Given an ip address 156.233.42.56 with a subnet mask of 7 bits.how many hosts and subnets are possible
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Class B network has the form N: N: H: H, the default subnet mask is 16 bits long.
There are additional 7 bits to the default subnet mask. the total number of bits in
the subnet are 16+7 = 23.
This leaves us with 32-23 = 9 bits for assigning to hosts.
7 bits of subnet mask corresponds to 2^7-2 = 128-2 = 126 subnets.
9 bits belonging to host addresses correspond to 2^9-2 = 512-2 = 510 hosts.
so, the answer will 510 hosts and 128 subnets.
There are additional 7 bits to the default subnet mask. the total number of bits in
the subnet are 16+7 = 23.
This leaves us with 32-23 = 9 bits for assigning to hosts.
7 bits of subnet mask corresponds to 2^7-2 = 128-2 = 126 subnets.
9 bits belonging to host addresses correspond to 2^9-2 = 512-2 = 510 hosts.
so, the answer will 510 hosts and 128 subnets.
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Answer:
7 bits of subnet mask corresponds to 2^7-2 = 128-2 = 126 subnets. 9 bits belonging to host addresses correspond to 2^9-2 = 512-2 = 510 hosts. so, the answer will 510 hosts and 128 subnets.
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