IP and different branches

[Onz]

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 IPV4 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

Number of subnet bits ( borrowed bits) (60 Host)
Borrowed bits is 2

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

New subnet mask (in dotted decimal notation) (60 Host)

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

Number of remaining host bits (60 Host)
Number of remaning host bits is 6

Number of host/subnet (60 Host)
Number of host 64/4

Useable number of host/subnet (60 Host)
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 being 32 - 2 = 30 which is lesser then any other hosts in the question.

55>30 x
40>30 x
31>30 x

Can anyone help me with this as i am stuck with it thanks.

[slowlyunruly]

6 months and have not any solutions here?