Another request for help...thanks for all so far.

OK, doing more subnetting practice questions: Here's the latest that has me a little stumped...they are from the www.************.net IP Subnetting Question Generator...

The question states: Assign server last valid host address on 4th subnet of 192.168.24.0 /28 network. The answer provided is 192.168.24.62.

I come up w/ the answer of 192.168.24.46...

0, 16, 32, 48 (4th subnet) .47 would be broadcast, 192.168.24.46 would be last valid host of 4th sn, right?

Wouldn't 62 be the last valid host of the 5th subnet?

Thanks for checking over my work...thanks for all the help often given...

Regards,

Me

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Comments

Nunya

I figured it out....48 is the 4th subnet..63 is the broadcast...62 is the last valid host of 4th subnet...thanks for making me figure it out my mistake on my own...

cpartin

.0 - .15 1st subnet
.16 - .31 2nd subnet
.32 - .47 3rd subnet
.48 - .63 4th subnet

andy4tech

Nunya,
its easy,you just have to use ip subnet zero,that means you start numbering from zero...........................................................................................

...0 16 32 48 64...................................
...1 17 33 49 65

...14 30 46 58 78
...15 31 47 59 79................................

Up above we have increment of 16 because /28 is 255.255.255.240,if we deduct 240 from the magic number 256 we get 16 which is the increment above.Looking at the arrangement counting from the 1st subnet which is 0,the fourth subnet will be 192.168.24.48 and the broadcast address for this subnet will be 192.168.24.63,the first available host on this subnet will be 192.168.24.49 and the last one will be 192.168.24.62.

mattlee09

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