Networking - Networking Basics - Discussion

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.

The subnet mask is 26, but the actual number of bits is 32 so 32-26 =6 bits for the one subnet. So 2^6 is 64 one subnet will have 64 addresses.

So, 0-63,64-127, 128-191 and 192-255 is the 4 subnets, in that 128 lies in 128-191 subnet.

How do know the subnet mask is /26? Please explain the concept in clear.

IP address is 192.168.168.188 .188 in binary is (1 0 1 1 1 1 0 0) =.188.
Subnet mask is 255.255.255.192 .192 in binary is (1 1 0 0 0 0 0 0) =.192.
as .255 is all ones, last octet of first network ID will be (1 0 0 0 0 0 0 0 )=.128.
with 256 - 192 = 64, N=(64) which is our magic number. N-2 = 62 hosts/network.
first N/W ID 192.168.168.128,

1st Host ID 192.168.168.129.
last Host ID 192.168.168.190.

Broadcast 192.168.168.191.
with (128+64=192), 2nd N/W ID will be 192.168.168. 192.

S1
192.168.168.188.
255.255.255.192.

Net id 192.168.168.128.
1011 0010.
1100 0000 AND LOGIC.
100000. = 128.

S2 255-192 = 63.
BROAD CAST 192.168.168.191.
.129 TO .190 is my valid host range.

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.

256-192 = 64 so increment by 64.

Networ - Broadcast
64 - 127
128 - 191 > so the usable host address range is 127 - 190 (answer is A).
192 - 255.

The answer is right they said "usable" ip address, so the network is 192.168.168.128 to .191, 128 is network address and 191 is broadcast address.

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.