GATE CSE Subnetting Questions: CIDR, Masks and Forwarding

Addressing questions in GATE CSE get solved like arithmetic, once you stop converting addresses to binary. A prefix length gives you a block size, a block size tells you where the boundaries fall, and everything else follows: the network address, the broadcast address, the host count, whether two machines can talk directly, and which forwarding-table entry wins. This guide gives the full method and the mistakes that cost marks.

Key takeaways

• A $/n$ prefix means $32 - n$ host bits, so the block holds $2^{32-n}$ addresses. That number is the only thing you need first.
• Blocks are aligned. A block of size $B$ starts at an address whose relevant octet is a multiple of $B$.
• The network address is the first address of the block, the directed broadcast is the last, and usable hosts are $B - 2$.
• Only the octet where the prefix falls ever needs any binary thinking, and usually not even that.
• Forwarding uses the most specific matching prefix, not the first matching row. The default route 0.0.0.0/0 wins only when nothing else matches.

What the 2027 syllabus covers

The clause that matters here is "IPv4 - Fragmentation, CIDR Notation". CIDR notation is the heading under which masks, network and broadcast addresses, host counts, subnetting into equal or unequal blocks, supernetting and forwarding-table lookup all sit. Routing appears separately as distance vector and link state. Classful addressing is not named in the 2027 text, although it is the backdrop for many older previous-year questions. The clause-by-clause comparison with 2026 is in the guide to what changed in the Computer Networks syllabus for 2027. Syllabus documents can be revised, so check the current official notification.

On weight: unofficial analyst compilations place Computer Networks at 6 to 11 marks per paper from 2009 to 2026, averaging about 8.2, with 8.3 across 2009 to 2017 and 8.0 across 2018 to 2026. The direction is flat. The recent high was 11 marks in 2022, and one analyst counted six Computer Networks questions in the 2026 second shift. Sources differ by 2 to 3 marks because borderline questions are classified differently. The subject ranking behind these numbers is in Computer Networks important topics.

Reading a prefix as a block size

Subtract the prefix from 32 and raise 2 to that power. A /26 has $32 - 26 = 6$ host bits and a block of $2^6 = 64$ addresses. A /22 has $32 - 22 = 10$ host bits and a block of $2^{10} = 1{,}024$. That single number drives everything else.

The mask is the same fact in another costume: $n$ ones followed by $32 - n$ zeros. The only octet that ever matters is the one where the prefix boundary falls, and there the mask value is $256 - \text{block size}$.

When the prefix is shorter than 24 the same table shifts one octet left. A /22 is mask 255.255.252.0 with a block of 4 in the third octet, so the network addresses are 192.168.0.0, 192.168.4.0, 192.168.8.0 and so on. A /18 is mask 255.255.192.0, block 64 in the third octet. Move the block size into the correct octet and binary conversion becomes unnecessary for almost every question in this family.

Network address, broadcast address, usable hosts

Take 192.168.10.100/26. The block size is 64, so the boundaries in the last octet are 0, 64, 128 and 192. The value 100 falls in the 64 to 127 range. Therefore:

• Network address: 192.168.10.64
• Directed broadcast: 192.168.10.127
• Usable host range: 192.168.10.65 to 192.168.10.126
• Usable hosts: $64 - 2 = 62$

The general formula is:

Two exceptions are worth carrying. A /31 holds two addresses and, on a point-to-point link, both are used as host addresses rather than as a network and broadcast pair, so it gives 2 usable. A /32 is a single host route and appears mainly inside forwarding tables. Unless a question sets up a point-to-point link explicitly, use the minus two form.

Running the other way is just as mechanical. If a department needs 300 hosts, you need at least $300 + 2 = 302$ addresses, the next power of two is $2^9 = 512$, so the prefix is $32 - 9 = 23$, a /23. Add 2 before rounding up, never after.

The same-subnet test in one step

Two hosts share a subnet if they have the same network address under the same mask, and you never need to compute both network addresses in full. Look at the octet where the prefix falls and compare the block index, which is the octet value divided by the block size, discarding the remainder:

Under a /26 mask, 192.168.10.100 and 192.168.10.130 give block indices $\lfloor 100/64 \rfloor = 1$ and $\lfloor 130/64 \rfloor = 2$, so they sit in different subnets and traffic between them must cross a router. Under a /24 mask the block size is 256 and both give index 0 ($\lfloor 100/256 \rfloor = \lfloor 130/256 \rfloor = 0$), so they share a subnet. Same addresses, different answer, purely because of the mask. Questions in this family often hand you two addresses and ask which masks are valid.

Variable-length subnet masking

When departments need different host counts, cutting a block into equal parts wastes addresses. Variable-length subnet masking cuts a block into pieces of different sizes, and the working rule is to allocate the largest requirement first. If you place a small block at the start of the space, it sits on a boundary that a later, larger block also needs, and the allocation fails alignment.

Take 192.168.20.0/24 and requirements of 60, 28, 12 and 5 hosts.

Every allocation starts on a multiple of its own block size, which is what makes it a legal CIDR block. Addresses 192.168.20.120 to 192.168.20.255 stay free. A common form of this question offers four allocations and asks which is valid, and the invalid ones usually fail on alignment rather than on the host arithmetic.

Supernetting and aggregation

Supernetting is the reverse operation: several blocks collapse into one advertised prefix. Three conditions must hold together. The blocks must be contiguous, their count must be a power of two, and the first block's address must be a multiple of the combined block size.

192.168.8.0/24, 192.168.9.0/24, 192.168.10.0/24 and 192.168.11.0/24 aggregate to 192.168.8.0/22. Four blocks, contiguous, and 8 is a multiple of 4.

Shift the same four blocks by one and they do not aggregate. 192.168.9.0/24 through 192.168.12.0/24 are still four contiguous /24s, but 9 is not a multiple of 4, so no single /22 covers exactly those four and nothing else. The alignment condition is the part candidates skip, and it is exactly what the question is testing.

Longest prefix match in a forwarding table

A router does not read its table top to bottom and stop at the first hit. It collects every entry whose prefix matches the destination and forwards using the longest one, because that is the most specific route. Row order carries no meaning.

The two prefixes that need care are /18, whose block is 64 in the third octet, covering 10.12.64.0 to 10.12.127.255, and /21, whose block is 8 in the third octet, covering 10.12.72.0 to 10.12.79.255.

• 10.12.73.5 matches /8, /16, /18 and /21. Longest is /21, so the next hop is R4.
• 10.12.80.1 falls outside the 72 to 79 range, so the /21 does not match. Longest remaining is /18, giving R3.
• 10.12.200.9 falls outside 64 to 127, so /18 drops out. Longest remaining is /16, giving R2.
• 10.30.1.1 matches only /8 and the default, giving R1.
• 192.168.4.9 matches nothing but 0.0.0.0/0, giving R0.

That last line is the role of the default route. A /0 prefix matches every destination with the lowest possible specificity, so it is chosen only when the table offers nothing else.

The traps in one list

• Forgetting to subtract the network and broadcast addresses, and answering 64 instead of 62.
• Mis-locating the block boundary when the prefix is not a multiple of 8. The block size belongs in the octet where the prefix falls, and only there.
• Assuming the forwarding table is scanned top to bottom. The most specific match wins regardless of order.
• Allocating the smallest block first in a variable-length split and breaking the alignment of a later, larger block.
• Aggregating blocks that are contiguous but not aligned, or whose count is not a power of two.
• Adding 2 to the host requirement after rounding to a power of two instead of before it.
• Treating the default route as a fallback the router reaches last in the list rather than as the shortest prefix.

The cross-subject version of this list is in common mistakes that cost marks in GATE CSE, and the mask, block-size and host-count relations are condensed in the GATE CSE formula sheet.

How to practise

Build one worksheet and reuse it. Take a single /24 from a private range, write every prefix from /25 to /30 with its mask, block size and host count, and check that you can produce any column from any other in under ten seconds. Then pick random addresses inside that block and write their network and broadcast addresses without touching binary.

After that, drill the forwarding table. Write five overlapping prefixes with a default route, place the most specific entry in the middle of the list, and answer six destinations. When that stops needing thought, the two-mark version in the paper takes under ninety seconds. The Computer Networks chapter of the GATE CSE 2027 book extends this method with a graded question bank on fragmentation, address translation and forwarding, each worked step by step.