Posted by rocky on Saturday, November 7, 2009 at 11:09pm.
Is Programming your School Subject?
I am not an expert on networking, but I will try to give an answer to the best of my knowledge.
Since the network part is 128.171, it is a class B subnet. Using 10 bits for the subnet means that it will use a 10 bit subnet-mask, namely 255.255.255.192, or in bits,
11111111.11111111.11111111.11000000.
Since subnets with all zeroes (address of network) and all ones (broadcast address) are not allowed, the first subnet begins with
128.171.0.64
or
10000000.10101011.00000000.01------.
Thus there can be 2^{10} = 1022 possible subnets, and each subnet can have 2^{6}-2 = 62 hosts.
The first host on the first subnet is therefore:
10000000.10101011.00000000.01000001.
or
128.171.0.65
Oops:
"Thus there can be 2^{10}-2 = 1022 possible subnets...."
a) Draw the bits for the four octets of the IP address of the first host on the first subnet.
10000000.10101011.00000000.00000001
b) Convert this answer into dotted decimal notation.
128.171.0.1
c) Draw the bits for the second host on the third subnet.
10000000.10101011.00000000.10000010
d) Convert this into dotted decimal notation.
128.171.0.130
e) Draw the bits for the last host on the third subnet.
10000000.10101011.00000000.10111110
f) Convert this answer into dotted decimal notation.
128.171.0.190