# 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 - 200.180.16.0/22B- 200.180.20.0/22C- 200.180.24.0/22D- 200.180.28.0/22 Is maximum usable host is 2^12 -2=4094 or is it 4096-8 = 4088I 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 range200.180.16.0 - 200.180.31.255which has4096 subnets 4094 max addresses If you were to divide this into 4 subnets the max subnet for each would be 1024200.180.16.0 - 200.180.19.255200.180.20.0 - 200.180.23.255200.180.24.0 - 200.180.27.255200.180.28.0 - 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=4088Answers 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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