Subnetting Help

I have a problem that I would appreciate help on.

Number of needed subnets = 2
Network Address = 195.223.50.0

Address Class - C
Default subnet mask - 255.255.255.0
Custom subnet mask - 255.255.255.128
Total number of subnets - 2
Total number of host addresses - 64
Number of useable addresses - 62
Number of bits borrowed - 1

What is the third subnet range?

Ok, I'm unsure about the answers starting with Total number of host addresses.

I think it should be 128 for that particular one. However.........

if that one bit is borrowed at the "128" doesn't that mean that the next one left hosts is 64?

Also, would appreciate any links to websites that give questions/answers on problems like this that has some explanation on why answers are right or wrong.

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Comments

MrBrian

An easy way to calculate the "number of subnets" and the "number of hosts" for a scenario is to use the 2^x method.. where x is the number of bits.

So we know there's 8 total bits in that last octet.. and then you say you want to borrow one for the subnet, which gives you one subnet bit, and 7 host bits.

Then you can just use 2^x, where x = bits, to calculate your totals..

One Subnet bit----> 2^1 = 2 possible subnets

7 Host bits----> (2^7) - 2 = 126 possible hosts

hodgey87

if you're only needing 2 subnets you can go with a /25 like you already have 255.255.255.128 - This give you an increment of 128 so the first subnet would be 192.223.50.1 (First Usable) - 192.223.50.126 with a broadcast of 192.223.50.127.

Second network would be 192.223.50.129 (First usable) - 192.223.50.254 with a broadcast of 192.223.50.255.

Ismaeljrp

I think you should practice subnetting in binary first, if you are lacking in binary, that's were the errors in subnetting will happen, especially if you use the speed methods, which, IMO does not teach you to troubleshoot subnetting mistakes.

Ismaeljrp

armycat23 wrote: »

I have a problem that I would appreciate help on.

Number of needed subnets = 2
Network Address = 195.223.50.0

Address Class - C
Default subnet mask - 255.255.255.0
Custom subnet mask - 255.255.255.128
Total number of subnets - 2
Total number of host addresses - 64
Number of useable addresses - 62
Number of bits borrowed - 1

What is the third subnet range?

Here is the giveaway buddy, " What is the THIRD subnet range? "

You only NEED 2 subnets, but that is not a strict requirement, you can have more than that.
Why? because it says so in the question. What is the third subnet range, which means, you actually need 3 subnets in order to answer the question. If not, you'll never be able to even have a 3rd subnet range.

You can take it from here.

armycat23

Ismaeljrp wrote: »

Here is the giveaway buddy, " What is the THIRD subnet range? "

You only NEED 2 subnets, but that is not a strict requirement, you can have more than that.
Why? because it says so in the question. What is the third subnet range, which means, you actually need 3 subnets in order to answer the question. If not, you'll never be able to even have a 3rd subnet range.

You can take it from here.

So, basically I would have to borrow another bit to have enough for a third subnet(4 total with 3 available). That would leave 6 host bits left. 2x2x2x2x2x2 = 64.

So, I have to modify the ranges depending on the number of networks added later.

So, the subnet ranges would be
195.223.50. 1 -195.223.50. 63
195. 223.50. 64 -192.223.50.127
195.223.50.128 - 192.223.50.191 .........so would this be the correct third subnet range?

If it is than does that mean that I have to change my custom subnet mask from earlier to 255.255.255. 192 and the following answers to the questions following to reflect that?

Thanks for all the help.

Ismaeljrp

armycat23 wrote: »

So, basically I would have to borrow another bit to have enough for a third subnet(4 total with 3 available). That would leave 6 host bits left. 2x2x2x2x2x2 = 64.

So, I have to modify the ranges depending on the number of networks added later.

So, the subnet ranges would be
195.223.50. 1 -195.223.50. 63
195. 223.50. 64 -192.223.50.127
195.223.50.128 - 192.223.50.191 .........so would this be the correct third subnet range?

If it is than does that mean that I have to change my custom subnet mask from earlier to 255.255.255. 192 and the following answers to the questions following to reflect that?

Thanks for all the help.

Yes. You would have to change you mask to a /26. The mask you were using before didn't provide enough subnets.That would be the actual third subnet range.

Remember the first Subnet ID is the same as the network ID. So your first subnet would be :
195.223.50.0 -195.223.50.63 ( 195.223.50.1 is the first usable host ip address )
.
.
. and so on, the rest is correct.