Networking - Networking Basics - Discussion

sbnet mask=255.255.255.{192} equiv to [11111111.11111111.11111111.11]00 0000 in binary. Now the binary values in [] represents the n/w portion i.e. first 26 bits is allocated for n/w portion and remaining 6 bits for the host portion.

Now take a look into the IP address.. 192.168.168.188 where the 1st 26 bits represent d N/W portion. So the binary representation of IP address is [11000000.10101000.10101000.10]111100,the bits in [] again representing the N/W portion.

There are four imp. things dat can be concluded from an IP address given with subnet mask,namely,
-> N/W ADDRESS
Represent d IP address in binary form and convert d HOST portion into ZEROS,and leave d N/W portion as it is. for eg. 11000000.10101000.10101000.10"111100" (192.168.168.188) is changed to 11000000.10101000.10101000.10"000000"(192.168.168.128;represents d n/w address). NOTICE THE CONVERSIONS REPRESENTED BY ".."

->1ST USABLE ADDRESS
The next IP address immediately after the N/W address is d 1st usable address. For eg. 192.168.168.129 is the 1st usable address in this case.

->LAST USABLE ADDRESS
Note that d host portion is of 6 bits i.e.2^6(=64) hosts can be connected. So d host portion can be divided in 64 blocks each i.e 0-63,64-127,128-191..and so on. Here in this case,since the n/w add. started from .128 so it must end on .191. So the LAST USABLE IP ADDRESS IS THE IP ADDRESS WHICH COMES IMMEDIATELY BEFORE THE BROADCAST ADDRESS. So the last usable IP address is 192.168.168.190.

->BROADCAST ADDRESS
It refers to d last IP address of the block. In this eg, the broadcast address is 192.168.168.191 since d block is 128-191 as seen in the previous step.

I hope u must have understood clearly.

This is the trick for this one.

The given is 192.168.188.188 with a subnet of 255.255.255.192. This subnet is /26. You should know the value of the increment of each subnet mask. In this case, it's 64.

Since it's in the fourth octet, we can now find the CLOSEST number which can be divided by the increment value. Only below the range of the given IP.

In this case, 128 is the CLOSEST number to the given 188 that CAN BE DIVIDED by the INCREMENT VALUE which is 64.

So we know that the IP will be 192.168.188.188 /26.

Since the increment value is 64 which is also the number of IP addresses, we will just ADD 64 to 128. Which will give us the last IP 192.168.188.192.

So, what do we have now?

First IP: 192.168.188.128 which is the NETWORK ADDRESS.
Last IP: 192.168.188.192 which is the BROADCAST ADDRESS.

These 2 IPs cannot be used.

So the usable IP range will be 192.168.188.129 to 192.168.188.191

Always remember that the first and last IP of the IP range is the Network and Broadcast addresses, respectively. And they are cannot be used.

I hope this will be a clear solution.

Everybody is confused because desimal binary conversation, here is the way to better understand this: subnet is all 1s but the last 6 digits determines hosts. You cant do anything for the fist 24 bits because they are fixed (255.255.255) you have play with the first 2 bits of the last octet, they can be:

00xxxxxx
01xxxxxx
10xxxxxx
11xxxxxx

In other word, you can create 4 different networks out of these bits, now convert them to decimal:
The first one's max value will be 63 (if all x were 1) which is not even close to 188.

Second posibility or network can be max 127 not close to 188.

Third one is good, it can be max 191 (don't forget you can't change the 0 on the second digit from left, it is belong to network portion not host, so max=1011111y. Here you cannot make y=1 because when it is 1 with other bits, it becomes network address.

11xxxxxx's min address is 192 so it will never contain 188 although other bits were 0.

Here are the answers using 255.255.255.192:

How many subnet bits are used in this mask? Answer: 2 2^2-2=2 subnets.

How many host bits are available per subnet? Answer: 6 2^6-2=62 hosts per subnet.

What are the subnet addresses?Answer: 256-192=64 (the first subnet)64+64=128 (the second subnet)64+128=192. However, although 192 is the subnet mask value, it's not a valid subnet. The valid subnets are 64 and 128.

What is the broadcast address of each subnet?Answer: 64 is the first subnet and 128 is the second subnet. The broadcast address is always the number before the next subnet. The broadcast address of the 64 subnet is 127. The broadcast address of the 128 subnet is 191.

What is the valid host range of each subnet? Answer: The valid hosts are the numbers between the subnet number and the mask. For the 64 subnet, the valid host range is 64-126. For the 128 subnet, the valid host range is 129-190.

The answer is correct.

Subnet Mask = 255.255.255.192 where. CIDR /26 means, 11111111.11111111.11111111.11000000.
The default subnet mask is 255.255.255.0, So, We can only change the last byte.

Subnet = 2^2 = 4 Total Subnets.
Valid IP Addresses = 2^6 = 64 Valid IP's.
Valid Host Ranges = 2^6 - 2 = 62 Host addresses.

Subnet Number ----- First network adrs ----- Valid Host Range ----- Broadcast adrs
1. 192.168.168.0 ----- 192.168.168.1 ----- 192.168.168.62 ----- 192.168.168.63
2. 192.168.168.64 ----- 192.168.168.65 - ----- 192.168.168.126 ----- 192.168.168.127
3. 92.168.168.128 ----- 192.168.168.129 - ----- 192.168.168.190 ----- 192.168.168.191( 192.168.168.188 is here)
4.192.168.168.192 ----- 192.168.168.191 ----- 192.168.168.254 ----- 192.168.168.255.

Hence, Ans is correct.

188 = 10111100.
192 = 11000000.
255.255.255.192 means that network id will be of 26 bits and host id will be of 6 bits
we know that 2^6=64 IP addresses can be generated.

10"111100" is the binary form of 188

Put all zeroes in the double quotes, we get the first IP address which is the network address i.e 192.168.168.128.

So the next IP will be 192.168.168.129
put all ones in the double quotes, we get the last IP address, which is the broadcast address i.e 10"111111", 192.168.168.191.

So the valid IP range will be excluding the network id and broadcast id.
Therefore, 192.168.168.128-192.168.168.190 is the required solution.

Subnet masks 255.255.255.192 means that it has 6 bits to represent hosts. which means in the last octet: first two bits represent network bits and other 6 host bits.

Now you are looking to fit 192.168.168.188.

Out of two bits you can have : either 10 or 11( 11 is not possible it will go beyond 192) so network bit is 10 , which makes it

128.( 255.255.255.128)

Now , possible values for hosts (for 6 bits) will start with: all zeroes to all 11s( but zero and all ones are meant for this network and broadcast)

Hence the possible values are: from 1 to 62

Which will be: 129 to 190

First of all you have to find the base number for that the formula:-.

256-mask = base number will be used.
Now that will be 256-192 = 64.

Now to find the number of valid subnet's do the following.

Subnet 1: 64.
Subnet 2: 64+64 = 128.
Subnet 3: 128+64 = 192 (not valid as it comes upto subnet 's mask address).

So 2 valid subnets.

Next find the broad cast address that would be 127 and 191.

So first and last valid address of the subnet would be.

65, 129 (first valid address).
And
126, 190 (last valid address).

So the range would be 129 to 190.
Hence the answer. :).

Here the subnet mask is 255.255.255.192 means only the last value is changing so subtract it from 255, you will get 64 means out of 255 IP's 64 is already used only 192 IP's are available.

256/4=64 means 0-63, 64-127, 128-191, 192-255 all 4 segments contains 64 IP's, out of which one segment is already used (lets not worry about it).

Now check the IP given which is 192. 168. 168. 188 which falls in the range 128-191 so here the valid host range will be from 129-190 because first one is always network IP, the last one is always the broadcast IP.

To find valid host range, we need subnet mask.

Here subnet mask 255.255.255.192 (11111111.11111111.11111111.1100000).

So for 192 (128+64+0+0+0+0+0+0). Here magic number is 64.

0-63 (192.168.168.0 (n/w address) & 192.168.168.63 (broadcast address). So we shouldn't use as IP address in between we can use these IP addresses).

64-127.
128-191.
192-255.

Our question where 192.168.168.188 address lies in these range?

Answer is 192.168.168.129-192.168.168.190.

I hope you understood well.

Thank you.

Ravi sankar.