
Minimum Time = Wrap around time = Total number of bits in sequence number / Bandwidth = 232 * 8 / 109 = 34.35 == 34 (closest integer)
Question 8

Match the following:

           Field                       Length in bits
P. UDP Header's Port Number               I.  48
Q. Ethernet MAC Address                   II.  8
R. IPv6 Next Header                       III.32
S. TCP Header's Sequence Number           IV. 16
A
P-III, Q-IV, R-II, S-I
B
P-II, Q-I, R-IV, S-III
C
P-IV, Q-I, R-II, S-III
D
P-IV, Q-I, R-III, S-II
       Computer-Networks       Header       GATE 2018       Video-Explanation
Question 8 Explanation: 
P. UDP Header’s Port Number – 16 bits
Q. Ethernet MAC Address – 48 bits
R. IPV6 Next Header – 8 bits
S. TCP Header’s Sequence Number – 32 bits
Question 9
Consider the following statements regarding the slow start phase of the TCP congestion control algorithm. Note that cwnd stands for the TCP congestion window and MSS denotes the Maximum Segment Size.
(i) The cwnd increase by 2 MSS on every successful acknowledgement.
(ii) The cwnd approximately doubles on every successful acknowledgement.
(iii) The cwnd increase by 1 MSS every round trip time.
(iv) The cwnd approximately doubles every round trip time.
Which one of the following is correct?
A
Only (ii) and (iii) are true
B
Only (i) and (iii) are true
C
Only (iv) is true
D
Only (i) and (iv) are true
       Computer-Networks       TCP-Congestion-Window       GATE 2018       Video-Explanation
Question 9 Explanation: 
In Slow-start, the value of the Congestion Window will be increased by 1 MSS with each acknowledgement (ACK) received, and effectively doubling the window size each round-trip time
Initially, TCP starts with cwnd of 1 MSS. On every ack, it increases cwnd by 1 MSS.
That is, cwnd doubles every RTT.
Initially sends 1 segment. On ack, sends 2 segments.
After these 2 acks come back, sends 4 segments etc.
TCP rate increases exponentially during slow start.
Slow start continues till cwnd reaches threshold.
After threshold is reached, cwnd increases more slowly, by one 1 MSS every RTT.
Question 10

Consider a simple communication system where multiple nodes are connected by a shared broadcast medium (like Ethernet or wireless). The nodes in the system use the following carrier-sense based medium access protocol. A node that receives a packet to transmit will carrier-sense the medium for 5 units of time. If the node does not detect any other transmission in this duration, it starts transmitting its packet in the next time unit. If the node detects another transmission, it waits until this other transmission finishes, and then begins to carrier-sense for 5 time units again. Once they start to transmit, nodes do not perform any collision detection and continue transmission even if a collision occurs. All transmissions last for 20 units of time. Assume that the transmission signal travels at the speed of 10 meters per unit time in the medium.

Assume that the system has two nodes P and Q, located at a distance d meters from each other. P starts transmitting a packet at time t=0 after successfully completing its carrier-sense phase. Node Q has a packet to transmit at time t=0 and begins to carrier-sense the medium.

The maximum distance d (in meters, rounded to the closest integer) that allows Q to successfully avoid a collision between its proposed transmission and P’s ongoing transmission is ___________.

A
50
B
51
C
52
D
53
       Computer-Networks       Ethernet       GATE 2018       Video-Explanation
Question 10 Explanation: 
Node senses the medium for 5 unit time. it means, any packet which arrives within 5 unit will be sensed and keep the channel busy.
Now signal travels at the speed of 10 meters per unit time.
Therefore, in 5 unit time, it can travel a maximum distance (d) of 50 m (5*10), which allows the receiver (Q) to sense that the channel is busy.
Question 11

Consider an IP packet with a length of 4,500 bytes that includes a 20-byte IPv4 header and a 40-byte TCP header. The packet is forwarded to an IPv4 router that supports a Maximum Transmission Unit (MTU) of 600 bytes. Assume that the length of the IP header in all the outgoing fragments of this packet is 20 bytes. Assume that the fragmentation offset value stored in the first fragment is 0.

The fragmentation offset value stored in the third fragment is __________.

A
144
B
145
C
146
D
147
       Computer-Networks       IPv4-an-Fragmentation       GATE 2018       Video-Explanation
Question 11 Explanation: 
MTU = 600 bytes, IP header = 20 bytes
Therefore Payload = 600 – 20 = 580 bytes.
As we know fragment size should be multiple of 8 but 580 bytes is not a multiple of 8, therefore fragment size is 576 bytes.
Offset value of kth fragment = Fragment size *( kth fragment – 1) / scaling factor
Offset value of third fragment = 576 * (3-1) / 8 = 144
Question 12

Consider a TCP client and a TCP server running on two different machines. After completing data transfer, the TCP client calls close to terminate the connection and a FIN segment is sent to the TCP server. Server-side TCP responds by sending an ACK, which is received by the client-side TCP. As per the TCP connection state diagram (RFC 793), in which state does the client-side TCP connection wait for the FIN from the server-side TCP?

A
LAST-ACK
B
TIME-WAIT
C
FIN-WAIT-1
D
FIN-WAIT-2
       Computer-Networks       TCP-Transition-Diagram       GATE 2017 [Set-1]       Video-Explanation
Question 12 Explanation: 
Client has sent FIN segment to the server and moves to FIN-WAIT-1,
i.e. waiting for the ACK for own FIN segment.
There are two possibilities here:
I. If Client receives ACK for its FIN then client will move to FIN-WAIT-2 and will wait for matching FIN from server side.
After receiving the FIN from server, client will send ACK and move to TIME-WAIT state.
II. Client has sent FIN segment but didn’t get ACK till the time.
Instead of ACK, client received FIN from server side.
Client will acknowledge this FIN and move to CLOSE state.
Here Client will wait for the ACK for its own FIN.
After receiving ACK, client will move to TIME-WAIT state.
Here we encounter First Case.
So, the solution is (D).
Refer this TCP state transition diagram:
Question 13
A sender S sends a message m to receiver R, which is digitally signed by S with its private key. In this scenario, one or more of the following security violations can take place.
(I) S can launch a birthday attack to replace m with a fraudulent message.
(II) A third party attacker can launch a birthday attack to replace m with a fraudulent message.
(III) R can launch a birthday attack to replace m with a fraudulent message.
Which of the following are possible security violations?
A
(I) and (II) only
B
(I) only
C
(II) only
D
(II) and (III) only
       Computer-Networks       Security       GATE 2017 [Set-1]       Video-Explanation
Question 13 Explanation: 
Birthday attack Problem is when sender replaces original message with fraud message having same message digest as the original message, along with the digital signature of the original message.
(I) Can the sender replace the message with a fraudulent message?
Yes, definitely because the sender will encrypt the message with its private key.
It can encrypt another message also with its private key.
(II) Can the third party send a fraudulent message?
No, because the third party doesn’t know about the private key of the sender.
(III) Can receiver send the fraudulent message?
No, the receiver also doesn’t know about the Private key of the sender.
So receiver also cannot send the fraudulent message.
Question 14

A computer network uses polynomials over GF(2) for error checking with 8 bits as information bits and uses x3 + x + 1 as the generator polynomial to generate the check bits. In this network, the message 01011011 is transmitted as

A
01011011010
B
01011011011
C
01011011101
D
01011011100
       Computer-Networks       CRC       GATE 2017 [Set-1]       Video-Explanation
Question 14 Explanation: 
Given CRC generator polynomial = x3+x+1
= 1∙x3+0∙x2+1∙x1+1∙x0
= 1011
Message = 01011011

So, the message 01011011 is transmitted as
Question 15

In a RSA cryptosystem, a participant A uses two prime numbers p = 13 and q = 17 to generate her public and private keys. If the public key of A is 35, then the private key of A is _________.

A
11
B
12
C
13
D
14
       Computer-Networks       RSA-Algorithm       GATE 2017 [Set-1]       Video-Explanation
Question 15 Explanation: 
Correct Answer is 11.
Given, p=13, q=17, e=35 (Public key), d=? (Private key)
As per RSA Algorithm; following steps:
Step 1: Find n = p×q = 13×17 = 221
Step 2: Find ∅(n) = (p-1)×(q-1) = 12×16 = 192
Step 3: d×e mod ∅(n) = 1 ⇒ (d = e(-1) mod ∅(n))
or
d×e = 1 mod ∅(n)
⇒ d×35 mod 192 = 1
Question 16

The value of parameters for the Stop-and-Wait ARQ protocol are as given below:

          Bit rate of the transmission channel = 1 Mbps.
          Propagation delay from sender to receiver = 0.75 ms.
          Time to process a frame = 0.25 ms.
          Number of bytes in the information frame = 1980.
          Number of bytes in the acknowledge frame = 20.
          Number of overhead bytes in the information frame = 20.

Assume that there are no transmission errors. Then, the transmission efficiency (expressed in percentage) of the Stop-and-Wait ARQ protocol for the above parameters is _________ (correct to 2 decimal places).

A
89.33%
B
89.34%
C
89.35%
D
89.36%
       Computer-Networks       Stop-and-Wait-ARQ-Protocol       GATE 2017 [Set-1]       Video-Explanation
Question 16 Explanation: 
Given Data:
B = 1Mbps, L = 1980Bytes, Overhead = 20Bytes
TProc = 0.25ms, LAck = 20Bytes
Tp=0.75ms
Total Data size(L) = (L + overhead) = 1980+20 = 2000Bytes
Efficiency of Stop & Wait ARQ?
Tt = L/B = 2000Bytes/1Mbps = (2000×8bits)/(106 b/s) = 16msec
TAck = LAck/B = (20×8bits)/(106 bits/sec) = 0.16msec
∴ In Stop and Wait ARQ, efficiency
ƞ = Tt/(Tt+TAck+2Tp+TProc) = 16ms/(16+0.16+2×0.75+0.25ms) = 16ms/17.91ms = 0.8933
Question 17

Consider the following statements about the routing protocols, Routing Information Protocol (RIP) and Open Shortest Path First (OSPF) in an IPv4 network.

    • I: RIP uses distance vector routing
    • II: RIP packets are sent using UDP
    • III: OSPF packets are sent using TCP
    IV: OSPF operation is based on link-state routing

Which of the statements above are CORRECT?

A
I and IV only
B
I, II and III only
C
I, II and IV only
D
II, III and IV only
       Computer-Networks       Routing       GATE 2017 [Set-2]       Video-Explanation
Question 17 Explanation: 
I: RIP uses distance vector routing. “TRUE”
RIP is one of the oldest DVR protocol which employ the hop count as a routing metric.
II: RIP packets are sent using UDP. “TRUE”
RIP uses the UDP as its transport protocol, and is assigned the reserved port no 520.
III: OSPF packets are sent using TCP. “FASLE”
OSPF encapsulates its data directly into IP Packets and does not use either TCP or UDP.
IV: OSPF operation is based on link state routing. “TRUE”
OSPF is a routing protocol which uses link state routing (LSR) and works within a single autonomous system.
Hence correct is answer “C”.
Question 18

Consider socket API on a Linux machine that supports connected UDP sockets. A connected UDP socket is a UDP socket on which connect function has already been called. Which of the following statement is/are CORRECT?

    I. A connected UDP socket can be used to communicate with multiple peers simultaneously.
    II. A process can successfully call connect function again for an already connected UDP socket.
A
I only
B
II only
C
Both I and II
D
Neither I nor II
       Computer-Networks       UDP       GATE 2017 [Set-2]       Video-Explanation
Question 18 Explanation: 
A connected UDP socket, the result of calling connection a UDP socket, Connect ( ) specifying the remote address.
I. A connected UDP socket can be used to communicate with only one peer.
A DNS client can be configured to use one or more servers, normally by listing the IP addresses of the servers in the file /etc/resolv.conf.
If a single server is listed, the client can call connect, but if multiple servers are listed the client cannot call connect.
II. A process with a connected UDP socket can call connect function again for that socket for one of two reasons:
(a) To specify a new IP address and port.
(b) To unconnect the socket.
Hence, the correct answer is (B).
Question 19

The maximum number of IPv4 router addresses that can be listed in the record route (RR) option field of an IPv4 header is _________.

A
9
B
10
C
11
D
12
       Computer-Networks       IPv4       GATE 2017 [Set-2]       Video-Explanation
Question 19 Explanation: 
A record route option is used to record the internet router that handles the datagram. It can be used for debugging and management purpose.
In IPv4 header, 40 bytes are reserved for OPTIONS.
For Record Route to stores, 1 byte is used to store type of option, 1 byte for length and 1 byte for pointer. Out of 40 bytes, 37 bytes are left.
Each IP4 address takes 32 bits or 4 bytes.
Therefore, it can store at most floor (37/4) = 9 router addresses.
Hence correct answer is 9 router address.
Question 20

Consider two hosts X and Y, connected by a single direct link of rate 106bits/sec. The distance between the two hosts is 10,000 km and the propagation speed along the link is 2×108m/sec. Host X sends a file of 50,000 bytes as one large message to host Y continuously. Let the transmission and propagation delays be p milliseconds and q milliseconds, respectively. Then the values of p and q are

A
p=50 and q=100
B
p=50 and q=400
C
p=100 and q=50
D
p=400 and q=50
       Computer-Networks       Network-Communication       GATE 2017 [Set-2]       Video-Explanation
Question 20 Explanation: 
Given Data:
B = 106 bits/sec
L = 50000 Bytes
d = 10000 Km = 107 m
v = 8×108 m/sec
Transmission time,
P = L/B = 50000×8bits/ 106 bits/sec = 0.4sec = 400msec
Propagation time,
q = d/v = 107m/ 2×108 m/s = 0.05sec = 50 msec
Question 21

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
       Computer-Networks       Protocols       GATE 2016 [Set-1]       Video-Explanation
Question 21 Explanation: 
DHCP is dynamic host configuration protocol: allocates one of the unused IP address.
Except DHCP, remaining all the protocols are used to resolve one form of address to another one.
I. DNS is going to convert hostname to IP address.
II. ARP is going to convert IP to MAC.
III. DHCP is going to assign IP dynamically.
IV. RARP is going to convert MAC to IP.
Question 22
Which of the following is/are example(s) of stateful application layer protocols?
(i) HTTP
(ii) FTP
(iii) TCP
(iv) POP3
A
(i) and (ii) only
B
(ii) and (iii) only
C
(ii) and (iv) only
D
(iv) only
       Computer-Networks       Application-Layer-Protocol       GATE 2016 [Set-1]       Video-Explanation
Question 22 Explanation: 
Stateless protocol is a communications protocol in which no information is retained by either sender or receiver.
A protocol that requires keeping of the internal state on the server is known as a stateful protocol.
Stateless – HTTP, IP
Stateful – FTP, SMTP, POP3, TCP
TCP is stateful as it maintains connection information across multiple transfers, but TCP is a Transport layer protocol.
FTP and POP3 is stateful Application layer protocol.
Question 23

Consider that B wants to send a message m that is digitally signed to A. Let the pair of private and public keys for A and B be denoted by Kx and Kx+ for x = A,B, respectively. Let Kx(m)  represent the operation of encrypting m with a key Kx and H(m) represent the message digest. Which one of the following indicates the CORRECT way of sending the message m along with the digital signature to A?

A
B
C
D
       Computer-Networks       Security       GATE 2016 [Set-1]       Video-Explanation
Question 23 Explanation: 
Digital signatures are electronic signatures which ensure the integrity, non-repudiation and authenticity of message.
Message digest is a hash value generated by applying a function on it.
Message digest is encrypted using private key of sender, so it can only be decrypted by public key of sender.
This ensures that the message was sent by the known sender.
Message digest is sent with the original message to the receiving end, where hash function is used on the original message and the value generated by that is matched with the message digest.
This ensures the integrity and thus, that the message was not altered.
Digital signature uses private key of the sender to sign message digest.
Question 24

An IP datagram of size 1000 bytes arrives at a router. The router has to forward this packet on a link whose MTU (maximum transmission unit) is 100 bytes. Assume that the size of the IP header is 20 bytes.

The number of fragments that the IP datagram will be divided into for transmission is _________.

A
13
B
14
C
15
D
16
       Computer-Networks       IPv4-and-Fragmentation       GATE 2016 [Set-1]       Video-Explanation
Question 24 Explanation: 

Size of Datagram (L) = 1000 bytes
MTU = 100 bytes
Size of IP header = 20 bytes
Size of Data that can be transmitted in one fragment (payload) = 100 – 20 = 80 bytes
Size of Data to be transmitted = Size of Datagram – size of header = 1000 – 20 = 980 bytes
No. of fragments required = ⌈980/80⌉ = 13
Question 25

For a host machine that uses the token bucket algorithm for congestion control, the token bucket has a capacity of 1 megabyte and the maximum output rate is 20 megabytes per second. Tokens arrive at a rate to sustain output at a rate of 10 megabytes per second. The token bucket is currently full and the machine needs to send 12 megabytes of data. The minimum time required to transmit the data is seconds _________.

A
1.1 sec
B
1.2 sec
C
1.3 sec
D
1.4 sec
       Computer-Networks       Token-Bucket       GATE 2016 [Set-1]       Video-Explanation
Question 25 Explanation: 
According to the token bucket algorithm, the minimum time required sending 1 MB of data or the maximum rate of data transmission is given by:
S = C / (M – P)
Where,
M = Maximum output rate,
C = capacity of the bucket,
P = Rate of arrival of a token,
Given, M=20 Mb, C=1Mbps, P=10 Mbps
Therefore, S= 1 Mb / (20-10) Mbps = 1/10 = 0.1 sec
Since, the bucket is initially full, it already has 1 Mb to transmit so it will be transmitted instantly.
So, we are left with only (12 – 1) Mb, i.e. 11 Mb of data to be transmitted.
Therefore, time required to send the 11 MB will be 11 * 0.1 = 1.1 sec
Question 26

A sender uses the Stop-and-Wait ARQ protocol for reliable transmission of frames. Frames are of size 1000 bytes and the transmission rate at the sender is 80 Kbps (1Kbps = 1000 bits/second). Size of an acknowledgement is 100 bytes and the transmission rate at the receiver is 8 Kbps. The one-way propagation delay is 100 milliseconds.

Assuming no frame is lost, the sender throughput is _________ bytes/second.

A
2500
B
2501
C
2502
D
2503
       Computer-Networks       Stop-and-Wait-ARQ       GATE 2016 [Set-1]       Video-Explanation
Question 26 Explanation: 
Given,
Frame size (L) =1000 bytes
Sender side bandwidth (BS) = 80 kbps = 10 * 103 bytes/sec
Acknowledgement size (LA) =100 bytes
Receiver side bandwidth (BR) = 8 kbps = 1 * 103 bytes/sec
Propagation delay (Tp) =100 ms
By formula:
Transmission delay (Tt ) = L/BS = 1000 bytes / 10 * 103 bytes/sec = 100 ms
Acknowledge delay (Tack ) = LA / BR = 100 bytes / 1 * 103 bytes/sec = 100 ms
Total cycle time = Tt + 2 * Tp + Tack = 100 ms + 2 * 100 ms + 100 ms = 400 ms
Efficiency (η) = Tt / Total cycle time = 100 ms / 400 ms = 1 / 4 = 0.25
Throughput = Efficiency (η) * Bandwidth (BS) = 0.25 * 10 *103 bytes/s = 2500 bytes/second
Question 27

Anarkali digitally signs a message and sends it to Salim. Verification of the signature by Salim requires

A
A station stops to sense the channel once it starts transmitting a frame.
B
The purpose of the jamming signal is to pad the frames that are smaller than the minimum frame size.
C
A station continues to transmit the packet even after the collision is detected.
D
The exponential backoff mechanism reduces the probability of collision on retransmissions.
       Computer-Networks       Security       GATE 2016 [Set-2]       Video-Explanation
Question 27 Explanation: 
An Ethernet is the most popularly and widely used LAN network for data transmission.
It is a protocol of data link layer and it tells how the data can be formatted to transmit and how to place the data on network for transmission.
Now considering the Ethernet protocol we will discuss all the options one by one
I. In Ethernet the station does not required, stops to sense the channel prior frame transmission.
II. A Jamming signal used to inform all the other devices or stations about collision that has occurred, so that further data transmission is stopped.
III. No, stations sends jamming signal if collusion is detected
IV. To reduce the probability of collision on retransmissions an exponential back off mechanism is used.
This is only True.
Question 28

In an Ethernet local area network, which one of the following statements is TRUE?

A
A station stops to sense the channel once it starts transmitting a frame.
B
The purpose of the jamming signal is to pad the frames that are smaller than the minimum frame size.
C
A station continues to transmit the packet even after the collision is detected.
D
The exponential backoff mechanism reduces the probability of collision on retransmissions.
       Computer-Networks       Ethernet       GATE 2016 [Set-2]       Video-Explanation
Question 28 Explanation: 
An Ethernet is the most popularly and widely used LAN network for data transmission.
It is a protocol of data link layer and it tells how the data can be formatted to transmit and how to place the data on network for transmission.
Now considering the Ethernet protocol we will discuss all the options one by one
I. In Ethernet the station does not required, stops to sense the channel prior frame transmission.
II. A Jamming signal used to inform all the other devices or stations about collision that has occurred, so that further data transmission is stopped.
III. No, stations sends jamming signal if collusion is detected
IV. To reduce the probability of collision on retransmissions an exponential back off mechanism is used.
This is only True.
Question 29

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
       Computer-Networks       TCP       GATE 2016 [Set-2]       Video-Explanation
Question 29 Explanation: 
When a browser requests a web page from a remote server then that requests (URL address) will be mapped to IP address using DNS query, then TCP synchronization takes place after that HTTP verify whether it is existed in the web server or not.
Question 30

A network has a data transmission bandwidth of 20 × 106 bits per second. It uses CSMA/CD in the MAC layer. The maximum signal propagation time from one node to another node is 40 microseconds. The minimum size of a frame in the network is _________ bytes.

A
200
B
201
C
202
D
203
       Computer-Networks       CSMA/CD       GATE 2016 [Set-2]       Video-Explanation
Question 30 Explanation: 
For frame size to be minimum, its transmission time should be equal to twice of one way propagation delay. i.e, Tt = 2 × TP
Given,
Bandwidth (B) = 20 × 106 bps
TP = 40 μs ⇒ 40 × 10– 6 sec
Suppose minimum frame size is L.
Tt = 2 × TP ⇒ L / B = 2 × TP
⇒ L = 2 × TP × B = 2 × 40 × 10-6 × 20 × 106 = 1600 bits ⇒ 200 bytes
Therefore, L = 200 bytes
Question 31
For the IEEE 802.11 MAC protocol for wireless communication, which of the following statements is/are TRUE?
I. At least three non-overlapping channels are available for transmissions.
II. The RTS-CTS mechanism is used for collision detection.
III. Unicast frames are ACKed.
A
All I, II and III
B
I and III only
C
II and III only
D
II only
       Computer-Networks       IEEE802.11       GATE 2016 [Set-2]       Video-Explanation
Question 31 Explanation: 
802.11 MAC = Wifi
I. This is true, maximum 3 overlapping channels are possible in Wifi.
II. The RTS (Request To Send) and CTS(Clear To Send) are control frames which is used for collision avoidance, not in collision detection, (so, II is False)
III. Every frame in Wifi is Acked, because Wifi stations do not use collusion detection. (True)
Question 32

Consider a 128 × 103 bits/ second satellite communication link with one way propagation delay of 150 milliseconds. Selective retransmission (repeat) protocol is used on this link to send data with a frame size of 1 kilobyte. Neglect the transmission time of acknowledgement. The minimum number of bits required for the sequence number field to achieve 100% utilization is __________.

A
4
B
5
C
6
D
7
       Computer-Networks       SR-Protocol       GATE 2016 [Set-2]       Video-Explanation
Question 32 Explanation: 
To achieve 100% efficiency, the number of frames that we should send N = 1 + 2 * a
a = Tp / Tt where Tp is propagation delay, and Tt is transmission delay.
Given, B = 128 kbps, Tp = 150 msec,
L = 1 KB = 1 * 8 * 210 bits
Tt = L / B ⇒ 1 * 8 * 210 bits / 128 * 103 bps ⇒ 0.064 sec = 64 msec
So, a = 150 msec / 64 msec = 2.343
Efficiency (η) = 100 % ⇒ 1 = N/ 1 + 2 * a
So, N = 1 + 2 * a ⇒ 1 + 2 * 2.343 = 5.686
No. of sequence numbers requires in SR is 2*N = 2 *5.686 = 11.375
Minimum No. of bits required in the sequence number = [ log2 (11.375) ] = 4
Question 33

Suppose two hosts use a TCP connection to transfer a large file. Which of the following statements is/are False with respect to the TCP connection?

    1. If the sequence number of a segment is m, then the sequence number of the subsequent segment is always m+1.
    2. If the estimated round trip time at any given point of time is t sec, the value of the retransmission timeout is always set to greater than or equal to t sec.
    3. The size of the advertised window never changes during the course of the TCP connection.
    4. The number of unacknowledged bytes at the sender is always less than or equal to the advertised window.
A
III only
B
I and III only
C
I and IV only
D
II and IV only
       Computer-Networks       TCP       GATE 2015 [Set-1]
Question 33 Explanation: 
I. False.
If the sequence no. of the segment is m, then the sequence number of the subsequent segment depends on the current segment size.
II. True.
If the estimated RTT at any given point of time is t second, then the value of the re-transmission timeout is always set to greater than or equal to t sec.
III. False.
The size of the advertised window may change during the course of the TCP connection depending on the processing capability at the receiver’s side and the network traffic.
IV. True.
The number of unacknowledged bytes at the sender is always less than or equal to the advertised window, because the sender never sends no. of bytes greater than advertised window.
Question 34

Which one of the following fields of an IP header is NOT modified by a typical IP router?

A
Checksum
B
Source address
C
Time to Live (TTL)
D
Length
       Computer-Networks       IPv4-Header       GATE 2015 [Set-1]
Question 34 Explanation: 
Option C (TTL) is decremented by each visited router. When it reaches to Zero, then Packet will be discarded.
Option A (Checksum) needs to be updated by each visited Router since TTL Value is modified.
Option D (Length) also modified whenever there is a need of performing the fragmentation process.
Option B (Source Address) can’t be modified by an IP router. Only NAT can modify it.
Question 35

Consider  a LAN with four nodes S1, S2, S3 and S4. Time is divided into fixed-size slots, and a node can begin its transmission only at the beginning of a slot. A collision is said to have occurred if more than one node transmit in the same slot. The probabilities of generation of a frame in a time slot by S1, S2, S3 and S4 are 0.1, 0.2, 0.3 and 0.4, respectively. The probability of sending a frame in the first slot without any collision by any of these four stations is _________.

A
0.4404
B
0.463
C
0.464
D
0.465
       Computer-Networks       Slotted-Channel-and-Probability       GATE 2015 [Set-1]
Question 35 Explanation: 
S1→0.1
S2→0.2
S3→0.3
S4→0.4
The probability of sending a frame without any collision by any of these stations is
Question 36

Suppose that the stop-and-wait protocol is used on a link with a bit rate of 64 kilobits per second and 20 milliseconds propagation delay. Assume that the transmission time for the acknowledgment and the processing time at nodes are negligible. Then the minimum frame size in bytes to achieve a link utilization of at least 50% is _________.

A
320
B
321
C
322
D
323
       Computer-Networks       Stop-and-Wait-protocol       GATE 2015 [Set-1]
Question 36 Explanation: 
Given B = 64 kbps
Tp = 20 ms
η ≥ 50%
For η≥50% ⇒ L≥BR
⇒ L = 64×103×2×20×10-3
= 2560bits
= 320bytes
Question 37

Identify the correct order in which a server process must invoke the function calls accept, bind, listen, and recv according to UNIX socket API.

A
listen, accept, bind recv
B
bind, listen, accept, recv
C
bind, accept, listen, recv
D
accept, listen, bind recv
       Computer-Networks       Sockets       GATE 2015 [Set-2]
Question 37 Explanation: 
Question 38

Which one of the following statements is NOT correct about HTTP cookies?

A
A cookie is a piece of code that has the potential to compromise the security of an internet user
B
A cookie gains entry to the user’s work area through an HTTP header
C
A cookie has an expiry date and time
D
Cookies can be used to track the browsing pattern of a user at a particular site
       Computer-Networks       HTTP-Cookies       GATE 2015 [Set-2]
Question 38 Explanation: 
An HTTP cookie (also called web cookie, Internet cookie, browser cookie, or simply cookie) is a small piece of data sent from a website and stored on the user’s computer by the user’s web browser while the user is browsing.
Cookies are not piece of code, they are just strings typically in the form of key value pairs.
Question 39

A link has a transmission speed of 106 bits/sec. It uses data packets of size 1000 bytes each. Assume that the acknowledgement has negligible transmission delay, and that its propagation delay is the same as the data propagation delay. Also assume that the processing delays at the nodes are negligible. The efficiency of the stop-and-wait protocol in this setup is exactly 25%. The value of the one-way propagation delay (in milliseconds) is ___________.

A
12
B
13
C
14
D
15
       Computer-Networks       Stop-and-Wait-ARQ       GATE 2015 [Set-2]
Question 39 Explanation: 
Given, B = 106bps
L = 1000
η = 25%
Tp = ?
In stop-and-wait, η = 1/1 + 2a
⇒1/4 = 1/1 + 2a ⇒ 1 + 2a = 4
2a = 3; a = 32
Tx = L/B = 8×103/106 = 8ms
Tp/Tx = 3/2; 2Tp = 3Tx
2Tp = 24ms
Tp = 12ms
Question 40

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 heard is 20 bytes. There is no option field in IP header. How many 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
       Computer-Networks       IPv4-an-Fragmentation       GATE 2015 [Set-2]
Question 40 Explanation: 
UDP data = 8880 bytes, UDP header = 8 bytes, IP Header = 20 bytes
Total Size excluding IP Header = 8888 bytes.
Number of fragments = ceil(8880+ UDP or TCP header /1500-IP header)
= ceil(8880+8 /1500-20)
= ceil(8888/1480)
= 7
Offset of last fragment = (MTU-IP header ) *( number of fragments -1) / scaling factor = 1110 (scaling factor of 8 is used in offset field).
= (1500-20)* (7-1)/8
= 1110
Question 41

Consider the following routing table at an IP router:

For each IP address in Group I identify the correct choice of the next hop from Group II using the entries from the routing table above.

     Group I                 Group II
(i) 128.96.171.92        (a) Interface 0 
(ii) 128.96.167.151      (b) Interface 1
(iii) 128.96.163.121     (c) R2
(iv) 128.96.165.121      (d) R3
                         (e) R4
A
i-a, ii-c, iii-e, iv-d
B
i-a, ii-d, iii-b, iv-e
C
i-b, ii-c, iii-d, iv-e
D
i-b, ii-c, iii-e, iv-d
       Computer-Networks       Subnetting       GATE 2015 [Set-2]
Question 41 Explanation: 
Do the AND operation of group I IP address with Netmask and compare the result with network number. if it match with network number, forward packet through that interface. Ex: 128.96.167.151 AND 255.255.254.0 = 128.96.166.0 Therefore packet will forward through R2.
Question 42

Consider the following statements.

    I. TCP connections are full duplex.
    II. TCP has no option for selective acknowledgment.
    III. TCP connections are message streams.
A
Only I is correct
B
Only I and III are correct
C
Only II and III are correct
D
All of I, II and III are correct
       Computer-Networks       TCP       GATE 2015 [Set-3]
Question 42 Explanation: 
In TCP, as sender and receiver can send segments at the same time, It is FULL-DUPLEX. TCP can use selective ACK and TCP uses byte streams that is every byte is send using TCP is numbered.
Question 43

Consider a CSMA/CD network that transmits data at a rate of 100 Mbps (108 bits second) over a 1 km (kilometer) cable with no repeaters. If the minimum frame size required for this network is 1250 bytes, what is the signal speed (km/sec) in the cable?  

A
8000
B
10000
C
16000
D
20000
       Computer-Networks       CSMA/CD       GATE 2015 [Set-3]
Question 43 Explanation: 
Given: L = 1250 Bytes
B = 100 Mbps
d = 1 km
v = ?
In CSMA/CD, L = 2×d/v×B
⇒ v = 2dB/L = 2×103×108/104
⇒ v = 20,000 km/sec
Question 44

Consider a network connected two systems located 8000 kilometers apart. The bandwidth of the network is 500 × 106 bits per second. The propagation speed of the media is 4 × 106 meters per second. It is needed to design a Go-Back-N sliding window protocol for this network. The average packet size is 107 bits. The network is to be used to its full capacity. Assume that processing delays at nodes are negligible. Then the minimum size in bits of the sequence number field has to be ___________.

A
8
B
7
C
6
D
5
       Computer-Networks       Sliding-Window-Protocol       GATE 2015 [Set-3]
Question 44 Explanation: 
η=100%
n = ?
∴a=Tp/Tn =2/0.02=100
Given Protocol, Go-back-N protocol, So η = w/(1+2a) where w = 2n-1
100/100 = w/(1+2a) ⇒ w = 1+2a
⇒ 2(n-1) = 1+2(100)
⇒ 2n – 1 = 201
⇒ 2n = 202 ⇒ 2n = 28
⇒ n = 8
Question 45

In the network 200.10.11.144/27, the fourth octet (in decimal) of the last IP address of the network which can be assigned to a host is ____________.

A
158
B
157
C
156
D
155
       Computer-Networks       Subnetting       GATE 2015 [Set-3]
Question 45 Explanation: 
No. of bit in HID part = 32-27 = 5 bits
Subnet mask is 255.255.255.224
Do AND with given IP and subnet mask then we get NID 200.10.11.128
In fourth octet first three bit will fixed for subnet and remaining 5 bits is for HID, so maximum value as 11111.
The address with all 1s in host part is broadcast address and can’t be assigned to a host.
So the maximum possible last octal in a host IP is 10011110 which is 158.
Question 46

Consider the following three statements about link state and distance vector routing protocols, for a large network with 500 network nodes and 4000 links.

    [S1] The computational overhead in link state protocols is higher than in distance vector protocols.
    [S2] A distance vector protocol (with split horizon) avoids persistent routing loops, but not a link state protocol.
    [S3] After a topology change, a link state protocol will converge faster than a distance vector protocol.

Which one of the following is correct about S1, S2, and S3 ?

A
S1, S2, and S3 are all true.
B
S1, S2, and S3 are all false.
C
S1 and S2 are true, but S3 is false.
D
S1 and S3 are true, but S2 is false.
       Computer-Networks       Routing       GATE 2014 [Set-1]
Question 46 Explanation: 
S1: The computational overhead in link state protocols is higher than in distance vector protocols. Because LSR is based upon global knowledge whereas DVR is based upon Local info.(True)
S2: A distance vector protocol with split horizon avoid persistent routing loops is true, but not a link state protocol is false because link state protocols do not have count to infinity problem.
S3: As Distance vector protocol has count to infinity problem and converges slower. (True)
Question 47

Which of the following are used to generate a message digest by the network security protocols?

    (P) RSA
    (Q) SHA-1
    (R) DES
    (S) MD5
A
P and R only
B
Q and R only
C
Q and S only
D
R and S only
       Computer-Networks       Security       GATE 2014 [Set-1]
Question 47 Explanation: 
RSA and DES are for Encryption where MD5 and SHA – 1 are used to generate Message Digest.
Question 48

Identify the correct order in which the following actions take place in an interaction between a web browser and a web server.

    2. The web browser establishes a TCP connection with the web server.
    3. The web server sends the requested webpage using HTTP.
    4. The web browser resolves the domain name using DNS.
A
4,2,1,3
B
1,2,3,4
C
4,1,2,3
D
2,4,1,3
       Computer-Networks       TCP       GATE 2014 [Set-1]
Question 48 Explanation: 
First of all the browser must now know what IP to connect to. For this purpose browser takes help of Domain name system (DNS) servers which are used for resolving hostnames to IP addresses. As browser is an HTTP client and as HTTP is based on the TCP/IP protocols, first it establishes a TCP connection with the web server and requests a web page using HTTP, and then the web server sends the requested web page using HTTP. Hence the order is 4,2,1,3.
Question 49

Consider a token ring network with a length of 2 km having 10 stations including a monitoring station. The propagation speed of the signal is 2×108 m/s and the token transmission time is ignored. If each station is allowed to hold the token for 2 µsec, the minimum time for which the monitoring station should wait (in µsec) before assuming that the token is lost is _______.

A
28μs to 30 μs
B
29μs to 31 μs
C
30μs to 32 μs
D
31μs to 33 μs
       Computer-Networks       Token-Ring       GATE 2014 [Set-1]
Question 49 Explanation: 
Given length (d) = 2 Km
No. of Stations (m) = 10
Propagation speed (v) = 2⨯108 m/s
THT = 2μs
So, Max, TRT = Tp in the ring + No. of Active Stations * THT
= 10 ⨯ 10-6 + 10 ⨯ 2 ⨯ 10-6
= 30 μs
Question 50

Let the size of congestion window of a TCP connection be 32 KB when a timeout occurs. The round trip time of the connection is 100 msec and the maximum segment size used is 2 KB. The time taken (in msec) by the TCP connection to get back to 32 KB congestion window is _________.

A
1100 to 1300
B
1101 to 1301
C
1102 to 1302
D
1103 to 1303
       Computer-Networks       TCP       GATE 2014 [Set-1]
Question 50 Explanation: 
Given that at the time of Time Out, Congestion Window Size is 32KB and RTT = 100ms
When Time Out occurs, for the next round of Slow Start, Threshold = (size of Cwnd) / 2
It means Threshold = 16KB
Slow Start
2KB
1RTT
4KB
2RTT
8KB
3RTT
16KB ———– Threshold reaches. So Additive Increase Starts
4RTT
18KB
5RTT
20KB
6RTT
22KB
7RTT
24KB
8RTT
26KB
9RTT
28KB
10RTT
30KB
11RTT
32KB
So, Total no. of RTTs = 11 → 11 * 100 = 1100
There are 50 questions to complete.

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