IPv4 Header, Packet Fragmentation & CIDR Subnetting
“Header fields, MTU fragmentation offsets, and VLSM network bit masks.”
1.IPv4 Header Fields: Crucial Exam Dimensions
§ 5.1.The IPv4 header has a variable length: 20 bytes (minimum, when Options field is empty) up to 60 bytes (maximum, when Options field is 40 bytes).
Key Header Fields:
1. HLEN (Header Length, 4 bits): Expressed in 4-byte (32-bit) words. Minimum value is 5 (5 4 = 20 bytes). Maximum value is 15 (15 4 = 60 bytes).
2. Total Length (16 bits): Entire IP datagram size (Header + Data) in bytes. Maximum theoretical datagram size = 2^16 - 1 = 65,535 bytes.
3. TTL (Time to Live, 8 bits): Hop counter decremented by 1 at each router. When TTL reaches 0, router drops the packet and sends an ICMP Time Exceeded (Type 11) message back to the sender.
4. Protocol (8 bits): Multiplexing identifier for higher layer: 1 = ICMP, 2 = IGMP, 6 = TCP, 17 = UDP, 89 = OSPF.
5. Fragmentation Fields: Identification (16 bits), Flags (3 bits: Reserved, DF, MF), Fragment Offset (13 bits).
Because the Fragment Offset field is only 13 bits (range 0 to 8191) while the datagram can be 65,535 bytes long, the offset is measured in units of 8-BYTE BLOCKS! Every fragment data payload MUST be an exact multiple of 8 bytes (except the final fragment).
2. GATE Step-by-Step Numerical: IP Datagram Fragmentation
Remember to subtract the 20-byte IP header from the link MTU to determine maximum payload per fragment, and ensure payload is divisible by 8!
Problem: An IPv4 datagram of Total Length = 4000 bytes (with 20-byte IP header) arrives at a router whose output interface has an MTU (Maximum Transmission Unit) of 1500 bytes. Calculate the number of fragments, their sizes, DF/MF flags, and fragment offsets.
Step 1: Total Data Payload = 4000 - 20 = 3980 bytes.
Step 2: Maximum payload per fragment = MTU - Header = 1500 - 20 = 1480 bytes. Since 1480 / 8 = 185 (exact integer), 1480 is valid.
Step 3: Fragment generation:
Fragment 1: Payload = 1480 bytes (bytes 0 to 1479). Total Length = 1500 bytes. MF = 1. Offset = 0 / 8 = 0.
Fragment 2: Payload = 1480 bytes (bytes 1480 to 2959). Total Length = 1500 bytes. MF = 1. Offset = 1480 / 8 = 185.
Fragment 3: Remaining Payload = 3980 - (1480 + 1480) = 1020 bytes (bytes 2960 to 3979). Total Length = 1020 + 20 = 1040 bytes. MF = 0 (last fragment!). Offset = 2960 / 8 = 370.
Summary: 3 fragments with Offsets [0, 185, 370] and MF flags [1, 1, 0].
3.CIDR (Classless Inter-Domain Routing) & Subnet Mathematics
§ 5.3.Classless addressing uses CIDR notation: a.b.c.d / n, where n is the prefix length (number of contiguous 1s in the subnet mask, 0 <= n <= 32).
Formulas for /n Block:
1. Subnet Mask: n bits of 1s followed by (32 - n) bits of 0s.
2. Host bits h = 32 - n.
3. Total IP addresses in block = 2^h = 2^(32 - n).
4. Usable Host addresses = 2^h - 2 (subtracting Network Address where host bits are all 0s, and Direct Broadcast Address where host bits are all 1s).
5. Network Address: IP address bitwise-AND Subnet Mask.
6. Broadcast Address: Network Address with all h host bits set to 1.
4. GATE Worked Numerical: Subnet Mask & Usable IP Range
Always check which octet the prefix split falls into. For /26, 26 = 8 + 8 + 8 + 2, so the boundary is in the 4th octet with 2 subnet bits and 6 host bits.
Problem: An ISP assigns host IP 198.51.100.142 / 27. Find the Subnet Mask, Network ID, Broadcast ID, and the range of usable host addresses.
Step 1: Prefix /27 means 27 ones: 11111111.11111111.11111111.11100000 -> Subnet Mask = 255.255.255.224.
Step 2: Host bits h = 32 - 27 = 5 bits. Block size = 2^5 = 32 addresses.
Step 3: Analyze 4th octet (142 in binary = 10001110):
142 / 32 = 4 with remainder 14.
Block begins at 4 * 32 = 128 (10000000).
Network Address = 198.51.100.128.
Step 4: Block ends at 128 + 32 - 1 = 159 (10011111).
Direct Broadcast Address = 198.51.100.159.
Step 5: Usable Host IP Range = 198.51.100.129 to 198.51.100.158 (Total = 32 - 2 = 30 usable hosts).
Primary Academic References & Exam Sources
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GATE-level numerical drilling and PYQ analysis.
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