A network engineer is configuring a default gateway for a VoIP phone subnet that has been … — CompTIA Network+

A network engineer is configuring a default gateway for a VoIP phone subnet that has been allocated the IPv4 address block $172.20.48.0/20$ . Which of the following IP addresses represents the last usable host address available for assignment within this subnet?

$172.20.63.254$ Cevap
B
$172.20.63.255$
C
$172.20.64.254$
D
$172.20.55.254$

Cevap

172.20.63.254

/20

prefix (

255.255.240.0

) provides a block size of 16 in the third octet. For the subnet starting at

172.20.48.0

, the full block extends up to

172.20.63.255

. Subtracting the network address (

172.20.48.0

) and the broadcast address (

172.20.63.255

) leaves usable host addresses ranging from

172.20.48.1

172.20.63.254

. Thus, the address ending in

.63.254

is the last usable host IP.

Adım Adım Çözüm

Determine the subnet mask and block size for a

/20

prefix.

/20

prefix corresponds to a subnet mask of

255.255.240.0

. The block size in the third octet is

256 - 240 = 16

The prefix length determines how many bits are allocated to the network portion.

Calculate the network address boundaries.

Starting at

172.20.48.0

, adding the third-octet block size of 16 yields the next subnet address, which is

172.20.64.0

Finding the next network boundary defines the upper limit of the current subnet.

Determine the broadcast address and last usable host IP.

The broadcast address is one position below the next network ID (

172.20.63.255

). The last usable host address is one below the broadcast address (

172.20.63.254

The network address (first) and broadcast address (last) are reserved and cannot be assigned to hosts.

Anahtar Kavram

IPv4 Subnet Boundaries and Host Address Allocation

Soru

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Konular

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