http://tftpd32.jounin.net/tftpd32_download.html
Thursday, October 2, 2014
Monday, June 2, 2014
Determining the Number of IP Subnets and Hosts
CREATED BY A VALID MASK. NO CIDR OR VLSM.
Step One
Use the first octet of the IP address to determine the class of address (A, B, or C).
169.199.109.137 = CLASS B
255.255.255.192
Step Two
Use the class of the address to determine which octets are available for hosts.
CLASS B = Network. Network. Host. Host
169.199.109.137
255.255.255.192
Step Three
Look at the host octet(s) in the subnet mask. Use the "Possible Masks" chart to determine which bits are set to one. If no bits are set to one, there are no subnets. If any bits are set to one, proceed to step four.
169.199.109.137
255.255.255.192 = 11111111 11000000 (host octets only)
Step Four
Count the total number of ones in the host octet(s) of the subnet mask. Call this number X. Raise 2 to the power of X. Use the "Powers of 2" chart if necessary. This is the number of potential subnets created by the mask. Two of these potential subnets are normally not usable.
11111111 11000000 = 10 ones. 210 = 1,024 - 2 = 1,022 usable subnets created.
Step Five
Count the total number of zeros in the host octet(s) of the subnet mask. Call this number Y. Raise 2 to the power of Y. Use the "Powers of 2" chart if necessary. This is the number of potential subnets created by the mask. Two of these numbers are never used to address hosts.
11111111 11000000 = 6 zeros. 26 = 64 - 2 = 62 usable host addresses created.
Source (http://www.chabotcollege.edu/faculty/netacad/prot/sem2/study_sheets/detIPadd.htm)
Subnetting - Quick Method
After looking around the internet over the past few months I found the following method works best:
There are 3 main classes of IP address that we are concerned with.
Class A Range 0 - 127 in the first octet (0 and 127 are reserved)
Class B Range 128 - 191 in the first octet
Class C Range 192 - 223 in the first octet
Below shows you how, for each class, the address is split in terms of network (N) and host (H) portions.
NNNNNNNN . HHHHHHHH . HHHHHHHH . HHHHHHHH Class A
NNNNNNNN . NNNNNNNN . HHHHHHHH . HHHHHHHH Class B
NNNNNNNN . NNNNNNNN . NNNNNNNN . HHHHHHHH Class C
At each dot I like to think that there is a boundary, therefore there are boundaries after bits 8, 16, 24, and 32. This is an important concept to remember.
We will now look at typical questions that you may see on subnetting. More often than not they ask what a host range is for a specific address or which subnet a certain address is located on. I shall run through examples of each, for each class of IP address.
What subnet does 192.168.12.78/29 belong to?
You may wonder where to begin. Well to start with let's find the next boundary of this address.
Our mask is a /29. The next boundary is 32. So 32 - 29 = 3. Now 2^3 = 8 which gives us our block size i.e. 2 to the power of 3 equals 8.
We have borrowed from the last octet as the 29th bit is in the last octet. We start from zero and count up in our block size. Therefore it follows that the subnets are:-
192.168.12.0
192.168.12.8
192.168.12.16
192.168.12.24
192.168.12.32
192.168.12.40
192.168.12.48
192.168.12.56
192.168.12.64
192.168.12.72
192.168.12.80
.............etc
Our address is 192.168.12.78 so it must sit on the 192.168.12.72 subnet.
What subnet does 172.16.116.4/19 sit on?
Our mask is /19 and our next boundary is 24. Therefore 24 - 19 = 5. The block size is 2^5 = 32.
We have borrowed into the third octet as bit 19 is in the third octet so we count up our block size in that octet. The subnets are:-
172.16.0.0
172.16.32.0
172.16.64.0
172.16.96.0
172.16.128.0
172.16.160.0
.............etc
Our address is 172.16.116.4 so it must sit on the 172.16.96.0 subnet. Easy eh?
What subnet does 10.34.67.234/12 sit on?
Our mask is 12. Our next boundary is 16. Therefore 16 - 12 = 4. 2^4 = 16 which gives us our block size.
We have borrowed from the second octet as bit 12 sits in the second octet so we count up the block size in that octet. The subnets are:-
10.0.0.0
10.16.0.0
10.32.0.0
10.48.0.0
.............etc
Our address is 10.34.67.234 which must sit on the 10.32.0.0 subnet.
Hopefully the penny is starting to drop and you are slapping the side of your head realising that you were a fool to think that subnetting was hard. We will now change the type of question so that we have to give a particular host range of a subnet.
What is the valid host range of of the 4th subnet of 192.168.10.0/28?
Easy as pie! The block size is 16 since 32 - 28 = 4 and 2^4 = 16. We need to count up in the block size in the last octet as bit 28 is in the last octet.
192.168.10.0
192.168.10.16
192.168.10.32
192.168.10.48
192.168.10.64
.................etc
Therefore the 4th subnet is 192.168.10.48 and the host range must be 192.168.10.49 to 192.168.10.62, remembering that the subnet and broadcast address cannot be used.
What is the valid host range of the 1st subnet of 172.16.0.0/17?
/17 tells us that the block size is 2^(24-17) = 2^7 = 128. We are borrowing in the 3rd octet as bit 17 is in the 3rd octet. Our subnets are:-
172.16.0.0
172.16.128.0
The first subnet is 172.16.0.0 and the valid host range is 172.16.0.1 to 172.16.127.254. You must remember not to include the subnet address (172.16.0.0) and the broadcast address (172.16.127.255).
What is the valid host range of the 7th subnet of address 10.0.0.0/14?
The block size is 4, from 16 - 14 = 2 then 22 = 4. We are borrowing in the second octet so count in the block size from 0 seven times to get the seventh subnet.
The seventh subnet is 10.24.0.0. Our valid host range must be 10.24.0.1 to 10.27.255.254 again remembering not to include our subnet (10.24.0.0) and the broadcast address (10.27.255.255).
What if they give me the subnet mask in dotted decimal?
If you're lucky and they give you a mask in dotted decimal format then you should have an even easier time. All you need again is your block size.
Let's say they have given a mask of 255.255.255.248 and you wish to know the block size. Here's the technique:
1. Starting from the left of the mask find which is the first octet to NOT have 255 in it.
2. Subtract the number in that octet from 256 to get your block size e.g. above it is 256 - 248 = block size of 8.
3. Count up from zero in your block size in the octet identified in step 1 as you have learned above (the example above would be in the last octet).
Another example is a mask of 255.255.192.0 - you would simply count up in 256 - 192 = 64 in the third octet.
One more example is 255.224.0.0 - block size is 256 - 224 = 32 in the second octet.
Class A Range 0 - 127 in the first octet (0 and 127 are reserved)
Class B Range 128 - 191 in the first octet
Class C Range 192 - 223 in the first octet
Below shows you how, for each class, the address is split in terms of network (N) and host (H) portions.
NNNNNNNN . HHHHHHHH . HHHHHHHH . HHHHHHHH Class A
NNNNNNNN . NNNNNNNN . HHHHHHHH . HHHHHHHH Class B
NNNNNNNN . NNNNNNNN . NNNNNNNN . HHHHHHHH Class C
At each dot I like to think that there is a boundary, therefore there are boundaries after bits 8, 16, 24, and 32. This is an important concept to remember.
We will now look at typical questions that you may see on subnetting. More often than not they ask what a host range is for a specific address or which subnet a certain address is located on. I shall run through examples of each, for each class of IP address.
What subnet does 192.168.12.78/29 belong to?
You may wonder where to begin. Well to start with let's find the next boundary of this address.
Our mask is a /29. The next boundary is 32. So 32 - 29 = 3. Now 2^3 = 8 which gives us our block size i.e. 2 to the power of 3 equals 8.
We have borrowed from the last octet as the 29th bit is in the last octet. We start from zero and count up in our block size. Therefore it follows that the subnets are:-
192.168.12.0
192.168.12.8
192.168.12.16
192.168.12.24
192.168.12.32
192.168.12.40
192.168.12.48
192.168.12.56
192.168.12.64
192.168.12.72
192.168.12.80
.............etc
Our address is 192.168.12.78 so it must sit on the 192.168.12.72 subnet.
What subnet does 172.16.116.4/19 sit on?
Our mask is /19 and our next boundary is 24. Therefore 24 - 19 = 5. The block size is 2^5 = 32.
We have borrowed into the third octet as bit 19 is in the third octet so we count up our block size in that octet. The subnets are:-
172.16.0.0
172.16.32.0
172.16.64.0
172.16.96.0
172.16.128.0
172.16.160.0
.............etc
Our address is 172.16.116.4 so it must sit on the 172.16.96.0 subnet. Easy eh?
What subnet does 10.34.67.234/12 sit on?
Our mask is 12. Our next boundary is 16. Therefore 16 - 12 = 4. 2^4 = 16 which gives us our block size.
We have borrowed from the second octet as bit 12 sits in the second octet so we count up the block size in that octet. The subnets are:-
10.0.0.0
10.16.0.0
10.32.0.0
10.48.0.0
.............etc
Our address is 10.34.67.234 which must sit on the 10.32.0.0 subnet.
Hopefully the penny is starting to drop and you are slapping the side of your head realising that you were a fool to think that subnetting was hard. We will now change the type of question so that we have to give a particular host range of a subnet.
What is the valid host range of of the 4th subnet of 192.168.10.0/28?
Easy as pie! The block size is 16 since 32 - 28 = 4 and 2^4 = 16. We need to count up in the block size in the last octet as bit 28 is in the last octet.
192.168.10.0
192.168.10.16
192.168.10.32
192.168.10.48
192.168.10.64
.................etc
Therefore the 4th subnet is 192.168.10.48 and the host range must be 192.168.10.49 to 192.168.10.62, remembering that the subnet and broadcast address cannot be used.
What is the valid host range of the 1st subnet of 172.16.0.0/17?
/17 tells us that the block size is 2^(24-17) = 2^7 = 128. We are borrowing in the 3rd octet as bit 17 is in the 3rd octet. Our subnets are:-
172.16.0.0
172.16.128.0
The first subnet is 172.16.0.0 and the valid host range is 172.16.0.1 to 172.16.127.254. You must remember not to include the subnet address (172.16.0.0) and the broadcast address (172.16.127.255).
What is the valid host range of the 7th subnet of address 10.0.0.0/14?
The block size is 4, from 16 - 14 = 2 then 22 = 4. We are borrowing in the second octet so count in the block size from 0 seven times to get the seventh subnet.
The seventh subnet is 10.24.0.0. Our valid host range must be 10.24.0.1 to 10.27.255.254 again remembering not to include our subnet (10.24.0.0) and the broadcast address (10.27.255.255).
What if they give me the subnet mask in dotted decimal?
If you're lucky and they give you a mask in dotted decimal format then you should have an even easier time. All you need again is your block size.
Let's say they have given a mask of 255.255.255.248 and you wish to know the block size. Here's the technique:
1. Starting from the left of the mask find which is the first octet to NOT have 255 in it.
2. Subtract the number in that octet from 256 to get your block size e.g. above it is 256 - 248 = block size of 8.
3. Count up from zero in your block size in the octet identified in step 1 as you have learned above (the example above would be in the last octet).
Another example is a mask of 255.255.192.0 - you would simply count up in 256 - 192 = 64 in the third octet.
One more example is 255.224.0.0 - block size is 256 - 224 = 32 in the second octet.
Source (http://www.techexams.net/forums/ccna-ccent/38772-subnetting-made-easy.html)
Spending time using this method you can quickly determine the Network ID and Broadcast Address.
What subnet does 172.16.116.4/19 sit on?
Our mask is /19 and our next boundary is 24. Therefore 24 - 19 = 5. The block size is 2^5 = 32.
We have borrowed into the third octet as bit 19 is in the third octet so we count up our block size in that octet. The subnets are:-
172.16.0.0
172.16.32.0
172.16.64.0
172.16.96.0
172.16.128.0
172.16.160.0
.............etc
Our address is 172.16.116.4 so it must sit on the 172.16.96.0 subnet.
172.16.128.0 would be the next subnet so subtract 1 from third octet, 128- 1 = 127. The Broadcast Address, for subnet 172.16.96.0 is 172.16.127.255
What subnet does 172.16.116.4/ 255.255.255.224 sit on?
Sweet we almost have to do no math to solve this one. First identify which number in the subnet mask is not 255. That's right it's 224.
Subtract 256- 224= 32 is our block size.
(Additional fun! 2^5 = 32. Looking at the 8 16 24 32
We would use 32-5= 17. Now we could rewrite the problem 172.16.116.4/17
Hopefully this looks familiar. This is unessary in solving the question, but it help if develop the thinking method)
Count up block size. The subnets are:-
172.16.0.0
172.16.32.0
172.16.64.0
172.16.96.0
172.16.128.0
172.16.160.0
.............etc
172.16.32.0
172.16.64.0
172.16.96.0
172.16.128.0
172.16.160.0
.............etc
Our address is 172.16.116.4 so it must sit on the 172.16.96.0 subnet.
172.16.128.0 would be the next subnet so subtract 1 from third octet, 128- 1 = 127. The Broadcast Address, for subnet 172.16.96.0 is 172.16.127.255
Thursday, September 5, 2013
KALI Raspberry PI
I have a raspberry Pi that has been sitting around my house gathering dust. Recently I saw the Kali Linux has a version for ARM architecture.
I set out to install Kali (a new BackTrack) on the device. In an attempt to have a box I can always access. The Kali Pi plugs into the back of a ASUS (DD-WRT) RTN-16 router. I have NOT tried to power the Pi from the USB port attached on the router.
While attempting to write this article (get the Kali PI up and running) I have started over from scratch more times then I would like to admit (about 5). I do not view these as failed attempts, but as a refresher Linux course. I have not had direct usage of a Linux based system in almost 4 years and make of my skills are rusty (or non existent). Also I'm pretty lazy.
1.) Follow these instructions
Use image for PI located here "http://cdimage.kali.org/"
Use WinRAR to extract folder "kali-linux-1.0-armel-raspberrypi.img" > Open folder
Using Fedora ARM "http://fedoraproject.org/wiki/Fedora_ARM_Installer#Download_.26_Install_-_1.1.1-8"
Select "kali-custom-rpi.img" and proceed to flash your card.
At this point your SD card will have used about 4 GB of space. It's important to notice that since we flashed the SC CARD your card will only be as large as the size of the file you flashed from. In other words, that 15GB SD card is going to have about 10 GB of unallocated space. DON'T GO INSTALLING EVERYTHING JUST YET!
2.) Hook up a keyboard to your PI and Boot the device (plug in the USB power). At this point we want to just start the SSH Server. Additionally this step requires some form of a Network Scanner. I suggest NMap just make a quick scan of your local network ("nmap -T4 -F 192.168.1.1/24")
root <WAIT>
toor <WAIT>
service ssh start
<Run your NMap Scan to locate the IP of your Kali PI>
You can now remove the keyboard attached to your PI. The only things plug into the little box should be Ethernet Cable, USB Power, and SD CARD.
3.) Installing 'raspi-config' (Access box using KiTTY or PuTTY)
Thank you internet!
[-- COPY THIS INTO YOUR TERMINAL-->
wget http://archive.raspberrypi.org/debian/pool/main/r/raspi-config/raspi-config_20121028_all.deb
wget http://http.us.debian.org/debian/pool/main/l/lua5.1/lua5.1_5.1.5-4_armel.deb
wget http://http.us.debian.org/debian/pool/main/t/triggerhappy/triggerhappy_0.3.4-2_armel.deb
dpkg -i triggerhappy_0.3.4-2_armel.deb
dpkg -i lua5.1_5.1.5-4_armel.deb
dpkg -i raspi-config_20121028_all.deb
<-- END --]
Enter the following Commands:
raspi-config
Select 'expand_rootfs' <~ Fill that card
Select 'ssh' <~ Enable ssh on startup
reboot
4.) Install xRDP
Scan your network for Kali Pi
apt-get install xrdp <~ Will need to do this from remote terminal
service xrdp start
reboot
Now all you do is need to connect to the box from your windows computer.
NOTES:
At this point in the the PI does not have a STATIC IP.
'service xrdp start' has not been configured to start on startup.
I set out to install Kali (a new BackTrack) on the device. In an attempt to have a box I can always access. The Kali Pi plugs into the back of a ASUS (DD-WRT) RTN-16 router. I have NOT tried to power the Pi from the USB port attached on the router.
While attempting to write this article (get the Kali PI up and running) I have started over from scratch more times then I would like to admit (about 5). I do not view these as failed attempts, but as a refresher Linux course. I have not had direct usage of a Linux based system in almost 4 years and make of my skills are rusty (or non existent). Also I'm pretty lazy.
1.) Follow these instructions
Use image for PI located here "http://cdimage.kali.org/"
Use WinRAR to extract folder "kali-linux-1.0-armel-raspberrypi.img" > Open folder
Using Fedora ARM "http://fedoraproject.org/wiki/Fedora_ARM_Installer#Download_.26_Install_-_1.1.1-8"
Select "kali-custom-rpi.img" and proceed to flash your card.
At this point your SD card will have used about 4 GB of space. It's important to notice that since we flashed the SC CARD your card will only be as large as the size of the file you flashed from. In other words, that 15GB SD card is going to have about 10 GB of unallocated space. DON'T GO INSTALLING EVERYTHING JUST YET!
2.) Hook up a keyboard to your PI and Boot the device (plug in the USB power). At this point we want to just start the SSH Server. Additionally this step requires some form of a Network Scanner. I suggest NMap just make a quick scan of your local network ("nmap -T4 -F 192.168.1.1/24")
root <WAIT>
toor <WAIT>
service ssh start
<Run your NMap Scan to locate the IP of your Kali PI>
You can now remove the keyboard attached to your PI. The only things plug into the little box should be Ethernet Cable, USB Power, and SD CARD.
3.) Installing 'raspi-config' (Access box using KiTTY or PuTTY)
Thank you internet!
[-- COPY THIS INTO YOUR TERMINAL-->
wget http://archive.raspberrypi.org/debian/pool/main/r/raspi-config/raspi-config_20121028_all.deb
wget http://http.us.debian.org/debian/pool/main/l/lua5.1/lua5.1_5.1.5-4_armel.deb
wget http://http.us.debian.org/debian/pool/main/t/triggerhappy/triggerhappy_0.3.4-2_armel.deb
dpkg -i triggerhappy_0.3.4-2_armel.deb
dpkg -i lua5.1_5.1.5-4_armel.deb
dpkg -i raspi-config_20121028_all.deb
<-- END --]
Enter the following Commands:
raspi-config
Select 'expand_rootfs' <~ Fill that card
Select 'ssh' <~ Enable ssh on startup
reboot
4.) Install xRDP
Scan your network for Kali Pi
apt-get install xrdp <~ Will need to do this from remote terminal
service xrdp start
reboot
Now all you do is need to connect to the box from your windows computer.
NOTES:
At this point in the the PI does not have a STATIC IP.
'service xrdp start' has not been configured to start on startup.
Wednesday, April 3, 2013
Acrobat failed to load its Core DLL
After three emails from my parents I decided to look into their Adobe issue.
The issue just started earlier in the week, I believe Windows automatically patched, and adobe started producing the error message of "...failed to load its Core DLL"
There are bunch of Adobe forums that attempt to resolve the issues with no clear fix.
Here's how I fixed it....
http://forums.adobe.com/message/5201386
"Copy all the FILES, not folders, from C:\Program Files (x86)\Adobe\Reader x.x\reader to C:\"
~NeemobeerX
On the back end it might not be the cleanest way to fix the issues.
But it resolves all their issues!
The issue just started earlier in the week, I believe Windows automatically patched, and adobe started producing the error message of "...failed to load its Core DLL"
There are bunch of Adobe forums that attempt to resolve the issues with no clear fix.
Here's how I fixed it....
http://forums.adobe.com/message/5201386
"Copy all the FILES, not folders, from C:\Program Files (x86)\Adobe\Reader x.x\reader to C:\"
~NeemobeerX
On the back end it might not be the cleanest way to fix the issues.
But it resolves all their issues!
Friday, January 11, 2013
Provisioning (RingCentral) Polycom on GetJive
Moving a Polycom IP phone from RingCentral to GetJive
Here at the office, we stopped pushing RingCentral and switched over to GetJive.
I have 5 hours to learn how to provision Polycom Phones and switch them over to our clients GetJive account. I've only been working with VoIP for a year, and all my previous experience is with Cisco phones. Here's what I learned today:
Polycom
provisioning instructions:
•
Plug the phone into the internet, and the power
•It
will say loading application – during that, press cancel press "setup"
•
Enter the password ‘456’, then hit ok
•
Scroll down to and select ‘Server Menu’ (Prov.Server/Provisioning)
•
Set ‘Server Type’ to HTTP (NOT https)
•
Set Server Address to (ADD PBX DOMAIN).onjive.com/p (note: this MUST be lower-case)
•
for (.) press *
•
for (/) press # three times
•
Remove everything from the Server User and Password Fields
•
Press exit a few times, choosing to save and reboot when prompted.
If you need to
Factory Reset, follow these instructions.
Polycom
provisioning instructions:
•
Reboot the phone
•
During the Count down hold (1 3 5 7 for the 3xx and 4xx model OR 4 6 8 * for the
5xx and 6xx model) until prompted for a password
•
Follow the steps above
456
is the password for a phone that is not provisioned to Jive or is factory reset
8647
(voip) is the password for a phone that
is provisioned to Jive
Because RingCentral locks down phones they sell you, it's essential to DISCONNECT the phone from the network until after you have applied the new provisioning URL.
Thursday, July 19, 2012
Falling back into Security
Ugh Summer, how I hate you.
I've fallen behind on InfoSec. Been too busy helping my job/ clients improve their relationships with us.
Need to get structure back in mylife.
I've fallen behind on InfoSec. Been too busy helping my job/ clients improve their relationships with us.
Need to get structure back in mylife.
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