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• Why is my RAID /dev/md1 showing up as /dev/md126? Is mdadm.conf being ignored?

  - by mmorris

  I created a RAID with: sudo mdadm --create --verbose /dev/md1 --level=mirror --raid-devices=2 /dev/sdb1 /dev/sdc1 sudo mdadm --create --verbose /dev/md2 --level=mirror --raid-devices=2 /dev/sdb2 /dev/sdc2 sudo mdadm --detail --scan returns: ARRAY /dev/md1 metadata=1.2 name=ion:1 UUID=aa1f85b0:a2391657:cfd38029:772c560e ARRAY /dev/md2 metadata=1.2 name=ion:2 UUID=528e5385:e61eaa4c:1db2dba7:44b556fb Which I appended it to /etc/mdadm/mdadm.conf, see below: # mdadm.conf # # Please refer to mdadm.conf(5) for information about this file. # # by default (built-in), scan all partitions (/proc/partitions) and all # containers for MD superblocks. alternatively, specify devices to scan, using # wildcards if desired. #DEVICE partitions containers # auto-create devices with Debian standard permissions CREATE owner=root group=disk mode=0660 auto=yes # automatically tag new arrays as belonging to the local system HOMEHOST <system> # instruct the monitoring daemon where to send mail alerts MAILADDR root # definitions of existing MD arrays # This file was auto-generated on Mon, 29 Oct 2012 16:06:12 -0500 # by mkconf $Id$ ARRAY /dev/md1 metadata=1.2 name=ion:1 UUID=aa1f85b0:a2391657:cfd38029:772c560e ARRAY /dev/md2 metadata=1.2 name=ion:2 UUID=528e5385:e61eaa4c:1db2dba7:44b556fb cat /proc/mdstat returns: Personalities : [raid1] [linear] [multipath] [raid0] [raid6] [raid5] [raid4] [raid10] md2 : active raid1 sdb2[0] sdc2[1] 208629632 blocks super 1.2 [2/2] [UU] md1 : active raid1 sdb1[0] sdc1[1] 767868736 blocks super 1.2 [2/2] [UU] unused devices: <none> ls -la /dev | grep md returns: brw-rw---- 1 root disk 9, 1 Oct 30 11:06 md1 brw-rw---- 1 root disk 9, 2 Oct 30 11:06 md2 So I think all is good and I reboot. After the reboot, /dev/md1 is now /dev/md126 and /dev/md2 is now /dev/md127????? sudo mdadm --detail --scan returns: ARRAY /dev/md/ion:1 metadata=1.2 name=ion:1 UUID=aa1f85b0:a2391657:cfd38029:772c560e ARRAY /dev/md/ion:2 metadata=1.2 name=ion:2 UUID=528e5385:e61eaa4c:1db2dba7:44b556fb cat /proc/mdstat returns: Personalities : [raid1] [linear] [multipath] [raid0] [raid6] [raid5] [raid4] [raid10] md126 : active raid1 sdc2[1] sdb2[0] 208629632 blocks super 1.2 [2/2] [UU] md127 : active (auto-read-only) raid1 sdb1[0] sdc1[1] 767868736 blocks super 1.2 [2/2] [UU] unused devices: <none> ls -la /dev | grep md returns: drwxr-xr-x 2 root root 80 Oct 30 11:18 md brw-rw---- 1 root disk 9, 126 Oct 30 11:18 md126 brw-rw---- 1 root disk 9, 127 Oct 30 11:18 md127 All is not lost, I: sudo mdadm --stop /dev/md126 sudo mdadm --stop /dev/md127 sudo mdadm --assemble --verbose /dev/md1 /dev/sdb1 /dev/sdc1 sudo mdadm --assemble --verbose /dev/md2 /dev/sdb2 /dev/sdc2 and verify everything: sudo mdadm --detail --scan returns: ARRAY /dev/md1 metadata=1.2 name=ion:1 UUID=aa1f85b0:a2391657:cfd38029:772c560e ARRAY /dev/md2 metadata=1.2 name=ion:2 UUID=528e5385:e61eaa4c:1db2dba7:44b556fb cat /proc/mdstat returns: Personalities : [raid1] [linear] [multipath] [raid0] [raid6] [raid5] [raid4] [raid10] md2 : active raid1 sdb2[0] sdc2[1] 208629632 blocks super 1.2 [2/2] [UU] md1 : active raid1 sdb1[0] sdc1[1] 767868736 blocks super 1.2 [2/2] [UU] unused devices: <none> ls -la /dev | grep md returns: brw-rw---- 1 root disk 9, 1 Oct 30 11:26 md1 brw-rw---- 1 root disk 9, 2 Oct 30 11:26 md2 So once again, I think all is good and I reboot. Again, after the reboot, /dev/md1 is /dev/md126 and /dev/md2 is /dev/md127????? sudo mdadm --detail --scan returns: ARRAY /dev/md/ion:1 metadata=1.2 name=ion:1 UUID=aa1f85b0:a2391657:cfd38029:772c560e ARRAY /dev/md/ion:2 metadata=1.2 name=ion:2 UUID=528e5385:e61eaa4c:1db2dba7:44b556fb cat /proc/mdstat returns: Personalities : [raid1] [linear] [multipath] [raid0] [raid6] [raid5] [raid4] [raid10] md126 : active raid1 sdc2[1] sdb2[0] 208629632 blocks super 1.2 [2/2] [UU] md127 : active (auto-read-only) raid1 sdb1[0] sdc1[1] 767868736 blocks super 1.2 [2/2] [UU] unused devices: <none> ls -la /dev | grep md returns: drwxr-xr-x 2 root root 80 Oct 30 11:42 md brw-rw---- 1 root disk 9, 126 Oct 30 11:42 md126 brw-rw---- 1 root disk 9, 127 Oct 30 11:42 md127 What am I missing here?

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• SPARC T5-4 LDoms for RAC and WebLogic Clusters

  - by Jeff Taylor-Oracle

  I wanted to use two Oracle SPARC T5-4 servers to simultaneously host both Oracle RAC and a WebLogic Server Cluster. I chose to use Oracle VM Server for SPARC to create a cluster like this: There are plenty of trade offs and decisions that need to be made, for example: Rather than configuring the system by hand, you might want to use an Oracle SuperCluster T5-8 My configuration is similar to jsavit's: Availability Best Practices - Example configuring a T5-8 but I chose to ignore some of the advice. Maybe I should have included an  alternate service domain, but I decided that I already had enough redundancy Both Oracle SPARC T5-4 servers were to be configured like this: Cntl 0.25  4  64GB                     App LDom                    2.75 CPU's                                        44 cores                                          704 GB              DB LDom      One CPU         16 cores         256 GB   The systems started with everything in the primary domain: # ldm list NAME             STATE      FLAGS   CONS    VCPU  MEMORY   UTIL  NORM  UPTIME primary          active     -n-c--  UART    512   1023G    0.0%  0.0%  11m # ldm list-spconfig factory-default [current] primary # ldm list -o core,memory,physio NAME              primary           CORE     CID    CPUSET     0      (0, 1, 2, 3, 4, 5, 6, 7)     1      (8, 9, 10, 11, 12, 13, 14, 15)     2      (16, 17, 18, 19, 20, 21, 22, 23) -- SNIP     62     (496, 497, 498, 499, 500, 501, 502, 503)     63     (504, 505, 506, 507, 508, 509, 510, 511) MEMORY     RA               PA               SIZE                 0x30000000       0x30000000       255G     0x80000000000    0x80000000000    256G     0x100000000000   0x100000000000   256G     0x180000000000   0x180000000000   256G # Give this memory block to the DB LDom IO     DEVICE                           PSEUDONYM        OPTIONS     pci@300                          pci_0                pci@340                          pci_1                pci@380                          pci_2                pci@3c0                          pci_3                pci@400                          pci_4                pci@440                          pci_5                pci@480                          pci_6                pci@4c0                          pci_7                pci@300/pci@1/pci@0/pci@6        /SYS/RCSA/PCIE1     pci@300/pci@1/pci@0/pci@c        /SYS/RCSA/PCIE2     pci@300/pci@1/pci@0/pci@4/pci@0/pci@c /SYS/MB/SASHBA0     pci@300/pci@1/pci@0/pci@4/pci@0/pci@8 /SYS/RIO/NET0        pci@340/pci@1/pci@0/pci@6        /SYS/RCSA/PCIE3     pci@340/pci@1/pci@0/pci@c        /SYS/RCSA/PCIE4     pci@380/pci@1/pci@0/pci@a        /SYS/RCSA/PCIE9     pci@380/pci@1/pci@0/pci@4        /SYS/RCSA/PCIE10     pci@3c0/pci@1/pci@0/pci@e        /SYS/RCSA/PCIE11     pci@3c0/pci@1/pci@0/pci@8        /SYS/RCSA/PCIE12     pci@400/pci@1/pci@0/pci@e        /SYS/RCSA/PCIE5     pci@400/pci@1/pci@0/pci@8        /SYS/RCSA/PCIE6     pci@440/pci@1/pci@0/pci@e        /SYS/RCSA/PCIE7     pci@440/pci@1/pci@0/pci@8        /SYS/RCSA/PCIE8     pci@480/pci@1/pci@0/pci@a        /SYS/RCSA/PCIE13     pci@480/pci@1/pci@0/pci@4        /SYS/RCSA/PCIE14     pci@4c0/pci@1/pci@0/pci@8        /SYS/RCSA/PCIE15     pci@4c0/pci@1/pci@0/pci@4        /SYS/RCSA/PCIE16     pci@4c0/pci@1/pci@0/pci@c/pci@0/pci@c /SYS/MB/SASHBA1     pci@4c0/pci@1/pci@0/pci@c/pci@0/pci@4 /SYS/RIO/NET2    Added an additional service processor configuration: # ldm add-spconfig split # ldm list-spconfig factory-default primary split [current] And removed many of the resources from the primary domain: # ldm start-reconf primary # ldm set-core 4 primary # ldm set-memory 32G primary # ldm rm-io pci@340 primary # ldm rm-io pci@380 primary # ldm rm-io pci@3c0 primary # ldm rm-io pci@400 primary # ldm rm-io pci@440 primary # ldm rm-io pci@480 primary # ldm rm-io pci@4c0 primary # init 6 Needed to add resources to the guest domains: # ldm add-domain db # ldm set-core cid=`seq -s"," 48 63` db # ldm add-memory mblock=0x180000000000:256G db # ldm add-io pci@480 db # ldm add-io pci@4c0 db # ldm add-domain app # ldm set-core 44 app # ldm set-memory 704G  app # ldm add-io pci@340 app # ldm add-io pci@380 app # ldm add-io pci@3c0 app # ldm add-io pci@400 app # ldm add-io pci@440 app Needed to set up services: # ldm add-vds primary-vds0 primary # ldm add-vcc port-range=5000-5100 primary-vcc0 primary Needed to add a virtual network port for the WebLogic application domain: # ipadm NAME              CLASS/TYPE STATE        UNDER      ADDR lo0               loopback   ok           --         --    lo0/v4         static     ok           --         ...    lo0/v6         static     ok           --         ... net0              ip         ok           --         ...    net0/v4        static     ok           --         xxx.xxx.xxx.xxx/24    net0/v6        addrconf   ok           --         ....    net0/v6        addrconf   ok           --         ... net8              ip         ok           --         --    net8/v4        static     ok           --         ... # dladm show-phys LINK              MEDIA                STATE      SPEED  DUPLEX    DEVICE net1              Ethernet             unknown    0      unknown   ixgbe1 net0              Ethernet             up         1000   full      ixgbe0 net8              Ethernet             up         10     full      usbecm2 # ldm add-vsw net-dev=net0 primary-vsw0 primary # ldm add-vnet vnet1 primary-vsw0 app Needed to add a virtual disk to the WebLogic application domain: # format Searching for disks...done AVAILABLE DISK SELECTIONS:        0. c0t5000CCA02505F874d0 <HITACHI-H106060SDSUN600G-A2B0-558.91GB>           /scsi_vhci/disk@g5000cca02505f874           /dev/chassis/SPARC_T5-4.AK00084038/SYS/SASBP0/HDD0/disk        1. c0t5000CCA02506C468d0 <HITACHI-H106060SDSUN600G-A2B0-558.91GB>           /scsi_vhci/disk@g5000cca02506c468           /dev/chassis/SPARC_T5-4.AK00084038/SYS/SASBP0/HDD1/disk        2. c0t5000CCA025067E5Cd0 <HITACHI-H106060SDSUN600G-A2B0-558.91GB>           /scsi_vhci/disk@g5000cca025067e5c           /dev/chassis/SPARC_T5-4.AK00084038/SYS/SASBP0/HDD2/disk        3. c0t5000CCA02506C258d0 <HITACHI-H106060SDSUN600G-A2B0-558.91GB>           /scsi_vhci/disk@g5000cca02506c258           /dev/chassis/SPARC_T5-4.AK00084038/SYS/SASBP0/HDD3/disk Specify disk (enter its number): ^C # ldm add-vdsdev /dev/dsk/c0t5000CCA02506C468d0s2 HDD1@primary-vds0 # ldm add-vdisk HDD1 HDD1@primary-vds0 app Add some additional spice to the pot: # ldm set-variable auto-boot\\?=false db # ldm set-variable auto-boot\\?=false app # ldm set-var boot-device=HDD1 app Bind the logical domains: # ldm bind db # ldm bind app At the end of the process, the system is set up like this: # ldm list -o core,memory,physio NAME             primary          CORE     CID    CPUSET     0      (0, 1, 2, 3, 4, 5, 6, 7)     1      (8, 9, 10, 11, 12, 13, 14, 15)     2      (16, 17, 18, 19, 20, 21, 22, 23)     3      (24, 25, 26, 27, 28, 29, 30, 31) MEMORY     RA               PA               SIZE                0x30000000       0x30000000       32G IO     DEVICE                           PSEUDONYM        OPTIONS     pci@300                          pci_0               pci@300/pci@1/pci@0/pci@6        /SYS/RCSA/PCIE1     pci@300/pci@1/pci@0/pci@c        /SYS/RCSA/PCIE2     pci@300/pci@1/pci@0/pci@4/pci@0/pci@c /SYS/MB/SASHBA0     pci@300/pci@1/pci@0/pci@4/pci@0/pci@8 /SYS/RIO/NET0   ------------------------------------------------------------------------------ NAME             app              CORE     CID    CPUSET     4      (32, 33, 34, 35, 36, 37, 38, 39)     5      (40, 41, 42, 43, 44, 45, 46, 47)     6      (48, 49, 50, 51, 52, 53, 54, 55)     7      (56, 57, 58, 59, 60, 61, 62, 63)     8      (64, 65, 66, 67, 68, 69, 70, 71)     9      (72, 73, 74, 75, 76, 77, 78, 79)     10     (80, 81, 82, 83, 84, 85, 86, 87)     11     (88, 89, 90, 91, 92, 93, 94, 95)     12     (96, 97, 98, 99, 100, 101, 102, 103)     13     (104, 105, 106, 107, 108, 109, 110, 111)     14     (112, 113, 114, 115, 116, 117, 118, 119)     15     (120, 121, 122, 123, 124, 125, 126, 127)     16     (128, 129, 130, 131, 132, 133, 134, 135)     17     (136, 137, 138, 139, 140, 141, 142, 143)     18     (144, 145, 146, 147, 148, 149, 150, 151)     19     (152, 153, 154, 155, 156, 157, 158, 159)     20     (160, 161, 162, 163, 164, 165, 166, 167)     21     (168, 169, 170, 171, 172, 173, 174, 175)     22     (176, 177, 178, 179, 180, 181, 182, 183)     23     (184, 185, 186, 187, 188, 189, 190, 191)     24     (192, 193, 194, 195, 196, 197, 198, 199)     25     (200, 201, 202, 203, 204, 205, 206, 207)     26     (208, 209, 210, 211, 212, 213, 214, 215)     27     (216, 217, 218, 219, 220, 221, 222, 223)     28     (224, 225, 226, 227, 228, 229, 230, 231)     29     (232, 233, 234, 235, 236, 237, 238, 239)     30     (240, 241, 242, 243, 244, 245, 246, 247)     31     (248, 249, 250, 251, 252, 253, 254, 255)     32     (256, 257, 258, 259, 260, 261, 262, 263)     33     (264, 265, 266, 267, 268, 269, 270, 271)     34     (272, 273, 274, 275, 276, 277, 278, 279)     35     (280, 281, 282, 283, 284, 285, 286, 287)     36     (288, 289, 290, 291, 292, 293, 294, 295)     37     (296, 297, 298, 299, 300, 301, 302, 303)     38     (304, 305, 306, 307, 308, 309, 310, 311)     39     (312, 313, 314, 315, 316, 317, 318, 319)     40     (320, 321, 322, 323, 324, 325, 326, 327)     41     (328, 329, 330, 331, 332, 333, 334, 335)     42     (336, 337, 338, 339, 340, 341, 342, 343)     43     (344, 345, 346, 347, 348, 349, 350, 351)     44     (352, 353, 354, 355, 356, 357, 358, 359)     45     (360, 361, 362, 363, 364, 365, 366, 367)     46     (368, 369, 370, 371, 372, 373, 374, 375)     47     (376, 377, 378, 379, 380, 381, 382, 383) MEMORY     RA               PA               SIZE                0x30000000       0x830000000      192G     0x4000000000     0x80000000000    256G     0x8080000000     0x100000000000   256G IO     DEVICE                           PSEUDONYM        OPTIONS     pci@340                          pci_1               pci@380                          pci_2               pci@3c0                          pci_3               pci@400                          pci_4               pci@440                          pci_5               pci@340/pci@1/pci@0/pci@6        /SYS/RCSA/PCIE3     pci@340/pci@1/pci@0/pci@c        /SYS/RCSA/PCIE4     pci@380/pci@1/pci@0/pci@a        /SYS/RCSA/PCIE9     pci@380/pci@1/pci@0/pci@4        /SYS/RCSA/PCIE10     pci@3c0/pci@1/pci@0/pci@e        /SYS/RCSA/PCIE11     pci@3c0/pci@1/pci@0/pci@8        /SYS/RCSA/PCIE12     pci@400/pci@1/pci@0/pci@e        /SYS/RCSA/PCIE5     pci@400/pci@1/pci@0/pci@8        /SYS/RCSA/PCIE6     pci@440/pci@1/pci@0/pci@e        /SYS/RCSA/PCIE7     pci@440/pci@1/pci@0/pci@8        /SYS/RCSA/PCIE8 ------------------------------------------------------------------------------ NAME             db               CORE     CID    CPUSET     48     (384, 385, 386, 387, 388, 389, 390, 391)     49     (392, 393, 394, 395, 396, 397, 398, 399)     50     (400, 401, 402, 403, 404, 405, 406, 407)     51     (408, 409, 410, 411, 412, 413, 414, 415)     52     (416, 417, 418, 419, 420, 421, 422, 423)     53     (424, 425, 426, 427, 428, 429, 430, 431)     54     (432, 433, 434, 435, 436, 437, 438, 439)     55     (440, 441, 442, 443, 444, 445, 446, 447)     56     (448, 449, 450, 451, 452, 453, 454, 455)     57     (456, 457, 458, 459, 460, 461, 462, 463)     58     (464, 465, 466, 467, 468, 469, 470, 471)     59     (472, 473, 474, 475, 476, 477, 478, 479)     60     (480, 481, 482, 483, 484, 485, 486, 487)     61     (488, 489, 490, 491, 492, 493, 494, 495)     62     (496, 497, 498, 499, 500, 501, 502, 503)     63     (504, 505, 506, 507, 508, 509, 510, 511) MEMORY     RA               PA               SIZE                0x80000000       0x180000000000   256G IO     DEVICE                           PSEUDONYM        OPTIONS     pci@480                          pci_6               pci@4c0                          pci_7               pci@480/pci@1/pci@0/pci@a        /SYS/RCSA/PCIE13     pci@480/pci@1/pci@0/pci@4        /SYS/RCSA/PCIE14     pci@4c0/pci@1/pci@0/pci@8        /SYS/RCSA/PCIE15     pci@4c0/pci@1/pci@0/pci@4        /SYS/RCSA/PCIE16     pci@4c0/pci@1/pci@0/pci@c/pci@0/pci@c /SYS/MB/SASHBA1     pci@4c0/pci@1/pci@0/pci@c/pci@0/pci@4 /SYS/RIO/NET2   Start the domains: # ldm start app LDom app started # ldm start db LDom db started Make sure to start the vntsd service that was created, above. # svcs -a | grep ldo disabled        8:38:38 svc:/ldoms/vntsd:default online          8:38:58 svc:/ldoms/agents:default online          8:39:25 svc:/ldoms/ldmd:default # svcadm enable vntsd Now use the MAC address to configure the Solaris 11 Automated Installation. Database Logical Domain # telnet localhost 5000 {0} ok devalias screen                   /pci@4c0/pci@1/pci@0/pci@c/pci@0/pci@7/display@0 disk7                    /pci@4c0/pci@1/pci@0/pci@c/pci@0/pci@c/scsi@0/disk@p3 disk6                    /pci@4c0/pci@1/pci@0/pci@c/pci@0/pci@c/scsi@0/disk@p2 disk5                    /pci@4c0/pci@1/pci@0/pci@c/pci@0/pci@c/scsi@0/disk@p1 disk4                    /pci@4c0/pci@1/pci@0/pci@c/pci@0/pci@c/scsi@0/disk@p0 scsi1                    /pci@4c0/pci@1/pci@0/pci@c/pci@0/pci@c/scsi@0 net3                     /pci@4c0/pci@1/pci@0/pci@c/pci@0/pci@4/network@0,1 net2                     /pci@4c0/pci@1/pci@0/pci@c/pci@0/pci@4/network@0 virtual-console          /virtual-devices/console@1 name                     aliases {0} ok boot net2 Boot device: /pci@4c0/pci@1/pci@0/pci@c/pci@0/pci@4/network@0  File and args: 1000 Mbps full duplex Link up Requesting Internet Address for xx:xx:xx:xx:xx:xx Requesting Internet Address for xx:xx:xx:xx:xx:xx WLS Logical Domain # telnet localhost 5001 {0} ok devalias hdd1                     /virtual-devices@100/channel-devices@200/disk@0 vnet1                    /virtual-devices@100/channel-devices@200/network@0 net                      /virtual-devices@100/channel-devices@200/network@0 disk                     /virtual-devices@100/channel-devices@200/disk@0 virtual-console          /virtual-devices/console@1 name                     aliases {0} ok boot net Boot device: /virtual-devices@100/channel-devices@200/network@0  File and args: Requesting Internet Address for xx:xx:xx:xx:xx:xx Requesting Internet Address for xx:xx:xx:xx:xx:xx Repeat the process for the second SPARC T5-4, install Solaris, RAC and WebLogic Cluster, and you are ready to go. Maybe buying a SuperCluster would have been easier.

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• How can I fix my corrupted RAID1 ext4 partition on a Synology DS212 NAS?

  - by Neil

  I have two identical 3 TB disks that were in a RAID1 array, where one disk crashed. I replaced the failed disk, but not after the RAID partitions got messed up. I need to figure out how to restore the RAID array and get at my ext4 partition. Here are the properties of the surviving disk: # fdisk -l /dev/sda fdisk: device has more than 2^32 sectors, can't use all of them Disk /dev/sda: 2199.0 GB, 2199023255040 bytes 255 heads, 63 sectors/track, 267349 cylinders Units = cylinders of 16065 * 512 = 8225280 bytes Device Boot Start End Blocks Id System /dev/sda1 1 267350 2147483647+ ee EFI GPT # parted /dev/sda print Model: ATA ST3000DM001-9YN1 (scsi) Disk /dev/sda: 3001GB Sector size (logical/physical): 512B/512B Partition Table: gpt Disk Flags: Number Start End Size File system Name Flags 1 131kB 2550MB 2550MB ext4 raid 2 2550MB 4698MB 2147MB linux-swap(v1) raid 5 4840MB 3001GB 2996GB raid I replaced the failed drive, and cloned the surviving drive to it so I have something to work with. I cloned the drives with dd if=/dev/sdb of=/dev/sda conv=noerror bs=64M, and now /dev/sda and /dev/sdb are identical. Here is the RAID information: # cat /proc/mdstat Personalities : [linear] [raid0] [raid1] [raid10] [raid6] [raid5] [raid4] md1 : active raid1 sdb2[1] 2097088 blocks [2/1] [_U] md0 : active raid1 sdb1[1] 2490176 blocks [2/1] [_U] unused devices: <none> It seems that md2 is missing. Here is what testdisk 6.14-WIP finds: Disk /dev/sda - 3000 GB / 2794 GiB - CHS 364801 255 63 Current partition structure: Partition Start End Size in sectors 1 P Linux Raid 256 4980735 4980480 [md0] 2 P Linux Raid 4980736 9175039 4194304 [md1] Invalid RAID superblock 5 P Linux Raid 9453280 5860519007 5851065728 5 P Linux Raid 9453280 5860519007 5851065728 # After a quick search Disk /dev/sda - 3000 GB / 2794 GiB - CHS 364801 255 63 Partition Start End Size in sectors D MS Data 256 4980607 4980352 [1.41.12-2197] D Linux Raid 256 4980735 4980480 [md0] D Linux Swap 4980736 9174895 4194160 D Linux Raid 4980736 9175039 4194304 [md1] >P MS Data 9481056 5858437983 5848956928 [1.41.12-2228] And listing the files on the last partition in the list shows all of my files intact. What should I do?

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• Fedora-13 not detecting USB HDD enclosure (with HDD)

  - by Ramy

  I recently purchased this enclosure: http://www.amazon.com/Inland-2-5-Inc.../dp/B003SZ2Y12 and this HDD: http://www.amazon.com/Seagate-Barrac...3811667&sr=8-1 Now, I let my brother in law use the enclosure with his 160GB disk to back some stuff up. He then gave me that disk in my enclosure and I backed up my computer and my fiances computer. So...obviously, i had no problem mounting that disk. I plan on keeping this disk as my "natural disaster backup" (in case my apartment building burns down, i still have that disk with my stuff backed up). I want to use the 1.5T disk as my regular/more frequent backup device, but it doesn't seem to be mounting to my F-13 machine. I searched through this forum and found someone advising to run the following: # mount -t vfat /dev/sda1 /mnt this is the output i get when I run that: mount: /dev/sda1 already mounted or /mnt busy mount: according to mtab, /dev/sda1 is mounted on /boot Thing is, shouldn't this disk automatically mount just like the LAST disk in the same enclosure with the same USB cable and power supply? Any help would be greatly appreciated. THANKS!

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• SQL 2K5 - Multiple databases vs. Multiple files

  - by Bob Palmer

  Hey all, quick question. Our current legacy system was built using multiple distinct databases (about ten of them). These are all part of the same discreet system, and a large number of SPs and functionalty span multiple databases. There are also key relationships that span (for example, a header table may be in database A with history, etc. in database B). When deploying multiple copies of our app to the same server therefore, we have to use multiple instances (because the database names are coded into so many sprocs). We're evaluating the idea of taking these ten databases (about 30gb total with individual sizes ranging from 100mb to 10gb) and merging them into a single database. Currently, we have our databases spread accross multiple spindles for better IO. The question I have is whether or not there is any performance loss or benefit of having 10 different databases vs. 10 different database files? i.e. rather than having three databases (A, B, and C) Disk D: A.mdf (1gb) Disk E: B.mdf (4gb) Disk F: C.mdf (10gb) Disk G: A_Log.ldf, B_Log.ldf, C_Log.ldf have one database (X) Disk D: X1.mdf (5gb) Disk E: X2.mdf (5gb) Disk F: X3.mdf (5gb) Disk G: X1_log.ldf,X2_log.ldf,X3_log.ldf Thanks! -Bob

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• Windows Server 2008 - RAID 5 Fails on Reboot

  - by Adam

  Hey, I've got an install of Windows Server 2008 Enterprise. It's running software RAID-5 with five disks. The disks were originally formatted under Windows Server 2003, but came up fine once I installed Windows Server 2008. The issue I'm having is that every time I reboot the server, the RAID comes up with a "Failed Redundancy" - the data stays available. I have 4 disks on a PCI SATA controller, and one of the disks connected to the motherboard's on-board SATA ports. (The other on-board port has the system disk connected.) I was having Disk #4 fail consistently, so I tried swapping the cables on the controller end. I swapped the on-board RAID disk with one on the PCI controller. Same issue now, expect with disk #1. Once the system's up, I can reactivate the RAID, it will resync for a while, then go to "Healthy", and will stay that way for an indefinite amount of time - until I reboot. As soon as I reboot, the disk drops again. I've ruled out disk + cable with the recabling. I don't believe it would be the controller as it seems to work fine most of the time - only failing on reboot, and the other port on the same controller connects the system disk - which is clearly working. I did look in the event log, but didn't see anything particularly relevant (although I didn't know what I was looking for - just looked for anything with a "Warning" or "Error" symbol that looked disk-related :)). I'm not particularly familiar with RAID on Windows, does anyone have any idea why this might be doing this? Any idea how to fix it? Any suggestions appreciated! -- Adam

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• How do I copy files between harddrives on Ubuntu CLI?

  - by ed209

  I have a dedicated server with a 120gb main ssd. The server happens to come with a couple of 3000GB hard drives. I'd like to use them to back up my main drive. Preferably, I'd like one as an exact copy of the main SSD and the other with incremental backups of the mysql database and a user uploads file. These are the drives I have Disk /dev/sda: 120.0 GB, 120034123776 bytes 255 heads, 63 sectors/track, 14593 cylinders, total 234441648 sectors Units = sectors of 1 * 512 = 512 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x000f2e18 Device Boot Start End Blocks Id System /dev/sda1 2048 4196352 2097152+ 83 Linux /dev/sda2 4198400 5246976 524288+ 83 Linux /dev/sda3 5249024 234441647 114596312 83 Linux Disk /dev/sdb: 3000.6 GB, 3000592982016 bytes 255 heads, 63 sectors/track, 364801 cylinders, total 5860533168 sectors Units = sectors of 1 * 512 = 512 bytes Sector size (logical/physical): 512 bytes / 4096 bytes I/O size (minimum/optimal): 4096 bytes / 4096 bytes Disk identifier: 0x00000000 Disk /dev/sdb doesn't contain a valid partition table Disk /dev/sdc: 3000.6 GB, 3000592982016 bytes 255 heads, 63 sectors/track, 364801 cylinders, total 5860533168 sectors Units = sectors of 1 * 512 = 512 bytes Sector size (logical/physical): 512 bytes / 4096 bytes I/O size (minimum/optimal): 4096 bytes / 4096 bytes Disk identifier: 0x00000000 Disk /dev/sdc doesn't contain a valid partition table The first problem I have, is that I have no idea how to copy from one drive to another. Kind of embarrassing I know, but I don't know where to start. I'm thinking of this in terms of Mac OS cli where I'm able to copy between /Volumes - is there an equivalent? (there is nothing under /mnt or /media)

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• How to fix missing RAID1 drive

  - by Sodved

  I had to do some fiddling about with my cables inside thebox and now I am getting a "Critical Error" about the RAID disks during startup. I have a gigabyte GA-MA770T-UD3 motherboard. Aparently the RAID controller is an AMD SB710 chip. I'm pretty sure I know what happenned. The first time I rebooted I had forgotten the power cable on one of the disks in the RAID1 (mirror) and let it boot up. So I shut down and put the power back in. So now when it boots up I go into the RAID admin interface (between the BIOS screen and the OS loading): it shows the RAID1 as in error the logical device has one disk and says the other is disconnected or missing the other physical disk shows up as a single disk If I boot to the OS (Windows 7 32 bit) the data all seems to be there. If I go into computer management it says my partition is on a disk and working OK. But the other disk is offline because: "The disk is offline because it has a signature collision with another disk that is online" So I am guessing because I STUPIDLY booted up with only one of the disks powered on, the other disk fell out of synch with the mirror and so now cannot rejoin the mirror. How do I fix this? I want to get the RAID1 mirror working again. There does not appear to be any "Repair" option in the basic RAID admin tool which I get into during startup before the OS boots. I have not made any explicit changes to the online one (but I guess the OS has probably written some admin data).

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• How to handle BL cache for multiple web applications?

  - by Eran Betzalel

  I recently received a project that contains multiple web applications with no MVC structure. For starters I've created a library (DLL) that will contain the main Business Logic. The problem is with Caching - If I use the current web context cache object than I might end up with duplicate caching (as the web context will be different for every application). I'm currently thinking about implementing a simple caching mechanism with a singleton pattern that will allow the different web sites (aka different application domains) to share their "caching wisdom". I'd like to know what is the best way to solve this problem.

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• Custom ASP.NET MVC cache controllers in a shared hosting environment?

  - by Daniel Crenna

  I'm using custom controllers that cache static resources (CSS, JS, etc.) and images. I'm currently working with a hosting provider that has set me up under a full trust profile. Despite being in full trust, my controllers fail because the caching strategy relies on the File class to directly open a resource file prior to treatment and storage in memory. Is this something that would likely occur in all full trust shared hosting environments or is this specific to my host? The static files live within my application's structure and not in an arbitrary server path. It seems to me that custom caching would require code to access the file directly, and am hoping someone else has dealt with this issue.

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• Is it possible to cache JSP bytecode to avoid recompiles w/ Tomcat?

  - by Computer Guru

  Hi, Is there any way of caching the bytecode for JSP webapps/ In particular, using Tomcat as the Java servlet? I'm getting really fed up of Tomcat taking up all the CPU for 10 minutes while it compiles 4 different webapps every time I restart it.... I'm already using Jikes to "speed up" the compiles, but it's really killing me. The code does not change unless the webapp is upgraded (very rarely), and I cannot believe that there is no way to cache the compiled java bytecode instead of recompiling it each and every time. I'd appreciate any advice on the matter!

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• ASP.Net: Is it possible to cache the js-proxies generated by scriptmanager?

  - by AndreasKnudsen

  We have the following code: <asp:ScriptManager runat="server"> ... <Services> <asp:ServiceReference Path="~/JSONServices/ProfileService.svc" /> </Services> ... This results in a Javascript proxy found in /JSONServices/ProfileService.svc/js. This Javascript has content expiry set to the same time it was called (so it is never cached on the client). Is it possible to have the clients cache these proxies for some time?

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• Cache an FTP connection via session variables for use via AJAX?

  - by Chad Johnson

  I'm working on a Ruby web Application that uses the Net::FTP library. One part of it allows users to interact with an FTP site via AJAX. When the user does something, and AJAX call is made, and then Ruby reconnects to the FTP server, performs an action, and outputs information. Every time the AJAX call is made, Ruby has to reconnect to the FTP server, and that's slow. Is there a way I could cache this FTP connection? I've tried caching in the session hash, but "We're sorry, but something went wrong" is displayed, and a TCP dump is outputted in my logs whenever I attempt to store it in the session hash. I haven't tried memcache yet. Any suggestions?

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• Is it possible to evaluate a JSP only once per session, and cache it after that?

  - by Bears will eat you

  My site has a nav menu that is dynamically built as a separate JSP, and included in most pages via <jsp:include />. The contents and styling of the menu are determined by which pages the user does and doesn't have access to. The set of accessible pages is retrieved from the database when a user logs in, and not during the course of a session. So, there's really no need to re-evaluate the nav menu code every time the user requests a page. Is there an easy way to generate the markup from the JSP only once per session, and cache/reuse it during the session?

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• How John Got 15x Improvement Without Really Trying

  - by rchrd

  The following article was published on a Sun Microsystems website a number of years ago by John Feo. It is still useful and worth preserving. So I'm republishing it here.  How I Got 15x Improvement Without Really Trying John Feo, Sun Microsystems Taking ten "personal" program codes used in scientific and engineering research, the author was able to get from 2 to 15 times performance improvement easily by applying some simple general optimization techniques. Introduction Scientific research based on computer simulation depends on the simulation for advancement. The research can advance only as fast as the computational codes can execute. The codes' efficiency determines both the rate and quality of results. In the same amount of time, a faster program can generate more results and can carry out a more detailed simulation of physical phenomena than a slower program. Highly optimized programs help science advance quickly and insure that monies supporting scientific research are used as effectively as possible. Scientific computer codes divide into three broad categories: ISV, community, and personal. ISV codes are large, mature production codes developed and sold commercially. The codes improve slowly over time both in methods and capabilities, and they are well tuned for most vendor platforms. Since the codes are mature and complex, there are few opportunities to improve their performance solely through code optimization. Improvements of 10% to 15% are typical. Examples of ISV codes are DYNA3D, Gaussian, and Nastran. Community codes are non-commercial production codes used by a particular research field. Generally, they are developed and distributed by a single academic or research institution with assistance from the community. Most users just run the codes, but some develop new methods and extensions that feed back into the general release. The codes are available on most vendor platforms. Since these codes are younger than ISV codes, there are more opportunities to optimize the source code. Improvements of 50% are not unusual. Examples of community codes are AMBER, CHARM, BLAST, and FASTA. Personal codes are those written by single users or small research groups for their own use. These codes are not distributed, but may be passed from professor-to-student or student-to-student over several years. They form the primordial ocean of applications from which community and ISV codes emerge. Government research grants pay for the development of most personal codes. This paper reports on the nature and performance of this class of codes. Over the last year, I have looked at over two dozen personal codes from more than a dozen research institutions. The codes cover a variety of scientific fields, including astronomy, atmospheric sciences, bioinformatics, biology, chemistry, geology, and physics. The sources range from a few hundred lines to more than ten thousand lines, and are written in Fortran, Fortran 90, C, and C++. For the most part, the codes are modular, documented, and written in a clear, straightforward manner. They do not use complex language features, advanced data structures, programming tricks, or libraries. I had little trouble understanding what the codes did or how data structures were used. Most came with a makefile. Surprisingly, only one of the applications is parallel. All developers have access to parallel machines, so availability is not an issue. Several tried to parallelize their applications, but stopped after encountering difficulties. Lack of education and a perception that parallelism is difficult prevented most from trying. I parallelized several of the codes using OpenMP, and did not judge any of the codes as difficult to parallelize. Even more surprising than the lack of parallelism is the inefficiency of the codes. I was able to get large improvements in performance in a matter of a few days applying simple optimization techniques. Table 1 lists ten representative codes [names and affiliation are omitted to preserve anonymity]. Improvements on one processor range from 2x to 15.5x with a simple average of 4.75x. I did not use sophisticated performance tools or drill deep into the program's execution character as one would do when tuning ISV or community codes. Using only a profiler and source line timers, I identified inefficient sections of code and improved their performance by inspection. The changes were at a high level. I am sure there is another factor of 2 or 3 in each code, and more if the codes are parallelized. The study’s results show that personal scientific codes are running many times slower than they should and that the problem is pervasive. Computational scientists are not sloppy programmers; however, few are trained in the art of computer programming or code optimization. I found that most have a working knowledge of some programming language and standard software engineering practices; but they do not know, or think about, how to make their programs run faster. They simply do not know the standard techniques used to make codes run faster. In fact, they do not even perceive that such techniques exist. The case studies described in this paper show that applying simple, well known techniques can significantly increase the performance of personal codes. It is important that the scientific community and the Government agencies that support scientific research find ways to better educate academic scientific programmers. The inefficiency of their codes is so bad that it is retarding both the quality and progress of scientific research. # cacheperformance redundantoperations loopstructures performanceimprovement 1 x x 15.5 2 x 2.8 3 x x 2.5 4 x 2.1 5 x x 2.0 6 x 5.0 7 x 5.8 8 x 6.3 9 2.2 10 x x 3.3 Table 1 — Area of improvement and performance gains of 10 codes The remainder of the paper is organized as follows: sections 2, 3, and 4 discuss the three most common sources of inefficiencies in the codes studied. These are cache performance, redundant operations, and loop structures. Each section includes several examples. The last section summaries the work and suggests a possible solution to the issues raised. Optimizing cache performance Commodity microprocessor systems use caches to increase memory bandwidth and reduce memory latencies. Typical latencies from processor to L1, L2, local, and remote memory are 3, 10, 50, and 200 cycles, respectively. Moreover, bandwidth falls off dramatically as memory distances increase. Programs that do not use cache effectively run many times slower than programs that do. When optimizing for cache, the biggest performance gains are achieved by accessing data in cache order and reusing data to amortize the overhead of cache misses. Secondary considerations are prefetching, associativity, and replacement; however, the understanding and analysis required to optimize for the latter are probably beyond the capabilities of the non-expert. Much can be gained simply by accessing data in the correct order and maximizing data reuse. 6 out of the 10 codes studied here benefited from such high level optimizations. Array Accesses The most important cache optimization is the most basic: accessing Fortran array elements in column order and C array elements in row order. Four of the ten codes—1, 2, 4, and 10—got it wrong. Compilers will restructure nested loops to optimize cache performance, but may not do so if the loop structure is too complex, or the loop body includes conditionals, complex addressing, or function calls. In code 1, the compiler failed to invert a key loop because of complex addressing do I = 0, 1010, delta_x IM = I - delta_x IP = I + delta_x do J = 5, 995, delta_x JM = J - delta_x JP = J + delta_x T1 = CA1(IP, J) + CA1(I, JP) T2 = CA1(IM, J) + CA1(I, JM) S1 = T1 + T2 - 4 * CA1(I, J) CA(I, J) = CA1(I, J) + D * S1 end do end do In code 2, the culprit is conditionals do I = 1, N do J = 1, N If (IFLAG(I,J) .EQ. 0) then T1 = Value(I, J-1) T2 = Value(I-1, J) T3 = Value(I, J) T4 = Value(I+1, J) T5 = Value(I, J+1) Value(I,J) = 0.25 * (T1 + T2 + T5 + T4) Delta = ABS(T3 - Value(I,J)) If (Delta .GT. MaxDelta) MaxDelta = Delta endif enddo enddo I fixed both programs by inverting the loops by hand. Code 10 has three-dimensional arrays and triply nested loops. The structure of the most computationally intensive loops is too complex to invert automatically or by hand. The only practical solution is to transpose the arrays so that the dimension accessed by the innermost loop is in cache order. The arrays can be transposed at construction or prior to entering a computationally intensive section of code. The former requires all array references to be modified, while the latter is cost effective only if the cost of the transpose is amortized over many accesses. I used the second approach to optimize code 10. Code 5 has four-dimensional arrays and loops are nested four deep. For all of the reasons cited above the compiler is not able to restructure three key loops. Assume C arrays and let the four dimensions of the arrays be i, j, k, and l. In the original code, the index structure of the three loops is L1: for i L2: for i L3: for i for l for l for j for k for j for k for j for k for l So only L3 accesses array elements in cache order. L1 is a very complex loop—much too complex to invert. I brought the loop into cache alignment by transposing the second and fourth dimensions of the arrays. Since the code uses a macro to compute all array indexes, I effected the transpose at construction and changed the macro appropriately. The dimensions of the new arrays are now: i, l, k, and j. L3 is a simple loop and easily inverted. L2 has a loop-carried scalar dependence in k. By promoting the scalar name that carries the dependence to an array, I was able to invert the third and fourth subloops aligning the loop with cache. Code 5 is by far the most difficult of the four codes to optimize for array accesses; but the knowledge required to fix the problems is no more than that required for the other codes. I would judge this code at the limits of, but not beyond, the capabilities of appropriately trained computational scientists. Array Strides When a cache miss occurs, a line (64 bytes) rather than just one word is loaded into the cache. If data is accessed stride 1, than the cost of the miss is amortized over 8 words. Any stride other than one reduces the cost savings. Two of the ten codes studied suffered from non-unit strides. The codes represent two important classes of "strided" codes. Code 1 employs a multi-grid algorithm to reduce time to convergence. The grids are every tenth, fifth, second, and unit element. Since time to convergence is inversely proportional to the distance between elements, coarse grids converge quickly providing good starting values for finer grids. The better starting values further reduce the time to convergence. The downside is that grids of every nth element, n > 1, introduce non-unit strides into the computation. In the original code, much of the savings of the multi-grid algorithm were lost due to this problem. I eliminated the problem by compressing (copying) coarse grids into continuous memory, and rewriting the computation as a function of the compressed grid. On convergence, I copied the final values of the compressed grid back to the original grid. The savings gained from unit stride access of the compressed grid more than paid for the cost of copying. Using compressed grids, the loop from code 1 included in the previous section becomes do j = 1, GZ do i = 1, GZ T1 = CA(i+0, j-1) + CA(i-1, j+0) T4 = CA1(i+1, j+0) + CA1(i+0, j+1) S1 = T1 + T4 - 4 * CA1(i+0, j+0) CA(i+0, j+0) = CA1(i+0, j+0) + DD * S1 enddo enddo where CA and CA1 are compressed arrays of size GZ. Code 7 traverses a list of objects selecting objects for later processing. The labels of the selected objects are stored in an array. The selection step has unit stride, but the processing steps have irregular stride. A fix is to save the parameters of the selected objects in temporary arrays as they are selected, and pass the temporary arrays to the processing functions. The fix is practical if the same parameters are used in selection as in processing, or if processing comprises a series of distinct steps which use overlapping subsets of the parameters. Both conditions are true for code 7, so I achieved significant improvement by copying parameters to temporary arrays during selection. Data reuse In the previous sections, we optimized for spatial locality. It is also important to optimize for temporal locality. Once read, a datum should be used as much as possible before it is forced from cache. Loop fusion and loop unrolling are two techniques that increase temporal locality. Unfortunately, both techniques increase register pressure—as loop bodies become larger, the number of registers required to hold temporary values grows. Once register spilling occurs, any gains evaporate quickly. For multiprocessors with small register sets or small caches, the sweet spot can be very small. In the ten codes presented here, I found no opportunities for loop fusion and only two opportunities for loop unrolling (codes 1 and 3). In code 1, unrolling the outer and inner loop one iteration increases the number of result values computed by the loop body from 1 to 4, do J = 1, GZ-2, 2 do I = 1, GZ-2, 2 T1 = CA1(i+0, j-1) + CA1(i-1, j+0) T2 = CA1(i+1, j-1) + CA1(i+0, j+0) T3 = CA1(i+0, j+0) + CA1(i-1, j+1) T4 = CA1(i+1, j+0) + CA1(i+0, j+1) T5 = CA1(i+2, j+0) + CA1(i+1, j+1) T6 = CA1(i+1, j+1) + CA1(i+0, j+2) T7 = CA1(i+2, j+1) + CA1(i+1, j+2) S1 = T1 + T4 - 4 * CA1(i+0, j+0) S2 = T2 + T5 - 4 * CA1(i+1, j+0) S3 = T3 + T6 - 4 * CA1(i+0, j+1) S4 = T4 + T7 - 4 * CA1(i+1, j+1) CA(i+0, j+0) = CA1(i+0, j+0) + DD * S1 CA(i+1, j+0) = CA1(i+1, j+0) + DD * S2 CA(i+0, j+1) = CA1(i+0, j+1) + DD * S3 CA(i+1, j+1) = CA1(i+1, j+1) + DD * S4 enddo enddo The loop body executes 12 reads, whereas as the rolled loop shown in the previous section executes 20 reads to compute the same four values. In code 3, two loops are unrolled 8 times and one loop is unrolled 4 times. Here is the before for (k = 0; k < NK[u]; k++) { sum = 0.0; for (y = 0; y < NY; y++) { sum += W[y][u][k] * delta[y]; } backprop[i++]=sum; } and after code for (k = 0; k < KK - 8; k+=8) { sum0 = 0.0; sum1 = 0.0; sum2 = 0.0; sum3 = 0.0; sum4 = 0.0; sum5 = 0.0; sum6 = 0.0; sum7 = 0.0; for (y = 0; y < NY; y++) { sum0 += W[y][0][k+0] * delta[y]; sum1 += W[y][0][k+1] * delta[y]; sum2 += W[y][0][k+2] * delta[y]; sum3 += W[y][0][k+3] * delta[y]; sum4 += W[y][0][k+4] * delta[y]; sum5 += W[y][0][k+5] * delta[y]; sum6 += W[y][0][k+6] * delta[y]; sum7 += W[y][0][k+7] * delta[y]; } backprop[k+0] = sum0; backprop[k+1] = sum1; backprop[k+2] = sum2; backprop[k+3] = sum3; backprop[k+4] = sum4; backprop[k+5] = sum5; backprop[k+6] = sum6; backprop[k+7] = sum7; } for one of the loops unrolled 8 times. Optimizing for temporal locality is the most difficult optimization considered in this paper. The concepts are not difficult, but the sweet spot is small. Identifying where the program can benefit from loop unrolling or loop fusion is not trivial. Moreover, it takes some effort to get it right. Still, educating scientific programmers about temporal locality and teaching them how to optimize for it will pay dividends. Reducing instruction count Execution time is a function of instruction count. Reduce the count and you usually reduce the time. The best solution is to use a more efficient algorithm; that is, an algorithm whose order of complexity is smaller, that converges quicker, or is more accurate. Optimizing source code without changing the algorithm yields smaller, but still significant, gains. This paper considers only the latter because the intent is to study how much better codes can run if written by programmers schooled in basic code optimization techniques. The ten codes studied benefited from three types of "instruction reducing" optimizations. The two most prevalent were hoisting invariant memory and data operations out of inner loops. The third was eliminating unnecessary data copying. The nature of these inefficiencies is language dependent. Memory operations The semantics of C make it difficult for the compiler to determine all the invariant memory operations in a loop. The problem is particularly acute for loops in functions since the compiler may not know the values of the function's parameters at every call site when compiling the function. Most compilers support pragmas to help resolve ambiguities; however, these pragmas are not comprehensive and there is no standard syntax. To guarantee that invariant memory operations are not executed repetitively, the user has little choice but to hoist the operations by hand. The problem is not as severe in Fortran programs because in the absence of equivalence statements, it is a violation of the language's semantics for two names to share memory. Codes 3 and 5 are C programs. In both cases, the compiler did not hoist all invariant memory operations from inner loops. Consider the following loop from code 3 for (y = 0; y < NY; y++) { i = 0; for (u = 0; u < NU; u++) { for (k = 0; k < NK[u]; k++) { dW[y][u][k] += delta[y] * I1[i++]; } } } Since dW[y][u] can point to the same memory space as delta for one or more values of y and u, assignment to dW[y][u][k] may change the value of delta[y]. In reality, dW and delta do not overlap in memory, so I rewrote the loop as for (y = 0; y < NY; y++) { i = 0; Dy = delta[y]; for (u = 0; u < NU; u++) { for (k = 0; k < NK[u]; k++) { dW[y][u][k] += Dy * I1[i++]; } } } Failure to hoist invariant memory operations may be due to complex address calculations. If the compiler can not determine that the address calculation is invariant, then it can hoist neither the calculation nor the associated memory operations. As noted above, code 5 uses a macro to address four-dimensional arrays #define MAT4D(a,q,i,j,k) (double *)((a)->data + (q)*(a)->strides[0] + (i)*(a)->strides[3] + (j)*(a)->strides[2] + (k)*(a)->strides[1]) The macro is too complex for the compiler to understand and so, it does not identify any subexpressions as loop invariant. The simplest way to eliminate the address calculation from the innermost loop (over i) is to define a0 = MAT4D(a,q,0,j,k) before the loop and then replace all instances of *MAT4D(a,q,i,j,k) in the loop with a0[i] A similar problem appears in code 6, a Fortran program. The key loop in this program is do n1 = 1, nh nx1 = (n1 - 1) / nz + 1 nz1 = n1 - nz * (nx1 - 1) do n2 = 1, nh nx2 = (n2 - 1) / nz + 1 nz2 = n2 - nz * (nx2 - 1) ndx = nx2 - nx1 ndy = nz2 - nz1 gxx = grn(1,ndx,ndy) gyy = grn(2,ndx,ndy) gxy = grn(3,ndx,ndy) balance(n1,1) = balance(n1,1) + (force(n2,1) * gxx + force(n2,2) * gxy) * h1 balance(n1,2) = balance(n1,2) + (force(n2,1) * gxy + force(n2,2) * gyy)*h1 end do end do The programmer has written this loop well—there are no loop invariant operations with respect to n1 and n2. However, the loop resides within an iterative loop over time and the index calculations are independent with respect to time. Trading space for time, I precomputed the index values prior to the entering the time loop and stored the values in two arrays. I then replaced the index calculations with reads of the arrays. Data operations Ways to reduce data operations can appear in many forms. Implementing a more efficient algorithm produces the biggest gains. The closest I came to an algorithm change was in code 4. This code computes the inner product of K-vectors A(i) and B(j), 0 = i < N, 0 = j < M, for most values of i and j. Since the program computes most of the NM possible inner products, it is more efficient to compute all the inner products in one triply-nested loop rather than one at a time when needed. The savings accrue from reading A(i) once for all B(j) vectors and from loop unrolling. for (i = 0; i < N; i+=8) { for (j = 0; j < M; j++) { sum0 = 0.0; sum1 = 0.0; sum2 = 0.0; sum3 = 0.0; sum4 = 0.0; sum5 = 0.0; sum6 = 0.0; sum7 = 0.0; for (k = 0; k < K; k++) { sum0 += A[i+0][k] * B[j][k]; sum1 += A[i+1][k] * B[j][k]; sum2 += A[i+2][k] * B[j][k]; sum3 += A[i+3][k] * B[j][k]; sum4 += A[i+4][k] * B[j][k]; sum5 += A[i+5][k] * B[j][k]; sum6 += A[i+6][k] * B[j][k]; sum7 += A[i+7][k] * B[j][k]; } C[i+0][j] = sum0; C[i+1][j] = sum1; C[i+2][j] = sum2; C[i+3][j] = sum3; C[i+4][j] = sum4; C[i+5][j] = sum5; C[i+6][j] = sum6; C[i+7][j] = sum7; }} This change requires knowledge of a typical run; i.e., that most inner products are computed. The reasons for the change, however, derive from basic optimization concepts. It is the type of change easily made at development time by a knowledgeable programmer. In code 5, we have the data version of the index optimization in code 6. Here a very expensive computation is a function of the loop indices and so cannot be hoisted out of the loop; however, the computation is invariant with respect to an outer iterative loop over time. We can compute its value for each iteration of the computation loop prior to entering the time loop and save the values in an array. The increase in memory required to store the values is small in comparison to the large savings in time. The main loop in Code 8 is doubly nested. The inner loop includes a series of guarded computations; some are a function of the inner loop index but not the outer loop index while others are a function of the outer loop index but not the inner loop index for (j = 0; j < N; j++) { for (i = 0; i < M; i++) { r = i * hrmax; R = A[j]; temp = (PRM[3] == 0.0) ? 1.0 : pow(r, PRM[3]); high = temp * kcoeff * B[j] * PRM[2] * PRM[4]; low = high * PRM[6] * PRM[6] / (1.0 + pow(PRM[4] * PRM[6], 2.0)); kap = (R > PRM[6]) ? high * R * R / (1.0 + pow(PRM[4]*r, 2.0) : low * pow(R/PRM[6], PRM[5]); < rest of loop omitted > }} Note that the value of temp is invariant to j. Thus, we can hoist the computation for temp out of the loop and save its values in an array. for (i = 0; i < M; i++) { r = i * hrmax; TEMP[i] = pow(r, PRM[3]); } [N.B. – the case for PRM[3] = 0 is omitted and will be reintroduced later.] We now hoist out of the inner loop the computations invariant to i. Since the conditional guarding the value of kap is invariant to i, it behooves us to hoist the computation out of the inner loop, thereby executing the guard once rather than M times. The final version of the code is for (j = 0; j < N; j++) { R = rig[j] / 1000.; tmp1 = kcoeff * par[2] * beta[j] * par[4]; tmp2 = 1.0 + (par[4] * par[4] * par[6] * par[6]); tmp3 = 1.0 + (par[4] * par[4] * R * R); tmp4 = par[6] * par[6] / tmp2; tmp5 = R * R / tmp3; tmp6 = pow(R / par[6], par[5]); if ((par[3] == 0.0) && (R > par[6])) { for (i = 1; i <= imax1; i++) KAP[i] = tmp1 * tmp5; } else if ((par[3] == 0.0) && (R <= par[6])) { for (i = 1; i <= imax1; i++) KAP[i] = tmp1 * tmp4 * tmp6; } else if ((par[3] != 0.0) && (R > par[6])) { for (i = 1; i <= imax1; i++) KAP[i] = tmp1 * TEMP[i] * tmp5; } else if ((par[3] != 0.0) && (R <= par[6])) { for (i = 1; i <= imax1; i++) KAP[i] = tmp1 * TEMP[i] * tmp4 * tmp6; } for (i = 0; i < M; i++) { kap = KAP[i]; r = i * hrmax; < rest of loop omitted > } } Maybe not the prettiest piece of code, but certainly much more efficient than the original loop, Copy operations Several programs unnecessarily copy data from one data structure to another. This problem occurs in both Fortran and C programs, although it manifests itself differently in the two languages. Code 1 declares two arrays—one for old values and one for new values. At the end of each iteration, the array of new values is copied to the array of old values to reset the data structures for the next iteration. This problem occurs in Fortran programs not included in this study and in both Fortran 77 and Fortran 90 code. Introducing pointers to the arrays and swapping pointer values is an obvious way to eliminate the copying; but pointers is not a feature that many Fortran programmers know well or are comfortable using. An easy solution not involving pointers is to extend the dimension of the value array by 1 and use the last dimension to differentiate between arrays at different times. For example, if the data space is N x N, declare the array (N, N, 2). Then store the problem’s initial values in (_, _, 2) and define the scalar names new = 2 and old = 1. At the start of each iteration, swap old and new to reset the arrays. The old–new copy problem did not appear in any C program. In programs that had new and old values, the code swapped pointers to reset data structures. Where unnecessary coping did occur is in structure assignment and parameter passing. Structures in C are handled much like scalars. Assignment causes the data space of the right-hand name to be copied to the data space of the left-hand name. Similarly, when a structure is passed to a function, the data space of the actual parameter is copied to the data space of the formal parameter. If the structure is large and the assignment or function call is in an inner loop, then copying costs can grow quite large. While none of the ten programs considered here manifested this problem, it did occur in programs not included in the study. A simple fix is always to refer to structures via pointers. Optimizing loop structures Since scientific programs spend almost all their time in loops, efficient loops are the key to good performance. Conditionals, function calls, little instruction level parallelism, and large numbers of temporary values make it difficult for the compiler to generate tightly packed, highly efficient code. Conditionals and function calls introduce jumps that disrupt code flow. Users should eliminate or isolate conditionls to their own loops as much as possible. Often logical expressions can be substituted for if-then-else statements. For example, code 2 includes the following snippet MaxDelta = 0.0 do J = 1, N do I = 1, M < code omitted > Delta = abs(OldValue ? NewValue) if (Delta > MaxDelta) MaxDelta = Delta enddo enddo if (MaxDelta .gt. 0.001) goto 200 Since the only use of MaxDelta is to control the jump to 200 and all that matters is whether or not it is greater than 0.001, I made MaxDelta a boolean and rewrote the snippet as MaxDelta = .false. do J = 1, N do I = 1, M < code omitted > Delta = abs(OldValue ? NewValue) MaxDelta = MaxDelta .or. (Delta .gt. 0.001) enddo enddo if (MaxDelta) goto 200 thereby, eliminating the conditional expression from the inner loop. A microprocessor can execute many instructions per instruction cycle. Typically, it can execute one or more memory, floating point, integer, and jump operations. To be executed simultaneously, the operations must be independent. Thick loops tend to have more instruction level parallelism than thin loops. Moreover, they reduce memory traffice by maximizing data reuse. Loop unrolling and loop fusion are two techniques to increase the size of loop bodies. Several of the codes studied benefitted from loop unrolling, but none benefitted from loop fusion. This observation is not too surpising since it is the general tendency of programmers to write thick loops. As loops become thicker, the number of temporary values grows, increasing register pressure. If registers spill, then memory traffic increases and code flow is disrupted. A thick loop with many temporary values may execute slower than an equivalent series of thin loops. The biggest gain will be achieved if the thick loop can be split into a series of independent loops eliminating the need to write and read temporary arrays. I found such an occasion in code 10 where I split the loop do i = 1, n do j = 1, m A24(j,i)= S24(j,i) * T24(j,i) + S25(j,i) * U25(j,i) B24(j,i)= S24(j,i) * T25(j,i) + S25(j,i) * U24(j,i) A25(j,i)= S24(j,i) * C24(j,i) + S25(j,i) * V24(j,i) B25(j,i)= S24(j,i) * U25(j,i) + S25(j,i) * V25(j,i) C24(j,i)= S26(j,i) * T26(j,i) + S27(j,i) * U26(j,i) D24(j,i)= S26(j,i) * T27(j,i) + S27(j,i) * V26(j,i) C25(j,i)= S27(j,i) * S28(j,i) + S26(j,i) * U28(j,i) D25(j,i)= S27(j,i) * T28(j,i) + S26(j,i) * V28(j,i) end do end do into two disjoint loops do i = 1, n do j = 1, m A24(j,i)= S24(j,i) * T24(j,i) + S25(j,i) * U25(j,i) B24(j,i)= S24(j,i) * T25(j,i) + S25(j,i) * U24(j,i) A25(j,i)= S24(j,i) * C24(j,i) + S25(j,i) * V24(j,i) B25(j,i)= S24(j,i) * U25(j,i) + S25(j,i) * V25(j,i) end do end do do i = 1, n do j = 1, m C24(j,i)= S26(j,i) * T26(j,i) + S27(j,i) * U26(j,i) D24(j,i)= S26(j,i) * T27(j,i) + S27(j,i) * V26(j,i) C25(j,i)= S27(j,i) * S28(j,i) + S26(j,i) * U28(j,i) D25(j,i)= S27(j,i) * T28(j,i) + S26(j,i) * V28(j,i) end do end do Conclusions Over the course of the last year, I have had the opportunity to work with over two dozen academic scientific programmers at leading research universities. Their research interests span a broad range of scientific fields. Except for two programs that relied almost exclusively on library routines (matrix multiply and fast Fourier transform), I was able to improve significantly the single processor performance of all codes. Improvements range from 2x to 15.5x with a simple average of 4.75x. Changes to the source code were at a very high level. I did not use sophisticated techniques or programming tools to discover inefficiencies or effect the changes. Only one code was parallel despite the availability of parallel systems to all developers. Clearly, we have a problem—personal scientific research codes are highly inefficient and not running parallel. The developers are unaware of simple optimization techniques to make programs run faster. They lack education in the art of code optimization and parallel programming. I do not believe we can fix the problem by publishing additional books or training manuals. To date, the developers in questions have not studied the books or manual available, and are unlikely to do so in the future. Short courses are a possible solution, but I believe they are too concentrated to be much use. The general concepts can be taught in a three or four day course, but that is not enough time for students to practice what they learn and acquire the experience to apply and extend the concepts to their codes. Practice is the key to becoming proficient at optimization. I recommend that graduate students be required to take a semester length course in optimization and parallel programming. We would never give someone access to state-of-the-art scientific equipment costing hundreds of thousands of dollars without first requiring them to demonstrate that they know how to use the equipment. Yet the criterion for time on state-of-the-art supercomputers is at most an interesting project. Requestors are never asked to demonstrate that they know how to use the system, or can use the system effectively. A semester course would teach them the required skills. Government agencies that fund academic scientific research pay for most of the computer systems supporting scientific research as well as the development of most personal scientific codes. These agencies should require graduate schools to offer a course in optimization and parallel programming as a requirement for funding. About the Author John Feo received his Ph.D. in Computer Science from The University of Texas at Austin in 1986. After graduate school, Dr. Feo worked at Lawrence Livermore National Laboratory where he was the Group Leader of the Computer Research Group and principal investigator of the Sisal Language Project. In 1997, Dr. Feo joined Tera Computer Company where he was project manager for the MTA, and oversaw the programming and evaluation of the MTA at the San Diego Supercomputer Center. In 2000, Dr. Feo joined Sun Microsystems as an HPC application specialist. He works with university research groups to optimize and parallelize scientific codes. Dr. Feo has published over two dozen research articles in the areas of parallel parallel programming, parallel programming languages, and application performance.

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• Remove partition for failed installation

  - by kapitanluffy

  i tried installing ubuntu alongside windows 7 Installing Ubuntu alongside Windows and i failed so i decided to go with wubi again. after installing it, i noticed a separate hard disk. i investigated and found out that this hard disk is actually for the failed installation. i don't know where to find it inside the windows system. can anyone please teach me how to remove the 'failed' hard disk. here's a screenshot the left side is the current filesystem. the right side on the other hand is the 'failed' harddisk. i verified that it is the failed one because the wubi installation will provide a 'host' folder for the current partition it is currently installed. i tried looking for the 'failed' one using the windows' commandline but i don't know where to look for the 'failed' disk. (i used the cmd coz i don't want root.disk to mysteriously disappear again.) see http://ubuntu-with-wubi.blogspot.com/2011/01/mystery-of-disappearing-rootdisk.html

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• remove failed ubuntu installation

  - by kapitanluffy

  i tried installing ubuntu alongside windows 7 Installing Ubuntu alongside Windows and i failed so i decided to go with wubi again. after installing it, i noticed a separate hard disk. i investigated and found out that this hard disk is actually for the failed installation. i don't know where to find it inside the windows system. can anyone please teach me how to remove the 'failed' hard disk. here's a screenshot the left side is the current filesystem. the right side on the other hand is the 'failed' harddisk. i verified that it is the failed one because the wubi installation will provide a 'host' folder for the current partition it is currently installed. i tried looking for the 'failed' one using the windows' commandline but i don't know where to look for the 'failed' disk. (i used the cmd coz i don't want root.disk to mysteriously disappear again.) see http://ubuntu-with-wubi.blogspot.com/2011/01/mystery-of-disappearing-rootdisk.html

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• Error MSB4019: The imported project "C:\Program Files\MSBuild\Microsoft\VisualStudio\v10.0\WebApplications\Microsoft.WebApplication.targets" was not found. Confirm that the path in the <Import> declaration is correct, and that the file exists on disk

  - by Tim Huffam

  This error occurred on our TFS2008 build server which we had upgraded to cater for VS2010 projects (by installing VS2010 on the build server - see this article). Error MSB4019: The imported project "C:\Program Files\MSBuild\Microsoft\VisualStudio\v10.0\WebApplications\Microsoft.WebApplication.targets" was not found. Confirm that the path in the <Import> declaration is correct, and that the file exists on disk. However - although we had installed VS2010 on the build server - we had not installed the web development components (Visual Web Developer) - this is what caused the error. To fix - simply add the web development components: Go into Control Panel - Add or Remove Programs Select Microsoft Visual Studio 2010, and click on Change/Remove In the VS Maintenance Mode screens, select Add or Remove Features In the Setup - Options page make sure 'Visual Web Developer' is checked. Click on Update.   You shouldn't need to restart your build service. HTH Tim

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• Installing Ubuntu 12.04.1 x64 with Fake RAID 1 [SOLVED]

  - by Arkadius

  I had: Software: Dual boot with Windows XP Ubuntu 10.04 LTS x32 Hardware Fake RAID 1 (mirroring) with 2x1 TB: Partition 1 - Windows Partition 2 - SWAP Partition 3 - / (root) Partition 4 - Extended Partition 5 - /home Partition 6 - /data arek@domek:/var/log/installer$ sudo fdisk -l Disk /dev/sda: 1000.2 GB, 1000204886016 bytes 255 heads, 63 sectors/track, 121601 cylinders, total 1953525168 sectors Units = sectors of 1 * 512 = 512 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x000de1b9 Device Boot Start End Blocks Id System /dev/sda1 * 63 524297339 262148638+ 7 HPFS/NTFS/exFAT /dev/sda2 524297340 528506369 2104515 82 Linux swap / Solaris /dev/sda3 528506370 570468149 20980890 83 Linux /dev/sda4 570468150 1953118439 691325145 5 Extended /dev/sda5 570468213 675340469 52436128+ 83 Linux /dev/sda6 675340533 1953118439 638888953+ 83 Linux Disk /dev/sdb: 1000.2 GB, 1000204886016 bytes 255 heads, 63 sectors/track, 121601 cylinders, total 1953525168 sectors Units = sectors of 1 * 512 = 512 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x000de1b9 Device Boot Start End Blocks Id System /dev/sdb1 * 63 524297339 262148638+ 7 HPFS/NTFS/exFAT /dev/sdb2 524297340 528506369 2104515 82 Linux swap / Solaris /dev/sdb3 528506370 570468149 20980890 83 Linux /dev/sdb4 570468150 1953118439 691325145 5 Extended /dev/sdb5 570468213 675340469 52436128+ 83 Linux /dev/sdb6 675340533 1953118439 638888953+ 83 Linux arek@domek:/var/log/installer$ ls -l /dev/mapper/ total 0 crw------- 1 root root 10, 236 Oct 7 20:17 control lrwxrwxrwx 1 root root 7 Oct 7 20:17 pdc_jhjbcaha -> ../dm-0 lrwxrwxrwx 1 root root 7 Oct 7 20:17 pdc_jhjbcaha1 -> ../dm-1 lrwxrwxrwx 1 root root 7 Oct 7 20:17 pdc_jhjbcaha2 -> ../dm-2 lrwxrwxrwx 1 root root 7 Oct 7 20:17 pdc_jhjbcaha3 -> ../dm-3 lrwxrwxrwx 1 root root 7 Oct 7 20:17 pdc_jhjbcaha4 -> ../dm-4 lrwxrwxrwx 1 root root 7 Oct 7 20:17 pdc_jhjbcaha5 -> ../dm-5 lrwxrwxrwx 1 root root 7 Oct 7 20:17 pdc_jhjbcaha6 -> ../dm-6 I wanted to upgrade from 10.04 x32 to 12.04 x64 using FRESH installation. So, run installation of Ubuntu 12.04.1 x64 LTS using alternate CD. During the installation I selected manual partitioning and to: - Use and Format / (root) - Use and Format SWAP - Use and Keep data on /home - Use and Keep data on /data After I clicked "Continue" I get error creating and formatting SWAP partition. I go to terminal with Alt + F2 (?) and hit enter. I discovered that there was visible RAID as only disk with NO partitions. Something like this: arek@domek:/var/log/installer$ ls -l /dev/mapper/ lrwxrwxrwx 1 root root 7 Oct 7 20:17 /dev/mapper/pdc_jhjbcaha -> ../dm-0 arek@domek:/var/log/installer$ ls -l /dev/dm* brw-rw---- 1 root disk 252, 0 Oct 7 20:17 /dev/dm-0 So I switched to log console Alt+F3 (?) and saw errors like below: Oct 7 14:02:45 check-missing-firmware: /dev/.udev/firmware-missing does not exist, skipping Oct 7 14:02:45 check-missing-firmware: /run/udev/firmware-missing does not exist, skipping Oct 7 14:02:45 check-missing-firmware: no missing firmware in /dev/.udev/firmware-missing /run/udev/firmware-missing Oct 7 14:02:45 anna-install: Installing dmraid-udeb Oct 7 14:02:45 anna[12599]: DEBUG: retrieving dmraid-udeb 1.0.0.rc16-4.1ubuntu8 Oct 7 14:02:49 anna[12599]: DEBUG: retrieving libdmraid1.0.0.rc16-udeb 1.0.0.rc16-4.1ubuntu8 Oct 7 14:02:49 anna[12599]: DEBUG: retrieving kpartx-udeb 0.4.9-3ubuntu5 Oct 7 14:02:49 disk-detect: Serial ATA RAID disk(s) detected. Oct 7 14:02:55 disk-detect: Enabling dmraid support. Oct 7 14:02:55 disk-detect: RAID set "pdc_jhjbcaha" was activated Oct 7 14:02:55 HERE --> dmraid-activate: ERROR: Cannot retrieve RAID set information for pdc_jhjbcaha Oct 7 14:02:56 check-missing-firmware: /dev/.udev/firmware-missing does not exist, skipping Oct 7 14:02:56 check-missing-firmware: /run/udev/firmware-missing does not exist, skipping Oct 7 14:02:56 check-missing-firmware: no missing firmware in /dev/.udev/firmware-missing /run/udev/firmware-missing Oct 7 14:02:57 main-menu[428]: DEBUG: resolver (libnewt0.52): package doesn't exist (ignored) Oct 7 14:02:57 main-menu[428]: DEBUG: resolver (ext2-modules): package doesn't exist (ignored) Oct 7 14:02:57 main-menu[428]: INFO: Menu item 'partman-base' selected Oct 7 14:02:57 kernel: [ 316.512999] NTFS driver 2.1.30 [Flags: R/O MODULE]. Oct 7 14:02:57 kernel: [ 316.523221] Btrfs loaded Oct 7 14:02:57 kernel: [ 316.534781] JFS: nTxBlock = 8192, nTxLock = 65536 Oct 7 14:02:57 kernel: [ 316.554749] SGI XFS with ACLs, security attributes, realtime, large block/inode numbers, no debug enabled Oct 7 14:02:57 kernel: [ 316.555336] SGI XFS Quota Management subsystem Oct 7 14:02:58 md-devices: mdadm: No arrays found in config file or automatically Oct 7 14:02:58 partman: No matching physical volumes found Oct 7 14:02:58 partman: No volume groups found Oct 7 14:02:58 partman: Reading all physical volumes. This may take a while... Oct 7 14:02:58 partman-lvm: No volume groups found Oct 7 14:02:58 partman: Error running 'tune2fs -l /dev/mapper/pdc_jhjbcaha' Oct 7 14:02:58 partman: Error running 'tune2fs -l /dev/mapper/pdc_jhjbcaha' Oct 7 14:02:58 partman: Error running 'tune2fs -l /dev/mapper/pdc_jhjbcaha' Oct 7 14:06:11 HERE --> partman: mkswap: can't open '/dev/mapper/pdc_jhjbcaha2': No such file or directory Oct 7 14:07:28 init: starting pid 401, tty '/dev/tty2': '-/bin/sh' Oct 7 14:15:00 net/hw-detect.hotplug: Detected hotpluggable network interface eth0 Oct 7 14:15:00 net/hw-detect.hotplug: Detected hotpluggable network interface lo As You can see there are 2 errors Oct 7 14:02:55 dmraid-activate: ERROR: Cannot retrieve RAID set information for pdc_jhjbcaha and Oct 7 14:06:11 partman: mkswap: can't open '/dev/mapper/pdc_jhjbcaha2': No such file or directory I looked in the internet and try to run command "dmraid -ay" and get something like that: dmraid -ay /dev/mapper/pdc_jhjbcaha -> Already activated /dev/mapper/pdc_jhjbcaha1 -> Successfully activated /dev/mapper/pdc_jhjbcaha2 -> Successfully activated /dev/mapper/pdc_jhjbcaha3 -> Successfully activated /dev/mapper/pdc_jhjbcaha4 -> Successfully activated /dev/mapper/pdc_jhjbcaha5 -> Successfully activated /dev/mapper/pdc_jhjbcaha6 -> Successfully activated Then I returned to installer with Alt+F1 (?) and click "Return" to return to partitioning menu. I did NOT change anything just selected again "Continue" and everything goes smoothly. I hope this will help someone. arkadius

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• How do I take an image/backup of Ubuntu partition and restore to VirtualBox VM

  - by whizkid

  I have Ubuntu 10.04 installed on an older hard disk. I recently bought a new disk and already installed Windows 7. I dont want to use the older disk anymore, and I would like to keep on using Ubuntu in a virtual machine on the new disk(to avoid the possible mess-ups of dual boot and I found VirtualBox is the best free tool for this). I wish to keep the exact same data\programs\configurations\settings I had been using in Ubuntu for so long, and avoid the tedious part of having to reconfigure so many things. How do I backup\restore Ubuntu to another disk? I would prefer a free tool to do the backup\restore.

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• How to use more than 3 virtual disks in Linux using CentOS and XenServer

  - by 010110110101

  I've attached 5 virtual disks to a Virtual Machine in Citrix XenServer. The VM has the xs-tools installed. Initially it said that it couldn't add so many disks. After I installed the xs-tools, it let me add all the disks. But /dev doesn't show all the disks. It shows these: /dev/xvda /dev/xvdb /dev/xvdc /dev/cdrom Perhaps it is bound by the limits of an IDE bus? (3 disks + CD-ROM) If so, how does one change the VM to use SCSI? Edit: According to the documentation: 2.6.3. VM Block Devices In the PV Linux case, block devices are passed through as PV devices. XenServer does not attempt to emulate SCSI or IDE, but instead provides a more suitable interface in the virtual environment in the form of xvd* devices. It is also possible to get an sd* device using the same mechanism, where the PV driver inside the VM takes over the SCSI device namespace. This is not desirable so it is best to use xvd* where possible for PV guests (this is the default for Debian and RHEL). For Windows or other fully virtualized guests, XenServer emulates an IDE bus in the form of an hd* device. When using Windows, installing the Citrix Tools for Virtual Machines installs a special PV driver that works in a similar way to Linux, except in the fully virtualized environment. Still, with 5 virtual disks attached, I don't see the other xvd devices. Edit #2: (attached requested info) Host Machine: XenServer 6.1 Linux version 2.6.32.43-0.4.1.xs1.6.10.777.170770xen (geeko@buildhost) (gcc version 4.1.2 20080704 (Red Hat 4.1.2-51)) #1 SMP Wed Apr 17 05:52:03 EDT 2013 Guest Machine: CentOS release 6.4 (Final) Linux version 2.6.32-358.6.2.el6.x86_64 ([email protected]) (gcc version 4.4.7 20120313 (Red Hat 4.4.7-3) (GCC) ) #1 SMP Thu May 16 20:59:36 UTC 2013 Output of 'fdisk -l' on Guest Machine: Note, the disk beyond the first 3 attached are not displaying -- there should be 4 100GB disks. (There are a total of 5 disks displayed in XenCenter -- 16GB, 100GB, 100GB, 100GB, 100GB) Disk /dev/xvdb: 107.4 GB, 107374182400 bytes 255 heads, 63 sectors/track, 13054 cylinders Units = cylinders of 16065 * 512 = 8225280 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0xfb6c95b9 Device Boot Start End Blocks Id System /dev/xvdb1 1 13054 104856223+ 83 Linux Disk /dev/xvda: 17.2 GB, 17179869184 bytes 255 heads, 63 sectors/track, 2088 cylinders Units = cylinders of 16065 * 512 = 8225280 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x000e5f41 Device Boot Start End Blocks Id System /dev/xvda1 * 1 64 512000 83 Linux Partition 1 does not end on cylinder boundary. /dev/xvda2 64 2089 16264192 8e Linux LVM Disk /dev/xvdc: 107.4 GB, 107374182400 bytes 255 heads, 63 sectors/track, 13054 cylinders Units = cylinders of 16065 * 512 = 8225280 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0xed249ced Device Boot Start End Blocks Id System /dev/xvdc1 1 13054 104856223+ 83 Linux Disk /dev/mapper/vg_blue-lv_root: 14.6 GB, 14571012096 bytes 255 heads, 63 sectors/track, 1771 cylinders Units = cylinders of 16065 * 512 = 8225280 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x00000000 Disk /dev/mapper/vg_blue-lv_swap: 2080 MB, 2080374784 bytes 255 heads, 63 sectors/track, 252 cylinders Units = cylinders of 16065 * 512 = 8225280 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x00000000 I see that the Linux versions say SMP. The Guest VM doesn't say "xen" in the name. However, I have already run yum install kernel-xen. Could be a clue?

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• Removing mdadm array and converting to regular disks while preserving data

  - by Jeffrey Kevin Pry

  I have a 6 disk (2TB each) mdadm RAID 5 volume created in Ubuntu 12.04 Server. However, I'm moving to a different solution and want to "unraid" my disks but keep the data. Only 50% is in use. From what I can surmise I basically have to do this recursively for each physical disk. Fail the disk Format the failed disk Move a portion of files to the new disk. Reshape the array Shrink the logical volume md0 This seems like a very time consuming process. Is there an easier way to do this (automatically perhaps) without buying new disks to temporarily hold the data? I am also aware that during this processing my RAID volume will be degraded and vulnerable the entire time. I am not too concerned about this and will be using battery backup and moving the most important files off first. Thank you for your help!

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• I/O errors are reported when I try to install Ubuntu, but the SMART data is good. Is my hard disk dying?

  - by James

  When I try to install linux, it tells me there is an input output error on dev sda. I have tried both Ubuntu and Mint on two different computers. So that narrows it down to the hdd. After hours of googling and trying different things I tried making the hardrive ext4 with gparted but that comes up with an error. This makes me think that the hdd is bad. There are a few reasons I think the hdd isn't bad. I can use the hdd in windows fully. Windows and gparted disk health checks both say it is fine. Its SMART data is all good. So... help?

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• Why don't I have the option ''Install Ubuntu alongside with them''

  - by almqgh

  Why dont I have this option? here are my disk sudo fdisk -l Disk /dev/sda: 640.1 GB, 640135028736 bytes 255 heads, 63 sectors/track, 77825 cylinders, total 1250263728 sectors Units = sectors of 1 * 512 = 512 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x5b53cc54 Device Boot Start End Blocks Id System /dev/sda1 * 2048 409599 203776 7 HPFS/NTFS/exFAT /dev/sda2 409600 1153767021 576678711 7 HPFS/NTFS/exFAT /dev/sda3 1216962560 1250050047 16543744 7 HPFS/NTFS/exFAT /dev/sda4 1250050048 1250261679 105816 c W95 FAT32 (LBA) Disk /dev/sdb: 4005 MB, 4005527552 bytes 32 heads, 63 sectors/track, 3880 cylinders, total 7823296 sectors Units = sectors of 1 * 512 = 512 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x20d8782d Device Boot Start End Blocks Id System /dev/sdb1 * 63 7822079 3911008+ c W95 FAT32 (LBA)

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• Best tool to understand source

  - by cache

  I have a source code for a project. I am working on porting it to another device as the current source code is for a linux environment. I am having some error on the newly ported code. So i was thinking it would be best to once again understand the whole source code and this will help me localise the errors. Now the problem is that i tried using 'gdb' for linux to debug the code but it does not help. So is there any tool that I can use to trace the program line by line ? By doing so i can understand the program flow. Please Help !

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