Having trouble memorizing power tables.
Raymond Mason
Member Posts: 74 ■■□□□□□□□□
in CCNA & CCENT
Hello! Not too long ago, I finally figured out how to (slowly) subnet. Although the problem I came across was memorizing the power tables. Going all the way up to class A networks, how far up the power table should I memorize? I can't seem to do it without using my figures but that takes time.
Here is what I know without using a calculator:
1^2 = 2 1^5 = 32
2^2 = 4 1^6 = 64
2^3 = 8 1^7 = 128
2^4 = 16 1^8 = 256
This is what has been really hindering me from subnetting faster. I tried to configure DHCP but I am having trouble learning how to configure that as well.
Here is what I know without using a calculator:
1^2 = 2 1^5 = 32
2^2 = 4 1^6 = 64
2^3 = 8 1^7 = 128
2^4 = 16 1^8 = 256
This is what has been really hindering me from subnetting faster. I tried to configure DHCP but I am having trouble learning how to configure that as well.
Comments
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OctalDump Member Posts: 1,722Well, if you know 8, then getting to 9 isn't hard, and if you can do 9, then 10 should be fairly easy and so on.
The useful ones for networking are probably 2^1 - 2^8, 2^10, 2^16, 2^24. You can then interpolate and extrapolate to get the others fairly easily.
2^8, 2^16 and 2^24 are the octet boundaries. I include 2^10 because knowing that 2^10 is ~1000, helps to get a feel for any others. Like 2^20 is 2^10 * 2^10, or ~1,000,000. So if you had 2^23, you know that it was about 2^20 x 2^3, or about 8,000,000.
It's one of those things that repetition will drive into your brain.
DHCP is relatively straight forward. You need to exclude the addresses you don't want to hand out. Define and name your pool. Add a network to your pool, default router and anything else you like. You also need to have at least one interface with an IP in the defined network (well, you don't have to, but that's not CCNA level).2017 Goals - Something Cisco, Something Linux, Agile PM -
pinkiaiii Member Posts: 216few months back when lying awake at night i would do subneting in my head i knew the rules magic number managed to sort of grasp concept of getting any subnets just by doing it on my fingers,to say the least once i thought i knew it i sort of left it at that,now im basically at square one.Not sure is there many questions in final exam for ccna but A class aside its sort of knowing top subnet amounts removing 2 for hosts then working on magic numbers.
That said people who pass ccna dont rely on subneting in their heads there's so many calculators to give you all info is way easier and most engineers would just do that,its just exam wants you to know how to do it - sort of like in school you learn formulas,but in real life its more of calculator.
what recently got me thou was the table of classfull as classless networks ,since seen class A=8 B=12 C=16,then all the labs you do its always 8-16-24 and i was like how the hell they get them masks,and which is which.
btw on subneting class c is easy-it goes 2-4-6,reverse is knowing how to multiply hosts past /23 510 1022 and so on until you hit the 65k for class b,no not following the 2^x i rather memorized max hosts for class b and if working with magical number its divide or multiply. -
fearlessfreap24 Member Posts: 10 ■□□□□□□□□□
Class B CIDR
17
18
19
20
21
22
23
24
Class C CIDR
25
26
27
28
29
30
31
32
Binary
128
64
32
16
8
4
2
1
Mask
128
192
224
240
248
252
254
255
this is the table I make when subnetting. Note that the binary is the increment of the subnet. I.e /27 is 32s. 0 32 64 96 128 160 192 224 It's a little more difficult to figure out for Class B, but this table makes it manageable.
I hope this helps -
AllwaysANewB Member Posts: 11 ■□□□□□□□□□I'm using the brute-force memorization method: Write them out over and over and over. You got 3 ones for networks, 2^3 = 8. And 5 ones for hosts? 2^5 = 32. Over and over again. I can do them in my head at this point. Just don't forget the -2 for the host IPs
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pinkiaiii Member Posts: 216AllwaysANewB wrote: »I'm using the brute-force memorization method: Write them out over and over and over. You got 3 ones for networks, 2^3 = 8. And 5 ones for hosts? 2^5 = 32. Over and over again. I can do them in my head at this point. Just don't forget the -2 for the host IPs
I guess that works if you cant do it out like that,but what would screw me over would be superneting going another direction.
since as person above said /27,i just know binary table and skip in my head for 32,if i see mask 240 i know thats fourth bit 16.
but Wonder with all that how many questions are simple like that in ccna,since dont believe they give easy stuff as how many subnets/hosts are in 172.16.0.0/19.
more like you got 172.18.20.10/21 and whats the br address or next available subnet if you know what i mean. -
AllwaysANewB Member Posts: 11 ■□□□□□□□□□I guess that works if you cant do it out like that,but what would screw me over would be superneting going another direction.
since as person above said /27,i just know binary table and skip in my head for 32,if i see mask 240 i know thats fourth bit 16.
but Wonder with all that how many questions are simple like that in ccna,since dont believe they give easy stuff as how many subnets/hosts are in 172.16.0.0/19.
more like you got 172.18.20.10/21 and whats the br address or next available subnet if you know what i mean.
With the brute-force memorization method it all starts to make sense. I use subnettingquestions.com - Free Subnetting Questions and Answers Randomly Generated Online to test my knowledge and that website hits you with plenty of "find the last available IP address for a host" and the like.
The other thing seems to be using the 256 - (whatever the subnetmask is) = number of IPs in a range.
Either way, Repetition is the Mother of Learning. Whatever method works for ya, do it over and over.