Networking - Networking Basics - Discussion
Discussion Forum : Networking Basics - Networking Basics (Q.No. 7)
7.
Which of the following is the valid host range for the subnet on which the IP address 192.168.168.188 255.255.255.192 resides?
Answer: Option
Explanation:
256 - 192 = 64. 64 + 64 = 128. 128 + 64 = 192. The subnet is 128, the broadcast address
is 191, and the valid host range is the numbers in between, or 129-190.
Discussion:
46 comments Page 5 of 5.
Jayanti moger said:
7 years ago
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.
So, 0-63,64-127, 128-191 and 192-255 is the 4 subnets, in that 128 lies in 128-191 subnet.
(3)
Ali Hasan said:
6 years ago
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.
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.
Karthikeyan said:
5 years ago
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.
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.
(8)
Suprio said:
4 years ago
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.
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.
(4)
Aditya Sharma said:
3 years ago
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.
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.
(1)
Anan said:
2 years ago
How do know the subnet mask is /26? Please explain the concept in clear.
(2)
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