Networking - Networking Basics - Discussion

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.

It's simple guys.

Given IP address : 192.168.168.188.
Given subnet mask address:- 255.255.255.192.

To find the N/W ID
step1: Convert both addresses into binary form.
step2: Perform AND operation b/w them.
step3: Result will be your network ID i.e(192.168.168.128).
step4: So range of the N/W starts from 192.168.168.129 (starting host ID).
step5: The last id of the N/w is 192.168.168.191 (which is broadcast domain).
step6: So the last id of the N/W will be 192.168.168.190.

Thank you.

Hello, all.

The answer is obtained as follows:

IP: 192.168.168.188
Subnet: 255.255.255.192

192 in binary :
1 1 0 0 0 0 0 0
188 in binary:
1 0 1 0 1 1 0 0

192 AND 188:
1 1 0 0 0 0 0 0
1 0 1 0 1 1 0 0
Ans: 1 0 0 0 0 0 0 0

In 192 first 2 digits are 1 therefore in Ans 1 0 will be part of subnet id.

While remaining 6 will give host address;

First Host addr: 1 0 0 0 0 0 0 1 = 129.
Last host addr: 1 0 1 1 1 1 1 0 = 190.

And Broad cast addr will be : 1 0 1 1 1 1 1 1.

192.168.168.188 255.255.255.192
mask 26.

And now have to find first address,
Applying AND operation with first address.

192.168.168.188
8 8 8 2

188 = 10111100.
And last 2 bit of mask,
11.

Hence first address,
192.168.168.128.

And last address,

Adding complement of mask into first address.

00000000 00000000 00000000 0011111111.
(63).

Since 1 address 192.168.168.128.

+ 0 0 0 63.

192.168.168.191.

This value is right not wrong.
First of all we need to be the see the address and now.

We take broadcast address 256 then in this we subtract the value of last gourd subnet mask (192). Then we add the same bit of subtracting then we find the 128 also and then we re add the value (64) then we find out the 192 this will be the broadcast address of segment.

Actually the value given is wrong. the value should be 192.168.168.128

then only we can get the correct value.

192-128=64

so that we can get 64 values.that valid host range value should be between 129 to 191.

i think correct answer is B.

does anybody refuse it....?

Hi, @All.

Subtract 192 from 256 = 64,

Means, the subnetting has been done by dividing into four parts, 64 * 4 = 256.
Now see the option.
64 * 2 = 128,
And 64 * 3 = 192.

So IP add should be in between.

192.168.168.129 to 192.168.168.191.

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.

Answer (A) is right.

Because valid host means usable ip.
The CIDR is;

Subnet=192.168.168.128
First Usable IP= 192.168.168.129
Last usable IP is= 192.168.168.190
Broadcast IP is= 192.168.168.191.
so, answer is correct.

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.