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       GATE 2019
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       GATE 2019
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       GATE 2019
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 Totient Function, then the prime factor of n which is greater than 50 is ______.

A
107
B
97
C
45
D
92
       Computer-Networks       GATE 2019
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       GATE 2019
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       GATE 2019
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
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
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 acknowledgedment.
    (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
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
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
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-01
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-01
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 x+ 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-01
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-01
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-01
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-02)
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-02)
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-02)
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-02)
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       2016 set-01
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       2016 set-01
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       2016 set-01
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       2016 set-01
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       2016 set-01
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       2016 set-01
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
Anarkali’s public key.
B
Salim’s public key.
C
Salim’s private key.
D
Anarkali’s private key.
       Computer-Networks       Security       GATE 2016 set-2
Question 27 Explanation: 
In digital signature generation process, Sender uses its own private key to digitally sign a document.
Receiver uses sender's public key to verify signature.
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
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
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
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
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
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-01)
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 retransmission timeout is always set to greater than or equal to t sec.
III. False.
The size of the advertized 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-01)
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-01)
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-01)
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:  
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-03)
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-03)
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-03)
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-03)
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-01)
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-01)
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.
  1. The web browser requests a webpage using HTTP.
  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-01)
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-01)
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-01)
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
Question 51
Consider a selective repeat sliding window protocol that uses a frame size of 1 KB to send data on a 1.5 Mbps link with a one-way latency of 50 msec. To achieve a link utilization of 60%, the minimum number of bits required to represent the sequence number field is ________.
A
5
B
6
C
7
D
8
       Computer-Networks       Sliding-Window-Protocol       GATE 2014(Set-01)
Question 51 Explanation: 
Given L=1 KB
B=1.5 Mbps
Tp=50 ms
η=60%
Efficiency formula for SR protocol is
η=W/(1+2a)⇒60/100=W/(1+2a) (∵a=Tp/Tx)
Tx=L/B=8×103/1.5×106=5.3ms
a=Tp/Tx=50/5.3=500/53=9.43
⇒ 60/100=W/19.86⇒W=11.9≈12
⇒ W=2n-1=12⇒2n=24⇒2n=24≈25⇒n=5
Question 52
Which one of the following is TRUE about the interior gateway routing protocols – Routing Information Protocol (RIP) and Open Shortest Path First (OSPF)?
A
RIP uses distance vector routing and OSPF uses link state routing
B
OSPF uses distance vector routing and RIP uses link state routing
C
Both RIP and OSPF use link state routing
D
Both RIP and OSPF use distance vector routing
       Computer-Networks       Routing       Gate 2014 Set -02
Question 52 Explanation: 
RIP Uses Distance Vector Routing and OSPF uses Link State Routing.
Question 53
Which one of the following socket API functions converts an unconnected active TCP socket into a passive socket?
A
connect
B
bind
C
listen
D
accept
       Computer-Networks       Sockets       Gate 2014 Set -02
Question 53 Explanation: 
(a) The connect function is used by a TCP client to establish a connection with a TCP server.
(b) The bind function assigns a local protocol address to a socket. With the Internet protocols, the protocol address is the combination of either a 32-bit IPv4 address or a 128-bit IPv6 address, along with a 16-bit TCP or UDP port number.
(c) The listen function converts an unconnected socket into a passive socket, indicating that the kernel should accept incoming connection requests directed to this socket.
(d) The accept function is called by a TCP server to return the next completed connection from the front of the completed connection queue. If the completed connection queue is empty, the process is put to sleep (assuming the default of a blocking socket).
Question 54
In the diagram shown below L1 is an Ethernet LAN and L2 is a Token-Ring LAN. An IP packet originates from sender S and traverses to R, as shown. The link within each ISP, and across two ISPs, are all point to point optical links. The initial value of TTL is 32. The maximum possible value of TTL field when R receives the datagram is
A
26
B
27
C
28
D
29
       Computer-Networks       Time-to-live       Gate 2014 Set -02
Question 54 Explanation: 
TTL field reduced at each router. There are 6 routers in the above diagram.
Initially TTL value was 32, so at the Receiver it will become 32 – 6 = 26.
Question 55
Consider the store and forward packet switched network given below. Assume that the bandwidth of each link is 106 bytes / sec. A user on host A sends a file of size 103 bytes to host B through routers R1 and R2 in three different ways. In the first case a single packet containing the complete file is transmitted from A to B. In the second case, the file is split into 10 equal parts, and these packets are transmitted from A to B. In the third case, the file is split into 20 equal parts and these packets are sent from A to B. Each packet contains 100 bytes of header information along with the user data. Consider only transmission time and ignore processing, queuing and propagation delays. Also assume that there are no errors during transmission. Let T1, T2 and T3 be the times taken to transmit the file in the first, second and third case respectively. Which one of the following is CORRECT?
A
T1 < T2 < T3
B
T1 > T2 > T3
C
T2 = T3, T3 < T1
D
T1 = T3, T3 > T2
       Computer-Networks       Packet-Switching       Gate 2014 Set -02
Question 55 Explanation: 
Given Bandwidth = 106 bytes/sec
L = 103 byte
Case: 1
L = 1000 bytes
Header size = 100 bytes
Total Frame size = 1000 + 100 = 1100 bytes
∴ Tx = 1100 × 8/ 106×8 = 1100μs
So, T1=3300μs
Case: 2
L = 100 bytes
Header size = 100 bytes
Total Frame size = 100 + 100 = 200 bytes
∴ Tx = 200×8/ 106×8 = 200μs for 1 packet
For 10 packets ⇒ Tx = 2000μs
So, T2 = 2000+200+200 = 2400μs
Case: 2
L = 50 bytes
Header size = 100 bytes
Total Frame size = 50 + 100 = 150 bytes
∴ Tx = 150×8/ 106×8 = 150μs for 1 packet
For 20 packets ⇒ Tx = 3000μs
So, T3 = 3000+150+150 = 3300μs
∴ T1 = T3
T3 > T2
Question 56
An IP machine Q has a path to another IP machine H via three IP routers R1, R2, and R3.
Q—R1—R2—R3—H
H acts as an HTTP server, and Q connects to H via HTTP and downloads a file. Session layer encryption is used, with DES as the shared key encryption protocol. Consider the following four pieces of information:
[I1] The URL of the file downloaded by Q
[I2] The TCP port numbers at Q and H
[I3] The IP addresses of Q and H
[I4] The link layer addresses of Q and H
Which of I1, I2, I3, and I4 can an intruder learn through sniffing at R2 alone?
A
Only I1 and I2
B
Only I1
C
Only I2 and I3
D
Only I3 and I4
       Computer-Networks       Security       Gate 2014 Set -02
Question 56 Explanation: 
An Intruder can’t learn [I1] through sniffing at R2 because URLs and Download are functioned at Application layer of OSI Model.
An Intruder can learn [I2] through sniffing at R2 because Port Numbers are encapsulated in the payload field of IP Datagram.
An Intruder can learn [I3] through sniffing at R2 because IP Addresses and Routers are functioned at network layer of OSI Model.
An Intruder can’t learn [I4] through sniffing at R2 because it is related to Data Link Layer of OSI Model.
Question 57
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
       Computer-Networks       OSI-Layers       Gate 2014 Set -03
Question 57 Explanation: 
(a) One of the main functionality of Network Layer is Routing. So Option (a) is CORRECT.
(b) Bit Synchronization is always handled by Physical Layer of OSI model but not Data Link Layer. So Option (b) is INCORRECT.
(c) End – to – End Process Communication is handled by Transport Layer. So Option (c) is CORRECT.
(d) MAC sub layer have 3 types of protocols (Random, Controlled and Channelized Access).
Question 58
A bit-stuffing based framing protocol uses an 8-bit delimiter pattern of 01111110. If the output bit-string after stuffing is 01111100101, then the input bit-string is
A
0111110100
B
0111110101
C
0111111101
D
0111111111
       Computer-Networks       Data-Link-Layer       Gate 2014 Set -03
Question 58 Explanation: 
Given 8-bit delimiter pattern of 01111110.
Output Bit string after stuffing is 01111100101.
Question 59
Host A (on TCP/IP v4 network A) sends an IP datagram D to host B (also on TCP/IP v4 network B). Assume that no error occurred during the transmission of D. When D reaches B, which of the following IP header field(s) may be different from that of the original datagram D? (i) TTL                   (ii) Checksum                 (iii) Fragment Offset
A
(i) only
B
(i) and (ii) only
C
(i) and (ii) only
D
(i), (ii) and (iii)
       Computer-Networks       TCP       Gate 2014 Set -03
Question 59 Explanation: 
While an IP Datagram is transferring from one host to another host, TTL, Checksum and Fragmentation Offset will be changed.
Question 60
An IP router implementing Classless Inter-domain Routing (CIDR) receives a packet with address 131.23.151.76. The router’s routing table has the following entries:  
Prefix Output Interface  
Identifier  
   
131.16.0.0/ 12 3  
131.28.0.0/ 14 5  
131.19.0.0/ 16 2  
131.22.0.0/ 15 1  
  The identifier of the output interface on which this packet will be forwarded is ______.
A
1
B
2
C
3
D
4
       Computer-Networks       Network-Layer       Gate 2014 Set -03
Question 60 Explanation: 
Lets take, 131.22.0.0/15 Its Net Mask is 255.254.0.0.
If we do AND operation between 255.254.0.0 and given IP 131.23.151.76, gives 131.22.0.0 which is matching with interface 1.
Question 61
Every host in an IPv4 network has a 1-second resolution real-time clock with battery backup. Each host needs to generate up to 1000 unique identifiers per second. Assume that each host has a globally unique IPv4 address. Design a 50-bit globally unique ID for this purpose. After what period (in seconds) will the identifiers generated by a host wrap around?
A
256
B
257
C
258
D
259
       Computer-Networks       IPv4-Protocol       Gate 2014 Set -03
Question 61 Explanation: 
Given that each host has a globally unique IPv4 Address and we have to design 50 bit unique Id. So, 50 bit in the sense (32 + 18). So, It is clearly showing that IP Address (32 bit) followed by 18 bits.
1000 unique Ids ⇒ 1Sec
218 unique Ids ⇒ 218/1000=28=256
Question 62
An IP router with a Maximum Transmission Unit (MTU) of 1500 bytes has received an IP packet of size 4404 bytes with an IP header of length 20 bytes. The values of the relevant fields in the header of the third IP fragment generated by the router for this packet are
A
MF bit: 0, Datagram Length: 1444; Offset: 370
B
MF bit: 1, Datagram Length: 1424; Offset: 185
C
MF bit: 1, Datagram Length: 1500; Offset: 370
D
MF bit: 0, Datagram Length: 1424; Offset: 2960
       Computer-Networks       IP-Routing-MTU       Gate 2014 Set -03
Question 62 Explanation: 
Number of packet fragments = ⌈ (total size of packet)/(MTU) ⌉
So Datagram with data 4404 byte fragmented into 3 fragments.
Question 63
Consider the transactions T1, T2, and T3 and the schedules S1 and S2 given below. T1: r1(X); r1(Z); w1(X); w1(Z) T2: r2(Y); r2(Z); w2(Z) T3: r3(Y); r3(X); w3(Y) S1: r1(X); r3(Y); r3(X); r2(Y); r2(Z); w3(Y); w2(Z); r1(Z); w1(X); w1(Z) S2: r1(X); r3(Y); r2(Y); r3(X); r1(Z); r2(Z); w3(Y); w1(X); w2(Z); w1(Z) Which one of the following statements about the schedules is TRUE?
A
Only S1 is conflict-serializable.
B
Only S2 is conflict-serializable.
C
Both S1 and S2 are conflict-serializable.
D
Neither S1 nor S2 is conflict-serializable.
       Computer-Networks       Transactions and concurrency control       Gate 2014 Set -03
Question 63 Explanation: 
S1:

No cycle, so schedule S1 is conflict serializable.
S2:

There is a cycle, so S2 is not conflict serializable.
Question 64
The transport layer protocols used for real time multimedia, file transfer, DNS and email, respectively are
A
TCP, UDP, UDP and TCP
B
UDP, TCP, TCP and UDP
C
UDP, TCP, UDP and TCP
D
TCP, UDP, TCP and UDP
       Computer-Networks       Transport Layer       Gate 2013
Question 64 Explanation: 
Real time multimedia needs connectionless service, so underlying transport layer protocol used is UDP. File transfer runs over TCP protocol with port no-21.
DNS runs over UDP protocol within port no-53.
Email needs, SMTP protocol which runs over TCP protocol within port no –25.
Question 65
Using public key cryptography, X adds a digital signature  to message M, encrypts <M, σ>, and sends it to Y, where it is decrypted. Which one of the following sequences of keys is used for the operations?
A
Encryption: X’s private key followed by Y’s private key; Decryption: X’s public key followed by Y’s public key
B
Encryption: X’s private key followed by Y’s public key; Decryption: X’s public key followed by Y’s private key
C
Encryption: X’s public key followed by Y’s private key; Decryption: Y’s public key followed by X’s private key
D
Encryption: X’s private key followed by Y’s public key; Decryption: Y’s private key followed by X’s public key
       Computer-Networks       Network-Security       Gate 2013
Question 65 Explanation: 

Encryption: Source has to encrypt with its private key for forming Digital signature for Authentication. Source has to encrypt the (M, σ) with Y’s public key to send it confidentially.
Decryption: Destination Y has to decrypt first with its private key, then decrypt using source public key.
Question 66
A
Network layer – 4 times and Data link layer – 4 times
B
Network layer – 4 times and Data link layer – 3 times
C
Network layer – 4 times and Data link layer – 6 times
D
Network layer – 2 times and Data link layer – 6 times
       Computer-Networks       Network-Layer       Gate 2013
Question 66 Explanation: 

From above given diagram, its early visible that packet will visit network layer 4 times, once at each node [S, R, R, D] and packet will visit Data Link layer 6 times. One time at S and one time at D, then two times for each intermediate router R as data link layer is used for link to link communication.
One at packet reaches R and goes up from physical –DL-Network and second time when packet coming out of router in order Network –DL-Physical.
Question 67
Determine the maximum length of the cable (in km) for transmitting data at a rate of 500 Mbps in an Ethernet LAN with frames of size 10,000 bits. Assume the signal speed in the cable to be 2,00,000 km/s.
A
1
B
2
C
2.5
D
5
       Computer-Networks       Data-Link-Layer       Gate 2013
Question 67 Explanation: 
500×106 bits ------ 1 sec
∴104 bits -----5×108/104=104/5×108sec=1/5×104sec
1 sec------2×105 km
1/5×104-----2×105/5×104= 4 km
Maximum length of cable=4/2=2 km
Question 68
In an IPv4 datagram, the M bit is 0, the value of HLEN is 10, the value of total length is 400 and the fragment offset value is 300. The position of the datagram, the sequence numbers of the first and the last bytes of the payload, respectively are
A
Last fragment, 2400 and 2789
B
First fragment, 2400 and 2759
C
Last fragment, 2400 and 2759
D
Middle fragment, 300 and 689
       Computer-Networks       Network-Layer       Gate 2013
Question 68 Explanation: 
M = 0 - Means there is no fragment after this, i.e. Last fragment.
HLEN = 10 - So header length is 4×10=40, as 4 is constant scale factor.
Total Length = 400 (40 Byte Header + 360 Byte Payload)
Fragment Offset = 300, that means 300×8 Byte = 2400 bytes are before this last fragment.
So the position of datagram is last fragment.
Sequence number of First Byte of Payload = 2400 (as 0 to 2399 Sequence no are used)
Sequence number of Last Byte of Payload = 2400+360-1=2759
Question 69
The protocol data unit (PDU) for the application layer in the Internet stack is
A
Segment
B
Datagram
C
Message
D
Frame
       Computer-Networks       Application-Layer-Protocol       Gate 2012
Question 69 Explanation: 
The PDU for Data Link layer, Network layer, Transport layer and Application layer are frame, datagram, segment and message respectively.
Question 70
Which of the following transport layer protocols is used to support electronic mail?
A
SMTP
B
IP
C
TCP
D
UDP
       Computer-Networks       Transport Layer Protocols       Gate 2012
Question 70 Explanation: 
TCP is used in transport layer to carry out mail which is initiated by application layer protocol SMTP.
Question 71
In the IPv4 addressing format, the number of networks allowed under Class C addresses is
A
214
B
27
C
221
D
224
       Computer-Networks       IPv4       Gate 2012
Question 71 Explanation: 
For class C address, size of network field is 24 bits. But first 3 bits are fixed as 110; hence total number of networks possible is 221.
Question 72
An Internet Service Provider (ISP) has the following chunk of CIDR-based IP addresses available with it: 245.248.128.0/20. The ISP wants to give half of this chunk of addresses to Organization A, and a quarter to Organization B, while retaining the remaining with itself. Which of the following is a valid allocation of addresses to A and B?
A
245.248.136.0/21 and 245.248.128.0/22
B
245.248.128.0/21 and 245.248.128.0/22
C
245.248.132.0/22 and 245.248.132.0/21
D
245.248.136.0/24 and 245.248.132.0/21
       Computer-Networks       IPv4       Gate 2012
Question 72 Explanation: 
Question 73
Consider a source computer (S) transmitting a file of size 106 bits to a destination computer (D) over a network of two routers (R1 and R2) and three links (L1, L2 and L3). L1 connects S to R1; L2 connects R1 to R2; and L3 connects R2 to D. Let each link be of length 100 km. Assume signals travel over each link at a speed of 108 meters per second. Assume that the link bandwidth on each link is 1Mbps. Let the file be broken down into 1000 packets each of size 1000 bits. Find the total sum of transmission and propagation delays in transmitting the file from S to D?
A
1005 ms
B
1010 ms
C
3000 ms
D
3003 ms
       Computer-Networks       Switching       Gate 2012
Question 73 Explanation: 

Propagation delay = (Distance) / (Velocity) = 3*105/108 = 3ms
Total transmission delay for 1 packet = 3 * L / B = 3*(1000/106) = 3ms. Because at source and 2 routers, we need to transmit the bits.
The first packet will reach destination = Tt + Tp = 6ms.
While the first packet was reaching to D, other packets must have been processing in parallel. So D will receive remaining packets 1 packet per 1 ms from R2. So remaining 999 packets will take 999 ms.
And total time will be 999 + 6 = 1005 ms
Question 74
Consider an instance of TCP’s Additive Increase Multiplicative Decrease (AIMD) algorithm where the window size at the start of the slow start phase is 2 MSS and the threshold at the start of the first transmission is 8 MSS. Assume that a timeout occurs during the fifth transmission. Find the congestion window size at the end of the tenth transmission.
A
8 MSS
B
14 MSS
C
7 MSS
D
12 MSS
       Computer-Networks       Congestion-Control       Gate 2012
Question 74 Explanation: 
Given initial threshold = 8
Time = 1 during 1st trans. , window size = 2 (Slow start),
Time = 2 congestion window size = 4 (double the no. of ack.)
Time = 3 congestion window = 8
Time = 4 congestion window size = 9, after threshold, increase by one additive increase.
Time = 5 transmit 10 MSS, but time out occur congestion window size = 10
Hence threshold = (congestion window size)/2 = 10/2 = 5
Time = 6 transmit 2(since in the question, they are saying ss is starting from 2)
Time = 7 transmit 4
Time = 8 transmit 5
Time =9 transmit 6
Time =10 transmit 7
Question 75
A layer-4 firewall (a device that can look at all protocol headers up to the transport layer) CANNOT
A
block entire HTTP traffic during 9:00PM and 5:00AM
B
block all ICMP traffic
C
stop incoming traffic from a specific IP address but allow outgoing traffic to the same IP address
D
block TCP traffic from a specific user on a multi-user system during 9:00PM and 5:00AM
       Computer-Networks       Network-Security       Gate 2011
Question 75 Explanation: 
(A) It is possible to block entire HTTP traffic by blocking port no.80.
(B) Possible because it is network layer protocol.
(C) Possible because SP address is present in Network layer.
(D) Not possible, because to block specific user, we need user id which is present in Application layer.
Question 76
Consider a network with five nodes, N1 to N5 as shown below. The network a distance vector protocol. What is the routes Have stabilised, the distance vector a different notes are as follows N1:(0,1,7,8,4) N2:(1,0,6,7,3) N3:(7,6,0,2,6) N4:(8,7,2,0,4) N5:(4,3,6,4,0) The network uses a Distance Vector Routing protocol. Once the routes have stabilized, the distance vectors at different nodes are as following. N1: (0, 1, 7, 8, 4) N2: (1, 0, 6, 7, 3) N3: (7, 6, 0, 2, 6) N4: (8, 7, 2, 0, 4) N5: (4, 3, 6, 4, 0) Each distance vector is the distance of the best known path at the instance to nodes, N1 to N5, where the distance to itself is 0. Also, all links are symmetric and the cost is identical in both directions. In each round, all nodes exchange their distance vectors with their respective neighbors. Then all nodes update their distance vectors. In between two rounds, any change in cost of a link will cause the two incident nodes to change only that entry in their distance vectors. 52. The cost of link N2-N3 reduces to 2(in both directions). After the next round of updates, what will be the new distance vector at node, N3.        
A
(3, 2, 0, 2, 5)
B
(3, 2, 0, 2, 6)
C
(7, 2, 0, 2, 5)
D
(7, 2, 0, 2, 6)
       Computer-Networks       Routing       Gate 2011
Question 76 Explanation: 
Question 77
Consider a network with five nodes, N1 to N5m as show below. The network a distance vector protocol. What is the routes Have stabilised, the distance vector a different notes are as follows N1:(0,1,7,8,4) N2:(1,0,6,7,3) N3:(7,6,0,2,6) N4:(8,7,2,0,4) N5:(4,3,6,4,0)      
A
3
B
9
C
10
D
       Computer-Networks       Routing       Gate 2011
Question 77 Explanation: 
N3 has neighbors N2 and N4
N2 has made entry ∞
N4 has the distance of 8 to N1
N3 has the distance of 2 to N4
So 2 + 8 = 10
Question 78
One of the header fields in an IP datagram is the Time to Live (TTL) field. Which of the following statements best explains the need for this field?
A
It can be used to prioritize packets
B
It can be used to reduce delays
C
It can be used to optimize throughput
D
It can be used to prevent packet looping
       Computer-Networks       IP-Packet       2010
Question 78 Explanation: 
Time to Live (TTL) is a limit on the period of time or transmissions in computer and computer network technology that a unit of data (e.g. a packet) can experience before it should be discarded. If the limit is not defined then the packets can go into an indefinite loop. The packet is discarded when the Time to Live field reaches 0 to prevent looping.
Question 79
Which one of the following is not a client server application?
A
Internet chat
B
Web browsing
C
E-mail
D
ping
       Computer-Networks       Client-Server-Application       2010
Question 79 Explanation: 
Ping is used for knowing status of a host by another host. it is not a client server application.
Question 80
Suppose computers A and B have IP addresses 10.105.1.113 and 10.105.1.91 respectively and they both use the same net mask N. Which of the values of N given below should not be used if A and B should belong to the same network?
A
255.255.255.0
B
255.255.255.128
C
255.255.255.192
D
255.255.255.224
       Computer-Networks       IP-Address       2010
Question 80 Explanation: 
When we perform bitwise AND operation between IP Address and Subnet Mask, it gives Network ID. If for both IP results is same Network ID. It means, both IP are belong to the same network else it's on different network.
When we perform AND operation between IP address 10.105.1.113 and 255.255.255.224 result is 10.105.1.96 and when we perform AND operation between IP address 10.105.1.91 and 255.255.255.224 result is 10.105.1.64.
Therefore, 10.105.1.96 and 10.105.1.64 are different network, so D is correct answer.
Question 81
A
4
B
3
C
2
D
1
       Computer-Networks       Routing       2010
Question 81 Explanation: 
Link R1- R2 will not be used because its cost is 6 and link R1-R3-R2 costs 5, which is lesser than R1-R2 link.
Similarly, link R4-R6 will not be used, instead this link we can use R4-R5-R6 link which costs only 5 unit.
Question 82
Consider a network with 6 routers R1 to R6 connected with links having weights as shown in the following diagram: All the routers use the distance vector based routing algorithm to update their routing tables. Each router starts with its routing table initialized to contain an entry for each neighbour with the weight of the respective connecting link. After all the routing tables stabilize, how many links in the network will never be used for carrying any data?  
A
0
B
1
C
2
D
4
       Computer-Networks       Routing       2010
Question 82 Explanation: 
Now Graph will look like

And only link that will be removed is R5-R6 link.
Question 83
In the RSA public key cryptosystem, the private and public keys are (e,n) and (d,n) respectively, where n=p*q and p and q are large primes. Besides, n is public and p and q are private. Let M be an integer such Which of the above equations correctly represent RSA cryptosystem?
A
I and II
B
I and III
C
II and IV
D
III and IV
       Computer-Networks       Network-Security       2009
Question 83 Explanation: 
To generate the encryption and decryption keys, we can proceed as follows.
1. Generate randomly two “large” primes p and q.
2. Compute n=pq and ∅=(p-1)(q-1).
3. Choose a number e so that
gcd(e,∅)=1
4. Find the multiplicative inverse of e modulo ∅, i.e., find d so that
ed≡1 (mod ∅)
This can be done efficiently using Euclid’s Extended Algorithm.
The encryption public key is KE=(n,e) and the decryption private key is KD=(n,d).
The encryption function is
E(M)=Me mod n
The decryption function is
D(M)=Md mod n
Question 84
While opening a TCP connection, the initial sequence number is to be derived using a time-of-day (ToD) clock that keeps running even when the host is down. The low order 32 bits of the counter of the ToD clock is to be used for the initial sequence numbers. The clock counter increments once per millisecond. The maximum packet lifetime is given to be Which one of the choices given below is closest to the minimum permissible rate at which sequence numbers used for packets of a connection can increase?
A
0.015/s
B
0.064/s
C
0.135/s
D
0.327/s
       Computer-Networks       TCP       2009
Question 84 Explanation: 
32 bits are used to represent a sequence number. So we have 232 different sequence number.
The maximum packet lifetime is given is given 64s.
Maximum data rate possible(bandwidth) to avoid the wraparound = 232/64 = 226 Byte/sec.
The clock counter increments once per milliseconds = That means when then counter increments next possible sequence number is generated. The packet lifetime is 64 seconds and after this 64 seconds next sequence number is come. So that means in this 64 seconds only 1 sequence number is generated.
Hence the minimum rate is = 1/64 = 0.015/sec.
Question 85
Let G(x) be the generator polynomial used for CRC checking. What is the condition that should be satisfied by G(x) to detect odd number of bits in error?
A
G(x) contains more than two terms
B
G(x) does not divide 1+xk, for any k not exceeding the frame length
C
1+x is a factor of G(x)
D
G(x) has an odd number of terms
       Computer-Networks       Error-Detection       2009
Question 85 Explanation: 
Odd number of bit errors can be detected if G(x) contains (x+1) as a factor.
Question 86
Frames of 1000 bits are sent over a 106 bps duplex link between two hosts. The propagation time is 25ms. Frames are to be transmitted into this link to maximally pack them in transit (within the link). What is the minimum number of bits (l) that will be required to represent the sequence numbers distinctly? Assume that no time gap needs to be given between transmission of two frames.
A
I = 2
B
I = 3
C
I = 4
D
I = 5
       Computer-Networks       Sliding-Window-Protocol       2009
Question 86 Explanation: 
Transmission time (Tt)=1000/106 seconds =1 ms
Maximum number of frames that can be transmit to maximally pack them is=(Tt+2Tp)/Tx = (25+1)/1=26 which is window size
Minimum sequence numbers required = 26
Minimum number of bits required for sequence number is 5.
Question 87

Frames of 1000 bits are sent over a 106 bps duplex link between two hosts. The propagation time is 25ms. Frames are to be transmitted into this link to maximally pack them in transit (within the link).

Suppose that the sliding window protocol is used with the sender window size of 2l, where l is the number of bits identified in the earlier part and acknowledgements are always piggy backed. After sending 2l frames, what is the minimum time the sender will have to wait before starting transmission of the next frame? (Identify the closest choice ignoring the frame processing time.)

A
16ms
B
18ms
C
20ms
D
22ms
       Computer-Networks       Sliding-Window-Protocol       2009
Question 87 Explanation: 

From above diagram we can say that Total RTT will be
1ms (transmitting time for first frame)
+ 25ms (propagation delay from S to R)
+ 1ms (transmitting time for piggybacked ACK)
+ 25ms (propagation delay from R to S)
= 52 ms
Also total time for sender needed to transmit 32 frames (22 = 25 = 32) is 32×1ms = 32ms
So sender has to wait for,
(52 - 32)ms
= 20ms
Question 88
What is the maximum size of data that the application layer can pass on to the TCP layer below?
A
Any size
B
216 bytes-size of TCP header
C
216 bytes
D
1500 bytes
       Computer-Networks       Application-Layer-Protocol       Gate-2008
Question 88 Explanation: 
Application Layer - Any size
Transport Layer - 65515 byte
Network layer - 65535 byte
Data link layer - 1500 byte
Question 89
Which of the following system calls results in the sending of SYN packets?
A
socket
B
bind
C
listen
D
connect
       Computer-Networks       Sockets       Gate-2008
Question 89 Explanation: 
When connect( ) is called by client, following three way handshake happens to establish the connection in TCP.
1) The client requests a connection by sending a SYN (synchronize) message to the server.
2) The server acknowledges this request by sending SYN-ACK back to the client.
3) The client responds with an ACK, and the connection is established.
Question 90
In the slow start phase of the TCP congestion control algorithm, the size of the congestion window
A
does not increase
B
increases linearly
C
increases quadratically
D
increases exponentially
       Computer-Networks       TCP       Gate-2008
Question 90 Explanation: 
In slow start phase, window size will grow exponentially. when the threshold is reached and congestion avoidance phase begins. In congestion avoidance phase, the window is increased linearly.
Question 91
If a class B network on the Internet has a subnet mask of 255.255.248.0, what is the maximum number of hosts per subnet?
A
1022
B
1023
C
2046
D
2047
       Computer-Networks       IP-Address       Gate-2008
Question 91 Explanation: 
255.255.248.0 can be written as 11111111.11111111.11111000.00000000
Number of bits assigned for host id is the number of zeros in subnet mask. Here 11 bits are used.
for host id so maximum possible hosts are= 211 -2=2046
Question 92
A computer on a 10Mbps network is regulated by a token bucket. The token bucket is filled at a rate of 2Mbps. It is initially filled to capacity with 16Megabits. What is the maximum duration for which the computer can transmit at the full 10Mbps?
A
1.6 seconds
B
2 seconds
C
5 seconds
D
8 seconds
       Computer-Networks       Token-Bucket       Gate-2008
Question 92 Explanation: 
Duration = C/x-y, where C is the initial capacity, x is outgoing rate and y is incoming rate.
Question 93

A client process P needs to make a TCP connection to a server process S. Consider the following situation: the server process S executes a socket (), a bind () and a listen () system call in that order, following which it is preempted. Subsequently, the client process P executes a socket () system call followed by connect () system call to connect to the server process S. The server process has not executed any accept() system call. Which one of the following events could take place?

A
connect ( ) system call returns successfully
B
connect ( ) system call blocks
C
connect ( ) system call returns an error
D
connect ( ) system call results in a core dump
       Computer-Networks       Sockets       Gate-2008
Question 93 Explanation: 
Connect() System call is not blocking system call but it blocks until connection is established or rejected. If accept() is not executed at server then connection will be rejected and an error statement is returned.
Question 94
Provider the best matching between the entries in the two columns given in the table below:
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
       Computer-Networks       Match-the-Following       Gate 2008-IT
Question 94 Explanation: 
(i) Proxy server: Proxy server and firewall are to be combined.
(ii) Kazaa and DC++ are P2P applications.
(iii) P2P slip is a processor of PPP.
(iv) DNS allows caching of entries at local server.
Question 95

A 1Mbps satellite link connects two ground stations. The altitude of the satellite is 36,504 km and speed of the signal is 3 × 108 m/s. What should be the packet size for a channel utilization of 25% for a satellite link using go-back-127 sliding window protocol? Assume that the acknowledgment packets are negligible in size and that there are no errors during communication.

A
120 bytes
B
60 bytes
C
240 bytes
D
90 bytes
       Computer-Networks       Sliding-Window-Protocol       Gate 2008-IT
Question 95 Explanation: 
Time to reach satellite = 36504000/3×108 = 0.121685
RTT = 4×Time to reach satellite (S1→S, S→S2, S2→S, S→S1)

∴ RTT = 0.48
Question 96
The minimum frame size required for a CSMA/CD based computer network running at 1 Gbps on a 200 m cable with a link speed of 2 × 10m/s is
A
125 bytes
B
250 bytes
C
500 bytes
D
None of these
       Computer-Networks       Ethernet       Gate 2008-IT
Question 96 Explanation: 
For CSMA/CD protocol to work, the transmission time of the frame must be more than minimum value. This minimum value is the RTT.
So,

Question 97
Data transmitted on a link uses the following 2D parity scheme for error detection: Each sequence of 28 bits is arranged in a 4×7 matrix (rows r0 through r3, and columns d7 through d1) and is padded with a column d0 and row r4 of parity bits computed using the Even parity scheme. Each bit of column d0 (respectively, row r4) gives the parity of the corresponding row (respectively, column). These 40 bits are transmitted over the data link. The table shows data received by a receiver and has n corrupted bits. What is the mini­mum possible value of n?    
A
1
B
2
C
3
D
4
       Computer-Networks       Error-Detection       Gate 2008-IT
Question 97 Explanation: 

Here, we need to change minimum 3-bits, and by doing it we get correct parity column wise and row wise (correction marked by boxed number).
Question 98
Two popular routing algorithms are Distance Vector(DV) and Link State (LS) routing. Which of the following are true? (S1) Count to infinity is a problem only with DV and not LS routing (S2) In LS, the shortest path algorithm is run only at one node (S3) In DV, the shortest path algorithm is run only at one node (S4) DV requires lesser number of network messages than LS
A
S1, S2 and S4 only
B
S1, S3 and S4 only
C
S2 and S3 only
D
S1 and S4 only
       Computer-Networks       Routing       Gate 2008-IT
Question 98 Explanation: 
→ Count to infinity problem only exist into the DVR.
→ In LSR shortest path is calculated at each every router. (B) is wrong.
→ In DVR also shortest path is calculated at each and every router. (C) is wrong.
→ Since DVR is based upon local knowledge whereas LSR is based upon global knowledge.
Question 99
Which of the following statements are TRUE? (S1) TCP handles both congestion and flow control (S2) UDP handles congestion but not flow control (S3) Fast retransmit deals with congestion but not flow control (S4) Slow start mechanism deals with both congestion and flow control
A
S1, S2 and S3 only
B
S1 and S3 only
C
S3 and S4 only
D
S1, S3 and S4 only
       Computer-Networks       Congestion-Control       Gate 2008-IT
Question 99 Explanation: 
S1:
TCP handles both congestion and flow control ⇒ True.
S2:
UDP handles congestion but not flow control ⇒ False, because UDP neither handles congestion control nor flow control.
S3:
Fast retransmits deals with congestion but not flow control ⇒ True.
S4:
Slow start mechanism deals with both congestion and flow control ⇒ False, because it has nothing to do with flow control. Flow control is taken care by Advertisement window.
Question 100
The three way handshake for TCP connection establishment is shown below. Which of the following statements are TRUE? (S1) Loss of SYN + ACK from the server will not establish a connection (S2) Loss of ACK from the client cannot establish the connection (S3) The server moves LISTEN → SYN_RCVD → SYN_SENT → ESTABLISHED in the state machine on no packet loss (S4) The server moves LISTEN → SYN_RCVD → ESTABLISHED in the state machine on no packet loss.  
A
S2 and S3 only
B
S1 and S4
C
S1 and S3
D
S2 and S4
       Computer-Networks       TCP       Gate 2008-IT
Question 100 Explanation: 
S1 → True.
S2 → False, because if after ACK client immediately sends data then everything goes on without worry.
S3 → False.
S4 → True.
Question 101
The total number of keys required for a set of n individuals to be able to communicate with each other using secret key and public key crypto-systems, respectively are:
A
B
C
D
       Computer-Networks       Network-Security       Gate 2008-IT
Question 101 Explanation: 
For private key crypto, a key used for encryption as well as decryption. So, no. of keys required for n individuals is same as no. of communication link between any two individuals which is
nC2 = n(n-1)/2
In case of public key, each sender has its own public key as well as private key. So, no. of keys are 2n.
Question 102
Host X has IP address 192.168.1.97 and is connected through two routers R1 and R2 to an­other host Y with IP address 192.168.1.80. Router R1 has IP addresses 192.168.1.135 and 192.168.1.110. R2 has IP addresses 192.168.1.67 and 192.168.1.155. The netmask used in the network is 255.255.255.224. Given the information above, how many distinct subnets are guaranteed to already exist in the network?
A
1
B
2
C
3
D
6
       Computer-Networks       IP-Address       Gate 2008-IT
Question 102 Explanation: 
Simply, Bitwise AND the bits of fourth octet for each of the following IP address with fourth octet of subnet mask. You will get only
XX.XX.XX.96, XX.XX.XX.64 and XX.XX.XX.128.
Question 103
Host X has IP address 192.168.1.97 and is connected through two routers R1 and R2 to an­other host Y with IP address 192.168.1.80. Router R1 has IP addresses 192.168.1.135 and 192.168.1.110. R2 has IP addresses 192.168.1.67 and 192.168.1.155. The netmask used in the network is 255.255.255.224. Which IP address should X configure its gateway as?  
A
192.168.1.67
B
192.168.1.110
C
192.168.1.135
D
192.168.1.155
       Computer-Networks       IP-Address       Gate 2008-IT
Question 103 Explanation: 
X must be able to reach the gateway using the net mask.
Subnet no. of host X is,

Now, the gateway must also have the same subnet number.
Let's take IP 192.168.1.110 of R1,

and hence this can be used by X.
Question 104
In Ethernet when Manchester encoding is used, the bit rate is:
A
Half the baud rate.
B
Twice the baud rate.
C
Same as the baud rate.
D
None of the above.
       Computer-Networks       Ethernet       Gate-2007
Question 104 Explanation: 
Bit rate is half the baud rate in Manchester encoding as bits are transferred only during a positive transition of the clock.
Question 105
Which one of the following uses UDP as the transport protocol?
A
HTTP
B
Telnet
C
DNS
D
SMTP
       Computer-Networks       Transport Layer Protocol       Gate-2007
Question 105 Explanation: 
DNS uses the UDP at the transport layer with port number 53 for name resolution.
HTTP, Telnet and SMTP uses TCP.
Question 106
There are n stations in a slotted LAN. Each station attempts to transmit with a probability p in each time slot. What is the probability that ONLY one station transmits in a given time slot?
A
np(1-p)n-1
B
(1-p)n-1
C
p(1-p)n-1
D
1-(1-p)n-1
       Computer-Networks       LAN-and-Probability       Gate-2007
Question 106 Explanation: 
This question can be solved by binomial distribution,
nC1P1(1-p)n-1
= nP(1-P)n-1
Question 107
In a token ring network the transmission speed is 107 bps and the propagation speed is 200 metres/μs. The 1-bit delay in this network is equivalent to:
A
500 metres of cable.
B
200 metres of cable.
C
20 metres of cable.
D
50 metres of cable.
       Computer-Networks       Token-Ring-Network       Gate-2007
Question 107 Explanation: 
Tt= L / B => 1/ 107 = 0.1 microsec
Given Tp= 200 m / microsec
In, 1 microsec it covers 200m
Therefore, in 0.1 microsec it is 200 * 0.1 = 20 meters
Question 108
The address of a class B host is to be split into subnets with a 6-bit subnet number. What is the maximum number of subnets and the maximum number of hosts in each subnet?
A
62 subnets and 262142 hosts.
B
64 subnets and 262142 hosts.
C
62 subnets and 1022 hosts.
D
64 subnets and 1024 hosts.
       Computer-Networks       IP-Address       Gate-2007
Question 108 Explanation: 
It is a class B address, so there 16-bits for NID and 16-bits for HID.
From HID, we took 6-bits for subnetting.
Then total subnets possible = ( 26 ) - 2 = 64
Total hosts possible for each subnet = (210) - 2 = 1022
Question 109
The message 11001001 is to be transmitted using the CRC polynomial x3+1 to protect it from errors. The message that should be transmitted is:
A
11001001000
B
11001001011
C
11001010
D
110010010011
       Computer-Networks       CRC       Gate-2007
Question 109 Explanation: 
CRC polynomial = x3+1 [∵ In data 3-zero’s need to be append to data]
= 1001

∴ Data transmitted is: 11001001011
Question 110
The distance between two stations M and N is L kilometers. All frames are K bits long. The propagation delay per kilometer is t seconds. Let R bits/second be the channel capacity. Assuming that processing delay is negligible, the minimum number of bits for the sequence number field in a frame for maximum utilization, when the sliding window protocol is used, is:  
A
⌈log2(2LtR+2K/K)⌉
B
⌈log2(2LtR/K)⌉
C
⌈log2(2LtR+K/K)⌉
D
⌈log2(2LtR+K/2K)⌉
       Computer-Networks       Sliding-Window-Protocol       Gate-2007
Question 110 Explanation: 
Maximum window size at sender side,
N = (Tk + 2Tp)/Tk
Tp = l × l sec
Tk = K/R sec
So, N = (K/R+2Lt)/(K/R) = K+2LtR/K
Note that when asked for general sliding window protocol (not GBN nor SR) then we do not care about receiver's window size.
So, no. of bits required,
⌈log2 K+2LtR/K⌉
Question 111
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
       Computer-Networks       Match-the-Following       Gate-2007
Question 111 Explanation: 
P) SMTP is used for email transmission.
Q) BGP is network layer protocol that manages low packets are routed across the network.
R) TCP is a transport layer protocol.
S) PPP is a data link layer protocol.
Question 112
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
       Computer-Networks       TCP       Gate 2007-IT
Question 112 Explanation: 
Statement-I: It is False.
The timeout value cannot be fixed for entire duration as it will turn timer to static timer, we need dynamic timer for timeout.
Statement-II: It is True.
Basic algorithm, Jacobson's algorithm, Karl's modification; these three algorithms are to be appropriate to RTT estimation algorithm used to set timeout value dynamically.
Statement-III: It 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.
Question 113

Consider a TCP connection in a state where there are no outstanding ACKs. The sender sends two segments back to back. The sequence numbers of the first and second segments are 230 and 290 respectively. The first segment was lost, but the second segment was received correctly by the receiver. Let X be the amount of data carried in the first segment (in bytes), and Y be the ACK number sent by the receiver. The values of X and Y (in that order) are

A
60 and 290
B
230 and 291
C
60 and 231
D
60 and 230
       Computer-Networks       TCP       Gate 2007-IT
Question 113 Explanation: 
In the 1st segment data is from byte number 230 to byte number 289, that is 60 bytes. As 1st segment is lost, so TCP will send ACK for the next-in-order segment receiver is to be expecting. So, it will be for 230.
Question 114
Consider the following two statements: i. A hash function (these are often used for computing digital signatures) is an injective function. A. encryption technique such as DES performs a permutation on the elements of its input alphabet. Which one of the following options is valid for the above two statements?  
A
Both are false
B
Statement (i) is true and the other is false
C
Statement (ii) is true and the other is false
D
Both are true
       Computer-Networks       Network-Security       Gate 2007-IT
Question 114 Explanation: 
i) Hash function is many to one function. It is not one-one (or) injective.
ii) It uses the P-Box permutation.
Statement-I is false, II is true.
Question 115

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
       Computer-Networks       TCP       Gate 2007-IT
Question 115 Explanation: 
Pushdown automata with two stacks which is the Turing machine.
Turing machine can do everything as the normal computer can do, so firewall can be created by the TM.
Question 116
An error correcting code has the following code words: 00000000, 00001111, 01010101, 10101010, 11110000. What is the maximum number of bit errors that can be corrected ?
A
0
B
1
C
2
D
3
       Computer-Networks       Error-Correction       Gate 2007-IT
Question 116 Explanation: 
For correction:
The no. of bits that can be corrected is
Question 117
For the network given in the figure below, the routing tables of the four nodes A, E, D and G are shown. Suppose that F has estimated its delay to its neighbors, A, E, D and G as 8, 10, 12 and 6 msecs respectively and updates its routing table using distance vector routing technique.  
A
B
C
D
       Computer-Networks       Routing       Gate 2007-IT
Question 117 Explanation: 
Distance from F to F is 0 which eliminates option (C).
Using distance vector routing protocol, F→D→B yeilds distance as 20 which eliminates option (B) and (D).
Question 118
In the waveform (a) given below, a bit stream is encoded by Manchester encoding scheme. The same bit stream is encoded in a different coding scheme in wave form (b). The bit stream and the coding scheme are  
A
1000010111 and Differential Manchester respectively
B
0111101000 and Differential Manchester respectively
C
1000010111 and Integral Manchester respectively
D
0111101000 and Integral Manchester respectively
       Computer-Networks       Manchester-Encoding       Gate 2007-IT
Question 118 Explanation: 
Both (A) & (B) is correct, it is just the convention which determine the correct answer.
As per IEEE (B) is correct.
As per GE Thomas (A) is correct.
As we usually follow IEEE thus (B) is correct.
Question 119

Let us consider a statistical time division multiplexing of packets. The number of sources is 10. In a time unit, a source transmits a packet of 1000 bits. The number of sources sending data for the first 20 time units is 6, 9, 3, 7, 2, 2, 2, 3, 4, 6, 1, 10, 7, 5, 8, 3, 6, 2, 9, 5 respectively. The output capacity of multiplexer is 5000 bits per time unit. Then the average number of backlogged of packets per time unit during the given period is

A
5
B
4.45
C
3.45
D
0
       Computer-Networks       Access-Control-Methods       Gate 2007-IT
Question 119 Explanation: 
The capacity of multiplexer is 5000 bits per time unit. This means there are 5 packets per unit time since each source transmits a packet of 1000 bits in a unit time.
If the no. of packets transmitted is larger than 5 then the extra packets are backlogged. This means gets added to the next number and further backlog is calculated.

Average no. of backlogged packets = 89/20 = 4.45
Question 120

A group of 15 routers are interconnected in a centralized complete binary tree with a router at each tree node. Router j communicates with router j by sending a message to the root of the tree. The root then sends the message back down to router j. The mean number of hops per message, assuming all possible router pairs are equally likely is

 
A
3
B
4.26
C
4.53
D
5.26
       Computer-Networks       Routing       Gate 2007-IT
Question 120 Explanation: 
Step 1:
Message goes up from sender to root,
Average hops message goes to root
= (3×8) + (2×4) + (1×2) + (0×1)/15
= 2.267
Here, (3×8) represents 3 hops and 8 routers for bottom most level and so on.
Step 2:
Similarly, average hops when message comes down
= (3×8) + (2×4) + (1×2) + (0×1)/15
= 2.267
So, Total hops = 2×2.267 = 4.53
Question 121
A broadcast channel has 10 nodes and total capacity of 10 Mbps. It uses polling for medium access. Once a node finishes transmission, there is a polling delay of 80 μs to poll the next node. Whenever a node is polled, it is allowed to transmit a maximum of 1000 bytes. The maximum throughput of the broadcast channel is
A
1 Mbps
B
100/11 Mbps
C
10 Mbps
D
100 Mbps
       Computer-Networks       Access-Control-Methods       Gate 2007-IT
Question 121 Explanation: 
Question 122
Consider the following clauses: i. Not inherently suitable for client authentication. ii. Not a state sensitive protocol. iii. Must be operated with more than one server. iv. Suitable for structured message organization. v. May need two ports on the serve side for proper operation. The option that has the maximum number of correct matches is
A
IMAP-(i), FTP-(ii), HTTP-(iii), DNS-(iv), POP3-(v)
B
FTP-(i), POP3-(ii), SMTP-(iii), HTTP-(iv), IMAP-(v)
C
POP3-(i), SMTP-(ii), DNS-(iii), IMAP-(iv), HTTP-(v)
D
SMTP-(i), HTTP-(ii), IMAP-(iii), DNS-(iv), FTP-(v)
       Computer-Networks       Network-Protocols       Gate 2007-IT
Question 122 Explanation: 
(ii) HTTP as it does not depend on state of device or operating system.
(v) FTP needs two ports, 20 for data and 21 for control.
Hence, only (D) matches.
Question 123
Your are given the following four bytes : 10100011            00110111            11101001            10101011 Which of the following are substrings of the base 64 encoding of the above four bytes ?  
A
zdp
B
fpq
C
qwA
D
oze
       Computer-Networks       Network-Security       Gate 2007-IT
Question 123 Explanation: 
You are given the following four bytes:
10100011 00110111 11101001 10101011
So, in total we have 32 bits. And for base 64we need 6 digits of binary no. to represent one digit of base 64 no.
So lets padd 4 bits on RHS, so that total digits will become 36 and we can separate then as group of 6 digits each.

Now, the longest substring will be from checking option is 'fpq'.
Question 124
Consider a token ring topology with N stations (numbered 1 to N) running token ring protocol where the stations are equally spaced. When a station gets the token it is allowed to send one frame of fixed size. Ring latency is tp, while the transmission time of a frame is tt. All other latencies can be neglected. The maximum utilization of the token ring when tt =3 ms, tp = 5 ms, N = 10 is
 
A
0.545
B
0.6
C
0.857
D
0.961
       Computer-Networks       Token-Ring-Topology       Gate 2007-IT
Question 124 Explanation: 
Note: Out of syllabus.
Question 125
Consider a token ring topology with N stations (numbered 1 to N) running token ring protocol where the stations are equally spaced. When a station gets the token it is allowed to send one frame of fixed size. Ring latency is tp, while the transmission time of a frame is tt. All other latencies can be neglected. The maximum utilization of the token ring when tt = 5 ms, tp = 3 ms, N = 15 is :
A
0.545
B
0.655
C
0.9375
D
0.961
       Computer-Networks       Token Ring Topology       Gate 2007-IT
Question 125 Explanation: 
Note: Out of syllabus.
Question 126
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 126 Explanation: 
It prevent infinite looping over network .Bcz each router decrease its value by one and when this(TTL) value become zero ,then it is discarded by router silently.
Question 127
Consider the diagram shown below where a number of LANs are connected by (transparent) bridges. In order to avoid packets looping through circuits in the graph, the bridges organize themselves in a spanning tree. First, the root bridge is identified as the bridge with the least serial number. Next, the root sends out (one or more) data units to enable the setting up of the spanning tree of shortest paths from the root bridge to each bridge. Each bridge identifies a port (the root port) through which it will forward frames to the root bridge. Port conflicts are always resolved in favour of the port with the lower index value. When there is a possibility of multiple bridges forwarding to the same LAN (but not through the root port), ties are broken as follows: bridges closest to the root get preference and between such bridges, the one with the lowest serial number is preferred. For the given connection of LANs by bridges, which one of the following choices represents the depth first traversal of the spanning tree of bridges?  
A
B
C
D
       Computer-Networks       Bridges       Gate-2006
Question 127 Explanation: 
Use Spanning tree generated in previous question.
Question 128
Consider the diagram shown below where a number of LANs are connected by (transparent) bridges. In order to avoid packets looping through circuits in the graph, the bridges organize themselves in a spanning tree. First, the root bridge is identified as the bridge with the least serial number. Next, the root sends out (one or more) data units to enable the setting up of the spanning tree of shortest paths from the root bridge to each bridge. Each bridge identifies a port (the root port) through which it will forward frames to the root bridge. Port conflicts are always resolved in favour of the port with the lower index value. When there is a possibility of multiple bridges forwarding to the same LAN (but not through the root port), ties are broken as follows: bridges closest to the root get preference and between such bridges, the one with the lowest serial number is preferred. For the given connection of LANs by bridges, which one of the following choices represents the depth first traversal of the spanning tree of bridges?
A
B1, B5, B3, B4, B2
B
B1, B3, B5, B2, B4
C
B1, B5, B2, B3, B4
D
B1, B3, B4, B5, B2
       Computer-Networks       Bridges       Gate-2006
Question 128 Explanation: 
Start Depth First traversal from B1 . B5 is connected to port 1 of B1 so B5 will be traversed after B1.
Port1 of B5 is connected to port1 of B2 and port2 of B3. B2 is connected with lower index port so B2 is traversed next.
Port 2 Both B3 and B4 is connected with port1 of B2, but B3 is Closer to root so B3 will be traversed next.
Depth First traversal is B1,B5,B2,B3,B4.
Question 129
Station A needs to send a message consisting of 9 packets to Station B using a sliding window (window size 3) and go-back-n error control strategy. All packets are ready and immediately available for transmission. If every 5th packet that A transmits gets lost (but no acks from B ever get lost), then what is the number of packets that A will transmit for sending the message to B?
A
12
B
14
C
16
D
18
       Computer-Networks       Sliding-Window-Protocol       Gate-2006
Question 129 Explanation: 
Window size is 3, so maximum 3 packets can be remained unacknowledged. In go back ‘n’ if acknowledge for a packet is not received then packets after that packet is also retransmitted.
Frame sequence for 9 frame is shown below. Frame with bold sequence number gets lost.
1 2 3 4 [5 6 7] 5 6 [7 8 9] 7 8 9 9 = 16
Question 130
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 130 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.
Question 131
Two computers C1 and C2 are configured as follows. C1 has IP address 203.197.2.53 and netmask 255.255.128.0. C2 has IP address 203.197.75.201 and netmask 255.255.192.0. Which one of the following statements is true?
A
C1 and C2 both assume they are on the same network
B
C2 assumes C1 is on same network, but C1 assumes C2 is on a different network
C
C1 assumes C2 is on same network, but C2 assumes C1 is on a different network
D
C1 and C2 both assume they are on different networks
       Computer-Networks       IP-Address       Gate-2006
Question 131 Explanation: 
From C1 side,
Subnet mask for C1 is 255.255.128.0.
So it finds the Network ID as,
C1 → 203.197.2.53 AND 255.255.128.0 = 203.197.0.0
C2 → 203.197.75.201 AND 255.255.128.0 = 203.197.0.0
Both same.
From C2 side,
Subnet mask for C2 is 255.255.192.0.
So it finds the network ID as,
C1 → 203.197.2.53 AND 255.255.192.0 = 203.197.0.0
C2 → 203.197.75.201 AND 255.255.192.0 = 203.197.64.0
Both different.
Hence, option 'C' is correct.
Question 132
HELO and PORT, respectively, are commands from the protocols
A
FTP and HTTP
B
TELNET and POP3
C
HTTP and TELNET
D
SMTP and FTP
       Computer-Networks       Network-Protocols       Gate 2006-IT
Question 132 Explanation: 
Note: Out of syllabus.
Question 133
Which of the following statements is TRUE?
A
Both Ethernet frame and IP packet include checksum fields
B
Ethernet frame includes a checksum field and IP packet includes a CRC field
C
Ethernet frame includes a CRC field and IP packet includes a checksum field
D
Both Ethernet frame and IP packet include CRC fields
       Computer-Networks       Ethernet       Gate 2006-IT
Question 133 Explanation: 
Ethernet frame:

IP packet:

IP Datagram:
Question 134
Which of the following statement(s) is TRUE?
  1. A hash function takes a message of arbitrary length and generates a fixed length code.
  2. A hash function takes a message of fixed length and generates a code of variable length.
  3. A hash function may give the same hash value for distinct messages.
A
1 only
B
2 and 3 only
C
1 and 3 only
D
2 only
       Computer-Networks       Network-Security       Gate 2006-IT
Question 134 Explanation: 
(1) A hash function takes a message of arbitary length and generates a fixed length code. So this is correct.
(2) Statement-2 is wrong, refer statement-1.
(3) Statement-3 is correct, for example hash function N%10, this will generate same values for 1 as well as 2!
Question 135
A router uses the following routing table: A packet bearing a destination address 144.16.68.117 arrives at the router. On which interface will it be forwarded?  
A
eth0
B
eth1
C
eth2
D
eth3
       Computer-Networks       IP-Address       Gate 2006-IT
Question 135 Explanation: 
Firstly start with longest mask.
144.16.68.117 = 144.16.68.01110101 AND 255.255.255.224 = 255.255.255.11100000
= 144.16.68.96(Not matching with destination)
Now, take 255.255.255.0
144.16.68.117 AND 255.255.255.0
= 144.16.68.0 (matched)
Hence, option (C) is correct.
Question 136

Suppose that it takes 1 unit of time to transmit a packet (of fixed size) on a communication link. The link layer uses a window flow control protocol with a window size of N packets. Each packet causes an ack or a nak to be generated by the receiver, and ack/nak transmission times are negligible. Further, the round trip time on the link is equal to N units. Consider time i > N. If only acks have been received till time i(no naks), then the goodput evaluated at the transmitter at time i(in packets per unit time) is

A
1 – N/i
B
i/(N + i)
C
1
D
1 – e(i/N)
       Computer-Networks       Sliding-Window-Protocol       Gate 2006-IT
Question 136 Explanation: 
Goodput is the application level throughtout, i.e., the no. of useful information bits delivered by the network to a certain destination per unit of time.
So, successful delivery of packet can be assured if ack has been received for it.
So till time 'i' we would have transmitted 'i' packets but only (i - N) can be acknowledged as minimum time for a packet to get acknowledged is N (since RTT is N which is equal to the window size, there is no waiting for the sender).
So, successfully delivered packets = (i - N)
Time for transmission = i
Goodput = Successfully delivered data/Time
= (i - N)/i
= 1 - N/i
Question 137

In the 4B/5B encoding scheme, every 4 bits of data are encoded in a 5-bit codeword. It is required that the codewords have at most 1 leading and at most 1 trailing zero. How many such codewords are possible?

A
14
B
16
C
18
D
20
       Computer-Networks       Encoding       Gate 2006-IT
Question 137 Explanation: 
It says we have 5 bit codeword such that "it can't have two consecutive zeros in first and second bit" and also "can't have two consecutive zeros in last two bits".
Codeword with first two bits '0'
= 0 0 x x x
= 23
= 8
Codeword with last two bits '0'
= x x x 0 0
= 23 = 8
Codeword with first two and last two bits '0'
= 0 0 x 0 0
= 2
Codeword with first or last two bits '0'
= 8 + 8 - 2
= 14
Therefore possible codewords
= 32 - 14
= 18
Question 138

A router has two full-duplex Ethernet interfaces each operating at 100 Mb/s. Ethernet frames are at least 84 bytes long (including the Preamble and the Inter-Packet-Gap). The maximum packet processing time at the router for wirespeed forwarding to be possible is (in micro­seconds)

A
0.01
B
3.36
C
6.72
D
8
       Computer-Networks       Ethernet       Gate 2006-IT
Question 138 Explanation: 
Let's first calculate transmission time Tt,
Tt = 84×8/10×106 = 6.72μs
But since a router has two full-duplex ethernet interfaces, so the maximum processing time should be,
6.72/2 μs = 3.36μs
Question 139

A link of capacity 100 Mbps is carrying traffic from a number of sources. Each source generates an on-off traffic stream; when the source is on, the rate of traffic is 10 Mbps, and when the source is off, the rate of traffic is zero. The duty cycle, which is the ratio of on-time to off-time, is 1:2. When there is no buffer at the link, the minimum number of sources that can be multiplexed on the link so that link capacity is not wasted and no data loss occurs is S1. Assuming that all sources are synchronized and that the link is provided with a large buffer, the maximum number of sources that can be multiplexed so that no data loss occurs is S2. The values of S1 and S2 are, respectively,

A
10 and 30
B
12 and 25
C
5 and 33
D
15 and 22
       Computer-Networks       Network-Layer       Gate 2006-IT
Question 139 Explanation: 
For S1,
Since there is no buffer and constraint given is there should not be any data lost, and no wastage of capacity as well.
Since data should not be lost, we calculate for the extreme case when all sources are ontime (that is transmitting).
10Mbps × n-station ≤ 100Mbps
n = 10 = S1
In the next part of question, it is given that the link is provided with large buffer and we are asked to find out large no. of stations.
For that we calculate expected value of bandwidth usage,
E = 1/3 × 10 + 1/3 × 10 + .......+ ....... n-station times ≤ 100Mbps
⇒ 1/3 × 10 × n-station ≤ 100Mbps ⇒ n-station = 30 = S2
So option (A) is answer.
Question 140

On a wireless link, the probability of packet error is 0.2. A stop-and-wait protocol is used to transfer data across the link. The channel condition is assumed to be independent from transmission to transmission. What is the average number of transmission attempts required to transfer 100 packets?

A
100
B
125
C
150
D
200
       Computer-Networks       Stop-and-Wait-protocol       Gate 2006-IT
Question 140 Explanation: 
Total no. of re-transmissions for one frame, in general, is 1/(1-P) where P is probability of error.
So here it would be for one frame = 1/(1-0.2) = 1/0.8
So for 100 frames = 100/0.8 = 125
Question 141

A program on machine X attempts to open a UDP connection to port 5376 on a machine Y, and a TCP connection to port 8632 on machine Z. However, there are no applications listening at the corresponding ports on Y and Z. An ICMP Port Unreachable error will be generated by

A
Y but not Z
B
Z but not Y
C
Neither Y nor Z
D
Both Y and Z
       Computer-Networks       TCP-and-UDP       Gate 2006-IT
Question 141 Explanation: 
When an IP packet is lost or discarded ICMP packet will be generated by receiver or the router. It doesn't matter whether its containing TCP or UDP inside it.
Question 142
A subnetted Class B network has the following broadcase address : 144.16.95.255. Its subnet mask  
A
is necessarily 255.255.224.0
B
is necessarily 255.255.240.0
C
is necessarily 255.255.248.0
D
could be any one of 255.255.224.0, 255.255.240.0, 255.255.248.0
       Computer-Networks       IP-Address       Gate 2006-IT
Question 142 Explanation: 
In the broadcast address for a subnet, all the host bits are set to 1. So as long as all the bits to the right are 1, bits left to it can be taken as possible subnet.
Broadcast address for subnet is
.95.255 or .01011111.11111111
(as in class B, 16 bits each are used for network and host)
So, we can take minimum 3 bits (from left) as subnet and make rest as host bits (as they are 1)
.224.0 → 11100000.00000000 (leftmost 3 bits for subnet)
.240.0 → 11110000.00000000 (leftmost 4 bits for subnet)
.248.0 → 11111000.00000000 (leftmost 5 bits for subnet)
Question 143
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 143 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 144
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 144 Explanation: 

Packet size have minimum header overhead.
Question 145
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 145 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 146
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 146 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 147

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 147 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 148
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 148 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 149
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 149 Explanation: 
Both TCP and UDP uses IP, which is a datagram service.
Question 150
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 150 Explanation: 
Address Resolution Protocol (ARP) is a request and reply protocol used to find MAC address from IP address.
Question 151
Traceroute reports a possible route that is taken by packets moving from some host A to some other host B. Which of the following options represents the technique used by traceroute to identify these hosts
A
By progressively querying routers about the next router on the path to B using ICMP packets, starting with the first router
B
By requiring each router to append the address to the ICMP packet as it is forwarded to B. The list of all routers en-route to B is returned by B in an ICMP reply packet
C
By ensuring that an ICMP reply packet is returned to A by each router en-route to B, in the ascending order of their hop distance from A
D
By locally computing the shortest path from A to B
       Computer-Networks       Network-Layer       Gate 2005-IT
Question 151 Explanation: 
Traceroute works by sending packets with gradually increasing TTL value, starting with TTL value of 1. The first router receives the packet, decrements the TTL value and drops the packet because it then has TTL value zero. The router sends an ICMP time exceeded message back to the source. The next set of packets are given a TTL value of 2.
So the first router forwards the packets, but the second router drops them and replies with ICMP time exceeded. Proceeding in this way, traceroute uses the returned ICMP time exceeded messages to build a list of routers that packets traverse, until the destination is reached and returns an ICMP echo reply message.
Question 152
Which of the following statements is TRUE about CSMA/CD  
A
IEEE 802.11 wireless LAN runs CSMA/CD protocol
B
Ethernet is not based on CSMA/CD protocol
C
CSMA/CD is not suitable for a high propagation delay network like satellite network
D
There is no contention in a CSMA/CD network
       Computer-Networks       CSMA/CD       Gate 2005-IT
Question 152 Explanation: 
For CSMA/CD requires that sender is to be transmitting atleast till the first bit reaches the receiver. So the collision will be eliminated in case if it is present.
For networks with high propagation delay this time becomes too long hence the minimum packet size required becomes too big to be feasible.
Question 153
Which of the following statements is FALSE regarding a bridge?
A
Bridge is a layer 2 device
B
Bridge reduces collision domain
C
Bridge is used to connect two or more LAN segments
D
Bridge reduces broadcast domain
       Computer-Networks       Network-Layer       Gate 2005-IT
Question 153 Explanation: 
Bridge devices works at the data link layer of the open system interconnected (OSI) model, connecting two different networks together and providing communication between them. So, option A, C are true.
The bridge acts as a interface between two networks and speed the traffic between them and there by reduces the collision domain.
So, option B also True.
Question 154
Count to infinity is a problem associated with
A
link state routing protocol.
B
distance vector routing protocol.
C
DNS while resolving host name.
D
TCP for congestion control.
       Computer-Networks       Network-Layer       Gate 2005-IT
Question 154 Explanation: 
Distance vector routing protocol uses the Bellman-Ford algorithm and Ford-Fulkerson algorithm.
The Bellman-Ford algorithm does not prevent routing loops from happening and suffers from the count-to-infinity problem.
Question 155

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

 
A
10000 bits
B
10000 bytes
C
5000 bits
D
5000 bytes
       Computer-Networks       CSMA/CD       Gate 2005-IT
Question 155 Explanation: 
For CSMA/CD protocol we know that minimum frame size required is,
L ≤ 2×Tp×B
L ≤ 2×(d/v)×B
d = 1Km = 1000m
v = 2×103 m/s
B = 109 bps
By solving the above equation we will set the value of L as,
10000 bits.
Question 156

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

A
80 bytes
B
80 bits
C
160 bytes
D
160 bits
       Computer-Networks       Data-Link-Layer       Gate 2005-IT
Question 156 Explanation: 
Question 157

On a TCP connection, current congestion window size is Congestion Window=4 KB. The window size advertised by the receiver is Advertise Window=6 KB. The last byte sent by the sender is LastByteSent=10240 and the last byte acknowledged by the receiver is LastByteAcked=8192. The current window size at the sender is

A
2048 bytes
B
4096 bytes
C
6144 bytes
D
8192 bytes
       Computer-Networks       Transport Layer       Gate 2005-IT
Question 157 Explanation: 
Corrent sender window
= min (4KB, 6KB)
= 4KB
Question 158

In a communication network, a packet of length L bits takes link L1 with a probability of por link L2 with a probability of p2. Link L1 and L2 have bit error probability of b1and brespectively. The probability that the packet will be received without error via either L1 or L2 is

A
(1 – b1)L p1 + (1 – b2)Lp2
B
[1 – (b1 + b2)L]p1p2
C
(1 – b1)L (1 – b2)Lp1p2
D
1 – (b1 Lp1 + b2 Lp2)
       Computer-Networks       Network-Layer       Gate 2005-IT
Question 158 Explanation: 
Probability of choosing link L1 = p1
Probability for no bit error for any single bit = (1 - b1)
Similarly, for link L2,
Probability of no bit error = (1 - b2)
Packet can go either through link L1 or L2, they are mutually exclusive events.
Probability packet will be received without any error = Probability of L1 being chosen and no errors in any of L bits + Probability of L2 being chosen and no error in any of the L bits
= (1 - b1)L p1 + (1 - b2)L p2
Hence, answer is option A.
Question 159

In a TDM medium access control bus LAN, each station is assigned one time slot per cycle for transmission. Assume that the length of each time slot is the time to transmit 100 bits plus the end-to-end propagation delay. Assume a propagation speed of 2 × 108 m/sec. The length of the LAN is 1 km with a bandwidth of 10 Mbps. The maximum number of stations that can be allowed in the LAN so that the throughput of each station can be 2/3 Mbps is

A
3
B
5
C
10
D
20
       Computer-Networks       Data-Link-Layer       Gate 2005-IT
Question 159 Explanation: 
Question 160

A company has a class C network address of 204.204.204.0. It wishes to have three subnets, one with 100 hosts and two with 50 hosts each. Which one of the following options represents a feasible set of subnet address/subnet mask pairs?

A
204.204.204.128/255.255.255.192
204.204.204.0/255.255.255.128
204.204.204.64/255.255.255.128
B
204.204.204.0/255.255.255.192
204.204.204.192/255.255.255.128
204.204.204.64/255.255.255.128
C
204.204.204.128/255.255.255.128
204.204.204.192/255.255.255.192
204.204.204.224/255.255.255.192
D
204.204.204.128/255.255.255.128
204.204.204.64/255.255.255.192
204.204.204.0/255.255.255.192
       Computer-Networks       Network-Layer       Gate 2005-IT
Question 160 Explanation: 
MSB in last 8 bits is in use to get subnet since it is class C IP.
10000000/128 (mask) - subnet id bit (1) (subnet 1)
01000000/192 (mask) - subnet id bit (01) (subnet 2)
0000000/192 (mask) - subnet id bit (00) (subnet 3)
Question 161

Assume that “host1.mydomain.dom” has an IP address of 145.128.16.8. Which of the following options would be most appropriate as a subsequence of steps in performing the reverse lookup of 145.128.16.8? In the following options “NS” is an abbreviation of “nameserver”.

A
Query a NS for the root domain and then NS for the “dom” domains
B
Directly query a NS for “dom” and then a NS for “mydomain.dom” domains
C
Query a NS for in-addr.arpa and then a NS for 128.145.in-addr.arpa domains
D
Directly query a NS for 145.in-addr.arpa and then a NS for 128.145.in-addr.arpa domains
       Computer-Networks       Network-Layer       Gate 2005-IT
Question 161 Explanation: 
We are performing reverse lookup of IP address to its hostname.
First we need to locate in-addr.apra, then perform reverse lookup of 8.16.128.145.in-addr.arpa which will point to host1.mydomain.com.
Question 162
Consider the following message M = 1010001101. The cyclic redundancy check (CRC) for this message using the divisor polynomial x5 + x4 + x2 + 1 is :
A
01110
B
01011
C
10101
D
10110
       Computer-Networks       Data-Link-Layer       Gate 2005-IT
Question 162 Explanation: 
Degree of generator polynomial is 5. Hence, 5 zeroes is appended before division.
M = 1010001101

append 5 zeroes = M = 101000110100000

∴ CRC = 01110
Question 163

Suppose that two parties A and B wish to setup a common secret key (D-H key) between themselves using the Diffie-Hellman key exchange technique. They agree on 7 as the modulus and 3 as the primitive root. Party A chooses 2 and party B chooses 5 as their respective secrets. Their D-H key is

A
3
B
4
C
5
D
6
       Computer-Networks       Data-Link-Layer       Gate 2005-IT
Question 163 Explanation: 
For Diffe-Hellaman the secret key is (pab)modn,
where p is the primitive root and n is the modulus and 'a' and 'b' are the secret values selected by parity A & B.
So answer is,
32×5 mod 7 = 310 mod 7 = 4
Question 164
 
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 164 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 165
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 165 Explanation: 
A bridge use MAC addresses (DLL layer) and router uses IP addresses (network layer).
Question 166
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 166 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 167

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 167 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 168
 
A
Eth1 and Eth2
B
Eth0 and Eth2
C
Eth0 and Eth3
D
Eth1 and Eth3
       Computer-Networks       IP-Address       Gate-2004
Question 168 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 169
 
A
200
B
220
C
240
D
260
       Computer-Networks       IPv4-and-TCP       Gate-2004
Question 169 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 170
 
A
325.5 Kbps
B
354.5 Kbps
C
409.6 Kbps
D
512.0 Kbps
       Computer-Networks       IPv4-and-TCP       Gate-2004
Question 170 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 171
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 172
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 172 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 173
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 173 Explanation: 
Note: Out of syllabus.
Question 174
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 174 Explanation: 
Sender can encrypts using the receiver public key and receiver decrypts it using his own private key.
Question 175
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 175 Explanation: 
Maximum no. of hosts = 2(no. of bits in HID) - 2
= 26- 2
= 64 - 2
= 62
Question 176

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 176 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 177
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 177 Explanation: 
In TCP, a unique sequence number is assigned to each byte.
Question 178
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 178 Explanation: 
Note: Out of syllabus.
Question 179
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 179 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 180

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 180 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 181

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 181 Explanation: 
For such a scheme to work properly, we will need a total of M+N distinct sequence numbers.
Question 182

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 182 Explanation: 
Note: Out of syllabus.
Question 183

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 183 Explanation: 
Question 184

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 184 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 185
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 185 Explanation: 
Destination port number is not present at IP header.
Question 186

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 186 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 187

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 187 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 188
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 188 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 189
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 189 Explanation: 
End to end connectivity is the required functionality provided by Transport protocol.
Question 190
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 190 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 191
A 2km long broadcast LAN has 107 bps bandwidth and uses CSMA/CD. The signal travels along the wire at 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 191 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 192
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 192 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 193
What is the distance of the following code 000000, 010101, 000111, 011001, 111111?
A
2
B
3
C
4
D
1
       Computer-Networks       Error-Detection       Gate-1995
Question 193 Explanation: 
Distance = Minimum hamming distance = 2
010101 ⊕ 011001 = 001100
Question 194
A simple and reliable data transfer can be accomplished by using the ‘handshake protocol’. It accomplishes reliable data transfer because for every data item sent by the transmitter __________.
A
in this case receiver has to respond that receiver can be able to receive the data item.
       Computer-Networks       Handshake-Protocol       Gate-1992
Question 195
Start and stop bits do not contain an ‘information’ but are used in serial communication for
A
Error detection
B
Error correction
C
Synchronization
D
Slowing down the communications
       Computer-Networks       Sequential-Circuits       Gate-1992
Question 195 Explanation: 
The start and stop bits are used to synchronize the serial receivers.
Question 196
 
A
Out of syllabus.
       Computer-Networks       General       Gate-1991
There are 196 questions to complete.
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