CIDR notation (/n), block size, alignment & route aggregation

1. What is the prefix length (number of 1s) of the subnet mask 255.255.248.0?
   Stuck? Peek at 3.1.3 Subnet masks, subnetting, network/broadcast & usable hosts

   • Borrow s bits from host field, leaving h bits → 2ˢ subnets, each with 2ʰ − 2 usable hosts.
   • Subnet address = IP AND Mask; Broadcast = host bits all 1s.
   • Stride Δ = 256 − Mask octet: blocks start at 0, Δ, 2Δ, … and end at Δ−1, 2Δ−1, ….
   • Valid mask octets: 128, 192, 224, 240, 248, 252, 254, 255.

   Read lesson 3.1.3

2. Why did Class B allocations exhaust the IPv4 address space so quickly?
   Stuck? Peek at 3.1.2 Classful addressing (A–E) and special IPv4 addresses

   • Classes by 1st octet: A = 0–127 (0, 8 net bits), B = 128–191 (10, 16 net bits), C = 192–223 (110, 24 net bits), D = 224–239 (1110, multicast), E = 240–255 (1111, reserved).
   • Usable hosts = 2ʰ − 2 (exclude host all-0s = network address, host all-1s = directed broadcast).
   • Specials: 0.0.0.0 = this host (src only); 255.255.255.255 = limited broadcast (local LAN only); 127.x.y.z = loopback.

   Read lesson 3.1.2

3. What is log₂(2048)?
   Stuck? Peek at 0.2.4 Exponents, logarithms and basic probability

   • 2ᵃ·2ᵇ = 2ᵃ⁺ᵇ; 2¹⁰ ≈ 10³ (estimates only).
   • log₂ x = power of 2 giving x; bits for N values = ⌈log₂ N⌉.
   • P(none) = (1−p)ⁿ · P(≥ 1) = 1 − (1−p)ⁿ · P(exactly 1) = n·p·(1−p)ⁿ⁻¹.
   • (1 − 1/n)ⁿ → 1/e ≈ 0.368; 1/(2e) ≈ 0.184.
   • 1 + 2 + … + 2ᵏ⁻¹ = 2ᵏ − 1; expected tries = 1/p.

   Read lesson 0.2.4

By 1993 the Internet faced two disasters at once: Class B blocks were nearly exhausted, and core routers were drowning in tens of thousands of individual Class C entries. The rescue was Classless Inter-Domain Routing (CIDR).

CIDR abolishes Classes A, B, and C completely. The network boundary can sit at any bit position from 0 to 32, written simply as a slash notation /n at the end of the address — and contiguous blocks can be glued together into a single router entry!

Can you form a valid 1024-address CIDR block (/22, which spans four 256-address octets) starting at 192.168.2.0 (covering .2.0 through .5.255)?

No! In a /22 block, the last 10 bits are host bits, and in the network address all 10 host bits must be 0. That means the 3rd octet must end in 00₂ — so it must be a multiple of 4 (.0.0, .4.0, .8.0, .12.0, …). Starting at .2.0 violates the alignment rule!

Slash Notation (/n) & Block Size

In CIDR, an address block is written as a.b.c.d/n, where prefix length n is the number of leftmost network bits (1s in the mask), and h = 32 − n is the number of host bits.

Supernetting & Route Aggregation (Summarisation)

Just as subnetting splits one block into smaller blocks by increasing n, supernetting (route aggregation) combines 2ᵏ contiguous, aligned blocks of size /n into a single block of size /(n − k).

Finding the smallest enclosing supernet in binary

When asked for the single CIDR prefix that covers a list of addresses or blocks, write the varying octet in binary from the lowest address to the highest address and count how many leading bits stay identical:

Putting it together

• CIDR replaces fixed classes with arbitrary prefix lengths /n, giving block size 2³²⁻ⁿ.
• Every CIDR block must start at an address whose rightmost 32−n bits are zero (the alignment rule).
• Route aggregation merges 2ᵏ aligned contiguous /n blocks into a single /(n−k) prefix, shrinking router tables.

• 4 contiguous blocks ≠ always 1 supernet! Always check that the starting octet is divisible by the number of /24s being merged (e.g., .1, .2, .3, .4 cannot form a single /22).
• "Exact aggregation" vs "Smallest single covering prefix": If a question asks for ONE prefix that covers .1.0/24 through .4.0/24 (even if extra addresses are included), find the first bit where 00000001₂ (1) and 00000100₂ (4) differ (the 4s bit) → 200.1.0.0/21.
• Don't forget: a /n block still loses 2 addresses (first = network, last = directed broadcast) if used as a single LAN, so usable hosts = 2³²⁻ⁿ − 2.

Size 2ʰ must start at a multiple of 2ʰ — you can't park a 4-slot bus starting at slot #1!

With CIDR and route aggregation, a router might have both a broad summary route like 200.1.0.0/16 AND a specific exception route like 200.1.2.0/24 in its table. Next question (3.1.5): When a packet matches BOTH entries, which one does the router choose?

1. NAT

   How many total IPv4 addresses are contained in a /19 CIDR block?

   Write your answer before opening the key.

2. MCQ

   Which of the following is a VALID CIDR block (satisfies the alignment rule)?

   1. 192.168.6.0/22
   2. 192.168.8.0/21
   3. 192.168.4.0/21
   4. 192.168.18.0/20
3. MCQ

   Four contiguous blocks 198.16.0.0/24, 198.16.1.0/24, 198.16.2.0/24, and 198.16.3.0/24 are aggregated into a single CIDR block. What is the aggregated prefix?

   1. 198.16.0.0/20
   2. 198.16.0.0/21
   3. 198.16.0.0/22
   4. 198.16.0.0/26
4. NAT

   What is the minimum number of CIDR prefixes needed to represent the exact set of addresses in the three blocks 10.0.1.0/24, 10.0.2.0/24, and 10.0.3.0/24?

   Write your answer before opening the key.

5. MSQ

   Which of the following statements about the CIDR block 172.16.32.0/20 are TRUE?

   1. Its subnet mask is 255.255.240.0.
   2. Its last address is 172.16.47.255.
   3. It contains 4096 total addresses.
   4. It contains the address 172.16.48.1.

• CIDR a.b.c.d/n: n prefix bits, h = 32 − n host bits, block size N = 2ʰ.
• Alignment rule: first address must be divisible by 2ʰ (for /(24−k), 3rd octet must be divisible by 2ᵏ).
• Supernetting: 2ᵏ contiguous, aligned /n blocks merge into one /(n − k) block.

• CIDR
• slash notation /n
• prefix
• prefix length
• block size 2³²⁻ⁿ
• alignment rule
• supernetting
• route aggregation (summarisation)

1. What are the 3 rules for a valid CIDR block /n?
   Reveal answer

   1. Contiguous addresses; 2. Size is 2³²⁻ⁿ; 3. Starting address is divisible by 2³²⁻ⁿ (last 32−n bits are 0).
2. When can 2ᵏ blocks of /n be aggregated into a single /(n−k) block?
   Reveal answer

   When they are contiguous AND the first block starts at an address aligned to /(n−k).
3. How many /24 blocks are inside a /20 block, and what is the 3rd-octet stride?
   Reveal answer

   2²⁴⁻²⁰ = 16 blocks of /24; 3rd-octet stride is 16.

• 3rd-octet alignment check for /16../24: For prefix /(24 − k), the 3rd octet must be a multiple of 2ᵏ (and 4th octet = 0). — checking CIDR alignment in 3 seconds on GATE.
• Pairwise merge rule: Two blocks X/n and Y/n merge into X/(n−1) iff X is an even multiple of 2³²⁻ⁿ and Y is the very next multiple. — simplifying a list of prefixes into the minimum number of routing entries.