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• Load average is have been high over some period

  - by user111196

  We have a dedicated MySQL server and below is the a snapshot of the top. The load average has been staying at nearly 100 for an hour plus ready. top - 20:54:28 up 7:31, 2 users, load average: 83.08, 96.88, 106.23 Tasks: 278 total, 2 running, 274 sleeping, 2 stopped, 0 zombie Cpu0 : 18.8%us, 10.2%sy, 0.0%ni, 70.9%id, 0.0%wa, 0.0%hi, 0.0%si, 0.0%st Cpu1 : 51.2%us, 4.3%sy, 0.0%ni, 44.2%id, 0.0%wa, 0.0%hi, 0.3%si, 0.0%st Cpu2 : 9.0%us, 10.3%sy, 0.0%ni, 80.6%id, 0.0%wa, 0.0%hi, 0.0%si, 0.0%st Cpu3 : 18.8%us, 7.4%sy, 0.0%ni, 73.8%id, 0.0%wa, 0.0%hi, 0.0%si, 0.0%st Cpu4 : 7.8%us, 8.8%sy, 0.0%ni, 83.4%id, 0.0%wa, 0.0%hi, 0.0%si, 0.0%st Cpu5 : 10.3%us, 8.4%sy, 0.0%ni, 81.4%id, 0.0%wa, 0.0%hi, 0.0%si, 0.0%st Cpu6 : 6.2%us, 7.5%sy, 0.0%ni, 86.2%id, 0.0%wa, 0.0%hi, 0.0%si, 0.0%st Cpu7 : 6.2%us, 6.2%sy, 0.0%ni, 87.3%id, 0.0%wa, 0.0%hi, 0.3%si, 0.0%st Cpu8 : 8.8%us, 10.4%sy, 0.0%ni, 80.5%id, 0.0%wa, 0.0%hi, 0.3%si, 0.0%st Cpu9 : 63.7%us, 4.6%sy, 0.0%ni, 12.2%id, 0.0%wa, 4.3%hi, 15.2%si, 0.0%st Cpu10 : 9.2%us, 10.2%sy, 0.0%ni, 80.6%id, 0.0%wa, 0.0%hi, 0.0%si, 0.0%st Cpu11 : 17.3%us, 5.9%sy, 0.0%ni, 76.8%id, 0.0%wa, 0.0%hi, 0.0%si, 0.0%st Cpu12 : 8.0%us, 8.7%sy, 0.0%ni, 83.3%id, 0.0%wa, 0.0%hi, 0.0%si, 0.0%st Cpu13 : 10.9%us, 7.4%sy, 0.0%ni, 81.7%id, 0.0%wa, 0.0%hi, 0.0%si, 0.0%st Cpu14 : 6.2%us, 6.9%sy, 0.0%ni, 86.9%id, 0.0%wa, 0.0%hi, 0.0%si, 0.0%st Cpu15 : 4.8%us, 6.1%sy, 0.0%ni, 89.0%id, 0.0%wa, 0.0%hi, 0.0%si, 0.0%st Mem: 33009800k total, 23174396k used, 9835404k free, 120604k buffers Swap: 35061752k total, 0k used, 35061752k free, 16459540k cached PID USER PR NI VIRT RES SHR S %CPU %MEM TIME+ COMMAND 3341 mysql 20 0 14.3g 4.6g 4240 S 417.8 14.5 1673:51 mysqld 24406 root 20 0 15008 1292 876 R 0.3 0.0 0:00.19 top 1 root 20 0 4080 852 608 S 0.0 0.0 0:01.92 init 2 root 15 -5 0 0 0 S 0.0 0.0 0:00.00 kthreadd 3 root RT -5 0 0 0 S 0.0 0.0 0:00.32 migration/0 4 root 15 -5 0 0 0 S 0.0 0.0 0:00.29 ksoftirqd/0 5 root RT -5 0 0 0 S 0.0 0.0 0:00.00 watchdog/0 6 root RT -5 0 0 0 S 0.0 0.0 0:03.21 migration/1 7 root 15 -5 0 0 0 S 0.0 0.0 0:00.07 ksoftirqd/1 8 root RT -5 0 0 0 S 0.0 0.0 0:00.00 watchdog/1 9 root RT -5 0 0 0 S 0.0 0.0 0:00.17 migration/2 10 root 15 -5 0 0 0 S 0.0 0.0 0:00.03 ksoftirqd/2 11 root RT -5 0 0 0 S 0.0 0.0 0:00.00 watchdog/2 12 root RT -5 0 0 0 S 0.0 0.0 0:00.32 migration/3 13 root 15 -5 0 0 0 S 0.0 0.0 0:00.02 ksoftirqd/3 14 root RT -5 0 0 0 S 0.0 0.0 0:00.00 watchdog/3 15 root RT -5 0 0 0 S 0.0 0.0 0:00.10 migration/4 16 root 15 -5 0 0 0 S 0.0 0.0 0:00.04 ksoftirqd/4 17 root RT -5 0 0 0 S 0.0 0.0 0:00.00 watchdog/4 18 root RT -5 0 0 0 S 0.0 0.0 0:00.35 migration/5 We have also tried to run this command. What else command can help us diagnose the exact problem of this high load? netstat -nat |grep 3306 | awk '{print $6}' | sort | uniq -c | sort -n 1 LISTEN 1 SYN_RECV 410 ESTABLISHED 964 TIME_WAIT Output of vmstat 1: ---------------memory--------------- --swap-- --io-- --system-- -----cpu------ r b swpd free buff cache si so bi bo in cs us sy id wa st 2 0 0 12978936 30944 15172360 0 0 259 3 184 265 6 6 77 12 0

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• « Si on veut aller sur Windows Phone c'est maintenant, après ce sera trop tard », entretien avec Nicolas Sorel, PDG de Magma Mobile

  « Si on veut aller sur Windows Phone c'est maintenant, après ce sera trop tard » , entretien avec Nicolas Sorel, DG de Magma Mobile Les cabinets d'analyse à l'instar d'IDC s'accordent à montrer que les dispositifs mobiles sont de plus en plus vendus de par le monde. Un signe peut-être pour les développeurs qui ne l'ont pas encore fait de songer à penser au mobile. D'ailleurs, certaines structures qui délivrent des services essayent de s'adapter aux réalités du marché en proposant désormais à...

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• Une entreprise devrait-elle payer moins d'impôts si elle investit beaucoup localement ? En Australie, Google suggère de « mettre l'accent dessus »

  Une entreprise devrait-elle payer moins d'impôts si elle investit beaucoup localement ? En Australie, le patron de Google suggère de « mettre l'accent dessus » Maile Carnegie, le chef Google Australie, s'attend à ce que le gouvernement clarifie les zones d'ombres dans son système global d'imposition. La directrice a reconnu que les critiques à l'endroit des impôts que verse son entreprise est compréhensible, mais ne tient cependant pas compte de l'impact de l'investissement de la compagnie dans...

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• Google met à jour reCAPTCHA et teste si vous êtes un humain avant, pendant et après que vous ayez interagi avec le CAPTCHA

  Google met à jour reCAPTCHA et teste si vous êtes un humain avant, pendant et après que vous ayez interagi avec le CAPTCHA Le CAPTCHA (Completely Automated Public Turing test to tell Computers and Humans Apart) est un moyen de sécurité sur le net permettant de différencier un humain des robots ou des spammeurs par la génération à l'écran des caractères dont le but est d'être difficilement déchiffrés par les bots.Plusieurs CAPTCHA irritent la plus part de temps les utilisateurs, les mots générés...

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• Low load average with plenty of cpu-intersive processes

  - by sds

  I see loadavg at about 1 with at least 3 processes running at full tile. How can that be? top - 11:48:32 up 147 days, 5:38, 8 users, load average: 1.08, 1.11, 1.05 Tasks: 416 total, 4 running, 410 sleeping, 2 stopped, 0 zombie Cpu0 : 43.3%us, 13.7%sy, 0.0%ni, 43.0%id, 0.0%wa, 0.0%hi, 0.0%si, 0.0%st Cpu1 : 48.8%us, 12.4%sy, 0.0%ni, 38.8%id, 0.0%wa, 0.0%hi, 0.0%si, 0.0%st Cpu2 : 0.7%us, 0.7%sy, 0.0%ni, 98.3%id, 0.0%wa, 0.0%hi, 0.3%si, 0.0%st Cpu3 : 99.3%us, 0.7%sy, 0.0%ni, 0.0%id, 0.0%wa, 0.0%hi, 0.0%si, 0.0%st Cpu4 : 0.0%us, 0.3%sy, 0.0%ni, 99.7%id, 0.0%wa, 0.0%hi, 0.0%si, 0.0%st Cpu5 : 5.7%us, 0.7%sy, 0.0%ni, 93.6%id, 0.0%wa, 0.0%hi, 0.0%si, 0.0%st Cpu6 : 2.3%us, 0.3%sy, 0.0%ni, 97.4%id, 0.0%wa, 0.0%hi, 0.0%si, 0.0%st Cpu7 : 0.3%us, 0.3%sy, 0.0%ni, 99.0%id, 0.0%wa, 0.0%hi, 0.3%si, 0.0%st Cpu8 : 38.4%us, 17.4%sy, 0.0%ni, 44.2%id, 0.0%wa, 0.0%hi, 0.0%si, 0.0%st Cpu9 : 43.4%us, 13.5%sy, 0.0%ni, 43.1%id, 0.0%wa, 0.0%hi, 0.0%si, 0.0%st Cpu10 : 0.0%us, 0.0%sy, 0.0%ni,100.0%id, 0.0%wa, 0.0%hi, 0.0%si, 0.0%st Cpu11 : 0.0%us, 0.0%sy, 0.0%ni,100.0%id, 0.0%wa, 0.0%hi, 0.0%si, 0.0%st Cpu12 : 0.0%us, 0.0%sy, 0.0%ni,100.0%id, 0.0%wa, 0.0%hi, 0.0%si, 0.0%st Cpu13 : 0.3%us, 0.3%sy, 0.0%ni, 99.3%id, 0.0%wa, 0.0%hi, 0.0%si, 0.0%st Cpu14 : 0.0%us, 0.0%sy, 0.0%ni,100.0%id, 0.0%wa, 0.0%hi, 0.0%si, 0.0%st Cpu15 : 1.0%us, 0.7%sy, 0.0%ni, 98.3%id, 0.0%wa, 0.0%hi, 0.0%si, 0.0%st Mem: 132145404k total, 88125080k used, 44020324k free, 516476k buffers Swap: 8388600k total, 620232k used, 7768368k free, 55729064k cached PID USER PR NI VIRT RES SHR S %CPU %MEM TIME+ COMMAND 25424 jonathan 20 0 4404m 4.1g 3268 R 99.7 3.3 212:58.17 python2.7 20939 sam 20 0 908m 733m 3376 R 81.2 0.6 603:08.07 python2.7 20987 sam 20 0 908m 732m 3376 R 79.8 0.6 598:49.18 python2.7 25428 jonathan 20 0 774m 164m 15m S 14.2 0.1 24:22.60 java 20996 sam 20 0 98.4m 7780 1880 S 4.3 0.0 17:48.15 vw 20941 sam 20 0 161m 70m 1880 S 3.0 0.1 18:10.03 vw 20940 sam 20 0 98.4m 8068 1880 S 2.6 0.0 18:06.28 vw 20942 sam 20 0 98.4m 8080 1880 S 2.6 0.0 17:39.45 vw 20944 sam 20 0 161m 71m 1880 S 2.6 0.1 17:29.29 vw 20947 sam 20 0 161m 71m 1880 S 2.6 0.1 17:25.58 vw 20959 sam 20 0 161m 70m 1880 S 2.6 0.1 17:28.00 vw 20962 sam 20 0 161m 70m 1880 S 2.6 0.1 17:26.96 vw 20963 sam 20 0 98.4m 8076 1880 S 2.6 0.0 18:07.19 vw 20965 sam 20 0 161m 71m 1880 S 2.6 0.1 18:08.13 vw 20995 sam 20 0 161m 71m 1880 S 2.6 0.1 17:38.67 vw 6399 root 20 0 558m 19m 5028 S 2.3 0.0 4329:56 BESClient 20945 sam 20 0 98.4m 8068 1880 S 2.3 0.0 17:35.38 vw 20948 sam 20 0 98.4m 8068 1880 S 2.3 0.0 17:26.01 vw 20950 sam 20 0 161m 70m 1880 S 2.3 0.1 17:25.79 vw 20952 sam 20 0 98.4m 8076 1880 S 2.3 0.0 17:32.94 vw 20955 sam 20 0 161m 70m 1880 S 2.3 0.1 17:26.61 vw 20956 sam 20 0 98.4m 8072 1880 S 2.3 0.0 17:34.76 vw 20960 sam 20 0 98.4m 8072 1880 S 2.3 0.0 17:34.04 vw Adding up CPU loads gives about 300%. The top process list also adds up to about 300%. Why is load average about 1?

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• Que se passerait-il si les clients payaient tout comme ils payent les développeurs ? Une vidéo humoristique pose la question

  Que se passerait-il si les clients décidaient de tout payer comme ils payent les développeurs ? Vous retrouvez-vous dans cette vidéo humoristique très bien vue La relation entre développeurs et clients a toujours été particulières. Souvent, le client fixe le budget qu'il « peut » dédier à un projet et c'est aux développeurs de s'adapter pour livrer le résultat escompté. Quite à faire des concessions sur la qualité du code et de l'architecture globale du projet... ou de carrément travailler à perte et de se contenter "d'égayer son portefolio" comme le suggère certains clients peu scrupuleux. Une situation aberrante qui fait grincer les dents dans les sociét...

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• Et si Apple rachetait le fournisseur de puces de ses principaux concurrents ? L'hypothèse est sérieu

  Et si Apple rachetait ARM, le fournisseur de puces de ses concurrents ? L'hypothèse est sérieusement envisagée dans les milieux financiers Soyons clairs. Il s'agit d'une rumeur. Mais une rumeur persistante qui ne cesse de faire le tour de la City de Londres. Résultat, ce « bruit qui court » a fait bondir le cours d'ARM, la joint-venture fondée entre Acorn Computers, Apple et VLSI Technology, de +8 points en une seule et unique journée. ARM est la holding qui possède la technologie des célèbres puces qui équipent les teminaux de Nokia, Samsung, HTC (avec lequel il est en procès), la Nintendo DS ou encore la PSP de Sony (pour ne citer que ceux-ci). Bref, autant de concurrents...

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• Les meilleurs programmeurs sont-ils ceux qui disent connaître C ++ ? Pas si sûr...

  Les meilleurs programmeurs sont-ils ceux qui disent connaître C ++ ? Pas si sûr... L'une des particularité de C++, est que les programmeurs travaillant avec ce langage se divisent en deux catégories. Ces deux groupes s'opposent frontalement et se considèrent chacun comme le seul ayant raison. Dans le premier, on trouve les développeurs qui viennent de C, ou d'autres qui pensent maitriser C++ très rapidement. Ce genre d'invidu vous dira "je connais C++". Mentent-ils ? Ce langage peut être vu comme l'ascension d'une montagne. Il y a ceux qui sont en bas et se croient déjà au sommet, et ceux qui onen ont déjà passé le col. Ceux-là souffrent. C'est la deuxième catégorie de programmeurs : ceux qui en ont déjà ...

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• Adobe : « Nous allons réaliser une super trousse à outils pour le HTML5 », et si Flash et HTML5 coha

  « Nous allons réaliser une super trousse à outils pour le HTML5 » Déclare Adobe : et si Flash et HTML5 cohabitaient en harmonie ? Le directeur de la technologie (CTO) d'Adobe, Kevin Lynch, vient de donner quelques réponses très intéressantes sur l'articulation entre Flash et le HTML 5 lors d'une intervention au Web 2.0 Expo qui se déroule actuellement à San Francisco. « Nous allons réaliser une super trousse à outils pour le HTML 5 », a-t-il annoncé, « Nous allons faire les meilleurs outils au monde pour HTML 5 ». Pour lui l'avenir du web passera certainement par cette nouvelle norme. Mais cela ne signifie nullement la fin de la technologie d'Ad...

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• Google menace de ne plus indexer la presse française si une taxe sur le référencement est votée : qu'en pensez-vous ?

  Google menace de ne plus indexer la presse française si une taxe sur le référencement venait à être votée Google vient de lancer un ultimatum à la presse française et au gouvernement en menaçant de déréférencer les contenus des éditeurs en cas d'adoption de la taxe Google. L'Association de la presse d'information politique et générale (IPG) et le Syndicat de la presse quotidienne nationale ont adressé en septembre dernier une ébauche d'un projet de loi au gouvernement français, visant à l'instauration d'une taxe Google. Ceux-ci estiment que le géant de la recherche tire profit de leurs contenus et actualités pour enrichir son moteur de recherche, permettant à celui-ci de générer de...

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• Jocuri friv pentru toti

  - by haioase

  Jocurile online sunt o modalitate foarte simpla de a te relaxa in timpul liber si spun acest lucru deoarece nu ai altceva de facut decat sa te asezi in fata calculatorului, sa cauti pe internet ceea ce iti place si sa te joci cat vrei - sau cat de mult timp ai la dispozitie.Poate si din cauza ca sunt atat de cautate, industria jocurilor de pe internet a devenit tot mai infloritoare in ultima decada.Un site unde poti sa gasesti o multitudine de jocuri flash este si friv.me.uk games - Play your favorite online game, site dedicat in exclusivitate jocurilor friv de tot felul. Dupa cum se vede din titlu este in limba engleza, dar acest lucru nu cred ca reprezinta un impediment pentru vizitatorii din lumea intreaga care viziteaza www.friv.me.uk , deoarece astazi pana si copiii de gradinita stiu semnificatia cuvintelor play this game sau click here to play.Daca va intrebati ce sa alegeti din multele jocuri de acolo, v-as sugera sa incercati noile jocuri friv de strategie, deoarece sunt atat haioase, cat si interesante si educative pentru cei mici. Nu iti trebuie dexteritate in apasarea tastelor, ci o minte organizata, deoarece trebuie sa iti faci un plan de aparare foarte bun pentru a putea castiga un joc de genul lui Bloons tower defense, de exemplu.Fetelor care vor sa se joace pe friv.me.uk as vrea sa le sugerez cateva jocuri speciale pentru ele, cum ar fi cele de gatit impreuna cu Dora. Se vor distra copios, preparand cea mai gustoasa pizza in bucataria virtuala a Dorei si, in acelasi timp, vor invata fractiile, deoarece trebuie sa imparta pizza in felii egale pentru toti cei aflati la masa.Acestea au fost doar cateva idei despre ce jocuri friv puteti sa va jucati in fiecare zi pe friv.me.uk. Voi alegeti orice va place si stati oricat vreti acolo, pentru ca este un site unde va puteti amuza foarte tare impreuna cu prietenii sau familia. Distractie placuta tuturor!

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• Problem with stack based implementation of function 0x42 of int 0x13

  - by IceCoder

  I'm trying a new approach to int 0x13 (just to learn more about the way the system works): using stack to create a DAP.. Assuming that DL contains the disk number, AX contains the address of the bootable entry in PT, DS is updated to the right segment and the stack is correctly set, this is the code: push DWORD 0x00000000 add ax, 0x0008 mov si, ax push DWORD [ds:(si)] push DWORD 0x00007c00 push WORD 0x0001 push WORD 0x0010 push ss pop ds mov si, sp mov sp, bp mov ah, 0x42 int 0x13 As you can see: I push the dap structure onto the stack, update DS:SI in order to point to it, DL is already set, then set AX to 0x42 and call int 0x13 the result is error 0x01 in AH and obviously CF set. No sectors are transferred. I checked the stack trace endlessly and it is ok, the partition table is ok too.. I cannot figure out what I'm missing... This is the stack trace portion of the disk address packet: 0x000079ea: 10 00 adc %al,(%bx,%si) 0x000079ec: 01 00 add %ax,(%bx,%si) 0x000079ee: 00 7c 00 add %bh,0x0(%si) 0x000079f1: 00 00 add %al,(%bx,%si) 0x000079f3: 08 00 or %al,(%bx,%si) 0x000079f5: 00 00 add %al,(%bx,%si) 0x000079f7: 00 00 add %al,(%bx,%si) 0x000079f9: 00 a0 07 be add %ah,-0x41f9(%bx,%si) I'm using qemu latest version and trying to read from hard drive (0x80), have also tried with a 4bytes alignment for the structure with the same result (CF 1 AH 0x01), the extensions are present.

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• Improving SAS multipath to JBOD performance on Linux

  - by user36825

  Hello all I'm trying to optimize a storage setup on some Sun hardware with Linux. Any thoughts would be greatly appreciated. We have the following hardware: Sun Blade X6270 2* LSISAS1068E SAS controllers 2* Sun J4400 JBODs with 1 TB disks (24 disks per JBOD) Fedora Core 12 2.6.33 release kernel from FC13 (also tried with latest 2.6.31 kernel from FC12, same results) Here's the datasheet for the SAS hardware: http://www.sun.com/storage/storage_networking/hba/sas/PCIe.pdf It's using PCI Express 1.0a, 8x lanes. With a bandwidth of 250 MB/sec per lane, we should be able to do 2000 MB/sec per SAS controller. Each controller can do 3 Gb/sec per port and has two 4 port PHYs. We connect both PHYs from a controller to a JBOD. So between the JBOD and the controller we have 2 PHYs * 4 SAS ports * 3 Gb/sec = 24 Gb/sec of bandwidth, which is more than the PCI Express bandwidth. With write caching enabled and when doing big writes, each disk can sustain about 80 MB/sec (near the start of the disk). With 24 disks, that means we should be able to do 1920 MB/sec per JBOD. multipath { rr_min_io 100 uid 0 path_grouping_policy multibus failback manual path_selector "round-robin 0" rr_weight priorities alias somealias no_path_retry queue mode 0644 gid 0 wwid somewwid } I tried values of 50, 100, 1000 for rr_min_io, but it doesn't seem to make much difference. Along with varying rr_min_io I tried adding some delay between starting the dd's to prevent all of them writing over the same PHY at the same time, but this didn't make any difference, so I think the I/O's are getting properly spread out. According to /proc/interrupts, the SAS controllers are using a "IR-IO-APIC-fasteoi" interrupt scheme. For some reason only core #0 in the machine is handling these interrupts. I can improve performance slightly by assigning a separate core to handle the interrupts for each SAS controller: echo 2 /proc/irq/24/smp_affinity echo 4 /proc/irq/26/smp_affinity Using dd to write to the disk generates "Function call interrupts" (no idea what these are), which are handled by core #4, so I keep other processes off this core too. I run 48 dd's (one for each disk), assigning them to cores not dealing with interrupts like so: taskset -c somecore dd if=/dev/zero of=/dev/mapper/mpathx oflag=direct bs=128M oflag=direct prevents any kind of buffer cache from getting involved. None of my cores seem maxed out. The cores dealing with interrupts are mostly idle and all the other cores are waiting on I/O as one would expect. Cpu0 : 0.0%us, 1.0%sy, 0.0%ni, 91.2%id, 7.5%wa, 0.0%hi, 0.2%si, 0.0%st Cpu1 : 0.0%us, 0.8%sy, 0.0%ni, 93.0%id, 0.2%wa, 0.0%hi, 6.0%si, 0.0%st Cpu2 : 0.0%us, 0.6%sy, 0.0%ni, 94.4%id, 0.1%wa, 0.0%hi, 4.8%si, 0.0%st Cpu3 : 0.0%us, 7.5%sy, 0.0%ni, 36.3%id, 56.1%wa, 0.0%hi, 0.0%si, 0.0%st Cpu4 : 0.0%us, 1.3%sy, 0.0%ni, 85.7%id, 4.9%wa, 0.0%hi, 8.1%si, 0.0%st Cpu5 : 0.1%us, 5.5%sy, 0.0%ni, 36.2%id, 58.3%wa, 0.0%hi, 0.0%si, 0.0%st Cpu6 : 0.0%us, 5.0%sy, 0.0%ni, 36.3%id, 58.7%wa, 0.0%hi, 0.0%si, 0.0%st Cpu7 : 0.0%us, 5.1%sy, 0.0%ni, 36.3%id, 58.5%wa, 0.0%hi, 0.0%si, 0.0%st Cpu8 : 0.1%us, 8.3%sy, 0.0%ni, 27.2%id, 64.4%wa, 0.0%hi, 0.0%si, 0.0%st Cpu9 : 0.1%us, 7.9%sy, 0.0%ni, 36.2%id, 55.8%wa, 0.0%hi, 0.0%si, 0.0%st Cpu10 : 0.0%us, 7.8%sy, 0.0%ni, 36.2%id, 56.0%wa, 0.0%hi, 0.0%si, 0.0%st Cpu11 : 0.0%us, 7.3%sy, 0.0%ni, 36.3%id, 56.4%wa, 0.0%hi, 0.0%si, 0.0%st Cpu12 : 0.0%us, 5.6%sy, 0.0%ni, 33.1%id, 61.2%wa, 0.0%hi, 0.0%si, 0.0%st Cpu13 : 0.1%us, 5.3%sy, 0.0%ni, 36.1%id, 58.5%wa, 0.0%hi, 0.0%si, 0.0%st Cpu14 : 0.0%us, 4.9%sy, 0.0%ni, 36.4%id, 58.7%wa, 0.0%hi, 0.0%si, 0.0%st Cpu15 : 0.1%us, 5.4%sy, 0.0%ni, 36.5%id, 58.1%wa, 0.0%hi, 0.0%si, 0.0%st Given all this, the throughput reported by running "dstat 10" is in the range of 2200-2300 MB/sec. Given the math above I would expect something in the range of 2*1920 ~= 3600+ MB/sec. Does anybody have any idea where my missing bandwidth went? Thanks!

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• Getting SQL table row counts via sysindexes vs. sys.indexes

  - by Bill Osuch

  Among the many useful SQL snippets I regularly use is this little bit that will return row counts in a table: SELECT so.name as TableName, MAX(si.rows) as [RowCount] FROM sysobjects so JOIN sysindexes si ON si.id = OBJECT_ID(so.name) WHERE so.xtype = 'U' GROUP BY so.name ORDER BY [RowCount] DESC This is handy to find tables that have grown wildly, zero-row tables that could (possibly) be dropped, or other clues into the data. Right off the bat you may spot some "non-ideal" code - I'm using sysobjects rather than sys.objects. What's the difference? In SQL Server 2005 and later, sysobjects is no longer a table, but a "compatibility view", meant for backward compatibility only. SELECT * from each and you'll see the different data that each returns. Microsoft advises that sysindexes could be removed in a future version of SQL Server, but this has never really been an issue for me since my company is still using SQL 2000. However, there are murmurs that we may actually migrate to 2008 some year, so I might as well go ahead and start using an updated version of this snippet on the servers that can handle it: SELECT so.name as TableName, ddps.row_count as [RowCount] FROM sys.objects so JOIN sys.indexes si ON si.OBJECT_ID = so.OBJECT_ID JOIN sys.dm_db_partition_stats AS ddps ON si.OBJECT_ID = ddps.OBJECT_ID  AND si.index_id = ddps.index_id WHERE si.index_id < 2  AND so.is_ms_shipped = 0 ORDER BY ddps.row_count DESC

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• La loi de Moore a-t-elle encore un avenir ? Oui, répond un ingénieur, si les développeurs se mettent

  La loi de Moore est elle encore pertinente ? Plus pour les CPU, répond un responsable de Nvidia qui appelle au développement de la programmation parallèle Bill Dally est un des ingénieurs les plus importants de Nvidia. Dans une tribune publiée dans le magasine Forbes, l'ingénieur en chef doute fortement que la loi de Moore puisse encore s'appliquer aux processeurs (CPU) : « Les performances des CPUs ne peuvent plus doubler tous les 18 mois », constate-t-il, « et cela va poser un grave problème à de nombreuses industries qui reposent sur cette croissance des performances ». Optimiste, Bill Dally en tire cependant une raison d'espérer « La bonne nouvelle, c'e...

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• La ripresa economica si sta consolidando, siete pronti a cogliere questa opportunità?

  - by antonella.buonagurio(at)oracle.com

  L'esclusiva ricerca IDC indica i percorsi strategici più innovativi a supporto delle Vendite, del Customer Service e del Marketing.        La ricerca basata su più di 300 interviste a executive, CIO e CEO di medie e grandi organizzazioni in tutta Europa, vi guiderà nel comprendere l'evoluzione e l'impatto dei trend più rilevanti sui processi che gestiscono ed indirizzano la relazione tra azienda e clienti.

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• "Nos petits trucs utiles de développeurs", et si on partageait nos astuces et nos trouvailles de programmation ?

  Bonjour J'espère que je poste dans le bon forum, sinon merci à la modération de rediriger ce thread. Depuis longtemps, j'avais envie de créer quelque chose de très utile pouvant servir à tout le monde et surtout soit au niveau débutant, mais puisse, pourquoi pas, également servir aux autres. Voici donc un thread entièrement consacré aux petits trucs sans prétention bêtes comme choux, mais très efficaces à l'usage. Il ne s'agit en aucun cas de code compliqué ni de manipulation de haut vol ; tout le contraire et accessible à tous. Je vous invite pour notre plus grand plaisir de développeurs d'ajouter v...

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• Creating independent process!

  - by Neha

  I am trying to create a process from a service in C++. This new process is creating as a child process. I want to create an independent process and not a child process... I am using CreateProcess function for the same. Since the new process i create is a child process when i try to kill process tree at the service level it is killing the child process too... I dont want this to happen. I want the new process created to run independent of the service. Please advice on the same.. Thanks.. Code STARTUPINFO si; PROCESS_INFORMATION pi; ZeroMemory( &si, sizeof(si) ); si.cb = sizeof(si); // Start the child process. ZeroMemory( &pi, sizeof(pi) ); si.dwFlags = STARTF_USESHOWWINDOW; if(bRunOnWinLogonDesktop) { if(csDesktopName.empty()) si.lpDesktop = _T("winsta0\\default"); else _tcscpy(si.lpDesktop, csDesktopName.c_str()); } if(bHide) si.wShowWindow = SW_HIDE; /* maybe even SW_HIDE */ else si.wShowWindow = SW_SHOW; /* maybe even SW_HIDE */ TCHAR szCmdLine[512]; _tcscpy(szCmdLine, csCmdLine.c_str()); if( !CreateProcess( NULL, szCmdLine, NULL, NULL, FALSE, CREATE_NEW_PROCESS_GROUP, NULL, NULL, &si, &pi ) )

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• What is required to use LODSB in assembly?

  - by Harvey

  What is the minimum set of steps required to use LODSB to load a relative address to a string in my code? I have the following test program that I'm using PXE to boot. I boot it two ways: via pxelinux.0 and directly. If I boot it directly, my program prints both strings. If I boot via pxelinux.0, it only prints the first string. Why? Working technique (for both): Set the direction flag to increment, cld Set ds to cs Put the address (from start) of string in si Add the starting offset to si Non-working technique (just for pxelinux): Calculate a new segment address based on (((cs << 4) + offset) >> 4) Set ds to that. (either A000 or 07C0) text here to fix bug in markdown // Note: If you try this code, don't forget to set // the "#if 0" below appropriately! .text .globl start, _start start: _start: _start1: .code16 jmp real_start . = _start1 + 0x1fe .byte 0x55, 0xAA // Next sector . = _start1 + 0x200 jmp real_start test1_str: .asciz "\r\nTest: 9020:fe00" test2_str: .asciz "\r\nTest: a000:0000" real_start: cld // Make sure %si gets incremented. #if 0 // When loaded by pxelinux, we're here: // 9020:fe00 ==> a000:0000 // This works. movw $0x9020, %bx movw %bx, %ds movw $(test1_str - _start1), %si addw $0xfe00, %si call print_message // This does not. movw $0xA000, %bx movw %bx, %ds movw $(test2_str - _start1), %si call print_message #else // If we are loaded directly without pxelinux, we're here: // 0000:7c00 ==> 07c0:0000 // This works. movw $0x0000, %bx movw %bx, %ds movw $(test1_str - _start1), %si addw $0x7c00, %si call print_message // This does, too. movw $0x07c0, %bx movw %bx, %ds movw $(test2_str - _start1), %si call print_message #endif // Hang the computer sti 1: jmp 1b // Prints string DS:SI (modifies AX BX SI) print_message: pushw %ax jmp 2f 3: movb $0x0e, %ah /* print char in AL */ int $0x10 /* via TTY mode */ 2: lodsb (%si), %al /* get token */ cmpb $0, %al /* end of string? */ jne 3b popw %ax ret .balign 0x200 Here's the compilation: /usr/bin/ccache gcc -Os -fno-stack-protector -fno-builtin -nostdinc -DSUPPORT_SERIAL=1 -DSUPPORT_HERCULES=1 -DSUPPORT_GRAPHICS=1 -DHAVE_CONFIG_H -I. -Wall -ggdb3 -Wmissing-prototypes -Wunused -Wshadow -Wpointer-arith -falign-jumps=1 -falign-loops=1 -falign-functions=1 -Wundef -g -c -o ds_teststart_exec-ds_teststart.o ds_test.S /usr/bin/ccache gcc -g -o ds_teststart.exec -nostdlib -Wl,-N -Wl,-Ttext -Wl,8000 ds_teststart_exec-ds_teststart.o objcopy -O binary ds_teststart.exec ds_teststart

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• Splitting a set of object into several subsets of 'similar' objects

  - by doublep

  Suppose I have a set of objects, S. There is an algorithm f that, given a set S builds certain data structure D on it: f(S) = D. If S is large and/or contains vastly different objects, D becomes large, to the point of being unusable (i.e. not fitting in allotted memory). To overcome this, I split S into several non-intersecting subsets: S = S1 + S2 + ... + Sn and build Di for each subset. Using n structures is less efficient than using one, but at least this way I can fit into memory constraints. Since size of f(S) grows faster than S itself, combined size of Di is much less than size of D. However, it is still desirable to reduce n, i.e. the number of subsets; or reduce the combined size of Di. For this, I need to split S in such a way that each Si contains "similar" objects, because then f will produce a smaller output structure if input objects are "similar enough" to each other. The problems is that while "similarity" of objects in S and size of f(S) do correlate, there is no way to compute the latter other than just evaluating f(S), and f is not quite fast. Algorithm I have currently is to iteratively add each next object from S into one of Si, so that this results in the least possible (at this stage) increase in combined Di size: for x in S: i = such i that size(f(Si + {x})) - size(f(Si)) is min Si = Si + {x} This gives practically useful results, but certainly pretty far from optimum (i.e. the minimal possible combined size). Also, this is slow. To speed up somewhat, I compute size(f(Si + {x})) - size(f(Si)) only for those i where x is "similar enough" to objects already in Si. Is there any standard approach to such kinds of problems? I know of branch and bounds algorithm family, but it cannot be applied here because it would be prohibitively slow. My guess is that it is simply not possible to compute optimal distribution of S into Si in reasonable time. But is there some common iteratively improving algorithm?

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• Top cpu% analyis

  - by user111196

  I would like the know in what cpus % can be considered in save range and also load average? Which indication will give signal something is wrong with the server? top - 22:55:51 up 3 days, 6:39, 1 user, load average: 0.53, 0.43, 0.37 Tasks: 229 total, 2 running, 227 sleeping, 0 stopped, 0 zombie Cpu0 : 16.2%us, 0.7%sy, 0.0%ni, 82.8%id, 0.0%wa, 0.0%hi, 0.3%si, 0.0%st Cpu1 : 10.5%us, 0.7%sy, 0.0%ni, 88.5%id, 0.0%wa, 0.0%hi, 0.3%si, 0.0%st Cpu2 : 9.0%us, 0.0%sy, 0.0%ni, 91.0%id, 0.0%wa, 0.0%hi, 0.0%si, 0.0%st Cpu3 : 0.3%us, 0.3%sy, 0.0%ni, 99.4%id, 0.0%wa, 0.0%hi, 0.0%si, 0.0%st Cpu4 : 1.0%us, 0.0%sy, 0.0%ni, 99.0%id, 0.0%wa, 0.0%hi, 0.0%si, 0.0%st Cpu5 : 44.8%us, 2.6%sy, 0.0%ni, 37.0%id, 0.0%wa, 9.4%hi, 6.2%si, 0.0%st Cpu6 : 3.0%us, 0.0%sy, 0.0%ni, 96.7%id, 0.0%wa, 0.0%hi, 0.4%si, 0.0%st Cpu7 : 0.0%us, 0.0%sy, 0.0%ni,100.0%id, 0.0%wa, 0.0%hi, 0.0%si, 0.0%st Mem: 16468596k total, 2423908k used, 14044688k free, 200172k buffers

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• Glume super amuzante

  - by haioase

  Un ardelean si un lepros stau in inchisoare. La un moment dat, leprosului ii cade o ureche... o ia si o arunca pe geam. Dupa un timp, leprosului ii cade nasul... il ia si il arunca pe geam. Dupa un timp, leprosului ii cade un deget... il ia si il arunca pe geam... Ardeleanul nu mai suporta: -No, dupa cum bag de sama, 'mneata vrei sa evadezi!– Nu va suparati, ce se da aici? - Nu stiu, domnule, stai sa întrebam. Întrebarea circula pâna la primul din rând. - Nu stiu. Mi s-a facut rau si m-am sprijinit de zid. Când mi-am revenit, era deja coada în spatele meu. - De ce nu pleci atunci? - Pai, acum ca sunt primu...

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• Where do all these messages come from?

  - by stacker

  This configuration works fine, but inbound-channel-adapter which is supposed to poll every 15 secs is running continously. Does anyone have an idea what I'm doning wrong? <si:channel id="msgChannel" /> <si:inbound-channel-adapter ref="jdbcInputAdapter" method="fetchData" channel="msgChannel"> <si:poller> <si:interval-trigger interval="15000" /> </si:poller> </si:inbound-channel-adapter> <si:outbound-channel-adapter ref="shouter" method="shout" channel="msgChannel"/>

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• Oracle Enterprise Innovation Days

  - by Lara Ermacora

  Si è tenuto lo scorso 10 e 11 novembre l'appuntamento con l'innovazione marcato Oracle. L' Oracle Enterprise Innovation Days, alla sua seconda edizione, ha portato a Bologna tutte le aziende che pensano all'innovazione come leva principale per difendere e rafforzare la propria competitività. All'interno di un panorama, come quello odierno, complesso ed eterogeneo si è discusso a lungo di approcci strategici, soluzioni possibili e sono state portate d'esempio alcune esperienze significative. Fra gli ospiti dell'evento Rajan Krishnan, Vice President, Applications Product Development and Product Management for EMEA, ha presentato le strategie applicative di Oracle aprendo così la discussione sulla tematica principale della sessione plenaria: Oracle Fusion Applications. Il suo intervento è stato subito seguito da Enrico Pagliarini, giornalista del sole 24 ore che ha intervistato 3 diverse coppie Partner / Cliente per approfondire con loro i progetti altamente innovativi a cui le loro aziende hanno collaborato.  Si è parlato di Enel Servizi Srl che grazie ad Accenture ha portato la soluzione Syebel Energy CRM alla sua attuale versione 8.0 per una migliore gestione dei clienti all'interno del mercato libero caratterizzato dalla sua alta competitività; Prysmian che, a fronte dell'acquisizione della società olandese Draka, insieme a Reply, ha deciso di rimodellare il processo di Reporting Civilistico e Gestionale di gruppo, creando una nuova applicazione che soddisfi i requisiti della nuova organizzazione nascente; Kinexia e Waste Italia precedentemente parte del gruppo Unendo e ora divisesi l'una nel mercato dei rinnovabili l'altra in quello dello smaltimento rifiuti che con l'aiuto di Deloitte si sono dotate della soluzione full outsourcing JDE, a seguito di  una sw selection tra JDE, SAP e altre soluzioni italiane.Durante la cena altri due momenti hanno attirato l'attenzione dei partecipanti: la presentazione di Michele Stroligo, giovanissimo  Designer Team Member Oracle Racing e i Reference Customer Award ovvero le premiazioni dei clienti che si sono contraddistinti come migliori referenze nei diversi mercati con diversi prodotti. I premi sono stati assegnati a: FIAT, Enel, Boiron Laboratoires, Champion Europe, Mediaset, Coeclerici. Il pomeriggio ha interessato invece vari percorsi di approfondimento declinati sulle diverse figure professionali concludendosi con la presentazione del Tenente Colonello Marco Lant delle Frecce Tricolori, esempio di eccellezza italiana noto in tutto il mondo. La giornata si è conclusa con la cena di gala nel famoso palazzo Re Enzo che troneggia sulla piazza principale della città.  La mattinata del secondo giorno è stata interamente dedicata all'approfondimento degli argomenti di maggior interesse attraverso tavoli interattivi e workshop a cura dei partner Oracle. L'evento si è poi concluso con una serie di iniziative culturali dedicate ai congressisti. A breve sarà disponibile il sito dedicato all'evento con tutte le foto della giornata, i video degli interventi più salienti, potrete inoltre scaricare tutte le presentazioni fatte durante i lavori. Rimani aggiornato sull'Oracle Enterprise Innovation Days 2011 visitando il blog! Strategie Applicative di Oracle - Rajan Krishnan bologna nov 2011 View more presentations from Oracle Apps - Italia .

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• reading from File in assembly

  - by Natasha

  i am trying to read a username and a password from a file in x86 assembly for the perpose of authentication obviously the file consists of two lines , the user name and the password how can i read the two lines seperately and compare them? My attempt: proc read_file mov ah,3dh lea dx,file_name int 21h mov bx, ax xor si,si repeat: mov ah, 3fh lea dx, buffer mov cx, 100 int 21h mov si, ax mov buffer[si], '$' mov ah, 09h int 21h ;print on screen cmp si, 100 je repeat jmp stop;jump to end stop: RET read_file ENDP

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