How to subent superneted IP ?

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January 11, 2011 at 13:00:06
Specs: Windows XP, 3.199 GHz / 1526 MB
My question is, if i have a block of IP as 200.180.16.0/20 and asked to divide into 4 subnets with equal hosts how should i do it ? I do understand subnetting clearly. This is bit tricky...

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#1
January 11, 2011 at 13:22:18
I don't think that you do understand subnetting clearly. Have a look at
the tutorial here: http://www.ralphb.net/IPSubnet/. What you are asking is just basic subnetting.

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#2
January 11, 2011 at 13:33:31
Ok. let's keep "my understanding" away from this quiz. Just give me the answer and i could try to understand.

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#3
January 11, 2011 at 13:44:13
Read the link I gave. It's a basic tutorial on subnetting.

Is this coursework for a college course by any chance?


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Related Solutions

#4
January 11, 2011 at 13:51:58
/22
     

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#5
January 11, 2011 at 13:57:33
I have read enough. This was an exam quiz.Come on man... I'm not going to get any marks :)
whenever I ask something in forum either some one will come and post a link or talk about something else like you did. I'll come to my answer now. i did like, Since four subents required i thought i should borrow 4bits and my answer was 200.180.16.0/22. Then I m confused of first host and last host of each subnet. Now would some1 explain this ?

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#6
January 11, 2011 at 14:03:22
Thanks fmwap. But most of the other students are saying that /20 will remain and the host IP will be divided or something similar to that. I couldn't think how we could do that....

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#7
January 11, 2011 at 14:05:59
Well, I'm sorry. I was trying to help your understanding rather than just give you an answer so that you could score highly in a test without understanding what you were doing.

It's not very fair on the other students if you just want to cheat, is it?


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#8
January 11, 2011 at 14:10:09
Thanks for your understanding. But agin off topic ;). I believe i got the answer correct. Answer would be,
A - 200.180.16.0/22
B- 200.180.20.0/22
C- 200.180.24.0/22
D- 200.180.28.0/22
Is maximum usable host is 2^12 -2=4094 or is it 4096-8 = 4088
I believe i made a mistake only in host number...

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#9
January 11, 2011 at 15:10:53
How come the range 200.180.160.0 - 200.180.175.255 ???
I'm bit confused now :( R u trying to say the question is wrong ?

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#10
January 11, 2011 at 15:14:30
Sorry had 160 instead of 16. Here is the corrected version.

the criteria was;
" 200.180.16.0/20 and asked to divide into 4 subnets with equal hosts"
this results in the following address range
200.180.16.0 - 200.180.31.255
which has
4096 subnets 4094 max addresses
If you were to divide this into 4 subnets the max subnet for each would be 1024
200.180.16.0 - 200.180.19.255
200.180.20.0 - 200.180.23.255
200.180.24.0 - 200.180.27.255
200.180.28.0 - 200.180.31.255

Yep you have it.

Answers are only as good as the information you provide.
How to properly post a question:
Sorry no tech support via PM's


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#11
January 11, 2011 at 15:23:45
So Do you agree total number of IP addresses available in all four subnets are 4088 ?

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#12
January 11, 2011 at 15:37:58
1022x4=4088

Answers are only as good as the information you provide.
How to properly post a question:
Sorry no tech support via PM's


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