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Difficulty: EasyIPv4 Addressing and Subnetting

A network technician is configuring an IPv4 interface on a server assigned a network prefix of /28/28. What is the corresponding subnet mask in dotted-decimal notation?

Answer: 255.255.255.240

Answer

The subnet mask in dotted-decimal notation for a /28/28 prefix is 255.255.255.240255.255.255.240.
A /28/28 CIDR prefix specifies that 2828 bits are allocated for the network portion of the address. The first three octets account for 2424 bits (255.255.255255.255.255). The remaining 44 network bits in the fourth octet (11110000211110000_2) sum to 128+64+32+16=240128 + 64 + 32 + 16 = 240, giving the dotted-decimal subnet mask 255.255.255.240255.255.255.240.

Step-by-Step Solution

1
Determine the distribution of network bits across the four 8-bit octets.
A /28/28 prefix allocates 2828 network bits out of 3232 total bits, leaving 44 bits for host addressing (32−28=432 - 28 = 4).
An IPv4 address consists of four 8-bit octets (3232 bits total).
2
Write the subnet mask in binary form.
The binary representation is 11111111.11111111.11111111.1111000011111111.11111111.11111111.11110000.
The first three octets are completely filled (8×3=248 \times 3 = 24 bits), and the fourth octet contains 44 network bits.
3
Convert each binary octet to decimal notation.
The fourth octet binary value 11110000211110000_2 equals 128+64+32+16=240128 + 64 + 32 + 16 = 240, resulting in 255.255.255.240255.255.255.240.
Converting binary bit positions (128,64,32,16128, 64, 32, 16) to decimal yields the final subnet mask.

Key Concept

CIDR prefix notation to dotted-decimal subnet mask conversion

Alternative Method

Calculate the host block size by using 2host bits=24=162^{\text{host bits}} = 2^4 = 16. Subtract the block size from 256256 in the interesting octet: 256−16=240256 - 16 = 240. Thus, the subnet mask is 255.255.255.240255.255.255.240.
Estimated Time:45s
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Topics

IPv4 Addressing and Subnetting

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