In my case I saw no symptoms of any issues before the BSOD, though one minute the system was working just fine and the next, it was hosed and wouldn’t boot.

The exception the BSOD had thrown indicated an SSD failure though I assumed perhaps a RAM hiccup, further troubleshooting confirmed my fears however so I went trawling for a solution.

I came across OCZ’s bootable Linux environment for flashing the drives firmware which I thought it was worth a shot since by this point I already had an RMA number in hand from OCZ themselves; this route proved fruitless however with “Internal Error” being returned constantly upon attempting to flash the device.  I phoned up Scan International (the place I bought the drive from) and they said they’ll replace the drive with a 60GB OCZ Agility 3 of which is a SATA 3 drive (not SATA 2 like the Vertex 2) and boasts on-par if not slightly better performance than the Vertex 2 equivalent according to OCZ’s website.

It turns out OCZ have since EOL’d the Vertex 2 series as well as a large chunk of other SATA 2 SSD drive ranges they had out at the same time; I suspect heads rolled for that screw up at OCZ’s HQ.

Main point, don’t bother trying to fix the bugger; contact the place where you bought the drive from and also OCZ themselves (remember: these drives have a 3 year warranty!), then go with the best option from there; any retailer worth it’s salt will sort out this very well-known and widespread issue.

Hi all,

I’ve received an email from one of your members asking me to post up my technique for subnetting as links to external blogs are not allowed due to forum rules. I know that he benefited from it and he wishes to help out others so here goes:

First of all I need you to get rid of all of the negative thoughts surrounding subnetting. Put down all of the books that you have read about the subject and navigate away from other sites claiming to provide an easy way to subnet. This technique requires no charts, just simply the know-how to work with the powers of 2.

We need to start with the fundamentals of IP addressing. An IP address is made up of 32 bits, split into 4 octets (oct = 8, yes?). Some bits are reserved for identifying the network and the other bits are left to identify the host.

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.

What now?

Now it’s time to go and pick up those books again and go straight to the practice questions, completely by-passing any of their techniques. Use my method and you will be laughing!

Happy subnetting!

You can now find him at the following channel; he appears to be posting the old content up as well as brand new stuff too:

http://www.youtube.com/user/photonvids

Let’s hope he’s here to stay this time!

http://www.bbc.co.uk/news/business-17931814

This is in an effort to make your searches secure as it becomes a bigger part of our daily lives and ever more a cognitive expansion of our brains, though could this have a negative impact on the quality of sites overall now that webmasters are essentially being left in the dark?

http://insidesearch.blogspot.com/2012/03/bringing-more-secure-search-around.html

Before now, only logged in users didnt have their keywords passed onto webmasters statistics/tracking software as the keyword was directly tied to a Google account (i.e. an identifiable individual) which was a violation of Google’s privacy policy, however some fear that with this data being suppressed across the board that the quality of sites will suffer as those creating content dont know truly if they’re going to be satisfying the audience they’re reaching.

I beg to differ however, this should also revolutionise the way SEO is carried out on.

Rather than the focus being purely on keywords, it’ll now be all about the bounce-rates, content quality and overall user experience instead of simply trying to rank in the top 3 spots for keywords with x,xxx searches per-month.

I suspect it’ll annoy the majority of sleasy snake-oil “SEO Experts” though the cream should rise to the top.  I’ll be grabbing some popcorn and watchin the fallout unfold for this one.

root@server [~]# php -i | grep

post_max_sizepost_max_size => 16M

Simply head into your php.ini file and raise this limit to something higher, or alternatively tweak your script to use less resources.

date.timezone = “Europe/London”

Supported timezones can be found here.

You can also place “date_default_timezone_set(‘Europe/London’)” within the script being called or mentioned within the error message.

• Open “./modules/mod_mainmenu/helper.php“
• Find “function buildXML(&$params)”
• Change to “function buildXML($params)”
• Save and Retry