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How many valid subnets do you get if you borrow 3 bits?
When you borrow 3 bits from the host portion of an IP address, you can create (2^3 = 8) valid subnets. However, one subnet is reserved for the network address and another for the broadcast address, leaving you with 6 usable subnets. Thus, by borrowing 3 bits, you obtain 6 valid subnets for use.
How many bits are borrowed from a class B host address to create 30 new subnets?
5 bits are necessary to create up to 30 subnets.
How many subnets of 300 hosts can be created on a network of IP address 172.16.1.0?
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How many subnets and hosts per subnts can you get from 172.30.0.0?
The IP address 172.30.0.0 is a private IP address in the Class B range, which has a default subnet mask of 255.255.0.0 (or /16). If you use the default mask, you can create 65,536 addresses (2^16), allowing for 65,534 usable hosts per subnet (subtracting 2 for the network and broadcast addresses). If you further subnet this address, the number of subnets and hosts per subnet will depend on the subnet mask you choose. For example, using a /24 subnet mask would give you 256 subnets with 254 usable hosts in each.
Assume all devices are using default configurations How many subnets are required to address the topology that is shown?
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How do you use 172.16.0.0 to create 6 subnets?
To create 6 subnets from the IP address 172.16.0.0, you need to determine how many bits to borrow from the host portion of the address. Since 2^3 = 8, borrowing 3 bits allows for 8 subnets, which is sufficient for your requirement of 6. This results in a subnet mask of 255.255.248.0 (or /21), providing each subnet with 2046 usable host addresses. The subnets would range from 172.16.0.0/21 to 172.16.7.0/21.
How many bits are borrowed for the subnet?
Subnets are created in powers of 2 due to the way netmasks work. To accomodate 10 addresses, you would need a /28 (255.255.255.240) netmask, which would provide 14 usable IP addresses. Here are the details of that mask in a private network address area (192.168.x.x) Output from the unix "ipcalc" program: Address: 192.168.1.0 11000000.10101000.00000001.0000 0000 Netmask: 255.255.255.240 = 28 11111111.11111111.11111111.1111 0000 Wildcard: 0.0.0.15 00000000.00000000.00000000.0000 1111 => Network: 192.168.1.0/28 11000000.10101000.00000001.0000 0000 HostMin: 192.168.1.1 11000000.10101000.00000001.0000 0001 HostMax: 192.168.1.14 11000000.10101000.00000001.0000 1110 Broadcast: 192.168.1.15 11000000.10101000.00000001.0000 1111 Hosts/Net: 14 Class C, Private Internet With the /28 netmask, 192.168.1.0 is the network address, and unusable. 192.168.1.15 is the broadcast address, and is also non-assignable. This gives 14 usable addresses. A /29 netmask (one bit less) gives 6 usable addresses. To create 10 subnets, the size of the subnets would need to be known, and the process is similar, but the subnets would have different start and end addresses.
An IP address of all ONES refers to?
A special type of IP address is the limited broadcast address 255.255.255.255. A broadcast involves delivering a message from one sender to many recipients. This broadcast is 'limited' in that it does not reach every node on the Internet, only nodes on the LAN.
How many /64 subnets are available in a /48 prefix?
There are 65,536 subnets available in an IPv6 /48 block.
Given class C address with a default subnet mask How many possible subnets and usable hosts if 4 bits were borrowed?
that gives you 16 subnets with 14 usable IPs for hosts that is because one is for subnet and one for broadcas in that subnet for example: 192.168.1.0/28 - subnet number 192.168.1.15 -broadcast number usable IPs for hosts - IPs between them that is 14
How many hosts and subnets are possible if you have an IP of 151.242.16.49 with a subnet mask of 7 bits?
151.242.16.49 would be class B in a classful environment with a network mask of 16 bits. An additional 7 bit subnet mask puts the total mask at 23 (i.e.: 255.255.254). Subnet ID would be: 151.242.16.0 Host address range: 151.242.16.1 through 151.242.17.254 Subnet Broadcast address would be: 151.242.17.255
How many hosts in 512 subnets?
To determine the number of hosts in 512 subnets, we first need to understand the subnetting. If you have 512 subnets, that means you need 9 bits (since 2^9 = 512). Assuming you are using a standard classful subnet mask, the remaining bits from a total of 32 bits (IPv4) can be used for hosts. This would typically leave you with 23 bits for hosts, allowing for 2^23 - 2 = 8,388,606 usable hosts per subnet, after accounting for network and broadcast addresses.
