Two subnetting problems:

1 - Given the following address: 10.100.48.27/19 what is the valid host range?

2 - Given the following address: 172.16.10.22 255.255.255.240 what is the valid host range?

Just trying to confirm my own approach and answers.

Just trying to confirm my own approach and answers.

1 - /19 indicates two things: interesting octet is the third, and increment size is 32.

So the network address should be 10.100.32.0 with usable host range of 10.100.32.1 - 10.100.63.254 and broadcast of 10.100.63.255

2 - Subnet mask of 255.255.255.240 is a /28 so interesting octes is 4th and increment size is 16.

Network address should be 172.16.10.16 with host range of 172.16.10.17 - 172.16.10.30 and broadcast of 172.16.10.31.

Is this correct, or am I missing something?

Thanks.

William

]]>If you can multiply by 2, you can write down a quick table to show you each segment.

I know that each octet is 8 bits. So, I write down an 8 bit decimal equivalent:

128 64 32 16 8 4 2 1

Now, when I see a /19, I know that there are all 1's in the first 2 octets (16 places), and all 0's in the last octet (places 25 through 32). The third octet is where I need to find the network number, and I do that by deciding how many bits to use, starting from the left (19 minus 16). That leaves me 3 bits for the third octet, which is 128 + 64 + 32 = 224. So, my decimal mask is 255.255.224.0 if I need that to answer an exam question. If I'm given the decimal mask instead of the network mask bitcount, I just convert it to binary as usual.

The subnet you gave, 10.100.48.27/19 uses the first 3 bits of the third octet as a mask. 48 is (0 0 1) (1 0 0 0 0) - you should already know how to convert a decimal number to binary using the decimal representation of 8 bits above. Just using those first 3 bits, I have a beginning network address of 32. The rest of the bits are assumed to be zero in the network address. 10.100.32.0/19 is the network. To find the end of that network segment, set the remaining 5 bits to 1 and add them to the beginning of the network address: 16 + 8 + 4 + 2 + 1 = 31 (I know this without adding because it will be 1 less than the next higher binary place (32)). I add 32 (the network octet) and 31 (the all 1's for the host addresses) and I get 63 (again, one less than the next higher binary place, so you can do this in your head without adding). The last possible address in the network (broadcast address) is 10.100.63.255/19 (all 1's after the last network bit, so the fourth octet is 255).

This shows you one process for finding the network and broadcast addresses. It's how I do it in my head. The host addresses are simply all addresses except the first and last within this range. As long as you can do a quick decimal to binary conversion using the power of 2 table: 128 64 32 16 8 4 2 1, then you're all set and can do each of these in mere seconds. Keep in mind the binary math trick that adding all bits to the right of any place in binary is equal to the next higher place minus 1 and you'll never have to add together the least significant bits of a binary number to convert it back to decimal.

]]>Both answers look good, remember the key for preparing for the exam is to look get these answers in about a minute per question.

]]>These problems were NOT taken from the CCENT EXAM. They do pertain to CCENT material.

Thanks,

William

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