GATE 2026 CS2 Q.33: solution and answer

GATE 2026 · CS2Q.33 · 1 mark · NAT

If an IP network uses a subnet mask of 255.255.240.0, the maximum number of IP addresses that can be assigned to network interfaces is ______. (answer in integer)

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Answer and solution

ANSWER

The maximum number of IP addresses that can be assigned to network interfaces is 4094.

Official answer: 4094Source: Official answer key, GATE 2026 CS2 (IIT Guwahati)

The mask is 255.255.240.0. In binary the third octet is $240 = 1111000 0_{2}$ , and the first two octets are all ones.
Count the 1-bits in the mask: $8 + 8 + 4 = 20$ , so the prefix length is /20.
The number of host bits is $32 - 20 = 12$ .
The network has $2^{12} = 4096$ addresses in total.
Two of these cannot be assigned to interfaces: the all-zeros host address (the network address) and the all-ones host address (the directed broadcast address).
Assignable addresses $= 2^{12} - 2 = 4094$ .

KEY IDEA

Count the 1-bits in the mask to get the prefix length, then take the host bits as 32 minus the prefix. The number of assignable addresses is $2^{h} - 2$ , because the network and broadcast addresses are reserved.

WATCH OUT FOR

Answering 4096 (all $2^{12}$ addresses), which forgets that the network address and the broadcast address cannot be given to interfaces. Another slip is reading 240 as 4 ones in the wrong octet and getting a prefix other than /20.

From the GATE 2026 CS2 paper. Every question of this paper · Download the full paper · All 20 IPv4 questions, with notes

GATE 2026 CS2 Q.33: solution and answer

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