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Zorluk: ZorIPv4 Addressing and Subnetting

A network administrator is provisioning IPv4 subnets from the block 172.16.96.0/21172.16.96.0/21 to support server clusters requiring at least 200200 usable host IP addresses per subnet. To maximize the number of subnets created while meeting this host requirement, which subnet mask must be configured, and what is the broadcast address of the third allocated subnet?

  1. Subnet mask 255.255.255.0255.255.255.0 and broadcast address 172.16.98.255172.16.98.255Cevap
  2. B
    Subnet mask 255.255.255.128255.255.255.128 and broadcast address 172.16.97.127172.16.97.127
  3. C
    Subnet mask 255.255.255.0255.255.255.0 and broadcast address 172.16.98.254172.16.98.254
  4. D
    Subnet mask 255.255.255.0255.255.255.0 and broadcast address 172.16.99.255172.16.99.255

Cevap

Subnet mask 255.255.255.0255.255.255.0 and broadcast address 172.16.98.255172.16.98.255
To support 200200 usable hosts, 88 host bits are required because 282=2542^8 - 2 = 254 usable IP addresses (77 host bits only provide 126126). Subtracting 88 host bits from 3232 yields a /24/24 prefix (255.255.255.0255.255.255.0). Starting from 172.16.96.0/21172.16.96.0/21, the subnets sequence as: Subnet 1 (172.16.96.0/24172.16.96.0/24), Subnet 2 (172.16.97.0/24172.16.97.0/24), and Subnet 3 (172.16.98.0/24172.16.98.0/24). The broadcast address of the third subnet is the highest IP in its range, 172.16.98.255172.16.98.255.

Adım Adım Çözüm

1
Determine the minimum host bits needed
8 host bits (282=2542002^8 - 2 = 254 \ge 200 usable hosts)
7 host bits only yield 126126 usable hosts (272=1262^7 - 2 = 126), which is insufficient for 200 hosts.
2
Calculate the CIDR prefix and dotted-decimal subnet mask
Prefix /24/24 (328=2432 - 8 = 24), subnet mask 255.255.255.0255.255.255.0
Subnetting a /21/21 block into /24/24 subnets maximizes the subnet count while guaranteeing at least 200200 usable hosts per segment.
3
Identify the network boundaries of the subnets
1st subnet: 172.16.96.0/24172.16.96.0/24, 2nd subnet: 172.16.97.0/24172.16.97.0/24, 3rd subnet: 172.16.98.0/24172.16.98.0/24
Each /24/24 subnet increments the third octet by 11.
4
Determine the broadcast address of the 3rd subnet
Broadcast address is 172.16.98.255172.16.98.255
The broadcast address is the last IP address within the 172.16.98.0/24172.16.98.0/24 network range.

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