Subnetting help!!!!

Hi everybody.

Just have few questions about Subnetting. Iam Using the shortcut method(magic no method) described in the Cisco press ICND book. However Iam stuck and cannot figure out how to solve this problem. Here goes...

1) Given the IP address 140.1.1.1 and the mask 255.255.255.248, what are the assignable
IP addresses in this subnet?

I cant make out which octet in the subnet mask is the so called "interesting" octet. The IP address given is a class B address, so is the interesting octet the 3rd octet? but the book at times says that the interesting octet is the octet with non 0's or 1's( as in other than 0 or 255)?

Another example to make myself clear---

2) IP 167.88.99.66 and the mask 255.255.255.192 or

even

3) 10.5.118.3 and the mask 255.255.255.0

In the last example even though there is no so called "difficult" octets, which octet do we use to find the "magic no". Although in the example the resultting magic no will be same i.e 1, but the end result would certainly be different.

Can someone pls help me, if iam not clear pls let me know.

Thanks

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Comments

EdTheLad

hshah12 wrote:

1) Given the IP address 140.1.1.1 and the mask 255.255.255.248, what are the assignable
IP addresses in this subnet?

hosts 140.1.1.1-> 140.1.1.6

When your subneting forget about the class of address,just look at the subnet mask.

hshah12 wrote:

Another example to make myself clear---

2) IP 167.88.99.66 and the mask 255.255.255.192
or even
3) 10.5.118.3 and the mask 255.255.255.0

167.88.99.65 -> 167.88.99.126
10.5.118.1 -> 10.5.118.254

hshah12

Sorry i still cannot understand what octet to consider an "interesting" octet when subnetting has been done on more than 8 bits.

I knew the answers, I just want to know how to solve the prob using the magic no method?

Thanks for your help

pannupandit

usable ip addersses will be

140.1.1.1 - 140.1.1.6

beoadcast address will be 140.1.1.7

and subnet is 140.1.1.0

there is no need to look for class of address.just look for the subnet mask

here the subnet mask is 255.255.255.248 so /29 bits will be used for the subnet mask.
i will explain more.
given ip address is 140.1.1.1 and subnet is 255.255.255.248
convert it into binary we get
ip- 10001100.00000001.00000001.00000001
subnet- 11111111.11111111.11111111.11111000
Anding- 10001100.00000001.00000001.00000000

anding will give the Required subnet address-144.1.1.0
subnet mask is telling out of given ip address,the last 3 bits are for host part.so last bits range from 000 t0 111 which is from 0 to 7.1st address is subnet address and last is broadcast address. and inbetween are usable addresses.