Simple subnetting help please, Class C network...

BefallenBefallen Posts: 1Registered Users ■□□□□□□□□□
So I'm going to be doing this skills exam next week, and I'm still not sure how things change if variables in the problem change. Basically, I have a network of 192.168.1.0/24, and I have worked through most of the questions (see attachment IMG_0018_ed.jpg) My main question is, and I'm sure I'm missing something simple; How am I choosing the IP Subnet, First IP address, and Last IP address of Subnet B with relation to the ones already defined in Subnet A? I'm not sure how to transition between the 2 subnets without overlapping ranges. What am I missing?

Comments

  • IsmaeljrpIsmaeljrp Posts: 480Member ■■■□□□□□□□
    read up on VLSM.
  • fadhilfadhil Posts: 200Member
    simple to answer this question
    first know the number of subnet needed and number host needed
    since on your question there,the thing to consider is the number of subnet because number of host are not mentioned
    we need to find power of 2(2^n) which is greater or equal to number of subnet needed
    where n is borrowing bits
    there, only two subnet are needed so it will be 2^1 =2 here 1 is borrowing bit

    hence from a network given 192.168.1.0/24 the subnet mask is 255.255.255.00000000 so put the borrowing bit into the lower bit(first 0 from a given subnet mask) that is 255.255.255.10000000 and the subnet now will be 255.255.255.128.
    find block size:a block size 256-increment no
    for the subnet mask the block size is 256-128=128
    hence the subnet will be 192.168.1.0/25 and the second subnet is 192.168.1.128/25 no more subnet here.

    to find number of valid host 2^n -2 where n here is the number of zeros from the subnet mask .In our subnet mask there are 7 zeros valid host per subnet are 2^7 -2=126
    so for eac subnet there are 126 host .
    for a network A
    lets assume that ip address is 192.168.1.0/25 hence the address is 192.168.1.1 and the last address is 192.168.1.126
    for network B
    ip address 192.168.1.128/25 hence first address is 192.168.1.129 the last address is 192.168.1.254
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