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  • ubuntu 12.04 can't find root partition (it doesn't look for btrfs partitions) end up with kernel-panic [closed]

    - by zalesz
    Possible Duplicate: There's an issue with an Alpha/Beta Release of Ubuntu, what should I do? I'm running Ubuntu 12.04 from kernel v. 3.2.0-17 with all partitions formatted as BTRFS. It was everything ok till kernel 3.2.0-18/19. Now system don't load, after trying to run it with recovery there is a msg that kernel panic occurred cause there is no partition with ext3/4 and some other partitions but I don't see any btrfs alike type. Any ideas how to fix it? Best

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  • How large should I make root, home, usr, var, and tmp partitions?

    - by Teddy Okidoki
    i install ubuntu server 10.04, have 64 Gb VHD. And want to separate partitions in this mode: /dev/xvda0 p on swap (2 Gb) /dev/xvda1a0 e on /boot (128 Mb) /dev/xvda1a1 e on / type ffs (local) /dev/xvda1a2 e on /usr type ffs (local, nodev) /dev/xvda1a3 e on /tmp type ffs (local, nodev) /dev/xvda1a4 e on /var/log type ffs (local, nodev) /dev/xvda1a5 e on /var type ffs (local, nodev, nosuid) /dev/xvda1a6 e on /home type ffs (local, nodev, nosuid, with quotas) /dev/xvda2 p on /new (local, nodev, nosuid, noexec) with rest of space ~50Gb. But i'ma stuck, and don't know what size to give to each folder. Also i want to encrypt partitions. Thank you for any tips. EDIT: System need minimum size, here will be installed about 10 apps like ufw, apache,mysql, chkrootkit and so on.

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  • How do I check whether partitions on my SSD are properly aligned?

    - by elementz
    I just installed ubuntu on my new intel SSD. Now I am not sure, whether paritions are properly aligned in respect to my specific SSD. Here's my fdisk output. $ fdisk -l Platte /dev/sda: 120.0 GByte, 120034123776 Byte 255 Köpfe, 63 Sektoren/Spur, 14593 Zylinder Einheiten = Zylinder von 16065 × 512 = 8225280 Bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x000a6294 Gerät boot. Anfang Ende Blöcke Id System /dev/sda1 * 1 1913 15360000 83 Linux /dev/sda2 1913 14058 97558528 83 Linux /dev/sda3 14058 14594 4300800 82 Linux Swap / Solaris Also, do I still need to align my SSD at all, since I am using TRIM on the ext4 partitions by mounting them with the discard flag. If it is the case, that my partitions are not properly aligned, what could I do to fix this without having to reinstall everything?

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  • Default mount options on auto-mounted NTFS partitions (how to add `noexec` and `fmask=0111`?)

    - by jetxee
    I use auto-mounting of external USB devices, and it works as expected, except that NTFS partitions are mounted with executability flag on. For example: /dev/sdb1 on /media/Elements type fuseblk (rw,nosuid,nodev,allow_other,blksize=4096,default_permissions) All normal files are -rwxrwxrwx on this partition. I am not happy with the xs. I know I can have it mounted the way I want if I pass the fmask=0111 option. Now I use Lucid, and suppose it uses some new auto-mounting mechanism (gvfs-mount?), but I don't really know how the default mounting options can be changed now. Gconf settings in /system/storage/default_options/ntfs/mount_options have no effect. So, how do I make fmask=0111 the default automounting option for all NTFS partitions? (I'd be grateful also if someone explains how the current automounting mechanism works, how to configure it, and if the default mounting options are hard-coded, what I have to recompile to change them). I know that I can put a line in the /etc/fstab and/or mount manually, but this is not the solution I want, because 1) I don't want to edit /etc/fstab for each and every external drive I use, 2) fstab records appear in the Places pane of Nautilus, even if the drives are not present. The questions is how to change the defaults.

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  • If I partition a drive connected via eSata will it show different partitions when connected via USB?

    - by jeffreypriebe
    I have an odd problem with an external drive. I'm formatting it connected to my laptop prior to connecting it to my router. The HDD enclosure has both an eSata and USB connections. Generally, I connect it via eSata to my laptop. I created my partitions and connected it to the router, but I see partition information that is different than what I created. After chasing leads concerning large HDD size, I mindlessly connected the HDD to my laptop with USB. Lo! I see the same partitions as the router. Attached are screenshots using the same program and the HDD in question. The only difference is the connection. For the first, I connected via eSata and hit "refresh" on the partition program. Then, turned off the HDD, disconnected the eSata cable, and connected via USB. Power and refresh. eSata: reports a total HDD size of 2328 GB, with four partitions (the third being 1.96TB) USB: reports a total HDD size of 280 GB, with three partitions (the third being 279 GB) Any idea why this is happening? It looks like it clearly is an issue of the 4K sector size and not playing nice with the USB enclosure. I tried it eSata and USB in Windows and Linux and it appears consistently that eSata is reporting correctly, USB incorrectly.

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  • Will deleting partitions affect my hard drive in any way?

    - by Portali5t
    I installed a Suse partition of around 200 Gigabytes on my hard drive, primarily running Windows 7. I am sick of Suse's crap, and just want to get rid of the OS and get that partition back for Windows' use. Is it as simple as that partition gets deleted,and I can choose what partition that space goes to, or is it communal that all partitions can access? I know next to nothing about partitions, so any help would be great. Also, if someone knows HOW to delete partitions, that would be a great help too. Thanks!

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  • Recover Data Like a Forensics Expert Using an Ubuntu Live CD

    - by Trevor Bekolay
    There are lots of utilities to recover deleted files, but what if you can’t boot up your computer, or the whole drive has been formatted? We’ll show you some tools that will dig deep and recover the most elusive deleted files, or even whole hard drive partitions. We’ve shown you simple ways to recover accidentally deleted files, even a simple method that can be done from an Ubuntu Live CD, but for hard disks that have been heavily corrupted, those methods aren’t going to cut it. In this article, we’ll examine four tools that can recover data from the most messed up hard drives, regardless of whether they were formatted for a Windows, Linux, or Mac computer, or even if the partition table is wiped out entirely. Note: These tools cannot recover data that has been overwritten on a hard disk. Whether a deleted file has been overwritten depends on many factors – the quicker you realize that you want to recover a file, the more likely you will be able to do so. Our setup To show these tools, we’ve set up a small 1 GB hard drive, with half of the space partitioned as ext2, a file system used in Linux, and half the space partitioned as FAT32, a file system used in older Windows systems. We stored ten random pictures on each hard drive. We then wiped the partition table from the hard drive by deleting the partitions in GParted. Is our data lost forever? Installing the tools All of the tools we’re going to use are in Ubuntu’s universe repository. To enable the repository, open Synaptic Package Manager by clicking on System in the top-left, then Administration > Synaptic Package Manager. Click on Settings > Repositories and add a check in the box labelled “Community-maintained Open Source software (universe)”. Click Close, and then in the main Synaptic Package Manager window, click the Reload button. Once the package list has reloaded, and the search index rebuilt, search for and mark for installation one or all of the following packages: testdisk, foremost, and scalpel. Testdisk includes TestDisk, which can recover lost partitions and repair boot sectors, and PhotoRec, which can recover many different types of files from tons of different file systems. Foremost, originally developed by the US Air Force Office of Special Investigations, recovers files based on their headers and other internal structures. Foremost operates on hard drives or drive image files generated by various tools. Finally, scalpel performs the same functions as foremost, but is focused on enhanced performance and lower memory usage. Scalpel may run better if you have an older machine with less RAM. Recover hard drive partitions If you can’t mount your hard drive, then its partition table might be corrupted. Before you start trying to recover your important files, it may be possible to recover one or more partitions on your drive, recovering all of your files with one step. Testdisk is the tool for the job. Start it by opening a terminal (Applications > Accessories > Terminal) and typing in: sudo testdisk If you’d like, you can create a log file, though it won’t affect how much data you recover. Once you make your choice, you’re greeted with a list of the storage media on your machine. You should be able to identify the hard drive you want to recover partitions from by its size and label. TestDisk asks you select the type of partition table to search for. In most cases (ext2/3, NTFS, FAT32, etc.) you should select Intel and press Enter. Highlight Analyse and press enter. In our case, our small hard drive has previously been formatted as NTFS. Amazingly, TestDisk finds this partition, though it is unable to recover it. It also finds the two partitions we just deleted. We are able to change their attributes, or add more partitions, but we’ll just recover them by pressing Enter. If TestDisk hasn’t found all of your partitions, you can try doing a deeper search by selecting that option with the left and right arrow keys. We only had these two partitions, so we’ll recover them by selecting Write and pressing Enter. Testdisk informs us that we will have to reboot. Note: If your Ubuntu Live CD is not persistent, then when you reboot you will have to reinstall any tools that you installed earlier. After restarting, both of our partitions are back to their original states, pictures and all. Recover files of certain types For the following examples, we deleted the 10 pictures from both partitions and then reformatted them. PhotoRec Of the three tools we’ll show, PhotoRec is the most user-friendly, despite being a console-based utility. To start recovering files, open a terminal (Applications > Accessories > Terminal) and type in: sudo photorec To begin, you are asked to select a storage device to search. You should be able to identify the right device by its size and label. Select the right device, and then hit Enter. PhotoRec asks you select the type of partition to search. In most cases (ext2/3, NTFS, FAT, etc.) you should select Intel and press Enter. You are given a list of the partitions on your selected hard drive. If you want to recover all of the files on a partition, then select Search and hit enter. However, this process can be very slow, and in our case we only want to search for pictures files, so instead we use the right arrow key to select File Opt and press Enter. PhotoRec can recover many different types of files, and deselecting each one would take a long time. Instead, we press “s” to clear all of the selections, and then find the appropriate file types – jpg, gif, and png – and select them by pressing the right arrow key. Once we’ve selected these three, we press “b” to save these selections. Press enter to return to the list of hard drive partitions. We want to search both of our partitions, so we highlight “No partition” and “Search” and then press Enter. PhotoRec prompts for a location to store the recovered files. If you have a different healthy hard drive, then we recommend storing the recovered files there. Since we’re not recovering very much, we’ll store it on the Ubuntu Live CD’s desktop. Note: Do not recover files to the hard drive you’re recovering from. PhotoRec is able to recover the 20 pictures from the partitions on our hard drive! A quick look in the recup_dir.1 directory that it creates confirms that PhotoRec has recovered all of our pictures, save for the file names. Foremost Foremost is a command-line program with no interactive interface like PhotoRec, but offers a number of command-line options to get as much data out of your had drive as possible. For a full list of options that can be tweaked via the command line, open up a terminal (Applications > Accessories > Terminal) and type in: foremost –h In our case, the command line options that we are going to use are: -t, a comma-separated list of types of files to search for. In our case, this is “jpeg,png,gif”. -v, enabling verbose-mode, giving us more information about what foremost is doing. -o, the output folder to store recovered files in. In our case, we created a directory called “foremost” on the desktop. -i, the input that will be searched for files. This can be a disk image in several different formats; however, we will use a hard disk, /dev/sda. Our foremost invocation is: sudo foremost –t jpeg,png,gif –o foremost –v –i /dev/sda Your invocation will differ depending on what you’re searching for and where you’re searching for it. Foremost is able to recover 17 of the 20 files stored on the hard drive. Looking at the files, we can confirm that these files were recovered relatively well, though we can see some errors in the thumbnail for 00622449.jpg. Part of this may be due to the ext2 filesystem. Foremost recommends using the –d command-line option for Linux file systems like ext2. We’ll run foremost again, adding the –d command-line option to our foremost invocation: sudo foremost –t jpeg,png,gif –d –o foremost –v –i /dev/sda This time, foremost is able to recover all 20 images! A final look at the pictures reveals that the pictures were recovered with no problems. Scalpel Scalpel is another powerful program that, like Foremost, is heavily configurable. Unlike Foremost, Scalpel requires you to edit a configuration file before attempting any data recovery. Any text editor will do, but we’ll use gedit to change the configuration file. In a terminal window (Applications > Accessories > Terminal), type in: sudo gedit /etc/scalpel/scalpel.conf scalpel.conf contains information about a number of different file types. Scroll through this file and uncomment lines that start with a file type that you want to recover (i.e. remove the “#” character at the start of those lines). Save the file and close it. Return to the terminal window. Scalpel also has a ton of command-line options that can help you search quickly and effectively; however, we’ll just define the input device (/dev/sda) and the output folder (a folder called “scalpel” that we created on the desktop). Our invocation is: sudo scalpel /dev/sda –o scalpel Scalpel is able to recover 18 of our 20 files. A quick look at the files scalpel recovered reveals that most of our files were recovered successfully, though there were some problems (e.g. 00000012.jpg). Conclusion In our quick toy example, TestDisk was able to recover two deleted partitions, and PhotoRec and Foremost were able to recover all 20 deleted images. Scalpel recovered most of the files, but it’s very likely that playing with the command-line options for scalpel would have enabled us to recover all 20 images. These tools are lifesavers when something goes wrong with your hard drive. If your data is on the hard drive somewhere, then one of these tools will track it down! 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  • Improving Partitioned Table Join Performance

    - by Paul White
    The query optimizer does not always choose an optimal strategy when joining partitioned tables. This post looks at an example, showing how a manual rewrite of the query can almost double performance, while reducing the memory grant to almost nothing. Test Data The two tables in this example use a common partitioning partition scheme. The partition function uses 41 equal-size partitions: CREATE PARTITION FUNCTION PFT (integer) AS RANGE RIGHT FOR VALUES ( 125000, 250000, 375000, 500000, 625000, 750000, 875000, 1000000, 1125000, 1250000, 1375000, 1500000, 1625000, 1750000, 1875000, 2000000, 2125000, 2250000, 2375000, 2500000, 2625000, 2750000, 2875000, 3000000, 3125000, 3250000, 3375000, 3500000, 3625000, 3750000, 3875000, 4000000, 4125000, 4250000, 4375000, 4500000, 4625000, 4750000, 4875000, 5000000 ); GO CREATE PARTITION SCHEME PST AS PARTITION PFT ALL TO ([PRIMARY]); There two tables are: CREATE TABLE dbo.T1 ( TID integer NOT NULL IDENTITY(0,1), Column1 integer NOT NULL, Padding binary(100) NOT NULL DEFAULT 0x,   CONSTRAINT PK_T1 PRIMARY KEY CLUSTERED (TID) ON PST (TID) );   CREATE TABLE dbo.T2 ( TID integer NOT NULL, Column1 integer NOT NULL, Padding binary(100) NOT NULL DEFAULT 0x,   CONSTRAINT PK_T2 PRIMARY KEY CLUSTERED (TID, Column1) ON PST (TID) ); The next script loads 5 million rows into T1 with a pseudo-random value between 1 and 5 for Column1. The table is partitioned on the IDENTITY column TID: INSERT dbo.T1 WITH (TABLOCKX) (Column1) SELECT (ABS(CHECKSUM(NEWID())) % 5) + 1 FROM dbo.Numbers AS N WHERE n BETWEEN 1 AND 5000000; In case you don’t already have an auxiliary table of numbers lying around, here’s a script to create one with 10 million rows: CREATE TABLE dbo.Numbers (n bigint PRIMARY KEY);   WITH L0 AS(SELECT 1 AS c UNION ALL SELECT 1), L1 AS(SELECT 1 AS c FROM L0 AS A CROSS JOIN L0 AS B), L2 AS(SELECT 1 AS c FROM L1 AS A CROSS JOIN L1 AS B), L3 AS(SELECT 1 AS c FROM L2 AS A CROSS JOIN L2 AS B), L4 AS(SELECT 1 AS c FROM L3 AS A CROSS JOIN L3 AS B), L5 AS(SELECT 1 AS c FROM L4 AS A CROSS JOIN L4 AS B), Nums AS(SELECT ROW_NUMBER() OVER (ORDER BY (SELECT NULL)) AS n FROM L5) INSERT dbo.Numbers WITH (TABLOCKX) SELECT TOP (10000000) n FROM Nums ORDER BY n OPTION (MAXDOP 1); Table T1 contains data like this: Next we load data into table T2. The relationship between the two tables is that table 2 contains ‘n’ rows for each row in table 1, where ‘n’ is determined by the value in Column1 of table T1. There is nothing particularly special about the data or distribution, by the way. INSERT dbo.T2 WITH (TABLOCKX) (TID, Column1) SELECT T.TID, N.n FROM dbo.T1 AS T JOIN dbo.Numbers AS N ON N.n >= 1 AND N.n <= T.Column1; Table T2 ends up containing about 15 million rows: The primary key for table T2 is a combination of TID and Column1. The data is partitioned according to the value in column TID alone. Partition Distribution The following query shows the number of rows in each partition of table T1: SELECT PartitionID = CA1.P, NumRows = COUNT_BIG(*) FROM dbo.T1 AS T CROSS APPLY (VALUES ($PARTITION.PFT(TID))) AS CA1 (P) GROUP BY CA1.P ORDER BY CA1.P; There are 40 partitions containing 125,000 rows (40 * 125k = 5m rows). The rightmost partition remains empty. The next query shows the distribution for table 2: SELECT PartitionID = CA1.P, NumRows = COUNT_BIG(*) FROM dbo.T2 AS T CROSS APPLY (VALUES ($PARTITION.PFT(TID))) AS CA1 (P) GROUP BY CA1.P ORDER BY CA1.P; There are roughly 375,000 rows in each partition (the rightmost partition is also empty): Ok, that’s the test data done. Test Query and Execution Plan The task is to count the rows resulting from joining tables 1 and 2 on the TID column: SET STATISTICS IO ON; DECLARE @s datetime2 = SYSUTCDATETIME();   SELECT COUNT_BIG(*) FROM dbo.T1 AS T1 JOIN dbo.T2 AS T2 ON T2.TID = T1.TID;   SELECT DATEDIFF(Millisecond, @s, SYSUTCDATETIME()); SET STATISTICS IO OFF; The optimizer chooses a plan using parallel hash join, and partial aggregation: The Plan Explorer plan tree view shows accurate cardinality estimates and an even distribution of rows across threads (click to enlarge the image): With a warm data cache, the STATISTICS IO output shows that no physical I/O was needed, and all 41 partitions were touched: Running the query without actual execution plan or STATISTICS IO information for maximum performance, the query returns in around 2600ms. Execution Plan Analysis The first step toward improving on the execution plan produced by the query optimizer is to understand how it works, at least in outline. The two parallel Clustered Index Scans use multiple threads to read rows from tables T1 and T2. Parallel scan uses a demand-based scheme where threads are given page(s) to scan from the table as needed. This arrangement has certain important advantages, but does result in an unpredictable distribution of rows amongst threads. The point is that multiple threads cooperate to scan the whole table, but it is impossible to predict which rows end up on which threads. For correct results from the parallel hash join, the execution plan has to ensure that rows from T1 and T2 that might join are processed on the same thread. For example, if a row from T1 with join key value ‘1234’ is placed in thread 5’s hash table, the execution plan must guarantee that any rows from T2 that also have join key value ‘1234’ probe thread 5’s hash table for matches. The way this guarantee is enforced in this parallel hash join plan is by repartitioning rows to threads after each parallel scan. The two repartitioning exchanges route rows to threads using a hash function over the hash join keys. The two repartitioning exchanges use the same hash function so rows from T1 and T2 with the same join key must end up on the same hash join thread. Expensive Exchanges This business of repartitioning rows between threads can be very expensive, especially if a large number of rows is involved. The execution plan selected by the optimizer moves 5 million rows through one repartitioning exchange and around 15 million across the other. As a first step toward removing these exchanges, consider the execution plan selected by the optimizer if we join just one partition from each table, disallowing parallelism: SELECT COUNT_BIG(*) FROM dbo.T1 AS T1 JOIN dbo.T2 AS T2 ON T2.TID = T1.TID WHERE $PARTITION.PFT(T1.TID) = 1 AND $PARTITION.PFT(T2.TID) = 1 OPTION (MAXDOP 1); The optimizer has chosen a (one-to-many) merge join instead of a hash join. The single-partition query completes in around 100ms. If everything scaled linearly, we would expect that extending this strategy to all 40 populated partitions would result in an execution time around 4000ms. Using parallelism could reduce that further, perhaps to be competitive with the parallel hash join chosen by the optimizer. This raises a question. If the most efficient way to join one partition from each of the tables is to use a merge join, why does the optimizer not choose a merge join for the full query? Forcing a Merge Join Let’s force the optimizer to use a merge join on the test query using a hint: SELECT COUNT_BIG(*) FROM dbo.T1 AS T1 JOIN dbo.T2 AS T2 ON T2.TID = T1.TID OPTION (MERGE JOIN); This is the execution plan selected by the optimizer: This plan results in the same number of logical reads reported previously, but instead of 2600ms the query takes 5000ms. The natural explanation for this drop in performance is that the merge join plan is only using a single thread, whereas the parallel hash join plan could use multiple threads. Parallel Merge Join We can get a parallel merge join plan using the same query hint as before, and adding trace flag 8649: SELECT COUNT_BIG(*) FROM dbo.T1 AS T1 JOIN dbo.T2 AS T2 ON T2.TID = T1.TID OPTION (MERGE JOIN, QUERYTRACEON 8649); The execution plan is: This looks promising. It uses a similar strategy to distribute work across threads as seen for the parallel hash join. In practice though, performance is disappointing. On a typical run, the parallel merge plan runs for around 8400ms; slower than the single-threaded merge join plan (5000ms) and much worse than the 2600ms for the parallel hash join. We seem to be going backwards! The logical reads for the parallel merge are still exactly the same as before, with no physical IOs. The cardinality estimates and thread distribution are also still very good (click to enlarge): A big clue to the reason for the poor performance is shown in the wait statistics (captured by Plan Explorer Pro): CXPACKET waits require careful interpretation, and are most often benign, but in this case excessive waiting occurs at the repartitioning exchanges. Unlike the parallel hash join, the repartitioning exchanges in this plan are order-preserving ‘merging’ exchanges (because merge join requires ordered inputs): Parallelism works best when threads can just grab any available unit of work and get on with processing it. Preserving order introduces inter-thread dependencies that can easily lead to significant waits occurring. In extreme cases, these dependencies can result in an intra-query deadlock, though the details of that will have to wait for another time to explore in detail. The potential for waits and deadlocks leads the query optimizer to cost parallel merge join relatively highly, especially as the degree of parallelism (DOP) increases. This high costing resulted in the optimizer choosing a serial merge join rather than parallel in this case. The test results certainly confirm its reasoning. Collocated Joins In SQL Server 2008 and later, the optimizer has another available strategy when joining tables that share a common partition scheme. This strategy is a collocated join, also known as as a per-partition join. It can be applied in both serial and parallel execution plans, though it is limited to 2-way joins in the current optimizer. Whether the optimizer chooses a collocated join or not depends on cost estimation. The primary benefits of a collocated join are that it eliminates an exchange and requires less memory, as we will see next. Costing and Plan Selection The query optimizer did consider a collocated join for our original query, but it was rejected on cost grounds. The parallel hash join with repartitioning exchanges appeared to be a cheaper option. There is no query hint to force a collocated join, so we have to mess with the costing framework to produce one for our test query. Pretending that IOs cost 50 times more than usual is enough to convince the optimizer to use collocated join with our test query: -- Pretend IOs are 50x cost temporarily DBCC SETIOWEIGHT(50);   -- Co-located hash join SELECT COUNT_BIG(*) FROM dbo.T1 AS T1 JOIN dbo.T2 AS T2 ON T2.TID = T1.TID OPTION (RECOMPILE);   -- Reset IO costing DBCC SETIOWEIGHT(1); Collocated Join Plan The estimated execution plan for the collocated join is: The Constant Scan contains one row for each partition of the shared partitioning scheme, from 1 to 41. The hash repartitioning exchanges seen previously are replaced by a single Distribute Streams exchange using Demand partitioning. Demand partitioning means that the next partition id is given to the next parallel thread that asks for one. My test machine has eight logical processors, and all are available for SQL Server to use. As a result, there are eight threads in the single parallel branch in this plan, each processing one partition from each table at a time. Once a thread finishes processing a partition, it grabs a new partition number from the Distribute Streams exchange…and so on until all partitions have been processed. It is important to understand that the parallel scans in this plan are different from the parallel hash join plan. Although the scans have the same parallelism icon, tables T1 and T2 are not being co-operatively scanned by multiple threads in the same way. Each thread reads a single partition of T1 and performs a hash match join with the same partition from table T2. The properties of the two Clustered Index Scans show a Seek Predicate (unusual for a scan!) limiting the rows to a single partition: The crucial point is that the join between T1 and T2 is on TID, and TID is the partitioning column for both tables. A thread that processes partition ‘n’ is guaranteed to see all rows that can possibly join on TID for that partition. In addition, no other thread will see rows from that partition, so this removes the need for repartitioning exchanges. CPU and Memory Efficiency Improvements The collocated join has removed two expensive repartitioning exchanges and added a single exchange processing 41 rows (one for each partition id). Remember, the parallel hash join plan exchanges had to process 5 million and 15 million rows. The amount of processor time spent on exchanges will be much lower in the collocated join plan. In addition, the collocated join plan has a maximum of 8 threads processing single partitions at any one time. The 41 partitions will all be processed eventually, but a new partition is not started until a thread asks for it. Threads can reuse hash table memory for the new partition. The parallel hash join plan also had 8 hash tables, but with all 5,000,000 build rows loaded at the same time. The collocated plan needs memory for only 8 * 125,000 = 1,000,000 rows at any one time. Collocated Hash Join Performance The collated join plan has disappointing performance in this case. The query runs for around 25,300ms despite the same IO statistics as usual. This is much the worst result so far, so what went wrong? It turns out that cardinality estimation for the single partition scans of table T1 is slightly low. The properties of the Clustered Index Scan of T1 (graphic immediately above) show the estimation was for 121,951 rows. This is a small shortfall compared with the 125,000 rows actually encountered, but it was enough to cause the hash join to spill to physical tempdb: A level 1 spill doesn’t sound too bad, until you realize that the spill to tempdb probably occurs for each of the 41 partitions. As a side note, the cardinality estimation error is a little surprising because the system tables accurately show there are 125,000 rows in every partition of T1. Unfortunately, the optimizer uses regular column and index statistics to derive cardinality estimates here rather than system table information (e.g. sys.partitions). Collocated Merge Join We will never know how well the collocated parallel hash join plan might have worked without the cardinality estimation error (and the resulting 41 spills to tempdb) but we do know: Merge join does not require a memory grant; and Merge join was the optimizer’s preferred join option for a single partition join Putting this all together, what we would really like to see is the same collocated join strategy, but using merge join instead of hash join. Unfortunately, the current query optimizer cannot produce a collocated merge join; it only knows how to do collocated hash join. So where does this leave us? CROSS APPLY sys.partitions We can try to write our own collocated join query. We can use sys.partitions to find the partition numbers, and CROSS APPLY to get a count per partition, with a final step to sum the partial counts. The following query implements this idea: SELECT row_count = SUM(Subtotals.cnt) FROM ( -- Partition numbers SELECT p.partition_number FROM sys.partitions AS p WHERE p.[object_id] = OBJECT_ID(N'T1', N'U') AND p.index_id = 1 ) AS P CROSS APPLY ( -- Count per collocated join SELECT cnt = COUNT_BIG(*) FROM dbo.T1 AS T1 JOIN dbo.T2 AS T2 ON T2.TID = T1.TID WHERE $PARTITION.PFT(T1.TID) = p.partition_number AND $PARTITION.PFT(T2.TID) = p.partition_number ) AS SubTotals; The estimated plan is: The cardinality estimates aren’t all that good here, especially the estimate for the scan of the system table underlying the sys.partitions view. Nevertheless, the plan shape is heading toward where we would like to be. Each partition number from the system table results in a per-partition scan of T1 and T2, a one-to-many Merge Join, and a Stream Aggregate to compute the partial counts. The final Stream Aggregate just sums the partial counts. Execution time for this query is around 3,500ms, with the same IO statistics as always. This compares favourably with 5,000ms for the serial plan produced by the optimizer with the OPTION (MERGE JOIN) hint. This is another case of the sum of the parts being less than the whole – summing 41 partial counts from 41 single-partition merge joins is faster than a single merge join and count over all partitions. Even so, this single-threaded collocated merge join is not as quick as the original parallel hash join plan, which executed in 2,600ms. On the positive side, our collocated merge join uses only one logical processor and requires no memory grant. The parallel hash join plan used 16 threads and reserved 569 MB of memory:   Using a Temporary Table Our collocated merge join plan should benefit from parallelism. The reason parallelism is not being used is that the query references a system table. We can work around that by writing the partition numbers to a temporary table (or table variable): SET STATISTICS IO ON; DECLARE @s datetime2 = SYSUTCDATETIME();   CREATE TABLE #P ( partition_number integer PRIMARY KEY);   INSERT #P (partition_number) SELECT p.partition_number FROM sys.partitions AS p WHERE p.[object_id] = OBJECT_ID(N'T1', N'U') AND p.index_id = 1;   SELECT row_count = SUM(Subtotals.cnt) FROM #P AS p CROSS APPLY ( SELECT cnt = COUNT_BIG(*) FROM dbo.T1 AS T1 JOIN dbo.T2 AS T2 ON T2.TID = T1.TID WHERE $PARTITION.PFT(T1.TID) = p.partition_number AND $PARTITION.PFT(T2.TID) = p.partition_number ) AS SubTotals;   DROP TABLE #P;   SELECT DATEDIFF(Millisecond, @s, SYSUTCDATETIME()); SET STATISTICS IO OFF; Using the temporary table adds a few logical reads, but the overall execution time is still around 3500ms, indistinguishable from the same query without the temporary table. The problem is that the query optimizer still doesn’t choose a parallel plan for this query, though the removal of the system table reference means that it could if it chose to: In fact the optimizer did enter the parallel plan phase of query optimization (running search 1 for a second time): Unfortunately, the parallel plan found seemed to be more expensive than the serial plan. This is a crazy result, caused by the optimizer’s cost model not reducing operator CPU costs on the inner side of a nested loops join. Don’t get me started on that, we’ll be here all night. In this plan, everything expensive happens on the inner side of a nested loops join. Without a CPU cost reduction to compensate for the added cost of exchange operators, candidate parallel plans always look more expensive to the optimizer than the equivalent serial plan. Parallel Collocated Merge Join We can produce the desired parallel plan using trace flag 8649 again: SELECT row_count = SUM(Subtotals.cnt) FROM #P AS p CROSS APPLY ( SELECT cnt = COUNT_BIG(*) FROM dbo.T1 AS T1 JOIN dbo.T2 AS T2 ON T2.TID = T1.TID WHERE $PARTITION.PFT(T1.TID) = p.partition_number AND $PARTITION.PFT(T2.TID) = p.partition_number ) AS SubTotals OPTION (QUERYTRACEON 8649); The actual execution plan is: One difference between this plan and the collocated hash join plan is that a Repartition Streams exchange operator is used instead of Distribute Streams. The effect is similar, though not quite identical. The Repartition uses round-robin partitioning, meaning the next partition id is pushed to the next thread in sequence. The Distribute Streams exchange seen earlier used Demand partitioning, meaning the next partition id is pulled across the exchange by the next thread that is ready for more work. There are subtle performance implications for each partitioning option, but going into that would again take us too far off the main point of this post. Performance The important thing is the performance of this parallel collocated merge join – just 1350ms on a typical run. The list below shows all the alternatives from this post (all timings include creation, population, and deletion of the temporary table where appropriate) from quickest to slowest: Collocated parallel merge join: 1350ms Parallel hash join: 2600ms Collocated serial merge join: 3500ms Serial merge join: 5000ms Parallel merge join: 8400ms Collated parallel hash join: 25,300ms (hash spill per partition) The parallel collocated merge join requires no memory grant (aside from a paltry 1.2MB used for exchange buffers). This plan uses 16 threads at DOP 8; but 8 of those are (rather pointlessly) allocated to the parallel scan of the temporary table. These are minor concerns, but it turns out there is a way to address them if it bothers you. Parallel Collocated Merge Join with Demand Partitioning This final tweak replaces the temporary table with a hard-coded list of partition ids (dynamic SQL could be used to generate this query from sys.partitions): SELECT row_count = SUM(Subtotals.cnt) FROM ( VALUES (1),(2),(3),(4),(5),(6),(7),(8),(9),(10), (11),(12),(13),(14),(15),(16),(17),(18),(19),(20), (21),(22),(23),(24),(25),(26),(27),(28),(29),(30), (31),(32),(33),(34),(35),(36),(37),(38),(39),(40),(41) ) AS P (partition_number) CROSS APPLY ( SELECT cnt = COUNT_BIG(*) FROM dbo.T1 AS T1 JOIN dbo.T2 AS T2 ON T2.TID = T1.TID WHERE $PARTITION.PFT(T1.TID) = p.partition_number AND $PARTITION.PFT(T2.TID) = p.partition_number ) AS SubTotals OPTION (QUERYTRACEON 8649); The actual execution plan is: The parallel collocated hash join plan is reproduced below for comparison: The manual rewrite has another advantage that has not been mentioned so far: the partial counts (per partition) can be computed earlier than the partial counts (per thread) in the optimizer’s collocated join plan. The earlier aggregation is performed by the extra Stream Aggregate under the nested loops join. The performance of the parallel collocated merge join is unchanged at around 1350ms. Final Words It is a shame that the current query optimizer does not consider a collocated merge join (Connect item closed as Won’t Fix). The example used in this post showed an improvement in execution time from 2600ms to 1350ms using a modestly-sized data set and limited parallelism. In addition, the memory requirement for the query was almost completely eliminated  – down from 569MB to 1.2MB. The problem with the parallel hash join selected by the optimizer is that it attempts to process the full data set all at once (albeit using eight threads). It requires a large memory grant to hold all 5 million rows from table T1 across the eight hash tables, and does not take advantage of the divide-and-conquer opportunity offered by the common partitioning. The great thing about the collocated join strategies is that each parallel thread works on a single partition from both tables, reading rows, performing the join, and computing a per-partition subtotal, before moving on to a new partition. From a thread’s point of view… If you have trouble visualizing what is happening from just looking at the parallel collocated merge join execution plan, let’s look at it again, but from the point of view of just one thread operating between the two Parallelism (exchange) operators. Our thread picks up a single partition id from the Distribute Streams exchange, and starts a merge join using ordered rows from partition 1 of table T1 and partition 1 of table T2. By definition, this is all happening on a single thread. As rows join, they are added to a (per-partition) count in the Stream Aggregate immediately above the Merge Join. Eventually, either T1 (partition 1) or T2 (partition 1) runs out of rows and the merge join stops. The per-partition count from the aggregate passes on through the Nested Loops join to another Stream Aggregate, which is maintaining a per-thread subtotal. Our same thread now picks up a new partition id from the exchange (say it gets id 9 this time). The count in the per-partition aggregate is reset to zero, and the processing of partition 9 of both tables proceeds just as it did for partition 1, and on the same thread. Each thread picks up a single partition id and processes all the data for that partition, completely independently from other threads working on other partitions. One thread might eventually process partitions (1, 9, 17, 25, 33, 41) while another is concurrently processing partitions (2, 10, 18, 26, 34) and so on for the other six threads at DOP 8. The point is that all 8 threads can execute independently and concurrently, continuing to process new partitions until the wider job (of which the thread has no knowledge!) is done. This divide-and-conquer technique can be much more efficient than simply splitting the entire workload across eight threads all at once. Related Reading Understanding and Using Parallelism in SQL Server Parallel Execution Plans Suck © 2013 Paul White – All Rights Reserved Twitter: @SQL_Kiwi

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  • The partition table is corrupt

    - by Tim
    I have a corrupt the partition table on the laptop that is running Ubunutu 10.4. Before the partition table was corrupt I had the following partitions: 2 primary partitions: 1st - NTFS 2nd - Extended 4 logical partitons that are built within 2nd extended: 1st NTFS (68 Gib) 2nd Linux (19 Gib) 3rd Swap (1.4 Gib) 4th Linux (24 Gib) The physical order of these partitions was the following: ( 4th Linux ) - ( 1st NTFS ) - ( 2nd Linux ) - ( 3rd Swap ) The logical order of the partition was different: ( 1st NTFS ) - ( 2nd Linux ) - ( 3rd Swap ) ( 4th Linux ) NTFS partition was big and it resided between 2 Linux partitions, neither of these partitions had enough space to install Oracle 11g. Therefore, I decided to a) either move the NTFS partion to the left or b) remove it completely and extend partition where Linux resides. As I tool I have chosen GParted. But unfortunately it was not able to move the partition because he found that in NTFS partition there are some blocks that are referenced multiple times. Also it was not able to remove the partition neither, because in this case the partitions that follow it ( 2nd Linux ) - ( 3rd Swap ) have to be in his opinion also removed, because the organization of extended partition is a linked list. Since GParted was not able to do such thing I was trying to find another tool. I found diskdrake tool on PSLinuxOS distribution of linux. That tool silently deleted ( 1st NTFS ) partition and I thought that everything was fine. But diskdrake has damaged the partition in a way that I am not able either to boot from the hard disk nor to see the partitions with GParted and even with diskdrake itself! Fortunately I have a live CD of Ubuntu 8.10 and I am able to boot and see hard disk. I have 2 ideas how I can solve the problem: 1) Manually change disk partitions and point them to the correct partitions. 2) Create partition table with GParted that as much as possible is the same with the previous one I find the 2nd approach less time consuming but some data will be lost because of it is not possible to place borders of the partitions exactly how it was before. And moreover I am not sure if such approach would work, for example, if the OS is able to locate files after repartitioning. I feel like that it will but not 100% sure. Are there some ideas how the problem may be solved?

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  • How should I generate the partitions / pairs for the Chinese Postman problem?

    - by Simucal
    I'm working on a program for class that involves solving the Chinese Postman problem. Our assignment only requires us to write a program to solve it for a hard-coded graph but I'm attempting to solve it for the general case on my own. The part that is giving me trouble is generating the partitions of pairings for the odd vertices. For example, if I had the following labeled odd verticies in a graph: 1 2 3 4 5 6 I need to find all the possible pairings / partitions I can make with these vertices. I've figured out I'll have i paritions given: n = num of odd verticies k = n / 2 i = ((2k)(2k-1)(2k-2)...(k+1))/2 So, given the 6 odd verticies above, we will know that we need to generate i = 15 partitions. The 15 partions would look like: 1 2 3 4 5 6 1 2 3 5 4 6 1 2 3 6 4 5 ... 1 6 ... Then, for each partition, I take each pair and find the shortest distance between them and sum them for that partition. The partition with the total smallest distance between its pairs is selected, and I then double all the edges between the shortest path between the odd vertices (found in the selected partition). These represent the edges the postman will have to walk twice. At first I thought I had worked out an appropriate algorithm for generating these partitions / pairs but it is flawed. I found it wasn't a simple permutation/combination problem. Does anyone who has studied this problem before have any tips that can help point me in the right direction for generating these partitions?

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  • What's the reason to break HDD to few partitions for MDADM+LVM2?

    - by archer
    I'm using 2 HDDs each 1TB in size. I'm going to create MDADM+LVM2 over them. Initially I though about this partition layout: /dev/sda1 - 1Gb (boot) /dev/sda2 - 500Gb (md0) /dev/sda3 - 499Gb (md1) /dev/sdb1 - 1Gb (boot) /dev/sdb2 - 500Gb (md0) /dev/sdb3 - 499Gb (md1) md0 is going to be raid0 and md1 is going to be raid1 however, I found some info that this would be better to break each drive to more partitions (lets say 10 partitions 100Gb in size each). What's the reason of doing that?

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  • I've two swap partitions, how can i delete one?

    - by Gp2mv3
    I've installed Ubuntu on my computer but I did a mistake during the installation and it created two swap's. In fact I had tree partitions (system, home, swap) but the installator crashed so I restarted the installation and it has installed everything in the system partition. So now I separated the home in the appropriate partition but I've two swap partitions. How can I delete one ? If I delete one, how will it go ?

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  • Partition Wise Joins

    - by jean-pierre.dijcks
    Some say they are the holy grail of parallel computing and PWJ is the basis for a shared nothing system and the only join method that is available on a shared nothing system (yes this is oversimplified!). The magic in Oracle is of course that is one of many ways to join data. And yes, this is the old flexibility vs. simplicity discussion all over, so I won't go there... the point is that what you must do in a shared nothing system, you can do in Oracle with the same speed and methods. The Theory A partition wise join is a join between (for simplicity) two tables that are partitioned on the same column with the same partitioning scheme. In shared nothing this is effectively hard partitioning locating data on a specific node / storage combo. In Oracle is is logical partitioning. If you now join the two tables on that partitioned column you can break up the join in smaller joins exactly along the partitions in the data. Since they are partitioned (grouped) into the same buckets, all values required to do the join live in the equivalent bucket on either sides. No need to talk to anyone else, no need to redistribute data to anyone else... in short, the optimal join method for parallel processing of two large data sets. PWJ's in Oracle Since we do not hard partition the data across nodes in Oracle we use the Partitioning option to the database to create the buckets, then set the Degree of Parallelism (or run Auto DOP - see here) and get our PWJs. The main questions always asked are: How many partitions should I create? What should my DOP be? In a shared nothing system the answer is of course, as many partitions as there are nodes which will be your DOP. In Oracle we do want you to look at the workload and concurrency, and once you know that to understand the following rules of thumb. Within Oracle we have more ways of joining of data, so it is important to understand some of the PWJ ideas and what it means if you have an uneven distribution across processes. Assume we have a simple scenario where we partition the data on a hash key resulting in 4 hash partitions (H1 -H4). We have 2 parallel processes that have been tasked with reading these partitions (P1 - P2). The work is evenly divided assuming the partitions are the same size and we can scan this in time t1 as shown below. Now assume that we have changed the system and have a 5th partition but still have our 2 workers P1 and P2. The time it takes is actually 50% more assuming the 5th partition has the same size as the original H1 - H4 partitions. In other words to scan these 5 partitions, the time t2 it takes is not 1/5th more expensive, it is a lot more expensive and some other join plans may now start to look exciting to the optimizer. Just to post the disclaimer, it is not as simple as I state it here, but you get the idea on how much more expensive this plan may now look... Based on this little example there are a few rules of thumb to follow to get the partition wise joins. First, choose a DOP that is a factor of two (2). So always choose something like 2, 4, 8, 16, 32 and so on... Second, choose a number of partitions that is larger or equal to 2* DOP. Third, make sure the number of partitions is divisible through 2 without orphans. This is also known as an even number... Fourth, choose a stable partition count strategy, which is typically hash, which can be a sub partitioning strategy rather than the main strategy (range - hash is a popular one). Fifth, make sure you do this on the join key between the two large tables you want to join (and this should be the obvious one...). Translating this into an example: DOP = 8 (determined based on concurrency or by using Auto DOP with a cap due to concurrency) says that the number of partitions >= 16. Number of hash (sub) partitions = 32, which gives each process four partitions to work on. This number is somewhat arbitrary and depends on your data and system. In this case my main reasoning is that if you get more room on the box you can easily move the DOP for the query to 16 without repartitioning... and of course it makes for no leftovers on the table... And yes, we recommend up-to-date statistics. And before you start complaining, do read this post on a cool way to do stats in 11.

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  • Are high office partitions as effective as private offices?

    - by CraigS
    I'm currently reading DeMarco and Lister's Peopleware, and I've been struck (as everyone is) by their comments about noise reduction via using private offices, and the effect this has on productivity. Private offices are probably not going to happen at my workplace, but I'm wondering if high cubicle partitions (say, 6 feet) might be nearly as good? I imagine they wouldn't deflect quite as much noise, but they would have some effect. One down-side is that the center cubicles would have less natural light. That seems quite a big downer to me. I'd be interested to hear what peoples experiences are.

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  • root and home are on seperate partitions, but home still uses up space on root?

    - by Void
    When installing Ubuntu 12.10 for the first time, I made sure /home had it's own partition, gave it about 185GB, and gave root ~10GB. Now root is almost full after a few days and I've noticed that root is as big as /home and some additional MB from some actual root files. Note that I deleted some unused files in /home to make sure it really affects how much space is used in root, and it clearly does. (checked in gparted) I've also made sure the files I am talking about are actually in /home/myname/ and not just in root's home. df -h spits out this information: df: `/root/.gvfs': Permission denied Filesystem Size Used Avail Use% Mounted on /dev/sdb2 9.2G 7.9G 908M 90% / udev 3.9G 12K 3.9G 1% /dev tmpfs 1.6G 932K 1.6G 1% /run none 5.0M 0 5.0M 0% /run/lock none 3.9G 6.4M 3.9G 1% /run/shm none 100M 112K 100M 1% /run/user /dev/sdb3 184G 4.9G 170G 3% /home /dev/sdb4 729G 87G 642G 12% /media/mave/Storage I cut out my windows partitions as I don't see any relevance. I hope this is enough information for someone to tell me what I did wrong

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  • Database Partitioning and Multiple Data Source Considerations

    - by Jeffrey McDaniel
    With the release of P6 Reporting Database 3.0 partitioning was added as a feature to help with performance and data management.  Careful investigation of requirements should be conducting prior to installation to help improve overall performance throughout the lifecycle of the data warehouse, preventing future maintenance that would result in data loss. Before installation try to determine how many data sources and partitions will be required along with the ranges.  In P6 Reporting Database 3.0 any adjustments outside of defaults must be made in the scripts and changes will require new ETL runs for each data source.  Considerations: 1. Standard Edition or Enterprise Edition of Oracle Database.   If you aren't using Oracle Enterprise Edition Database; the partitioning feature is not available. Multiple Data sources are only supported on Enterprise Edition of Oracle   Database. 2. Number of Data source Ids for partitioning during configuration.   This setting will specify how many partitions will be allocated for tables containing data source information.  This setting requires some evaluation prior to installation as       there are repercussions if you don't estimate correctly.   For example, if you configured the software for only 2 data sources and the partition setting was set to 2, however along came a 3rd data source.  The necessary steps to  accommodate this change are as follows: a) By default, 3 partitions are configured in the Reporting Database scripts. Edit the create_star_tables_part.sql script located in <installation directory>\star\scripts   and search for partition.  You’ll see P1, P2, P3.  Add additional partitions and sub-partitions for P4 and so on. These will appear in several areas.  (See P6 Reporting Database 3.0 Installation and Configuration guide for more information on this and how to adjust partition ranges). b) Run starETL -r.  This will recreate each table with the new partition key.  The effect of this step is that all tables data will be lost except for history related tables.   c) Run starETL for each of the 3 data sources (with the data source # (starETL.bat "-s2" -as defined in P6 Reporting Database 3.0 Installation and Configuration guide) The best strategy for this setting is to overestimate based on possible growth.  If during implementation it is deemed that there are atleast 2 data sources with possibility for growth, it is a better idea to set this setting to 4 or 5, allowing room for the future and preventing a ‘start over’ scenario. 3. The Number of Partitions and the Number of Months per Partitions are not specific to multi-data source.  These settings work in accordance to a sub partition of larger tables with regard to time related data.  These settings are dataset specific for optimization.  The number of months per partition is self explanatory, optimally the smaller the partition, the better query performance so if the dataset has an extremely large number of spread/history records, a lower number of months is optimal.  Working in accordance with this setting is the number of partitions, this will determine how many "buckets" will be created per the number of months setting.  For example, if you kept the default for # of partitions of 3, and select 2 months for each partitions you would end up with: -1st partition, 2 months -2nd partition, 2 months -3rd partition, all the remaining records Therefore with records to this setting, it is important to analyze your source db spread ranges and history settings when determining the proper number of months per partition and number of partitions to optimize performance.  Also be aware the DBA will need to monitor when these partition ranges will fill up and when additional partitions will need to be added.  If you get to the final range partition and there are no additional range partitions all data will be included into the last partition. 

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  • After installing ubuntu, all my partitions are gone. Boot-repair log

    - by user211079
    I have an HP Pavilion Sleekbook that came with windows 8 pre-installed. I had trouble dual booting after installing ubuntu, so I disabled safeboot on bios and proceeded to try boot-repair, nothing happened. No dual boot yet. So I tried to reinstall ubuntu, but without the manual partitioning. So I chose to erase ubuntu 13 and reinstall it. Instead it deleted all my HP recovery partitions and windows as well. Here is the log of boot-repair. http://paste.ubuntu.com/6354919/ Gparted and fdisk only show one partition: /dev/sda1 I am wondering if you could suggest any way of recovering my windows partition and have a working windows 8 again? I need some information there with urgency. If you could help me I will be welcome. I am desperated. Thanks

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  • How can I recover XFS partitions from a formatted HD?

    - by giuprivite
    I deleted the partition table of my HD. I wanted to format another one, but by mistake, I formatted the wrong one. Then I also created some new partition on it. Now I would like, if possible, to recover my old data. The old configuration was this: A primary NTFS partition with Windows, and a secondary partition with four logical partitions: a swap and three XFS partitions (two for Ubuntu and OpenSuSE, and one with the home for both systems). This is the output I get when I run gpart in a terminal: ubuntu@ubuntu:~$ sudo gpart /dev/sdb Begin scan... Possible partition(Windows NT/W2K FS), size(39997mb), offset(0mb) Possible extended partition at offset(39997mb) Possible partition(Linux swap), size(8189mb), offset(39997mb) Possible partition(SGI XFS filesystem), size(40942mb), offset(48187mb) Possible partition(SGI XFS filesystem), size(40942mb), offset(89149mb) Possible partition(SGI XFS filesystem), size(175044mb), offset(130112mb) End scan. Checking partitions... Partition(OS/2 HPFS, NTFS, QNX or Advanced UNIX): primary Partition(Linux swap or Solaris/x86): logical Partition(Linux ext2 filesystem): logical Partition(Linux ext2 filesystem): orphaned logical Partition(Linux ext2 filesystem): orphaned logical Ok. Guessed primary partition table: Primary partition(1) type: 007(0x07)(OS/2 HPFS, NTFS, QNX or Advanced UNIX) size: 39997mb #s(81915360) s(63-81915422) chs: (0/1/1)-(1023/254/63)d (0/1/1)-(5098/254/51)r Primary partition(2) type: 015(0x0F)(Extended DOS, LBA) size: 265245mb #s(543221849) s(81915435-625137283) chs: (1023/254/63)-(1023/254/63)d (5099/0/1)-(38912/254/2)r Primary partition(3) type: 000(0x00)(unused) size: 0mb #s(0) s(0-0) chs: (0/0/0)-(0/0/0)d (0/0/0)-(0/0/0)r Primary partition(4) type: 000(0x00)(unused) size: 0mb #s(0) s(0-0) chs: (0/0/0)-(0/0/0)d (0/0/0)-(0/0/0)r Looking the first eight lines, it seems the data are still there... but I don't know how to recover them. I have a free second HD of about 500 GB (the formatted one is 320 GB) that I can use for the recovery process.

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  • How can I recover XFS partitions from a formatted HD?

    - by giuprivite
    I deleted the partition table of my HD. I wanted to format another one, but by mistake, I formatted the wrong one. Then I also created some new partition on it. Now I would like, if possible, to recover my old data. The old configuration was this: A primary NTFS partition with Windows, and a secondary partition with four logical partitions: a swap and three XFS partitions (two for Ubuntu and OpenSuSE, and one with the home for both systems). This is the output I get when I run gpart in a terminal: ubuntu@ubuntu:~$ sudo gpart /dev/sdb Begin scan... Possible partition(Windows NT/W2K FS), size(39997mb), offset(0mb) Possible extended partition at offset(39997mb) Possible partition(Linux swap), size(8189mb), offset(39997mb) Possible partition(SGI XFS filesystem), size(40942mb), offset(48187mb) Possible partition(SGI XFS filesystem), size(40942mb), offset(89149mb) Possible partition(SGI XFS filesystem), size(175044mb), offset(130112mb) End scan. Checking partitions... Partition(OS/2 HPFS, NTFS, QNX or Advanced UNIX): primary Partition(Linux swap or Solaris/x86): logical Partition(Linux ext2 filesystem): logical Partition(Linux ext2 filesystem): orphaned logical Partition(Linux ext2 filesystem): orphaned logical Ok. Guessed primary partition table: Primary partition(1) type: 007(0x07)(OS/2 HPFS, NTFS, QNX or Advanced UNIX) size: 39997mb #s(81915360) s(63-81915422) chs: (0/1/1)-(1023/254/63)d (0/1/1)-(5098/254/51)r Primary partition(2) type: 015(0x0F)(Extended DOS, LBA) size: 265245mb #s(543221849) s(81915435-625137283) chs: (1023/254/63)-(1023/254/63)d (5099/0/1)-(38912/254/2)r Primary partition(3) type: 000(0x00)(unused) size: 0mb #s(0) s(0-0) chs: (0/0/0)-(0/0/0)d (0/0/0)-(0/0/0)r Primary partition(4) type: 000(0x00)(unused) size: 0mb #s(0) s(0-0) chs: (0/0/0)-(0/0/0)d (0/0/0)-(0/0/0)r Looking the first eight lines, it seems the data are still there... but I don't know how to recover them. I have a free second HD of about 500 GB (the formatted one is 320 GB) that I can use for the recovery process.

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  • Messed up partitions... system will not boot!

    - by someguy
    I did a really dumb thing. cfdisk threw an error at me saying "FATAL ERROR: Bad primary partition 3: Partition ends in the final partial cylinder", so I installed Partition Table Doctor to see if I could fix the problem. When the program started up, it told me there were problems with my partitions, and asked if I wanted them fixed (cannot remember real message, but I believe it had something to do with the cylinder boundaries), so, blindly, without thinking of the consequences, I did. Now, my system will not boot. I tried booting from the Windows 7 installation CD. I went to install a fresh copy, but it said that "No drives were found". I then opened up diskpart. According to diskpart, there is only one partition, containing one volume, assigned the letter "C". Before, I had four partitions! It is also saying that the file system is RAW. Is there any way I can fix this? I have important data that I do not want to lose. Later on... I tried fdisk with the option -l, which lists the partition table(s), and this is what I got: Ignoring extra extended partition 4 Disk /dev/sda: 250.1 GB, 250059350016 bytes 255 heads, 64 sectors/track, 30401 cylinders Units = cylinders of 16065 * 512 = 8225280 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x163df116 Device Boot Start End Blocks Id System /dev/sda1 6 18 102400 7 HPFS/NTFS Partition 1 does not end on cylinder boundary. /dev/sda2 18 7851 62918572+ 7 HPFS/NTFS /dev/sda3 13073 30402 139196416 f W95 Ext'd (LBA) /dev/sda3 13073 30402 139196416 f W95 Ext'd (LBA) /dev/sda3 13073 30403 139203193 7 HPFS/NTFS I don't know if this will help, but it's extra information, at least. Also, this is how I had my partitions: 40MB (Unallocated) 100MB (System Reserved) 60GB (Windows, C:) 40GB (Was reserved for secondary OS) ~132GB (Home, E:)

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  • Gparted resize of an extended partition fails with error "can't have overlapping partitions".

    - by Marcus
    I just decided to install Ubuntu 12.04 alongside Windows 7 on my Dell laptop. However I didn't do this manually but instead used the "Install Ubuntu alongside Windows 7" option during the installation. Now the partition that Ubuntu runs in has very little space and I am getting warning messages. I'm trying to use gparted 0.12.1-5 (via a live CD) to give Windows less space and give Ubuntu more. I've managed to remove 100GB from the Windows partition so I now have some unallocated space between Windows and Ubuntu. This is what it looks like inside Ubuntu (not using the live CD, since it won't let me mount a USB to save a screenshot): So first I take sda4 (extended?) and resize it to the left so it takes up all the unallocated space. Then I resize sda5 (ext4) as well so it takes up all the new space. However, when I hit apply, it fails on the first action (resizing sd4) with the error message can't have overlapping partitions. Any ideas as to why this happens? I also tried resizing sda4 by just a few MB so that it definitely didn't overlap anything, but I still got the same error message. To clarify, I am using gparted from the LiveCD, I just took the screenshot from Ubuntu. I couldn't attach the details file containing the error information from gparted because I can't mount a USB drive when I'm running from the LiveCD. I'm tried following the guide on the gparted website but it says Invalid argument or something like that. If the gparted details are needed, I may need some hints on how to solve the USB issue as well. :)

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  • Problem in multi booting Ubuntu 12.04 with existing Windows XP, 7 and 8 in 500GB HDD with 5 Partitions

    - by Dhruva
    Here's my case. I have 500GB HDD with 5 Partitions with XP, Windows7 and Windows8 RP in the first three. As per one of the instruction I've seen in this forum, I did shrink my 4th Partition to create a 30GB unallocated free space to install Ubuntu 12.04. But, when next I'm trying to boot the Ubuntu CD and choosing "Something Else", its only recognizing my 500GB HDD in whole as "/sda" and not reading the free 30GB space separately to install Ubuntu in it as suggested in the instruction mentioned in this forum. I've also tried to install in from within Windows7, by mounting the Ubuntu ISO file and using the .exe file and instruction thereupon (choosing free drive, user name, installation size, etc.), but that also failed after the PC restarted to continue the installation, showing as error for file extension, partition something error. One thing to be noted that the PC I'm trying to install Ubuntu in it is my Home PC and doesn't have any internet connection. Hence, no updates or otherwise online help. What shall I do?? Kindly suggest. Sorry if I made some grammatical mistakes as English is not my first language. Thanks in advance.

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  • How to set individual NTFS partitions permissions behaviour for each user account?

    - by ryniek
    I have two NTFS partitions (DOWNLOADS for downloaded files and VM for my VirtualBox .vdi file) for which i must have full permissions for my allday use account. They should be also automounting when i login to this account. But i've also set Guest account for guests. For Guest account, i want make VM partition fully disabled and invisible (and thus it mustn't automount) but DOWNLOADS partition should be shared with limited privileges with Guest account. Editing fstab i'm able to share DOWNLOADS partition with Guest on limited privileges but VM can be only set to limited and have disabled automounting - so guest can't mount it but it still can be seen in Nautilus, plus I must always mount it manually when i login to allday account. Is there some trick to make what i want? Here's my fstab config: # /etc/fstab: static file system information. # # <file system> <mount point> <type> <options> <dump> <pass> proc /proc proc nodev,noexec,nosuid 0 0 #Entry for /dev/sda1 : UUID=35e66658-5ee9-40cf-bf56-8204959e3df0 / xfs defaults 01 #Entry for /dev/sda2 : UUID=26c714cf-4236-45e7-9c46-cfcf91a215ae /home xfs defaults 02 #Entry for /dev/sda5 : UUID=1315BCB027C44639 /media/DOWNLOADS ntfs-3g auto,uid=1000,gid=1000,umask=0022,nodev,locale=pl_PL.utf8 0 0 #Entry for /dev/sda6 : UUID=60FF39EB72B72264 /media/VM ntfs noauto,uid=1000,gid=1000,umask=0077,nodev,locale=pl_PL.utf8 0 0 #Entry for /dev/sda7 : UUID=c52411f5-105c-45d1-971f-412f962c350e none swap sw 0 0 /dev/fd0 /media/floppy0 auto rw,user,noauto,exec,utf8 0 0 Thanks

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  • many partitions on a single filegroup?¿ does it make sense?

    - by river0
    Hi, I'm designing a datawarehouse solution and I'm a newbie in disk configuration issues, let me explain you. Our storage is spread over 6 storage enlosures having each of them 5 raid-1 disk arrays, and having 2 LUNS defined per each disk array, which makes a total 48 LUNS (this is following Microsoft fast track recommendations for datawarehouse architectures). I would like to partition my data, on other projects I have worked before, we always followed a 1 partition - 1 filegroup rule. On the microsoft fast track recomendations it is advised to create a filegroup and then for that filegroup a data file per each lun... but I pretend to have a week level partitioning... if I apply that rule I think that I'll get too many files and a complex layout. I'm thinking of just creating just one filegroup (with the 48 lun data files), but still create the partitions since I want to keep soem of the benefits of partitions like partition switching... Is this scenario not recommended? What would you suggest?

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  • Dual HDD dual boot Win7/Win7 hibernate corrupts partitions?

    - by Ivan Zlatev
    I have two SSDs in my laptop. Both have 2 partitions each one with Windows 7 64bit and another one with OEM stuff on it (SYSTEM_DRV). I use the BIOS boot menu to boot between the drives. I have removed the drive letters so that no partitions of drive 1 are mapped in windows 2 on drive 2 and wise-versa. What I have observered however is that when I hibernate windows 1, boot windows 2 do some work and shutdown then resume windows 1 - the windows 2 partition will get corrupt quite often. Just happened actually - it is shown as "RAW" in the Disk manager instead of NTFS and windows will blue screen at boot. Alternatively I've seen cases where chkdsk will run and will wipe out all security descriptors making the partition completly unbootable. Why am I seeing these corruptions and what can I do to prevent them?

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