Current broadcast address+1 for next subnet network address?

[shivajikobardan]

Homework Statement

subnetting problem

Relevant Equations

none

https://www.cisco.com/c/dam/en_us/training-events/netacad/demos/CCNA3v3Demo/PDF/lab_1_1_4.pdf

This is the pdf I am studying from. This is not exactly what I need for my exams but it’s pretty clearly written and concept giving thing.

192.168.10.0/24 is given IP address to subnet.
We need 4 subnets
So
192.168.10.00 00 0000
This is an ip address that has 3 octets of network address and 1 octet of host address. So a class C address.
For 4 subnets, i need 2 bits of host to be borrowed.
00 00 0000
01
10
11

So I get these IP address for subnets-:
Subnet #0->192.168.10.0/26
Subnet #1->192.168.10.64/26
Subnet #2->192.168.10.128/26
Subnet #3->192.168.10.192/26

A) Take subnet #0. Perth 60 hosts
The start address here is 192.168.10.0/26
192.168.10.00 000000

192.168.10.00 1111111

This gives

Network address=192.168.10.0 /26

Broadcast address=192.168.10.63 /26

B) Take subnet #1-:
Kuala Lampur 28 hosts.

Next subnet IP address was 192.168.10.64/26

Question-: Why was this the next subnet IP address?

1. Is it becuase of this?

So I get these IP address for subnets-:

“Subnet #0->192.168.10.0/26
Subnet #1->192.168.10.64/26
Subnet #2->192.168.10.128/26
Subnet #3->192.168.10.192/26”

1. Is it because the next consecutive ip address after the perth’s broadcast address is 192.168.10.64 /26?

28 hosts required ie 5 host bits. I.e 27 network bits.
192.168.10.010 00000
192.168.10.010 11111

I.e
Network address=192.168.10.64/27

Broadcast address=192.168.10.95/27

C) Syndey 12 hosts.
Now what will be the ip address to use? The answer is 192.168.10.96/27
Question I’ve is-:
Why did we calculate those 4 subnets

“So I get these IP address for subnets-:
Subnet #0->192.168.10.0/26
Subnet #1->192.168.10.64/26
Subnet #2->192.168.10.128/26
Subnet #3->192.168.10.192/26”

If we are just calculating the next consecutive ip address and using it for the next subnet? I don’t really understand this.
So for sydney we require 4 host bits.
192.168.10.1000 0000
192.168.10.1000 1111

Network address=192.168.10.96/28

Broadcast address=192.168.10.111/28

D) Singapore 12 hosts

The ip we are using is the next consecutive IP address 192.168.10.112/28. It does make some sense as I feel this would save a lot of IP space than the firstly calculated 4 subnets. But I am still confuse because if we ignore that step I have quoted twice, how would we know if we could make the required number of subnets with this ip address, i know i am not asking clear question but that is confusing me. do we really need to calculate that quoted part? or just randomly keep on adding 1 to broadcast address and keep solving? Because I assume that it is assuming same number of host bits for everyone thus wasting lots of IP addresses.
192.168.10.0111 0000
192.168.10.0111 1111

Network address=192.168.10.112/28

Broadcast address=192.168.10.127/28

Finally for WANs-:

I don’t need to learn for the WAN according to my requirements. But just for knowledge, why just only 3 subnets?
It says perth,syndey and singapore has a WAN connection to kuala lampur.
Why 3 subnets?

 

[Andrew Mason]

Conceptually, there would be four subnets created, each able to serve 62 hosts. Because 3 of the locations can fit into one subnet (52 hosts required), only two of these subnets are used (192.168.10.0/26 and 192.168.10.64/26). The first is used for Perth and the second is divided into two sub-subnets (192.168.10.64/27 and 192.168.10.96/27). The first of those is used for KL and the second sub-subnet is divided into two sub-sub-subnets (192.168.10.96/28 and 192.168.10.112/28). The first sub-sub-subnet is allocated to Sydney and the second to Singapore.

I am not sure why they did it this way. It may be to accommodate more locations. But if any location needs another 3 host addresses you have to reconfigure the other locations.

To accommodate future expansions, it might be better to divide into two subnets ie. 192.168.10.0/25 and 192.168.10.128/25 and give the first to Perth (using only half eg. 192.168.10.0/26). Then divide the second subnet into two sub-subnets ie. 192.168.10.128/26 and 192.168.10.192/26 and give the first to KL (using only half eg. 192.168.10.128/27). Then divide the second into two sub-sub-subnets ie. 192.168.10.192/27 and 192.168.10.224/27 for each of Sydney and Singapore (using only half for each eg. 192.168.10.192/28 for Sydney and 192.168.224/28 for Singapore)

Regardless of how you allocated the subnets, you would use the mask 255.255.255.192 for the Perth subnetwork, the mask 255.255.255.224 for the Kuala Lumpur sub-subnetwork and the mask 255.255.255.240 for each of the Sydney and Singapore sub-sub-subnetworks.

The final part (3 not 4 WAN links) is not explained and expects you to know that each WAN router has to be on its own subnet. The KL router is the top level.

 

[shivajikobardan]

Conceptually, there would be four subnets created, each able to serve 62 hosts. Because 3 of the locations can fit into one subnet (52 hosts required), only two of these subnets are used (192.168.10.0/26 and 192.168.10.64/26). The first is used for Perth and the second is divided into two sub-subnets (192.168.10.64/27 and 192.168.10.96/27). The first of those is used for KL and the second sub-subnet is divided into two sub-sub-subnets (192.168.10.96/28 and 192.168.10.112/28). The first sub-sub-subnet is allocated to Sydney and the second to Singapore.

I am not sure why they did it this way. It may be to accommodate more locations. But if any location needs another 3 host addresses you have to reconfigure the other locations.

To accommodate future expansions, it might be better to divide into two subnets ie. 192.168.10.0/25 and 192.168.10.128/25 and give the first to Perth (using only half eg. 192.168.10.0/26). Then divide the second subnet into two sub-subnets ie. 192.168.10.128/26 and 192.168.10.192/26 and give the first to KL (using only half eg. 192.168.10.128/27). Then divide the second into two sub-sub-subnets ie. 192.168.10.192/27 and 192.168.10.224/27 for each of Sydney and Singapore (using only half for each eg. 192.168.10.192/28 for Sydney and 192.168.224/28 for Singapore)

Regardless of how you allocated the subnets, you would use the mask 255.255.255.192 for the Perth subnetwork, the mask 255.255.255.224 for the Kuala Lumpur sub-subnetwork and the mask 255.255.255.240 for each of the Sydney and Singapore sub-sub-subnetworks.

The final part (3 not 4 WAN links) is not explained and expects you to know that each WAN router has to be on its own subnet. The KL router is the top level.

this is confusing. but i value your input. thanks for the response. i got a feel of this. my conclusion now is that subnet bits hold no significance is vlsm/cidr compared to flsm.

 

[Andrew Mason]

this is confusing. but i value your input. thanks for the response. i got a feel of this. my conclusion now is that subnet bits hold no significance is vlsm/cidr compared to flsm.

The trailing bits in the mask, whether in VLSM or FLSM, have signficance. The sizes of the subnets in VLSN/CIDR or FLSM are determined by the number of trailing 0s (8- no of leading 1s) in the subnet mask (in binary). For a class C network (first 3 Octets fixed) if the number of trailing 0s in the last Octet is n, the subnet size is 2ⁿ (with a capacity of 2ⁿ-2 hosts).

So, for example, if the mask is 255.255.255.224 (e.g. Kuala Lumpur sub-subnet 192.168.10.96/27) there are five trailing zeros in the last octet in the binary mask (11111111.11111111.11111111.11100000) so the subnet size is 2⁵=32.

If the mask is 255.255.255.240 (e.g the Sydney sub-sub-subnet 192.168.10.96/28) there are four trailing zeros in the binary mask (11111111.11111111.11111111.11110000) so the subnet size is 2^4=16.

In FLSM, there is only one subnet mask, which determines the size of all subnets, which must be large enough to accommodate the largest subnet required. With FLSM there would be 4 subnets of 64 each. You would use subnet mask 255.255.255.192 for all sub-networks (in binary: 11111111.11111111.11111111.11000000) ie. subnet size = 2⁶=64 (six trailing zeros).