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The answer is yes. This all depends on the type of IP address.That is the class it belongs to.All that is needed to be done is to subnet the given IP adress. Depending on the address given (Class A, Class B, or Class C) you can determine how many hosts are available. This is also subject to whether the address is classless or classful. If it is classful the a Class C address can have254 valid hosts, a Class B can have64,534 and a Class A can allow over16 million hosts! If the address is classless then more focus is on the subnet mask. To illustrate a quick example, if you had the subnet address255.255.255.240, you know that it is a Class C address, but the last octet has been subnetted. Converting240 into binary gives. This means that4 bits were used for subnetting (1) and four bits are left for addresses (0). If you did all the possible ranges in binary, starting with0 and working all the way til1 you see that you get16 binary instances. Knowing that you cannot use the first and last address you are left with14, and since the .240 address splits the octet down the middle, you get14 subnets and14 valid hosts
I always answer the "Four" big questions in order to do any subnetting homework on any given subnet mask and address:
You're given this
192.168.10.0
255.255.255.240
1- How many valid subnets?
Answer:16.
Explanation:240 is in binary, therefore,2^4=16
2- How many hosts per subnet?
Answer:14
Explanation:4 bits left for host addressing, therefore,2^4-2=14. (Note:2 bits must be left for host addressing that is why we subtract2 bits)
3- What are those valid subnets?
Answer:256-240=16. Start at16 then do the incrementing job by16.
For example,16,32,48,64,80,96,112,128,144,160,176,192,208, 224,240. (Note: You could start by subnet "Zero" if it is enabled on the router)
Explanation: These are the16 valid subnets that we just calculated in the first question but we are more curious about what are they. :)
4- What is the broadcast address for each subnet and what are the valid hosts?
Answer: Let's say for subnet192.168.1.64, the broadcast address is79 and valid hosts are65 through78. The same applies to all subnets.
Explanation:
Subnet 64
First Host 65
Last Host 78
Broadcast 79
As you see, this method is easier than the binary way.
Cheers
Please vote if it helps.
An example is 192.168.0.0/24 means you can effectively connect 253 machines with 2 addreses reserved for the network and broadcast i.e 255.255.255.255
The answer is yes and you need to dertemine which class the IP block falls under(A,B and C).
Yes, By Subnetting and Supernetting
Yes, of course. Depending on the CLASS on which the IP ADD belongs, then the MASK is known. My knowlege of a Classless Inter-Domain Routing (CIDR) make it easy to determine the usable IP addresses base on the Subetmask.
yes. you either look at the subnet mask to determine how many bits were borrowed to the network portion and how many are left for the host portion.
if subnet mask tell me according to this.
Each IP Address has 2 segments
IP Address and Subnet Mask e.g. Unique Street Name (Subnet Mask) and Unique House Number (IP Address)
Once we have both - we can find how many computers can be available
Of course - Broadcast ID and NW id has to be omitted from the total numbers i.e. -2
Subnet mask should be known to know the IP range,Subnet mask calculate ip range
Subnet mask should be known to calculate ip range
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