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TCP/IP and Subnetting

Tarrant64

Diamond Member
I'm trying to understand a couple things about subnetting at the moment. I'm taking a network comm. class and we have 1 week dedicated to TCP/IP. It's crucial I get most(if not all) understanding of it, as once I'm done with this class I have one whole class dedicated to TCP/IP. Anyways, to the problem.

I've been given an IP address, 202.16.105.27 with a subnet mask of 255.255.255.248. I'm asked questions as to what I know about the class, default subnet, etc. about it.

This is what I have so far. Class C, default subnet of 255.255.255.0(for class C).
Now, what has thrown me off is 202.16.105.27 /29.

I ran across it in my reading but only for a paragraph or so. I wanted to say that it supports 30 sub networks, and supports 2,046 hosts.

I was thinking the network address and broadcast address would be easy too, just 202.16.105.0 and the broadcast address is 202.16.105.255

Can someone help me out?
 
The /29 is shorthand for 255.255.255.248 (29 consecutive bits flipped on)

The network is 202.16.105.24, broadcast is 202.16.105.31 and there are only 6 addresses for hosts.

Sounds like you need to learn more about subnetting in binary before you move forward, if you do this will make a lot more sense.

Erik
 
Thanks a lot for the link. When I did go back through and concentrated on the binary aspect of subnetting it made more sense. "/29" wasn't covered in the reading, but it was online in a few places but it was never really explained to be 'shorthand'. After going through the binary, it was a "duh" moment.

11111111 11111111 11111111 11111000 - >29 bits, or 255.255.255.248, right?

The subnet is 3 bits long, so 2^N gives me 8. Subtract 2 and we get 6, the number of hosts.

Using AND I got 24, which is as you say the network ID, and 31 is as you say the broadcast address. Makes a lot more sense now. The website linked had a few practice examples which helps too. It was mainly the AND that wasn't covered in my book and a couple of websites didn't have it either, but that one did the trick!

Thanks.
 
yup...and instead of using powers, just subtract the last subnet octet from 256 to get the total number of hosts.

256 - 252 = 4
256 - 248 = 8
256 - 240 - 16
256 - 224 = 32
256 - 192 = 64
256 - 128 = 128
 
Originally posted by: jlazzaro
yup...and instead of using powers, just subtract the last subnet octet from 256 to get the total number of hosts.

256 - 252 = 3
256 - 248 = 8
256 - 240 - 16
256 - 224 = 32
256 - 192 = 64
256 - 128 = 128

😕
 
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