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IP addressing

Posted on 2010-09-15
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Last Modified: 2013-01-20
I have very little knowledge of IP addressing and I would appreciate some help here

I have an assigned range of 75.137.45.144/28. How many machines can I connect to it and can I make it into two equal sized subnets? How?  Is there a Math formula to figure this stuff out??
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Question by:dawber39
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Expert Comment

by:mccarl
ID: 33687455
You can have up to 14 devices in this range. For your info the subnet mask would be 255.255.255.240

If you want two subnets, then they could have only 6 devices each. They would have ranges of 75.137.45.144/29 and 75.137.45.152/29 and the subnet mask for these two subnets would be 255.255.255.248
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Qlemo earned 500 total points
ID: 33687474
with a /28 you have 32-28 = 4 bits = 16 addresses left for your hosts. You need to subtract two for the reserved addresses of all-zero and all-one bits, so in fact you can use 14 addresses. If you subnet that, you loose 2 addresses per subnet for the same reason. Your subnets would then be
75.137.45.144/29 with .145 - .150
75.137.45..152/29 with .153 - .158
that are 6 addresses per subnet = 12 overall.

The /28 part tells us how much bits from the left to the right are assigned to the network. Each part of the address is a single byte = 8 bit, 4 parts = 32 bits, and /28 is 32-4, which means 4 bits are left for the hosts.
We can only add bits on the highest part of the host addresses for implementing subnets, so the next bit to use for a subnet is bit 29, which has a value of 8 (2 power 3).That is why the next subnet starts at 152 (144+8).
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by:rfc1180
ID: 33687544
>I have an assigned range of 75.137.45.144/28. How many machines can I connect to it and can I make it into two equal sized subnets?

This depends on whether or not you are assigning this range to an interface; if all you are doing is defining static one to one NAT, then you can use all 16 addresses; if you are however, assigning this to a physical interface on a firewall/router, you will then need 1 IP address for the physical interface then you will have 13 usable addresses for your hosts on this LAN.

Billy
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Author Comment

by:dawber39
ID: 33687976
I appreciate the efforts of all of you - but Qlemo gave me a better explanation of how - thank you all
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Author Closing Comment

by:dawber39
ID: 33687981
Excellent
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by:mccarl
ID: 33687997
If you "appreciate the efforts of all" you should really be sharing the points... but whatever!
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by:rfc1180
ID: 33688030
too bad you can't block authors!
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Author Comment

by:dawber39
ID: 33688221
Wow - all I did was ask how - and Qlemo was quite definitive in his explanation - I am sorry if I hurt your feelings
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