ITPEC FE Morning April 2019 Question 34

30 — the mask is /27, which leaves 5 host bits: 32 addresses, 2 of them reserved.

• •Last octet of the mask: 224 = 11100000, so 3 more network bits (27 in all) and 5 host bits
• •Addresses per subnet: 2⁵ = 32
• •Host addresses: 32 − 2 = 30, since the all-0s host part is the network address and the all-1s host part is the broadcast address

Answer (d)

Why not others:

• •(a) 14 — the count for mask 255.255.255.240 (/28, 4 host bits): 2⁴ − 2 = 14; it takes one host bit too few
• •(b) 20 and (c) 26 — neither has the form 2ⁿ − 2, so no subnet mask gives these host counts

Key rule: Count the 0 bits n in the mask; the number of assignable host addresses is 2ⁿ − 2.

Networks · AI-written, may contain errors