Question
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The default (Clssful) subnet mask forthis network is: Usingthe default subnet mask for this network, can you support how many hosts or clients? Subnet this network into 2 subnets and provide the following information: Whatisthe new subnet mask for this network? List the two subnets (Network IDs): How many hosts can be supported by each subnet? How many total hosts can be supported? List the address range each subnet and the directed broadcast address: Given IP 140.150.67.117 and mask 255.255.248.0 Specify the subnet that contains this IP host address and the range of addresses within this subnet: # hosts per subnet:2n - 2: where 'n' is the number of host bits (# of '0' in the subnet mask). Network ID: N = H & M (N = Network ID, H = Host Address, M = Subnet Mask) # required subnets: 2n-2> = X; where 'n' is the number of host bits (what we are solving for) 3nd 'X' is the numberof required subnets (i.e. ifwe require 25 subnets;then 2n - 2 > = 25, 'n' should be 5 since: 24 -2 = 22 which is not greater or equal to 25. In this case, 'n' represents the number of BORROWED bits and not the number of host bits. M (Mega) and k (kilo) should not be taken literally. Since we are dealing with binary numbers (powers of two) lk is not 1,000 (or 103) but instead 210 (or 1024). 1M is not 1,000,000 (or 10s); instead, it means 2Jvor 1,048,576. Numbers in between these 21'and 220 could be easily calculated simply by multiplyingthese base numberstimesamultiplier. Forinstance, 16k could be calculated 16 * 1k, or 16 * 210. Alternatively, 16k could be also calculated by applyingthe proper power of two. In the case of 16k equals 214, or 1k (210), 2k (211), 4k (212), 8k (213), 16k (214).
Explanation / Answer
class B network = 150.200.0.0
default subnet mask is = 255.255.0.0
can you support how many hosts / clients = 32766.
no of subnets needed = 2;
new subnet mask for this network = 255.255.192.0
List two subnets (Network ID's)
= 150.200.0.64 and 150.200.64.0
how many hosts supported by each subnet = 2^14-2 = 16382
total hosts supported = 16382 + 16382 = 32764
subnet identifier valid host range broadcast address for subnet.
150.200.0.0/18 150.200.0.1 to 150.200.63.254 150.200.63.255
150.200.64.0/18 150.200.64.1 to 150.200.127.254 150.200.127.255
IP = 140.150.67.117 mask is 255.255.248.0
Subnet Containing IP start range - end range
140.150.64.0 140.150.64.1 - 140.150.71.254
