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subnetting question

Posted on 2010-09-20
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Last Modified: 2012-05-10
Given address 202.56.64.134/27, what is the network and broadcast address?

Solution
consider only the 4th octet, where the subnet mask (27bits) boundary occurs.
step1. how many host bits? number of host bits = 5

step2. what is the starting address for the address block containing given address?
possible starting addresses in the 4th octet are: 0, 32, 64, 96, 128, 160,....

Ans: network address: 202.56.64.128 (first address)
         broadcast address: 202.56.64.159 (last address)

on the question above I can understand step1 but  I'm kind of confused with step2 where it says "possible starting addresses in the 4th octet are:  0, 32, 64, 96, 128, 160,...."

Need help with understanding step2.
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Question by:bt1942
6 Comments
 
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Expert Comment

by:duggie264
ID: 33720930
on a /27 subnet you have 256/32 subnets. therefore you have 8 subnets. each of those has a network and broadcast address (first and last) with the bits in between being the host address's.
Therefore in the example above the starting address would be 202.56.64.128/27 (the network address) host addresses .129-.158 and the broadcast address of .159

Hope this helps

Duggie
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Author Comment

by:bt1942
ID: 33721038
how did you get the number 128 at the end of the starting address?
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Expert Comment

by:jimmyray7
ID: 33721172
Because  202.56.64.134 is between  202.56.64.129 and  202.56.64.158
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Expert Comment

by:jimmyray7
ID: 33721182
If the problem had used  202.56.64.12 instead, the last octet would be 0.  Hope that helps.
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Expert Comment

by:kuoh
ID: 33721932
To be able to do it in your head, it really requires a solid understanding of binary, and a good memory doesn't hurt either.  However, the easy way to get the answers is just to use the calculator in Windows.  Switch to BIN mode, in the scientific or programmer view depending on version, enter the number of host bits (5), 00011111 in this case.  Then click on DEC and you have 31, so each block contains 32 addresses because 0 also counts.  That's where they come up with 0,32,64,96,etc..., but you want to know the beginning of the block for a random host.  This is also simple, just put the inverse 11100000 in, click DEC and you have 224, or you could also just take 255 - 31 = 224.  That's your mask to figure out the beginning of the block.  So enter the example value of 134, then click on AND, then enter the mask 224, click = and you get the answer 128.  Try 197 AND 224 and you get 192 as the start of the block.  The end of the block (broadcast) is just the start + block size, or 128 + 31 = 159.  Ok, so maybe it's not all that easy.
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Accepted Solution

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bijal7612 earned 2000 total points
ID: 33809325
u wil find the best practice on youtube for subnet.Please review that and will make u understand about subnetting and VLSM. Watch all 6 parts avaialble.
http://www.youtube.com/watch?v=UXN5XrmsaV8 
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