Can this be made a sticky, spent 25 minutes looking for it....
every time I forget subnetting this page help re-plant the seed...
Can this be made a sticky, spent 25 minutes looking for it....
every time I forget subnetting this page help re-plant the seed...
Ty for bumping that Deathmage! got my exam in a few days and still needed to work on my subnetting. I really like this explanation of it to!!
Definitely should be stickied
Been using this workbook the past 24 hours and I'm half way thru it and it's really helping me breeze through subnetting; I learned subnetting before I switched gears onto the VCP5-DCV track and lost touch with it but from my superb notes and this work book that I noted in my 5-star massive 6 subject notebook to do in-case I forgot (and see I knew I would) I'm picking it up again.. Right now I can do one in about 45 seconds; there is still 45 pages left of worksheets so maybe I'll get better at it.
Here is a link: Let me google that for you
Last edited by Deathmage; 03-03-2015 at 03:34 PM.
Has anyone found a way to remember subnetting for class A?
See I can subnet Class B and C fine in binary but once I get to Class A remembering those large A** numbers is daunting to say the least. Curious if the CCENT/CCNA will have Class A's on it or if it's really just tons and tons of practice to just remember them?
The past 3 weeks I've been doing on average 20 or so subnetting practices a day (when I have down-time at work) between my workbook above and subnetting.net; and watch every subnetting video on youtube.
I don't expect you will need anything past a /22.
If you do then you can revert to the math quickly on the whiteboard they give you. If you are familiar with binary up to like 4096 (/20) I don't think you will have any problems.
Thanks Jon, I can do Class A but with a calculator. I've been just doing Class B and C's and really honing them. I mean If I had to do a Class A I could do it but with long math, just it's tedious to remember those numbers and I suck with multiplication. I haven't used math is like 6 years since college....
making sure I got this Class A address right:
number of needed hosts: 29
Network Address: 23.0.0.0
so.....
Address Class: A
Default Mask: 255.0.0.0
Custom Mask: 255.255.255.224
Total number of Subnets: 524,288
Total number of Host addresses: 32
Total usable addresses: 30
Number of bits borrowed: 19
Does that look right; I can remember up to 65,538 but need a calculator to go any higher. Really hope the exam doesn't have anything higher than 65,538.
I go with the
(Left to Right)
Subnets: X. 2 4 8 16 32 64 128 256 . 512 1024 2048 4096 8192 16384 32,768 65,536 . 131,072 262,144 524,288 1,048,576 (you all know the rest)
(Right to Left)
Hosts: X. (the big number for 8 blocks) .65,536 32,768 16,384 8,192 4096 2048 1025 512 . 256 128 64 32 16 8 4 2 1
now since I only need up to 32 hosts (32-2 would equal 30 and I need 29) and that's 3 bits into the 4th octet that would be 128+64+32=224 as the custom mask, so it would be 255.255.255.224.
been subnetting with this workbook day and night for weeks and I think it finally clicked. now I just reprinted it and going to do it again and again until I get really fast at it...
Maybe I'll try the power of 2 method but I got binary in my head right now, is binary good enough for the exam or is it still too long?
been finding that 5 to 10 minutes at work doing a few subnetting problems really helps with keep the skillz in my head; i feel like I'm dreaming binary anymore...
just to add to this here is a Class B:
number of needed subnets: 2000
number of needed hosts: 15
Network Address: 178.100.0.0
so.....
Address Class: B
Default Mask: 255.255.0.0
Custom Mask: 255.255.224.0
Total number of Subnets: 2048
Total number of Host addresses: 32
Total usable addresses: 30
Number of bits borrowed: 11
am I on the right track?
been using this: Let me google that for you
Last edited by Deathmage; 03-10-2015 at 04:06 AM.
Hi guys,
I understood how to do the example originally posted and went through most of the examples posted by others , but i could not find answer to this type of question and this is the tricky ones that I am not able to get my head around
What is the broadcast address of the network 172.28.74.192 255.255.255.192?
Answer: 172.28.74.255
What i did was took the block size as 64
and my subnets were
172.28.0.0
172.28.64.0
172.28.128.0
so going by this logic the address specified in the question belongs to 172.28.64.0 and the broadcast address will be 172.28.127.255 .
But the answer says its 172.28.74.255 where the 3rd octet remained unchanged. But if it is a Class B Address (starting with 172) then should the 3rd octet change ?
You are designing a subnet mask for the 172.16.0.0 network. You want 120 subnets with up to 300 hosts on each subnet. What subnet mask should you use?
Answer: 255.255.254.0
How you do work out the subnet mask in this iteration.
I know for 300 hosts you will be taking 9bits ( or on the 2*9 = 512 ).
I know that its a Class B address ( by default first two octets are network )
But cannot make the translation to equate- 254 in the third octet
There are 7 bits set in the third octet, adding them together results in 254.
Here's a cheat sheet I made up: Subnet Cheat Sheet
Thank you for your sheet but am not clear on how your table translates. I know that by default that a CLass B would be a /16. With getting 300 hosts this would now become a /25 ( Am thinkinh... not sure ) BUT how do you take /25 to make up the netmask- 255.255.254.0
Edit. No answers required any more. Worked it out, albeit using another method. Throwing into questions now to get it all nailed down.
Thanks
Last edited by Phara0h; 04-23-2015 at 01:18 PM.
hi , can someone calculate step by step this ip 10.0.0.0/8 . i need only 100000 host.
please help me
Thanks
Dude there is plenty of material here to help with that, give it a go and learn from the process. We have all had to do it the hardway!
For anyone reading this, where he went wrong is that you add subnets to the subnet mask or subtract host from /32, so 32 - 9 = /23 OR /16 + 7 bits of subnets (128 subnets) = /23
Again for anyone reading... I have a few power of two "anchor points" if you will, and think of them as MegaBytes for thinking.
So
2^3 = 8MB
2^6 = 64MB
2^10 = 1GB (1024)
For 2^ 10, 11, 12, 13, it's really easy, it's 1024, 1024*2, *4, or *8
For mental math, this is how I do the bigger numbers, it seems to work best for my mind, though I'll switch it up with different techniques at times. Useful for interviews I guess? I tend to do mental over calculator by habit
For 14 to 23, you do -10 to the exponent. Let's call it x. We want to multiply it by 1024 (2^10). x*1000 + 2x*10 + 2*2x
Let's take 4 examples, 14 16 19 and 23:
For 2^14:
2^4 = 16, so 16,000 + 320 + 64 = 16,384
for 2^16
2^6 = 64, so 64,000 + 1,280 + 256 = 65,536
for 2^19
2^9 = 512, so 512,000 + 10,240 + 2048 = 524,288
and for the hardest, 2^23
2^13 = 8,192 so 8,192,000 + 163,840 + 32,768
I usually calculate it backwards. So say for 2^15, my thought process is, 2^5 = 32, so 32,64,128. 128+640=768, +32000 = 32,768. For 2^21, 2^11 = 2048. 2048;4096;8192. 8,192+ 40,960 = 49,152 + 2,048,000 = 2,096,000 + 1,152 = 2,097,152
Last edited by nster; 08-26-2015 at 09:10 PM.
Thanks for this! This has made my life so much easier. I do have a question though. Is there an easier way to figure out the type of questions below other than just counting up in your block size???
Question: How many subnets and hosts per subnet can you get from the network 172.29.0.0/21?
Answer: 32 subnets and 2046 hosts
Hello, can someone done this task for me?
An ISP is granted a block of addresses starting with 158.78.0.0/16. The ISP wants to distribute
these blocks to 3200 customers as follows.
a. The first group has 400 medium-size businesses; each needs 64 addresses.
b. The second group has 600 small businesses; each needs 16 addresses.
c. The third group has 80 households; each needs 256 addresses.
Design the subblocks and give the slash notation for each subblock. Find out how many
addresses are still available after these allocations
Oh my! This made things so easy. Been away from networking for a while and this made it seem like a walk in the park. Thanks!
Hi all, can you please check if below answers are correct, as I have been given below for next coming interview questions.
Thanks in advance for the assistance.
1. Enter the last valid host on the network 192.168.92.0 255.255.255.224?
Increment 32, so the last valid host is 192.168.92.30
2. What is the maximum number of valid subnets one will have from the network 10.29.252.144/13? Assume this is a class A address.
2^5 = 32 maximum number of valid subnets, since borrowing 5 bits
3. What is the most efficent subnet if you need 110686 usable hosts on a subnet? Present your answer as a subnet mask?
2^17 - 2 = 131070. So the subnet mask will be 255.254.0.0, as we need to have 17 zero bits for the host.
4. What is the broadcast address of the network 10.121.108.13/14?
Network ID: 10.120.0.0, 10.124.0.0
So, the broadcast is 10.123.255.255
5.What is the maximum number of valid hosts one will have from the network 192.168.124.0 255.255.255.224?
5 bits zero for the host, so the answer is 30
6. What is the number of network IDs in a Class C network? 3, because of 255.255.255.0
Wow, so glad I came across this post from 2008. Brilliant.
Lordflasheart, I don't know if you came up with this on your own? But it's amazing. I'm not as advanced as some (I'm working on my net+ cert) but this is very helpful and getting the basics down. Thanks, I really appreciate it!
Question: What is the first valid host on the subnetwork that the node 172.18.186.50/23 belongs to?
I'm confused a little - you've asked the question that is clearly copied and from a website, on a forum that is about technology. AND on a subnetting thread that TEACHES you how to subnet. This one is easy to answer. How about you have a little stab at it. If you get it wrong, we can help you.
Can't thank you enough for this. It's cleared up so much confusion. Even the cisco netacad class I'm in at the community college didn't teach it this well.
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