Student question stumped the class, unable to drive the nail home for this question.

Okay, let's imagine this like a big game of organizing candy!


You have 8 different boxes of candy. Each box has its special number, like 192.168.0, 192.168.1, and so on, up to 192.168.7.


Now, your mom wants to put all these boxes into one big container. She needs to find a way to label this container so that it includes all the smaller boxes inside.


The trick is to look at the numbers and see what they all have in common. In this case, they all start with 192.168, right? That's like saying all the candy is chocolate!


But we need to be a bit more specific. We notice that the last number changes from 0 to 7. In computer language, this is like saying we need the first 21 bits to be the same (because 21 bits can describe all numbers from 0 to 7).


So, the label for the big container (what grown-ups call a "supernet route") would be:

• Network prefix: 192.168.0.0
• Subnet mask: 255.255.248.0

This is like saying "This big box contains all the chocolate candy numbered from 0 to 7!"

Brolen Gumb said:

In CompTIA Network+ N10-008 Student Guide, Pg. 163. or Instructor Guide Pg. 168.
Question 5 states:
5. A company has eight networks, using the subnet addresses 192.168.0.0/24,
192.168.1.0/24 … 192.168.7.0/24. What network prefix and subnet mask can be used to summarize a
supernet route to these networks?

Answer:
It takes 3 bits to summarize eight networks (23 = 8). Subtracting 3 bits from the existing network
mask makes the supernet network prefix /21. The third octet of the mask will use 5 bits, which is
248 in decimal (25 = 248), so the full mask is 255.255.248.0.

This class is unable to grasp this concept, & I am running out of ways to explain & time to stay on track to finish class where I need to for the day. Any help explaining is greatly appreciated.

Click to expand...

For my classes, they understand it better when they see the bits in a table.
I hope this helps.

In addition to Jarrel's table I also explain it as
00000000
00000|000 (borrowed the 3 digits)
128+64+32+16+8|4+2+1 or 0
248 | 7+(the zero network) or 2^3
248 | 8

When I am teaching subnetting and binary, I use/build the same chart for them. I use WORD, and build a table for only 8 bits at a time.
https://filebin.net/9r05cq1z4uuc4xor I uploaded the file to filebin, that I usually use (its a .docx) I will build my chart in front of them. I will explain binary to them and color coordinate which is the Network bits, the Host bits, and the subnet (borrowed) bits. I have been able to get my point across to everyone so far.
There is some left over writings on their from my last class, you should be able to piece together, what I was doing from those remnants.
I hope this at least helps in some small way.
FYI, that filebin link is only good for 7 days?
Regards,
Curtis

Brolen Gumb said:

In CompTIA Network+ N10-008 Student Guide, Pg. 163. or Instructor Guide Pg. 168.
Question 5 states:
5. A company has eight networks, using the subnet addresses 192.168.0.0/24,
192.168.1.0/24 … 192.168.7.0/24. What network prefix and subnet mask can be used to summarize a
supernet route to these networks?

Answer:
It takes 3 bits to summarize eight networks (23 = 8). Subtracting 3 bits from the existing network
mask makes the supernet network prefix /21. The third octet of the mask will use 5 bits, which is
248 in decimal (25 = 248), so the full mask is 255.255.248.0.

This class is unable to grasp this concept, & I am running out of ways to explain & time to stay on track to finish class where I need to for the day. Any help explaining is greatly appreciated.

Click to expand...

I am impressed about this

Brolen Gumb said:

In CompTIA Network+ N10-008 Student Guide, Pg. 163. or Instructor Guide Pg. 168.
Question 5 states:
5. A company has eight networks, using the subnet addresses 192.168.0.0/24,
192.168.1.0/24 … 192.168.7.0/24. What network prefix and subnet mask can be used to summarize a
supernet route to these networks?

Answer:
It takes 3 bits to summarize eight networks (23 = 8). Subtracting 3 bits from the existing network
mask makes the supernet network prefix /21. The third octet of the mask will use 5 bits, which is
248 in decimal (25 = 248), so the full mask is 255.255.248.0.

This class is unable to grasp this concept, & I am running out of ways to explain & time to stay on track to finish class where I need to for the day. Any help explaining is greatly appreciated.

Click to expand...

This is how my brain works it out --

A /23 has two /24's
A /22 has two /23's (4 /24s)
A /21 has two /22 (8 /24's)

I make it a point to say you will only 255, 254, 248, 240, 224, 192, 128, or 0 in a subnet mask (a fast way to eliminate wrong answers on a Cisco exam).

For students, I use the full dotted quad, for me I just use the last 9 values

/32 255.255.255.255
/31 255.255.255.254
/30 255.255.255.252
/29 255.255.255.248
/28 255.255.255.240
/27 255.255.255.224
/26 255.255.255.192
/25 255.255.255.128
/24 255.255.255.000

/23 255.255.254.000
/22 255.255.252.000
/21 255.255.248.000
/20 255.255.240.000

/32 255
/31 254
/30 252
/29 248
/28 240
/27 224
/26 192
/25 128
/24 0

/23 254.0
/22 252.0
/21 248.0
/20 240.0

Andrew H said:

This is how my brain works it out --

A /23 has two /24's
A /22 has two /23's (4 /24s)
A /21 has two /22 (8 /24's)

I make it a point to say you will only 255, 254, 248, 240, 224, 192, 128, or 0 in a subnet mask (a fast way to eliminate wrong answers on a Cisco exam).

For students, I use the full dotted quad, for me I just use the last 9 values

/32 255.255.255.255
/31 255.255.255.254
/30 255.255.255.252
/29 255.255.255.248
/28 255.255.255.240
/27 255.255.255.224
/26 255.255.255.192
/25 255.255.255.128
/24 255.255.255.000

/23 255.255.254.000
/22 255.255.252.000
/21 255.255.248.000
/20 255.255.240.000

/32 255
/31 254
/30 252
/29 248
/28 240
/27 224
/26 192
/25 128
/24 0

/23 254.0
/22 252.0
/21 248.0
/20 240.0

Click to expand...

What did you what is to learn from this
So that we can achieve from it.

I would try to do it a bit more visually & more from the bottom up:
1 - Ask what is in common between the 8 networks - hopefully get 195.168.
2-then Ask where (which octet) is the change taking place? Hopefully get answered octet 3.
Then ask what are the values in octet 3? Hopefully get answered 0 thru 7.
3- NOW ask: Ok, can you count in Binary from 0 to 7 using 8 bits?
Hopefully get this response:

• 0: 00000000
• 1: 00000001
• 2: 00000010
• 3: 00000011
• 4: 00000100
• 5: 00000101
• 6: 00000110
• 7: 00000111

4-Now ask: What do you see about the bits in these binary numbers?
--HINT: How many bits are common across all 7 addresses/numbers?
Hopefully you hear the answer "5 zeros in common".
Now ask so how many of those 5 zeros are part of the network address? -Hopefully hear the answer of 5
so then how many 1's would that be if the subnet is in binary? hopefully also hear the number 5

5- Exactly - So how many ones were in the 192.168 portion already? 16 -- and how many now when you add 5 more? 21
Perfect - THAT is your subnet mask in CIDR: /21

6-Now ask, -what does that means in BINARY for the submask in octet 3? hopefully someone says "5 ones & 3 zeros".
(or if you have beeen showing the entire 11111111. 11111111. 11111000 .00000000
- so the subnet mask bits for the 3rd octet are 1111 1000.
Now ask, what is that value in Dotted decimal notation of 11111000? 128+64+32+16+8=248
So the subnet mask is 192.168.248..0
7- But what about the summary network number? Look at the 5 network bits - since we only used up to 7 networks,
all of the first 5 network bits are 0 (since even for network 7 it would be 0000 0111 - 7.
so 0+0+0+0+0 = 0 that wan't too bad.
Your final answer so your summary network is 192.168.0.0/24