Ok they have given us the IP address which is: 10.11.12.13
Next. At first glance, it just looks like all we have is an IP address but, they provided the subnet for us. The subnet has 5 bits. The IP address is a class A network (look at the bottom for a reference) which is 10.0.0.0 | 255.0.0.0 or 11111111.00000000.00000000.00000000. Since the subnet has 5 bits, that would make it 255.248.0.0 or 11111111.11111000.00000000.00000000 (I underlined the 5 bits that were added. Also scroll to the bottom for a quick review on binary). So now we have our subnet mask. Now to find out what the network address is, we can do a couple things. We can just subtract 248 (255.248.0.0) from 256 which = 8 or you can do that ANDing process where you take the IP address and subnet in binary form and you stack them on top of each other and wherever there is a matching 1, you copy that down as a 1. Anything else would equal a 0. For example:
10.11.12.13 = 00001010.00001011.00001100.00001101
255.248.0.0 = 11111111.11111000.00000000.00000000
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10.8.0.0 .....= 00001010.00001000.00000000.00000000
Here are the network ranges:
10.0.0.0 - 10.7.255.255
10.8.0.0 - 10.15.255.255
10.16.0.0 - 10.23.255.255
10.24.0.0 - 10.31.255.255
etc...
If you notice, there is a pattern. They are going up by 8 (the number we got earlier). Since the network on the exam is 10.11.12.13, there is only 1 network from that list that fits this, which is: 10.8.0.0. (we already know this from earlier)
Blue = Network bits
Red = Subnet bits
Green = Host bits
11111111.11111000.00000000.00000000 (8 network bits, 5 subnet bits, 19 host bits = 32 bits, which is what an IP (version 4) is made up of)
Subnet mask = 255.248.0.0
Subnet Bits = 5
Network bits = 8
Network address = 10.8.0.0
Broad cast address = 10.15.255.255
Network range = 10.8.0.0 - 10.15.255.255
Number of addresses = 524288 (we get that by doing: 2^19. Remember that the 0's represent host address. Since there are 19 0's, we do 2^19.)
Host bits = 19
Host range? 10.8.0.1 - 10.15.255.254 (We can't use the first and last IP address in the network. 10.8.0.0 and 10.15.255.255)
Default gateway = Anything between 10.8.0.1 - 10.15.255.254 (as long as it is not being used)
If you aren't familiar with the binary, here is how I got the values. Remember how an IP address is setup:
128|64|32|16|8|4|2|1 . 128|64|32|16|8|4|2|1 . 128|64|32|16|8|4|2|1 . 128|64|32|16|8|4|2|1
In binary, which ever slot has a 1 in it, then you take that value and add it with any other value within the same octet. For example:
10.8.0.0 | 00001010.00001011.00000000.00000000
10 = 00001010 (There is a 1 in the 8 and 2 slot so you add that and it makes 10)
8 = 00001000 (just an 8. No addition needed)
0 = 00000000 (no 1's)
0 = 00000000 (no 1's)
Another example: 192.168.1.1 | 11000000.1010100.00000001.00000001
192 = 11000000 (128 + 64 = 192)
168 = 10101000 (128 + 32 + 8 = 168)
1 = 00000001 (No addition needed)
1 = 00000001 (No addition needed)
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Class A
0.0.0.0 - 127.255.255.255
Class B
128.0.0.0 - 191.255.255.255
Class C
192.0.0.0 - 223.255.255.255
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