VLSM & Summarization help

Can anyone refer a resource to help me understand VLSM & Summarization? I already have Sybex 5th Edition & ExamCram...I'm just not getting it! Help!

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NightShade1

ill post you 2 tutorials for this in the next messages
One tutorial of subnettnig made by Spacyfreak and another of router summerazing tutorial made by myself.

NightShade1

Subnetting by Spacyfreak

You think Subnetting is a beast?
You think you have to be Superbrain to understand it?

You are wrong!

Here the step-by-step course.
After reading and some self-training, you should be able to fix Subnetting-Questions in CCNA Exam
without any problems in a snatch.
Relax!

What is a Subnetmask?

With Subnetmasks, we can divide an IP-Address in network-part and in host-part.
A given IP-Network can be divided in smaller parts. Each of this smaller parts is called a "Subnet".

If we for example have the network

192.168.10.0 255.255.255.0
We have here ONE Class C - network, with 253 useable IPs for Client-PCs.

The useable IP Range of this network is

192.168.10.1 - 192.168.10.254

The very last IP of each Subnet is called Broadcast-Address.
This address is in that example 192.168.10.255 and its NOT useable for host-pcs.

If we want to divide this network in two parts, we must use subnetting.

With Subnetmask 255.255.255.128 we would divide the network in two parts.

192.168.10.1 - 192.168.10.127

192.168.10.128 - 192.168.10.255

So in this example, BEFORE we had one big Network.
With the change of the Subnetmask we did divide it in two smaller networks.

First with Subnetmask 255.255.255.0 we had THIS network:
192.168.10.0 >>> This is the "Network-IP" which is NOT useable for Host-PCs

192.168.10.1
192.168.10.2
192.168.10.3
192.168.10.4
192.168.10.5
...
...
...
192.168.10.253
192.168.10.254
192.168.10.255 >>>This is the Broadcast-IP, which is NOT useable for Host-PCs

Now with Subnetmask 255.255.255.128 we have THIS two networks:

First Subnet:

192.168.10.0 >>> This is the "Network-IP" which is NOT useable for Host-PCs

192.168.10.1
192.168.10.2
192.168.10.3
192.168.10.4
192.168.10.5
...
...
...
192.168.10.125
192.168.10.126
192.168.10.127 >>>This is the Broadcast-IP, which is NOT useable for Host-PCs

Second Subnet:

192.168.10.128 >>> This is the "Network-IP" which is NOT useable for Host-PCs

192.168.10.129
192.168.10.130
192.168.10.131
192.168.10..132
192.168.10.133
...
...
...
192.168.10.253
192.168.10.254
192.168.10.255 >>>This is the Broadcast-IP, which is NOT useable for Host-PCs

The Subnetmask defines how big the subnet is.
That means - how many Client-PCs will have place in that subnetwork.

A Subnetmask of 255.255.255.0 means in binary

11111111.11111111.11111111.00000000

So, what do we see?

4 Blocks, divided with a ".". Each of these blocks is also called "octett". Because - each Block has 8 bits.

To be able to do subnet-calculation, we first must understand binary calculation.

Lets take the first block.

The first "1" stands for a 128.

The second "1" stands for a 64.

The third "1" stands for a 32.

The fourth "1" stands for a 16.

The fifth "1" stands for a 8.

And so on. That means:

11111111=255

11110000=240

11100000=224

If we see something like "/24", that means that 24 bits are set to "1", from the left side.

Examples:

/16 = 255.255.0.0 = 11111111.11111111.00000000.00000000

/20 = 255.255.240.0 = 11111111.11111111.11110000

If we would take a subnetmask of 255.255.255.255 that would be

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

and in binary it would be

11111111.11111111.11111111.11111111

Calculation of Subnetmask big enough for a specified number of Hosts

If they ask..

"create a subnet with minimum 10 host IPs"

than

1. calculate a power of two, that is minimum 10

2^3=8. That is not enough
2^4=16 That is higher than 10. Good.

2. Now put the LAST 4 Bits of your subnetmask to 0.

11111111.11111111.11111111.11110000

That is in decimal

255.255.255.240

With THIS Subnetmask, you have minimum 10 Host-Ips in the Subnet, without wasting to much IP-Addresses.

Other example

If they ask

Create a subnet with minimum 70 Host-IPs

1. Calculate a Power of 2 that is MINIMUM 70

2^6=64. Not enough.
2^7=128. Thats higher than 70. Good.

2. Put the LAST 7 Bits of your Subnetmask to 0.

11111111.11111111.11111111.10000000

That is in decimal

255.255.255.128

You have a Subnetmask, with more than 70 Host-IPs.

Calculation what is the Broadcast-IP of a Subnet

When they ask
"There is subnet 172.16.64.0/20. What is the BROADCAST ADDRESS of that Subnet, dude?"

1. Step

/20 meens 255.255.240.0

2. Step

Now analyze the Subnet Oktett to find out the "network-jumps"

240 means 11110000

The LAST of the 1s is under decimal 16. That are our "network jumps"
(128/64/32/16/8/4/2/1)
3. Step

Write down the network-jumps

172.16.64.0 - 172.16.79.255
+16 172.16.80.0 - 172.16.95.255
+16 172.16.96.0 - 172.16.111.255
+16 172.16.112.0 - 172.16.127.255

Because the NEXT Subnet in the example is 172.16.80.0, the broadcast must be 172.16.79.255, cause THAT is the IP BEFORE the next Subnet starts = the BroadcastAddress.

Other example of Broadcast-IP calculation:

If it would be 172.16.64.0 /26

Same procedure

/26 means 255.255.255.192

192 is binary 11000000
The LAST 1 stands under the 64. That are in that example our "net-jumps".

172.16.64.0 - 172.16.64.63 <<<in this example THIS is the broadcastaddress of first subnet
172.16.64.64 - 172.16.64.127
172.16.64.128 - 172.16.64.191

Calculating first and last possible IP for a Host

You have Network 192.168.20.32 /27
The very first IP is reserved for Default Gateway!
What is the first and last valid IP for a Host-PC?

1.Step

/27 is 255.255.255.224

2.Step

224 means 11100000

The LAST 1 is under the 32. That are our "network-jumps" in this example
(128/64/32/16/8/4/2/1)

Valid IPs in that subnet:
192.168.20.33 - 192.168.20.62
(192.168.20.63 is NOT useable, this is the very last IP and so the BROADCAST-IP).
192.168.20.64 <<<this is the network-IP of the NEXT Subnet!

So, because the very first IP is reserved for Default Gateway, our first Host-PC IP would be
192.168.20.34
The very last Host-PC IP would be
192.168.20.62

###################################################################

Here some examples and step by step solutions:

Given that you have a class B IP address network range, which of the subnet masks
below will allow for 100 subnets with 500 usable host addresses per subnet?
A. 255.255.0.0
B. 255.255.224.0
C. 255.255.254.0
D. 255.255.255.0
E. 255.255.255.224

Solution:

Allways the same game...

Power of 2 that is minimum 500?
2^7=128
2^8=256
2^9=512 >>voila!

Now - put the last 9 Bits of your Subnetmask to "0"

11111111.11111111.11111110.00000000

That is in decimal
255.255.254.0

=====================================================================
If a host on a network has the address 172.16.45.14/30, what is the address of the
subnetwork to which this host belongs?
A. 172.16.45.0
B. 172.16.45.4
C. 172.16.45.8
D. 172.16.45.12
E. 172.16.45.18

Solution:

172.16.45.14/30

/30 means 11111111.11111111.11111111.11111100

The last of the ones stands under the "4". That is our increment or network jump.
172.16.45.0 - 172.16.45.3
172.16.45.4 - 172.16.45.7
172.16.45.8 - 172.16.45.11
172.16.45.12 - 172.16.45.15
172.16.45.16 - 172.16.45.19

As we see, the Ip is in the Range of 172.16.45.12 - 172.16.45.15.
So the network Address is 172.16.45.12

=================================================
QUESTION NO: 9
Which two of the addresses below are available for host addresses on the subnet
192.168.15.19/28? (Select two answer choices)
A. 192.168.15.17
B. 192.168.15.14
C. 192.168.15.29
D. 192.168.15.16
E. 192.168.15.31
F. None of the above

Solution:

/28 means 11111111.11111111.11111111.11110000

The last 1 stands under the 16. This is increment or network-jumps.

192.168.15.0 - 192.168.15.15
192.168.15.16 - 192.168.15.31
192.168.15.32 - 192.168.15.47

Only A and C are IPs in the right range.
Also E is in the right range. But - this is not useable for hosts, cause its broadcastaddress.
So answer is A and C.

################################################################

Calculation of Wildcard-Masks (Needed for Access Lists and OSPF Configuration)

You have Network 192.168.32.0 /28
Only THIS network should be denied of accessing a network or server.

1. Step
calculate the wildcard mask

/28 means 255.255.255.240

binary this is

11111111.11111111.11111111.11110000

For wildcard-mask only the ZEROS are interesting.

11110000 Make a addition of all the fields, that are set to zero

128/64/32/16/8/4/2/1
That is 8+4+2+1=15

So the wildcard-mask will be
0.0.0.15

access-list will be

access-list 1 deny 192.168.32.0 0.0.0.15
access-list 1 permit ip any any

now, we have to bind that access-list to a routerinterface. In the example, this is e0.

interface e0
ip access-group 1 out (or in!)
exit

PS.

Its good to write on a BIG paper the powers of 2

2^2=4
2^3=8
2^4=16
2^5=32
2^6=64
2^7=128
2^8=256
2^9=512
2^10=1024
2^11=2048
2^12=4096

And write on that paper the numbers

128 192 224 240 248 252 254

Cause this are the Numbers, you will allways need in calculating Subnets.

Burn them in your mind! Hang the paper in front of your eyes to never forget them.
Then you will be able to calculate Subnets in your head in a half second!

Spacyfreak, all Rights reserved hehehe

NightShade1

Router Summarization by NightShade
This is a mini tutorial i did myself of summerazing)

Ripv1 and IGRP
those protocols do auto summary to the classfull network which means
if you try to add a network of 10.1.1.1 it will auto summary to 10.0.0.0

You cant turn off this auto summary in Ripv1 and IGRP

you can turn it off on Ripv2 and EIGRP summarizing

OSPF and IS IS does not auto summery

Note:
Classful routing protocols (RIPv1 and IGRP) automatically summarize routes on the classful network boundary and do not support summarization on any other bit boundaries. Classless routing protocols support summarization on any bit boundary.

Route Summarization Calculation example

suppose you got this networks attached to your router like different LAns

172.16.12.0/24

172.16.13.0/24

172.16.14.0/24

172.16.15.0/24

1) first step:
You will convert everything to binary

10101100.00010000.00001100.00000000 -->172.16.12.0
10101100.00010000.00001101.00000000 --->172.16.13.0
10101100.00010000.00001110.00000000 -->172.16.14.0
10101100.00010000.00001111.00000000 -->172.16.15.0

2) Second Step

You need to locate the bits till where the bits are equals
What i mean with this ?

10101100.00010000.00001100.00000000 -->172.16.12.0
10101100.00010000.00001101.00000000 --->172.16.13.0
10101100.00010000.00001110.00000000 -->172.16.14.0
10101100.00010000.00001111.00000000 -->172.16.15.0

As you can see all the bits are equal till 6th bit of the 3rd octet(can you can see it above i BOLDED all the equals bits)

Now you find this we will go to step 3

3)Third step

You need to count those bits, the summary route number will be the first IP address in the block, and the subnet mask will be the number of common bits.

so the answer is
172.16.12.0/22

Well i think this is all you need to know about summarizing If anyone find any mistake on my explanation they are welcome but im pretty sure its like this.

NightShade, All right reserved

bperiod

Wow, thank you VERY MUCH...I really appreciate your time and effort. As soon as I get off work..I will study these and let you know if my thick head got it!

rock360

wow thank you

Kaminsky

Technote vote for summerisation from me.

Lot of work there Nightshade. Thanks.

malcybood

Nightshade,

Great informative post, thanks....I'll second Kaminsky on this being a potential addition to the technotes.

Good job

Malc

hectorjhrdz

LOL !!! what a BIG POST!!!

but very helpful of course

JNeko

Hello NightShade!

Awesome post! Your calculation has helped me understand Router Summarization and how it works in detail, interesting. I have a question for you in regards to VLSM implementation, topology, packet routing, and not calculation. Sound stupid? Well...that's me, haha. I understand all these calculations and scenerio ip address schemes, and why its good to save on ip address space to a decent extent, and often answer correctly on any practice test, but I want to look at this from a configuration aspect. I know this is probably a stupid question, but I hope someone can help me. Thanks!

I want to use an example I previously saw in another post:

Corporate network: 202.2.2.0

Corporate site: 50 hosts
Site 1: 48 hosts
Site 2: 30 hosts
Site 3: 23 hosts
Site 4: 12 hosts
Serial links: 4

"Give the IP subnet addy, 1st usable host, last usable host, broadcast, and subnet mask for each."

I actually did the calculations and got the answer (mostly) right, but I want to talk about some other things here, what's all this looking like when doing the configurations from ur telnet screen? And what kind of topology are we looking at visually? So here are some quick questions I have:

This address 202.2.2.0 is assigned to the network by an ISP, right?

Is each site represented by a seperate router interface basically? For instance,
Site 1: 48 hosts
Subnet=.0, 1st host=.1, last host=.62, broadcast=.63, mask=255.255.255.192
So when configuring an ethernet interface on whatever router, it would be like 202.2.2.1 with a 255.255.255.192 prefix mask, so when a packet arrives at this router, the router parses the table, looking for the packets destination addresses, sees it is valid address between .1 - .62 and forwards it out the corresponding interface. But I thought the router would parse the table and send the packet out through whatever interface due to the summarazation of the address (for quicker parsing and reduction of router memory blah blah)? I am confused here.

And then it would be the same concept for the other interfaces on the same router, like on another ethernet interface it would be site 2 with 30 hosts and a 255.255.255.224 mask? and so on and so on, I don't know, just sounds like I am reaching with this one...

Thanks for any help!

Darthn3ss

with summarys you can take a short cut - say you need to s ummarize the 192.168.1.0/24 network....(okay not hard i know)... instead of writing out all of your addresses in binary, you can write the first address of the subnet andt he last address of the subnet, and summarize those two..

Darthn3ss

why its good to save on ip address space to a decent extent

Because IPv4 is very limited in the number of IPs availble... say a link between 2 routers.. with an ip like in your exapmle, 202.2.2.0. if the subnet mask was the default mask IP, you would be wasting 252 out of 254 available addresses. doesn't seem like much but it adds up. With VLSM and classless routing protocols, you could make that serial link between 2 routers 202.2.2.0/252, which would only use 2 addresses and waste none.

I want to use an example I previously saw in another post:

Corporate network: 202.2.2.0

Corporate site: 50 hosts
Site 1: 48 hosts
Site 2: 30 hosts
Site 3: 23 hosts
Site 4: 12 hosts
Serial links: 4

"Give the IP subnet addy, 1st usable host, last usable host, broadcast, and subnet mask for each."

I actually did the calculations and got the answer (mostly) right, but I want to talk about some other things here, what's all this looking like when doing the configurations from ur telnet screen? And what kind of topology are we looking at visually? So here are some quick questions I have:

This address 202.2.2.0 is assigned to the network by an ISP, right?

Is each site represented by a separate router interface basically? For instance,
Site 1: 48 hosts

Subnet=.0, 1st host=.1, last host=.62, broadcast=.63, mask=255.255.255.192
So when configuring an Ethernet interface on whatever router, it would be like 202.2.2.1 with a 255.255.255.192 prefix mask, so when a packet arrives at this router, the router parses the table, looking for the packets destination addresses, sees it is valid address between .1 - .62 and forwards it out the corresponding interface. But I thought the router would parse the table and send the packet out through whatever interface due to the summarazation of the address (for quicker parsing and reduction of router memory blah blah)? I am confused here.

And then it would be the same concept for the other interfaces on the same router, like on another ethernet interface it would be site 2 with 30 hosts and a 255.255.255.224 mask? and so on and so on, I don't know, just sounds like I am reaching with this one...

well with route summarization i believe that an external router would summarize routes to your networks...

say you subnetted your 202.2.2.0 to 9 networks.... instead of that external router (say the isp) having 9 different routes to your network, i beleive they could summarize it to 202.2.2.0. i'm still kind of shady on this area myself so maybe someone else can chrip in. i haven't practiced it much yet.

for your example above:

200.2.2.0/24
8 subnets (4 sites and then 4 serial links...
first: 202.0.0.0/25 like you said .1-.62
second: /27 (could even go a size bigger to account for expansion .65-.94
third: /27 .97-.126
fourth: /28 .129-.142
fifth (serial link): /30 .145-.146
sixth (Serial link): /30 .149-150
seventh (serial link): /30 .153-.154
eighth (serial link): /30 .157-.158

if you had 4 routers..

R1:
E0: 202.2.2.1 255.255.255.192
S0: 202.2.2.145 255.255.255.252

R2:
E0: 202.2.2.65 255.255.255.224
S0: 202.2.2.149 255.255.255.252 (connected to R2)
S1: 202.2.2.146 255.255.255.252 (connected to R1)

R3:
E0: 202.2.2.97 255.255.255.224
S0: 202.2.2.153 255.255.255.252 (connected to R4)
S1: 202.2.2.150 255.255.255.252 (connected to R2)

R4:
E0: 202.2.2.129 255.255.255.240
S0: 202.2.2.154 255.255.255.252 (To R3)

heres what your routing table looks like:

Gateway of last resort is not set

70.0.0.0/8 is variably subnetted, 2 subnets, 2 masks
D 70.0.0.0/8 is a summary, 00:12:07, Null0
C 70.5.5.0/24 is directly connected, Loopback0
202.2.2.0/24 is variably subnetted, 9 subnets, 5 masks
D 202.2.2.128/28 [90/2297856] via 202.2.2.157, 00:10:20, Serial0
D 202.2.2.152/30 [90/2681856] via 202.2.2.157, 00:10:20, Serial0
C 202.2.2.156/30 is directly connected, Serial0
D 202.2.2.144/30 [90/3705856] via 202.2.2.157, 00:07:51, Serial0
D 202.2.2.148/30 [90/3193856] via 202.2.2.157, 00:08:38, Serial0
D 202.2.2.64/27 [90/3321856] via 202.2.2.157, 00:07:51, Serial0
D 202.2.2.96/27 [90/2809856] via 202.2.2.157, 00:08:38, Serial0
D 202.2.2.0/26 [90/3833856] via 202.2.2.157, 00:02:11, Serial0
D 202.2.2.0/24 is a summary, 00:12:07, Null0

did this from the 5th router, in my setup it was the "ISP" router. the 70.0.0.0 network i just made up to simulate an internet connection or something.

phatsoundz

great post, thanks alot, it makes way more sense than any of my books.

JNeko

Thank you Darth, for your replies.

markzab

Hey, had to bring this thread back up real quick to make sure I understood properly. In your example you had...

10101100.00010000.00001100.00000000 -->172.16.12.0
10101100.00010000.00001101.00000000 --->172.16.13.0
10101100.00010000.00001110.00000000 -->172.16.14.0
10101100.00010000.00001111.00000000 -->172.16.15.0

With the summary route being the location where all the matchi9ng bits stopped. In this case 172.16.12.0/22.

But for my understanding, lets change the first route from 12.0 to 10.0, which would be...

10101100.00010000.00001010.00000000 -->172.16.10.0
10101100.00010000.00001101.00000000 --->172.16.13.0
10101100.00010000.00001110.00000000 -->172.16.14.0
10101100.00010000.00001111.00000000 -->172.16.15.0

In this situation, am I correct in saying that the summary route would be 172.16.8.0/21?

Am I getting this?

rjbarlow

markzab wrote:

Hey, had to bring this thread back up real quick to make sure I understood properly. In your example you had...

10101100.00010000.00001100.00000000 -->172.16.12.0
10101100.00010000.00001101.00000000 --->172.16.13.0
10101100.00010000.00001110.00000000 -->172.16.14.0
10101100.00010000.00001111.00000000 -->172.16.15.0

With the summary route being the location where all the matchi9ng bits stopped. In this case 172.16.12.0/22.

But for my understanding, lets change the first route from 12.0 to 10.0, which would be...

10101100.00010000.00001010.00000000 -->172.16.10.0
10101100.00010000.00001101.00000000 --->172.16.13.0
10101100.00010000.00001110.00000000 -->172.16.14.0
10101100.00010000.00001111.00000000 -->172.16.15.0

In this situation, am I correct in saying that the summary route would be 172.16.8.0/21?

Am I getting this?

This is totally right!

This topic go straight into my favorites...

wait2dominate

Another more 'visual/hands-on' way of learning VSLM(I've used this technique to teach people who didn't understand with just writing) is to use a sheet of paper as the network.
192.168.1.0-192.168.0.255 SM-255.255.255.0 is the full sheet with 256 nodes, 254 usable addresses.
Full sheet 1
192.168.1.0-192.168.1.255 SM 255.255.255.0

When you add the next bit, to make the mask 255.255.255.128, you tear the sheet of paper in half, and write the new information on each half.
Full sheet has been ripped in half
half 1(half of original network)
192.168.1.0-192.168.1.127 SM 255.255.255.128
half 2(half of original network)
192.168.1.128-192.168.1.255 SM 255.255.255.128

If you needed smaller pieces, you repeat the process.
half 1(half of original network)
192.168.1.0-192.168.1.127 SM 255.255.255.128

half 2 now becomes quarters with the subnet mask 255.255.255.192
quarter 3 (as this would be the third quarter if all were split)
192.168.1.128-192.168.1.191 SM 255.255.255.192
quarter 4 (as this would be the fourth quarter is all were split)
192.168.1.192-192.168.1.255 SM 255.255.255.192

Say we needed Q4 further split
half 1(half of original network)
192.168.1.0-192.168.1.127 SM 255.255.255.128
quarter 3 (as this would be the third quarter if all were split)
192.168.1.128-192.168.1.191 SM 255.255.255.192

Eight 7 (Half of Q4) W/ SM 255.255.255.224
192.168.1.192.192-192.168.1.223 SM 255.255.255.224
Eight 8 (Half of Q4)
192.168.1.224-192.168.1.255 SM 255.255.255.224

Repeat as needed.

Now, if you had two addresses that you wanted to summarize, you find the two pieces, and put together all the pieces needed to make that one piece.

IE Addresses 192.168.1.250 and 192.168.1.130
First address is from Eight 8, second is from Quarter 3
The only piece that encompasses both originally was half 2

half 2's address was 192.168.1.128 SM 255.255.255.128

It's a lot easier when you have the paper in front of you and actually do this than it is to explain online, but I tried my best

antonio banderas

NightShade1 and others, thanks because you are helping me out with these posts.

Im taking the test pretty soon, so this is great help...

Someone, sticky this post...

live_wire

thanx a lot guys.

..really good post.

edit:
can any1 tell me the difference between VLSM and FLSM

mattipler

markzab wrote:

Hey, had to bring this thread back up real quick to make sure I understood properly. In your example you had...

10101100.00010000.00001100.00000000 -->172.16.12.0
10101100.00010000.00001101.00000000 --->172.16.13.0
10101100.00010000.00001110.00000000 -->172.16.14.0
10101100.00010000.00001111.00000000 -->172.16.15.0

With the summary route being the location where all the matchi9ng bits stopped. In this case 172.16.12.0/22.

But for my understanding, lets change the first route from 12.0 to 10.0, which would be...

10101100.00010000.00001010.00000000 -->172.16.10.0
10101100.00010000.00001101.00000000 --->172.16.13.0
10101100.00010000.00001110.00000000 -->172.16.14.0
10101100.00010000.00001111.00000000 -->172.16.15.0

In this situation, am I correct in saying that the summary route would be 172.16.8.0/21?

Am I getting this?

Why summary route 172.16.8.0 /21 and not 172.16.10.0 /21??? Perfectly understand Subnetting and VLSM but I really don't understand route summarization at all. I can't get my head around it.

mikearama

Wow... what an awesome tutorial. Very well written.

Hey matt... summarization uses all the bits that are identical. Identical!

In this example:

10101100.00010000.00001010.00000000 -->172.16.10.0
10101100.00010000.00001101.00000000 --->172.16.13.0
10101100.00010000.00001110.00000000 -->172.16.14.0
10101100.00010000.00001111.00000000 -->172.16.15.0

the last common bit that is identical is the fifth bit (in third octect), which reflects the group of subnets counting by 8's.

There is no such thing as subnets that count by 10.

So just remember that summarized routes MUST always fall on bit positions... after the fifth bit (with all subnets counting by

, the next option is the sixth bit (with all subnets counting by 4).

Using the sixth bit, as in the previous example, 172.16.12.0 worked, and it would cover all subnets between it and 172.16.15.x.

Using the fifth bit, as above, covers everything between 172.16.8.0 and 172.16.15.x. Since the first subnet in question was the 172.16.10.0 subnet, you have to go a step backward and use the 172.16.8.0 /21 summary.

Hth.

Stotic

I wish the books would teach it like this. Straight to the point, without babbling about senseless nonsense.

r_durant

Ditto...I thought this was a great post as well...It's actually from this post that I understood summarization...

Stotic

Can this be added to the FAQ thread? I keep looking for it when I want to reference it.

bohra_ajay

This Post is Gem of a Work n as every1 is saying i wish this to be added to Technote.

I agree that the books would have gone straight n simple like this after going thru all this Summarization (which i was breaking my head with since start) seems like 5th Grade Maths.

Thanx u have helped me earn more marks in my Score