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...

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.

Ok. let's keep "my understanding" away from this quiz. Just give me the answer and i could try to understand.

Read the link I gave. It's a basic tutorial on subnetting. Is this coursework for a college course by any chance?

/22

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 ?

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....

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?

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...

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 ?

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 - 200.180.31.255Yep you have it.

How to properly post a question:

Sorry no tech support via PM's

So Do you agree total number of IP addresses available in all four subnets are 4088 ?

1022x4=4088 How to properly post a question:

Sorry no tech support via PM's

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