Test: Introduction & IP Addressing- 1 - Computer Science Engineering (CSE) MCQ

In class A, 24 bits are reserved for Host Id and 8 bits are reserved for Network Id.

In class B, 16 bits are reserved for Host Id and 16 bits are reserved for Network Id.

In class C, 8 bits are reserved for Host Id and 24 bits are reserved for Network Id.

Concept:

In class B, the first two octets(16 bits) represent network address and out of these 16 bits first two bits are fixed(10) to specify that its class B and the remaining 14 bits represent 2¹⁴ networks, and each network has 2¹⁶ - 2 host.

So,

IPv4 address: 32 bits

Network Id = 16 bits

Since there are 600 Host ID

∴ Host ID = ⌈log₂ 600⌉= 10

IP address = Network ID + Subnet ID + Host ID

32 = 16 + Subnet ID + 10

∴ Subnet ID = 6

In subnet mask, all host id bits are 0 and network id, subnet id are 1’s

Network ID + Subnet ID = 6 + 16 = 22 bits

subnet mask: 11111111. 11111111. 11111100.00000000

subnet mask: 255. 255.252.0

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:

1) Class A

2) Class B

3) Class C

4) Class D

5) 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.

The correct answer is A - III, B - IV, C - II, D - I

10.20.30.40 (A):

• This IP address falls within the range of 1.0.0.0 to 126.0.0.0, which is designated for Class A. So, A matches with III.

210.20.30.3 (B):

• This IP address falls within the range of 192.0.1.0 to 223.255.254.0, which is designated for Class C. So, B matches with IV.

180.30.100.10 (C):

• This IP address falls within the range of 128.1.0.0 to 191.255.0.0, which is designated for Class B. So, C matches with II.

252.5.15.11 (D):

• This IP address falls within the range of 240.0.0.0 to 255.255.255.255, which is designated for Class E. So, D matches with I.

Therefore, the correct option is A - III, B - IV, C - II, D - I. So, the correct option is 4).

The correct answer is option A.
In IPv4 0.0.0.0 is used for the default route. IP address 0.0.0.0 is used by hosts when they are being booted. If the packet's destination address is unknown, and the default route is unknown, but the default route is present in the router's routing table, the packet is not discarded but forwarded to the next router.
∴ Hence the correct answer is 0.0.0.0.

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:

1) Class A

2) Class B

3) Class C

4) Class D

5) 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 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.

Note: IPv6 is a128 bits address

IPv4 address: 32 bits

IP address = Network ID + Host ID

Network ID + Host ID =32

PREFIX is the part of the Network ID

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 6015=4 is introduced

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.

The correct answer is A assumes, B is on same network, but B assumes, A is on a different network

Given Information:

• Computer A:
  • IP Address: 203.197.2.53
  • Subnet Mask: 255.255.128.0
• Computer B:
  • IP Address: 203.197.75.201
  • Subnet Mask: 255.255.192.0

Step 1: Calculate the Network Address for A

1. Convert the Subnet Mask to Binary:

• Subnet Mask A: 255.255.128.0 
• In binary: 11111111.11111111.10000000.00000000

2. Convert the IP Address to Binary:

• IP Address A: 203.197.2.53 
• In binary: 11001011.11000101.00000010.00110101

3. Perform Bitwise AND:

• Network Address A:
  

Step 2: Calculate the Network Address for B

1. Convert the Subnet Mask to Binary:

• Subnet Mask B: 255.255.192.0
• In binary: 11111111.11111111.11000000.00000000

2. Convert the IP Address to Binary:

• IP Address B: 203.197.75.201 
• In binary: 11001011.11000101.01001011.11001001

​3. Perform Bitwise AND:

• Network Address B:
  

Summary of Network Addresses:

• Network Address A: 203.197.0.0
• Network Address B: 203.197.64.0

Step 3: Determine Network Assumptions

• For Computer A (203.197.2.53):
  • It calculates its network address as 203.197.0.0.
  • It considers Computer B (203.197.75.201) as being on the same network if the IP falls within the same network address range.
• For Computer B (203.197.75.201):
  • It calculates its network address as 203.197.64.0.
  • Since 203.197.2.53 does not fall within the range of 203.197.64.0, it assumes that A is on a different network.

Conclusion

• Computer A assumes that Computer B is on the same network.
• Computer B assumes that Computer A is on a different network.

Thus, the correct option is indeed: c) A assumes B is on the same network, but B assumes A is on a different network.

PAT (Port Address Translation):

• Allows a one to many approaches to network address translation (NAT)
• Permits multiple devices on a LAN to be mapped to a single public IP address.
• Goal of PAT is to conserve IP address
• Most home networks use PAT.
• Thousands of users can be connected to the internet by using one public address.

Static NAT:

• A single private IP address is mapped with single public IP address.
• Used in web hosting.
• Private IP address is translated to a public IP address.

Dynamic NAT:

• Multiple private IP address are mapped to a pool of public IP address.
• Used when we know the number of fixed user wants to access the internet at any time.

So,
From above concepts we can say that PAT is used can be used to have one IP address allow many users to connect to the global internet.