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withhelds
@singnet.com.sg

withhelds

Anon

Subnet class B address

I have the following question which I'm not sure if my calculation is correct or not because the usual question I did has indication on how many HOST it support.

Question A company is granted the block of Class B address 130.55.0.0. The network admin wants to create 50 subnets.

My Answers

a) Design an suitable subnet address scheme.

50 subnets requires 6 bits (2^6 * -2) = 62 subnets

| 1 0 (2 bits) | Network ID (14 bits) | Subnet ID (6 bits) | Host ID (10 bits)

b) What is the suitable subnet mask?

255.255.252.0 (252 is based on binary 11111100 to decimal)

c) What is the number of hosts in each subnet? Do not include network and broadcast address (all 0s & 1s)

2^10 - 2 = 1022 hosts

d) Calculate the range of subnet addresses? Do not include all 0s & 1s subnet addresses

First address: 130.57.4.0
...
Last address: 130.57.248.0

e) What is the first (network address) & the last address (broadcast address) in the last subnet?

I'm not too sure if it is:
First network address = 130.57.248.0
Last broadcast address = 130.57.248.255 ?
HELLFIRE
MVM
join:2009-11-25

HELLFIRE

MVM

Agree with you except on answers d and e

d) how did you get .57 as your 2nd octet if you were given 130.55.0.0 as the address block?

e1) how did you get .57 as your 2nd octet if you were given 130.55.0.0 as the address block?

e2) i get .252.0 and .255.255 myself.

Regards

withhelds
@220.255.4.x

withhelds

Anon

my mistake on d & e on the 2nd octet, it should be: 55 instead of 57.
therefore

d)
first address: 130.55.4.0
...
last/63th address 130.55.248.0

e)
First network address = 130.55.248.0
Last broadcast address = 130.55.248.255

Can you explain why your answer for e is 130.55.252.0 & 130.55.255.255 instead of 130.55.248.0 & 130.55.248.255?

I have worked out based on:
i) the last subnet address from answer d is 130.55.248.0. I did not use 130.55.252.0 because it mentioned to exclude all 0s and 1s subnet address
ii) from my understanding, first network address is ends with 0 and last broadcast address ends with 255.
iii) therefore, I select the last subnet address from question d and change the 3rd octet to 248 and 4th octet to 0 and 255 respectively.

Hope you can enlighten me on this.
Thanks.
HELLFIRE
MVM
join:2009-11-25

HELLFIRE to withhelds

MVM

to withhelds
My answer to e was done INDEPENDENT of the requirements of d's "Do not include all 0s & 1s subnet addresses."
In which case :

130.55. _ _ _ _ _ _ | _ _ . _ _ _ _ _ _ _ _
130.55. 1 1 1 1 1 1 | 0 0 . 0 0 0 0 0 0 0 0 (last subnet)
130.55. 1 1 1 1 1 1 | 0 0 . 0 0 0 0 0 0 0 0 (last subnet, network address) == 130.55.252.0
133.55. 1 1 1 1 1 1 | 1 1 . 1 1 1 1 1 1 1 1 (last subnet, broadcast address) == 130.55.255.255
 

If you were to do it for last subnet, but under the condition "do not include all 0s & 1s subnet addresses,"
I get the following result :

130.55. _ _ _ _ _ _ | _ _ . _ _ _ _ _ _ _ _
130.55. 1 1 1 1 1 0 | 0 0 . 0 0 0 0 0 0 0 0 (last subnet, do not include all 1's)
130.55. 1 1 1 1 1 0 | 0 0 . 0 0 0 0 0 0 0 0 (last subnet, network address) == 130.55.248.0
130.55. 1 1 1 1 1 0 | 1 1 . 1 1 1 1 1 1 1 1 (last subnet, broadcast address) == 130.55.251.255
 

Rememeber that the | is the divisor between the NETWORK and HOST part of the address -- so you've two more host
bits to use in the 3rd octet that you forgot about. Put another way, anything to the right of the | CAN change,
anything to the left cannot.

Either case, I'd clarify with your instructor about question e. This is why I hate questions that are very
imprecise as to a) what preconditions exist, and b) what they are looking for.

Regards
HELLFIRE

HELLFIRE to withhelds

MVM

to withhelds
I should add that there are many subnetting resources around the internet, and many ways to do subnetting.
I'd take some time to go around to learn the different methods, and see which one works best for you.

Regards

withhelds
@singnet.com.sg

withhelds

Anon

Thanks for the clear explanation once again.