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I am new to subnet masking. I have a question. If network address is 10.10.10.0/23, what will be the first and last usable IP address of this network?

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2 Answers 2

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I don't believe that 10.10.10.10/23 is a network address, it would be a host address on a network; the network address would be 10.10.10.0, the subnet mask would be 255.255.254.0.

Regarding the calculations: I always convert to binary - 10.10.10.10 is

00001010 00001010 00001010 00001010

With a 23 bit subnet mask, that's

11111111 11111111 11111110 00000000

Using it as a mask

11111111 11111111 11111110 00000000
00001010 00001010 00001010 00001010

the network address is the part below the 1's, followed by zeros:

00001010 00001010 00001010 00000000

which is 10.10.10.0

The hosts are determined by the rest of the bits. Since the mask is 23 bits, there are 9 bits for the host portion of the addresses. (32-23=9)

so they range from

00000000 00000000 00000000 00000001 

to

00000000 00000000 00000001 11111111

Adding those values to the network address - the low value:

 00001010 00001010 00001010 00000000
+00000000 00000000 00000000 00000001
____________________________________
 00001010 00001010 00001010 00000001 

which is 10.10.10.1 - the high value:

 00001010 00001010 00001010 00000000
+00000000 00000000 00000001 11111111
____________________________________
 00001010 00001010 00001011 11111111 

which is 10.10.11.255 - the broadcast address.

So the network address is 10.10.10.0 and the broadcast address is 10.10.11.255 and the usable addresses are 10.10.10.1 through 10.10.11.254

its easy in binary.

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  • edited. The network address is 10.10.10.0/23 @batsplatsterson Dec 14, 2015 at 13:35
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The range of usable hosts will be 10.10.10.1 - 10.10.11.254 with a subnet of 255.255.254.0. The 10.10.10.255 and 10.10.11.0 address will be usable in this situation.

You can run these questions through a subnet calcualtor. I use Wintelguy

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