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Help on figuring out VLSM
fourmyskull
Member Posts: 6 ■□□□□□□□□□
in CCNA & CCENT
I need to set up 2 routers with 12 host on each router, subnetting 192.168.0.0, Can someone tell me the steps to take on subnetting this out using VLSM
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Optionscaptobvious
Member Posts: 648
fourmyskull wrote: »I need to set up 2 routers with 12 host on each router, subnetting 192.168.0.0, Can someone tell me the steps to take on subnetting this out using VLSM
Start with the biggest subnets needed, 12 hosts on two subnets. Ok, this means we need to use 4 bits for hosts ie. 2^4 - 2 = 14. So our subnet mask is 255.255.255.240 or /28. So the first two subnets, assuming you can use the zero subnet, would be:
192.168.0.0/28 -> host ip range 192.168.0.1 - 192.168.0.14
192.198.0.16/28 -> host ip range 192.168.0.17 - 192.168.0.30
Now we need to connect the routers. But we only have a need for two host addresses. You would use the next available subnet from above 192.168.0.32 as to not overlap host addresses. So with only two hosts needed we can use 255.255.255.254 or /30 ie 2^2 - 2 = 2. This makes the subnet for the router connection:
192.168.0.32/30 -> host ip range 192.168.0.33 - 192.168.0.34
Hope this makes sense!
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Optionsfourmyskull
Member Posts: 6 ■□□□□□□□□□
Homework?
Yes kind of just getting ready for a practical on the final, this was an example of one that will be on the final. have to set up 2 routers on packet tracer -
Optionsfourmyskull
Member Posts: 6 ■□□□□□□□□□
captobvious wrote: »Start with the biggest subnets needed, 12 hosts on two subnets. Ok, this means we need to use 4 bits for hosts ie. 2^4 - 2 = 14. So our subnet mask is 255.255.255.240 or /28. So the first two subnets, assuming you can use the zero subnet, would be:
192.168.0.0/28 -> host ip range 192.168.0.1 - 192.168.0.14
192.198.0.16/28 -> host ip range 192.168.0.17 - 192.168.0.30
Now we need to connect the routers. But we only have a need for two host addresses. You would use the next available subnet from above 192.168.0.32 as to not overlap host addresses. So with only two hosts needed we can use 255.255.255.254 or /30 ie 2^2 - 2 = 2. This makes the subnet for the router connection:
192.168.0.32/30 -> host ip range 192.168.0.33 - 192.168.0.34
Hope this makes sense!
Thank you I will go over it and see if I can figure it out, -
Optionscaptobvious
Member Posts: 648
fourmyskull wrote: »Yes kind of just getting ready for a practical on the final, this was an example of one that will be on the final. have to set up 2 routers on packet tracer
Some final!
JK!
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blackninja
Member Posts: 385
fourmyskull wrote: »Yes kind of just getting ready for a practical on the final, this was an example of one that will be on the final. have to set up 2 routers on packet tracer
I wish they done this when I was at uni - imagine the extra drinkin' that could have been done
Currently studying:
CCIE R&S - using INE workbooks & videos
Currently reading:
Everything. Twice -
Optionscaptobvious
Member Posts: 648
Who needed an excuse?blackninja wrote: »I wish they done this when I was at uni - imagine the extra drinkin' that could have been done
happened every night. Heck, some of my best exams I didn't remember even taking! -
Optionsfourmyskull
Member Posts: 6 ■□□□□□□□□□
captobvious wrote: »Start with the biggest subnets needed, 12 hosts on two subnets. Ok, this means we need to use 4 bits for hosts ie. 2^4 - 2 = 14. So our subnet mask is 255.255.255.240 or /28. So the first two subnets, assuming you can use the zero subnet, would be:
192.168.0.0/28 -> host ip range 192.168.0.1 - 192.168.0.14
192.198.0.16/28 -> host ip range 192.168.0.17 - 192.168.0.30
Now we need to connect the routers. But we only have a need for two host addresses. You would use the next available subnet from above 192.168.0.32 as to not overlap host addresses. So with only two hosts needed we can use 255.255.255.254 or /30 ie 2^2 - 2 = 2. This makes the subnet for the router connection:
192.168.0.32/30 -> host ip range 192.168.0.33 - 192.168.0.34
Hope this makes sense!
To check my answer if I had 20 hosts what would be the range on the routers, I would be setting PPP on two routers.
Thanks again for all your help in advance!! -
Optionscaptobvious
Member Posts: 648
My suggestion would be to drink heavily!
After all, there always is next semester! -
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miller811
Member Posts: 897
fourmyskull wrote: »Does anyone have any suggestions?
here is the results from a previous post I left for someone else.. Plug your network address in and read through and hopefully this will click for you. You should be able to follow this, with your scheme and the number of networks you need to figure this out. BTW, you really need to know how to do this hands down.
Post your results when you figure it out.
The IP scheme that they gave you was a single subnet with 254 host possible.
You need to then break it up into smaller subnets by subnetting......
204.15.5.0 /24
could be broken into two segments of 128 bits by changing the mask to a /25 = 126 usable hosts per subnet
204.15.5.0 /25 204.15.5.0 - 204.15.5.127
204.15.5.128/25 204.15.128-204.15.5.255
it could also be broken into 4 subnets by starting out with a /26 mask = 62 usable host/subnet
204.15.5.0 /26 204.15.5.0-204.15.5.63
204.15.5.64/26 204.15.5.64-204.15.5.127
204.15.5.128/26 204.15.5.128 - 204.15.5.191
204.15.5.192/27 204.15.5.192-204.15.5.255
it could also be broken into 8 subnets with a /27 mask = 30 usable hosts/subnet
204.15.5.0 /27
204.15.5.32/27
204.15.5.64/27
204.15.5.96/27
204.15.5.128/27
204.15.5.160/27
204.15.5.192/27
204.15.5.224/27
it could also be broken into 16 subnets with a /28 mask = 14 usable hosts
it could also be broken into 32 subnets with a /29 mask = 6 usable hosts
it could also be broken into 64 subnets with a /30 mask = 2 usable hosts
So you take the highest number of host you need per subnet first, 2 of the networks required 28 hosts, so the first 2 subnets would be /27
204.15.5.0 /27
204.15.5.32/27
since the second subnet extends to 204.15.5.63, the starting point for the next subnet would be 204.15.5.64
The next subnet requires 14 hosts, we would use a /28 mask
so 204.15.5.64 /28 = 204.15.5.64-204.15.5-79
the next subnet requires 7 hosts, so we would need to use a /28 mask again
so we would start with the next IP
204.15.5.80 /28 = 204.15.5.80 - 204.15.5.95
the last subnet only requires 2 IP address so we would use a /30 mask
the next IP address is
204.15.5.96 /30 = 204.15.5.96 - 204.15.5.99
Hope that helpsI don't claim to be an expert, but I sure would like to become one someday.
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