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spacyfreak

Subnetting

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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

192.168.32.62

 

 

 

-------------------------------------------------------------------------------------------------------------------------

 

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 the network 132.43.56.0 /24

 

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 0.0.0.15 132.43.56.0 0.0.0.255

access-list 1 ip any any

 

-------------------------------------------------------------------------------------------------------------------------

 

 

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!

 

Isnt live easy?

Edited by rainbow9810

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:ph34r:

This is great, you didnt learn this on a windows 2000 course by any chance did you?

:ph34r:

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:ph34r:

This is great, you didnt learn this on a windows 2000 course by any chance did you?

:ph34r:

 

Well, i had three different teachers, which TRIED to make me understand the way to solve this beast "subnet-calculation". But they explained so complicated, that no "normal" person with an IQ under 160 could understand.

 

I was near to give it up to understand this sh*t EVER.

 

Than a shy and silent Engineer, who did sit behind me in school, explained it to me the way i explained it above. That is the simplest and very logical Way to fix the questions, i think.

But everyone has his own "best way" to do it. This is my best way. Hope it helpes other lost souls, too! Hehe.

Edited by lancewoodson

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I use a couple of tables that who the powers of two and network-jumps/bits borrowed, and subnet masks and wildcard masks/bits borrowed.

 

I have the numbers all columned up always the same way:

In the first table:

First line the powers, second line the values, third line the network jumps.

 

In the second table:

First column the bits borrowed, second column the network masks, third column the wild card masks.

 

Allows me to do the subnetting without even thinking of binary equivalents.

 

For instance:

Last address in 192.16.10.135/27

 

/27 = I borrowed 3 in the 4th octet.

First table tells me: network jump: 32

So subnet address: 192.16.10.128

First address one more: 192.16.10.129

Broadcast: 192.16.10.159

(159=128+32-1)

Last address: 192.16.10.158

 

If I am asked for a wildcard, I can find it in two seconds from the second table, on the line that corresponds to the number of bits borrowed, or on the line that corresponds to the subnet mask.

 

I am not saying it would work for everyone, but it sure works for me.

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/27 = I borrowed 3 in the 4th octet.

First table tells me: network jump: 32

So subnet address: 192.16.10.128

First address one more: 192.16.10.129

Broadcast: 192.16.10.159

(159=128+32-1)

Last address: 192.16.10.128

 

Last Address should be 192.16.10.158, right?

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Of course, you're right. Sorry...

 

So much for the demo :lol::lol::lol:

 

I'll blame it on the fact that it was soooo freaking early on a Sunday morning. I never take certs in the morning cause my brains usually sleep until noon, even though my body is at work at 8am :lol:

 

 

OK, so now I just corrected it, and both are posts look silly :lol::lol::lol:

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I made a table (for myself) and I think it is very easy. ;)

 

msk -- mbt --   sbn  -- hst -- wmk

255 -- /32 -- 0

254 -- /31 -- 2(128) -- 0   -- 1

252 -- /30 -- 4(64)  -- 2   -- 3

248 -- /29 -- 8(32)  -- 6   -- 7

240 -- /28 -- 16(16) -- 14  -- 15

224 -- /27 -- 32(8)  -- 30  -- 31

192 -- /26 -- 64(4)  -- 62  -- 63

128 -- /25 -- 128(2) -- 126 -- 127

 

I failed today, and I want revenge!

I was stupid and I didn't wrote this table on paper BEFORE the test. :blush:

Everybody can see relations.

msk=last octet of sub.mask ex. 255.255.255.192

mbt=mask bit

sbn=1st subnet address(how many on network)

hst=number of hosts

wmk=wildcard mask (last octet) ex. 0.0.0.63

Using binary notation it is easy to wrote this table upside-down.

sbn = 256 - msk

number of subnets = 256 / sbn

hst = sbn - 2

Wildcard mask = sbn - 1

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Nenadx,

 

What is the Wildcard mask ?. It's something I have never heard before ?!

 

Am I correct to say that it's the same as broadcast ??

 

Thanks to everyone who showed us the different techniques.

 

I am thinking of proposing this for the post of the month !!

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What is the Wildcard mask ?. It's something I have never heard before ?!

 

You put wildcard or inverse mask in Access Control List (in router). You can find a lot examples in TK. Just put "wildcard mask" in search.

 

With my method you dont need any tables, just 5 fingers of your hand...

I agree wit you. You pass, I didn't :angry:

I also using fingers method, but you need to count every time. Before the exam you can put that on paper and later just read it. I will prove that next week. :rolleyes:

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!<snip>

If we for example have the network

 

192.168.10.0 255.255.255.0

We have here ONE Class C - network, with 254 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 - 192.168.10.254

 

The Subnetmask defines how big the subnet is.

That means - how many Client-PCs will have place in that subnetwork.

!<snip>

 

 

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

 

i)192.168.10.1 - 192.168.10.127

 

ii)192.168.10.128 - 192.168.10.254

 

spacy according to me you aint right here ..infact the decimal equivalent of 128 is 1000 0000, which gets me 2^0 ie 0 no. of subnets ..whereas i require two subnets ..which can be achieved by 128+64 = 192 in dec 1100 0000 or which gives me 2^1(or 2^2-2=4-2=2)= 2 no. of subnets ..infact as far as ma math goes 128 in subnets is never used..! ie you will always require two subnit bits for a minimum.

 

also the formula used for calculating the no. of subnets is 2^subnet bits used-2 ie 2^1-2=0 ..which yields me 0 no. of subnets ..doesnt fit correctly with your solution.

 

ps: correct me if i have got it wrong..!?

Edited by MKD

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spacy according to me you aint right here ..infact the decimal equivalent of 128 is 1000 0000, which gets me 2^0 ie 0 no. of subnets ..whereas i require two subnets ..which can be achieved by 128+64 = 192 in dec 1100 0000 or which gives me 2^1(or 2^2-2=4-2=2)= 2 no. of subnets ..infact as far as ma math goes 128 in subnets is never used..! ie you will always require two subnit bits for a minimum.

 

also the formula used for calculating the no. of subnets is 2^subnet bits used-2 ie 2^1-2=0 ..which yields me 0 no. of subnets ..doesnt fit correctly with your solution.

 

ps: correct me if i have got it wrong..!?

 

 

subnet0uw.jpg

 

From mathematics side, i am right.

In my example the network is divided in two parts.

 

Normally, the first and last subnet of a network is not useable.

In that case the formula 2^n-2 is right.

With command "ip subnet-zero", they ARE useable.

Any other opinions??? ;)

So i dont think that i am wrong.

Edited by spacyfreak

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spacy according to me you aint right here ..infact the decimal equivalent of 128 is 1000 0000, which gets me 2^0 ie 0 no. of subnets ..whereas i require two subnets ..which can be achieved by 128+64 = 192 in dec 1100 0000 or which gives me 2^1(or 2^2-2=4-2=2)= 2 no. of subnets ..infact as far as ma math goes 128 in subnets is never used..! ie you will always require two subnit bits for a minimum.

 

also the formula used for calculating the no. of subnets is 2^subnet bits used-2 ie 2^1-2=0 ..which yields me 0 no. of subnets ..doesnt fit correctly with your solution.

 

ps: correct me if i have got it wrong..!?

 

 

subnet0uw.jpg

 

From mathematics side, i am right.

In my example the network is divided in two parts.

 

Normally, the first and last subnet of a network is not useable.

In that case the formula 2^n-2 is right.

With command "ip subnet-zero", they ARE useable.

Any other opinions??? ;)

So i dont think that i am wrong.

 

Spacy honey,

 

A pointer would have been good ..btw if you look it from the examinations point of view you have to use the minus two in the formula to get it correct ..of course, 'm aware of the "ip-subnet zero" command which can be used in the new versions of ios!

 

Regards,

mkd.

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hi.. and tnx for usefull guide.. yesterday in my network+ class i didnt undrestand anyhting about subnet :unsure: .. i was so confused .. but now i think it's not that hard... and my question:

how can we calculate subnet for a class b ip address? for example we have these 2 ip:182.18.8.12 and 182.18.63.12.. what is the smallest subnet for these ips? danke

 

 

So, the subnet must minimum have the given range 182.18.2.12 - 182.18.63.12

 

182.18.0.0 is Class-B Network, Standard Subnetmask is 255.255.0.0

 

If we change the Subnetmask to 255.255.192.0 (11111111.11111111.11000000.0000000) the increment is 64, because the LAST of the ones stands under 64.

 

So the subnets ranges are

 

182.18.0.0 - 182.18.63.255 <<<first subnet

+64 182.18.64.0 - 182.18.127.255

+64 182.18.128.0 - 182.18.191.255

...

 

The given IPs 182.18.2.12 and 182.18.63.12 are in the range of the first subnet.

Edited by spacyfreak

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Spacyfreak i could give you a hug!!! Thank you thank you thank you!!! I have read so many subnetting examples and guides but nothing was making sense for me. Until i met you...and your fourm post here. A light blub went off in my head and i can officially say i understand subnetting now! I just need to practice a little more to retain it, but i really understand it and its a great feeling.

 

THANK YOU!!!

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i cant get it why u in 1 of your examples add net jumps on 3d octet(from left)

172.16.64.0 - 172.16.79.255

+16 172.16.80.0 - 172.16.95.255

and here u add network jumps on last octet(from left)

172.16.64.0 - 172.16.64.63 +64

172.16.64.64 - 172.16.64.127

tnx.

  • Upvote 1

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Why not?

You can also do

 

174.0.0.0 with 255.240.0.0.

 

Subnets would be

 

 

174.0.0.0 - 174.15.255.255

174.16.0.0 - 174.31.255.255

174.32.0.0 - 174.63.255.255

 

 

and you can also do

 

 

174.64.64.0 with 255.255.255.240

 

Subnets would be

 

174.64.64.0 - 174.64.64.15

174.64.64.16 - 174.64.64.31

174.64.64.32 - 174.64.64.47

 

The SUBNETMASK defines what ranges your networks will have. Thats all.

  • Downvote 1

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Hi Spacy,

 

Sincerely i dont know how to thank u guys in this forum its one of the best i have ever joined! People takes subnetting like mini god but your example proof its just one of the materials use to make a semi god and not a god on its own. May God bless you all and continue to increase your knowledge.

cheers

babydoll

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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

 

subnettingdg5.jpg

 

 

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.

 

 

Look here, for some step by step examples, how to calculate subnets..

 

http://www.sadikhov.com/forum/index.php?showtopic=44874

 

-------------------------------------------------------------------------

 

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 of real exam questions, 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... Like in 5 minute course..

 

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!

 

Isnt live easy?

 

 

Here is a table that I used for myself to get the hang of it and to remember it better.

 

# of users per Net 128 64 32 16 8 4 2 1

# of users per Net -2 126 62 30 14 6 2 0 0

 

 

 

# of Subnets 2 4 8 16 32 64 128 256

 

 

 

 

Subnet Masks 128 192 224 240 248 252 254 255

 

 

 

 

# of Bits 1 2 3 4 5 6 7 8

 

______|______

8 4 2 1 | 8 4 2 1

 

8+4+2+1= 15 1 2 3 4 5 6 7 8 9 A(10)B(11)C(12)D(13)E(14)F(15)

 

IEE Private IP Addresses

10.0.0.0 – 10.255.255.255

172.16.0.0 – 172.31.255.255

192.168.0.0 – 192.168.255.255

 

Default Administrative Distance

E - 90

I - 100

O - 110

R – 120

 

N – NVRAM

F – Flash

T – TFTP

R – ROM

 

Physical

Data

Network

Transport

Session

Presentation

Application

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Here is a table that I used for myself to get the hang of it and to remember it better.

 

# of users per Net 128 64 32 16 8 4 2 1

# of users per Net -2 126 62 30 14 6 2 0 0

 

 

 

# of Subnets 2 4 8 16 32 64 128 256

 

 

 

 

Subnet Masks 128 192 224 240 248 252 254 255

 

 

 

 

# of Bits 1 2 3 4 5 6 7 8

 

______|______

8 4 2 1 | 8 4 2 1

 

8+4+2+1= 15 1 2 3 4 5 6 7 8 9 A(10)B(11)C(12)D(13)E(14)F(15)

 

IEE Private IP Addresses

10.0.0.0 – 10.255.255.255

172.16.0.0 – 172.31.255.255

192.168.0.0 – 192.168.255.255

 

Default Administrative Distance

E - 90

I - 100

O - 110

R – 120

 

N – NVRAM

F – Flash

T – TFTP

R – ROM

 

Physical

Data

Network

Transport

Session

Presentation

Application

 

 

 

 

 

 

 

is it ok to use a calculator in the ccna ex :unsure: am and is it ok to bring a piece of paper with the list above in it?? :unsure:

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Hi All,

 

How do I calculate the number of usable networks?

 

For example, how many usable Class C networks are created with a subnet mask of 255.255.255.240?

 

Thanks,

Ryno

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Hi All,

 

How do I calculate the number of usable networks?

 

For example, how many usable Class C networks are created with a subnet mask of 255.255.255.240?

 

Thanks,

Ryno

 

That depends on the network you have.

 

If you for example have B-class network

 

140.20.0.0 with 255.255.0.0 Subnetmask, and you change the Subnetmask to 255.25.255.240, than simply count the bits you did change from zero to one.

 

Before= 255.255.0.0 = 11111111.11111111.00000000.00000000

After= 255.255.255.240 = 11111111.11111111.11111111.11110000

 

So you have 12 bits changed from zero to one.

 

2^12=4096 Subnets you did create

 

But if you had before network

 

198.99.99.0 with subnetmask 255.255.255.0, and you change subnetmask to 255.255.255.240, than you changed only 4 bits from zero to one.

So in this case, you created 2^4=16 Subnets.

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Also, what I found as a neat trick is to divide by the number of ips that are provided by the mask. For example:

 

2^H=Ip address (H=number of 0's in subnet mask)

 

2^4=16

 

So for example if you have 192.168.0.34/28 and you wanted to find out the broadcast and the network for this ip address and the range

 

34/16=2 remaider 2

 

So at this point to find the network address all you need to do is subtract 2 from the provided ip address:

 

34-2=32

 

To find the broadcast add the number of IPs given minus -1:

 

32+16-1=47

 

That gives you the range of your ip addresses. This works best for Class C's but I have adaptations for Class B and Class A

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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

 

subnettingdg5.jpg

 

 

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.

 

 

Look here, for some step by step examples, how to calculate subnets..

 

http://www.sadikhov.com/forum/index.php?showtopic=44874

 

-------------------------------------------------------------------------

 

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 of real exam questions, 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... Like in 5 minute course..

 

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!

 

Isnt live easy?

 

 

subnettingdg5.jpg

 

Well, i think you may be missing something.

 

The default gateway ip address is a usable ip address even if it is used by the default gateway. this is because you assigned it to the default gateway. the only unusable ip address is that which identify the network and the broadcast. so you should have about 254 usable ip addresses.

 

thanks and correct me if i am wrong.

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Well, if i use ONE of these 254 USEABLE IPs for the default gateway, then there ARE only 253 IPs left for the Host-PCs. Or not?

I think thats not a mistake - its just a question of the specific point of view.

 

But now to something completely different - WHY THE HELL IS IT SO COLD TODAY? I thought its SUMMER===????

 

Cheer.

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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

 

subnettingdg5.jpg

 

 

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.

 

 

Look here, for some step by step examples, how to calculate subnets..

 

http://www.sadikhov.com/forum/index.php?showtopic=44874

 

-------------------------------------------------------------------------

 

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 of real exam questions, 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... Like in 5 minute course..

 

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!

 

Isnt live easy?

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