tricky vlsm question

DANMOH009

I got the following practice question whilst revising and found it extremely a hard, wanted to know if there were any quik ways of doing this.


Network Address = 87.6.0.0
Mask/Prefix = 15


Subnet A
8690

Subnet B
37134


Subnet C
18183


Subnet D
8122


Subnet E
2

Now i was suppose to work out the vlsm network. i started with the highest which is subnet b and i know subnet b needs 65,536 hosts which will give it a prefix of /16 so the answer for subnet b will be 87.6.0.0/16

now this is the tricky part, is there any easy way of me finding out where the next network starts????

i know i have to count up 65,536 hosts, but is there a quick way to do it??


Thanks

Dan

Mutata

The way I usually do this (now I could be wrong because i'm just working through this again myself) is.

87.6.0.0 /16 = 87.6.0.0 255.255.0.0

Last addy (Broadcast) is 87.6.255.255. After that would be the start of your next subnet.

That would make it 87.7.0.0

Anyone with more experience is more than welcome to chime in here on faster ways, I definitely am no expert.

r_durant

I'm by no means an expert either, but I think Mutata is correct...

The next subnet would start at 87.7.0.0, as you would have used up the entire 87.6.0.0 subnet (87.6.0.0 - 87.6.255.255).

oli356

Extreme numbers much? Considering Cisco recommends subnets of <500 IRC. Going to read about VLSM tomorrow

fadhil

another way
/16 means 255.255.0.0
find the block size
block size = 256-x where x is the value(number) present on an octet which was subnetted.
256-255 =1 block size is 1
then subnets
86.6.0.0/16
86.7.0.0/16
86.8.0.0/16
........

r_durant

oli356 wrote: »

Extreme numbers much? Considering Cisco recommends subnets of <500 IRC. Going to read about VLSM tomorrow

Extreme numbers indeed, especially when Subnet B is wasting 28400 IPs (65534- 37134)

DANMOH009

Thanks guys,

bit late for me now but gonna put this into practice tommoroqw.

cheers

Mutata

@fadhil yea we did that as well in school, its also a good way to do it