please i need help
i have an ip 22.214.171.124. and i need four subnets for this production network without using calculator. and i want to Complete the following:
Subnet the IP using 3 bits
Identify the subnet mask
Determine the usable blocks of addresses for each network
Establish and label IP addresses for each side of the routers (E0/0 and E0/1)
Establish and label the IP addresses for each workstation and server
List the information for all four subnets using the following format:
Network Address Mask Size Host Range Broadcast Address
so this is what i did
since the ip 126.96.36.199 is calss c the default subnet mask is 255.255.255.0
but they want us to subnet using 3 bits which means we eill borrow 3 bits of 8 or exactly from 6 bits 11100000=224
So are subnet mask is 255.255.255.224
now we can find the host/subnet variables by looking for the lowest bits turned on which is 2 power 5 = 32 that is mean we can have 8 subnet with 30host in each subnet but since we are looking for 4 subnet are subnet will be
is that correct or not
any help will be appreciated
First off, is this a homework problem? Please just say so, we like to help guide you to the right answer and not give it to you for homework.
Otherwise, yes, that looks right to me. Just don't forget about the network and broadcast addresses and that they are unusable.
Thank you for your replay but i am confused because he want us to use 3bits and he want he s network to be subnetted to 4 subnets
Originally Posted by AdamPI
so if i borrow 3 bits which means we will ended up with 8 subnets
but he wants just 4 subnet i am really confused
Could be an error in his logic. It could also be that he wants you to use 4 32-node subnets. I'm not quite sure. It's a good question to ask him.
July 10th, 2012, 04:12 AM
From the wording of the question, it sounds like the instructor wants four /27 networks out of your original 188.8.131.52/24 range.
If that's the case, there are multiple answers. Some more "right" than others when you consider future expansion or future use of the left over IPs.
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