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March 24, 2016 | Views: 19811

Let’s begin.

Steps:

- Write powers of 2 left to right
- Write CIDR subnet masks from 1 to 32. This will take 4 lines, one for each octet in the IP addresses
- Write the subnet mask: add powers of 2 from left to current column

You can invert 2 and 3 if you think it’s more logical (I’m used to this way).

You an add the number of usable IP by doing (power of 2) -2

A more mathematical way to express the the subnet mask calculation is: current value = value on the left (previous value) + power of 2 of this column.

It should take about 1 minute to do this when the exam starts. It will be extremely useful all through the exam to double check any IP and subnet. The only missing part would be a list of possible subnet values (i.e 0, 16, 32, … for a /28 CIDR), but I found it’s best to do this when you need it rather than pre-calculate it. It’s not needed that often during the exam, as most of the exam is being “nice” with you and gives you easy values such as 0, 128, 64, …

You should be able to generate this table:

2^{x} |
128 |
64 | 32 | 16 | 8 | 4 |
2 |
1 |

1st octet | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |

2nd | 9 | 10 | 11 | 12 |
13 | 14 | 15 | 16 |

3rd | 17 | 18 | 19 | 20 | 21 | 22 | 23 | 24 |

4th | 25 |
26 | 27 | 28 | 29 | 30 |
31 | 32 |

mask | 128 |
192 | 224 | 240 |
248 | 252 | 254 |
255 |

A few examples of how to use this table and concept:

*– Find the last usable IP in the subnet 172.16.1.0/30*

/30 gives a 4 total IP’s (0 to 3), meaning 2 usable IPs (1 and 2), 0 is the subnet and 3 is the broadcast. Answer is 172.16.1.2

*– Can 192.168.15.23 use 192.168.16.35 as default gateway if subnet is 255.255.254.0?*

254 as a mask gives us 2 IP in the third octet, so:

- 192.168.0.x and 192.168.1.x are in the same subnet
- 2.x and 3.x
- 4.x and 5.x
- 6.x and 7.x

The answer is “no,” as both IP’s are not in the same subnet.

*– What is the CIDR subnet mask for /12?*

The answer is 255.240.0.0

*– What mask do you need to have 64 hosts in a subnet?*

64 total IP’s won’t be enough (don’t forget broadcast and subnet IP’s remove usable IP’s from the total). You need to choose a subnet that can accommodate 126 hosts (128-2) hence a mask of 255.255.255.128, or /25

Thanks and I hope this is helpful to you.

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Of all the charts I have seen, this one is the most practical. You essentially have all 4 octets which is perfect. Thanks!

Of all the charts I have seen, this one is the most practical. Thanks!