How many IP addresses can be used for unique hosts inside the IPv4 subnet 192.168.2.128/26?
How many IP addresses can be used for unique hosts inside the IPv4 subnet 192.168.2.128/26?
The given subnet mask of /26 indicates that there are 26 bits dedicated to the network portion, leaving 6 bits for the host portion (32-26=6). The number of possible IP addresses for the host portion is calculated as 2^6, which equals 64. However, each subnet reserves 1 IP address for the network itself and 1 IP address for the broadcast address. Therefore, the number of usable IP addresses for unique hosts within the 192.168.2.128/26 subnet is 64 - 2 = 62.
Each subnet mask has 32 bits which are sperate in network and host bits. In this case we have 28 network bits which means 32 - 28 = 6 Host bits remain. 2^6 = 64.. BUT every subnet has 1 network adress (the first IP address in that subnet) and 1 broadcast address (the last one in that subnet) so 64 -2 = 62 hosts in each subnet
EDIT: Sorry it's 32 - 26 of course
Correct. IP for this subnet is 192.168.2.128 ~ .191. However, you need to remove the subnet address (.128) and broadcast address (.191) ...and you are left with 62 unique host addresses.
2^(32-26) - 2 = 62 is the formula to determine the number of hosts available on a given subnet.
D is correct, 64 available, but network and broadcast address are not available.
ambecilli
Ma fateve na pera
correct B
Perdon 64
64 no, because the broadcast address and the net address doesn't count
2^6-2 because 2 ip for broadcast and subnet
correct