Subnetting Question

# of workstations = 100
Network address = 192.168.1.0/24
Subnets required = 5

On this exercise, I wrote out:

192.168.1.0 - 7
.8 - 15
.16 - 23
.24 - 31
.32 - 39....96 - 103

But I got a different answer from this link (Subnetting Exercises « Cisco Networking Tutorial), why?

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Comments

Ltat42a

Need to remember your block (or range) sizes. 3 bits were used to create the subnets (2^3 = 6). Using the below chart, your subnets will increment by 32.

Alhaji265

So where do you get 3 from?

MAC_Addy

You said you needed 5 subnets. So it'll be the last 3 bits used.

Ltat42a

Also need to study up on binary. 2 "to the what power" will give you at least 5 subnets?
2^1=2
2^2=4
2^3=8

Alhaji265

MAC_Addy wrote: »

You said you needed 5 subnets. So it'll be the last 3 bits used.

So by using 3 (2^3), I will have at least 8 subnets enough for 5 subnets? Pardon me if this is a stupid question.

Ltat42a

Subnetting requires knowledge in binary math. To calculate a required number of subnets, the formula is 2^. ^ is the number between 1 and 32 you use to calculate subnets and/or hosts.

This thread will help also -

http://www.techexams.net/forums/ccna-ccent/38772-subnetting-made-easy.html

theodoxa

Alhaji265 wrote: »

# of workstations = 100
Network address = 192.168.1.0/24
Subnets required = 5

On this exercise, I wrote out:

192.168.1.0 - 7
.8 - 15
.16 - 23
.24 - 31
.32 - 39....96 - 103

But I got a different answer from this link (Subnetting Exercises « Cisco Networking Tutorial), why?

Is that 100 workstations per subnet or total?

If per subnet, then you will have to use 4 Class C Blocks (192.168.0.0, 192.168.1.0, 192.168.2.0, and 192.168.3.0).

If OTOH [as is most likely,] it is 100 workstations total that need to be divided among 5 subnets, then the answer is:

192.168.1.0/27

Subnet 0) 192.168.1.0
Subnet 1) 192.168.1.32
Subnet 2) 192.168.1.64
Subnet 3) 192.168.1.96
Subnet 4) 192.168.1.128
Subnet 5) 192.168.1.160
Subnet 6) 192.168.1.192
Subnet 7) 192.168.1.224

This will give you 8 subnets (6 if subnet zero is disabled) that can support up to 30 hosts each.

Jason R

Alhaji265 wrote: »

So by using 3 (2^3), I will have at least 8 subnets enough for 5 subnets? Pardon me if this is a stupid question.

You want 5 subnets, so in binary that is 11100000. The equation to 2^m gives you the number of subnets where M stands for masked bits. Therefore, 2^3 gives you 8 subnets. 2^2 does not give you enough subnets as you would only get 4 subnets. There is no problem in having more than what you need, but you want to be as close as possible. The 5 zeros are your non-masked bits which gives you the amount of hosts you can have. The equation is (2^N)-2. In this case, it is (2^5)-2. This equation tell you that you can have 30 addresses that you can utilize per subnet. You are left with 8 subnets that can have 30 usable addresses in each subnet. The 100 hosts would simply be distributed into each subnet. For example, 30 hosts would go in one subnet, 30 in the second, 30 in the third, and 10 in the fourth. I hope this helps. Let me know if you have any questions.