vlsm issue

hello. i have a question on vlsm and i just answered of the questions but got it wrong. i looked at the answer but it doesnt really make sense to me. can someone plz explain this. the parts that i dont understand is BOLDED and UNDERLINED. i dont understand how they came about to this answer.

You are configuring the IP addresses on the SFO router FastEthernet interfaces. Subnet 1 (connected to FastEthernet0/0) has 672 hosts and Subnet 2 (connected to FastEthernet0/1) has 258 hosts. Your task in this lab is to subnet the 172.20.0.0 network to provide sufficient host addresses for both subnets while minimizing wasted IP addresses.

Complete the following tasks:

Choose one subnet mask for each subnet. The masks should provide for enough host addresses without wasting IP addresses.
Assign IP addresses and correct subnet masks to the SFO interfaces connected to the switches (in this order):
1. For FastEthernet0/0, use the third possible subnet and the last address on the subnet.
2. For FastEthernet0/1, use the next available subnet that does not overlap and the last address on the subnet.
Enable both interfaces.
Save your changes to the startup-config file.

and here is the answer

to determine which subnet mask will work for the 172.20.0.0 network, first look at the number of hosts required for each subnet:

Subnet1 (connected to FastEthernet0/0) has 672 hosts. To support 672 hosts, a subnet mask of /22 is required (10 host bits in the 2n-2 formula will afford 1022 host addresses in the subnet).
Subnet2 (connected to FastEthernet0/1) has 258 hosts. To support 258 hosts, a subnet mask of /23 is required (9 host bits in the 2n-2 formula will afford 510 host addresses in the subnet).
With a network address of 172.20.0.0 and the masks needed to fit the requirements, you need to configure the following IP address and subnet masks:

For the FastEthernet0/0 connection:
172.20.8.0/22 is the third possible subnet. (172.20.0.0/22 is the first possible subnet and 172.20.4.0/22 is the second possible subnet.)
172.20.11.254 is the last possible IP address in the subnet.
255.255.252.0 is the decimal version of a 22-bit mask.
For the FastEthernet0/1 connection:
172.20.12.0/23 is the next available subnet that does not overlap.
172.20.13.254 is the last possible IP address in the subnet.
255.255.254.0 is the decimal version of a 23-bit mask.

how is 172.20.8.0 / 22 the third possible subnet. isnt it 172.20.6.0 the third possible subnet?

thanks

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Comments

CodeBlox

The third subnet is indeed 172.20.8.0 /22

Your "special octet" increments by 4 at a time so the subnets are as follows:

172.20.0.0
172.20.4.0
172.20.8.0
172.20.12.0
172.20.16.0
172.20.20.0
172.20.24.0
...
...
...
172.20.248.0
172.20.252.0

boredgamelad

A good question to ask here would be: how did you come up with your answer of 172.20.6.0? Maybe we can help you understand the flaw in your reasoning.

saied

wow im an idiot and now i know why. for some reason i counted 22 ending in the number 2 and not 4. i guess i studied too much today.
confusion fixed. thanks guys.

RouteMyPacket

It's a good feeling to see a "problem" like this and immediately see how easy it is to VLSM.

It's easy simply because it breaks down like this

672 hosts needed? Ok, well a /24 won't work, a /23 won't either, well a /22 will but wait I am technically "wasting" addresses but there are no other options so /22 it is. 258 hosts next? Crap, can;t they just fire a few people and let me use a /24? No? Ok, well a /23 it is then.

VLSM is awesome and following this basic rule and you're golden, ALWAYS start with the largest network first. Repetition will have you answering questions like this one without even thinking. Cheers!