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Subnetting

cisconoval
Beginner
Beginner

Hello,

How the following IP subnets pinging even its in different subnets?

172.16.35.20 255.255.192.0

172.16.40.25 255.255.224.0

PC1 having 172.16.35.0/18

PC2 having 172.16.40.0/19

both PCs connected on same switch. Please tell me how its working?

Thanks in Advance

8 Replies 8

network.king
Enthusiast
Enthusiast

Hi

Both the subnets fall under the supernet segment of 172.16.0.0/18 , so PC1 would broadcast and find out the mac-address and communicate .

When PC2 wants to communicate , ip segment falls in 172.16.32.0/19 segment , here again both the ips fall under the supernet , so it would do a broadcast and would communicate.

Hope this helps.

regards

vanesh k

If so then what is the uses of subnetting....

it shouldn't accept the either of the subnet on same lan right?

Hi, cisconoval

A good subnetting design will not allow addresses overlapping which is happenning in your case.The subnets 172.16.0.0/18 (not 172.16.35.0/18) and 172.16.32.0/19 (not 172.16.40.0/19) have overlapping addresses ranging from 172.16.32.1 to 172.16.63.254 which both PC1 and PC2 fall under.

You think PC1 and PC2 are in different subnets, but they don't think so by using their own algorithm.

PC1's will apply its own mask (255.255.192.0) to PC2's address (172.16.40.25) and find out it's on the same subnet(172.16.0.0/18) as its own. PC2's will apply it own mask (255.255.224.0) to PC1's address (172.16.35.20) and find out it's on the same subnet(172.16.32.0/19) as its own. That's the reason why they think they can talk to each other.

Try to change the subnet mask to 24 on both PCs, then you will have 2 totally different subnets: 172.16.35.0/24 and 172.16.40.0/24 without any overlapping addresses. You can put them on the same lan, but they will not be able to talk to each other anymore.

Hope this help

SSLIN

Hi All,

Now am clear on this..thanks for your suggestions...

CSCO10892433
Enthusiast
Enthusiast

Hi, cisconoval

The subnets you provided are not real subnets. In fact, they are host addresses. Let's analyze them as following:

172.16.35.0/18

network address 172.16.0.0

broadcast address 172.16.63.255

host address range: 172.16.0.1 ~ 172.16.63.254 (PC1 and PC2 are both in this range)

172.16.40.0/19

network address 172.16.32.0

broadcast address 172.16.63.255

host address range 172.16.32.1 ~ 172.16.63.254 (PC1 and PC2 are also in this range)

So, actually PC1 and PC2 are in the same subnet. You can say they are both in 172.16.0.0/18 (from PC1's ponit of view) or you can say they are both in 172.16.32.0/19 (from PC2's ponit of view)

Hope this help.

SSLIN

keduncan
Beginner
Beginner

PC1 subnet range is 172.16.32.0 - 172.16.96.255

PC2 subnet range is 172.16.32.0 - 172.16.64.255

IP addresses 172.16.35.20 and 172.16.40.25 fall into the ranges of both. As long as both PC's are using 255.255.255.255 as their broadcast address, the ARP packets will be answered and the PC's will communicate.

If they are using 172.16.64.255 and 172.16.96.255 as their broadcast addresses you might have problems

Hi, Kennith

Thank you for your revise, but I have to say I disagree with you about the host range and ARP.

The PC1's address is 172.16.35.20 which has binary form of

10101100 00010000 00100011 00010100.

The subnet mask is /18, means keeping the first 18 bits unchanged and filling 14 0s to the last 14 bits will derive the subnet address. We can derive it here as

10101100 00100000 00000000 00000000

which has decimal form of 172.16.0.0. The broadcast address is the same subnet address filled with 14 1s at the last 14 bits , we can derive it here as

10101100 00100000 00111111 11111111

which has decimal form of 172.16.63.255. The host range in this subnet is thus 172.16.0.1~172.16.63.254

The PC2's address is 172.16.40.25 which has binary form

10101100 00010000 00101000 00011001.

The subnet mask is /19, means keeping the first 19 bits unchanged and filling 13 0s to the last 13 bits will derive the subnet address. We can derive it here as

10101100 00100000 00100000 00000000

which has decimal form of 172.16.32.0. The broadcast address is the same subnet address filled with 13 1s at the last 13 bits , we can derive it here as

10101100 00100000 00111111 11111111

which has decimal form of 172.16.63.255. The host range in this subnet is thus 172.16.32.1~172.16.63.254

As for the ARP, I'm afraid it does not use the Layer3 broadcast address 255.255.255.255 but the Layer2 address ffff.ffff.ffff. So, whether they have the same Layer3 brocast address is not relevant for this case

Please check them again and let me know I'm wrong.

Thanks

SSLIN

Yes, I messed it up

PC1 subnet range is 172.16.0.0 - 172.16.63.255

PC2 subnet range is 172.16.32.0 - 172.16.63.255

Both are on the same subnet.. You will have problems if you use addresses 172.16.0.1 -172.16.31.255..

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