Subnet design

Ok with a theoretical question like that you have to assume a closed network unless they say otherwise. No ~single~ network on the planet has 5mil+ nodes, the electrical cost alone are astronomical. So what their really asking you to do is not realistic, instead just subnet out the IP range so that it can support that many nodes.

But there is a slight problem, 200.10.80.32/25 is too restricted to hold that many nodes, so please reply with the proper question limits.

What exactly do u mean by " 200.10.80.32/25 is too restricted to hold that many nodes"?
Do u mean that it is impossible to adjust that number of hosts in the subnet (Y/n)?
Would it be possible if there were only 2 departments with 40 & 28 hosts?

And regarding the question limits, I can't specify any further limits because that was the question given in exam & u find a solution following what it states.

5,40,28,12,6 hosts

I read that as 54,028,126 (54 million) hosts haha. Be careful with running numbers and comma's together like that.

5
40
28
12
6

Start with the biggest and work your way down, then find out the smallest subnet that can contain each and split accordingly. Remember to add +3 for each subnet as your need a network number, a router and a broadcast address.

40 -> /26 for 64 address, 61 nodes
28 -> /27 for 32 address, 29 nodes
12 -> /28 for 16 address, 13 nodes
6 -> /28 for 16 address, 13 nodes
5 -> /29 for 8 address, 5 nodes

Now I leave the rest up to you, we won't do your homework for you but we will show you how to do things.

Well then let me try.
200.10.80.32/26 to 200.10.80.95/26 1st subnet
200.10.80.96/27 to 200.10.80.127/27 2nd subnet
200.10.80.128/28 to 200.10.80.143/28 3rd subnet
& so on.

But still my doubt is unanswered. The original IP address is /25 meaning that 7 bits are for the hosts i.e. a maximum of (128-3) hosts. So, the 3rd, 4th & 5th subnet should be invalid since they need the 8th which belongs to n/w address.

Where am I misunderstanding?

200.10.80.32 /25 implies that the your network is 0-127. It doesn't start at 32 (The network address for what he gave you is 200.10.80.0 - 127.

40 -> /26 for 64 address, 61 nodes
28 -> /27 for 32 address, 29 nodes
12 -> /28 for 16 address, 13 nodes
6 -> /29 for 8 address, 6 nodes
5 -> /29 for 8 address, 5 nodes

I corrected the line that uses 6 hosts that palladin added. There's no need to use a /28 for that subnet. There's exactly enough room for all the subnets and the required amount of hosts with that mask he gave you.

Hope this helps.

Read what he wrote very carefully. You can't fit 6 real hosts inside a /29 as you wouldn't be able to route to it at all.
You were thinking of a /29 as

Bit order
0 -> Network
1
2
3
4
5
6
7 -> Broadcast

Which is entirely correct from a pure networking point of view but wouldn't actually work as you'd run 1 address short.
What you actually need is

Bit order
0 -> Network
1 -> Router / Switch / Vlan Interface
2
3
4
5
6
7 -> Broadcast

Which only gives you 5 address's for machines on the network. The teacher threw it in to see who would catch that. He can ask his teacher to give more guidance on whether he meant to include the Router/Switch/Vlan IP address inside hosts or not. Using classless routing you can break an IP range into as many pieces as you need, just ensure you include the additional three required ones or you'll catch yourself short.

I see what you're saying and I understand. However, when I was in school, we never took the router interface into consideration. I'm not sure which is the case here, but based on your allotted subnets and hosts, it would not work with a .128 subnet mask as the OP stated, you'd go into the 128 network.

Also, on the Cisco certifications, when they ask you to assign a certain number of hosts to a subnet, it's simply that. You leave out the network and broadcast address but the router interface isn't taken into consideration. The more I think about it, I'm quite certain this is the case. Good luck to you both!

And this is the case from a networking point of view, but from a systems point of view you must include the "router" as one of the hosts or you'll find yourself short. All too often I've had a network admin cut a subnet for me and ended up being out by one because he didn't think to add the interface IP. Its also why I said to ask the teacher to clarify, whenever I see a teacher list a network needing 5 and 6 "hosts" it's always been as a practice to see if the students are actually thinking real world implementation. You are right that you can't fit all those nodes practically into 5 subnets inside of a /25.

128
->64 (40)
->64 SN
->32 (28)
->32 SN
->16 (12)
->16 SN
->8 (5)
->8 (6) <=Short 1 or is unreachable=>

Is most likely what the teacher wants, but it wouldn't actually work in the real world. Maybe get bonus points for bringing that up in class?

Agreed! And good point... bonus points are always good!

I think I got my answer. Thank you all for the discussion.