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If an Ethernet port on a router were assigned an IP address of 172.16.112.1/25, what would be the valid subnet address of this host?

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If an Ethernet port on a router were assigned an IP address of 172.16.112.1/25, what would be the valid subnet address of this host?

(a) 172.16.112.0

(b) 172.16.0.0

(c) 172.16.96.0

(d) 172.16.255.0

The question was asked in examination.

Question is from Designing Subnet Masks in chapter TCP/IP Protocol Suite of Computer Network

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Correct option is (a) 172.16.112.0

For explanation I would say: A /25 mask is 255.255.255.128. Used with a Class B network, the third and fourth octets are used for subnetting with a total of 9 subnet bits, 8 bits in the third octet and 1 bit in the fourth octet. Since there is only 1 bit in the fourth octet, the bit is either off or on-which is a value of 0 or 128. The host in the question is in the 0 subnet, which has a broadcast address of 127 since 128 is the next subne

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