Could one of you advise on this please? Got an IP address of 192.168.128.0 and per question need 1022 hosts (no more allowed) Possible answers are /10, /20, /22 and /24.

I didn't really know the answer on the test exam. I eliminated /24 as I know that gives you 254. I can now see (looking at my notes) that it's not /10 (didn't think it was anyway) because that can only be used with a Class A IP address (1-127 - 127 reserved for loopback) So, the answer is coming from /20 and /22.

I know the formula now is 2 to the power of x minus 2.

So 2 to the power of 10 minus 2 = 1022. Therefore I need 10 bits for the hosts portion leaving 22 bits left over for the network address = /22. Is this the right answer though? Does the fact that the IP address is 192.168.128.0 (and not 192.168.0.0) have any bearing on the answer?

I've tried working it out by looking at the IP address 192.168.128.0 and subnet mask 255.255.252.0 to determine the range of IP addresses that could be used but it baffles me TBH. I know if the subnet mask was 255.255.255.0 then that would be 254 hosts and each host has to have its own unique IP address (ie 192.168.0.1 through 192.168.0.254) If you could ONLY use 192.168.0.128 upwards then you'd only have 126 unique IP addresses/hosts.

Am I just overthinking this and the answer is 192.168.128.0/22? Cheers!

]]>Could one of you advise on this please? Got an IP address of 192.168.128.0 and per question need 1022 hosts (no more allowed) Possible answers are /10, /20, /22 and /24.

I didn't really know the answer on the test exam. I eliminated /24 as I know that gives you 254. I can now see (looking at my notes) that it's not /10 (didn't think it was anyway) because that can only be used with a Class A IP address (1-127 - 127 reserved for loopback) So, the answer is coming from /20 and /22.

I know the formula now is 2 to the power of x minus 2.

So 2 to the power of 10 minus 2 = 1022. Therefore I need 10 bits for the hosts portion leaving 22 bits left over for the network address = /22. Is this the right answer though? Does the fact that the IP address is 192.168.128.0 (and not 192.168.0.0) have any bearing on the answer?

I've tried working it out by looking at the IP address 192.168.128.0 and subnet mask 255.255.252.0 to determine the range of IP addresses that could be used but it baffles me TBH. I know if the subnet mask was 255.255.255.0 then that would be 254 hosts and each host has to have its own unique IP address (ie 192.168.0.1 through 192.168.0.254) If you could ONLY use 192.168.0.128 upwards then you'd only have 126 unique IP addresses/hosts.

Am I just overthinking this and the answer is 192.168.128.0/22? Cheers!

]]>1024 (possibilities)

1022 (addressable, minus Net ID and broadcast)

So 2 to the power of 10 minus 2 = 1022. Therefore I need 10 bits for the hosts portion leaving 22 bits left over for the network address = /22. Is this the right answer though?

So this just looks weird because it's a class C address. Rest assured when dealing with a /22, the ranges will look a little strange. so for 192.168.128.0/22:

- The range itself : 192.168.128.1 - 192.168.131.255
- The broadcast: 192.168.131.254

Does the fact that the IP address is 192.168.128.0 (and not 192.168.0.0) have any bearing on the answer?

Not really. If anything it's just strange where the range looks weird when it crosses beyond a single number (e.g., 128) to extend beyond that (e.g., 131).

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