Find out the SUbnetwork Id, Broadcast Address and Maximum Number of host range for the following?

1. xxx.xx.xx.5/25
2. xxx.xx.xx.33/27
3. xxx.xx.xx.65/26
4. xxx.xx.xx.17/30

1. xxx.xx.xx.5/25
No. of subnets : 2
No. of valid host per subnets : 126
N/w ID : xxx.xx.xx.0
Broadcast ID : xxx.xx.xx.255

2. xxx.xx.xx.33/27
No. of subnets : 8
No. of valid host per subnets : 30
N/w ID : xxx.xx.xx.0
Broadcast ID : xxx.xx.xx.255

3. xxx.xx.xx.65/26
No. of subnets : 4
No. of valid host per subnets : 62
N/w ID : xxx.xx.xx.0
Broadcast ID : xxx.xx.xx.255

4. xxx.xx.xx.17/30
No. of subnets : 64
No. of valid host per subnets : 2
N/w ID : xxx.xx.xx.0
Broadcast ID : xxx.xx.xx.255

Cheers,
Sharme

@above answer...


for the first ip given ...
all the details about no of hosts and the no of subnets is correct
the network ID is ok!!
but then the Broadcast address is not correct the following ip lies in the first subnet hence the broadcast address must be the broadcast address for the first subnet... ie .. xxx.xx.xx.127(the last ipaddress in the first subnet)
the ip given in the question is xxx.xx.xx.5 which lies in the first subnet xxx.xx.xx.0-xxx.xx.xx.127

similarly for the rest of the ip the networkID and the Broadcast ID needs to be checked... all the rest of the other details provided by the person above answering the question is correct!!

for the second ip given...
the Network ID is xxx.xx.xx.32
and the Broadcast address is the xxx.xx.xx.63
because the ip address given lies in the 2nd subnet whose range lies from xxx.xx.xx.32-xxx.xx.xx.63

the correct procedure for finding the network addressing is to mask the subnet with the ipaddress bits to obtain the network address.. this applies to the CIDR concept also for the question above ... in case u want details for the topic ... post a reply and i would give a detailed xplaination!!!

in the third case the network ID is xxx.xx.xx.64
the broadcast ID is xxx.xx.xx.127
the ip lies in the 2nd subnet!!

in the last case the network ID is xxx.xx.xx.16
Broadcast ID is -xxx.xx.xx.20
the Ip given in the question lies in the 4rth subnet!!!

1. xxx.xx.xx.5/25
2. xxx.xx.xx.33/27
3. xxx.xx.xx.65/26
4. xxx.xx.xx.17/30
..how u people r calculating subnet mask..network ID Broadcast ID

IS THESE ANSWERS R TRUE...
1. xxx.xx.xx.5/25
No. of subnets : 2
No. of valid host per subnets : 126
N/w ID : xxx.xx.xx.0
Broadcast ID : xxx.xx.xx.255

2. xxx.xx.xx.33/27
No. of subnets : 8
No. of valid host per subnets : 30
N/w ID : xxx.xx.xx.0
Broadcast ID : xxx.xx.xx.255

3. xxx.xx.xx.65/26
No. of subnets : 4
No. of valid host per subnets : 62
N/w ID : xxx.xx.xx.0
Broadcast ID : xxx.xx.xx.255

4. xxx.xx.xx.17/30
No. of subnets : 64
No. of valid host per subnets : 2
N/w ID : xxx.xx.xx.0
Broadcast ID : xxx.xx.xx.255

Actually, the bcast is 19 here, not 20.

Regards,

The easiest way for me to find these if you know the number of hosts a subnet has (Very Simple). On the left hand side of your paper, write the 0 and each line thereafter increment by (4 in this case). Since you already know that there are only two usable hosts for this mask, you simply write in the two hosts, and the host after that will be the broadcast address. Continue this for each line. See below.

net hosts bcast
0 1-2 3
4 5-6 7
8 9-10 11
12 13-14 15
16 17-18 19
20 21-22 23
24 25-26

Good Questions.

The simple logic is in network address all the host bit must be 0 and in broadcast address all the host bit will be 1....find required number of subnet...and first address of your subnet will be network addresss and last ip address will be broadcast address...
Answer:

1. xxx.xx.xx.5/25
No. of subnets : 2
No. of valid host per subnets : 126
Sub N/w ID : xxx.xx.xx.0
sub Broadcast ID : xxx.xx.xx.127

Here ip lies in first subnet...so first valid ip of this subnet will be subnetwork address and last will be broadcast address...

2. xxx.xx.xx.33/27
No. of subnets : 8
No. of valid host per subnets : 30
sub N/w ID : xxx.xx.xx.32
sub Broadcast ID : xxx.xx.xx.63



Here ip lies in second subnet...so first valid ip of this subnet will be subnetwork address and last will be broadcast address...

3:3. xxx.xx.xx.65/26
No. of subnets : 4
No. of valid host per subnets : 62
sub N/w ID : xxx.xx.xx.64
sub Broadcast ID : xxx.xx.xx.127

Here ip lies in second subnet...so first valid ip of this subnet will be subnetwork address and last will be broadcast address...

4. xxx.xx.xx.17/30
No. of subnets : 64
No. of valid host per subnets : 2
sub N/w ID : xxx.xx.xx.16
sub Broadcast ID : xxx.xx.xx.19


Here ip lies in fifth subnet...so first valid ip of this subnet will be subnetwork address and last will be broadcast address...