Right, im going over subnets again after doing them last year, i thought i had remembered it until the Cisco testerererer cd hit me with this: (I have put this in code tags as it allows spacing)
Could the Cisco CD have a little error, or a i doing this the wrong way?
Given the IP address 22.214.171.124 & a subnet mask of 255.255.255.224 what is the broadcast address:
So I have done my working like this:
IP = 126.96.36.199 /27
. | N | N | N | S | H |
Broadcast: 011 11111 = 188.8.131.52
But the correct answer is D: 184.108.40.206??
I have tried to work through the example given again using ANDing:
Subnet mask = 220.127.116.11
IP = 18.104.22.168
Only showing 'The intresting Octet':
Subent = 224 = 11100000
IP = 100 = 01100100
Network address = 01100000 = 96
Broadcast = 01111111 = 127
I have worked through another example from the same CD:
What is the broadcast for IP subnet 192.168.100.32, given a mask of 255.255.255.244.
So, 192.168.100.32 /27
Which, according to the CD is correct. Is the CD wrong? This is confusing me a lil.
| S | H |
224 111 00000
001 00000 - Network
001 11111 - Broadcast = 192.168.100.63
May be you are right.There will be some mistake in cisco cd.
Put your IP address and Subnet Mask.The result will be the same.
Yes, the CD is wrong. The broadcast address for 22.214.171.124/27 is 126.96.36.199... 188.8.131.52 is the network address for the following subnet.