IPv4 - Free MCQ Practice Test with solutions, GATE CSE (CSE) Computer Networks

Concept:
CIDR stands for Classless Inter Domain Routing. CIDR denotes the number of bits in the Netid.  Upon subnetting, any IP address consist of three subparts: Netid (bits for identifying the network), subnetid (bits for subnetting), and Hostid (bits for address allocation to hosts)

Calculation:
Given CIDR is 17, that means Netid=17.
There are 1500 devices, so the number of bits in Hostid= 11 (because 211 ≥ 1500)
Therefore, subnet bits= 32- (17+11) = 4
Hence, in any IP address in the given network, first 17 bits be part of network, next 4 of subnet ans last 11 bits would be for host address. Host bits should be all 0s

Address I: Cannot be allotted
202.61.84.0/21= 202.61. 0 1010 100.00000000. This address is not possible because the all the 11 host bits should be zero, but it contains a 1.

Address II: Can be allotted
202.61.104.0/21= 202.61. 0 1101 000.00000000. This address is possible as all the host bits are 0 and network bits are also satisfied.

Address III: Can be allotted
202.61.64.0/21= 202.61.0 1000 000.0000000. This address is possible as all the host bits are 0 and network bits are also satisfied

Address IV: Cannot be allotted
202.61.144.0/21= 202.61. 1 0010 000. 00000000. This address is not possible because bit in Netid has become 1.

Concept
Pre-IPv6 is also known as IPv4
Pre-IPv6 address consists of the Network part and Host part

Explanation
IPV4 size consists of 32 bits
<---------------32 bit------------>

Network ID + Host ID = 32 bits
Calculation
8 bit =  1 byte
× by 4 on both sides
32 bits = 4 bytes
The length of an IP address in bytes (pre-IPv6) is 4.

There is a total of 32 bits in the IPv4 address space.
For example, if a network has the address “192.0. 2.0/24”, the number “24” refers to how many bits are contained in the network. From this, the number of bits left for address space can be calculated.
Hence the correct answer is 32.

The correct option is (a)
172.16.112.0
Concept:
Subnetting is the process of creating smaller networks called subnets by taking bits from the HOST portion of an IP address.
Explanation:
Given IP IS 172.16.112.1/25 ⇒ NID (Network ID) is 25 bits.
We know that a subnet mask is obtained by replacing all NID(Net ID) bits with 1 and HID(Host ID) bits with 0.
So SM is 255.255.255.10000000 (the First 25 bits are NID and the last 7 bits are HID).
i.e, SM is 255.255.255.128.
Now perform Bitwise AND operation between subnet address and given IP address.
⇒ (255.255.255.128) and (172.16.112.1) = 172.16.112.0

Concept:
Private IP addresses:
Private IP addresses are ones that only function on the local network. On the Internet, these addresses are not routable. The network router simply assigns the address to your specific device. Every device on the same network is assigned a unique private IP address.
Private IP address ranges are

• 192.168.0.0 - 192.168.255.255 (65,536 IP addresses)
• 172.16.0.0 - 172.31.255.255 (1,048,576 IP addresses)
• 10.0.0.0 - 10.255.255.255 (16,777,216 IP addresses)

Explanation:

• 172.30.255.255  is a valid private IP Address.
• 10.210.220.254 is a valid private IP Address.
• 172.16.21.240 is a valid private IP Address.
• 172.32.250.254 is not a valid private IP Address.

​Hence the correct answer is 3.

Concept:

• Private IP address:
• A private IP address range is a collection of non-internet-facing IP addresses that are used in an internal network.
• Network equipment, such as routers, uses network address translation to give private IP addresses.
• IP addresses are used to identify devices on the internet or a local network.
• IP addresses also allow information to be transferred between network devices.
• Private IP address ranges:
  • 192.168.0.0 - 192.168.255.255 (65,536 IP addresses)
  • 172.16.0.0 - 172.31.255.255 (1,048,576 IP addresses)
  • 10.0.0.0 - 10.255.255.255 (16,777,216 IP addresses)

Hence option A and option B and option C are valid private IP Addresses.
Hence the correct answer is option A, option B, and option C.

IP address:
IP address is an address having information about how to reach a specific host. The 32 bit IP address is divided into five sub classes. These are:

• Class A
• Class B
• Class C
• Class D
• Class E

Class D and E are reserved for multicast and experimental purposes.

IPv4 address:

An IPv4 address is a 32 bits address, which is categorized into different IP classes.

IPv4 addresses are divided into two parts: a) Network ID b) Host ID

• Class A: This class uses 8 bits for network ID part and 24 bit for host ID.
• Class B: This class uses 16 bits for network ID and 16 bits for host ID.
• Class C: This class uses 24 bits for network ID and 8 bits for host ID.
  

Note:

• IPv6 is a128 bits address

Concept:
Minimum header size of IPv4 = 20 byte
Maximum header size of IPv4 = 60 byte
Header length field in the IPv4 header is 4 bits.
Maximum possible value (1111) = 15
The scaling factor of 60/15 = 4 is introduced

Explanation:
IP header: 

 

But the first 8 bits in the Question are 0100 0010.
The first 4 bits indicate the version which is IPv4 (0100)
Now the next four bit indicates Header length (bits decimal value × 4) which is 2 × 4 = 8 bytes
which is not possible since the minimum header length for IPv4 is 20 bytes.
Hence Receiver will reject the packet.
Therefore option d is the correct answer.

Concept:
a.b.c.d/NID
Therefore 23 bits are in NID and only 9 bits for Host id.
Explanation:
Last two-byte of IP address is represented in binary
Ethernet: 172.16.2.1/23
Range of IP address: 172.16.00000010.0000000 to 172.16.0000011.11111111
Range of IP address: 172.16.2.0 to 172.16.3.255
First and last IP address is reserved
Valid Host IP address: 172.16.2.1 to 172.16.3.254
Statement 1: 172.16.1.100
It is not valid since it does not fall in the range
Statement 2: 172.16.1.198
It is not valid since it does not fall in the range
Statement 3: 172.16.2.255
It is valid since it does fall in the valid range
Statement 4:172.16.3.0
It is valid since it does fall in the valid range
Hence option 3 is the correct answer.
Important Point:
NID stands for network ID.

IP address is an address having information about how to reach a specific host. The 32 bit IP address is divided into five sub classes. These are:

• Class A
• Class B
• Class C
• Class D
• Class E

Class D and E are reserved for multicast and experimental purposes.
IPv4 addresses are divided into two parts: a) Network ID b) Host ID

• Class A: This class uses 24 bits for network ID part and 8 bit for host ID.
• Class B: This class uses 16 bits for network ID and 16 bits for host ID.
• Class C: This class uses 8 bits for network ID and 16 bits for host ID.