Q3IP ServicesMultiple answers
Refer to the exhibit. A network associate configured OSPF with the command: `City(config-router)# network 192.168.12.64 0.0.0.63 area 0` After finishing the configuration, the associate finds that not all interfaces are participating in OSPF. Based on this configuration statement, which three interfaces shown in the exhibit will participate in OSPF? **(Choose three.)**
Select 3 answers.
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Discussion · 19
50
OSPF will match IP addresses using 192.168.12.64 0.0.0.63.
11000000.10101000.00001100.01000000 => 192.168.12.64
00000000.00000000.00000000.00111111 => 0.0.0.63
Matches will be made on the IP only for the 1's not 0's above. We can flip the bits to make it more familiar as a network mask. This becomes:
11000000.10101000.00001100.01000000 => 192.168.12.64
11111111.11111111.11111111.11000000 => /26 or 255.255.255.192
This therefore gives a match of IPs in the network 192.168.12.64 (the next network is 192.168.12.128) so broadcast is 192.168.12.127 and usable IPs are .65 to 126.
We now match IPs in this range which are:
FastEthernet0/1 (192.168.12.65) - ANSWER B
Serial0/0 (192.168.12.121) - ANSWER C
Serial0/1/102 (192.168.12.125) - ANSWER D
If you are having problems understanding this one the key to write out 0.0.0.63 in binary and then invert the bits.
14
Use the magic number technique. Wildcard mask minus 255 so 255 minus 63 = 192.
192 gives a block size of 64 (256-192)
network 0 -63
network 64 -127
network 128 - 192
6
It has already been explained. The network 192.168.12.64 0.0.0.63 covers IP addresses ranging from .64 to .127 (it's using a /25 subnet mask).
So in this case, that's what you're looking for.
6
To identify what a wildcard mask actually represents you have to subtract the amount from 255 so in this case 255-63=192. So the subnet mask is actually 255.255.255.192 which is a /26.
Use my subnet calculator below to help you master subnetting
1st OCT /1 /2 /3 /4 /5 /6 /7 /8
2nd OCT /9 /10 /11 /12 /13 /14 /15 /16
3rd OCT /17 /18 /19 /20 /21 /22 /23 /24
4th OCT /25 /26 /27 /28 /29 /30 /31 /32
HOSTS 128 64 32 16 8 4 2 1
SUBNET 1 2 4 8 16 32 64 128
PREFIX 128 192 224 240 248 252 254 255
3
Network 192.168.12.64 0.0.0.63 covers the addresses from 192.168.12.64 to 127
192.168.12.64 - Network
192.168.12.65 to 126 - Gives 62 Usable host addresses
192.168.12.127 - Broadcast
Only Int F0/1, S0/0 and S0/1.102 fall inside this host range
S0/1.103 is in the next network
3
I see that nobody knows the explanation
3
It's /26 actually
3
sorry. typo. i meant /26
3
Given answer is correct.
I converted the wildcard bits to subnet mask and I used the fist subnet 192.168.12.65 to 192.168.12.126 , any Ip address between this interval are valid
2
why network 192.168.12.64 0.0.0.63 equals to network 192.168.12.64/26? can anyone help me explain please
2
excellent explanation, chr, that's the way
2
@Carter_Milk why use the right hand to reach the left ear?
It's simpler to add 63 (from wildcard mask) to 64 (the ip of network) and it will give you 127 (the upper limit of the network). Results the range x.x.x.64-127. Regards,
1
Can anyone explain this for me please? Thanks
1
I got it thanks.
1
So net id’$ Are 0/64/128 pick which ever fall in .64-127.
1
Why Serial 1.102 and not Serial 1.103?
192.168.12.128 is the next network, isn't it?
1
But why 192.168.12.64 0.0.0.63 --> IP address ranging .64 to .127 (why /25 subnet)
1
becuse it's a wildcard mask, which is the inverse of the subnet mask which in this case is 255.255.255.192
1
Thank you