Binary question

[mikesgroovin]

I made a similar post a while ago that might be of some help in addition to what oldskool is saying. May help you to understand subnetting limits and using the AND statements to get a results of which nodes in a subnet will talk to each other...

...

ok, not sure if this is going to help understand or not... but essentially, you are asking the same questions (just the reserve logic of each other)

http://www.techist.com/forums/f44/subnet-would-i-use-188682/#post1487128

 

[harnar123]

yeah haven't really tried to focus on supernetting. Trying to master subnetting first.

 

[mikesgroovin]

The logic is still the same. I figured that it may seem a little confusing, but once you understand subnetting, you'll understand supernetting.

Subnetting is talking a single network and dividing it up into smaller networks or "sub"networks for secure, concise and more efficient propagation methods.

Take the first question

1. For the Class C network address 192.168.10.0, which of the following subnet masks delivers 30 subnets?
a. 255.255.255.252
b. 255.255.255.248
c. 255.255.255.240
d. 255.255.255.224

You have a lot of work cut out for you.....
First of all... what does class C mean?
or better yet, what is the class C subnet mask of the 192.168.10.0. Thats where you are starting from ... then you have to subnet that particular network mask into 30 more.

 

[S0ULphIRE]

I'll solve the first question for you, just so you can get and idea of how to do it then.
The mask that best fits is 255.255.255.248
Here's how I get that answer:
1.You can't get exactly 30 subnets, so you have to go for the closest alternative, which is 32.
2.Ip Addresses range from .0 to .255, effectively 256 different addresses
3. 256 divided by 32 gives 8, which is the total number of addresses in each subnet
4. now we know the subnets addresses range from 192.168.10.0 - 192.168.10.7 (8 addresses, counting the network address, .0, and broadcast address, .7)
5. take the number you got from dividing 256 by your wanted number of subnets (8 in this case) and subtract 1 off it (8-1=7), then work out how many binary bits you'd need to make that number.
6. In this case, you'd need 3 bits to make 7 (which is 111 in binary)
7. 32 binary ip address bits total. 32 - 3 bits = 29
8. 29 is your mask. you can get the decimal values (255.255.255.248) by simply subtracting the 3 bits you got from right to left from the mask. 255 in binary= 11111111. take 3 bits from the right gives you 11111000. which=248

 

[harnar123]

thanks a lot everybody. I now have a much better understanding of binary. And soulphire, ya really made it click. Because I understand by seeing every step, I am a hands on learner.
so I am starting to get how to figure out but I had one more question.

if we created thirty subnets then, how do I know how many hosts are on each network?

is it just 8 hosts per a subnet. Then minus 2 so 6? if so it would always be six correct?

 

[Apokalipse]

I still dont get it.
24/2=0
12/2=0
6/2=0
3=1
1=1
that equals 00011 so wtf?

You placed the digits the wrong way around. You start from right to left*, as with any number system.

Binary is just a number system. It essentially works the same as decimal, just using a lot fewer digits

The highest digit is 1, and not 9. So after a number that has all 1's, the next number has an extra digit (eg 111 goes to 1000)

In decimal, it just happens after a number with all 9's (eg 999 goes to 1000)

in decimal, you count like this:
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

in binary, you count like this:
0, 1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011, 1100...

* I made a noob mistake, I initially said left to right, though that's not what I meant. It goes that (<---) way

 

[harnar123]

yeah I edited my last post include more of what I am still not understanding.

 

[S0ULphIRE]

as to your other question about the number of hosts, yes you were right. 8 addresses - 2 (because of the broadcast and network addresses) leaves you 6 host addresses per subnet.

 

[harnar123]

3. For the Class C network 192.168.220.0, what subnet mask supports up to 14 subnets?
a. 255.255.255.252
b. 255.255.255.248
c. 255.255.255.240
d. 255.255.255.224

So then the answer would be c correct?

 

[S0ULphIRE]

yup
you got it!