subnetting

Pffff I had never played with such big numbers for subnetting. I will not explain the process itself, there is enough written on the net, how the net mask calculates in which position they move 1 and 0 and from there it is already clear where the network part ends and the host part begins. I had a very interesting teasing task 2 /16-that (that is 255.255.0.0) or for the utterly unenlightened 2 x 65534 host address to split them into 2 region with several networks – one with requirements for 32k hosts, 16k host and 8k hosts which in turn had to be divided into more- 4 equals subnets. The second region had a requirement for 4k hosts, 2k host and 1k host, and again like the previous zone each network of 4 equal subnets 😆 In general I'm pretty good at subnetting but I've never played with networks on such a scale. There was a beautiful big calculation, but at least I gained experience with much larger sizes than dividing / 24-ki, which is done calmly in mind. Now all that is left is to scatter the networks on the devices to make the routations and networks to work 😀 hahaha. The diagram with my devices numbered above 30 as I stopped fighting them – beauty 😎

2 comments

  1. 1
    1x /17 = 4x/19
    1x / 18 = 4x /20
    1x /19 = 4x 21

    2
    1x / 20 = 4x / 22
    1x / 21 = 4x /23
    1x /22 = 4x/24
    Yes, it's easy when you do it all day, but look at me, 10 line code I read it for two days….
    🙂
    Welcome to subnetting. To do it quickly and easily, you have to learn to work only on decimal numbers.
    Let me explain the above
    32 000 the host is actually a network with 2 of degree 15 = 32768 (or half of all known numbers 65536), supposedly we want 15 host bits, 8 go to the last octet, remain 7 for the third => 24-7=/17.
    Another option for bills – /16 there is 65536 address, to have 32k we need one bit less – /16 => /17
    And now how to divide /17 on 4 Equal parts. My logic goes to the last line: 4 Equal parts – therefore, the 17th is divided into 4….so on 2 on the second. The degree gives us the number of bits we take from the host part – Veil 17 + 2 = 19. Demek 4x / 19. Why this is so I leave to your imagination.

    1. ehehe well done and my answers are like that and they are correct :D.
      Well, I generally do quite well, but not with such large numbers 😀
      This is because by shifting each bit we get x2 networks of the bits we have shifted by shifting by 2 we get bits 4 networks 😉 If they had to 6 we would move with 3 bits and others. To be honest, subnetting is fun. These examples are generally more elementary because they are in the CCNA separately, that they are deliberately such that it is easier for students to summarize the routines to one larger supernet 😉 I just liked the numbers as they are 😀 😉

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