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How many bits would you borrow to subnet 172.16.100.0 to have at least 500 hosts?
we need to borrow 7 bits to subnet 172.16.100.0 to have at least 500 hosts and the subnet mask will be 255.255.254.0
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
What is host for this 255.255.255.192?
You have given a subnet mask; it has nothing to do with hosts. You would need to supply the network number.
A company is planning to subnet its network for a maximum of 27 host .which subnet mask would provide the needed hosts and leave the fewest unused addresses in each subnet?
255.255.255.224 would give 32 per subnet or for ex. 192.168.1.1/27 leaving 5 bits or 2-5th power =32
What subnet mask would a network administrator assign to a network address of 192.30.5.0 if it were possible to have up to 254 hosts?
yes
If you need a to have a class b network address subnetted into exactly 510 hosts what subnet mask would you assign?
255.255.254.0
If you have the following Network ID 131.112.0.0 You need at least 500 hosts per network How many networks can you create What subnet mask will you use?
If you need to divide it up into the maximum number of subnets containing at least 500 hosts each, you should use a /23 subnet mask. This will provide you with 128 networks of 510 hosts each. If you used a /24 mask, you would be limited to 254 hosts. Similarly, a /22 mask would be wasteful, allowing you 1022 hosts.
What is the subnetting of 192.168.1.0 24 to provide for 100 hosts?
the /24 at the end of your IP is stating that your subnet is going to be 255.255.255.0 wich would give you 254 hosts so you have alot of room to expand if you are only looks for 100 hosts
What are necessary procedure in calculating subnet mask or simply subneting show an example forr calculation?
To calculate a subnet mask, first determine the number of subnets needed and the number of hosts per subnet. Use the formula (2^n \geq \text{number of subnets}) for subnetting and (2^h - 2 \geq \text{number of hosts}) for host calculation, where (n) is the number of bits borrowed for subnets and (h) is the number of bits left for hosts. For example, if you start with a Class C address like 192.168.1.0 and want 4 subnets, you would borrow 2 bits (since (2^2 = 4)), resulting in a subnet mask of 255.255.255.192 (or /26), which allows for 62 hosts per subnet.
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
As network administrator what is the subnet mask that allows 510 hosts given the IP address?
N - network bitsH - host bitsIf you are following the first octet rule, this is a class B network with a subnet mask of 255.255.0.0 (N.N.H.H). We can subnet this by "borrowing" some bits from the host portion. There are 16 network bits and 16 host bits. There is a simple formula to calculate the proper subnet mask.2 to the power of what equals at least 510(2^X)? We have a total of 16 host bits to borrow from. 2^1...2^2...2^3...Etc2^9= 512 - 2 = 510 host addressesWe subtract two because the network and broadcast address are not usable addresses. As we can see we need at least 9 host bits to get 510 hosts per subnet.Take 32 and subtract it from the host bits you need. So 32-9=23. Your subnet mask now has 23 network bits instead of 16.In binary the original subnet mask would be 11111111.11111111.00000000.00000000.In binary the new subnet mask is 11111111.11111111.11111110.00000000.If you convert this into dotted decimal form you get 255.255.254.0.TLDR: 172.30.0.0 - 172.30.1.255255.255.254.0
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.
