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IP Addressing (IPv4)

Short Notes

IPv4 addresses are 32-bit unique identifiers for network interfaces.

Classful Addressing

  • Class A: 0... (1-126). 8-bit Network ID.
  • Class B: 10... (128-191). 16-bit Network ID.
  • Class C: 110... (192-223). 24-bit Network ID.
  • Class D: 1110... (Multicast).
  • Class E: 1111... (Experimental).

Key Theories & Formulas

1. CIDR (Classless Inter-Domain Routing)

Represented as IP/Prefix.

  • Number of addresses in /n block = \(2^{32-n}\).
  • Netmask: \(n\) bits of 1 followed by \((32-n)\) bits of 0.

2. Subnetting

Borrowing bits from Host ID to create Subnet ID.

  • Number of subnets = \(2^{\text{borrowed bits}}\).
  • Usable hosts per subnet = \(2^{\text{remaining bits}} - 2\). (Exclude Network ID and Directed Broadcast Address).

Example Problems

Problem: How many hosts in a 192.168.1.0/24 network?

  • \(32 - 24 = 8\) bits for hosts.
  • \(2^8 - 2 = 254\) hosts.

Hardest GATE Questions

Topic: Variable Length Subnet Masking (VLSM) and Aggregation Tricky Question (GATE 2014/2015/2019): An ISP has block 10.1.1.0/24. It needs to divide this for 3 companies needing 100, 50, and 50 addresses respectively. Give the prefixes.

  • Analysis:
  • 100 needs 7 bits (\(2^7=128\)). Prefix: /25. Range: 10.1.1.0 - 10.1.1.127.
  • 50 needs 6 bits (\(2^6=64\)). Prefix: /26. Range: 10.1.1.128 - 10.1.1.191.
  • 50 needs 6 bits (\(2^6=64\)). Prefix: /26. Range: 10.1.1.192 - 10.1.1.255.
  • The "Trap": Route Aggregation (Supernetting).
  • Given four /26 blocks, can they be summarized into one /24? (Only if they are contiguous and start at a /24 boundary).
  • Hard Aspect: Fragment size calculation in IPv4.
  • Offset = (Bytes from start) / 8.
  • Data must be a multiple of 8 (except last fragment).
  • Complexity: Private IP ranges (10.x, 172.16-31.x, 192.168.x) and Loopback (127.0.0.1)

References