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  • Apache suddenly very slow on http and faster on https

    - by hsnm
    Background: I have Apache 2 running on ubuntu. There is a low usage on it and mostly being accessed for a web service URL from mobile apps. It was working fine until I installed SSL certificates. I now have both http and https. When I access the server using https, I get a fairly quick response (but probably not as fast as before). When I use http, it's so slow. What I tried: From this post: I curl localhost from the host and it takes some time, meaning there is no routing issue. The server runs on Amazon EC2 instance and is managed by me only. Also: I see that Apache once running, creates the maximum number of processes it is allowed to, which was not the case before. I lowered the MaxClients to 20 and I think I'm getting faster responses but it still takes over a minute and I always have MaxClients Apache processes. dmesg returns many [ 1953.655703] TCP: Possible SYN flooding on port 80. Sending cookies. When I netstat I get many entries with SYN_RECV. Possibly a DDoS attack? From EC2's monitoring diagrams I see a pattern of high "Maximum Network In (Bytes)" since 2 days ago. By the way the server is still being tested, the actual traffic is very low and not consistent. I tried to go with this solution to limit incoming connections using iptables, still no luck, but I'm trying. Question: What could be the problem? Is this a DDoS attack?

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  • MysqlTunner and query_cache_size dilemma

    - by wbad
    On a busy mysql server MySQLTuner 1.2.0 always recommends to add query_cache_size no matter how I increase the value (I tried up to 512MB). On the other hand it warns that : Increasing the query_cache size over 128M may reduce performance Here are the last results: >> MySQLTuner 1.2.0 - Major Hayden <[email protected]> >> Bug reports, feature requests, and downloads at http://mysqltuner.com/ >> Run with '--help' for additional options and output filtering -------- General Statistics -------------------------------------------------- [--] Skipped version check for MySQLTuner script [OK] Currently running supported MySQL version 5.5.25-1~dotdeb.0-log [OK] Operating on 64-bit architecture -------- Storage Engine Statistics ------------------------------------------- [--] Status: +Archive -BDB -Federated +InnoDB -ISAM -NDBCluster [--] Data in InnoDB tables: 6G (Tables: 195) [--] Data in PERFORMANCE_SCHEMA tables: 0B (Tables: 17) [!!] Total fragmented tables: 51 -------- Security Recommendations ------------------------------------------- [OK] All database users have passwords assigned -------- Performance Metrics ------------------------------------------------- [--] Up for: 1d 19h 17m 8s (254M q [1K qps], 5M conn, TX: 139B, RX: 32B) [--] Reads / Writes: 89% / 11% [--] Total buffers: 24.2G global + 92.2M per thread (1200 max threads) [!!] Maximum possible memory usage: 132.2G (139% of installed RAM) [OK] Slow queries: 0% (2K/254M) [OK] Highest usage of available connections: 32% (391/1200) [OK] Key buffer size / total MyISAM indexes: 128.0M/92.0K [OK] Key buffer hit rate: 100.0% (8B cached / 0 reads) [OK] Query cache efficiency: 79.9% (181M cached / 226M selects) [!!] Query cache prunes per day: 1033203 [OK] Sorts requiring temporary tables: 0% (341 temp sorts / 4M sorts) [OK] Temporary tables created on disk: 14% (760K on disk / 5M total) [OK] Thread cache hit rate: 99% (676 created / 5M connections) [OK] Table cache hit rate: 22% (1K open / 8K opened) [OK] Open file limit used: 0% (49/13K) [OK] Table locks acquired immediately: 99% (64M immediate / 64M locks) [OK] InnoDB data size / buffer pool: 6.1G/19.5G -------- Recommendations ----------------------------------------------------- General recommendations: Run OPTIMIZE TABLE to defragment tables for better performance Reduce your overall MySQL memory footprint for system stability Increasing the query_cache size over 128M may reduce performance Variables to adjust: *** MySQL's maximum memory usage is dangerously high *** *** Add RAM before increasing MySQL buffer variables *** query_cache_size (> 192M) [see warning above] The server has 76GB ram and dual E5-2650. The load is usually below 2. I appreciate your hints to interpret the recommendation and optimize the database configs.

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  • What does the 'Burst Rate' stat mean in HDTune?

    - by UpTheCreek
    I recently upgraded my laptop's v slow hard drive to a seagate momentus 7200. Everything is working fine, but I'm a bit confused by these benchmark results: The burst rate is significantly less than the Maximim transfer rate, and not much higher than the normal minimum (if you ignore the spikes). What's going on here? On the HDtune website it defines Burst Rate as: ...the highest speed (in megabytes per second) at which data can be transferred from the drive interface (IDE or SCSI for example) to the operating system. Which begs some questions... e.g. if this is the highest, then how did the bechmarking tool record the 103MB/sec maximum? And if this really is the true maximum, then where is the bottleneck? The laptops SATA interface is on an Intel 82801GBM southbridge controller. When I check in hardware manager, I see that it's driver is iaStor.sys from 2005. Maybe that's the issue? I'll look for a newever version, but any insights would be appreciated. Thanks

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  • Which RAM is faster (or, is Crucial's Memory Advisor giving non-optimal advice)?

    - by adpe
    In general, if a PC's motherboard is only specified for RAM up to a given core speed x, will that PC be faster with: RAM of latency y capable of running at a maximum core speed >x or RAM of latency <y capable of running at a maximum core speed of exactly x ? I would have thought the latter, but Crucial's Memory Adviser tool advises the former. So, which of us is correct - me, or the machine? (Here is a concrete example: I wish to upgrade a Toshiba Satellite Pro L300-155 laptop from its current 1GB RAM to 2GB Crucial RAM. The laptop's specifications are given here. I see from those specifications that the laptop is designed for DDR2-667 Ram. Crucial sells two compatible 2GB kits, priced exactly the same as each other: DDR2-667, CL=5; DDR2-800, CL=6. It seems to me that of these two upgrade kits, the first kit would run slightly faster on the L300-155 than the second, because both will presumably be capped at DDR2-667 core speed (see laptop specs), but the second kit has more latency. However, Crucial's Memory Advisor tool recommends the second kit.)

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  • What are the advantages of registered memory?

    - by odd parity
    I'm browsing for a few low-end servers for a startup and I'm a bit confused about the different memory types. The advantage of ECC is clear - single-bit error correction. When it comes to registered memory it seems more vague, especially in systems that support both registered and unbuffered memory. A Google search mostly finds copies of the Wikipedia article, which states that registered memory chips "...place less electrical load on the memory controller and allow single systems to remain stable with more memory modules than they would have otherwise". However I can't find any quantification of this. What I'm wondering about is: Is registered memory an improvement over unbuffered when it comes to soft error rate, or is it purely about the maximum number of modules supported? If yes, at what point (amount of modules or GB of memory) do these improvements start to become noticeable? For a specific example, the HP ProLiant DL 120 G6 server manual states that maximum supported memory configuration is 16 GB unbuffered (4x4GB) or 12 GB registered (6x2GB). In this case I'd rather have the extra 4GB of memory if the reliability difference is negligible.

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  • Throughput and why do ISPs sell too much bandwidth?

    - by jonescb
    I hope the question made sense how I worded it. :) I've been wondering, maximum theoretical bandwidth is measured as RWIN/RTT (Window size / round trip time) Source 1 and Souce 2 So if a major city only 100 miles away gives me a ping of 50ms, and I have the default 64kb TCP window size then my maximum throughput will be 12.5Mb/s. Everything further away would give me a higher ping and therefore a lower throughput. Is there any reason to buy something like FiOS with a 50Mb/s or greater connection? Will you ever be able to reach that kind of speed? I know you can increase the TCP window size to increase throughput, but it has to be at both ends which is a deal breaker because you can't control the server. I'm assuming other network protocols like UDP aren't quite as affected by latency as TCP is, but how much of overall network traffic does non-TCP make up vs TCP. Am I just misguided about how throughput works? But if the above is correct, then why should a consumer like me buy way more bandwidth than can be realistically used. Maybe the only reason is for downloading multiple things at once, or one thing from multiple servers/peers?

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  • Using multiple USB webcams in Linux

    - by rachelderp
    Running more than one USB webcam in Debian/Linux results in the the following error: libv4l2: error turning on stream: No space left on device VIDIOC_STREAMON: No space left on device What initially seemed to be a programming issue in OpenCV turned into a quest for a mysterious hardware/software problem after the same errors were produced by running cheese and xawtv. Apparently it's caused by webcams requesting all the available bandwidth on the USB host controller. With that in mind I decided to run wireshark and capinfos to see just how much bandwidth a single camera used. 4 megabits per second at 320x240 14 megabits per second at 640x480 32 megabits per second at 1920x1080 Interesting! That might explain why two cameras at 320x240 work but any higher resolution fails. It's as if my USB controller is only operating at USB 1 speeds, yet lsusb shows both webcams belonging to a device which supposedly supports 480 megabits per second. One solution proposed forcing the webcams to calculate their bandwidth usage instead of requesting their maximum by running the following commands: sudo rmmod uvcvideo sudo modprobe uvcvideo quirks=128 Unfortunately that made no difference, so I decided to try another solution. A post on StackOverflow suggested telling my webcams to use a lower FPS or compressed video format like MJPEG, but after running v4lctl list it doesn't appear either of my webcams support changing their video mode. And that's where I'm stuck. Why would two webcams operating well below the maximum speed of USB 2 would produce this error? ps: It's not a disk space issue, df displays no change when the webcams are started. pps: If it makes a difference, here's the output of lsusb

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  • Shell script to block proftp failled attempt

    - by Saif
    Hello, I want to filter and block failed attempt to access my proftp server. Here is an example line from the /var/log/secure file: Jan 2 18:38:25 server1 proftpd[17847]: spy1.XYZ.com (93.218.93.95[93.218.93.95]) - Maximum login attempts (3) exceeded There are several lines like this. I would like to block any attempts like this from any IP twice. Here's a script I'm trying to run to block those IPs. tail -1000 /var/log/secure | awk '/proftpd/ && /Maximum login/ { if (/attempts/) try[$7]++; else try[$11]++; } END { for (h in try) if (try[h] > 4) print h; }' | while read ip do /sbin/iptables -L -n | grep $ip > /dev/null if [ $? -eq 0 ] ; then # echo "already denied ip: [$ip]" ; true else logger -p authpriv.notice "*** Blocking ProFTPD attempt from: $ip" /sbin/iptables -I INPUT -s $ip -j DROP fi done how can I select the IP with "awk". with the current script it's selecting "(93.218.93.95[93.218.93.95])" this line completely. But i only want to select the IP.

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  • Is 40+ Logons on Exchange 2003 per user normal?

    - by cbsch
    Hello! We've had a problem at work where users sometimes randomly can't connect to exchange. I've found out that it's because they reached the limit of 32 concurrent logons. I increased the maximum allowed connections by adding the key "Maximum Allowed Sessions Per User" in HKLM\SYSTEM\CurrentControlSet\Services\MSExchangeIS\ParametersSystem. But I'm not sure if this is a real good fix. Looking at the logons some users has as many as 15 logons with the exact same logon time. I know for sure that Outlook 2007 does this, as I was watching them while a user connected with Outlook after a restart on the Exchange service. Every user also has an iPhone connected to exchange, I don't know if these cause the same thing. Is this normal? Could there be a bug in the software? (The Outlook 2007 has nothing configured, except added the user, pure vanilla installs). The users are mobile, and when Outlook generates up to 15 connection every time it connects, and I've read (no sources, sorry) that Outlook doesn't time out connections before 2 hours. I might have to set this number real high to prevent it from being a problem.

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  • Duplicate monitor on highest resolution in Windows 7

    - by AlexanderMP
    I have a monitor with a native resolution of 2560x1440, connected through display port. I also have an AV Receiver connected to the video card via HDMI, to have surround sound in games. All using Radeon HD 5670 (will upgrade soon to HD 7850). The problem is that my computer detects the receiver as a separate monitor, with the highest available resolution of 1920x1080. I have 3 options: Disconnect the second display. But then the sound (digital audio output through video card) also disappears. Duplicate displays. But then my primary monitor resolution is reduced to a maximum of just 1920x1080, that being the maximum of the second monitor. Extend desktop. This is the solution I picked so far, it being the least evil. The problems I face in this situations are 2: I have a blank part of the desktop where I sometimes lose my mouse pointer, so I made the extension small, 640x480, and placed it in a corner; when I turn off the main display, all windows resize to 640x480. In Kubuntu I had the option to duplicate the displays, while keeping the higher resolution. Which was great. I tried overriding using the Win7 netbook hack, but it's not available on non-netbooks. Is there a similar solution for this problem in Windows 7?

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  • Using R to Analyze G1GC Log Files

    - by user12620111
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  Using R to Analyze G1GC Log Files   Using R to Analyze G1GC Log Files Introduction Working in Oracle Platform Integration gives an engineer opportunities to work on a wide array of technologies. My team’s goal is to make Oracle applications run best on the Solaris/SPARC platform. When looking for bottlenecks in a modern applications, one needs to be aware of not only how the CPUs and operating system are executing, but also network, storage, and in some cases, the Java Virtual Machine. I was recently presented with about 1.5 GB of Java Garbage First Garbage Collector log file data. If you’re not familiar with the subject, you might want to review Garbage First Garbage Collector Tuning by Monica Beckwith. The customer had been running Java HotSpot 1.6.0_31 to host a web application server. I was told that the Solaris/SPARC server was running a Java process launched using a commmand line that included the following flags: -d64 -Xms9g -Xmx9g -XX:+UseG1GC -XX:MaxGCPauseMillis=200 -XX:InitiatingHeapOccupancyPercent=80 -XX:PermSize=256m -XX:MaxPermSize=256m -XX:+PrintGC -XX:+PrintGCTimeStamps -XX:+PrintHeapAtGC -XX:+PrintGCDateStamps -XX:+PrintFlagsFinal -XX:+DisableExplicitGC -XX:+UnlockExperimentalVMOptions -XX:ParallelGCThreads=8 Several sources on the internet indicate that if I were to print out the 1.5 GB of log files, it would require enough paper to fill the bed of a pick up truck. Of course, it would be fruitless to try to scan the log files by hand. Tools will be required to summarize the contents of the log files. Others have encountered large Java garbage collection log files. There are existing tools to analyze the log files: IBM’s GC toolkit The chewiebug GCViewer gchisto HPjmeter Instead of using one of the other tools listed, I decide to parse the log files with standard Unix tools, and analyze the data with R. Data Cleansing The log files arrived in two different formats. I guess that the difference is that one set of log files was generated using a more verbose option, maybe -XX:+PrintHeapAtGC, and the other set of log files was generated without that option. Format 1 In some of the log files, the log files with the less verbose format, a single trace, i.e. the report of a singe garbage collection event, looks like this: {Heap before GC invocations=12280 (full 61): garbage-first heap total 9437184K, used 7499918K [0xfffffffd00000000, 0xffffffff40000000, 0xffffffff40000000) region size 4096K, 1 young (4096K), 0 survivors (0K) compacting perm gen total 262144K, used 144077K [0xffffffff40000000, 0xffffffff50000000, 0xffffffff50000000) the space 262144K, 54% used [0xffffffff40000000, 0xffffffff48cb3758, 0xffffffff48cb3800, 0xffffffff50000000) No shared spaces configured. 2014-05-14T07:24:00.988-0700: 60586.353: [GC pause (young) 7324M->7320M(9216M), 0.1567265 secs] Heap after GC invocations=12281 (full 61): garbage-first heap total 9437184K, used 7496533K [0xfffffffd00000000, 0xffffffff40000000, 0xffffffff40000000) region size 4096K, 0 young (0K), 0 survivors (0K) compacting perm gen total 262144K, used 144077K [0xffffffff40000000, 0xffffffff50000000, 0xffffffff50000000) the space 262144K, 54% used [0xffffffff40000000, 0xffffffff48cb3758, 0xffffffff48cb3800, 0xffffffff50000000) No shared spaces configured. } A simple grep can be used to extract a summary: $ grep "\[ GC pause (young" g1gc.log 2014-05-13T13:24:35.091-0700: 3.109: [GC pause (young) 20M->5029K(9216M), 0.0146328 secs] 2014-05-13T13:24:35.440-0700: 3.459: [GC pause (young) 9125K->6077K(9216M), 0.0086723 secs] 2014-05-13T13:24:37.581-0700: 5.599: [GC pause (young) 25M->8470K(9216M), 0.0203820 secs] 2014-05-13T13:24:42.686-0700: 10.704: [GC pause (young) 44M->15M(9216M), 0.0288848 secs] 2014-05-13T13:24:48.941-0700: 16.958: [GC pause (young) 51M->20M(9216M), 0.0491244 secs] 2014-05-13T13:24:56.049-0700: 24.066: [GC pause (young) 92M->26M(9216M), 0.0525368 secs] 2014-05-13T13:25:34.368-0700: 62.383: [GC pause (young) 602M->68M(9216M), 0.1721173 secs] But that format wasn't easily read into R, so I needed to be a bit more tricky. I used the following Unix command to create a summary file that was easy for R to read. $ echo "SecondsSinceLaunch BeforeSize AfterSize TotalSize RealTime" $ grep "\[GC pause (young" g1gc.log | grep -v mark | sed -e 's/[A-SU-z\(\),]/ /g' -e 's/->/ /' -e 's/: / /g' | more SecondsSinceLaunch BeforeSize AfterSize TotalSize RealTime 2014-05-13T13:24:35.091-0700 3.109 20 5029 9216 0.0146328 2014-05-13T13:24:35.440-0700 3.459 9125 6077 9216 0.0086723 2014-05-13T13:24:37.581-0700 5.599 25 8470 9216 0.0203820 2014-05-13T13:24:42.686-0700 10.704 44 15 9216 0.0288848 2014-05-13T13:24:48.941-0700 16.958 51 20 9216 0.0491244 2014-05-13T13:24:56.049-0700 24.066 92 26 9216 0.0525368 2014-05-13T13:25:34.368-0700 62.383 602 68 9216 0.1721173 Format 2 In some of the log files, the log files with the more verbose format, a single trace, i.e. the report of a singe garbage collection event, was more complicated than Format 1. Here is a text file with an example of a single G1GC trace in the second format. As you can see, it is quite complicated. It is nice that there is so much information available, but the level of detail can be overwhelming. I wrote this awk script (download) to summarize each trace on a single line. #!/usr/bin/env awk -f BEGIN { printf("SecondsSinceLaunch IncrementalCount FullCount UserTime SysTime RealTime BeforeSize AfterSize TotalSize\n") } ###################### # Save count data from lines that are at the start of each G1GC trace. # Each trace starts out like this: # {Heap before GC invocations=14 (full 0): # garbage-first heap total 9437184K, used 325496K [0xfffffffd00000000, 0xffffffff40000000, 0xffffffff40000000) ###################### /{Heap.*full/{ gsub ( "\\)" , "" ); nf=split($0,a,"="); split(a[2],b," "); getline; if ( match($0, "first") ) { G1GC=1; IncrementalCount=b[1]; FullCount=substr( b[3], 1, length(b[3])-1 ); } else { G1GC=0; } } ###################### # Pull out time stamps that are in lines with this format: # 2014-05-12T14:02:06.025-0700: 94.312: [GC pause (young), 0.08870154 secs] ###################### /GC pause/ { DateTime=$1; SecondsSinceLaunch=substr($2, 1, length($2)-1); } ###################### # Heap sizes are in lines that look like this: # [ 4842M->4838M(9216M)] ###################### /\[ .*]$/ { gsub ( "\\[" , "" ); gsub ( "\ \]" , "" ); gsub ( "->" , " " ); gsub ( "\\( " , " " ); gsub ( "\ \)" , " " ); split($0,a," "); if ( split(a[1],b,"M") > 1 ) {BeforeSize=b[1]*1024;} if ( split(a[1],b,"K") > 1 ) {BeforeSize=b[1];} if ( split(a[2],b,"M") > 1 ) {AfterSize=b[1]*1024;} if ( split(a[2],b,"K") > 1 ) {AfterSize=b[1];} if ( split(a[3],b,"M") > 1 ) {TotalSize=b[1]*1024;} if ( split(a[3],b,"K") > 1 ) {TotalSize=b[1];} } ###################### # Emit an output line when you find input that looks like this: # [Times: user=1.41 sys=0.08, real=0.24 secs] ###################### /\[Times/ { if (G1GC==1) { gsub ( "," , "" ); split($2,a,"="); UserTime=a[2]; split($3,a,"="); SysTime=a[2]; split($4,a,"="); RealTime=a[2]; print DateTime,SecondsSinceLaunch,IncrementalCount,FullCount,UserTime,SysTime,RealTime,BeforeSize,AfterSize,TotalSize; G1GC=0; } } The resulting summary is about 25X smaller that the original file, but still difficult for a human to digest. SecondsSinceLaunch IncrementalCount FullCount UserTime SysTime RealTime BeforeSize AfterSize TotalSize ... 2014-05-12T18:36:34.669-0700: 3985.744 561 0 0.57 0.06 0.16 1724416 1720320 9437184 2014-05-12T18:36:34.839-0700: 3985.914 562 0 0.51 0.06 0.19 1724416 1720320 9437184 2014-05-12T18:36:35.069-0700: 3986.144 563 0 0.60 0.04 0.27 1724416 1721344 9437184 2014-05-12T18:36:35.354-0700: 3986.429 564 0 0.33 0.04 0.09 1725440 1722368 9437184 2014-05-12T18:36:35.545-0700: 3986.620 565 0 0.58 0.04 0.17 1726464 1722368 9437184 2014-05-12T18:36:35.726-0700: 3986.801 566 0 0.43 0.05 0.12 1726464 1722368 9437184 2014-05-12T18:36:35.856-0700: 3986.930 567 0 0.30 0.04 0.07 1726464 1723392 9437184 2014-05-12T18:36:35.947-0700: 3987.023 568 0 0.61 0.04 0.26 1727488 1723392 9437184 2014-05-12T18:36:36.228-0700: 3987.302 569 0 0.46 0.04 0.16 1731584 1724416 9437184 Reading the Data into R Once the GC log data had been cleansed, either by processing the first format with the shell script, or by processing the second format with the awk script, it was easy to read the data into R. g1gc.df = read.csv("summary.txt", row.names = NULL, stringsAsFactors=FALSE,sep="") str(g1gc.df) ## 'data.frame': 8307 obs. of 10 variables: ## $ row.names : chr "2014-05-12T14:00:32.868-0700:" "2014-05-12T14:00:33.179-0700:" "2014-05-12T14:00:33.677-0700:" "2014-05-12T14:00:35.538-0700:" ... ## $ SecondsSinceLaunch: num 1.16 1.47 1.97 3.83 6.1 ... ## $ IncrementalCount : int 0 1 2 3 4 5 6 7 8 9 ... ## $ FullCount : int 0 0 0 0 0 0 0 0 0 0 ... ## $ UserTime : num 0.11 0.05 0.04 0.21 0.08 0.26 0.31 0.33 0.34 0.56 ... ## $ SysTime : num 0.04 0.01 0.01 0.05 0.01 0.06 0.07 0.06 0.07 0.09 ... ## $ RealTime : num 0.02 0.02 0.01 0.04 0.02 0.04 0.05 0.04 0.04 0.06 ... ## $ BeforeSize : int 8192 5496 5768 22528 24576 43008 34816 53248 55296 93184 ... ## $ AfterSize : int 1400 1672 2557 4907 7072 14336 16384 18432 19456 21504 ... ## $ TotalSize : int 9437184 9437184 9437184 9437184 9437184 9437184 9437184 9437184 9437184 9437184 ... head(g1gc.df) ## row.names SecondsSinceLaunch IncrementalCount ## 1 2014-05-12T14:00:32.868-0700: 1.161 0 ## 2 2014-05-12T14:00:33.179-0700: 1.472 1 ## 3 2014-05-12T14:00:33.677-0700: 1.969 2 ## 4 2014-05-12T14:00:35.538-0700: 3.830 3 ## 5 2014-05-12T14:00:37.811-0700: 6.103 4 ## 6 2014-05-12T14:00:41.428-0700: 9.720 5 ## FullCount UserTime SysTime RealTime BeforeSize AfterSize TotalSize ## 1 0 0.11 0.04 0.02 8192 1400 9437184 ## 2 0 0.05 0.01 0.02 5496 1672 9437184 ## 3 0 0.04 0.01 0.01 5768 2557 9437184 ## 4 0 0.21 0.05 0.04 22528 4907 9437184 ## 5 0 0.08 0.01 0.02 24576 7072 9437184 ## 6 0 0.26 0.06 0.04 43008 14336 9437184 Basic Statistics Once the data has been read into R, simple statistics are very easy to generate. All of the numbers from high school statistics are available via simple commands. For example, generate a summary of every column: summary(g1gc.df) ## row.names SecondsSinceLaunch IncrementalCount FullCount ## Length:8307 Min. : 1 Min. : 0 Min. : 0.0 ## Class :character 1st Qu.: 9977 1st Qu.:2048 1st Qu.: 0.0 ## Mode :character Median :12855 Median :4136 Median : 12.0 ## Mean :12527 Mean :4156 Mean : 31.6 ## 3rd Qu.:15758 3rd Qu.:6262 3rd Qu.: 61.0 ## Max. :55484 Max. :8391 Max. :113.0 ## UserTime SysTime RealTime BeforeSize ## Min. :0.040 Min. :0.0000 Min. : 0.0 Min. : 5476 ## 1st Qu.:0.470 1st Qu.:0.0300 1st Qu.: 0.1 1st Qu.:5137920 ## Median :0.620 Median :0.0300 Median : 0.1 Median :6574080 ## Mean :0.751 Mean :0.0355 Mean : 0.3 Mean :5841855 ## 3rd Qu.:0.920 3rd Qu.:0.0400 3rd Qu.: 0.2 3rd Qu.:7084032 ## Max. :3.370 Max. :1.5600 Max. :488.1 Max. :8696832 ## AfterSize TotalSize ## Min. : 1380 Min. :9437184 ## 1st Qu.:5002752 1st Qu.:9437184 ## Median :6559744 Median :9437184 ## Mean :5785454 Mean :9437184 ## 3rd Qu.:7054336 3rd Qu.:9437184 ## Max. :8482816 Max. :9437184 Q: What is the total amount of User CPU time spent in garbage collection? sum(g1gc.df$UserTime) ## [1] 6236 As you can see, less than two hours of CPU time was spent in garbage collection. Is that too much? To find the percentage of time spent in garbage collection, divide the number above by total_elapsed_time*CPU_count. In this case, there are a lot of CPU’s and it turns out the the overall amount of CPU time spent in garbage collection isn’t a problem when viewed in isolation. When calculating rates, i.e. events per unit time, you need to ask yourself if the rate is homogenous across the time period in the log file. Does the log file include spikes of high activity that should be separately analyzed? Averaging in data from nights and weekends with data from business hours may alias problems. If you have a reason to suspect that the garbage collection rates include peaks and valleys that need independent analysis, see the “Time Series” section, below. Q: How much garbage is collected on each pass? The amount of heap space that is recovered per GC pass is surprisingly low: At least one collection didn’t recover any data. (“Min.=0”) 25% of the passes recovered 3MB or less. (“1st Qu.=3072”) Half of the GC passes recovered 4MB or less. (“Median=4096”) The average amount recovered was 56MB. (“Mean=56390”) 75% of the passes recovered 36MB or less. (“3rd Qu.=36860”) At least one pass recovered 2GB. (“Max.=2121000”) g1gc.df$Delta = g1gc.df$BeforeSize - g1gc.df$AfterSize summary(g1gc.df$Delta) ## Min. 1st Qu. Median Mean 3rd Qu. Max. ## 0 3070 4100 56400 36900 2120000 Q: What is the maximum User CPU time for a single collection? The worst garbage collection (“Max.”) is many standard deviations away from the mean. The data appears to be right skewed. summary(g1gc.df$UserTime) ## Min. 1st Qu. Median Mean 3rd Qu. Max. ## 0.040 0.470 0.620 0.751 0.920 3.370 sd(g1gc.df$UserTime) ## [1] 0.3966 Basic Graphics Once the data is in R, it is trivial to plot the data with formats including dot plots, line charts, bar charts (simple, stacked, grouped), pie charts, boxplots, scatter plots histograms, and kernel density plots. Histogram of User CPU Time per Collection I don't think that this graph requires any explanation. hist(g1gc.df$UserTime, main="User CPU Time per Collection", xlab="Seconds", ylab="Frequency") Box plot to identify outliers When the initial data is viewed with a box plot, you can see the one crazy outlier in the real time per GC. Save this data point for future analysis and drop the outlier so that it’s not throwing off our statistics. Now the box plot shows many outliers, which will be examined later, using times series analysis. Notice that the scale of the x-axis changes drastically once the crazy outlier is removed. par(mfrow=c(2,1)) boxplot(g1gc.df$UserTime,g1gc.df$SysTime,g1gc.df$RealTime, main="Box Plot of Time per GC\n(dominated by a crazy outlier)", names=c("usr","sys","elapsed"), xlab="Seconds per GC", ylab="Time (Seconds)", horizontal = TRUE, outcol="red") crazy.outlier.df=g1gc.df[g1gc.df$RealTime > 400,] g1gc.df=g1gc.df[g1gc.df$RealTime < 400,] boxplot(g1gc.df$UserTime,g1gc.df$SysTime,g1gc.df$RealTime, main="Box Plot of Time per GC\n(crazy outlier excluded)", names=c("usr","sys","elapsed"), xlab="Seconds per GC", ylab="Time (Seconds)", horizontal = TRUE, outcol="red") box(which = "outer", lty = "solid") Here is the crazy outlier for future analysis: crazy.outlier.df ## row.names SecondsSinceLaunch IncrementalCount ## 8233 2014-05-12T23:15:43.903-0700: 20741 8316 ## FullCount UserTime SysTime RealTime BeforeSize AfterSize TotalSize ## 8233 112 0.55 0.42 488.1 8381440 8235008 9437184 ## Delta ## 8233 146432 R Time Series Data To analyze the garbage collection as a time series, I’ll use Z’s Ordered Observations (zoo). “zoo is the creator for an S3 class of indexed totally ordered observations which includes irregular time series.” require(zoo) ## Loading required package: zoo ## ## Attaching package: 'zoo' ## ## The following objects are masked from 'package:base': ## ## as.Date, as.Date.numeric head(g1gc.df[,1]) ## [1] "2014-05-12T14:00:32.868-0700:" "2014-05-12T14:00:33.179-0700:" ## [3] "2014-05-12T14:00:33.677-0700:" "2014-05-12T14:00:35.538-0700:" ## [5] "2014-05-12T14:00:37.811-0700:" "2014-05-12T14:00:41.428-0700:" options("digits.secs"=3) times=as.POSIXct( g1gc.df[,1], format="%Y-%m-%dT%H:%M:%OS%z:") g1gc.z = zoo(g1gc.df[,-c(1)], order.by=times) head(g1gc.z) ## SecondsSinceLaunch IncrementalCount FullCount ## 2014-05-12 17:00:32.868 1.161 0 0 ## 2014-05-12 17:00:33.178 1.472 1 0 ## 2014-05-12 17:00:33.677 1.969 2 0 ## 2014-05-12 17:00:35.538 3.830 3 0 ## 2014-05-12 17:00:37.811 6.103 4 0 ## 2014-05-12 17:00:41.427 9.720 5 0 ## UserTime SysTime RealTime BeforeSize AfterSize ## 2014-05-12 17:00:32.868 0.11 0.04 0.02 8192 1400 ## 2014-05-12 17:00:33.178 0.05 0.01 0.02 5496 1672 ## 2014-05-12 17:00:33.677 0.04 0.01 0.01 5768 2557 ## 2014-05-12 17:00:35.538 0.21 0.05 0.04 22528 4907 ## 2014-05-12 17:00:37.811 0.08 0.01 0.02 24576 7072 ## 2014-05-12 17:00:41.427 0.26 0.06 0.04 43008 14336 ## TotalSize Delta ## 2014-05-12 17:00:32.868 9437184 6792 ## 2014-05-12 17:00:33.178 9437184 3824 ## 2014-05-12 17:00:33.677 9437184 3211 ## 2014-05-12 17:00:35.538 9437184 17621 ## 2014-05-12 17:00:37.811 9437184 17504 ## 2014-05-12 17:00:41.427 9437184 28672 Example of Two Benchmark Runs in One Log File The data in the following graph is from a different log file, not the one of primary interest to this article. I’m including this image because it is an example of idle periods followed by busy periods. It would be uninteresting to average the rate of garbage collection over the entire log file period. More interesting would be the rate of garbage collect in the two busy periods. Are they the same or different? Your production data may be similar, for example, bursts when employees return from lunch and idle times on weekend evenings, etc. Once the data is in an R Time Series, you can analyze isolated time windows. Clipping the Time Series data Flashing back to our test case… Viewing the data as a time series is interesting. You can see that the work intensive time period is between 9:00 PM and 3:00 AM. Lets clip the data to the interesting period:     par(mfrow=c(2,1)) plot(g1gc.z$UserTime, type="h", main="User Time per GC\nTime: Complete Log File", xlab="Time of Day", ylab="CPU Seconds per GC", col="#1b9e77") clipped.g1gc.z=window(g1gc.z, start=as.POSIXct("2014-05-12 21:00:00"), end=as.POSIXct("2014-05-13 03:00:00")) plot(clipped.g1gc.z$UserTime, type="h", main="User Time per GC\nTime: Limited to Benchmark Execution", xlab="Time of Day", ylab="CPU Seconds per GC", col="#1b9e77") box(which = "outer", lty = "solid") Cumulative Incremental and Full GC count Here is the cumulative incremental and full GC count. When the line is very steep, it indicates that the GCs are repeating very quickly. Notice that the scale on the Y axis is different for full vs. incremental. plot(clipped.g1gc.z[,c(2:3)], main="Cumulative Incremental and Full GC count", xlab="Time of Day", col="#1b9e77") GC Analysis of Benchmark Execution using Time Series data In the following series of 3 graphs: The “After Size” show the amount of heap space in use after each garbage collection. Many Java objects are still referenced, i.e. alive, during each garbage collection. This may indicate that the application has a memory leak, or may indicate that the application has a very large memory footprint. Typically, an application's memory footprint plateau's in the early stage of execution. One would expect this graph to have a flat top. The steep decline in the heap space may indicate that the application crashed after 2:00. The second graph shows that the outliers in real execution time, discussed above, occur near 2:00. when the Java heap seems to be quite full. The third graph shows that Full GCs are infrequent during the first few hours of execution. The rate of Full GC's, (the slope of the cummulative Full GC line), changes near midnight.   plot(clipped.g1gc.z[,c("AfterSize","RealTime","FullCount")], xlab="Time of Day", col=c("#1b9e77","red","#1b9e77")) GC Analysis of heap recovered Each GC trace includes the amount of heap space in use before and after the individual GC event. During garbage coolection, unreferenced objects are identified, the space holding the unreferenced objects is freed, and thus, the difference in before and after usage indicates how much space has been freed. The following box plot and bar chart both demonstrate the same point - the amount of heap space freed per garbage colloection is surprisingly low. par(mfrow=c(2,1)) boxplot(as.vector(clipped.g1gc.z$Delta), main="Amount of Heap Recovered per GC Pass", xlab="Size in KB", horizontal = TRUE, col="red") hist(as.vector(clipped.g1gc.z$Delta), main="Amount of Heap Recovered per GC Pass", xlab="Size in KB", breaks=100, col="red") box(which = "outer", lty = "solid") This graph is the most interesting. The dark blue area shows how much heap is occupied by referenced Java objects. This represents memory that holds live data. The red fringe at the top shows how much data was recovered after each garbage collection. barplot(clipped.g1gc.z[,c("AfterSize","Delta")], col=c("#7570b3","#e7298a"), xlab="Time of Day", border=NA) legend("topleft", c("Live Objects","Heap Recovered on GC"), fill=c("#7570b3","#e7298a")) box(which = "outer", lty = "solid") When I discuss the data in the log files with the customer, I will ask for an explaination for the large amount of referenced data resident in the Java heap. There are two are posibilities: There is a memory leak and the amount of space required to hold referenced objects will continue to grow, limited only by the maximum heap size. After the maximum heap size is reached, the JVM will throw an “Out of Memory” exception every time that the application tries to allocate a new object. If this is the case, the aplication needs to be debugged to identify why old objects are referenced when they are no longer needed. The application has a legitimate requirement to keep a large amount of data in memory. The customer may want to further increase the maximum heap size. Another possible solution would be to partition the application across multiple cluster nodes, where each node has responsibility for managing a unique subset of the data. Conclusion In conclusion, R is a very powerful tool for the analysis of Java garbage collection log files. The primary difficulty is data cleansing so that information can be read into an R data frame. Once the data has been read into R, a rich set of tools may be used for thorough evaluation.

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  • Nginx + Haproxy + Thin + Rails - 503 Service Unavailable -

    - by Luca G. Soave
    I don't know how troubleshoot this. I get "503 Service Unavailable" http error for all "nginx upstreams" proxy passing calls to haproxy fast_thin and slow_thin ( server 127.0.0.1:3100 and server 127.0.0.1:3200 ), which loadbalance on 6 Thin servers ( 127.0.0.1:3000 .. 3005 ). Static files like /blog are currently fine. The falldown is: nginx on port 80 - haproxy on 3100 and 3200 - thin on 3000 .. 3005 and then Rails. Here it is /etc/nginx/nginx.conf : user nginx; worker_processes 2; pid /var/run/nginx.pid; events { worker_connections 1024; } http { include /etc/nginx/mime.types; default_type application/octet-stream; log_format main '$remote_addr - $remote_user [$time_local] "$request" ' '$status $body_bytes_sent "$http_referer" ' '"$http_user_agent" "$http_x_forwarded_for"'; sendfile on; tcp_nopush on; keepalive_timeout 65; tcp_nodelay on; include /etc/nginx/conf.d/*.conf; } then /etc/nginx/conf.d/default.conf upstream fast_thin { server 127.0.0.1:3100; } upstream slow_thin { server 127.0.0.1:3200; } server { listen 80; server_name www.gitwatcher.com; rewrite ^/(.*) http://gitwatcher.com/$1 permanent; } server { listen 80; server_name gitwatcher.com; access_log /var/www/gitwatcher/log/access.log; error_log /var/www/gitwatcher/log/error.log; root /var/www/gitwatcher/public; # index index.html; location /about { proxy_pass http://fast_thin; break; } location /trends { proxy_pass http://slow_thin; break; } location /categories { proxy_pass http://slow_thin; break; } location /signout { proxy_pass http://slow_thin; break; } location /auth/github { proxy_pass http://slow_thin; break; } location / { proxy_set_header X-Real-IP $remote_addr; proxy_set_header X-Forwarded-For $proxy_add_x_forwarded_for; proxy_set_header Host $http_host; proxy_redirect off; if (-f $request_filename/index.html) { rewrite (.*) $1/index.html break; } if (-f $request_filename.html) { rewrite (.*) $1.html break; } if (!-f $request_filename) { proxy_pass http://slow_thin; break; } } } then haproxy config file /etc/haproxy/haproxy.cfg : global log 127.0.0.1 local0 log 127.0.0.1 local1 notice #log loghost local0 info maxconn 4096 #chroot /usr/share/haproxy user haproxy group haproxy daemon #debug #quiet nbproc 1 # number of processing cores defaults log global retries 3 maxconn 2000 contimeout 5000 mode http clitimeout 60000 # maximum inactivity time on the client side srvtimeout 30000 # maximum inactivity time on the server side timeout connect 4000 # maximum time to wait for a connection attempt to a server to succeed option httplog option dontlognull option redispatch option httpclose # disable keepalive (HAProxy does not yet support the HTTP keep-alive mode) option abortonclose # enable early dropping of aborted requests from pending queue option httpchk # enable HTTP protocol to check on servers health option forwardfor # enable insert of X-Forwarded-For headers balance roundrobin # each server is used in turns, according to assigned weight stats enable # enable web-stats at /haproxy?stats stats auth haproxy:pr0xystats # force HTTP Auth to view stats stats refresh 5s # refresh rate of stats page listen rails_proxy 127.0.0.1:3100 # - equal weights on all servers # - maxconn will queue requests at HAProxy if limit is reached # - minconn dynamically scales the connection concurrency (bound my maxconn) depending on size of HAProxy queue # - check health every 20000 microseconds server web1 127.0.0.1:3000 weight 1 minconn 3 maxconn 6 check inter 20000 server web1 127.0.0.1:3001 weight 1 minconn 3 maxconn 6 check inter 20000 server web1 127.0.0.1:3002 weight 1 minconn 3 maxconn 6 check inter 20000 listen slow_proxy 127.0.0.1:3200 # cluster for slow requests, lower the queues, check less frequently server slow1 127.0.0.1:3003 weight 1 minconn 1 maxconn 3 check inter 40000 server slow2 127.0.0.1:3004 weight 1 minconn 1 maxconn 3 check inter 40000 server slow3 127.0.0.1:3005 weight 1 minconn 1 maxconn 3 check inter 40000 and the Thin config file /etc/thin/gitwatcher.yml : --- chdir: /var/www/gitwatcher environment: production address: 0.0.0.0 port: 3000 timeout: 30 log: log/thin.log pid: tmp/pids/thin.pid max_conns: 1024 max_persistent_conns: 100 require: [] wait: 30 servers: 6 daemonize: true if I look into open listen ports, I got the following : root@fullness:/var/www/gitwatcher# lsof | grep TCP | egrep "nginx|haproxy|thin" nginx 834 root 8u IPv4 921 0t0 TCP *:http (LISTEN) nginx 835 nginx 8u IPv4 921 0t0 TCP *:http (LISTEN) nginx 837 nginx 8u IPv4 921 0t0 TCP *:http (LISTEN) haproxy 1908 haproxy 4u IPv4 11699 0t0 TCP localhost:3100 (LISTEN) haproxy 1908 haproxy 6u IPv4 11701 0t0 TCP localhost:3200 (LISTEN) root@fullness:/var/www/gitwatcher# iptables -L get me the following : Chain INPUT (policy DROP) target prot opt source destination ACCEPT all -- anywhere anywhere state RELATED,ESTABLISHED ACCEPT all -- anywhere anywhere state RELATED,ESTABLISHED ACCEPT tcp -- anywhere anywhere tcp dpt:22222 ACCEPT tcp -- anywhere anywhere tcp dpt:http ACCEPT tcp -- anywhere anywhere tcp dpt:https ACCEPT all -- anywhere anywhere DROP all -- anywhere anywhere Chain FORWARD (policy ACCEPT) target prot opt source destination Chain OUTPUT (policy ACCEPT) target prot opt source destination ACCEPT all -- anywhere anywhere Any help ?

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  • Ubuntu 12.04 does not see windows already install on my computer (dual installation)

    - by jacinta
    I was trying to install the ubuntu 12.4 along side windows 7 on my new HP Pavilion 64k desktop with windows 7 computer but Ubuntu said that ( This computer has no detected operating system) and some one said (I suggest you chkdsk your Windows partition. I also suggest you resize the NTFS in WIndows then install Ubuntu to the free space.) Therefore I did (To shrink a simple or spanned volume using the Windows interface In Disk Management, right-click the simple or spanned volume you want to shrink. Click Shrink Volume…. Follow the instructions on your screen.) Then When I try to install ubuntu 12.4 after doing this, I received the same error. I was going to undo what I did but I see that I lose 1g when I do that so now what do I do? it says I can do a new simple volume and maybe then the space will no longer be unallocated. Please help me. I think I have a bad cd (ubuntu 12.4) cause from my research I see that I am not suppose to get a screen saying that (The computer has no detected operating system) I think this is a bad cd and I hope I did not mess up my computer. Please help. .................................................................................... O k I think I am following what you said about how to edit my question irrational john. I did chkdsk as you and actionparsnip (andrew-woodhead666) told me to AND ALSO did a lot of other things before I found out how to chkdsk. No problems. Thank you. Then I put back the space (extended) I took from system. I still was only able to put back 15 and not 16 so it is up to 99mb not back to 100mb. Then I shrank HP (C) as you told me, to 10 13,240 mb which is (12.93gb Unallocated). I did not change it into NTSF by doing the (New Simple Volume Action) I just left it. Then I tried to install UBUNTU 12.04 live CD amd64 and it gave me the results it was sometimes giving me before which is result (THAT Ubuntu) does not tell me weather I have or have not an already installed windows7. It just goes to a window that would have showed me information on what I have and on the bottom (DEVICE FOR BOOT LOADER INSTALLATION /dev/sda ) and the option to go BACK, QUIT, or INSTALL. (I think it is the INSTALLATION TYPE window). Therefore I do what I have been doing and I QUIT. What do I do now? Sorry that it seems like I cannot do anything on my own. On the Youtube video how to install ubuntu dual-boot alongside windows UBUNTU is installed so easy. The installation option page gives 3 options including dual instillation and the disk even lets you use a slider to slide to the size of the partition size you want. Yet my UBUNTU live cd is a mess and I checked it as one of you guys told me and got back information that it is good. Oh well this guy says you should press a control key to tell which device you are using to install ubuntu before the screen comes up. I guess cause it is old. This page also shows you easy stuff that do not show up on my cd. how to dual-boot UBUNTU and windows 7 P.S.. I saw this on the windows 7 website windows.microsoft.com/en-US/windows7/Formatting-disks-and-drives-frequently-asked-questions CREATE A BOOT PARTITION I HAD TO LEAVE OUT THE HTTP STUFF CAUSE I AM ONLY ALLOWED 2 ON A PAGE IT SAID To create a boot partition Warning Warning If you are installing different versions of Windows, you must install the earliest version first. If you don't do this, your computer may become inoperable. Open Computer Management by clicking the Start button Picture of the Start button, clicking Control Panel, clicking System and Security, clicking Administrative Tools, and then double-clicking Computer Management.? Administrator permission required If you're prompted for an administrator password or confirmation, type the password or provide confirmation. In the left pane, under Storage, click Disk Management. Right-click an unallocated region on your hard disk, and then click New Simple Volume. In the New Simple Volume Wizard, click Next. Type the size of the volume you want to create in megabytes (MB) or accept the maximum default size, and then click Next. Accept the default drive letter or choose a different drive letter to identify the volume, and then click Next. In the Format Partition dialog box, do one of the following: If you don't want to format the volume right now, click Do not format this volume, and then click Next. To format the volume with the default settings, click Next. For more information about formatting, see Formatting disks and drives: frequently asked questions. Review your choices, and then click Finish. AND THIS ON ANOTHER PAGE. Formatting disks and drives: frequently asked questions Hard disks, the primary storage devices on your computer, need to be formatted before you can use them. When you format a disk, you configure it with a file system so that Windows can store information on the disk. Hard disks in new computers running Windows are already formatted. If you buy an additional hard disk to expand the storage of your computer, you might need to format it. Storage devices such as USB flash drives and flash memory cards usually come preformatted by the manufacturer, so you probably won't need to format them. CDs and DVDs, on the other hand, use different formats from hard disks and removable storage devices. For information about formatting CDs and DVDs, see Which CD or DVD format should I use? Warning Warning Formatting erases any existing files on a hard disk. If you format a hard disk that has files on it, the files will be deleted. WHAT I DID WAS I GOT TO COMPUTER MANAGEMENT SECTION THEN I CLICKED ON DRIVE HP(C) (it put stripes on to show it is selected) Then I click on ACTION selected ALL TASKS AND THEN selected SHRINK VOLUME and then chose how much space from what it was giving me that I wanted. (12.93gb) AND THAT WAS ALL I DID. THEN I TRIED TO INSTALL UBUNTU i NEVER GOT THE 3RD SCREEN THAT IS IN THE VIDEO I INCLUDED (THE YOUTUBE WITH THE ENGLISH GUY) INSTALLATION TYPE I ALSO DID NOT GET THE 4TH SCREEN THAT ALLOWS YOU TO SELECT PARTITION SIZE what i got next was the 2nd INSTILLATION TYPE window shown on the (LINUX BS DOS.COM) PAGE THAT I INCLUDED and it showed no information about any drives (no drives /partition or stuff was shown) only the Boot Loader statement and the dev/sda bar and that's why i did not press install but chose to QUIT. SORRY I JUST NOW SAW YOUR ANSWER IRRATIONAL JOHN. I SHRANK HP(C) BY 12.93GB MY UNALLOCATED SPACE IS NOW 12.93GB HP(C) = 907.17gb NTSF...YOU ARE CORRECT WITH EVERYTHING YOU SAID This is what i read on (http://)windows.microsoft.com/en-US/windows7/Create-a-boot-partition I am only allowed 2 links Create a boot partition You must be logged on as an administrator to perform these steps. A boot partition is a partition that contains the files for the Windows operating system. If you want to install a second operating system on your computer (called a dual-boot or multiboot configuration), you need to create another partition on the hard disk, and then install the additional operating system on the new partition. Your hard disk would then have one system partition and two boot partitions. (A system partition is the partition that contains the hardware-related files. These tell the computer where to look to start Windows.) To create a partition on a basic disk, there must be unallocated disk space on your hard disk. With Disk Management, you can create a maximum of three primary partitions on a hard disk. You can create extended partitions, which include logical drives within them, if you need more partitions on the disk. Picture of disk space in Computer ManagementUnallocated disk space If there is no unallocated space, you will either need to create space by shrinking or deleting an existing partition or by using a third-party partitioning tool to repartition your hard disk. For more information, see Can I repartition my hard disk? To create a boot partition Warning Warning If you are installing different versions of Windows, you must install the earliest version first. If you don't do this, your computer may become inoperable. Open Computer Management by clicking the Start button Picture of the Start button, clicking Control Panel, clicking System and Security, clicking Administrative Tools, and then double-clicking Computer Management.? Administrator permission required If you're prompted for an administrator password or confirmation, type the password or provide confirmation. In the left pane, under Storage, click Disk Management. Right-click an unallocated region on your hard disk, and then click New Simple Volume. In the New Simple Volume Wizard, click Next. Type the size of the volume you want to create in megabytes (MB) or accept the maximum default size, and then click Next. Accept the default drive letter or choose a different drive letter to identify the volume, and then click Next. In the Format Partition dialog box, do one of the following: If you don't want to format the volume right now, click Do not format this volume, and then click Next. To format the volume with the default settings, click Next. For more information about formatting, see Formatting disks and drives: frequently asked questions. Review your choices, and then click Finish. I did what you told me @irrational john and this is the screen shot. I ENTERED ubuntu@ubuntu:~$ sudo os-prober computer did not respond so I entered ubuntu@ubuntu:~$ sudo apt-get -y remove dmraid computer responded with Reading package lists... Done Building dependency tree Reading state information... Done The following packages will be REMOVED: dmraid 0 upgraded, 0 newly installed, 1 to remove and 0 not upgraded. After this operation, 141 kB disk space will be freed. (Reading database ... 147515 files and directories currently installed.) Removing dmraid ... update-initramfs is disabled since running on read-only media Processing triggers for man-db ... I entered ubuntu@ubuntu:~$ sudo os-prober Computer Responded with /dev/sda1:Windows 7 (loader):Windows:chain /dev/sda3:Windows Recovery Environment (loader):Windows1:chain ubuntu@ubuntu:~$ ............... @obsessiveFOSS I don't know what is a Grub menu and I do not know what is the Ubuntu boot option The answer you gave to me was correct. This one {This apparently removes the dmraid metadata. After doing that, you can use the desktop icon Install Ubuntu 12.04 LTS to start the Ubuntu installer. This time the Installation Type window should contain the option to Install Ubuntu alongside Windows 7.} This is what I decided to do. I did not see the rest of your help 'till now. Never the less. I think the best thing for me to do now is to get a cheap used laptop and either do a dual installation or just install Ubuntu on to it. This way if I have any issues that I cannot solve like the one I had here, at least I will still have a usable computer to work on and to use to get answers with because I am not an expert like the people on this forum. Thanks a lot I will try to keep learning and do research enough to some day help someone else.

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  • Salesforce/PHP - Bulk Outbound message (SOAP), Time out issue - See update #2

    - by Phill Pafford
    Salesforce can send up to 100 requests inside 1 SOAP message. While sending this type of Bulk Ooutbound message request my PHP script finishes executing but SF fails to accept the ACK used to clear the message queue on the Salesforce side of things. Looking at the Outbound message log (monitoring) I see all the messages in a pending state with the Delivery Failure Reason "java.net.SocketTimeoutException: Read timed out". If my script has finished execution, why do I get this error? I have tried these methods to increase the execution time on my server as I have no access on the Salesforce side: set_time_limit(0); // in the script max_execution_time = 360 ; Maximum execution time of each script, in seconds max_input_time = 360 ; Maximum amount of time each script may spend parsing request data memory_limit = 32M ; Maximum amount of memory a script may consume I used the high settings just for testing. Any thoughts as to why this is failing the ACK delivery back to Salesforce? Here is some of the code: This is how I accept and send the ACK file for the imcoming SOAP request $data = 'php://input'; $content = file_get_contents($data); if($content) { respond('true'); } else { respond('false'); } The respond function function respond($tf) { $ACK = <<<ACK <?xml version = "1.0" encoding = "utf-8"?> <soapenv:Envelope xmlns:soapenv="http://schemas.xmlsoap.org/soap/envelope/" xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"> <soapenv:Body> <notifications xmlns="http://soap.sforce.com/2005/09/outbound"> <Ack>$tf</Ack> </notifications> </soapenv:Body> </soapenv:Envelope> ACK; print trim($ACK); } These are in a generic script that I include into the script that uses the data for a specific workflow. I can process about 25 requests (That are in 1 SOAP response) but once I go over that I get the timeout error in the Salesforce queue. for 50 requests is usually takes my PHP script 86.77 seconds. Could it be Apache? PHP? I have also tested just accepting the 100 request SOAP response and just accepting and sending the ACK the queue clears out, so I know it's on my side of things. I show no errors in the apache log, the script runs fine. I did find some info on the Salesforce site but still no luck. Here is the link. Also I'm using the PHP Toolkit 11 (From Salesforce). Other forum with good SF help Thanks for any insight into this, --Phill UPDATE: If I receive the incoming message and print the response, should this happen first regardless if I do anything else after? Or does it wait for my process to finish and then print the response? UPDATE #2: okay I think I have the problem: PHP uses the single thread processing approach and will not send back the ACK file until the thread has completed it's processing. Is there a way to make this a mutli thread process? Thread #1 - accept the incoming SOAP request and send back the ACK Thread #2 - Process the SOAP request I know I could break it up into like a DB table or flat file, but is there a way to accomplish this without doing that? I'm going to try to close the socket after the ACK submission and continue the processing, cross my fingers it will work.

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  • Getting java.lang.ClassNotFoundException: com.mysql.jdbc.Driver Exception

    - by Yashwant Chavan
    Hi , I am getting Following Exception while configuring the Connection Pool in Tomcat This is Context.xml <Context path="/DBTest" docBase="DBTest" debug="5" reloadable="true" crossContext="true"> <!-- maxActive: Maximum number of dB connections in pool. Make sure you configure your mysqld max_connections large enough to handle all of your db connections. Set to -1 for no limit. --> <!-- maxIdle: Maximum number of idle dB connections to retain in pool. Set to -1 for no limit. See also the DBCP documentation on this and the minEvictableIdleTimeMillis configuration parameter. --> <!-- maxWait: Maximum time to wait for a dB connection to become available in ms, in this example 10 seconds. An Exception is thrown if this timeout is exceeded. Set to -1 to wait indefinitely. --> <!-- username and password: MySQL dB username and password for dB connections --> <!-- driverClassName: Class name for the old mm.mysql JDBC driver is org.gjt.mm.mysql.Driver - we recommend using Connector/J though. Class name for the official MySQL Connector/J driver is com.mysql.jdbc.Driver. --> <!-- url: The JDBC connection url for connecting to your MySQL dB. --> <Resource name="jdbc/TestDB" auth="Container" type="javax.sql.DataSource" maxActive="100" maxIdle="30" maxWait="10000" username="root" password="password" driverClassName="com.mysql.jdbc.Driver" url="jdbc:mysql:///BUSINESS"/> </Context> This is Bean Entry <bean id="dataSource" class="org.springframework.jndi.JndiObjectFactoryBean"> <property name="jndiName" value="jdbc/TestDB"></property> <property name="resourceRef" value="true"></property> </bean> org.springframework.jdbc.CannotGetJdbcConnectionException: Could not get JDBC Co nnection; nested exception is org.apache.tomcat.dbcp.dbcp.SQLNestedException: Ca nnot load JDBC driver class 'com.mysql.jdbc.Driver' at org.springframework.jdbc.datasource.DataSourceUtils.getConnection(Dat aSourceUtils.java:82) at org.springframework.jdbc.core.JdbcTemplate.execute(JdbcTemplate.java: 382) at org.springframework.jdbc.core.JdbcTemplate.query(JdbcTemplate.java:45 8) at org.springframework.jdbc.core.JdbcTemplate.query(JdbcTemplate.java:46 6) at com.businesscaliber.dao.Dao.getQueryForListMap(Dao.java:66) at com.businesscaliber.dao.MiscellaneousDao.getDefaultSucessStory(Miscel laneousDao.java:109) at com.businesscaliber.listeners.BusinessContextLoader.contextInitialize d(BusinessContextLoader.java:40) at org.apache.catalina.core.StandardContext.listenerStart(StandardContex t.java:3795) at org.apache.catalina.core.StandardContext.start(StandardContext.java:4 252) at org.apache.catalina.core.ContainerBase.addChildInternal(ContainerBase .java:760) at org.apache.catalina.core.ContainerBase.addChild(ContainerBase.java:74 0) at org.apache.catalina.core.StandardHost.addChild(StandardHost.java:544) at org.apache.catalina.startup.HostConfig.deployWAR(HostConfig.java:831) at org.apache.catalina.startup.HostConfig.deployWARs(HostConfig.java:720 ) at org.apache.catalina.startup.HostConfig.deployApps(HostConfig.java:490 ) at org.apache.catalina.startup.HostConfig.start(HostConfig.java:1150) at org.apache.catalina.startup.HostConfig.lifecycleEvent(HostConfig.java :311) at org.apache.catalina.util.LifecycleSupport.fireLifecycleEvent(Lifecycl eSupport.java:120) at org.apache.catalina.core.ContainerBase.start(ContainerBase.java:1022) at org.apache.catalina.core.StandardHost.start(StandardHost.java:736) at org.apache.catalina.core.ContainerBase.start(ContainerBase.java:1014) at org.apache.catalina.core.StandardEngine.start(StandardEngine.java:443 ) at org.apache.catalina.core.StandardService.start(StandardService.java:4 48) at org.apache.catalina.core.StandardServer.start(StandardServer.java:700 ) at org.apache.catalina.startup.Catalina.start(Catalina.java:552) at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method) at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl. java:39) at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAcces sorImpl.java:25) at java.lang.reflect.Method.invoke(Method.java:585) at org.apache.catalina.startup.Bootstrap.start(Bootstrap.java:295) at org.apache.catalina.startup.Bootstrap.main(Bootstrap.java:433) Caused by: org.apache.tomcat.dbcp.dbcp.SQLNestedException: Cannot load JDBC driv er class 'com.mysql.jdbc.Driver' at org.apache.tomcat.dbcp.dbcp.BasicDataSource.createDataSource(BasicDat aSource.java:1136) at org.apache.tomcat.dbcp.dbcp.BasicDataSource.getConnection(BasicDataSo urce.java:880) at org.springframework.jdbc.datasource.DataSourceUtils.doGetConnection(D ataSourceUtils.java:113) at org.springframework.jdbc.datasource.DataSourceUtils.getConnection(Dat aSourceUtils.java:79) ... 30 more Caused by: java.lang.ClassNotFoundException: com.mysql.jdbc.Driver at java.net.URLClassLoader$1.run(URLClassLoader.java:200) at java.security.AccessController.doPrivileged(Native Method) at java.net.URLClassLoader.findClass(URLClassLoader.java:188) at java.lang.ClassLoader.loadClass(ClassLoader.java:306) at java.lang.ClassLoader.loadClass(ClassLoader.java:251) at java.lang.ClassLoader.loadClassInternal(ClassLoader.java:319) at java.lang.Class.forName0(Native Method) at java.lang.Class.forName(Class.java:164) at org.apache.tomcat.dbcp.dbcp.BasicDataSource.createDataSource(BasicDat aSource.java:1130)

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  • Explain the Peak and Flag Algorithm

    - by Isaac Levin
    EDIT Just was pointed that the requirements state peaks cannot be ends of Arrays. So I ran across this site http://codility.com/ Which gives you programming problems and gives you certificates if you can solve them in 2 hours. The very first question is one I have seen before, typically called the Peaks and Flags question. If you are not familiar A non-empty zero-indexed array A consisting of N integers is given. A peak is an array element which is larger than its neighbours. More precisely, it is an index P such that 0 < P < N - 1 and A[P - 1] < A[P] A[P + 1] . For example, the following array A: A[0] = 1 A[1] = 5 A[2] = 3 A[3] = 4 A[4] = 3 A[5] = 4 A[6] = 1 A[7] = 2 A[8] = 3 A[9] = 4 A[10] = 6 A[11] = 2 has exactly four peaks: elements 1, 3, 5 and 10. You are going on a trip to a range of mountains whose relative heights are represented by array A. You have to choose how many flags you should take with you. The goal is to set the maximum number of flags on the peaks, according to certain rules. Flags can only be set on peaks. What's more, if you take K flags, then the distance between any two flags should be greater than or equal to K. The distance between indices P and Q is the absolute value |P - Q|. For example, given the mountain range represented by array A, above, with N = 12, if you take: two flags, you can set them on peaks 1 and 5; three flags, you can set them on peaks 1, 5 and 10; four flags, you can set only three flags, on peaks 1, 5 and 10. You can therefore set a maximum of three flags in this case. Write a function that, given a non-empty zero-indexed array A of N integers, returns the maximum number of flags that can be set on the peaks of the array. For example, given the array above the function should return 3, as explained above. Assume that: N is an integer within the range [1..100,000]; each element of array A is an integer within the range [0..1,000,000,000]. Complexity: expected worst-case time complexity is O(N); expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments). Elements of input arrays can be modified. So this makes sense, but I failed it using this code public int GetFlags(int[] A) { List<int> peakList = new List<int>(); for (int i = 0; i <= A.Length - 1; i++) { if ((A[i] > A[i + 1] && A[i] > A[i - 1])) { peakList.Add(i); } } List<int> flagList = new List<int>(); int distance = peakList.Count; flagList.Add(peakList[0]); for (int i = 1, j = 0, max = peakList.Count; i < max; i++) { if (Math.Abs(Convert.ToDecimal(peakList[j]) - Convert.ToDecimal(peakList[i])) >= distance) { flagList.Add(peakList[i]); j = i; } } return flagList.Count; } EDIT int[] A = new int[] { 7, 10, 4, 5, 7, 4, 6, 1, 4, 3, 3, 7 }; The correct answer is 3, but my application says 2 This I do not get, since there are 4 peaks (indices 1,4,6,8) and from that, you should be able to place a flag at 2 of the peaks (1 and 6) Am I missing something here? Obviously my assumption is that the beginning or end of an Array can be a peak, is this not the case? If this needs to go in Stack Exchange Programmers, I will move it, but thought dialog here would be helpful. EDIT

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  • Steganography Experiment - Trouble hiding message bits in DCT coefficients

    - by JohnHankinson
    I have an application requiring me to be able to embed loss-less data into an image. As such I've been experimenting with steganography, specifically via modification of DCT coefficients as the method I select, apart from being loss-less must also be relatively resilient against format conversion, scaling/DSP etc. From the research I've done thus far this method seems to be the best candidate. I've seen a number of papers on the subject which all seem to neglect specific details (some neglect to mention modification of 0 coefficients, or modification of AC coefficient etc). After combining the findings and making a few modifications of my own which include: 1) Using a more quantized version of the DCT matrix to ensure we only modify coefficients that would still be present should the image be JPEG'ed further or processed (I'm using this in place of simply following a zig-zag pattern). 2) I'm modifying bit 4 instead of the LSB and then based on what the original bit value was adjusting the lower bits to minimize the difference. 3) I'm only modifying the blue channel as it should be the least visible. This process must modify the actual image and not the DCT values stored in file (like jsteg) as there is no guarantee the file will be a JPEG, it may also be opened and re-saved at a later stage in a different format. For added robustness I've included the message multiple times and use the bits that occur most often, I had considered using a QR code as the message data or simply applying the reed-solomon error correction, but for this simple application and given that the "message" in question is usually going to be between 10-32 bytes I have plenty of room to repeat it which should provide sufficient redundancy to recover the true bits. No matter what I do I don't seem to be able to recover the bits at the decode stage. I've tried including / excluding various checks (even if it degrades image quality for the time being). I've tried using fixed point vs. double arithmetic, moving the bit to encode, I suspect that the message bits are being lost during the IDCT back to image. Any thoughts or suggestions on how to get this working would be hugely appreciated. (PS I am aware that the actual DCT/IDCT could be optimized from it's naive On4 operation using row column algorithm, or an FDCT like AAN, but for now it just needs to work :) ) Reference Papers: http://www.lokminglui.com/dct.pdf http://arxiv.org/ftp/arxiv/papers/1006/1006.1186.pdf Code for the Encode/Decode process in C# below: using System; using System.Collections.Generic; using System.Linq; using System.Text; using System.Drawing.Imaging; using System.Drawing; namespace ImageKey { public class Encoder { public const int HIDE_BIT_POS = 3; // use bit position 4 (1 << 3). public const int HIDE_COUNT = 16; // Number of times to repeat the message to avoid error. // JPEG Standard Quantization Matrix. // (to get higher quality multiply by (100-quality)/50 .. // for lower than 50 multiply by 50/quality. Then round to integers and clip to ensure only positive integers. public static double[] Q = {16,11,10,16,24,40,51,61, 12,12,14,19,26,58,60,55, 14,13,16,24,40,57,69,56, 14,17,22,29,51,87,80,62, 18,22,37,56,68,109,103,77, 24,35,55,64,81,104,113,92, 49,64,78,87,103,121,120,101, 72,92,95,98,112,100,103,99}; // Maximum qauality quantization matrix (if all 1's doesn't modify coefficients at all). public static double[] Q2 = {1,1,1,1,1,1,1,1, 1,1,1,1,1,1,1,1, 1,1,1,1,1,1,1,1, 1,1,1,1,1,1,1,1, 1,1,1,1,1,1,1,1, 1,1,1,1,1,1,1,1, 1,1,1,1,1,1,1,1, 1,1,1,1,1,1,1,1}; public static Bitmap Encode(Bitmap b, string key) { Bitmap response = new Bitmap(b.Width, b.Height, PixelFormat.Format32bppArgb); uint imgWidth = ((uint)b.Width) & ~((uint)7); // Maximum usable X resolution (divisible by 8). uint imgHeight = ((uint)b.Height) & ~((uint)7); // Maximum usable Y resolution (divisible by 8). // Start be transferring the unmodified image portions. // As we'll be using slightly less width/height for the encoding process we'll need the edges to be populated. for (int y = 0; y < b.Height; y++) for (int x = 0; x < b.Width; x++) { if( (x >= imgWidth && x < b.Width) || (y>=imgHeight && y < b.Height)) response.SetPixel(x, y, b.GetPixel(x, y)); } // Setup the counters and byte data for the message to encode. StringBuilder sb = new StringBuilder(); for(int i=0;i<HIDE_COUNT;i++) sb.Append(key); byte[] codeBytes = System.Text.Encoding.ASCII.GetBytes(sb.ToString()); int bitofs = 0; // Current bit position we've encoded too. int totalBits = (codeBytes.Length * 8); // Total number of bits to encode. for (int y = 0; y < imgHeight; y += 8) { for (int x = 0; x < imgWidth; x += 8) { int[] redData = GetRedChannelData(b, x, y); int[] greenData = GetGreenChannelData(b, x, y); int[] blueData = GetBlueChannelData(b, x, y); int[] newRedData; int[] newGreenData; int[] newBlueData; if (bitofs < totalBits) { double[] redDCT = DCT(ref redData); double[] greenDCT = DCT(ref greenData); double[] blueDCT = DCT(ref blueData); int[] redDCTI = Quantize(ref redDCT, ref Q2); int[] greenDCTI = Quantize(ref greenDCT, ref Q2); int[] blueDCTI = Quantize(ref blueDCT, ref Q2); int[] blueDCTC = Quantize(ref blueDCT, ref Q); HideBits(ref blueDCTI, ref blueDCTC, ref bitofs, ref totalBits, ref codeBytes); double[] redDCT2 = DeQuantize(ref redDCTI, ref Q2); double[] greenDCT2 = DeQuantize(ref greenDCTI, ref Q2); double[] blueDCT2 = DeQuantize(ref blueDCTI, ref Q2); newRedData = IDCT(ref redDCT2); newGreenData = IDCT(ref greenDCT2); newBlueData = IDCT(ref blueDCT2); } else { newRedData = redData; newGreenData = greenData; newBlueData = blueData; } MapToRGBRange(ref newRedData); MapToRGBRange(ref newGreenData); MapToRGBRange(ref newBlueData); for(int dy=0;dy<8;dy++) { for(int dx=0;dx<8;dx++) { int col = (0xff<<24) + (newRedData[dx+(dy*8)]<<16) + (newGreenData[dx+(dy*8)]<<8) + (newBlueData[dx+(dy*8)]); response.SetPixel(x+dx,y+dy,Color.FromArgb(col)); } } } } if (bitofs < totalBits) throw new Exception("Failed to encode data - insufficient cover image coefficients"); return (response); } public static void HideBits(ref int[] DCTMatrix, ref int[] CMatrix, ref int bitofs, ref int totalBits, ref byte[] codeBytes) { int tempValue = 0; for (int u = 0; u < 8; u++) { for (int v = 0; v < 8; v++) { if ( (u != 0 || v != 0) && CMatrix[v+(u*8)] != 0 && DCTMatrix[v+(u*8)] != 0) { if (bitofs < totalBits) { tempValue = DCTMatrix[v + (u * 8)]; int bytePos = (bitofs) >> 3; int bitPos = (bitofs) % 8; byte mask = (byte)(1 << bitPos); byte value = (byte)((codeBytes[bytePos] & mask) >> bitPos); // 0 or 1. if (value == 0) { int a = DCTMatrix[v + (u * 8)] & (1 << HIDE_BIT_POS); if (a != 0) DCTMatrix[v + (u * 8)] |= (1 << HIDE_BIT_POS) - 1; DCTMatrix[v + (u * 8)] &= ~(1 << HIDE_BIT_POS); } else if (value == 1) { int a = DCTMatrix[v + (u * 8)] & (1 << HIDE_BIT_POS); if (a == 0) DCTMatrix[v + (u * 8)] &= ~((1 << HIDE_BIT_POS) - 1); DCTMatrix[v + (u * 8)] |= (1 << HIDE_BIT_POS); } if (DCTMatrix[v + (u * 8)] != 0) bitofs++; else DCTMatrix[v + (u * 8)] = tempValue; } } } } } public static void MapToRGBRange(ref int[] data) { for(int i=0;i<data.Length;i++) { data[i] += 128; if(data[i] < 0) data[i] = 0; else if(data[i] > 255) data[i] = 255; } } public static int[] GetRedChannelData(Bitmap b, int sx, int sy) { int[] data = new int[8 * 8]; for (int y = sy; y < (sy + 8); y++) { for (int x = sx; x < (sx + 8); x++) { uint col = (uint)b.GetPixel(x,y).ToArgb(); data[(x - sx) + ((y - sy) * 8)] = (int)((col >> 16) & 0xff) - 128; } } return (data); } public static int[] GetGreenChannelData(Bitmap b, int sx, int sy) { int[] data = new int[8 * 8]; for (int y = sy; y < (sy + 8); y++) { for (int x = sx; x < (sx + 8); x++) { uint col = (uint)b.GetPixel(x, y).ToArgb(); data[(x - sx) + ((y - sy) * 8)] = (int)((col >> 8) & 0xff) - 128; } } return (data); } public static int[] GetBlueChannelData(Bitmap b, int sx, int sy) { int[] data = new int[8 * 8]; for (int y = sy; y < (sy + 8); y++) { for (int x = sx; x < (sx + 8); x++) { uint col = (uint)b.GetPixel(x, y).ToArgb(); data[(x - sx) + ((y - sy) * 8)] = (int)((col >> 0) & 0xff) - 128; } } return (data); } public static int[] Quantize(ref double[] DCTMatrix, ref double[] Q) { int[] DCTMatrixOut = new int[8*8]; for (int u = 0; u < 8; u++) { for (int v = 0; v < 8; v++) { DCTMatrixOut[v + (u * 8)] = (int)Math.Round(DCTMatrix[v + (u * 8)] / Q[v + (u * 8)]); } } return(DCTMatrixOut); } public static double[] DeQuantize(ref int[] DCTMatrix, ref double[] Q) { double[] DCTMatrixOut = new double[8*8]; for (int u = 0; u < 8; u++) { for (int v = 0; v < 8; v++) { DCTMatrixOut[v + (u * 8)] = (double)DCTMatrix[v + (u * 8)] * Q[v + (u * 8)]; } } return(DCTMatrixOut); } public static double[] DCT(ref int[] data) { double[] DCTMatrix = new double[8 * 8]; for (int v = 0; v < 8; v++) { for (int u = 0; u < 8; u++) { double cu = 1; if (u == 0) cu = (1.0 / Math.Sqrt(2.0)); double cv = 1; if (v == 0) cv = (1.0 / Math.Sqrt(2.0)); double sum = 0.0; for (int y = 0; y < 8; y++) { for (int x = 0; x < 8; x++) { double s = data[x + (y * 8)]; double dctVal = Math.Cos((2 * y + 1) * v * Math.PI / 16) * Math.Cos((2 * x + 1) * u * Math.PI / 16); sum += s * dctVal; } } DCTMatrix[u + (v * 8)] = (0.25 * cu * cv * sum); } } return (DCTMatrix); } public static int[] IDCT(ref double[] DCTMatrix) { int[] Matrix = new int[8 * 8]; for (int y = 0; y < 8; y++) { for (int x = 0; x < 8; x++) { double sum = 0; for (int v = 0; v < 8; v++) { for (int u = 0; u < 8; u++) { double cu = 1; if (u == 0) cu = (1.0 / Math.Sqrt(2.0)); double cv = 1; if (v == 0) cv = (1.0 / Math.Sqrt(2.0)); double idctVal = (cu * cv) / 4.0 * Math.Cos((2 * y + 1) * v * Math.PI / 16) * Math.Cos((2 * x + 1) * u * Math.PI / 16); sum += (DCTMatrix[u + (v * 8)] * idctVal); } } Matrix[x + (y * 8)] = (int)Math.Round(sum); } } return (Matrix); } } public class Decoder { public static string Decode(Bitmap b, int expectedLength) { expectedLength *= Encoder.HIDE_COUNT; uint imgWidth = ((uint)b.Width) & ~((uint)7); // Maximum usable X resolution (divisible by 8). uint imgHeight = ((uint)b.Height) & ~((uint)7); // Maximum usable Y resolution (divisible by 8). // Setup the counters and byte data for the message to decode. byte[] codeBytes = new byte[expectedLength]; byte[] outBytes = new byte[expectedLength / Encoder.HIDE_COUNT]; int bitofs = 0; // Current bit position we've decoded too. int totalBits = (codeBytes.Length * 8); // Total number of bits to decode. for (int y = 0; y < imgHeight; y += 8) { for (int x = 0; x < imgWidth; x += 8) { int[] blueData = ImageKey.Encoder.GetBlueChannelData(b, x, y); double[] blueDCT = ImageKey.Encoder.DCT(ref blueData); int[] blueDCTI = ImageKey.Encoder.Quantize(ref blueDCT, ref Encoder.Q2); int[] blueDCTC = ImageKey.Encoder.Quantize(ref blueDCT, ref Encoder.Q); if (bitofs < totalBits) GetBits(ref blueDCTI, ref blueDCTC, ref bitofs, ref totalBits, ref codeBytes); } } bitofs = 0; for (int i = 0; i < (expectedLength / Encoder.HIDE_COUNT) * 8; i++) { int bytePos = (bitofs) >> 3; int bitPos = (bitofs) % 8; byte mask = (byte)(1 << bitPos); List<int> values = new List<int>(); int zeroCount = 0; int oneCount = 0; for (int j = 0; j < Encoder.HIDE_COUNT; j++) { int val = (codeBytes[bytePos + ((expectedLength / Encoder.HIDE_COUNT) * j)] & mask) >> bitPos; values.Add(val); if (val == 0) zeroCount++; else oneCount++; } if (oneCount >= zeroCount) outBytes[bytePos] |= mask; bitofs++; values.Clear(); } return (System.Text.Encoding.ASCII.GetString(outBytes)); } public static void GetBits(ref int[] DCTMatrix, ref int[] CMatrix, ref int bitofs, ref int totalBits, ref byte[] codeBytes) { for (int u = 0; u < 8; u++) { for (int v = 0; v < 8; v++) { if ((u != 0 || v != 0) && CMatrix[v + (u * 8)] != 0 && DCTMatrix[v + (u * 8)] != 0) { if (bitofs < totalBits) { int bytePos = (bitofs) >> 3; int bitPos = (bitofs) % 8; byte mask = (byte)(1 << bitPos); int value = DCTMatrix[v + (u * 8)] & (1 << Encoder.HIDE_BIT_POS); if (value != 0) codeBytes[bytePos] |= mask; bitofs++; } } } } } } } UPDATE: By switching to using a QR Code as the source message and swapping a pair of coefficients in each block instead of bit manipulation I've been able to get the message to survive the transform. However to get the message to come through without corruption I have to adjust both coefficients as well as swap them. For example swapping (3,4) and (4,3) in the DCT matrix and then respectively adding 8 and subtracting 8 as an arbitrary constant seems to work. This survives a re-JPEG'ing of 96 but any form of scaling/cropping destroys the message again. I was hoping that by operating on mid to low frequency values that the message would be preserved even under some light image manipulation.

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  • SQL SERVER – Reducing CXPACKET Wait Stats for High Transactional Database

    - by pinaldave
    While engaging in a performance tuning consultation for a client, a situation occurred where they were facing a lot of CXPACKET Waits Stats. The client asked me if I could help them reduce this huge number of wait stats. I usually receive this kind of request from other client as well, but the important thing to understand is whether this question has any merits or benefits, or not. Before we continue the resolution, let us understand what CXPACKET Wait Stats are. The official definition suggests that CXPACKET Wait Stats occurs when trying to synchronize the query processor exchange iterator. You may consider lowering the degree of parallelism if a conflict concerning this wait type develops into a problem. (from BOL) In simpler words, when a parallel operation is created for SQL Query, there are multiple threads for a single query. Each query deals with a different set of the data (or rows). Due to some reasons, one or more of the threads lag behind, creating the CXPACKET Wait Stat. Threads which came first have to wait for the slower thread to finish. The Wait by a specific completed thread is called CXPACKET Wait Stat. Note that CXPACKET Wait is done by completed thread and not the one which are unfinished. “Note that not all the CXPACKET wait types are bad. You might experience a case when it totally makes sense. There might also be cases when this is also unavoidable. If you remove this particular wait type for any query, then that query may run slower because the parallel operations are disabled for the query.” Now let us see what the best practices to reduce the CXPACKET Wait Stats are. The suggestions, with which you will find that if you search online through the browser, would play a major role as and might be asked about their jobs In addition, might tell you that you should set ‘maximum degree of parallelism’ to 1. I do agree with these suggestions, too; however, I think this is not the final resolutions. As soon as you set your entire query to run on single CPU, you will get a very bad performance from the queries which are actually performing okay when using parallelism. The best suggestion to this is that you set ‘the maximum degree of parallelism’ to a lower number or 1 (be very careful with this – it can create more problems) but tune the queries which can be benefited from multiple CPU’s. You can use query hint OPTION (MAXDOP 0) to run the server to use parallelism. Here is the two-quick script which helps to resolve these issues: Change MAXDOP at Server Level EXEC sys.sp_configure N'max degree of parallelism', N'1' GO RECONFIGURE WITH OVERRIDE GO Run Query with all the CPU (using parallelism) USE AdventureWorks GO SELECT * FROM Sales.SalesOrderDetail ORDER BY ProductID OPTION (MAXDOP 0) GO Below is the blog post which will help you to find all the parallel query in your server. SQL SERVER – Find Queries using Parallelism from Cached Plan Please note running Queries in single CPU may worsen your performance and it is not recommended at all. Infect this can be very bad advise. I strongly suggest that you identify the queries which are offending and tune them instead of following any other suggestions. Reference: Pinal Dave (http://blog.sqlauthority.com) Filed under: SQL, SQL Authority, SQL Optimization, SQL Performance, SQL Query, SQL Server, SQL Tips and Tricks, SQL White Papers, SQLAuthority News, T SQL, Technology

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  • Sun Fire X4800 M2 Posts World Record x86 SPECjEnterprise2010 Result

    - by Brian
    Oracle's Sun Fire X4800 M2 using the Intel Xeon E7-8870 processor and Sun Fire X4470 M2 using the Intel Xeon E7-4870 processor, produced a world record single application server SPECjEnterprise2010 benchmark result of 27,150.05 SPECjEnterprise2010 EjOPS. The Sun Fire X4800 M2 server ran the application tier and the Sun Fire X4470 M2 server was used for the database tier. The Sun Fire X4800 M2 server demonstrated 63% better performance compared to IBM P780 server result of 16,646.34 SPECjEnterprise2010 EjOPS. The Sun Fire X4800 M2 server demonstrated 4% better performance than the Cisco UCS B440 M2 result, both results used the same number of processors. This result used Oracle WebLogic Server 12c, Java HotSpot(TM) 64-Bit Server 1.7.0_02, and Oracle Database 11g. This result was produced using Oracle Linux. Performance Landscape Complete benchmark results are at the SPEC website, SPECjEnterprise2010 Results. The table below compares against the best results from IBM and Cisco. SPECjEnterprise2010 Performance Chart as of 3/12/2012 Submitter EjOPS* Application Server Database Server Oracle 27,150.05 1x Sun Fire X4800 M2 8x 2.4 GHz Intel Xeon E7-8870 Oracle WebLogic 12c 1x Sun Fire X4470 M2 4x 2.4 GHz Intel Xeon E7-4870 Oracle Database 11g (11.2.0.2) Cisco 26,118.67 2x UCS B440 M2 Blade Server 4x 2.4 GHz Intel Xeon E7-4870 Oracle WebLogic 11g (10.3.5) 1x UCS C460 M2 Blade Server 4x 2.4 GHz Intel Xeon E7-4870 Oracle Database 11g (11.2.0.2) IBM 16,646.34 1x IBM Power 780 8x 3.86 GHz POWER 7 WebSphere Application Server V7 1x IBM Power 750 Express 4x 3.55 GHz POWER 7 IBM DB2 9.7 Workgroup Server Edition FP3a * SPECjEnterprise2010 EjOPS, bigger is better. Configuration Summary Application Server: 1 x Sun Fire X4800 M2 8 x 2.4 GHz Intel Xeon processor E7-8870 256 GB memory 4 x 10 GbE NIC 2 x FC HBA Oracle Linux 5 Update 6 Oracle WebLogic Server 11g Release 1 (10.3.5) Java HotSpot(TM) 64-Bit Server VM on Linux, version 1.7.0_02 (Java SE 7 Update 2) Database Server: 1 x Sun Fire X4470 M2 4 x 2.4 GHz Intel Xeon E7-4870 512 GB memory 4 x 10 GbE NIC 2 x FC HBA 2 x Sun StorageTek 2540 M2 4 x Sun Fire X4270 M2 4 x Sun Storage F5100 Flash Array Oracle Linux 5 Update 6 Oracle Database 11g Enterprise Edition Release 11.2.0.2 Benchmark Description SPECjEnterprise2010 is the third generation of the SPEC organization's J2EE end-to-end industry standard benchmark application. The SPECjEnterprise2010 benchmark has been designed and developed to cover the Java EE 5 specification's significantly expanded and simplified programming model, highlighting the major features used by developers in the industry today. This provides a real world workload driving the Application Server's implementation of the Java EE specification to its maximum potential and allowing maximum stressing of the underlying hardware and software systems. The workload consists of an end to end web based order processing domain, an RMI and Web Services driven manufacturing domain and a supply chain model utilizing document based Web Services. The application is a collection of Java classes, Java Servlets, Java Server Pages, Enterprise Java Beans, Java Persistence Entities (pojo's) and Message Driven Beans. The SPECjEnterprise2010 benchmark heavily exercises all parts of the underlying infrastructure that make up the application environment, including hardware, JVM software, database software, JDBC drivers, and the system network. The primary metric of the SPECjEnterprise2010 benchmark is jEnterprise Operations Per Second ("SPECjEnterprise2010 EjOPS"). This metric is calculated by adding the metrics of the Dealership Management Application in the Dealer Domain and the Manufacturing Application in the Manufacturing Domain. There is no price/performance metric in this benchmark. Key Points and Best Practices Sixteen Oracle WebLogic server instances were started using numactl, binding 2 instances per chip. Eight Oracle database listener processes were started, binding 2 instances per chip using taskset. Additional tuning information is in the report at http://spec.org. See Also Oracle Press Release -- SPECjEnterprise2010 Results Page Sun Fire X4800 M2 Server oracle.com OTN Sun Fire X4270 M2 Server oracle.com OTN Sun Storage 2540-M2 Array oracle.com OTN Oracle Linux oracle.com OTN Oracle Database 11g Release 2 Enterprise Edition oracle.com OTN WebLogic Suite oracle.com OTN Disclosure Statement SPEC and the benchmark name SPECjEnterprise are registered trademarks of the Standard Performance Evaluation Corporation. Sun Fire X4800 M2, 27,150.05 SPECjEnterprise2010 EjOPS; IBM Power 780, 16,646.34 SPECjEnterprise2010 EjOPS; Cisco UCS B440 M2, 26,118.67 SPECjEnterprise2010 EjOPS. Results from www.spec.org as of 3/27/2012.

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  • Big Data – Beginning Big Data – Day 1 of 21

    - by Pinal Dave
    What is Big Data? I want to learn Big Data. I have no clue where and how to start learning about it. Does Big Data really means data is big? What are the tools and software I need to know to learn Big Data? I often receive questions which I mentioned above. They are good questions and honestly when we search online, it is hard to find authoritative and authentic answers. I have been working with Big Data and NoSQL for a while and I have decided that I will attempt to discuss this subject over here in the blog. In the next 21 days we will understand what is so big about Big Data. Big Data – Big Thing! Big Data is becoming one of the most talked about technology trends nowadays. The real challenge with the big organization is to get maximum out of the data already available and predict what kind of data to collect in the future. How to take the existing data and make it meaningful that it provides us accurate insight in the past data is one of the key discussion points in many of the executive meetings in organizations. With the explosion of the data the challenge has gone to the next level and now a Big Data is becoming the reality in many organizations. Big Data – A Rubik’s Cube I like to compare big data with the Rubik’s cube. I believe they have many similarities. Just like a Rubik’s cube it has many different solutions. Let us visualize a Rubik’s cube solving challenge where there are many experts participating. If you take five Rubik’s cube and mix up the same way and give it to five different expert to solve it. It is quite possible that all the five people will solve the Rubik’s cube in fractions of the seconds but if you pay attention to the same closely, you will notice that even though the final outcome is the same, the route taken to solve the Rubik’s cube is not the same. Every expert will start at a different place and will try to resolve it with different methods. Some will solve one color first and others will solve another color first. Even though they follow the same kind of algorithm to solve the puzzle they will start and end at a different place and their moves will be different at many occasions. It is  nearly impossible to have a exact same route taken by two experts. Big Market and Multiple Solutions Big Data is exactly like a Rubik’s cube – even though the goal of every organization and expert is same to get maximum out of the data, the route and the starting point are different for each organization and expert. As organizations are evaluating and architecting big data solutions they are also learning the ways and opportunities which are related to Big Data. There is not a single solution to big data as well there is not a single vendor which can claim to know all about Big Data. Honestly, Big Data is too big a concept and there are many players – different architectures, different vendors and different technology. What is Next? In this 31 days series we will be exploring many essential topics related to big data. I do not claim that you will be master of the subject after 31 days but I claim that I will be covering following topics in easy to understand language. Architecture of Big Data Big Data a Management and Implementation Different Technologies – Hadoop, Mapreduce Real World Conversations Best Practices Tomorrow In tomorrow’s blog post we will try to answer one of the very essential questions – What is Big Data? Reference: Pinal Dave (http://blog.sqlauthority.com) Filed under: Big Data, PostADay, SQL, SQL Authority, SQL Query, SQL Server, SQL Tips and Tricks, T SQL

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  • Dual Screen will only mirror after 12.04 upgrade

    - by Ne0
    I have been using Ubuntu with a dual screen for years now, after upgrading to 12.04 LTS i cannot get my dual screen working properly Graphics: 01:00.0 VGA compatible controller: Advanced Micro Devices [AMD] nee ATI RV350 AR [Radeon 9600] 01:00.1 Display controller: Advanced Micro Devices [AMD] nee ATI RV350 AR [Radeon 9600] (Secondary) I noticed i was using open source drivers and attempted to install official binaries using the methods in this thread. Output: liam@liam-desktop:~$ sudo apt-get install fglrx fglrx-amdcccle Reading package lists... Done Building dependency tree Reading state information... Done The following packages will be upgraded: fglrx fglrx-amdcccle 2 upgraded, 0 newly installed, 0 to remove and 12 not upgraded. Need to get 45.1 MB of archives. After this operation, 739 kB of additional disk space will be used. Get:1 http://gb.archive.ubuntu.com/ubuntu/ precise/restricted fglrx i386 2:8.960-0ubuntu1 [39.2 MB] Get:2 http://gb.archive.ubuntu.com/ubuntu/ precise/restricted fglrx-amdcccle i386 2:8.960-0ubuntu1 [5,883 kB] Fetched 45.1 MB in 1min 33s (484 kB/s) (Reading database ... 328081 files and directories currently installed.) Preparing to replace fglrx 2:8.951-0ubuntu1 (using .../fglrx_2%3a8.960-0ubuntu1_i386.deb) ... Removing all DKMS Modules Error! There are no instances of module: fglrx 8.951 located in the DKMS tree. Done. Unpacking replacement fglrx ... Preparing to replace fglrx-amdcccle 2:8.951-0ubuntu1 (using .../fglrx-amdcccle_2%3a8.960-0ubuntu1_i386.deb) ... Unpacking replacement fglrx-amdcccle ... Processing triggers for ureadahead ... ureadahead will be reprofiled on next reboot Setting up fglrx (2:8.960-0ubuntu1) ... update-alternatives: warning: forcing reinstallation of alternative /usr/lib/fglrx/ld.so.conf because link group i386-linux-gnu_gl_conf is broken. update-alternatives: warning: skip creation of /etc/OpenCL/vendors/amdocl64.icd because associated file /usr/lib/fglrx/etc/OpenCL/vendors/amdocl64.icd (of link group i386-linux-gnu_gl_conf) doesn't exist. update-alternatives: warning: skip creation of /usr/lib32/libaticalcl.so because associated file /usr/lib32/fglrx/libaticalcl.so (of link group i386-linux-gnu_gl_conf) doesn't exist. update-alternatives: warning: skip creation of /usr/lib32/libaticalrt.so because associated file /usr/lib32/fglrx/libaticalrt.so (of link group i386-linux-gnu_gl_conf) doesn't exist. update-alternatives: warning: forcing reinstallation of alternative /usr/lib/fglrx/ld.so.conf because link group i386-linux-gnu_gl_conf is broken. update-alternatives: warning: skip creation of /etc/OpenCL/vendors/amdocl64.icd because associated file /usr/lib/fglrx/etc/OpenCL/vendors/amdocl64.icd (of link group i386-linux-gnu_gl_conf) doesn't exist. update-alternatives: warning: skip creation of /usr/lib32/libaticalcl.so because associated file /usr/lib32/fglrx/libaticalcl.so (of link group i386-linux-gnu_gl_conf) doesn't exist. update-alternatives: warning: skip creation of /usr/lib32/libaticalrt.so because associated file /usr/lib32/fglrx/libaticalrt.so (of link group i386-linux-gnu_gl_conf) doesn't exist. update-initramfs: deferring update (trigger activated) update-initramfs: Generating /boot/initrd.img-3.2.0-25-generic-pae Loading new fglrx-8.960 DKMS files... Building only for 3.2.0-25-generic-pae Building for architecture i686 Building initial module for 3.2.0-25-generic-pae Done. fglrx: Running module version sanity check. - Original module - No original module exists within this kernel - Installation - Installing to /lib/modules/3.2.0-25-generic-pae/updates/dkms/ depmod....... DKMS: install completed. update-initramfs: deferring update (trigger activated) Processing triggers for bamfdaemon ... Rebuilding /usr/share/applications/bamf.index... Setting up fglrx-amdcccle (2:8.960-0ubuntu1) ... Processing triggers for initramfs-tools ... update-initramfs: Generating /boot/initrd.img-3.2.0-25-generic-pae Processing triggers for libc-bin ... ldconfig deferred processing now taking place liam@liam-desktop:~$ sudo aticonfig --initial -f aticonfig: No supported adapters detected When i attempt to get my settings back to what they were before upgrading i get this message requested position/size for CRTC 81 is outside the allowed limit: position=(1440, 0), size=(1440, 900), maximum=(1680, 1680) and GDBus.Error:org.gtk.GDBus.UnmappedGError.Quark._gnome_2drr_2derror_2dquark.Code3: requested position/size for CRTC 81 is outside the allowed limit: position=(1440, 0), size=(1440, 900), maximum=(1680, 1680) Any idea's on what i need to do to fix this issue?

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  • Partner Webcast: Innovation in Products - October 1st, 2012 at 04:00 PM CET (03:00 PM GMT) Program

    - by Richard Lefebvre
    I am pleased to invite you to join the Innovations in Products – webcast. Innovations in Products will present Oracle Product's new functions and features including sales positioning. The key objectives of these webcasts are to inspire System Integrator's implementation personnel to conduct successful after sales in their Customer projects. Innovations in Products will be presented on the 1st Monday of each quarter after the billable day (4:00 to 5:00 PM CET). The webcast is intended for System Integrator's Implementation Certified Specialists but Innovations in Products is open for other system Integrator's personnel as well. At first, two Oracle representatives will discuss Oracle's contribution to partners. Then you will see product breakout session followed by Q&A with Oracle Experts. Each session will last for maximum 1 hour. A Q&A document covering all questions and answers will be made available after the webcast. What are the Benefits for partners? Find out how Innovations in Products helps you to improve your after sales Discover new functions and features so you can enrich your Customers's solution Learn more about Oracle products, especially sales positioning Hear crucial questions raised by colleague alike, learn from their interest Engage and present your questions to subject experts Be inspired of the richness of Oracle's product portfolio – for your and your customer's benefit. Note: Should you already be familiar with a specific Product, then choose another one. Doing so you would expand your knowledge of the overall product portfolio. Some presentations contain product demonstration, although these presentations are not intended to be extremely detailed technical presentations. Product breakout sessions available on October 1st:   Topics Speaker To Register Fusion HCM Social Capabilities, Enterprise social capabilities embedded in how you run your business Anca Dumitru, HCM Presales Consultant, EMEA Presales Center CLICK HERE Oracle Fusion Applications Security Concepts, Overview Alexandra Dan, Applications Technology Presales Consultant, EMEA Presales Centre CLICK HERE Fusion Financials Overview - Focus on Fusion Payables, Meeting the Payables' Challenges Elena Nita, Senior ERP Sales Consultant, EMEA Presales Center CLICK HERE Introduction to Oracle RightNow CX, Empowering companies to engage directly with their customers through great Social, Web, Chat and Contact Center experiences. Cais Champsi, Presales Consultant, EMEA Presales Centre CLICK HERE Oracle Endeca Information Discovery, Product Overview Emma Palii, BI Sales Consultant, EMEA Presales Center CLICK HERE To access previously presented 23 Applications Products presentations and 6 Public Sector Value Proposition presentations, please click here. You might want to bookmark the overall registration page Innovations in Products October 1st and the global event calendar page events.oracle.com. Delivery Format Innovations in Products – program is a series of FREE prerecorded Oracle product presentations followed by Q&A. It will be delivered over the Web. Participants have the opportunity to submit questions during the web cast via chat and subject matter experts will provide verbal answers live. Innovations in Products consists of several parallel prerecorded product breakout sessions, each lasting for max. 1 hour. At first, two Oracle representatives will discuss Oracle's contribution to Partners. Then you'll see the product breakout sessions followed by Q&A with Oracle Experts. A Q&A document covering all questions and answers will be made available after the webcast. You can also see Innovations in Products afterwards as its content will be available online for the next 6-12 months. The next Innovations in Products web casts will be presented as follows: October 1st 2012 January 14th 2013 April 8th 2013. Note: Depending on local network bandwidth please allow some seconds time the presentations to download. You might want to refresh your screen by pressing F5. Duration: Maximum 1 hour For further information please contact me Markku Rouhiainen. Best regards Markku Rouhiainen Director, Applications Partner Enablement EMEA

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  • Where is the value of OEA

    - by [email protected]
    In a room full of architects, if you were to ask for the definition of enterprise architecture, or the importance thereof,  you are likely to get a number of varying view points ranging from,  a complete analysis of the digital assets of an organization,  to, a strategic alignment of business goals/objectives to IT initiatives.  Similiarily in a room full of senior business executives,  if you asked them how they see their IT groups and their effectiveness to align to business strategy,  you would get a myriad of responses,  ranging from, “a huge drain on our bottom line”, “always more expensive than budgeted”, “lack of agility,  by the time IT is ready,  my business strategy has changed”, and on the rare occurrence, “ a leader of innovation,  that is lock step with my business strategy”. However does this necessarily demonstrate the overall value of enterprise architecture.  Having a framework, and process is of critical importance to help produce a number of the artefacts that ultimately align technology goals and initiatives to business strategy,  however,  is that really where the value is?  I believe that first we need to understand the concept of value.  Value typically is a measure of sorts,  when we purchase a product it’s value is equivalent to the maximum amount that someone is willing to pay for the product,  however,  is the same equation valid in terms of the business value of enterprise architecture? Is the library of artefacts generated through a process/framework, inclusive of a strategic roadmap to realize the enterprise architecture where the value is? If we agree that enterprise architecture is the alignment of IT and IT assets to support business strategy, and by achieving our business strategy, we have we have increased the business value of the enterprise then;  it seems that, in order to really identify the true value of an enterprise architecture,  we need to understand how we measure business value .  A number of formal measurement methodologies exist for this purpose, business models, balanced scorecards, etc   After we have an understanding on how to measure the business value of each of the organizational units within an enterprise, then we understand how the enterprise architecture contributes to the success of business strategy,  and EXECUTE on the roadmap to implement, and deliver the IT initiatives that provide MEASUREABLE returns, As we analyse the value chain of each of the individual organizational units within the enterprise we may identify how that unit has performed by quantitatively measuring it proximity to achieving the goals defined by the business for each unit. However, It would appear that true business value (the aggregate of all of the business units in the value chain), is to some degree subjectively measured  as for public companies this lies in shareholder value,  as the true value, or be it, the maximum amount that someone would pay for shares of an organization.

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  • Is this simple XOR encrypted communication absolutely secure?

    - by user3123061
    Say Alice have 4GB USB flash memory and Peter also have 4GB USB flash memory. They once meet and save on both of memories two files named alice_to_peter.key (2GB) and peter_to_alice.key (2GB) which is randomly generated bits. Then they never meet again and communicate electronicaly. Alice also maintains variable called alice_pointer and Peter maintains variable called peter_pointer which is both initially set to zero. Then when Alice needs to send message to Peter they do: encrypted_message_to_peter[n] = message_to_peter[n] XOR alice_to_peter.key[alice_pointer + n] Where n i n-th byte of message. Then alice_pointer is attached at begining of the encrypted message and (alice_pointer + encrypted message) is sent to Peter and then alice_pointer is incremented by length of message (and for maximum security can be used part of key erased) Peter receives encrypted_message, reads alice_pointer stored at beginning of message and do this: message_to_peter[n] = encrypted_message_to_peter[n] XOR alice_to_peter.key[alice_pointer + n] And for maximum security after reading of message also erases used part of key. - EDIT: In fact this step with this simple algorithm (without integrity check and authentication) decreases security, see Paulo Ebermann post below. When Peter needs to send message to Alice they do analogical steps with peter_to_alice.key and with peter_pointer. With this trivial schema they can send for next 50 years each day 2GB / (50 * 365) = cca 115kB of encrypted data in both directions. If they need more data to send, they simple use larger memory for keys for example with today 2TB harddiscs (1TB keys) is possible to exchange next 50years 60MB/day ! (thats practicaly lots of data for example with using compression its more than hour of high quality voice communication) It Seems to me there is no way for attacker to read encrypted message without keys even if they have infinitely fast computer. because even with infinitely fast computer with brute force they get ever possible message that can fit to length of message, but this is astronomical amount of messages and attacker dont know which of them is actual message. I am right? Is this communication schema really absolutely secure? And if its secure, has this communication method its own name? (I mean XOR encryption is well-known, but whats name of this concrete practical application with use large memories at both communication sides for keys? I am humbly expecting that this application has been invented someone before me :-) ) Note: If its absolutely secure then its amazing because with today low cost large memories it is practicaly much cheeper way of secure communication than expensive quantum cryptography and with equivalent security! EDIT: I think it will be more and more practical in future with lower a lower cost of memories. It can solve secure communication forever. Today you have no certainty if someone succesfuly atack to existing ciphers one year later and make its often expensive implementations unsecure. In many cases before comunication exist step where communicating sides meets personaly, thats time to generate large keys. I think its perfect for military communication for example for communication with submarines which can have installed harddrive with large keys and military central can have harddrive for each submarine they have. It can be also practical in everyday life for example for control your bank account because when you create your account you meet with bank etc.

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  • Laptop screen blank after login when external monitor is not connected

    - by Ramon Suarez
    Ubuntu does not switch back automatically to only monitor present when booting after disconnecting external monitor. Here's a video showing what happens. I get to the login window and everything looks ok, then I type my password, the desktop image shows up and... everything goes blank. It does not happen when I just login as a guest. When possible I work with my laptop connected to an external screen via the VGA port. The problem comes when I boot the computer without that secondary screen connected: The login screen comes out ok. After login the screen goes black, but I can hear the login sound. If I hit ctr + alt + backwards-delete and login again sometimes it is fixed, but not all. If I log in as a different user everything is OK. Then I log in as my user and sometimes it works. To have a screen I have to plug a monitor. Although I have turned on the laptop display with that monitor on, if I reboot it goes blank again after login, even if I turn off the external monitor before turning off the computer. I've managed to get my screen back with my username after going into recovery mode, but only sometimes. Failsafe would not load after second screen asking me what I wanted to do (no mouse to click nor keyboard working). My computer is a LDLC Aurore BB1-i5 -8 -S1. Which is the configuration file that keeps the information about the monitors using Displays under lightgdm and where is it? I guess if I could edit it I may have a chance :) One of the things I tried following a solution in another post was removing my monitors.xml file, but it does not work and I don't know how to create a good one that I could use now. When doing DISPLAY=:0 xrandrI get: Screen 0: minimum 320 x 200, current 320 x 200, maximum 8192 x 8192 LVDS1 connected (normal left inverted right x axis y axis) 1366x768 60.0 + 1360x768 59.8 60.0 1024x768 60.0 800x600 60.3 56.2 640x480 59.9 VGA1 disconnected (normal left inverted right x axis y axis) HDMI1 disconnected (normal left inverted right x axis y axis) DP1 disconnected (normal left inverted right x axis y axis) This is the full dmesg after activating sudo xdiagnoseas Bryce sugested. (If you tell me the relevant parts I will paste them here) When conecting the external monitor, only the external will work, although I can see using Displays that the computer thinks that both are working. I've asked the question in Launchpad but have it keeps on expiring without any feedback. In my opinion Ubuntu should be able to detect automatically that there is no external monitor present and switch to the laptop monitor. There's a similar question here, but it does not apply to my case External monitor set as primary even when disconnected from laptop Update: For clarification, the problem happens only with my user and once I log in. I even get to see the screensaver for about a second, and then it goes blank. Tried Bryce's example (see his answer below), but it did not work. This is the info I get from tty1 with Display=:0 xrandr: – Ramon Suarez Jul 9 at 16:36 Screen 0: minimum 320 x 200, current 320 x 200, maximum 8192 x 8192 LVDS1 connected (normal left inverted right x axis y axis) 1366x768 60.0 + 1360x768 59.8 60.0 1024x768 60.0 800x600 60.3 56.2 640x480 59.9 VGA1 disconnected (normal left inverted right x axis y axis) HDMI1 disconnected (normal left inverted right x axis y axis) DP1 disconnected (normal left inverted right x axis y axis)

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