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An organization is granted a block   172.89.0.0/16 . The administrator wants to create 1024 subnets using 10 bits. The first and last address of any host in subnet 1024 respectively are –

A. 172.89.255.0/26 and 172.89.255.255/26

​​​​​​​B. 172.89.255.1/26 and 172.89.255.245/26

​​​​​​​C. 172.89.0.1/26 and 164.76.255.255/26

​​​​​​​D. 172.89.0.1/26 and 172.89.255.245/26
asked in Computer Networks by Junior (977 points) | 58 views
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Is it d ?
0
yes , explain
0
The question mentions first and last address of any subnet, in option d the first address is the address for first subnet (subnet #1) and the second address is the last address for last subnet (subnet #1024) . at first you might be tempted to go with option b , but there the number of hosts are 245, which is not possible for (32-26=6) host Id bits.

* but a thing I don't understand is, the second part ,according to my calculation should be 172.89.255.254/26, because subnet #1024 net Id would be 172.89.255.192/26.

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