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IPV4 Addressing

Tags: Microsoft, Networking, Windows

Hi everyone, I am very new to IT in general got a question in my hand and i would love to get information from more experienced IT people. So i have this 200.168.100.0/24 IP and i got 4 different branches which branch 1 has 60 host, branch 2 has 55 host, branch 3 has 40 host and branch 5 has 31 hosts. I am not supposed to use VLSM ( variable lenght subnet mask) I need to calculate the following.

• Number of subnet bits (borrowed bits) required to accommodate the addressing requirements
• Total number of subnet
• The new subnet mask of the IP subnetting scheme
• Number of remaining host bits
• Number of hosts per subnet
• Useable number of hosts per subnet

This is what i have done so far

Branch 1 60 Host

Base network address 200.168.100.0 /24
Subnet mask 255.255.255.192 / 26 – 1111 1111 . 1111 1111 . 1111 1111 . 1100 0000
Number of subnet bits 2
number of subnets (22) = 4
number of host bits 6
number of useable hosts (26-2) = 62

[b]Number of subnet bits ( borrowed bits) (60 Host)[/b]
Borrowed bits is 2

Total number of subnets (60 Host)
Total number of subnets are 4

[b]New subnet mask (in dotted decimal notation) (60 Host)[/b]

0th subnet Address: 200.168.100.0000 0000 / 200.168.100.0
Mask: 1111 1111 . 111 1111 . 1111 1111 . 1100 0000 / 255.255.255.192

1st Subnet Address: 200.168.100.0100 0000 / 200.168.100.64
Mask: 1111 1111 . 1111 1111 . 1111 1111 . 1100 0000 /255.255.255.192

2nd Subnet Address: 200.168.100.1000 0000 / 200.168.100.128
Mask: 1111 1111 . 1111 1111 . 1111 1111 . 1100 000 / 255. 255.255.192

3rd Subnet Address: 200.168.100.1100 0000 / 200.168.100.192
Mask: 1111 1111 . 1111 1111 . 1111 1111 . 1100 0000 / 255.255.255.192

The answer is 255.255.255.192 / 1111 1111 . 1111 1111 . 1111 1111 . 1100 0000
[b]
Number of remaining host bits (60 Host)[/b]
Number of remaning host bits is 6
[b]
Number of host/subnet (60 Host)[/b]
Number of host 64/4

[b]Useable number of host/subnet (60 Host)[/b]
62/4

And the other branches with 55,40 and 31 is the same answer literally as 60 host ?? According to formula (2H-2) = 64-2 = 62 they all have to be 2 to the power of 6
62>55 /
62>40 /
62>31 /

because if i do 2 to the power of 5 which ends up

Topic

[kees_b]

Reply To: IPV4 Addressing

In reply to IPV4 Addressing

As you see, your post was longer than the allowed maximum.

And this seem perfect questions for your teacher. I’m sure he can answer them.

Finally, may I ask why you used Windows and Microsoft as tags? Even if you’re very new to IT, in this phase of your study you should know that the IP-protocol has nothing to do with Windows and Microsoft.

• This reply was modified 2 years, 2 months ago by kees_b.
• This reply was modified 2 years, 2 months ago by kees_b.

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