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Help... how do I figure this sub problem.
SurferdudeHB
Member Posts: 199 ■■■□□□□□□□
in CCNA & CCENT
I got the subnetting basics down but can't put it together to solve this, thank you.
Network ID: 213 214 150 0
Subnets Required: 13
What is the subnet mask?
1st Available Host
Address of Subnet 1?
Max # of hosts/subnet?
Network ID: 213 214 150 0
Subnets Required: 13
What is the subnet mask?
1st Available Host
Address of Subnet 1?
Max # of hosts/subnet?
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miller811
Member Posts: 897
The IP scheme that they gave you was a single subnet with 254 hosts possible. It is a Class C address which is a /24 or 255.255.255.0
Class C Range 192 - 223 in the first octet
it could be broken into two segments of 128 bits by changing the mask to a /25 = 126 usable hosts per subnet
it could also be broken into 4 subnets of 64 bits by changing the mask to a /26 mask = 62 usable host/subnet
it could also be broken into 8 subnets of 32 bits with a /27 mask = 30 usable hosts/subnet
it could also be broken into 16 subnets of 16 bits with a /28 mask = 14 usable hosts
it could also be broken into 32 subnets of 8 bits with a /29 mask = 6 usable hosts
it could also be broken into 64 subnets of 4 bits with a /30 mask = 2 usable hosts
That should be enough to get you started....I don't claim to be an expert, but I sure would like to become one someday.
Quest for 11K pages read in 2011
Page Count total to date - 1283 -
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phoeneous
Member Posts: 2,333 ■■■■■■■□□□
Start by asking yourself:
"How many binary bits make up 13?" -
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Haruna Umar Adoga
Member Posts: 24 ■□□□□□□□□□
The given network (213.214.150.0) should be subnetted to meet the requirements of 13 hosts right ?
First you have to know that the above address is a class c address which also means that it has a subnet mask of 255.255.255.0.
Then you can use the block size method of solving the problem (VLSM)
with your powers of 2 block sizes,
i.e 128 64 32 16 8 4 2 1 = 255 (for all ones)
from the above block sizes, its obvious that to meet the requirements of 13 hosts, you will need to use a block size of 16 which means that the first 4 bits will be ON (borrowed or added to your given subnet mask which is /24 in this case)
Therefore, using a block size of 16 gives us a new subnet mask of /28 i.e 255.255.255.240 (adding 4 bits to the given mask of /24) which will meet the above requirements of 13 hosts
since 4 bits were borrowed (added), it implies that :
number of subnets = 2^4 = 16
number of valid hosts/subnet = 2^4 - 2 = 16-2 = 14 hosts per subnet.
your subnets are :
(213.214.150.0 to 213.214.150.150) i.e first subnet
(213.214.150.16 to 213.214.150.31) 2nd subnet
(213.214.150.32 to 213.214.150.47) 3rd subnet
. . . and so on.
which means that the first valid host on the first subnet is :
213.214.150.1 with a broadcast address of 213.214.150.15 and a subnet mask of /28 (255.255.255.0)
and max # of hosts/ subnet = 14.
Hope you get to understand this method. -
Optionstech-airman
Member Posts: 953
SurferdudeHB wrote: »I got the subnetting basics down but can't put it together to solve this, thank you.
Network ID: 213 214 150 0
Subnets Required: 13
What is the subnet mask?
1st Available Host
Address of Subnet 1?
Max # of hosts/subnet?
SurferdudeHB,
What class is the network address? -
OptionsSurferdudeHB
Member Posts: 199 ■■■□□□□□□□
tech-airman wrote: »SurferdudeHB,
What class is the network address?
Class C -
Optionsmiller811
Member Posts: 897
did it click for you yet Surferdude?I don't claim to be an expert, but I sure would like to become one someday.
Quest for 11K pages read in 2011
Page Count total to date - 1283 -
OptionsSurferdudeHB
Member Posts: 199 ■■■□□□□□□□
tech-airman wrote: »surferdudehb,
what is the default mask for a class c network?
255.255.255.0 -
Optionsmella060
Member Posts: 198 ■■■□□□□□□□
SurferdudeHB wrote: »I got the subnetting basics down but can't put it together to solve this, thank you.
Network ID: 213 214 150 0
Subnets Required: 13
What is the subnet mask?
Max # of hosts/subnet?
hmmm lets see...this is a class C address so the default subnet mask is 255.255.255.0
You require 13 subnets right ? So then what power of 2 will give you the required amount of subnets ?
2^3 will only give you 8 subnets...2^4 will give you 16 subnets...so then you require 4 subnet bits.
So your subnet mask would be 255.255.255.240 /28...does that sound alright ?
That leaves you with 4 host bits so 2^4-2 = 14 hosts per subnet -
OptionsSurferdudeHB
Member Posts: 199 ■■■□□□□□□□
Ok I think I got it guys. I reviewed my CBT Nuggets videos and can now put it all together. Thank you all!
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Optionstech-airman
Member Posts: 953
SurferdudeHB wrote: »255.255.255.0
SurferdudeHB,
If you borrowed 1 subnet bit, how many subnets can you have?
Does it meet the design requirement of "Subnets Required: 13?" -
OptionsSurferdudeHB
Member Posts: 199 ■■■□□□□□□□
tech-airman wrote: »SurferdudeHB,
If you borrowed 1 subnet bit, how many subnets can you have?
Does it meet the design requirement of "Subnets Required: 13?"
No you need to borrow 4 bits.
