Subnet Masking - Is the following calculations right

Question

Had my own self notes which has gone missing in the worst of times. Desperately trying to figure out if I have got the calculations right for subnet masking in the following questions. I have answered them all. Just wish to be sure I got it right.

It is not a help with some outstanding homework. It is for my upcoming exams. Really appreciate any help. Thanks.

Need 5 subnets for class C network. starting ip address is 192.168.0.0
subnet mask is 255.255.255.0

Questions:
1. How bits are borrowed from to make the subnet? 
ANS: Borrowed from the host portion of the IP address. In this case we need 5 subnets so it 
will be 2 to the power of 3 (Cos 2 to the power of 2 only gives 4 subnets - not enough).

2. How many subnets are now available?  
ANS: 8 subnets are available. 2 to the power of 3 is 8.

3.What is the new subnet mask? 
ANS: 255.255.255.224

4.How many host addresses available in each subnet?
ANS: 8

5.How many host addresses can be used in each subnet and why?
ANS:   7.      192.168.0.0 is reserved. 

6.What is the IP address of each subnet?
192.168.0.0
192.168.0.32
192.168.0.64
192.168.0.96
192.168.0.128
192.168.0.160
192.168.0.192
192.168.0.224

Answer 1

This is really more suited for other StackExchange sites (like ServerFault), but I can help.

1) Correct. Although you may want to specifically mention that since 2^3 (2 to the power of 3) gives you enough subnets (8), then you borrow 3 bits from the host portion of the address. So the new subnet mask is /27 (24+3), or, as you correctly mention in Q3, 255.255.255.224.

2) Correct.

3) Correct. May also be noted as /27.

4) Wrong. Since from the last octet you borrowed 3 bits for subnetting, you only have the last 5 bits for host addresses. That gives you 2^5 = 32 host addresses in each subnet.

5) Wrong. Out of the 32 addresses available, the first from each subnet is reserved as the subnet's network address, and the last one from each subnet is reserved as the subnet's broadcast address. Therefore, you're left with 30 addresses (2^5 - 2) you can actually use in each subnet. For example, in the first of the subnets, 192.168.0.0 is the network address and 192.168.0.31 is the broadcast address. 192.168.0.1 through 192.168.0.30 are usable.

6) Correct.

Let me know if that helps!