Cisco Training CCNA IP Addressing - Part 6

Comments on 'Cisco Training CCNA IP Addressing - Part 6'

1. In many cases you can get away with using .48 as a
   host. The routers will just forward the packets to .48 to
   the right network, so it's a matter of whether the host
   itself is fussy or not. If the host is fussy and refuses to
   allow you to give it a "network address", you might be able
   to sacrifice connectivity with "neighbouring" addresses by
   setting an incorrect 255.255.255.224 mask.
2. Not E, because it is part of 32 subnet, not the 48 subnet
   segment that 172.16.12.54 is from.
3. Host Range is: 48 --> 63 with : 1. 48 ... as Network base
   Adress 2. 63 ... as Broadcast Adress anynumber above belongs
   to Network & between is Host ! right ? or .. left ?? =)
4. that's amazing ! =) guys you may have to use the calc a
   paper & practice it ;) thank you again for these vids ;)
5. Ah, much thanks. I understand what he means now. True, its a
   good trick.
6. His "Trick" is ever simpler, just subtract your last octet
   - .240 from the Magic number "256" and you get 16, this
   number represents the number of "Steps" in which the subnets
   increment. So subnets would be 16,32,48,64 etc... However we
   know that the numbers 16, 32, 48, 64 etc... are the Subnet
   which means the number preceding it will be the broadcast
   address for the preceding subnet, so .16's broadcast addy is
   .32, and we know neither of these are valid as hosts. So .54
   is in the range .49-.62
7. drakezen & DnSavior , .240 gives you a Subnet of .48, here's
   how long way - When you "Logical AND" 172.16.12.54 &
   255.255.255.240 you get 172.16.12.48, that is you "logical
   AND" 10101100 00010000
   00001100 0011 0110 & 11111111 11111111 11111111 1111 0000
   - 10101100 00010000 00001100 0011 0000 which
   translates to decimal to 172.16.12.48. Put all 1's in for
   the Host portion and it gives you -172.16.12.63 (Broadcast)
   so range is .49 - .62
8. I am confused as well. Is not te answer, B, C, and E?
9. If the mask is 240 and the ip address is 172.16.12.54
   wouldn't that give a subnet host range of 172.16.12.48 to
   172.16.12.54? with 172.16.12.55 being the broadcast
   address. So how would 172.16.12.64 be a valid host address
   with this range using that mask? The question is what are
   the valid hosts with that mask? I dont think I understand
   why you are showing what possible subnets are possible.
10. I listened to a guy who said we shouldn't use "tricks"
    because of blah blah....but on the test i just took,
    converting to binary really screwed me time wise. This trick
    is good, and thanks to this guy i wont spend a ridiculous
    amount of time converting crap during the test.
11. OMFG YOU ARE MY HERO! Thank you!
12. thanks for taking the time to do these..especially the
    "trick". I understand it now!! 5 Stars!
13. thanks a lot to post exmaples. it really helps me to
    understand much better. I wish my teacher could explain
    subneting like you..
14. This is great. Makes me want to learn more from your videos.
    Everything sinks in after the second watch.
15. Thanks for the post.
16. Thanks for your videos! Your explanations actuallly make
    sense! This method is great!
17. Great tip; kudos!
18. It would because 256 is the highest amount of address
    possibilities in one octet: (0 to 255 from all 0s to all
    1s). As in the vid, same deal goes with any masking (check
    equation below): possible octet bits - masked bits = host
    bits. e.g. 256 - 248 = 8 256 - 240 = 16 256 - 224 = 32 256
    - 192 = 64 256 - 128 = 128 etc...
19. will 256 work with all subnet masks to determine the subnet
    host ranges and broadcast addresses?
20. thank you very much keep going u make a great job /
21. Thanks for the tip. I'm reading a book and it says to use
    the magic number where you draw a line between the network
    bits and the host bits and look at the network bit next to
    the line. You look at the place value of that bit to see
    what the increments of the subnets will be. It works, but
    you still have to write out the binary value of the octet.
    Your way is much faster.