Previous GATE questions with solutions on Computer Networks (TCP/IP) - CS/IT
GATE-2003
1. 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 & 172.56.87.23
(b) 10.35.28.2 & 10.35.29.4
(c) 191.203.31.87 & 192.234.31.88
(d) 128.8.129.43 & 128.8.161.55
Ans: option (d)
Explanation:
Here is the question is: Which of the following pair of IP addresses belongs to the given network. Applying a subnet mask to an IP address separates network address from host address.
So you have to find the network-id from the IP address using the given subnet mask. Below shows the example of how to find a network-id from given Subnet mask and IP address
The network bits are represented by the 1's in the subnet mask, and the host bits are represented by 0's. Performing a bitwise logical AND operation on the IP address with the subnet mask produces the network address. For example, applying the Class C subnet mask to our IP address 216.3.128.12 produces the following network address:
IP: 1101 1000 . 0000 0011 . 1000 0000 . 0000 1100 (216.003.128.012)
Mask: 1111 1111 . 1111 1111 . 1111 1111 . 0000 0000 (255.255.255.000)
------------------------------------------------------
1101 1000 . 0000 0011 . 1000 0000 . 0000 0000 (216.003.128.000)
Therefore the network-id is: 216.003.128.000
Hence doing the above for all the options, you will find that option (d) belongs to the same network.
GATE-2004
2. The routing table of a router is shown below
On which interface will router forward packets addressed to destinations 128.75.43.16 and 192.12.17.10 respectively?
a)Eth1 and Eth2 b)Eth0 and Eth2
c)Eth0 and Eth3 d)Eth1 and Eth3
Ans: option(a)
Explanation:
Perform AND operation between incoming IP address and Subnet-mask and then compare the result with the destination. Note that if there is a match between multiple Destinations, then select the destination with longest length subnet mask. For example,
128.75.43.16, matches with 128.75.43.0 and 128.75.43.0. But the packet's addressed to 128.75.43.16 will be forwarded to Eth1.
If a result is not matching with any of the given destinations then the packet is forwarded to the default interface (here Eth2). Therefore the packet's addressed to 192.12.17.10 will be forwarded to Eth2.
GATE-2010
3. 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 priortize 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
Ans: (d)
GATE-2010
4. Suppose computers A and B have IP addresses 10.105.1.113 and 10.105.1.91 respectively and they both use the same netmask 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
Ans: option (d)
Explanation:
Apply subnet mask to the IP address, i.e. perform AND operation between IP address and subnet mask. If the result is equal for both IP address A & B then we can use that subnet mask else not. Applying subnet mask to all options we find that option (d) cannot be used.
GATE-2012
5. 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 time out 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
Ans: option (c)
Explanation:
Read Slow Start Algorithm: http://www.btechonline.org/2012/12/slow-start-algorithm-in-tcp.html
Read AIMD Algorithm: http://en.wikipedia.org/wiki/Additive_increase/multiplicative_decrease
Note: Got through the above two links before you get into the solution.
Its given that initial Threshold is: 8MSS
Hence the answer is: 7MSS (since it transmitted 7 segments during 10th transmission)
GATE-2008
6. 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
Ans: option (d)
Explanation:
Read Slow Start Algorithm: http://www.btechonline.org/2012/12/slow-start-algorithm-in-tcp.html
GATE-2008
7. 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
Ans: option (c)
Explanation:
Convert the subnet mask into binary format.
255.255.248.0 = 11111111.11111111.11111000.00000000
Note:
Number of 1's in the subnet mask indicates the Network-ID and the Subnet-ID part
Number of 0's in the subnet mask indicates the Host-ID part
Maximum number of Hosts per subnet = 211 = 2048
where 11 = Number of 0's in the Subnet Mask.
Out of 2048 values, 2 addresses are reserved, hence we remove them (2048-2 = 2046).
Note: In the host part of the address:- all bits as 1 is reserved as broadcast address and all bits as 0 is used as network address of subnet.
Hence option (c) is the answer.
GATE-2008
8. Which of the following system calls results in the sending of SYN packets?
(a) socket (b) bind (c) listen (d) connect
Ans: option (d)
Explanation:
A typical TCP client and server application issues a sequence of TCP system calls to attain certain functions.
The socket system call creates a new socket and assigns the protocol and resources to the created socket descriptor.
The bind system call associates a local network transport address with a socket. For a client process, it is not mandatory to issue a bind call. The kernel takes care of doing an implicit binding when the client process issues the connect system call. It is often necessary for a server process to issue an explicit bind request before it can accept connections or start communication with clients.
The listen call indicates to the protocol that the server process is ready to accept any new incoming connections on the socket. There is a limit on the number of connections that can be queued up, after which any further connection requests are ignored.
The connect system call is normally called by the client process to connect to the server process.
To know more on TCP system call sequences: http://www.ibm.com/developerworks/aix/library/au-tcpsystemcalls/index.html
GATE-2008
9. 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
Ans: option (a)
Explanation:
Application layer can forward any size of data to the transport layer. Transport layer in turn divides the data into several data packets.
GATE-2012
10. The protocol data unit(PDU) for the application layer in the Internet stack is
(a) Segment (b) Datagram (c) Message (d) Frame
Ans: (c)
Note:
* Having all 0's or all 1's either in the network-id or in the host-id is not allowed, because they are meant for special purposes and are reserved.
GATE-2003
1. 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 & 172.56.87.23
(b) 10.35.28.2 & 10.35.29.4
(c) 191.203.31.87 & 192.234.31.88
(d) 128.8.129.43 & 128.8.161.55
Ans: option (d)
Explanation:
Here is the question is: Which of the following pair of IP addresses belongs to the given network. Applying a subnet mask to an IP address separates network address from host address.
So you have to find the network-id from the IP address using the given subnet mask. Below shows the example of how to find a network-id from given Subnet mask and IP address
The network bits are represented by the 1's in the subnet mask, and the host bits are represented by 0's. Performing a bitwise logical AND operation on the IP address with the subnet mask produces the network address. For example, applying the Class C subnet mask to our IP address 216.3.128.12 produces the following network address:
IP: 1101 1000 . 0000 0011 . 1000 0000 . 0000 1100 (216.003.128.012)
Mask: 1111 1111 . 1111 1111 . 1111 1111 . 0000 0000 (255.255.255.000)
------------------------------------------------------
1101 1000 . 0000 0011 . 1000 0000 . 0000 0000 (216.003.128.000)
Therefore the network-id is: 216.003.128.000
Hence doing the above for all the options, you will find that option (d) belongs to the same network.
GATE-2004
2. The routing table of a router is shown below
DESTINATION
|
SUBNETMASK
|
ETHERNET
|
128.75.43.0
|
255.255.255.0
|
Eth0
|
128.75.43.0
|
255.255.255.128
|
Eth1
|
192.12.17.5
|
255.255.255.255
|
Eth3
|
Default
|
Eth2
|
a)Eth1 and Eth2 b)Eth0 and Eth2
c)Eth0 and Eth3 d)Eth1 and Eth3
Ans: option(a)
Explanation:
Perform AND operation between incoming IP address and Subnet-mask and then compare the result with the destination. Note that if there is a match between multiple Destinations, then select the destination with longest length subnet mask. For example,
128.75.43.16, matches with 128.75.43.0 and 128.75.43.0. But the packet's addressed to 128.75.43.16 will be forwarded to Eth1.
If a result is not matching with any of the given destinations then the packet is forwarded to the default interface (here Eth2). Therefore the packet's addressed to 192.12.17.10 will be forwarded to Eth2.
GATE-2010
3. 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 priortize 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
Ans: (d)
GATE-2010
4. Suppose computers A and B have IP addresses 10.105.1.113 and 10.105.1.91 respectively and they both use the same netmask 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
Ans: option (d)
Explanation:
Apply subnet mask to the IP address, i.e. perform AND operation between IP address and subnet mask. If the result is equal for both IP address A & B then we can use that subnet mask else not. Applying subnet mask to all options we find that option (d) cannot be used.
GATE-2012
5. 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 time out 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
Ans: option (c)
Explanation:
Read Slow Start Algorithm: http://www.btechonline.org/2012/12/slow-start-algorithm-in-tcp.html
Read AIMD Algorithm: http://en.wikipedia.org/wiki/Additive_increase/multiplicative_decrease
Note: Got through the above two links before you get into the solution.
Its given that initial Threshold is: 8MSS
Transmission
|
Window Size
|
Reason
|
First
|
2
|
Slow Start Phase
|
Second
|
4
|
“
|
Third
|
8
|
“
|
Fourth
|
9
|
Threshold reaches; According to AIMD increase by 1MSS
|
Fifth
|
10
|
Timeout occurs. New Threshold = 10/2 = 5MSS
|
Sixth
|
2
|
Reset to Slow Start State
|
Seventh
|
4
|
Slow Start Phase
|
Eight
|
5
|
Threshold reaches.
|
Ninth
|
6
|
According to AIMD increase by 1MSS
|
Tenth
|
7
|
“
|
Hence the answer is: 7MSS (since it transmitted 7 segments during 10th transmission)
GATE-2008
6. 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
Ans: option (d)
Explanation:
Read Slow Start Algorithm: http://www.btechonline.org/2012/12/slow-start-algorithm-in-tcp.html
GATE-2008
7. 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
Ans: option (c)
Explanation:
Convert the subnet mask into binary format.
255.255.248.0 = 11111111.11111111.11111000.00000000
Note:
Number of 1's in the subnet mask indicates the Network-ID and the Subnet-ID part
Number of 0's in the subnet mask indicates the Host-ID part
Maximum number of Hosts per subnet = 211 = 2048
where 11 = Number of 0's in the Subnet Mask.
Out of 2048 values, 2 addresses are reserved, hence we remove them (2048-2 = 2046).
Note: In the host part of the address:- all bits as 1 is reserved as broadcast address and all bits as 0 is used as network address of subnet.
Hence option (c) is the answer.
GATE-2008
8. Which of the following system calls results in the sending of SYN packets?
(a) socket (b) bind (c) listen (d) connect
Ans: option (d)
Explanation:
A typical TCP client and server application issues a sequence of TCP system calls to attain certain functions.
The socket system call creates a new socket and assigns the protocol and resources to the created socket descriptor.
The bind system call associates a local network transport address with a socket. For a client process, it is not mandatory to issue a bind call. The kernel takes care of doing an implicit binding when the client process issues the connect system call. It is often necessary for a server process to issue an explicit bind request before it can accept connections or start communication with clients.
The listen call indicates to the protocol that the server process is ready to accept any new incoming connections on the socket. There is a limit on the number of connections that can be queued up, after which any further connection requests are ignored.
The connect system call is normally called by the client process to connect to the server process.
To know more on TCP system call sequences: http://www.ibm.com/developerworks/aix/library/au-tcpsystemcalls/index.html
GATE-2008
9. 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
Ans: option (a)
Explanation:
Application layer can forward any size of data to the transport layer. Transport layer in turn divides the data into several data packets.
GATE-2012
10. The protocol data unit(PDU) for the application layer in the Internet stack is
(a) Segment (b) Datagram (c) Message (d) Frame
Ans: (c)
Note:
PDU for Application Layer - Message
PDU for Internet Layer - Datagram
PDU for Transport Layer - Segment
PDU for Link Layer - Frame
GATE-2012
11. Which of the following transport layer protocols is used to support electronic mail?
11. Which of the following transport layer protocols is used to support electronic mail?
(a) SMTP (b) IP (c) TCP (d) UDP
Ans: (c)
Explanation:
Application Layer protocol for sending email is SMTP (Simple Mail Transfer Protocol). SMTP is a connection oriented protocol, therefore it uses TCP at Transport Layer.
Note:
IP - It is a network layer protocol.
UDP - is a connection-less transport layer protocol.
GATE-2012
12. In the IPv4 addressing format, the number of networks allowed under Class C addresses is
12. In the IPv4 addressing format, the number of networks allowed under Class C addresses is
(a) 214 (b) 27 (c) 221 (d) 224
Ans: option (c)
Note:
Class
|
Leading
bits |
Number
of networks |
Number of Hosts
|
Start address
|
End address
|
Class A
|
0
|
27-2=126
|
16,777,214=224-2
|
0.0.0.0
|
126.255.255.255
|
Class B
|
10
|
214=16,384
|
65,534=216-2
|
128.0.0.0
|
191.255.255.255
|
Class C
|
110
|
2,097,152=221
|
254=28-2
|
192.0.0.0
|
223.255.255.255
|
Class D (multicast)
|
1110
|
not defined
|
not defined
|
224.0.0.0
|
239.255.255.255
|
Class E (reserved)
|
1111
|
not defined
|
not defined
|
240.0.0.0
|
255.255.255.255
|
GATE-2005
13. An organization has a class B network and wishes to form subnets for 64 departments. The subnet mask would be:
13. 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
Ans: option (d)
Explanation:
Number of 1's in the subnet mask indicates the Network-ID and the Subnet-ID part
Number of 0's in the subnet mask indicates the Host-ID part
Since Class B network, network-ID consists of 16 bits, and host-id consists of 16 bits. Since we need 64 subnetworks, we have to borrow 6 bits for subnet-ID. Therefore, subnet mask will be 255.255.252.0
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