The pace is picking up. This week is Chapter 6 of the Cisco courseware. It starts with binary math (which I already knew), then quickly moves on to subnetting and the /24, /27, etc. stuff.
The chapter 5 test was pretty difficult, with next-hop questions that were tricky.
Wow....the cert exams are gonna suck!
Subnetting is a difficult subject for many people to grasp. There are many sources of this information, but I encourage you continue searching until you find a description that helps you understand the best.
Think "computer". Look at IP addresses and imaging how they look in binary like the computer and networking devices see them, not in our decimal numbering system. As you compare a host IP address through its subnet mask, you can see what binary bits sepecify the network portion of an address and the host specific portion.
Quiz yourself with various IP addresses and subnet masks in your spare time. Be able to answer these four questions for every IP address and subnet mask you make up: (1) What class is the IP address (2) What is the size of the subnet which it is in (number of IPs & number of usable hosts) (3) What is the lowest IP address in that subnet - the subnet IP (or subnet ID, depending on the source) (4) What is the highest IP address in that subnet - the broadcast IP Once you can answer those four questions rapidly for any IP address and subnet mask given, notably the addresses with subnet masks using octets like
254, 252, 248, 240, 224, 192, and 128, then you have conquered IP address subnetting. Also, you need to know how to covert numbers from decimal to binary and binary to decimal rather quickly for 8 bit binary numbers and decimal numbers between 0 and 255. Once on the job, you will still be doing this but not notice because you are so quick calculating it in your head that you do not covert to binary any longer.********** BINARY METHOD:
********** Given an IP address and subnet mask: 192.168.50.3 255.255.240.0 You convert to binary: 11000000.10101000.00110010.00000011 11111111.11111111.11110000.00000000 If you break the IP address into its network portion host portion based on where the 1's and 0's in the subnet mask are seperating the address: 11000000.10101000.00110000.00000000 = 192.168.48.0 00000000.00000000.00000010.00000011 = 0.0.2.3 CHECK: Add these two decimal format IP address portions together, and you end up with the original address: 192.168.48.0 0.0.2.3 +_____________ 192.168.50.3 Now that you have verified your binary math, convert all of the host bits in 192.168.48.0 from 0's to 1's: 11000000.10101000.00111111.11111111 = 192.168.63.255 Figure up the number of binary bits in the host portion and figure up 2 to the nth power for that value: 2^12 = 4096 Now you have determined that the IP address is in a subnet with the following information: Lowest IP address: 192.168.48.0 Lowest IP address: 192.168.63.255 Number of IP addresses in the subnet: 4096 with only 4094 host addresses available********** MY METHOD:
********** Given an IP address and subnet mask: 192.168.50.3 255.255.240.0 Perform some odd subnet math subtracting each octet from 256 and then multiplying the differences together: 256 256 256 256 255 255 240 0 -_________________ 1 x 1 x 16 x 256 = 4096 Now take that "funny number" in the subnet mask subtracted from 256 and start counting ranges from 0 by that number and then look for that the value in the corresponding octet of the IP address: 256 - 240 (3rd octet of subnet mask) = 16 Couting by 16s, looking for 50 (3rd octet of IP address): 0...15, 16...31, 32...47, 48...63 FOUND 50! Insert those low and high ranges with a 0 and 255 after them, respectively: 192.168.*48*.*0* 192.168.*63*.*255* That is it. You just determined the low and high IP address in a range and the size of the range - no binary math!Slash format subnet masks keep us from having to say two-fifty-five over and over. Just take the slash value and subtract it from 32 then figure up 2 to the nth power. For example, 192.168.50.3//20 would require finding 32-20 which is 12 then figure out 2^12 which is 4096. I do not do exponent math, I just memorize the powers of 2 and count on my fingers. It is just like the sizes of computer RAM we look for deals on all of the time. Two (1), four (2) , eight (3), sixteen (4), thirty-two (5), sixty-four (6), one-twenty-eight (7), two-fifty-six (8), five-twelve (9), one-thousand-twenty-four (10), two-thousand-fourty-eight (11), four-thousand-ninety-six (12). Done!
----- Scott Perry Indianapolis, IN
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This is the method presented to us last night in a video by Odom. Very quick and easy. However, it was presented to us BEFORE the binary method, so it was sort of "Follow this formula and you'll get the answer."
I'm not like that. I need to know why something works before I blindly follow it. So once I learned about how to AND the addresses together, and then made the logical jump to the "magic number" method...very easy.
But this math is all well and good...we still haven't been given context. Why do this? What is the situation in which you have this information and you need to do these calculations?
I'd love to have a part-time internship and see this stuff in action.
And this is just the first of 4 courses to cover off the CCNA :-)
But subnetting is an essential skill.
At least with the current courses you weren't first hit with all that "Class" stuff that others before you have to suffer. Now's it's just a prefix or mask.
/22 means you can't touch the first (leftmost) 22 bits, so you have 10 to work with. From these 10 you work out how many you need to create your required number of subnets and how many to provide the number of hosts you need in each. This way is easier to teach and it's great to see students actually getting the concept sooner.
But like all things maths related you must practice and practice and practice - you won't become proficient at it by theorising about it.. When those minutes are ticking by in the cert exam this is one process you need to be just able to do with the minimum of effort.
Aubrey
When you have time, can you give an example of that tricky next-hop questions? It is probbaly the same one that I saw. I think that once we understandd the exact function of each application layer, it wouldn't look as tricky. When I took Chapter 6, I didn't get to read the material as much as I would have wanted because I was working and was just tight with time. But later, I understood what was what.
My instructor said that the first class covers 50% of CCNA material.
Subnetting is fun.
I know how to do those with a few reading and exmaples the instructor showed but I kow that I have to practise to practise a lot more to be efficient time-wise.
This is probably the only type of uestions where one needs to be fast (aside from knowing the concepts), right? Most other questions would be either you know it or you don't, I think.
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