Last subnet verification

[news2010a]

Hi, can you just confirm the below is right ?

For 10.0.0.0/19 ip address assignment, can you confirm my

calculation below for the last subnet is correct ?

Interesting because if I use an IP calculator, when I

get to subnet "10.255.224.0" I hit "next subnet" and then

I get 11.0.0.0. I know that may be a silly observation,

but I just want to make sure I am not overlooking anything.

First subnet Broadcast host range

10.0.0.0 (zero) 10.0.31.255 10.0.0.1 - 10.0.31.254

Last subnet Broadcast host range

10.255.224.0 10.255.255.255 10.255.224.1 - 10.255.255.254

Number of bits in the subnet identifier:11

Number of possible IP addresses in each subnet:2^11 = 2048

[sourabhagarwal]

you are having 10.0.0.0/19 subnet and I guess you are further trying to subnet this network. For this you need to tell us that what is the new subnet mask you are using to further subnet this network. hope my understanding is correct else correct me.

say, you are going to further subnet this /19 to /24 network. In this case you will have 32 /24 network each having 254 hosts. remember you can further subnet these /24s into /25 ... /30 depending on your requirement.

hope to helps ... rate if it does ...

[jolmo]

Hi Marlon

For network 10.0.0.0/19

Subnet Address is 10.0.0.0

Broadcast Address is 10.0.31.255

Host Range is 10.0.0.1 - 10.0.31.254

If you are trying to "subnet" this network with 2048 address per subnet, then you will have:

No. Subnet Address Hosts From Hosts To Broadcast Address

0 10.0.0.0 10.0.0.1 10.0.7.254 10.0.7.255

1 10.0.8.0 10.0.8.1 10.0.15.254 10.0.15.255

2 10.0.16.0 10.0.16.1 10.0.23.254 10.0.23.255

3 10.0.24.0 10.0.24.1 10.0.31.254 10.0.31.255

Hope this helps

[news2010a]

I also got confused by the question.

The book just tells this:

Given the "IP address assignment 10.0.0.0/19, what is the first and last subnet? What is the range of hosts for the first and last subnet? ".

Since thet 10.0.0.0 is a class A address, I figured that Subnet Mask bit = 11.

Subnet Mask = 255.255.224.0.

From there I found the data I posted.

If you think my interpretation from the question is wrong, please let me know.

[adrian.harte]

I don't know where you got the Subnet Mask bit=11 from... a /19 network has the subnet mask of 255.255.224.0

Your network is 10.0.0.0/19, which means each subnet has 2^13 usable addresses (32-19), minus wire/broadcast addresses. This equates to 8190 usable per /19 network.

First subnet (assuming ip subnet-zero), will be the 10.0.0.0/19, consisting of 10.0.0.0 through 10.0.31.255.

Last subnet will be 10.0.224.0 through 10.0.255.255.

The point to remember is for a /19 network, the first 19 bits cannot change. This makes the first two octets (10.0) easy to calculate, and the third octet just goes up in 32's.

[rajinikanth]

hi,

caluclations for subnet mask of 10.0.0.0 /19

is

8 bits + 8bits + 3bits =19 bits

11111111 + 11111111 + 11100000 + 00000000

255 . 255. 128+64+32 . 0

255.255.224.0

subnet mask for 10.0.0.0 /19 = 255.255.224.0

to caluclate the different subnets use this simple method

256 -224 = 32

subnet 1 = 0 31 :host range usable(1-30)

subnet 2 = 32 63

subnet 3 = 64 95

subnet 4 = 96 127

subnet 5 =128 159

subnet 6 =160 191

subnet 7 =192 223

subnet 8 =224 255

:host range usable(225-254)

Hope this helps you,

Thanks

Raj