Search Results

Search found 3392 results on 136 pages for 'average joe'.

Page 5/136 | < Previous Page | 1 2 3 4 5 6 7 8 9 10 11 12  | Next Page >

  • Sum and average over null values SQL Server 2008 Analysis Services

    - by Jonathan
    I have a simple problem, I think, but I have googled and can't find the solution. I have a cube that has MeasureA, MeasureB and MeasureC. Not all three measures have values for each record, sometimes they can be null, it's depending if it was applicable. Now for my totals, I need to average but the average must not take nulls into account. Any help will be much appreciated. When I view the measures, the null values show as zeros.

    Read the article

  • High load average due to high system cpu load (%sys)

    - by Nick
    We have server with high traffic website. Recently we moved from 2 x 4 core server (8 cores in /proc/cpuinfo), 32 GB RAM, running CentOS 5.x, to 2 x 4 core server (16 cores in /proc/cpuinfo), 32 GB RAM, running CentOS 6.3 Server running nginx as a proxy, mysql server and sphinx-search. Traffic is high, but mysql and sphinx-search databases are relatively small, and usually everything works blazing fast. Today server experienced load average of 100++. Looking at top and sar, we noticed that (%sys) is very high - 50 to 70%. Disk utilization was less 1%. We tried to reboot, but problem existed after the reboot. At any moment server had at least 3-4 GB free RAM. Only message shown by dmesg was "possible SYN flooding on port 80. Sending cookies.". Here is snippet of sar 11:00:01 CPU %user %nice %system %iowait %steal %idle 11:10:01 all 21.60 0.00 66.38 0.03 0.00 11.99 We know that this is traffic issue, but we do not know how to proceed future and where to check for solution. Is there a way we can find where exactly those "66.38%" are used. Any suggestions would be appreciated.

    Read the article

  • Average performance of binary search algorithm?

    - by Passionate Learner
    http://en.wikipedia.org/wiki/Binary_search_algorithm#Average_performance BinarySearch(int A[], int value, int low, int high) { int mid; if (high < low) return -1; mid = (low + high) / 2; if (A[mid] > value) return BinarySearch(A, value, low, mid-1); else if (A[mid] < value) return BinarySearch(A, value, mid+1, high); else return mid; } If the integer I'm trying to find is always in the array, can anyone help me write a program that can calculate the average performance of binary search algorithm? I know I can do this by actually running the program and counting the number of calls, but what I'm trying to do here is to do it without calling the function. I'm not asking for a time complexity, I'm trying to calculate the average number of calls. For example, the average number of calls to find a integer in A[2], it would be 1.67 (5/3).

    Read the article

  • how to Compute the average probe length for success and failure - Linear probe (Hash Tables)

    - by fang_dejavu
    hi everyone, I'm doing an assignment for my Data Structures class. we were asked to to study linear probing with load factors of .1, .2 , .3, ...., and .9. The formula for testing is: The average probe length using linear probing is roughly Success-- ( 1 + 1/(1-L)**2)/2 or Failure-- (1+1(1-L))/2. we are required to find the theoretical using the formula above which I did(just plug the load factor in the formula), then we have to calculate the empirical (which I not quite sure how to do). here is the rest of the requirements **For each load factor, 10,000 randomly generated positive ints between 1 and 50000 (inclusive) will be inserted into a table of the "right" size, where "right" is strictly based upon the load factor you are testing. Repeats are allowed. Be sure that your formula for randomly generated ints is correct. There is a class called Random in java.util. USE it! After a table of the right (based upon L) size is loaded with 10,000 ints, do 100 searches of newly generated random ints from the range of 1 to 50000. Compute the average probe length for each of the two formulas and indicate the denominators used in each calculationSo, for example, each test for a .5 load would have a table of size approximately 20,000 (adjusted to be prime) and similarly each test for a .9 load would have a table of approximate size 10,000/.9 (again adjusted to be prime). The program should run displaying the various load factors tested, the average probe for each search (the two denominators used to compute the averages will add to 100), and the theoretical answers using the formula above. .** how do I calculate the empirical success?

    Read the article

  • Total Average Week using a Parameter

    - by Jose
    I have a crystal report that shows sales volumes called week to date volume. It shows current week, previous week, and average week. The report prompts for a date parameter and I extract the week number to get current week and previous week volumes. Did it this way because Mngmt wants to be able to run report whenever. My problem is for Average Week I cant figure out how to get the number of weeks to divide by for my average. Report originates from June 1st, 2010. Right now I have: DATEPART("ww", {?date}) - DATEPART("ww", DATE(2010, 6, 1)) This returns 2 right now which is perfect, so i divide my total by 2. This code will work until the end of the year then I'm hooped. Any idea how I can make this a little more dynamic. I was thinking a counter somehow, just can't get the logic down because the date parameter will keep changing, meaning I cant increase my counter by 1 after each week??? Cheers.

    Read the article

  • How can I determine which processes are using up the most CPU?

    - by Rob
    05:54:17 up 6 days, 9:54, 1 user, load average: 15.70, 8.04, 4.56 Load average is a LOT higher than it should be. It was HALF of that a moment ago, I used ps aux to see which processes were using a lot of cpu, and the httpd was using a TON. It had several processes running with 0.9 each. So I restarted the httpd, and now those processes are gone, but the load averages have doubled. So my question is: is there a way I can list the processes that are consuming the most cpu?

    Read the article

  • About Load average in htop, how to decide if it's still doing ok?

    - by Joe Huang
    I use 'htop' to monitor my web server. It's recently quite loaded and the Load average is showing something like this: Load average: 3.10 2.56 1.63 I searched the web about these numbers and I found an article about it: http://blog.scoutapp.com/articles/2009/07/31/understanding-load-averages In the article, it says if I have 2 CPUs, 2.0 means 100% CPU utilization. And my VPS has two CPUs, so what does 3.1 mean? How could it exceed 100% CPU utilization? And from these numbers, does it mean I should be wary about the loading now? But the performance seems totally fine, and this is a managed VPS, the hosting company has not notified me any warning about it. During day time, Load average always show these high numbers... here is another snapshot while writing. Load average: 3.03 2.77 1.97 Load average: 0.41 1.29 1.60 <---- 5 more minutes later So I am wondering how much room left for this site to grow in current configurations? What kind of proactive actions I should take in advance? I don't want to wait until the server bursts. Thanks.

    Read the article

  • How could Load average numbers from 'htop' exceed 100% CPU utlization?

    - by Joe Huang
    I use 'htop' to monitor my web server. It's recently quite loaded and the Load average is showing something like this: Load average: 3.10 2.56 1.63 I searched the web about these numbers and I found an article about it: http://blog.scoutapp.com/articles/2009/07/31/understanding-load-averages In the article, it says if I have 2 CPUs, 2.0 means 100% CPU utilization. And my VPS has two CPUs, so what does 3.1 mean? How could it exceed 100% CPU utilization? And from these numbers, does it mean I should be wary about the loading now? But the performance seems totally fine, and this is a managed VPS, the hosting company has not notified me any warning about it. During day time, Load average always show these high numbers... here is another snapshot while writing. Load average: 3.03 2.77 1.97 Load average: 0.41 1.29 1.60 <---- 5 more minutes later So I am wondering how much room left for this site to grow in current configurations? What kind of proactive actions I should take in advance? I don't want to wait until the server bursts. Thanks.

    Read the article

  • high load average, high wait, dmesg raid error messages (debian nfs server)

    - by John Stumbles
    Debian 6 on HP proliant (2 CPU) with raid (2*1.5T RAID1 + 2*2T RAID1 joined RAID0 to make 3.5T) running mainly nfs & imapd (plus samba for windows share & local www for previewing web pages); with local ubuntu desktop client mounting $HOME, laptops accessing imap & odd files (e.g. videos) via nfs/smb; boxes connected 100baseT or wifi via home router/switch uname -a Linux prole 2.6.32-5-686 #1 SMP Wed Jan 11 12:29:30 UTC 2012 i686 GNU/Linux Setup has been working for months but prone to intermittently going very slow (user experience on desktop mounting $HOME from server, or laptop playing videos) and now consistently so bad I've had to delve into it to try to find what's wrong(!) Server seems OK at low load e.g. (laptop) client (with $HOME on local disk) connecting to server's imapd and nfs mounting RAID to access 1 file: top shows load ~ 0.1 or less, 0 wait but when (desktop) client mounts $HOME and starts user KDE session (all accessing server) then top shows e.g. top - 13:41:17 up 3:43, 3 users, load average: 9.29, 9.55, 8.27 Tasks: 158 total, 1 running, 157 sleeping, 0 stopped, 0 zombie Cpu(s): 0.4%us, 0.4%sy, 0.0%ni, 49.0%id, 49.7%wa, 0.0%hi, 0.5%si, 0.0%st Mem: 903856k total, 851784k used, 52072k free, 171152k buffers Swap: 0k total, 0k used, 0k free, 476896k cached PID USER PR NI VIRT RES SHR S %CPU %MEM TIME+ COMMAND 3935 root 20 0 2456 1088 784 R 2 0.1 0:00.02 top 1 root 20 0 2028 680 584 S 0 0.1 0:01.14 init 2 root 20 0 0 0 0 S 0 0.0 0:00.00 kthreadd 3 root RT 0 0 0 0 S 0 0.0 0:00.00 migration/0 4 root 20 0 0 0 0 S 0 0.0 0:00.12 ksoftirqd/0 5 root RT 0 0 0 0 S 0 0.0 0:00.00 watchdog/0 6 root RT 0 0 0 0 S 0 0.0 0:00.00 migration/1 7 root 20 0 0 0 0 S 0 0.0 0:00.16 ksoftirqd/1 8 root RT 0 0 0 0 S 0 0.0 0:00.00 watchdog/1 9 root 20 0 0 0 0 S 0 0.0 0:00.42 events/0 10 root 20 0 0 0 0 S 0 0.0 0:02.26 events/1 11 root 20 0 0 0 0 S 0 0.0 0:00.00 cpuset 12 root 20 0 0 0 0 S 0 0.0 0:00.00 khelper 13 root 20 0 0 0 0 S 0 0.0 0:00.00 netns 14 root 20 0 0 0 0 S 0 0.0 0:00.00 async/mgr 15 root 20 0 0 0 0 S 0 0.0 0:00.00 pm 16 root 20 0 0 0 0 S 0 0.0 0:00.02 sync_supers 17 root 20 0 0 0 0 S 0 0.0 0:00.02 bdi-default 18 root 20 0 0 0 0 S 0 0.0 0:00.00 kintegrityd/0 19 root 20 0 0 0 0 S 0 0.0 0:00.00 kintegrityd/1 20 root 20 0 0 0 0 S 0 0.0 0:00.02 kblockd/0 21 root 20 0 0 0 0 S 0 0.0 0:00.08 kblockd/1 22 root 20 0 0 0 0 S 0 0.0 0:00.00 kacpid 23 root 20 0 0 0 0 S 0 0.0 0:00.00 kacpi_notify 24 root 20 0 0 0 0 S 0 0.0 0:00.00 kacpi_hotplug 25 root 20 0 0 0 0 S 0 0.0 0:00.00 kseriod 28 root 20 0 0 0 0 S 0 0.0 0:04.19 kondemand/0 29 root 20 0 0 0 0 S 0 0.0 0:02.93 kondemand/1 30 root 20 0 0 0 0 S 0 0.0 0:00.00 khungtaskd 31 root 20 0 0 0 0 S 0 0.0 0:00.18 kswapd0 32 root 25 5 0 0 0 S 0 0.0 0:00.00 ksmd 33 root 20 0 0 0 0 S 0 0.0 0:00.00 aio/0 34 root 20 0 0 0 0 S 0 0.0 0:00.00 aio/1 35 root 20 0 0 0 0 S 0 0.0 0:00.00 crypto/0 36 root 20 0 0 0 0 S 0 0.0 0:00.00 crypto/1 203 root 20 0 0 0 0 S 0 0.0 0:00.00 ksuspend_usbd 204 root 20 0 0 0 0 S 0 0.0 0:00.00 khubd 205 root 20 0 0 0 0 S 0 0.0 0:00.00 ata/0 206 root 20 0 0 0 0 S 0 0.0 0:00.00 ata/1 207 root 20 0 0 0 0 S 0 0.0 0:00.14 ata_aux 208 root 20 0 0 0 0 S 0 0.0 0:00.01 scsi_eh_0 dmesg suggests there's a disk problem: .............. (previous episode) [13276.966004] raid1:md0: read error corrected (8 sectors at 489900360 on sdc7) [13276.966043] raid1: sdb7: redirecting sector 489898312 to another mirror [13279.569186] ata4.00: exception Emask 0x0 SAct 0x1 SErr 0x0 action 0x0 [13279.569211] ata4.00: irq_stat 0x40000008 [13279.569230] ata4.00: failed command: READ FPDMA QUEUED [13279.569257] ata4.00: cmd 60/08:00:00:6a:05/00:00:23:00:00/40 tag 0 ncq 4096 in [13279.569262] res 41/40:00:05:6a:05/00:00:23:00:00/40 Emask 0x409 (media error) <F> [13279.569306] ata4.00: status: { DRDY ERR } [13279.569321] ata4.00: error: { UNC } [13279.575362] ata4.00: configured for UDMA/133 [13279.575388] ata4: EH complete [13283.169224] ata4.00: exception Emask 0x0 SAct 0x1 SErr 0x0 action 0x0 [13283.169246] ata4.00: irq_stat 0x40000008 [13283.169263] ata4.00: failed command: READ FPDMA QUEUED [13283.169289] ata4.00: cmd 60/08:00:00:6a:05/00:00:23:00:00/40 tag 0 ncq 4096 in [13283.169294] res 41/40:00:07:6a:05/00:00:23:00:00/40 Emask 0x409 (media error) <F> [13283.169331] ata4.00: status: { DRDY ERR } [13283.169345] ata4.00: error: { UNC } [13283.176071] ata4.00: configured for UDMA/133 [13283.176104] ata4: EH complete [13286.224814] ata4.00: exception Emask 0x0 SAct 0x1 SErr 0x0 action 0x0 [13286.224837] ata4.00: irq_stat 0x40000008 [13286.224853] ata4.00: failed command: READ FPDMA QUEUED [13286.224879] ata4.00: cmd 60/08:00:00:6a:05/00:00:23:00:00/40 tag 0 ncq 4096 in [13286.224884] res 41/40:00:06:6a:05/00:00:23:00:00/40 Emask 0x409 (media error) <F> [13286.224922] ata4.00: status: { DRDY ERR } [13286.224935] ata4.00: error: { UNC } [13286.231277] ata4.00: configured for UDMA/133 [13286.231303] ata4: EH complete [13288.802623] ata4.00: exception Emask 0x0 SAct 0x1 SErr 0x0 action 0x0 [13288.802646] ata4.00: irq_stat 0x40000008 [13288.802662] ata4.00: failed command: READ FPDMA QUEUED [13288.802688] ata4.00: cmd 60/08:00:00:6a:05/00:00:23:00:00/40 tag 0 ncq 4096 in [13288.802693] res 41/40:00:05:6a:05/00:00:23:00:00/40 Emask 0x409 (media error) <F> [13288.802731] ata4.00: status: { DRDY ERR } [13288.802745] ata4.00: error: { UNC } [13288.808901] ata4.00: configured for UDMA/133 [13288.808927] ata4: EH complete [13291.380430] ata4.00: exception Emask 0x0 SAct 0x1 SErr 0x0 action 0x0 [13291.380453] ata4.00: irq_stat 0x40000008 [13291.380470] ata4.00: failed command: READ FPDMA QUEUED [13291.380496] ata4.00: cmd 60/08:00:00:6a:05/00:00:23:00:00/40 tag 0 ncq 4096 in [13291.380501] res 41/40:00:05:6a:05/00:00:23:00:00/40 Emask 0x409 (media error) <F> [13291.380577] ata4.00: status: { DRDY ERR } [13291.380594] ata4.00: error: { UNC } [13291.386517] ata4.00: configured for UDMA/133 [13291.386543] ata4: EH complete [13294.347147] ata4.00: exception Emask 0x0 SAct 0x1 SErr 0x0 action 0x0 [13294.347169] ata4.00: irq_stat 0x40000008 [13294.347186] ata4.00: failed command: READ FPDMA QUEUED [13294.347211] ata4.00: cmd 60/08:00:00:6a:05/00:00:23:00:00/40 tag 0 ncq 4096 in [13294.347217] res 41/40:00:06:6a:05/00:00:23:00:00/40 Emask 0x409 (media error) <F> [13294.347254] ata4.00: status: { DRDY ERR } [13294.347268] ata4.00: error: { UNC } [13294.353556] ata4.00: configured for UDMA/133 [13294.353583] sd 3:0:0:0: [sdc] Unhandled sense code [13294.353590] sd 3:0:0:0: [sdc] Result: hostbyte=DID_OK driverbyte=DRIVER_SENSE [13294.353599] sd 3:0:0:0: [sdc] Sense Key : Medium Error [current] [descriptor] [13294.353610] Descriptor sense data with sense descriptors (in hex): [13294.353616] 72 03 11 04 00 00 00 0c 00 0a 80 00 00 00 00 00 [13294.353635] 23 05 6a 06 [13294.353644] sd 3:0:0:0: [sdc] Add. Sense: Unrecovered read error - auto reallocate failed [13294.353657] sd 3:0:0:0: [sdc] CDB: Read(10): 28 00 23 05 6a 00 00 00 08 00 [13294.353675] end_request: I/O error, dev sdc, sector 587557382 [13294.353726] ata4: EH complete [13294.366953] raid1:md0: read error corrected (8 sectors at 489900544 on sdc7) [13294.366992] raid1: sdc7: redirecting sector 489898496 to another mirror and they're happening quite frequently, which I guess is liable to account for the performance problem(?) # dmesg | grep mirror [12433.561822] raid1: sdc7: redirecting sector 489900464 to another mirror [12449.428933] raid1: sdb7: redirecting sector 489900504 to another mirror [12464.807016] raid1: sdb7: redirecting sector 489900512 to another mirror [12480.196222] raid1: sdb7: redirecting sector 489900520 to another mirror [12495.585413] raid1: sdb7: redirecting sector 489900528 to another mirror [12510.974424] raid1: sdb7: redirecting sector 489900536 to another mirror [12526.374933] raid1: sdb7: redirecting sector 489900544 to another mirror [12542.619938] raid1: sdc7: redirecting sector 489900608 to another mirror [12559.431328] raid1: sdc7: redirecting sector 489900616 to another mirror [12576.553866] raid1: sdc7: redirecting sector 489900624 to another mirror [12592.065265] raid1: sdc7: redirecting sector 489900632 to another mirror [12607.621121] raid1: sdc7: redirecting sector 489900640 to another mirror [12623.165856] raid1: sdc7: redirecting sector 489900648 to another mirror [12638.699474] raid1: sdc7: redirecting sector 489900656 to another mirror [12655.610881] raid1: sdc7: redirecting sector 489900664 to another mirror [12672.255617] raid1: sdc7: redirecting sector 489900672 to another mirror [12672.288746] raid1: sdc7: redirecting sector 489900680 to another mirror [12672.332376] raid1: sdc7: redirecting sector 489900688 to another mirror [12672.362935] raid1: sdc7: redirecting sector 489900696 to another mirror [12674.201177] raid1: sdc7: redirecting sector 489900704 to another mirror [12698.045050] raid1: sdc7: redirecting sector 489900712 to another mirror [12698.089309] raid1: sdc7: redirecting sector 489900720 to another mirror [12698.111999] raid1: sdc7: redirecting sector 489900728 to another mirror [12698.134006] raid1: sdc7: redirecting sector 489900736 to another mirror [12719.034376] raid1: sdc7: redirecting sector 489900744 to another mirror [12734.545775] raid1: sdc7: redirecting sector 489900752 to another mirror [12734.590014] raid1: sdc7: redirecting sector 489900760 to another mirror [12734.624050] raid1: sdc7: redirecting sector 489900768 to another mirror [12734.647308] raid1: sdc7: redirecting sector 489900776 to another mirror [12734.664657] raid1: sdc7: redirecting sector 489900784 to another mirror [12734.710642] raid1: sdc7: redirecting sector 489900792 to another mirror [12734.721919] raid1: sdc7: redirecting sector 489900800 to another mirror [12734.744732] raid1: sdc7: redirecting sector 489900808 to another mirror [12734.779330] raid1: sdc7: redirecting sector 489900816 to another mirror [12782.604564] raid1: sdb7: redirecting sector 1242934216 to another mirror [12798.264153] raid1: sdc7: redirecting sector 1242935080 to another mirror [13245.832193] raid1: sdb7: redirecting sector 489898296 to another mirror [13261.376929] raid1: sdb7: redirecting sector 489898304 to another mirror [13276.966043] raid1: sdb7: redirecting sector 489898312 to another mirror [13294.366992] raid1: sdc7: redirecting sector 489898496 to another mirror although the arrays are still running on all disks - they haven't given up on any yet: # cat /proc/mdstat Personalities : [raid1] [raid0] md10 : active raid0 md0[0] md1[1] 3368770048 blocks super 1.2 512k chunks md1 : active raid1 sde2[2] sdd2[1] 1464087824 blocks super 1.2 [2/2] [UU] md0 : active raid1 sdb7[0] sdc7[2] 1904684920 blocks super 1.2 [2/2] [UU] unused devices: <none> So I think I have some idea what the problem is but I am not a linux sysadmin expert by the remotest stretch of the imagination and would really appreciate some clue checking here with my diagnosis and what do I need to do: obviously I need to source another drive for sdc. (I'm guessing I could buy a larger drive if the price is right: I'm thinking that one day I'll need to grow the size of the array and that would be one less drive to replace with a larger one) then use mdadm to fail out the existing sdc, remove it and fit the new drive fdisk the new drive with the same size partition for the array as the old one had use mdadm to add the new drive into the array that sound OK?

    Read the article

  • Java - Highest, Lowest and Average

    - by Emily
    Hello, I've just started studying and I need help on one of my exercises. I need the end user to input a rain fall number for each month. I then need to out put the average rainfall, highest month and lowest month and the months which rainfall was above average. I keep getting the same number in the highest and lowest and I have no idea why. I am seriously pulling my hair out. Any help would be greatly appreciated. This is what I have so far: public class rainfall { /** * @param args */ public static void main(String[] args) { int[] numgroup; numgroup = new int [13]; ConsoleReader console = new ConsoleReader(); int highest; int lowest; int index; int tempVal; int minMonth; int minIndex; int maxMonth; int maxIndex; System.out.println("Welcome to Rainfall"); for(index = 1; index < 13; index = index + 1) { System.out.println("Please enter the rainfall for month " + index); tempVal = console.readInt(); while (tempVal>100 || tempVal<0) { System.out.println("The rating must be within 0...100. Try again"); tempVal = console.readInt(); } numgroup[index] = tempVal; } lowest = numgroup[0]; for(minIndex = 0; minIndex < numgroup.length; minIndex = minIndex + 1); { if (numgroup[0] < lowest) { lowest = numgroup[0]; minMonth = minIndex; } } highest = numgroup[1]; for(maxIndex = 0; maxIndex < numgroup.length; maxIndex = maxIndex + 1); { if (numgroup[1] > highest) { highest = numgroup[1]; maxMonth = maxIndex; } } System.out.println("The average monthly rainfall was "); System.out.println("The lowest monthly rainfall was month " + minIndex); System.out.println("The highest monthly rainfall was month " + maxIndex); System.out.println("Thank you for using Rainfall"); } private static ConsoleReader ConsoleReader() { return null; } } Thanks, Emily

    Read the article

  • SQL Average Data Based on Distance

    - by jsmith
    I'm pretty new to SQL. I have a database with records based on road/milepoints. My goal is to get an average value every 52.8 ft along the road. My related table has data every 15 ft, this table of course has a foreign key relating it to the primary table. If I wanted to pull out the average value every 52.8 ft, along a given milepost, how would I go about this? Example Data: RecID Begin_MP End_MP 100 0 0.56 RecID MP Value1 Value2 100 0 159 127.7 100 0.003 95.3 115.3 100 0.006 82.3 107 100 0.009 56.5 74.5 100 0.011 58.1 89.1 100 0.014 95.2 78.8 100 0.017 108.9 242.5 100 0.02 71.8 73.3 100 0.023 84.1 80.2 100 0.026 65.5 66.1 100 0.028 122 135.8 100 0.031 99.9 230.7 100 0.034 95.7 111.5 100 0.037 127.3 74.3 100 0.04 140.7 543.1 The first Data is an example of a Road. The second subset of data are the values I need to query out every 52.8 ft. Thank you

    Read the article

  • average velocity, as3

    - by VideoDnd
    Hello, I need something accurate I can plug equations in to if you can help. How would you apply the equation bellow? Thanks guys. AVERAGE VELOCITY AND DISPLACEMENT average velocity V=X/T displacement x=v*T more info example I have 30 seconds and a field that is 170 yards. What average velocity would I need my horse to travel at to reach the end of the field in 30 seconds. I moved the decimal places around and got this. Here's what I tried 'the return value is close, but not close enough' FLA here var TIMER:int = 10; var T:int = 0; var V:int = 5.6; var X:int = 0; var Xf:int = 17000/10*2; var timer:Timer = new Timer(TIMER,Xf); timer.addEventListener(TimerEvent.TIMER, incrementCounter); timer.start(); function formatCount(i:int):String { var fraction:int = Math.abs(i % 100); var whole:int = Math.abs(i / 100); return ("0000000" + whole).substr(-7, 7) + "." + (fraction < 10 ? "0" + fraction : fraction); } function incrementCounter(event:TimerEvent) { T++; X = Math.abs(V*T); text.text = formatCount(X); } tests TARGET 5.6yards * 30seconds = 168yards INTEGERS 135.00 in 30 seconds MATH.ROUND 135.00 in 30 seconds NUMBERS 140.00 in 30 seconds control timer 'I tested with this and the clock on my desk' var timetest:Timer = new Timer(1000,30); var Dplus:int = 17000; timetest.addEventListener(TimerEvent.TIMER, cow); timetest.start(); function cow(evt:TimerEvent):void { tx.text = String("30 SECONDS: " + timetest.currentCount); if(timetest.currentCount> Dplus){ timetest.stop(); } } //far as I got...couldn't get delta to work... T = (V*timer.currentCount); X += Math.round(T);

    Read the article

  • Algorithm Question Maximize Average of Functions

    - by paradoxperfect
    Hello, I have a set of N functions each denoted by Fi(h). Each function returns some value when given an h. I'm trying to figure out a way to maximize the average of all of the functions given some total H value. For example, say each function represents a grade on an assignment. If I spend h hours on assignment i, I will get g = Fi(h) as my grade. I'm given H hours to finish all of the assignments. I want to maximize my average grade for all assignments. Can anyone point me in the right direction to figure this out?

    Read the article

  • Average Rating script

    - by MILESMIBALERR
    I have asked this once before but i didnt get a very clear answer. I need to know how to make a rating script for a site. I have a form that submits a rating out of ten to mysql. How would you get the average rating to be displayed from the mysql column using php? One person suggested having two tables; one for all the ratings, and one for the average rating of each page. Is there a simpler method than this?

    Read the article

  • Find the min max and average of one column of data in python

    - by user1440194
    I have a set of data that looks like this 201206040210 -3461.00000000 -8134.00000000 -4514.00000000 -4394.00000000 0 201206040211 -3580.00000000 -7967.00000000 -4614.00000000 -7876.00000000 0 201206040212 -3031.00000000 -9989.00000000 -9989.00000000 -3419.00000000 0 201206040213 -1199.00000000 -6961.00000000 -3798.00000000 -5822.00000000 0 201206040214 -2940.00000000 -5524.00000000 -5492.00000000 -3394.00000000 0 I want to take the second to last column and find the min, max, and average. Im a little confused on how to use split when the columns are delimited by a space and -. i Figure once i do that i can use min() and max function. I have written a shell script to do the same here #!/bin/ksh awk '{print substr($5,2);}' data' > /data1 sort -n data1 > data2 tail -1 data2 head -1 data2 awk '{sum+=$1} END {print "average = ",sum/NR}' data2 Im just not sure how to do this in python. Thanks

    Read the article

  • MySQL Rating system (calculating average from two tables).

    - by MussuR
    I have two tables, videos and videos_ratings. The videos table has an int videoid field (and many others but those fields are not important I think) and many records. The videos_ratings table has 3 int fields: videoid, rating, rated_by which has many records (multiple records for each fields from the videos table) but not for all records from the videos table. Currently I have the following mysql query: SELECT `videos`.*, avg(`videos_ratings`.`vote`) FROM `videos`, `videos_ratings` WHERE `videos_ratings`.`videoid` = `videos`.`videoid` GROUP BY `videos_ratings`.`videoid` ORDER BY RAND() LIMIT 0, 12 It selects all the records from table videos that have a rating in table video_ratings and calculates the average correctly. But what I need is to select all records from the videos table, no matter if there is a rating for that record or not. And if there aren't any records in the videos_ratings table for that particular videos record, the average function should show 0. Hope someone could understand what I want... :) Thanks!

    Read the article

  • Calculate average in LINQ C# with string representation of property name

    - by Paul
    I need to calculate a whole bunch of averages on an List of Surveys. The surveys have lots of properties that are int and double valued. I am creating a business object to handle all the calculations (there are like 100) and I'd rather not code 100 different methods for finding the average for a particular property. I'd like to be able to have the UI pass a string (representing the property) and have the the business object return an average for that property. So, like... int AverageHeightInInches = MyObject.GetIntAverage("HeightInInches"); . . . Then have linq code to calculate the result. Thanks!

    Read the article

  • average temperature of CPU

    - by lego 69
    I downloaded the program named speedFan(for measuring the temperature of the hardware), and I have parameters like these GPU: 60C HDO: 38C Temp1: 50C Core 0: 44C Core 1: 44C Core: 59C Ambient: 0C I know that it shows me temperature of the different parts of my hardware, but I have no idea which ones, can You please explain it and also, is it something wrong with my laptop (because temperature of GPU is 60C) thanks in advance

    Read the article

  • How to get average volume level of a mp3 file

    - by Stan
    I have to normalize volume level of some speech recording. But every recording seems to be at difference volume level. Is there any tool that can get some statistics of volume level? So I know how much dB I have to increase/decrease during normalization. Thanks.

    Read the article

  • ASP.NET retrieve Average CPU Usage

    - by Sam
    Last night I did a load test on a site. I found that one of my shared caches is a bottleneck. I'm using a ReaderWriterLockSlim to control the updates of the data. Unfortunately at one point there are ~200 requests trying to update the data at approximately the same time. This also coincided with CPU usage spikes. The data being updated is in the ASP.NET Cache. What I'd like to do is if the CPU usage is around 75%, I'd like to just skip the cache and hit the database on another machine. My problem is that I don't know how expensive it is to create a new performance counter to check the cpu usage. Also, if I would probably like the average cpu usage over the last 2 or 3 seconds. However, I can't sit there and calculate the cpu time as that would take longer than it's taking to update the cache currently. Is there an easy way to get the average CPU usage? Are there any drawbacks to this? I'm also considering totaling the wait count for the lock and then at a certain threshold switch over to the database. The concern I had with this approach would be that changing hardware might allow more locks with less of a strain on the system. And also finding the right balance for the threshold would be cumbersome and it doesn't take into account any other load on the machine. But it's a simple approach, and simple is 99% of the time better.

    Read the article

  • SQL: Daily Average of Logins Per Hour

    - by jerrygarciuh
    This query is producing counts of logins per hour: SELECT DATEADD(hour, DATEDIFF(hour, 0, EVENT_DATETIME), 0), COUNT(*) FROM EVENTS_ALL_RPT_V1 WHERE EVENT_NAME = 'Login' AND EVENT_DATETIME >= CONVERT(DATETIME, '2010-03-17 00:00:00', 120) AND EVENT_DATETIME <= CONVERT(DATETIME, '2010-03-24 00:00:00', 120) GROUP BY DATEADD(hour, DATEDIFF(hour, 0, EVENT_DATETIME), 0) ORDER BY DATEADD(hour, DATEDIFF(hour, 0, EVENT_DATETIME), 0) ...with lots of results like this: Datetime COUNT(*) ---------------------------------- 2010-03-17 12:00:00.000 135 2010-03-17 13:00:00.000 129 2010-03-17 14:00:00.000 147 What I need to figure out is how to query the average logins per hour for a given day. Any help?

    Read the article

< Previous Page | 1 2 3 4 5 6 7 8 9 10 11 12  | Next Page >