Computer Networks And Internet Protocol - Week 6 Assignment 6 | NPTEL | JAN 2023

Computer Networks And Internet Protocol - Week 6 | Assignment 6 : Answer with Explanation

What is the maximum number of hosts under class B addresses?

a. 254
b. 65534
c. 65535
d. 65536

Answer

b. 65534
Class B Networks (/16 Prefixes) : This network is a 16-bit network prefix; its highest bit order is set to 1-0. It is a 14-bit network number with a 16-bit host number. This class B defines 16,384 (2 to the power 14) /16 networks, and supports a maximum of 65,534 (2 to the power 16 -2) hosts per network. Class B /16 block address is (1,073,741,824) = 2 to the power 30; therefore it represent 25% of the total IPV4.

What is the maximum number of networks possible in class C IP addresses?

a. 2^14
b. 2^24
c. 2^21
d. 2^22

Answer

c. 2^21
This class defines a maximum of 2,097,152 (2 ^21 ) /24 networks. And each network supports up to 254 (2^8 -2) hosts.

Suppose computers P and Q have IP addresses 10.20.5.25 and 10.20.5.9, respectively, and they both use the same netmask N. Which of the values of N given below should not be used if P and Q should belong to the same network?

a. 255.255.255.128
b. 255.255.255.192
c. 255.255.255.224
d. 255.255.255.240

Answer

d. 255.255.255.240
To determine which netmask should not be used, we need to find out the network address of P and Q using each of the netmasks provided and then check whether they belong to the same network or not.

The network address of an IP address can be obtained by performing a bitwise AND operation between the IP address and the netmask.

Using the given IP addresses of P and Q, and the first netmask, we get:

P: 10.20.5.25 = 00001010.00010100.00000101.00011001
Q: 10.20.5.9 = 00001010.00010100.00000101.00001001
Netmask: 255.255.255.128 = 11111111.11111111.11111111.10000000

Performing a bitwise AND operation between each IP address and the netmask, we get:
P network address: 00001010.00010100.00000101.00000000 = 10.20.5.0
Q network address: 00001010.00010100.00000101.00000000 = 10.20.5.0
Since both P and Q have the same network address using the first netmask, they belong to the same network. We repeat this process for the remaining netmasks:

Using the second netmask, we get:
Netmask: 255.255.255.192 = 11111111.11111111.11111111.11000000
P network address: 00001010.00010100.00000101.00000000 = 10.20.5.0
Q network address: 00001010.00010100.00000101.00000000 = 10.20.5.0

Again, both P and Q have the same network address using the second netmask, so they belong to the same network.

Using the third netmask, we get:
Netmask: 255.255.255.224 = 11111111.11111111.11111111.11100000
P network address: 00001010.00010100.00000101.00000000 = 10.20.5.0
Q network address: 00001010.00010100.00000101.00000000 = 10.20.5.0
Once again, both P and Q have the same network address using the third netmask, so they belong to the same network.
Using the fourth netmask, we get:
Netmask: 255.255.255.240 = 11111111.11111111.11111111.11110000
P network address: 00001010.00010100.00000101.00010000 = 10.20.5.16
Q network address: 00001010.00010100.00000101.00001000 = 10.20.5.8
Using the fourth netmask, P and Q have different network addresses, so they do not belong to the same network.
Therefore, the netmask that should not be used if P and Q should belong to the same network is 255.255.255.240.

What is supernetting?

a. Split a large network into multiple small networks.
b. Combine multiple small networks in a single large network.
c. Enable the network to accept more hosts.
d. Connect a new network to an existing network.

Answer

b. Combine multiple small networks in a single large network.

Which of the following is/are not a valid IPv6 address?

a. AE82::1:800:23E7:F5DB
b. FC80:2:7:1:800:23E7:A:F5DB
c. DE62:6A42:1:5AC::800:23E7:F5DB
d. FE80:2030:31:24

Answer

d. FE80:2030:31:24

Which is true for class E?

a. It is used for multicasting
b. It is reserved for experimental purposes
c. It is used for a network with a large number of hosts
d. It is used for a network with less number of host

Answer

b. It is reserved for experimental purposes

What is the possible ip address range for class B?

a. from 0.0.0.0 to 127.255.255.255
b. from 192.0.0.0 to 223.255.255.255
c. from 128.0.0.0 to 191.255.255.255
d. from 240.0.0.0 to 247.255.255.255

Answer

c. from 128.0.0.0 to 191.255.255.255

You need 500 subnets, each with about 100 usable host addresses per subnet. What network mask will you assign using a class B network address?

a. 255.255.255.252
b. 255.255.255.128
c. 255.255.255.0
d. 255.255.254.0

Answer

b. 255.255.255.128

The header length of the ipv6 datagram is

a. 10 bytes
b. 20 bytes
c. 30 bytes
d. 40 bytes

Answer

d. 40 bytes

We write the IP address as 191.180.83.235/12 in CIDR notation. What is the subnet mask?

a. 255.240.0.0
b. 255.255.255.0
c. 255.255.240.0
d. 255.0.0.0

Answer

a. 255.240.0.0
To determine the subnet mask from an IP address in CIDR notation, we need to determine the number of bits in the network portion of the address.
The IP address 191.180.83.235/12 indicates that the first 12 bits of the address represent the network portion, and the remaining 20 bits represent the host portion.
To find the subnet mask, we need to set all the bits in the network portion to 1 and all the bits in the host portion to 0.

Starting with the leftmost bit of the network portion, we have: 11111111 11110000 00000000 00000000
Converting this to dotted decimal notation gives:
255.240.0.0
Therefore, the subnet mask for the IP address 191.180.83.235/12 is (a) 255.240.0.0.

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Computer Networks And Internet Protocol, Week 6 : Assignment 6,noc23_cs48