Hi, I'm studying to pass the 70-291 Microsoft certification. In the following exercise are the ip addresses class C ? Should we use supernetting or subnetting? thanks , Bruno Your ISP has assigned you 2 Class C network addresses, 131.107.10.0 and 131.107.11.0, to accommodate your network’s 400 hosts. Which network address and subnet mask (expressed as a network prefix) can you assign to this address space so that your routers and hosts view these 2 networks as a single network? a. 131.107.10.0/23 b. 131.107.11.0/24 c. 131.107.10.0/22 d. 131.107.11.0/22 Correct answer is a

These are two Class B network addresses . It may be a fault in your book. In this case you should use subnetting (create subnets for 131.107.0.0 /16. 131.107.11.0 /24 does not include the two network addresses = it is not the true response. For 131.107.10.0 /23, it includes the two network addresses so it should be the right answer. 131.107.10.0 /22 and 131.107.11.0 /22 = 131.107.8.0 /22. So as you see 131.107.10.0 /23 is enough to use and it is for that it is the correct response. This posting is provided "AS IS" with no warranties or guarantees , and confers no rights.

These are two Class B network addresses . It may be a fault in your book. In this case you should use supernetting. 131.107.11.0 /24 does not include the two network addresses = it is not the true response. For 131.107.10.0 /23, it includes the two network addresses so it should be the right answer. 131.107.10.0 /22 and 131.107.11.0 /22 = 131.107.8.0 /22. So as you see 131.107.10.0 /23 is enough to use and it is for that it is the correct response. This posting is provided "AS IS" with no warranties or guarantees , and confers no rights.

Hi, thanks for you reply. I was certain that there was a fault in the book! but thanks for your confirmation. in this case we use subnetting and not supernetting I think: default subnet mask for Class B address is 255.255.0.0 /16 then we are subtracting 6 bits from bits reserved to Host ID! then , answer a is the right answer not because 131.107.10.22/22 and 131.107.11.22 /22 aren't in the same subnet: 131.107.10.22 and 131.107.11.22 seem are on the same subnet: If you make an "and" between 252 and 10 and 252 and 11 the result is the same: 8 11111100 subnet mask 255.255.252 /22 OK 00001010 00001000 11111100 00001011 00001000 but with a subnet mask at 23 every subnet will have : 400 hosts : 400+1=401=100000001 then we need 9 bits for 400 hosts: 2^9-2=510 because a subnet mask at /22 should make: 10 bits left for ID host: 2^10-2= 1022 What do you thing about that? Regards,

Yes it is soubnetting. I made a fault when I was writing that I corrected now. We should create subnet for the 131.107.0.0 /16 network address. You're right for the calculation that you've done. When using /23, you've got 9 bits for hosts = 2^9-2 = 510 (510 > 400) When using /22, you've got 10 bits for hosts = 2^10-2 = 1022 (1022 > 400) So, here you can choose /22 or /23. The most optimized forumule is to use /23 so that there will not be a large part of addresses that is not used. This is why I said /23 is enough to use so you don't need /22 and that is why the correct response is a. For MOC, you may find some errors because when I was a trainee like you I found some of them. This posting is provided "AS IS" with no warranties or guarantees , and confers no rights.

Thanks for your reply! regards,