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New Member

subnet question help

Hi iam kinda confused with this question.

1) Given 192.168.100.0 /24, subnet it so as to give 5 subnet

    show the table.

1.1) What is the first and last usable ip address of the zeroth subnet?

1.2) What is the first and last usable ip address of the 2nd subnet?

I would be very appreciated to who can answer to this questions.

1 ACCEPTED SOLUTION

Accepted Solutions
Bronze

subnet question help

You can't get exactly 5 subnets without using VLSM. You can get 8 subnets by using /27 (255.255.255.224) mask.

Zeroth subnet is 192.168.100.0/27, first usable address is 192.168.100.1, last - 192.168.100.30

Second subnet is 192.168.100.64/27, first address is 192.168.100.65, last - 192.168.100.90

If you use VLSM you may get any result you want. It would not be deterministic.

3 REPLIES
Bronze

subnet question help

You can't get exactly 5 subnets without using VLSM. You can get 8 subnets by using /27 (255.255.255.224) mask.

Zeroth subnet is 192.168.100.0/27, first usable address is 192.168.100.1, last - 192.168.100.30

Second subnet is 192.168.100.64/27, first address is 192.168.100.65, last - 192.168.100.90

If you use VLSM you may get any result you want. It would not be deterministic.

New Member

Re: subnet question help

omg thank you i thought my answer was wrong

but questions 1.3) how do solve it? what i get is 65~94 usable ip

New Member

Re: subnet question help

As Fer said to get 5 subnet u use a /27 with the no of valid subnets as 8 and a block size of 32 with 30 being the valid hosts ( remember one goes for n/w address and one for broadcast ) so now.. calculate the range

Valid subnets --> 0,32,64,96,128,160, etc etc.. till 224

so to answer ur 1.3) how to solve it? the usuable IPs which u will get is from 192.168.100.65 - 192.168.100.94 ( 192.168.100.64 as the n/w address and 192.168.100.95 as the broadcast address) .. and that makes your answer right

      

Note: You still got 3 more usuable subnets which u can have fun with

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