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  • John Burke's Weclome to the Applications Strategy Blog

    - by Tony Ouk
    Hi I'm John Burke and I'm the group Vice President of Oracle's Applications Business Unit.  Thanks for stopping by our Applications blog today.  The purpose of this site is to provide you, our customers, with timely, relevant, and balanced information about the state of the applications business, both here at Oracle and industry-wide. So on this site, you'll find information about Oracle's application products, how our customers have used those products to transform their businesses, and general industry trends which might help you craft YOUR applications roadmap.  So right now I'm walking to meet with one of Oracle's development executives.  I also plan to talk to Oracle customers and leading industry analysts.  I plan to provide a complete and balanced view of the total applications landscape.  I hope you check back often and view our updates.

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  • How to mount a blu-ray drive?

    - by Stephan Schielke
    Maybe it is for the best to close this question. This has nothing to do with a bluray drive in general anymore. Probably a hardware defect. I will try to test it with a windows system and different cables again... Thx so far. I have a bluray/dvd/cdrom drive with SATA. Ubuntu wont find it under /dev/sd wodim --devices wodim: Overview of accessible drives (1 found) : ------------------------------------------------------------------------- 0 dev='/dev/sg2' rwrw-- : 'HL-DT-ST' 'BDDVDRW CH08LS10' ------------------------------------------------------------------------- cdrecord -scanbus scsibus2: 2,0,0 200) 'HL-DT-ST' 'BDDVDRW CH08LS10' '2.00' Removable CD-ROM fdisk dont even lists it. Ubuntu only automounts blank DVDs, but neither CDROM nor Blurays. I also changed the sata slot, sata cable and the power cable. The drive works with a windows system. This happens when I try to mount: sudo mount -t auto /dev/scd0 /media/bluray mount: you must specify the filesystem type I tried all filesystems there are. I also installed makemkv. It finds the drive but not the disc. Here is my /dev ls -al /dev total 12 drwxr-xr-x 17 root root 4420 2011-11-25 19:36 . drwxr-xr-x 28 root root 4096 2011-11-25 07:12 .. crw------- 1 root root 10, 235 2011-11-25 19:28 autofs -rw-r--r-- 1 root root 630 2011-11-25 19:28 .blkid.tab -rw-r--r-- 1 root root 630 2011-11-25 19:28 .blkid.tab.old drwxr-xr-x 2 root root 700 2011-11-25 19:27 block drwxr-xr-x 2 root root 100 2011-11-25 19:27 bsg crw------- 1 root root 10, 234 2011-11-25 19:28 btrfs-control drwxr-xr-x 3 root root 60 2011-11-25 19:27 bus drwxr-xr-x 2 root root 3820 2011-11-25 19:28 char crw------- 1 root root 5, 1 2011-11-25 19:28 console lrwxrwxrwx 1 root root 11 2011-11-25 19:28 core -> /proc/kcore drwxr-xr-x 2 root root 60 2011-11-25 19:28 cpu crw------- 1 root root 10, 60 2011-11-25 19:28 cpu_dma_latency drwxr-xr-x 7 root root 140 2011-11-25 19:27 disk crw------- 1 root root 10, 61 2011-11-25 19:28 ecryptfs crw-rw---- 1 root video 29, 0 2011-11-25 19:28 fb0 lrwxrwxrwx 1 root root 13 2011-11-25 19:28 fd -> /proc/self/fd crw-rw-rw- 1 root root 1, 7 2011-11-25 19:28 full crw-rw-rw- 1 root fuse 10, 229 2011-11-25 19:28 fuse crw------- 1 root root 251, 0 2011-11-25 19:28 hidraw0 crw------- 1 root root 251, 1 2011-11-25 19:28 hidraw1 crw------- 1 root root 10, 228 2011-11-25 19:28 hpet lrwxrwxrwx 1 root root 14 2011-11-25 19:27 .initramfs -> /run/initramfs drwxr-xr-x 4 root root 220 2011-11-25 19:28 input crw------- 1 root root 1, 11 2011-11-25 19:28 kmsg srw-rw-rw- 1 root root 0 2011-11-25 19:28 log brw-rw---- 1 root disk 7, 0 2011-11-25 19:28 loop0 brw-rw---- 1 root disk 7, 1 2011-11-25 19:28 loop1 brw-rw---- 1 root disk 7, 2 2011-11-25 19:28 loop2 brw-rw---- 1 root disk 7, 3 2011-11-25 19:28 loop3 brw-rw---- 1 root disk 7, 4 2011-11-25 19:28 loop4 brw-rw---- 1 root disk 7, 5 2011-11-25 19:28 loop5 brw-rw---- 1 root disk 7, 6 2011-11-25 19:28 loop6 brw-rw---- 1 root disk 7, 7 2011-11-25 19:28 loop7 drwxr-xr-x 2 root root 60 2011-11-25 19:27 mapper crw------- 1 root root 10, 227 2011-11-25 19:28 mcelog crw-r----- 1 root kmem 1, 1 2011-11-25 19:28 mem drwxr-xr-x 2 root root 60 2011-11-25 19:27 net crw------- 1 root root 10, 59 2011-11-25 19:28 network_latency crw------- 1 root root 10, 58 2011-11-25 19:28 network_throughput crw-rw-rw- 1 root root 1, 3 2011-11-25 19:28 null crw-rw-rw- 1 root root 195, 0 2011-11-25 19:28 nvidia0 crw-rw-rw- 1 root root 195, 255 2011-11-25 19:28 nvidiactl crw------- 1 root root 1, 12 2011-11-25 19:28 oldmem crw-r----- 1 root kmem 1, 4 2011-11-25 19:28 port crw------- 1 root root 108, 0 2011-11-25 19:28 ppp crw------- 1 root root 10, 1 2011-11-25 19:28 psaux crw-rw-rw- 1 root tty 5, 2 2011-11-25 20:00 ptmx drwxr-xr-x 2 root root 0 2011-11-25 19:27 pts brw-rw---- 1 root disk 1, 0 2011-11-25 19:28 ram0 brw-rw---- 1 root disk 1, 1 2011-11-25 19:28 ram1 brw-rw---- 1 root disk 1, 10 2011-11-25 19:28 ram10 brw-rw---- 1 root disk 1, 11 2011-11-25 19:28 ram11 brw-rw---- 1 root disk 1, 12 2011-11-25 19:28 ram12 brw-rw---- 1 root disk 1, 13 2011-11-25 19:28 ram13 brw-rw---- 1 root disk 1, 14 2011-11-25 19:28 ram14 brw-rw---- 1 root disk 1, 15 2011-11-25 19:28 ram15 brw-rw---- 1 root disk 1, 2 2011-11-25 19:28 ram2 brw-rw---- 1 root disk 1, 3 2011-11-25 19:28 ram3 brw-rw---- 1 root disk 1, 4 2011-11-25 19:28 ram4 brw-rw---- 1 root disk 1, 5 2011-11-25 19:28 ram5 brw-rw---- 1 root disk 1, 6 2011-11-25 19:28 ram6 brw-rw---- 1 root disk 1, 7 2011-11-25 19:28 ram7 brw-rw---- 1 root disk 1, 8 2011-11-25 19:28 ram8 brw-rw---- 1 root disk 1, 9 2011-11-25 19:28 ram9 crw-rw-rw- 1 root root 1, 8 2011-11-25 19:28 random crw-rw-r--+ 1 root root 10, 62 2011-11-25 19:28 rfkill lrwxrwxrwx 1 root root 4 2011-11-25 19:28 rtc -> rtc0 crw------- 1 root root 254, 0 2011-11-25 19:28 rtc0 lrwxrwxrwx 1 root root 3 2011-11-25 19:38 scd0 -> sr0 brw-rw---- 1 root disk 8, 0 2011-11-25 19:28 sda brw-rw---- 1 root disk 8, 1 2011-11-25 19:28 sda1 brw-rw---- 1 root disk 8, 2 2011-11-25 19:28 sda2 brw-rw---- 1 root disk 8, 3 2011-11-25 19:28 sda3 brw-rw---- 1 root disk 8, 5 2011-11-25 19:28 sda5 brw-rw---- 1 root disk 8, 6 2011-11-25 19:28 sda6 brw-rw---- 1 root disk 8, 16 2011-11-25 19:28 sdb brw-rw---- 1 root disk 8, 17 2011-11-25 19:28 sdb1 crw-rw---- 1 root disk 21, 0 2011-11-25 19:28 sg0 crw-rw---- 1 root disk 21, 1 2011-11-25 19:28 sg1 crw-rw----+ 1 root cdrom 21, 2 2011-11-25 19:28 sg2 lrwxrwxrwx 1 root root 8 2011-11-25 19:28 shm -> /run/shm crw------- 1 root root 10, 231 2011-11-25 19:28 snapshot drwxr-xr-x 4 root root 280 2011-11-25 19:28 snd brw-rw----+ 1 root cdrom 11, 0 2011-11-25 19:38 sr0 lrwxrwxrwx 1 root root 15 2011-11-25 19:28 stderr -> /proc/self/fd/2 lrwxrwxrwx 1 root root 15 2011-11-25 19:28 stdin -> /proc/self/fd/0 lrwxrwxrwx 1 root root 15 2011-11-25 19:28 stdout -> /proc/self/fd/1 crw-rw-rw- 1 root tty 5, 0 2011-11-25 19:35 tty crw--w---- 1 root tty 4, 0 2011-11-25 19:28 tty0 crw------- 1 root root 4, 1 2011-11-25 19:28 tty1 crw--w---- 1 root tty 4, 10 2011-11-25 19:28 tty10 crw--w---- 1 root tty 4, 11 2011-11-25 19:28 tty11 crw--w---- 1 root tty 4, 12 2011-11-25 19:28 tty12 crw--w---- 1 root tty 4, 13 2011-11-25 19:28 tty13 crw--w---- 1 root tty 4, 14 2011-11-25 19:28 tty14 crw--w---- 1 root tty 4, 15 2011-11-25 19:28 tty15 crw--w---- 1 root tty 4, 16 2011-11-25 19:28 tty16 crw--w---- 1 root tty 4, 17 2011-11-25 19:28 tty17 crw--w---- 1 root tty 4, 18 2011-11-25 19:28 tty18 crw--w---- 1 root tty 4, 19 2011-11-25 19:28 tty19 crw------- 1 root root 4, 2 2011-11-25 19:28 tty2 crw--w---- 1 root tty 4, 20 2011-11-25 19:28 tty20 crw--w---- 1 root tty 4, 21 2011-11-25 19:28 tty21 crw--w---- 1 root tty 4, 22 2011-11-25 19:28 tty22 crw--w---- 1 root tty 4, 23 2011-11-25 19:28 tty23 crw--w---- 1 root tty 4, 24 2011-11-25 19:28 tty24 crw--w---- 1 root tty 4, 25 2011-11-25 19:28 tty25 crw--w---- 1 root tty 4, 26 2011-11-25 19:28 tty26 crw--w---- 1 root tty 4, 27 2011-11-25 19:28 tty27 crw--w---- 1 root tty 4, 28 2011-11-25 19:28 tty28 crw--w---- 1 root tty 4, 29 2011-11-25 19:28 tty29 crw------- 1 root root 4, 3 2011-11-25 19:28 tty3 crw--w---- 1 root tty 4, 30 2011-11-25 19:28 tty30 crw--w---- 1 root tty 4, 31 2011-11-25 19:28 tty31 crw--w---- 1 root tty 4, 32 2011-11-25 19:28 tty32 crw--w---- 1 root tty 4, 33 2011-11-25 19:28 tty33 crw--w---- 1 root tty 4, 34 2011-11-25 19:28 tty34 crw--w---- 1 root tty 4, 35 2011-11-25 19:28 tty35 crw--w---- 1 root tty 4, 36 2011-11-25 19:28 tty36 crw--w---- 1 root tty 4, 37 2011-11-25 19:28 tty37 crw--w---- 1 root tty 4, 38 2011-11-25 19:28 tty38 crw--w---- 1 root tty 4, 39 2011-11-25 19:28 tty39 crw------- 1 root root 4, 4 2011-11-25 19:28 tty4 crw--w---- 1 root tty 4, 40 2011-11-25 19:28 tty40 crw--w---- 1 root tty 4, 41 2011-11-25 19:28 tty41 crw--w---- 1 root tty 4, 42 2011-11-25 19:28 tty42 crw--w---- 1 root tty 4, 43 2011-11-25 19:28 tty43 crw--w---- 1 root tty 4, 44 2011-11-25 19:28 tty44 crw--w---- 1 root tty 4, 45 2011-11-25 19:28 tty45 crw--w---- 1 root tty 4, 46 2011-11-25 19:28 tty46 crw--w---- 1 root tty 4, 47 2011-11-25 19:28 tty47 crw--w---- 1 root tty 4, 48 2011-11-25 19:28 tty48 crw--w---- 1 root tty 4, 49 2011-11-25 19:28 tty49 crw------- 1 root root 4, 5 2011-11-25 19:28 tty5 crw--w---- 1 root tty 4, 50 2011-11-25 19:28 tty50 crw--w---- 1 root tty 4, 51 2011-11-25 19:28 tty51 crw--w---- 1 root tty 4, 52 2011-11-25 19:28 tty52 crw--w---- 1 root tty 4, 53 2011-11-25 19:28 tty53 crw--w---- 1 root tty 4, 54 2011-11-25 19:28 tty54 crw--w---- 1 root tty 4, 55 2011-11-25 19:28 tty55 crw--w---- 1 root tty 4, 56 2011-11-25 19:28 tty56 crw--w---- 1 root tty 4, 57 2011-11-25 19:28 tty57 crw--w---- 1 root tty 4, 58 2011-11-25 19:28 tty58 crw--w---- 1 root tty 4, 59 2011-11-25 19:28 tty59 crw------- 1 root root 4, 6 2011-11-25 19:28 tty6 crw--w---- 1 root tty 4, 60 2011-11-25 19:28 tty60 crw--w---- 1 root tty 4, 61 2011-11-25 19:28 tty61 crw--w---- 1 root tty 4, 62 2011-11-25 19:28 tty62 crw--w---- 1 root tty 4, 63 2011-11-25 19:28 tty63 crw--w---- 1 root tty 4, 7 2011-11-25 19:28 tty7 crw--w---- 1 root tty 4, 8 2011-11-25 19:28 tty8 crw--w---- 1 root tty 4, 9 2011-11-25 19:28 tty9 crw------- 1 root root 5, 3 2011-11-25 19:28 ttyprintk crw-rw---- 1 root dialout 4, 64 2011-11-25 19:28 ttyS0 crw-rw---- 1 root dialout 4, 65 2011-11-25 19:28 ttyS1 crw-rw---- 1 root dialout 4, 74 2011-11-25 19:28 ttyS10 crw-rw---- 1 root dialout 4, 75 2011-11-25 19:28 ttyS11 crw-rw---- 1 root dialout 4, 76 2011-11-25 19:28 ttyS12 crw-rw---- 1 root dialout 4, 77 2011-11-25 19:28 ttyS13 crw-rw---- 1 root dialout 4, 78 2011-11-25 19:28 ttyS14 crw-rw---- 1 root dialout 4, 79 2011-11-25 19:28 ttyS15 crw-rw---- 1 root dialout 4, 80 2011-11-25 19:28 ttyS16 crw-rw---- 1 root dialout 4, 81 2011-11-25 19:28 ttyS17 crw-rw---- 1 root dialout 4, 82 2011-11-25 19:28 ttyS18 crw-rw---- 1 root dialout 4, 83 2011-11-25 19:28 ttyS19 crw-rw---- 1 root dialout 4, 66 2011-11-25 19:28 ttyS2 crw-rw---- 1 root dialout 4, 84 2011-11-25 19:28 ttyS20 crw-rw---- 1 root dialout 4, 85 2011-11-25 19:28 ttyS21 crw-rw---- 1 root dialout 4, 86 2011-11-25 19:28 ttyS22 crw-rw---- 1 root dialout 4, 87 2011-11-25 19:28 ttyS23 crw-rw---- 1 root dialout 4, 88 2011-11-25 19:28 ttyS24 crw-rw---- 1 root dialout 4, 89 2011-11-25 19:28 ttyS25 crw-rw---- 1 root dialout 4, 90 2011-11-25 19:28 ttyS26 crw-rw---- 1 root dialout 4, 91 2011-11-25 19:28 ttyS27 crw-rw---- 1 root dialout 4, 92 2011-11-25 19:28 ttyS28 crw-rw---- 1 root dialout 4, 93 2011-11-25 19:28 ttyS29 crw-rw---- 1 root dialout 4, 67 2011-11-25 19:28 ttyS3 crw-rw---- 1 root dialout 4, 94 2011-11-25 19:28 ttyS30 crw-rw---- 1 root dialout 4, 95 2011-11-25 19:28 ttyS31 crw-rw---- 1 root dialout 4, 68 2011-11-25 19:28 ttyS4 crw-rw---- 1 root dialout 4, 69 2011-11-25 19:28 ttyS5 crw-rw---- 1 root dialout 4, 70 2011-11-25 19:28 ttyS6 crw-rw---- 1 root dialout 4, 71 2011-11-25 19:28 ttyS7 crw-rw---- 1 root dialout 4, 72 2011-11-25 19:28 ttyS8 crw-rw---- 1 root dialout 4, 73 2011-11-25 19:28 ttyS9 d rwxr-xr-x 3 root root 60 2011-11-25 19:28 .udev crw-r----- 1 root root 10, 223 2011-11-25 19:28 uinput crw-rw-rw- 1 root root 1, 9 2011-11-25 19:28 urandom drwxr-xr-x 2 root root 60 2011-11-25 19:27 usb crw------- 1 root root 252, 0 2011-11-25 19:28 usbmon0 crw------- 1 root root 252, 1 2011-11-25 19:28 usbmon1 crw------- 1 root root 252, 2 2011-11-25 19:28 usbmon2 crw------- 1 root root 252, 3 2011-11-25 19:28 usbmon3 crw------- 1 root root 252, 4 2011-11-25 19:28 usbmon4 crw------- 1 root root 252, 5 2011-11-25 19:28 usbmon5 crw------- 1 root root 252, 6 2011-11-25 19:28 usbmon6 crw------- 1 root root 252, 7 2011-11-25 19:28 usbmon7 crw------- 1 root root 252, 8 2011-11-25 19:28 usbmon8 drwxr-xr-x 4 root root 80 2011-11-25 19:28 v4l crw------- 1 root root 10, 57 2011-11-25 19:28 vboxdrv crw------- 1 root root 10, 56 2011-11-25 19:28 vboxnetctl drwxr-x--- 4 root vboxusers 80 2011-11-25 19:28 vboxusb crw-rw---- 1 root tty 7, 0 2011-11-25 19:28 vcs crw-rw---- 1 root tty 7, 1 2011-11-25 19:28 vcs1 crw-rw---- 1 root tty 7, 2 2011-11-25 19:28 vcs2 crw-rw---- 1 root tty 7, 3 2011-11-25 19:28 vcs3 crw-rw---- 1 root tty 7, 4 2011-11-25 19:28 vcs4 crw-rw---- 1 root tty 7, 5 2011-11-25 19:28 vcs5 crw-rw---- 1 root tty 7, 6 2011-11-25 19:28 vcs6 crw-rw---- 1 root tty 7, 128 2011-11-25 19:28 vcsa crw-rw---- 1 root tty 7, 129 2011-11-25 19:28 vcsa1 crw-rw---- 1 root tty 7, 130 2011-11-25 19:28 vcsa2 crw-rw---- 1 root tty 7, 131 2011-11-25 19:28 vcsa3 crw-rw---- 1 root tty 7, 132 2011-11-25 19:28 vcsa4 crw-rw---- 1 root tty 7, 133 2011-11-25 19:28 vcsa5 crw-rw---- 1 root tty 7, 134 2011-11-25 19:28 vcsa6 crw------- 1 root root 10, 63 2011-11-25 19:28 vga_arbiter crw-rw----+ 1 root video 81, 0 2011-11-25 19:28 video0 crw-rw-rw- 1 root root 1, 5 2011-11-25 19:28 zero sg_scan -i gives me: sudo sg_scan -i /dev/sg0: scsi0 channel=0 id=0 lun=0 [em] ATA ST31000524NS SN12 [rmb=0 cmdq=0 pqual=0 pdev=0x0] /dev/sg1: scsi0 channel=0 id=1 lun=0 [em] ATA WDC WD15EADS-00S 01.0 [rmb=0 cmdq=0 pqual=0 pdev=0x0] /dev/sg2: scsi2 channel=0 id=0 lun=0 [em] HL-DT-ST BDDVDRW CH08LS10 2.00 [rmb=1 cmdq=0 pqual=0 pdev=0x5] sg_map gives me: /dev/sg0 /dev/sda /dev/sg1 /dev/sdb /dev/sg2 /dev/scd0 lsscsi -l gives me: [0:0:0:0] disk ATA ST31000524NS SN12 /dev/sda state=running queue_depth=1 scsi_level=6 type=0 device_blocked=0 timeout=30 [0:0:1:0] disk ATA WDC WD15EADS-00S 01.0 /dev/sdb state=running queue_depth=1 scsi_level=6 type=0 device_blocked=0 timeout=30 [2:0:0:0] cd/dvd HL-DT-ST BDDVDRW CH08LS10 2.00 /dev/sr0 state=running queue_depth=1 scsi_level=6 type=5 device_blocked=0 timeout=30 my udf mod is: filename: /lib/modules/3.0.0-14-generic/kernel/fs/udf/udf.ko license: GPL description: Universal Disk Format Filesystem author: Ben Fennema srcversion: 6ABDE012374D96B9685B8E5 depends: crc-itu-t vermagic: 3.0.0-14-generic SMP mod_unload modversions Do I need special drivers or mods enabled? Do I need to change some BIOS settings? edit: Somehow I am now able to fire the mount command without any filesystem errors, but now I get: mount: no medium found on /dev/sr0

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  • (and a new ray smith equipment webpage)

    - by raysmithequip
    Originally posted on: http://geekswithblogs.net/raysmithequip/archive/2013/10/15/154351.aspxPlease bear with me, apparently we lost jabry.com to what I am not sure.  I have yet another webpage coming to http://www.raysmithequip.netai.net/ . Right now it is pretty bare, I just spent an hour configuring web matrix 2.0 (3 no likey like vista!!).  I should have the shoppers corner sub page back up intime for black friday though, soo keep your eyes posted.To keep you busy meantime, be sure to check out inmoov, a really cool open source 3d printed diy robot.I chanced upon it from the dangerous prototypes web site some time ago and consider it the one project that will rock the world in the immediate future.inmoov.blogspot.com/ raysmithn3twu

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  • Employee Engagement Q&A with John Brunswick

    - by Kellsey Ruppel
    As we are focusing this week on Employee Engagement, I recently sat down with industry expert and thought leader John Brunswick on the topic. Here is the Q&A dialogue we shared.  Q: How do you effectively engage employees to drive business value?A: Motivation, both extrinsic and intrinsic, combined with the relevancy of various channels to support it.  Beyond chaining business strategies like compensation models within an organization, engagement ultimately is most successful when driven by employee's motivations.  Business value derived from engagement through technical capabilities can be objectively measured through metrics like the rate and accuracy of problem solving for a given business function or frequency of innovation created.  Providing employees performing "knowledge work" with capabilities that allow them to perform work with a higher degree of accuracy in the same or ideally less time, adds value for that individual and in turn, drives their level of engagement to drive business value. Q: Organizations with high levels of employee engagement outperform the total stock market index by 22%. Can you comment on why you think this might be? A: Alignment through shared purpose.  Zappos is an excellent example of a culture that arguably has higher than average levels of employee engagement and it permeates every aspect of their organization – embodied externally through their customer experience.  I recently made my first purchase with them and it was obvious through their web experience, visual design, communication style, customer service and attention to detail down to green packaging, that they have an amazingly strong shared purpose.  The Zappos.com ‘About page’ outlines their "Family Core Values", the first three being "Deliver WOW Through Service, Embrace and Drive Change & Create Fun and A Little Weirdness" – all reflected externally in my interaction with them.  Strong shared purpose enables higher product and service experience, equating to a dedicated customer base, repeat purchases and expanded marketshare. Q: Have you seen any trends in the market regarding employee engagement? A: Some companies now see offering a form of social engagement similar to Facebook and LinkedIn as standard communication infrastructure like email or instant messaging.  Originally offered as standalone tools, the value is now seen when these capabilities are offered in an integrated fashion in the context of business entities.  An emerging area of focus is around employee activities related to their organization on external social platforms, implicitly creating external communities with employees acting on behalf of the brand and interacting with each other (e.g. Twitter).  Companies have reached a formal understand that this now established communication medium requires strategies allowing employees to engage.  I have personally met colleagues from Oracle, like Oracle User Experience Director Ultan O'Broin (@ultan), via Twitter before meeting first through internal channels. Q: Employee engagement is important, but what about engaging customers and partners? A: The last few years we have witnessed an interesting evolution from the novelty of self-service to expectations of "intelligent" self-service.  From a consumer standpoint, engagement can end up being a key differentiator, especially in mature markets.  Customers that perform some level of interaction with a brand develop greater affinity for the brand and have a greater probability of acting as an advocate.  As organizations move toward a model of deeper engagement, they must ensure that their business is positioned to support deeper relationships, offering potentially greater transparency. From a partner standpoint greater engagement can lead to new types of business opportunities, much in the way that Amazon.com offers a unified shopping experience that can potentially span various vendors.  This same model can be extended to blending services and product delivery models, based on a closeness not easily possible before increased capability of engagement mechanisms. Q: What types of solutions are available to successfully deliver employee engagement? A: Solutions enabling higher levels of engagement do so on the basis of relevancy.  This relevancy is generally supported by aspects of content management, social collaboration, business intelligence, portal and process management technologies.  These technologies can help deliver an experience tailored to a given role or process within an organization that applies equally to work that is structured or unstructured, appearing in the form of functionality as simple as an online employee directory search, knowledge communities supported by social collaboration, as well as more feature rich business intelligence dashboards and portals. Looking to learn more about how to effectively engage your employees? Check out this webcast, or read more from John Brunswick. 

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  • I don't have permission to access other drives

    - by mcjohnalds45
    After messing with the user accounts & names, I found I can't access my external drives without using sudo. So when I access one normally with cd "/media/john/FreeAgent Drive" I receive bash: cd: /media/john/FreeAgent Drive: Permission denied However, using sudo: sudo cd /media/john sudo ls -l It gives: drwx------ 1 john john 20480 Sep 24 10:45 FreeAgent Drive/ And id returns uid=1003(john) gid=1003(john) groups=1003(john), ... So I'm interpreting this is as "you are john, only john can access this drive, however, you cannot access this drive." I have tried sudo chown john:john "FreeAgent Drive" and sudo chmod o+rw "john/FreeAgent Drive"but I still can't access it.

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  • Does RVM "failover" to another ruby instance on error?

    - by JohnMetta
    Have a strange problem in that I have a Rake task that seems to be using multiple versions of Ruby. When one fails, it seems to try another one. Details MacBook running 10.6.5 rvm 1.1.0 Rubies: 1.8.7-p302, ree-1.8.7-2010.02, ruby-1.9.2-p0 Rake 0.8.7 Gem 1.3.7 Veewee (provisioning Virtual Machines using Opcode.com, Vagrant and Chef) I'm not entirely sure the specific details of the error matter, but since it might be an issue with Veewee itself. So, what I'm trying to do is build a new box base on a veewee definition. The command fails with an error about a missing method- but what's interesting is how it fails. Errors I managed to figure out that if I only have one Ruby installed with RVM, it just fails. If I have more than one Ruby install, it fails at the same place, but execution seems to continue in another interpreter. Here are two different clipped console outputs. I've clipped them for size. The full outputs of each error are available as a gist. One Ruby version installed Here is the command run when I only have a single version of Ruby (1.8.7) available in RVM boudica:veewee john$ rvm rake build['mettabox'] --trace rvm 1.1.0 by Wayne E. Seguin ([email protected]) [http://rvm.beginrescueend.com/] (in /Users/john/Work/veewee) ** Invoke build (first_time) ** Execute build … creating new harddrive rake aborted! undefined method `max_vdi_size' for #<VirtualBox::SystemProperties:0x102d6af80> /Users/john/.rvm/gems/ruby-1.8.7-p302/gems/virtualbox-0.8.3/lib/virtualbox/abstract_model/dirty.rb:172:in `method_missing' <------ stacktraces cut ----------> /Users/john/.rvm/gems/ruby-1.8.7-p302/gems/rake-0.8.7/bin/rake:31 /Users/john/.rvm/gems/ruby-1.8.7-p302@global/bin/rake:19:in `load' /Users/john/.rvm/gems/ruby-1.8.7-p302@global/bin/rake:19 Multiple Ruby Versions Here is the same command run with three versions of Ruby available in RVM. Prior to doing this, I used "rvm use 1.8.7." Again, I don't know how important the details of the specific errors are- what's interesting to me is that there are three separate errors- each with it's own stacktrace- and each in a different Ruby interpreter. Look at the bottom of each stacktrace and you'll see that they are all sourced from different interpreter locations- First ree-1.8.7, then ruby-1.8.7, then ruby-1.9.2: boudica:veewee john$ rvm rake build['mettabox'] --trace rvm 1.1.0 by Wayne E. Seguin ([email protected]) [http://rvm.beginrescueend.com/] (in /Users/john/Work/veewee) ** Invoke build (first_time) ** Execute build … creating new harddrive rake aborted! undefined method `max_vdi_size' for #<VirtualBox::SystemProperties:0x1059dd608> /Users/john/.rvm/gems/ree-1.8.7-2010.02/gems/virtualbox-0.8.3/lib/virtualbox/abstract_model/dirty.rb:172:in `method_missing' … /Users/john/.rvm/gems/ree-1.8.7-2010.02/gems/rake-0.8.7/bin/rake:31 /Users/john/.rvm/gems/ree-1.8.7-2010.02@global/bin/rake:19:in `load' /Users/john/.rvm/gems/ree-1.8.7-2010.02@global/bin/rake:19 (in /Users/john/Work/veewee) ** Invoke build (first_time) ** Execute build isofile ubuntu-10.04.1-server-amd64.iso is available ["a1b857f92eecaf9f0a31ecfc39dee906", "30b5c6fdddbfe7b397fe506400be698d"] [] Last good state: -1 Current step: 0 last good state -1 destroying machine+disks (re-)executing step 0-initial-a1b857f92eecaf9f0a31ecfc39dee906 VBoxManage: error: Machine settings file '/Users/john/VirtualBox VMs/mettabox/mettabox.vbox' already exists VBoxManage: error: Details: code VBOX_E_FILE_ERROR (0x80bb0004), component Machine, interface IMachine, callee nsISupports Context: "CreateMachine(bstrSettingsFile.raw(), name.raw(), osTypeId.raw(), Guid(id).toUtf16().raw(), FALSE , machine.asOutParam())" at line 247 of file VBoxManageMisc.cpp rake aborted! undefined method `memory_size=' for nil:NilClass /Users/john/Work/veewee/lib/veewee/session.rb:303:in `create_vm' /Users/john/Work/veewee/lib/veewee/session.rb:166:in `build' /Users/john/Work/veewee/lib/veewee/session.rb:560:in `transaction' /Users/john/Work/veewee/lib/veewee/session.rb:163:in `build' /Users/john/Work/veewee/Rakefile:87 /Users/john/.rvm/gems/ruby-1.8.7-p302/gems/rake-0.8.7/lib/rake.rb:636:in `call' /Users/john/.rvm/gems/ruby-1.8.7-p302/gems/rake-0.8.7/lib/rake.rb:636:in `execute' /Users/john/.rvm/gems/ruby-1.8.7-p302/gems/rake-0.8.7/lib/rake.rb:631:in `each' … /Users/john/.rvm/gems/ruby-1.8.7-p302/gems/rake-0.8.7/bin/rake:31 /Users/john/.rvm/gems/ruby-1.8.7-p302@global/bin/rake:19:in `load' /Users/john/.rvm/gems/ruby-1.8.7-p302@global/bin/rake:19 (in /Users/john/Work/veewee) ** Invoke build (first_time) ** Execute build isofile ubuntu-10.04.1-server-amd64.iso is available ["a9c4ab3257e1da3479c984eae9905c2a", "30b5c6fdddbfe7b397fe506400be698d"] [] Last good state: -1 Current step: 0 last good state -1 (re-)executing step 0-initial-a9c4ab3257e1da3479c984eae9905c2a VBoxManage: error: Machine settings file '/Users/john/VirtualBox VMs/mettabox/mettabox.vbox' already exists VBoxManage: error: Details: code VBOX_E_FILE_ERROR (0x80bb0004), component Machine, interface IMachine, callee nsISupports Context: "CreateMachine(bstrSettingsFile.raw(), name.raw(), osTypeId.raw(), Guid(id).toUtf16().raw(), FALSE , machine.asOutParam())" at line 247 of file VBoxManageMisc.cpp rake aborted! undefined method `memory_size=' for nil:NilClass /Users/john/Work/veewee/lib/veewee/session.rb:303:in `create_vm' /Users/john/Work/veewee/lib/veewee/session.rb:166:in `block in build' /Users/john/Work/veewee/lib/veewee/session.rb:560:in `transaction' /Users/john/Work/veewee/lib/veewee/session.rb:163:in `build' /Users/john/Work/veewee/Rakefile:87:in `block in <top (required)>' /Users/john/.rvm/rubies/ruby-1.9.2-p0/lib/ruby/1.9.1/rake.rb:634:in `call' /Users/john/.rvm/rubies/ruby-1.9.2-p0/lib/ruby/1.9.1/rake.rb:634:in `block in execute' … /Users/john/.rvm/rubies/ruby-1.9.2-p0/lib/ruby/1.9.1/rake.rb:2013:in `top_level' /Users/john/.rvm/rubies/ruby-1.9.2-p0/lib/ruby/1.9.1/rake.rb:1992:in `run' /Users/john/.rvm/rubies/ruby-1.9.2-p0/bin/rake:35:in `<main>' It isn't until we reach the last installed version of Ruby that execution halts. Discussion Does anyone have any idea what's going on here? Has anyone seen this "failover"-like behavior before? It seems strange to me that the first exception would not halt execution as it did with one interpreter, but I wonder if there are things happening when RVM is installed that we Ruby developers are not considering.

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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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  • Raycasting tutorial / vector math question

    - by mattboy
    I'm checking out this nice raycasting tutorial at http://lodev.org/cgtutor/raycasting.html and have a probably very simple math question. In the DDA algorithm I'm having trouble understanding the calcuation of the deltaDistX and deltaDistY variables, which are the distances that the ray has to travel from 1 x-side to the next x-side, or from 1 y-side to the next y-side, in the square grid that makes up the world map (see below screenshot). In the tutorial they are calculated as follows, but without much explanation: //length of ray from one x or y-side to next x or y-side double deltaDistX = sqrt(1 + (rayDirY * rayDirY) / (rayDirX * rayDirX)); double deltaDistY = sqrt(1 + (rayDirX * rayDirX) / (rayDirY * rayDirY)); rayDirY and rayDirX are the direction of a ray that has been cast. How do you get these formulas? It looks like pythagorean theorem is part of it, but somehow there's division involved here. Can anyone clue me in as to what mathematical knowledge I'm missing here, or "prove" the formula by showing how it's derived?

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  • Wavefront mesh: determine which face a point belongs to?

    - by Mina Samy
    I have a 3D mesh Wavefront .obj file. Is there any algorithm that takes an arbitrary point coordinates as input and determines which face of the mesh that point belongs to ?? The mesh is rendered on the screen, then the user clicks on it, I want to determine which part of the mesh the user has clicked on ? Here's the code using LibGDX: Vector3 intersection=new Vector3(); Ray ray=camera.getPickRay(x, y); //vertices is an array that hold the coordinates of the mesh boolean ok=Intersector.intersectRayTriangles(ray, vertices, intersection); Thanks

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  • Web Experience Management: Segmentation & Targeting - Chalk Talk with John

    - by Michael Snow
    Today's post comes from our WebCenter friend, John Brunswick.  Normal 0 false false false EN-US X-NONE X-NONE MicrosoftInternetExplorer4 /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-priority:99; mso-style-qformat:yes; mso-style-parent:""; mso-padding-alt:0in 5.4pt 0in 5.4pt; mso-para-margin:0in; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:11.0pt; font-family:"Calibri","sans-serif"; mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin; mso-fareast-font-family:"Times New Roman"; mso-fareast-theme-font:minor-fareast; mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin; mso-bidi-font-family:"Times New Roman"; mso-bidi-theme-font:minor-bidi;} Having trouble getting your arms around the differences between Web Content Management (WCM) and Web Experience Management (WEM)?  Told through story, the video below outlines the differences in an easy to understand manner. By following the journey of Mr. and Mrs. Smith on their adventure to find the best amusement park in two neighboring towns, we can clearly see what an impact context and relevancy play in our decision making within online channels.  Just as when we search to connect with the best products and services for our needs, the Smiths have their grandchildren coming to visit next week and finding the best park is essential to guarantee a great family vacation.  One town effectively Segments and Targets visitors to enhance their experience, reducing the effort needed to learn about their park. Have a look below to join the Smiths in their search.    Learn MORE about how you might measure up: Deliver Engaging Digital Experiences Drive Digital Marketing SuccessAccess Free Assessment Tool

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  • Ray-box Intersection Theory

    - by Myx
    Hello: I wish to determine the intersection point between a ray and a box. The box is defined by its min 3D coordinate and max 3D coordinate and the ray is defined by its origin and the direction to which it points. Currently, I am forming a plane for each face of the box and I'm intersecting the ray with the plane. If the ray intersects the plane, then I check whether or not the intersection point is actually on the surface of the box. If so, I check whether it is the closest intersection for this ray and I return the closest intersection. The way I check whether the plane-intersection point is on the box surface itself is through a function bool PointOnBoxFace(R3Point point, R3Point corner1, R3Point corner2) { double min_x = min(corner1.X(), corner2.X()); double max_x = max(corner1.X(), corner2.X()); double min_y = min(corner1.Y(), corner2.Y()); double max_y = max(corner1.Y(), corner2.Y()); double min_z = min(corner1.Z(), corner2.Z()); double max_z = max(corner1.Z(), corner2.Z()); if(point.X() >= min_x && point.X() <= max_x && point.Y() >= min_y && point.Y() <= max_y && point.Z() >= min_z && point.Z() <= max_z) return true; return false; } where corner1 is one corner of the rectangle for that box face and corner2 is the opposite corner. My implementation works most of the time but sometimes it gives me the wrong intersection. I was wondering if the way I'm checking whether the intersection point is on the box is correct or if I should use some other algorithm. Thanks.

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  • What can I do with Java for Blu Ray or BD-J?

    - by Jay Askren
    I have a Blu Ray player which can connect to the internet to play media from netflix and youtube. I am intrigued by the possibilities of BD-J and wondering just how far the technology can be taken. For instance: Could I write a twitter, facebook, rss reader, or email client? Can I write a game which would allows people to play each other over the web from their own tv? Could I write a DVR app which stored tv shows on the thumbdrive plugged into the player. Can I run my applications from a thumbdrive or do I need to put them on a Blu Ray disk? Does anyone have real experience with BD-J? How do you like it as a development platform? How would you recommend getting started? Can I develop in BD-J using open source tools like Eclipse, Maven, etc...

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  • Ray-triangle intersetion

    - by gamemaker
    Hello! How can I test intersesion ray and triangle, and if it exist how to get distance from ray origin to intersection point?? What optimization I can use, if in my program I've got to check 1 ray to ~10000 triangles ??

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  • How can I draw an arrow at the edge of the screen pointing to an object that is off screen?

    - by Adam Henderson
    I am wishing to do what is described in this topic: http://www.allegro.cc/forums/print-thread/283220 I have attempted a variety of the methods mentioned here. First I tried to use the method described by Carrus85: Just take the ratio of the two triangle hypontenuses (doesn't matter which triagle you use for the other, I suggest point 1 and point 2 as the distance you calculate). This will give you the aspect ratio percentage of the triangle in the corner from the larger triangle. Then you simply multiply deltax by that value to get the x-coordinate offset, and deltay by that value to get the y-coordinate offset. But I could not find a way to calculate how far the object is away from the edge of the screen. I then tried using ray casting (which I have never done before) suggested by 23yrold3yrold: Fire a ray from the center of the screen to the offscreen object. Calculate where on the rectangle the ray intersects. There's your coordinates. I first calculated the hypotenuse of the triangle formed by the difference in x and y positions of the two points. I used this to create a unit vector along that line. I looped through that vector until either the x coordinate or the y coordinate was off the screen. The two current x and y values then form the x and y of the arrow. Here is the code for my ray casting method (written in C++ and Allegro 5) void renderArrows(Object* i) { float x1 = i->getX() + (i->getWidth() / 2); float y1 = i->getY() + (i->getHeight() / 2); float x2 = screenCentreX; float y2 = ScreenCentreY; float dx = x2 - x1; float dy = y2 - y1; float hypotSquared = (dx * dx) + (dy * dy); float hypot = sqrt(hypotSquared); float unitX = dx / hypot; float unitY = dy / hypot; float rayX = x2 - view->getViewportX(); float rayY = y2 - view->getViewportY(); float arrowX = 0; float arrowY = 0; bool posFound = false; while(posFound == false) { rayX += unitX; rayY += unitY; if(rayX <= 0 || rayX >= screenWidth || rayY <= 0 || rayY >= screenHeight) { arrowX = rayX; arrowY = rayY; posFound = true; } } al_draw_bitmap(sprite, arrowX - spriteWidth, arrowY - spriteHeight, 0); } This was relatively successful. Arrows are displayed in the bottom right section of the screen when objects are located above and left of the screen as if the locations of the where the arrows are drawn have been rotated 180 degrees around the center of the screen. I assumed this was due to the fact that when I was calculating the hypotenuse of the triangle, it would always be positive regardless of whether or not the difference in x or difference in y is negative. Thinking about it, ray casting does not seem like a good way of solving the problem (due to the fact that it involves using sqrt() and a large for loop). Any help finding a suitable solution would be greatly appreciated, Thanks Adam

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  • How do I make my blu-ray/hd-dvd/dvd drive multi-region?

    - by jammus
    I have a combined dvd/hd-dvd/bd drive on my machine (yeah, I'm very much the big man) which is all great and everything. However, I've got quite a few region 1 DVDs (on import, yeah?) that it won't play. I know I can switch my drive's region but this is limited to 5 times only. Is there a way I can make the drive permanently multi-region or lift this limit, the same way you can with most DVD players?

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  • are blue ray disks the cheapest storage medium per Gb?

    - by oshirowanen
    The question is as simple as that really. Are bluray disks the cheapest storage medium per gb? I am recording video which is using about 32gb per day. So a month of that would be almost 1Tb. A year around 12tb. I want to store at last a years worth with the possibility of more if needed. To me it seems that cheap bluray disks world be the cheapest solution. But I wanted to get this confirmed.

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  • Blender/POV-Ray differences

    - by Magnap
    I am currently following a tutorial for Blender, but came across POV-Ray mentioned as a renderer. After having researched it a bit, i took a look at it's scene description (scripting) language, which kind of fascinated me. But, after even more googling of the topic, i am still wondering: What are the main and key differences between working with 3D in resp. Blender and POV-Ray? PS: I suspect this might not be the best place for a question such as this, but it seems to be the most suitable.

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  • blu-ray archiving in vmware ESXi 4

    - by spacecadet77
    Hi, I need some advice about using blu-ray writer for archiving data on vmware ESXi 4. At office we have IBM System x3400 Tower server with ESXi 4 hipervisor and OpenSuse and CentOS GNU/Linux system as guests. Will blu-ray writer work in this setup, and if it will is there any particular model you can suggest. Best regards IBM System x3400 Tower server specification: 1x Intel Quad-Core Xeon E5410 2.33GHz/ 12MB/ 1333MHz (2x CPU max) Intel 5000P chipset, 2x 1GB PC2-5300 DDR2 667MHz SDRAM ECC Chipkill (32GB max) 2x4GB (2x2GB) PC2-5300 CL5 ECC DDR2 FBDIMM (x3400, x3550, x3650) SAS/SATA Hot-Swap Open Bay (0xHDD std, 4xHDD max, 8xHDD optional) ServeRAID 8K dual channel SAS/SATA controller (RAID 0,1,1E,10,5,6, 256MB, Battery Backup) Graphics ATI® RN50(ES1000) 16MB DDR, CD-RW/DVD Combo no FDD GigaEthernet, Tower with Power Supply 835W (opt Redudant) Slot 1: half-length, PCI-Express x8(x4 electrical) Slot 2: full, PCI-Express x8 Slot 3: full, PCI-Express x8 Slot 4: full, 64-bit 133MHz 3.3v PCI-X Slot 5: full, 64-bit 133MHz 3.3v PCI-X , Slot 6: half-length, 32-bit 33MHz 5.0v PCI ports: 4x USB (Vers 2.0), 2x PS/2, parallel, 2x serial (9-pin), VGA, RJ-45 (ethernet ), RJ-45 (sys mgm) HDD 4 x TB 7200rpm / Serial ATA II 3.0Gb/s / 16MB, RoHS

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  • Installing XAMPP in Xubuntu 13.10

    - by illage2
    I downloaded the XAMPP .run file from Apacheandfriends but the installation isn't working for me. I can't seem to navigate to my downloads folder and it just keeps saying command not found all the time. root@john-Aspire-V3-531:/home/john# cd ~/downloads bash: cd: /root/downloads: No such file or directory root@john-Aspire-V3-531:/home/john# cd ~/Downloads bash: cd: /root/Downloads: No such file or directory root@john-Aspire-V3-531:/home/john# /downloads bash: /downloads: No such file or directory root@john-Aspire-V3-531:/home/john# cd /downloads bash: cd: /downloads: No such file or directory root@john-Aspire-V3-531:/home/john# cd downloads bash: cd: downloads: No such file or directory root@john-Aspire-V3-531:/home/john# downloads downloads: command not found What do I need to do? Apacheandfriends says to: chmod 755 xampp-linux-1.8.2-0-installer.run and then ./xampp-linux-1.8.2-0-installer.run but it doesn't seem to think that the file exists. Can anyone help me?

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  • Spotlight on Claims: Serving Customers Under Extreme Conditions

    - by [email protected]
    Oracle Insurance's director of marketing for EMEA, John Sinclair, recently attended the CII Spotlight on Claims event in London. Bad weather and its implications for the insurance industry have become very topical as the frequency and diversity of natural disasters - including rains, wind and snow - has surged across Europe this winter. On England's wettest day on record, the county of Cumbria was flooded with 12 inches of rain within 24 hours. Freezing temperatures wreaked havoc on European travel, causing high speed TVG trains to break down and stranding hundreds of passengers under the English Chanel in a tunnel all night long without heat or electricity. A storm named Xynthia thrashed France and surrounding countries with hurricane force, flooding ports and killing 51 people. After the Spring Equinox, insurers may have thought the worst had past. Then came along Eyjafjallajökull, spewing out vast quantities of volcanic ash in what is turning out to be one of most costly natural disasters in history. Such extreme events challenge insurance companies' ability to service their customers just when customers need their help most. When you add economic downturn and competitive pressures to the mix, insurers are further stretched and required to continually learn and innovate to meet high customer expectations with reduced budgets. These and other issues were hot topics of discussion at the recent "Spotlight on Claims" seminar in London, focused on how weather is affecting claims and the insurance industry. The event was organized by the CII (Chartered Insurance Institute), a group with 90,000 members. CII has been at the forefront in setting professional standards for the insurance industry for over a century. Insurers came to the conference to hear how they could better serve their customers under extreme weather conditions, learn from the experience of their peers, and hear about technological breakthroughs in climate modeling, geographic intelligence and IT. Customer case studies at the conference highlighted the importance of effective and constant communication in handling the overflow of catastrophe related claims. First and foremost is the need to rapidly establish initial communication with claimants to build their confidence in a positive outcome. Ongoing communication then needs to be continued throughout the claims cycle to mange expectations and maintain ownership of the process from start to finish. Strong internal communication to support frontline staff was also deemed critical to successful crisis management, as was communication with the broader insurance ecosystem to tap into extended resources and business intelligence. Advances in technology - such web based systems to access policies and enter first notice of loss in the field - as well as customer-focused self-service portals and multichannel alerts, are instrumental in improving customer satisfaction and helping insurers to deal with the claims surge, which often can reach four or more times normal workloads. Dynamic models of the global climate system can now be used to better understand weather-related risks, and as these models mature it is hoped that they will soon become more accurate in predicting the timing of catastrophic events. Geographic intelligence is also being used within a claims environment to better assess loss reserves and detect fraud. Despite these advances in dealing with catastrophes and predicting their occurrence, there will never be a substitute for qualified front line staff to deal with customers. In light of pressures to streamline efficiency, there was debate as to whether outsourcing was the solution, or whether it was better to build on the people you have. In the final analysis, nearly everybody agreed that in the future insurance companies would have to work better and smarter to keep on top. An appeal was also made for greater collaboration amongst industry participants in dealing with the extreme conditions and systematic stress brought on by natural disasters. It was pointed out that the public oftentimes judged the industry as a whole rather than the individual carriers when it comes to freakish events, and that all would benefit at such times from the pooling of limited resources and professional skills rather than competing in silos for competitive advantage - especially the end customer. One case study that stood out was on how The Motorists Insurance Group was able to power through one of the most devastating catastrophes in recent years - Hurricane Ike. The keys to Motorists' success were superior people, processes and technology. They did a lot of upfront planning and invested in their people, creating a healthy team environment that delivered "max service" even when they were experiencing the same level of devastation as the rest of the population. Processes were rapidly adapted to meet the challenge of the catastrophe and continually adapted to Ike's specific conditions as they evolved. Technology was fundamental to the execution of their strategy, enabling them anywhere access, on the fly reassigning of resources and rapid training to augment the work force. You can learn more about the Motorists experience by watching this video. John Sinclair is marketing director for Oracle Insurance in EMEA. He has more than 20 years of experience in insurance and financial services.

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