ITPEC FE Morning October 2020 Question 33

192.168.59.255 — a /22 network spans four values of the third octet, and 57 lies in the block 56–59.

• •/22 leaves 10 host bits: the last 2 bits of the third octet and all of the fourth
• •Mask in the third octet: 252, binary 11111100
• •57 is 00111001; AND with 11111100 gives 00111000 = 56, so the network is 192.168.56.0
• •Broadcast: set all 10 host bits to 1, third octet 00111011 = 59, fourth octet 255

The network is 192.168.56.0/22, with broadcast address 192.168.59.255.

Answer (c)

Why not others:

• •(a) 192.168.55.255 — the broadcast address of the preceding /22 block, 52–55, which does not contain 57
• •(b) 192.168.57.255 — treats the prefix as /24, setting only the fourth octet to 255
• •(d) 192.168.63.255 — the broadcast address of the /21 block 56–63, one prefix bit too short

Key rule: For the broadcast address, AND the address with the mask to find the network, then set every host bit to 1; with /22 the third octet moves in steps of 4.

Networks · AI-written, may contain errors