Search Results

Search found 19975 results on 799 pages for 'disk queue length'.

facebook twitter digg google delicious technorati stumbleupon myspace wordpress linkedin gmail igoogle windows live tumblr viadeo yahoo buzz yahoo mail yahoo bookmarks favorites email print

Page 11/799 | < Previous Page | 7 8 9 10 11 12 13 14 15 16 17 18  | Next Page >

  • Content-Length header not returned from Pylons response

    - by Evgeny
    I'm still struggling to Stream a file to the HTTP response in Pylons. In addition to the original problem, I'm finding that I cannot return the Content-Length header, so that for large files the client cannot estimate how long the download will take. I've tried response.content_length = 12345 and I've tried response.headers['Content-Length'] = 12345 In both cases the HTTP response (viewed in Fiddler) simply does not contain the Content-Length header. How do I get Pylons to return this header? (Oh, and if you have any ideas on making it stream the file please reply to the original question - I'm all out of ideas there.)

    Read the article

  • TCP RST right after FIN/ACK

    - by Nitzan Shaked
    I am having the weirdest issue: I have a web server which sometimes, only on very specific requests, will send a RST to the client after having sent the FIN datagram. First, a description of the setup: The server runs on an Ubuntu 12.04.1 LTS, which itself is a VM guest inside a Win7 x64 host, in bridged mode. ufw is disabled on the host The client runs on a iOS simulator, which runs on OS X Mountain Lion, which is a VM guest (hackintosh) inside a Win7 x64 host, in bridged mode. Both client and server are on the same LAN, one is connected to the home router via an Ethernet cable, and then other thru WiFi. I happened to glimpse over the server's http logs and found that the client sometimes issuing multiple subsequent identical requests. Further investigation led me to discover that this happens when the server sends a RST, and that the client is simply re-trying. I am attaching several tcpdump's: Good1 is the server-side tcpdump of a good session ("good" meaning no RST was generated). Good3 is another sever-side tcpdump of a good session. (The difference between Good1 and Good3 is the order in which ACK's were sent from the server to the client, ACK'ing the client's request. The client's request arives in 2 segements (specifically: one for the http headers, and another for a body containing an empty json object, "{}"). In Good1, the server ACK's both request segments, using 2 ACK segments, after the second request has arrived. In Good3, the server ACK's each request segment with an ACK segment as soon as the request segment arrives. Not that it should make a difference.) Bad1 is a dump, both client- and server-side, of a bad session. Bad2 is another bad session, this time server-side only. Note that in all "bad" sessions, the server ACK's each request segments immediately after having received it. I've looked at a few other bad sessions, and the situation is the same in all of them. But this is also the behavior in "Good3", so I don't see how that observation helps me, of for that matter why it should matter. I can't find any difference between good and bad sessions, or at least one that I think should matter. My question is: why are those RST's being generated? Or at least: how do I go about debugging this, or providing more info here that'll help? Edit 2 new facts that I have learned: Section 4.2.2.13 of the RFC (1122) (and Wikipedia, in the article "TCP", under "Connection Termination") says that a TCP application on one host may close the connection before it has read all of the data in its socket buffer, and in such a case the TCP on the host will sent a RST to the other side, to let it know that not all the data it has sent has been read. I'm not sure I completely understand this, since closing my side of the connection still allows me to read, no? It also means that I can't write any more. I am not sure this is relevant, though, since I see a RST after FIN. There are multiple complaints of this happening with wsgiref (Python's dev-mode HTTP server), which is exactly what I'm using. I'll keep updating as I find out more. Thanks! ~~~~~~~~~~~~~~~~~~~~ Good1 -- Server Side ~~~~~~~~~~~~~~~~~~~~ 13:28:02.308319 IP 192.168.1.51.51479 > 192.168.1.132.5000: Flags [S], seq 94268074, win 65535, options [mss 1460,nop,wscale 4,nop,nop,TS val 943308864 ecr 0,sackOK,eol], length 0 13:28:02.308336 IP 192.168.1.132.5000 > 192.168.1.51.51479: Flags [S.], seq 1726304574, ack 94268075, win 14480, options [mss 1460,sackOK,TS val 326480982 ecr 943308864,nop,wscale 3], length 0 13:28:02.309750 IP 192.168.1.51.51479 > 192.168.1.132.5000: Flags [.], ack 1, win 8235, options [nop,nop,TS val 943308865 ecr 326480982], length 0 13:28:02.310744 IP 192.168.1.51.51479 > 192.168.1.132.5000: Flags [P.], seq 1:351, ack 1, win 8235, options [nop,nop,TS val 943308865 ecr 326480982], length 350 13:28:02.310766 IP 192.168.1.51.51479 > 192.168.1.132.5000: Flags [P.], seq 351:353, ack 1, win 8235, options [nop,nop,TS val 943308865 ecr 326480982], length 2 13:28:02.310841 IP 192.168.1.132.5000 > 192.168.1.51.51479: Flags [.], ack 351, win 1944, options [nop,nop,TS val 326480983 ecr 943308865], length 0 13:28:02.310918 IP 192.168.1.132.5000 > 192.168.1.51.51479: Flags [.], ack 353, win 1944, options [nop,nop,TS val 326480983 ecr 943308865], length 0 13:28:02.315931 IP 192.168.1.132.5000 > 192.168.1.51.51479: Flags [P.], seq 1:18, ack 353, win 1944, options [nop,nop,TS val 326480984 ecr 943308865], length 17 13:28:02.316107 IP 192.168.1.132.5000 > 192.168.1.51.51479: Flags [FP.], seq 18:684, ack 353, win 1944, options [nop,nop,TS val 326480984 ecr 943308865], length 666 13:28:02.317651 IP 192.168.1.51.51479 > 192.168.1.132.5000: Flags [.], ack 18, win 8234, options [nop,nop,TS val 943308872 ecr 326480984], length 0 13:28:02.318288 IP 192.168.1.51.51479 > 192.168.1.132.5000: Flags [.], ack 685, win 8192, options [nop,nop,TS val 943308872 ecr 326480984], length 0 13:28:02.318640 IP 192.168.1.51.51479 > 192.168.1.132.5000: Flags [F.], seq 353, ack 685, win 8192, options [nop,nop,TS val 943308872 ecr 326480984], length 0 13:28:02.318651 IP 192.168.1.132.5000 > 192.168.1.51.51479: Flags [.], ack 354, win 1944, options [nop,nop,TS val 326480985 ecr 943308872], length 0 ~~~~~~~~~~~~~~~~~~~~ Good3 -- Server Side ~~~~~~~~~~~~~~~~~~~~ 13:28:03.311143 IP 192.168.1.51.51486 > 192.168.1.132.5000: Flags [S], seq 1982901126, win 65535, options [mss 1460,nop,wscale 4,nop,nop,TS val 943309853 ecr 0,sackOK,eol], length 0 13:28:03.311155 IP 192.168.1.132.5000 > 192.168.1.51.51486: Flags [S.], seq 2245063571, ack 1982901127, win 14480, options [mss 1460,sackOK,TS val 326481233 ecr 943309853,nop,wscale 3], length 0 13:28:03.312671 IP 192.168.1.51.51486 > 192.168.1.132.5000: Flags [.], ack 1, win 8235, options [nop,nop,TS val 943309854 ecr 326481233], length 0 13:28:03.313330 IP 192.168.1.51.51486 > 192.168.1.132.5000: Flags [P.], seq 1:351, ack 1, win 8235, options [nop,nop,TS val 943309855 ecr 326481233], length 350 13:28:03.313337 IP 192.168.1.132.5000 > 192.168.1.51.51486: Flags [.], ack 351, win 1944, options [nop,nop,TS val 326481234 ecr 943309855], length 0 13:28:03.313342 IP 192.168.1.51.51486 > 192.168.1.132.5000: Flags [P.], seq 351:353, ack 1, win 8235, options [nop,nop,TS val 943309855 ecr 326481233], length 2 13:28:03.313346 IP 192.168.1.132.5000 > 192.168.1.51.51486: Flags [.], ack 353, win 1944, options [nop,nop,TS val 326481234 ecr 943309855], length 0 13:28:03.327942 IP 192.168.1.132.5000 > 192.168.1.51.51486: Flags [P.], seq 1:18, ack 353, win 1944, options [nop,nop,TS val 326481237 ecr 943309855], length 17 13:28:03.328253 IP 192.168.1.132.5000 > 192.168.1.51.51486: Flags [FP.], seq 18:684, ack 353, win 1944, options [nop,nop,TS val 326481237 ecr 943309855], length 666 13:28:03.329076 IP 192.168.1.51.51486 > 192.168.1.132.5000: Flags [.], ack 18, win 8234, options [nop,nop,TS val 943309868 ecr 326481237], length 0 13:28:03.329688 IP 192.168.1.51.51486 > 192.168.1.132.5000: Flags [.], ack 685, win 8192, options [nop,nop,TS val 943309868 ecr 326481237], length 0 13:28:03.330361 IP 192.168.1.51.51486 > 192.168.1.132.5000: Flags [F.], seq 353, ack 685, win 8192, options [nop,nop,TS val 943309869 ecr 326481237], length 0 13:28:03.330370 IP 192.168.1.132.5000 > 192.168.1.51.51486: Flags [.], ack 354, win 1944, options [nop,nop,TS val 326481238 ecr 943309869], length 0 ~~~~~~~~~~~~~~~~~~~~ Bad1 -- Server Side ~~~~~~~~~~~~~~~~~~~~ 13:28:01.311876 IP 192.168.1.51.51472 > 192.168.1.132.5000: Flags [S], seq 920400580, win 65535, options [mss 1460,nop,wscale 4,nop,nop,TS val 943307883 ecr 0,sackOK,eol], length 0 13:28:01.311896 IP 192.168.1.132.5000 > 192.168.1.51.51472: Flags [S.], seq 3103085782, ack 920400581, win 14480, options [mss 1460,sackOK,TS val 326480733 ecr 943307883,nop,wscale 3], length 0 13:28:01.313509 IP 192.168.1.51.51472 > 192.168.1.132.5000: Flags [.], ack 1, win 8235, options [nop,nop,TS val 943307884 ecr 326480733], length 0 13:28:01.315614 IP 192.168.1.51.51472 > 192.168.1.132.5000: Flags [P.], seq 1:351, ack 1, win 8235, options [nop,nop,TS val 943307886 ecr 326480733], length 350 13:28:01.315727 IP 192.168.1.132.5000 > 192.168.1.51.51472: Flags [.], ack 351, win 1944, options [nop,nop,TS val 326480734 ecr 943307886], length 0 13:28:01.316229 IP 192.168.1.51.51472 > 192.168.1.132.5000: Flags [P.], seq 351:353, ack 1, win 8235, options [nop,nop,TS val 943307886 ecr 326480733], length 2 13:28:01.316242 IP 192.168.1.132.5000 > 192.168.1.51.51472: Flags [.], ack 353, win 1944, options [nop,nop,TS val 326480734 ecr 943307886], length 0 13:28:01.321019 IP 192.168.1.132.5000 > 192.168.1.51.51472: Flags [P.], seq 1:18, ack 353, win 1944, options [nop,nop,TS val 326480735 ecr 943307886], length 17 13:28:01.321294 IP 192.168.1.132.5000 > 192.168.1.51.51472: Flags [FP.], seq 18:684, ack 353, win 1944, options [nop,nop,TS val 326480736 ecr 943307886], length 666 13:28:01.321386 IP 192.168.1.132.5000 > 192.168.1.51.51472: Flags [R.], seq 685, ack 353, win 1944, options [nop,nop,TS val 326480736 ecr 943307886], length 0 13:28:01.322727 IP 192.168.1.51.51472 > 192.168.1.132.5000: Flags [.], ack 18, win 8234, options [nop,nop,TS val 943307891 ecr 326480735], length 0 13:28:01.322733 IP 192.168.1.132.5000 > 192.168.1.51.51472: Flags [R], seq 3103085800, win 0, length 0 13:28:01.323221 IP 192.168.1.51.51472 > 192.168.1.132.5000: Flags [.], ack 685, win 8192, options [nop,nop,TS val 943307892 ecr 326480736], length 0 13:28:01.323231 IP 192.168.1.132.5000 > 192.168.1.51.51472: Flags [R], seq 3103086467, win 0, length 0 ~~~~~~~~~~~~~~~~~~~~ Bad1 -- Client Side ~~~~~~~~~~~~~~~~~~~~ 13:28:11.374654 IP 192.168.1.51.51472 > 192.168.1.132.5000: Flags [S], seq 920400580, win 65535, options [mss 1460,nop,wscale 4,nop,nop,TS val 943307883 ecr 0,sackOK,eol], length 0 13:28:11.375764 IP 192.168.1.132.5000 > 192.168.1.51.51472: Flags [S.], seq 3103085782, ack 920400581, win 14480, options [mss 1460,sackOK,TS val 326480733 ecr 943307883,nop,wscale 3], length 0 13:28:11.376352 IP 192.168.1.51.51472 > 192.168.1.132.5000: Flags [.], ack 1, win 8235, options [nop,nop,TS val 943307884 ecr 326480733], length 0 13:28:11.378252 IP 192.168.1.51.51472 > 192.168.1.132.5000: Flags [P.], seq 1:351, ack 1, win 8235, options [nop,nop,TS val 943307886 ecr 326480733], length 350 13:28:11.379027 IP 192.168.1.51.51472 > 192.168.1.132.5000: Flags [P.], seq 351:353, ack 1, win 8235, options [nop,nop,TS val 943307886 ecr 326480733], length 2 13:28:11.379732 IP 192.168.1.132.5000 > 192.168.1.51.51472: Flags [.], ack 351, win 1944, options [nop,nop,TS val 326480734 ecr 943307886], length 0 13:28:11.380592 IP 192.168.1.132.5000 > 192.168.1.51.51472: Flags [.], ack 353, win 1944, options [nop,nop,TS val 326480734 ecr 943307886], length 0 13:28:11.384968 IP 192.168.1.132.5000 > 192.168.1.51.51472: Flags [P.], seq 1:18, ack 353, win 1944, options [nop,nop,TS val 326480735 ecr 943307886], length 17 13:28:11.385044 IP 192.168.1.51.51472 > 192.168.1.132.5000: Flags [.], ack 18, win 8234, options [nop,nop,TS val 943307891 ecr 326480735], length 0 13:28:11.385586 IP 192.168.1.132.5000 > 192.168.1.51.51472: Flags [FP.], seq 18:684, ack 353, win 1944, options [nop,nop,TS val 326480736 ecr 943307886], length 666 13:28:11.385743 IP 192.168.1.51.51472 > 192.168.1.132.5000: Flags [.], ack 685, win 8192, options [nop,nop,TS val 943307892 ecr 326480736], length 0 13:28:11.385966 IP 192.168.1.132.5000 > 192.168.1.51.51472: Flags [R.], seq 685, ack 353, win 1944, options [nop,nop,TS val 326480736 ecr 943307886], length 0 13:28:11.387343 IP 192.168.1.132.5000 > 192.168.1.51.51472: Flags [R], seq 3103085800, win 0, length 0 13:28:11.387344 IP 192.168.1.132.5000 > 192.168.1.51.51472: Flags [R], seq 3103086467, win 0, length 0 ~~~~~~~~~~~~~~~~~~~~ Bad2 -- Server Side ~~~~~~~~~~~~~~~~~~~~ 13:28:01.319185 IP 192.168.1.51.51473 > 192.168.1.132.5000: Flags [S], seq 1631526992, win 65535, options [mss 1460,nop,wscale 4,nop,nop,TS val 943307889 ecr 0,sackOK,eol], length 0 13:28:01.319197 IP 192.168.1.132.5000 > 192.168.1.51.51473: Flags [S.], seq 2524685719, ack 1631526993, win 14480, options [mss 1460,sackOK,TS val 326480735 ecr 943307889,nop,wscale 3], length 0 13:28:01.320692 IP 192.168.1.51.51473 > 192.168.1.132.5000: Flags [.], ack 1, win 8235, options [nop,nop,TS val 943307890 ecr 326480735], length 0 13:28:01.322219 IP 192.168.1.51.51473 > 192.168.1.132.5000: Flags [P.], seq 1:351, ack 1, win 8235, options [nop,nop,TS val 943307890 ecr 326480735], length 350 13:28:01.322336 IP 192.168.1.132.5000 > 192.168.1.51.51473: Flags [.], ack 351, win 1944, options [nop,nop,TS val 326480736 ecr 943307890], length 0 13:28:01.322689 IP 192.168.1.51.51473 > 192.168.1.132.5000: Flags [P.], seq 351:353, ack 1, win 8235, options [nop,nop,TS val 943307890 ecr 326480735], length 2 13:28:01.322700 IP 192.168.1.132.5000 > 192.168.1.51.51473: Flags [.], ack 353, win 1944, options [nop,nop,TS val 326480736 ecr 943307890], length 0 13:28:01.326307 IP 192.168.1.132.5000 > 192.168.1.51.51473: Flags [P.], seq 1:18, ack 353, win 1944, options [nop,nop,TS val 326480737 ecr 943307890], length 17 13:28:01.326614 IP 192.168.1.132.5000 > 192.168.1.51.51473: Flags [FP.], seq 18:684, ack 353, win 1944, options [nop,nop,TS val 326480737 ecr 943307890], length 666 13:28:01.326710 IP 192.168.1.132.5000 > 192.168.1.51.51473: Flags [R.], seq 685, ack 353, win 1944, options [nop,nop,TS val 326480737 ecr 943307890], length 0 13:28:01.328499 IP 192.168.1.51.51473 > 192.168.1.132.5000: Flags [.], ack 18, win 8234, options [nop,nop,TS val 943307896 ecr 326480737], length 0 13:28:01.328509 IP 192.168.1.132.5000 > 192.168.1.51.51473: Flags [R], seq 2524685737, win 0, length 0 13:28:01.328514 IP 192.168.1.51.51473 > 192.168.1.132.5000: Flags [.], ack 685, win 8192, options [nop,nop,TS val 943307896 ecr 326480737], length 0 13:28:01.328517 IP 192.168.1.132.5000 > 192.168.1.51.51473: Flags [R], seq 2524686404, win 0, length 0

    Read the article

  • Disk performance below expectations

    - by paulH
    this is a follow-up to a previous question that I asked (Two servers with inconsistent disk speed). I have a PowerEdge R510 server with a PERC H700 integrated RAID controller (call this Server B) that was built using eight disks with 3Gb/s bandwidth that I was comparing with an almost identical server (call this Server A) that was built using four disks with 6Gb/s bandwidth. Server A had much better I/O rates than Server B. Once I discovered the difference with the disks, I had Server A rebuilt with faster 6Gbps disks. Unfortunately this resulted in no increase in the performance of the disks. Expecting that there must be some other configuration difference between the servers, we took the 6Gbps disks out of Server A and put them in Server B. This also resulted in no increase in the performance of the disks. We now have two identical servers built, with the exception that one is built with six 6Gbps disks and the other with eight 3Gbps disks, and the I/O rates of the disks is pretty much identical. This suggests that there is some bottleneck other than the disks, but I cannot understand how Server B originally had better I/O that has subsequently been 'lost'. Comparative I/O information below, as measured by SQLIO. The same parameters were used for each test. It's not the actual numbers that are significant but rather the variations between systems. In each case D: is a 2 disk RAID 1 volume, and E: is a 4 disk RAID 10 volume (apart from the original Server A, where E: was a 2 disk RAID 0 volume). Server A (original setup with 6Gpbs disks) D: Read (MB/s) 63 MB/s D: Write (MB/s) 170 MB/s E: Read (MB/s) 68 MB/s E: Write (MB/s) 320 MB/s Server B (original setup with 3Gpbs disks) D: Read (MB/s) 52 MB/s D: Write (MB/s) 88 MB/s E: Read (MB/s) 112 MB/s E: Write (MB/s) 130 MB/s Server A (new setup with 3Gpbs disks) D: Read (MB/s) 55 MB/s D: Write (MB/s) 85 MB/s E: Read (MB/s) 67 MB/s E: Write (MB/s) 180 MB/s Server B (new setup with 6Gpbs disks) D: Read (MB/s) 61 MB/s D: Write (MB/s) 95 MB/s E: Read (MB/s) 69 MB/s E: Write (MB/s) 180 MB/s Can anybody suggest any ideas what is going on here? The drives in use are as follows: Dell Seagate F617N ST3300657SS 300GB 15K RPM SAS Dell Hitachi HUS156030VLS600 300GB 3.5 inch 15000rpm 6GB SAS Hitachi Hus153030vls300 300GB Server SAS Dell ST3146855SS Seagate 3.5 inch 146GB 15K SAS

    Read the article

  • Store Varnish cache in hard disk

    - by Great Kuma
    Hello, The situation is: Im building PHP application, and need http caching. Varnish is great, and lots of people tell me that Varnish store the cached data in RAM. But I want it cached in hard disk. Is there any way to store the Varnish cached data in hard disk? thanks.

    Read the article

  • Why does storage's performance change at various queue depths?

    - by Mxx
    I'm in the market for a storage upgrade for our servers. I'm looking at benchmarks of various PCIe SSD devices and in comparisons I see that IOPS change at various queue depths. How can that be and why is that happening? The way I understand things is: I have a device with maximum (theoretical) of 100k IOPS. If my workload consistently produces 100,001 IOPS, I'll have a queue depth of 1, am I correct? However, from what I see in benchmarks some devices run slower at lower queue depths, then speedup at depth of 4-64 and then slow down again at even larger depths. Isn't queue depths a property of OS(or perhaps storage controller), so why would that affect IOPS?

    Read the article

  • Why is my mail stuck in the queue with the status "retry" in Exchange 2003?

    - by Mike C
    I have been having a problem recently where some of our mail is not getting to clients. I looked into the Message queue on my SBS2k3 server and noticed that several recipients are showing up in the queue with a state of "Retry". When I highlight the recipient, the additional queue information says "An SMTP protocol error occured". How can I find out more specifically what error occured, and how could I then go about correcting it? Thanks, Mike

    Read the article

  • Why is the disk making my motherboard beep?

    - by Mark Ransom
    Whenever I let my PC do heavy disk accesses for a long time, the speaker on the motherboard starts making a continuous chirping sound. Thankfully it doesn't happen often, but it drives me nuts when it does. Anybody know where this sound might be coming from, or have any hints as to how to track it down? Edit: The problem appears to be with the processor, the correlation with disk access was coincidental. Thanks for all the answers.

    Read the article

  • Clean install vs disk image

    - by Thanos
    Once a year I am making a clean install on windows, in order to keep my system fast. After posting a question on making a bootable windows usb with exe programs where I was adviced to make a disk image, a new question rose. What is the difference in making a disk image and performing a clean install on windows? Which is better in terms of speed, general performance, value for time and transfering between different computers?

    Read the article

  • How do I Change a damaged Disk in a Raid 5 array

    - by Egakagoc2xI
    Hi, I have a server with a 4-drives Raid 5 array; one of the disks is damaged. All the disks are hot pluggable. My Question is, I want to replace the damaged disk with a new one, do I have to shutdown the server or should I just change the hard disk with the server on and it will rebuild the array? There is a procedure to follow? My Server is a HP. Regards.

    Read the article

  • Queued Loadtest to remove Concurrency issues using Shared Data Service in OpenScript

    - by stefan.thieme(at)oracle.com
    Queued Processing to remove Concurrency issues in Loadtest ScriptsSome scripts act on information returned by the server, e.g. act on first item in the returned list of pending tasks/actions. This may lead to concurrency issues if the virtual users simulated in a load test scenario are not synchronized in some way.As the load test cases should be carried out in a comparable and straight forward manner simply cancel a transaction in case a collision occurs is clearly not an option. In case you increase the number of virtual users this approach would lead to a high number of requests for the early steps in your transaction (e.g. login, retrieve list of action points, assign an action point to the virtual user) but later steps would be rarely visited successfully or at all, depending on the application logic.A way to tackle this problem is to enqueue the virtual users in a Shared Data Service queue. Only the first virtual user in this queue will be allowed to carry out the critical steps (retrieve list of action points, assign an action point to the virtual user) in your transaction at any one time.Once a virtual user has passed the critical path it will dequeue himself from the head of the queue and continue with his actions. This does theoretically allow virtual users to run in parallel all steps of the transaction which are not part of the critical path.In practice it has been seen this is rarely the case, though it does not allow adding more than N users to perform a transaction without causing delays due to virtual users waiting in the queue. N being the time of the total transaction divided by the sum of the time of all critical steps in this transaction.While this problem can be circumvented by allowing multiple queues to act on individual segments of the list of actions, e.g. per country filter, ends with 0..9 filter, etc.This would require additional handling of these additional queues of slots for the virtual users at the head of the queue in order to maintain the mutually exclusive access to the first element in the list returned by the server at any one time of the load test. Such an improved handling of multiple queues and/or multiple slots is above the subject of this paper.Shared Data Services Pre-RequisitesStart WebLogic Server to host Shared Data ServicesYou will have to make sure that your WebLogic server is installed and started. Shared Data Services may not work if you installed only the minimal installation package for OpenScript. If however you installed the default package including OLT and OTM, you may follow the instructions below to start and verify WebLogic installation.To start the WebLogic Server deployed underneath of Oracle Load Testing and/or Oracle Test Manager you can go to your Start menu, Oracle Application Testing Suite and select the Restart Oracle Application Testing Suite Application Service entry from the Tools submenu.To verify the service has been started you can run the Microsoft Management Console for Services by Selecting Run from the Start Menu and entering services.msc. Look for the entry that reads Oracle Application Testing Suite Application Service, once it has changed it status from Starting to Started you can proceed to verify the login. Please note that this may take several minutes, I would say up to 10 minutes depending on the strength of your CPU horse-power.Verify WebLogic Server user credentialsYou will have to make sure that your WebLogic Server is installed and started. Next open the Oracle WebLogic Server Adminstration Console on http://localhost:8088/console.It may take a while until the application is deployed and started. It may display the following until the Administration Console has been deployed on the fly.Afterwards you can login using the username oats and the password that you selected during install time for your Application Testing Suite administrative purposes.This will bring up the Home page of you WebLogic Server. You have actually verified that you are able to login with these credentials already. However if you want to check the details, navigate to Security Realms, myrealm, Users and Groups tab.Here you could add users to your WebLogic Server which could be used in the later steps. Details on the Groups required for such a custom user to work are exceeding this quick overview and have to be selected with the WebLogic Server Adminstration Guide in mind.Shared Data Services pre-requisites for Load testingOpenScript Preferences have to be set to enable Encryption and provide a default Shared Data Service Connection for Playback.These are pre-requisites you want to use for load testing with Shared Data Services.Please note that the usage of the Connection Parameters (individual directive in the script) for Shared Data Services did not playback reliably in the current version 9.20.0370 of Oracle Load Testing (OLT) and encryption of credentials still seemed to be mandatory as well.General Encryption settingsSelect OpenScript Preferences from the View menu and navigate to the General, Encryption entry in the tree on the left. Select the Encrypt script data option from the list and enter the same password that you used for securing your WebLogic Server Administration Console.Enable global shared data access credentialsSelect OpenScript Preferences from the View menu and navigate to the Playback, Shared Data entry in the tree on the left. Enable the global shared data access credentials and enter the Address, User name and Password determined for your WebLogic Server to host Shared Data Services.Please note, that you may want to replace the localhost in Address with the hosts realname in case you plan to run load tests with Loadtest Agents running on remote systems.Queued Processing of TransactionsEnable Shared Data Services Module in Script PropertiesThe Shared Data Services Module has to be enabled for each Script that wants to employ the Shared Data Service Queue functionality in OpenScript. It can be enabled under the Script menu selecting Script Properties. On the Script Properties Dialog select the Modules section and check Shared Data to enable Shared Data Service Module for your script. Checking the Shared Data Services option will effectively add a line to your script code that adds the sharedData ScriptService to your script class of IteratingVUserScript.@ScriptService oracle.oats.scripting.modules.sharedData.api.SharedDataService sharedData;Record your scriptRecord your script as usual and then add the following things for Queue handling in the Initialize code block, before the first step and after the last step of your critical path and in the Finalize code block.The java code to be added at individual locations is explained in the following sections in full detail.Create a Shared Data Queue in InitializeTo create a Shared Data Queue go to the Java view of your script and enter the following statements to the initialize() code block.info("Create queueA with life time of 120 minutes");sharedData.createQueue("queueA", 120);This will create an instantiation of the Shared Data Queue object named queueA which is maintained for upto 120 minutes.If you want to use the code for multiple scripts, make sure to use a different queue name for each one here and in the subsequent steps. You may even consider to use a dynamic queueName based on filters of your result list being concurrently accessed.Prepare a unique id for each IterationIn order to keep track of individual virtual users in our queue we need to create a unique identifier from the virtual user id and the used username right after retrieving the next record from our databank file.getDatabank("Usernames").getNextDatabankRecord();getVariables().set("usernameValue1","VU_{{@vuid}}_{{@iterationnum}}_{{db.Usernames.Username}}_{{@timestamp}}_{{@random(10000)}}");String usernameValue = getVariables().get("usernameValue1");info("Now running virtual user " + usernameValue);As you can see from the above code block, we have set the OpenScript variable usernameValue1 to VU_{{@vuid}}_{{@iterationnum}}_{{db.Usernames.Username}}_{{@timestamp}}_{{@random(10000)}} which is a concatenation of the virtual user id and the iterationnumber for general uniqueness; as well as the username from our databank, the timestamp and a random number for making it further unique and ease spotting of errors.Not all of these fields are actually required to make it really unique, but adding the queue name may also be considered to help troubleshoot multiple queues.The value is then retrieved with the getVariables.get() method call and assigned to the usernameValue String used throughout the script.Please note that moving the getDatabank("Usernames").getNextDatabankRecord(); call to the initialize block was later considered to remove concurrency of multiple virtual users running with the same userid and therefor accessing the same "My Inbox" in step 6. This will effectively give each virtual user a userid from the databank file. Make sure you have enough userids to remove this second hurdle.Enqueue and attend Queue before Critical PathTo maintain the right order of virtual users being allowed into the critical path of the transaction the following pseudo step has to be added in front of the first critical step. In the case of this example this is right in front of the step where we retrieve the list of actions from which we select the first to be assigned to us.beginStep("[0] Waiting in the Queue", 0);{info("Enqueued virtual user " + usernameValue + " at the end of queueA");sharedData.offerLast("queueA", usernameValue);info("Wait until the user is the first in queueA");String queueValue1 = null;do {// we wait for at least 0.7 seconds before we check the head of the// queue. This is the time it takes one user to move through the// critical path, i.e. pass steps [5] Enter country and [6] Assign// to meThread.sleep(700);queueValue1 = (String) sharedData.peekFirst("queueA");info("The first user in queueA is currently: '" + queueValue1 + "' " + queueValue1.getClass() + " length " + queueValue1.length() );info("The current user is '"+ usernameValue + "' " + usernameValue.getClass() + " length " + usernameValue.length() + ": indexOf " + usernameValue.indexOf(queueValue1) + " equals " + usernameValue.equals(queueValue1) );} while ( queueValue1.indexOf(usernameValue) < 0 );info("Now the user is the first in queueA");}endStep();This will enqueue the username to the tail of our Queue. It will will wait for at least 700 milliseconds, the time it takes for one user to exit the critical path and then compare the head of our queue with it's username. This last step will be repeated while the two are not equal (indexOf less than zero). If they are equal the indexOf will yield a value of zero or larger and we will perform the critical steps.Dequeue after Critical PathAfter the virtual user has left the critical path and complete its last step the following code block needs to dequeue the virtual user. In the case of our example this is right after the action has been actually assigned to the virtual user. This will allow the next virtual user to retrieve the list of actions still available and in turn let him make his selection/assignment.info("Get and remove the current user from the head of queueA");String pollValue1 = (String) sharedData.pollFirst("queueA");The current user is removed from the head of the queue. The next one will now be able to match his username against the head of the queue.Clear and Destroy Queue for FinishWhen the script has completed, it should clear and destroy the queue. This code block can be put in the finish block of your script and/or in a separate script in order to clear and remove the queue in case you have spotted an error or want to reset the queue for some reason.info("Clear queueA");sharedData.clearQueue("queueA");info("Destroy queueA");sharedData.destroyQueue("queueA");The users waiting in queueA are cleared and the queue is destroyed. If you have scripts still executing they will be caught in a loop.I found it better to maintain a separate Reset Queue script which contained only the following code in the initialize() block. I use to call this script to make sure the queue is cleared in between multiple Loadtest runs. This script could also even be added as the first in a larger scenario, which would execute it only once at very start of the Loadtest and make sure the queues do not contain any stale entries.info("Create queueA with life time of 120 minutes");sharedData.createQueue("queueA", 120);info("Clear queueA");sharedData.clearQueue("queueA");This will create a Shared Data Queue instance of queueA and clear all entries from this queue.Monitoring QueueWhile creating the scripts it was useful to monitor the contents, i.e. the current first user in the Queue. The following code block will make sure the Shared Data Queue is accessible in the initialize() block.info("Create queueA with life time of 120 minutes");sharedData.createQueue("queueA", 120);In the run() block the following code will continuously monitor the first element of the Queue and write an informational message with the current username Value to the Result window.info("Monitor the first users in queueA");String queueValue1 = null;do {queueValue1 = (String) sharedData.peekFirst("queueA");if (queueValue1 != null)info("The first user in queueA is currently: '" + queueValue1 + "' " + queueValue1.getClass() + " length " + queueValue1.length() );} while ( true );This script can be run from OpenScript parallel to a loadtest performed by the Oracle Load Test.However it is not recommend to run this in a production loadtest as the performance impact is unknown. Accessing the Queue's head with the peekFirst() method has been reported with about 2 seconds response time by both OpenScript and OTL. It is advised to log a Service Request to see if this could be lowered in future releases of Application Testing Suite, as the pollFirst() and even offerLast() writing to the tail of the Queue usually returned after an average 0.1 seconds.Debugging QueueWhile debugging the scripts the following was useful to remove single entries from its head, i.e. the current first user in the Queue. The following code block will make sure the Shared Data Queue is accessible in the initialize() block.info("Create queueA with life time of 120 minutes");sharedData.createQueue("queueA", 120);In the run() block the following code will remove the first element of the Queue and write an informational message with the current username Value to the Result window.info("Get and remove the current user from the head of queueA");String pollValue1 = (String) sharedData.pollFirst("queueA");info("The first user in queueA was currently: '" + pollValue1 + "' " + pollValue1.getClass() + " length " + pollValue1.length() );ReferencesOracle Functional Testing OpenScript User's Guide Version 9.20 [E15488-05]Chapter 17 Using the Shared Data Modulehttp://download.oracle.com/otn/nt/apptesting/oats-docs-9.21.0030.zipOracle Fusion Middleware Oracle WebLogic Server Administration Console Online Help 11g Release 1 (10.3.4) [E13952-04]Administration Console Online Help - Manage users and groupshttp://download.oracle.com/docs/cd/E17904_01/apirefs.1111/e13952/taskhelp/security/ManageUsersAndGroups.htm

    Read the article

  • Using a priority queue in Java

    - by Bharat
    Forgive me if this is a tried question, but I'm having a little difficulty figuring it out. I currently have a class Node, and each 'node' is a square in a maze. I'm trying to implement the A* algorithm, so each of these nodes will have an f-cost (int) data member inside of it. I was wondering if there's a way that I can create a priority queue of these nodes, and set up the f-cost variable as the comparator? I've looked at examples online, but all I can find are String priority queues. Can I implement Comparator for the Node class? Would this allow me to access the data member stored inside it? Many Thanks!

    Read the article

  • Lock Free Queue -- Single Producer, Multiple Consumers

    - by Shirish
    Hello, I am looking for a method to implement lock-free queue data structure that supports single producer, and multiple consumers. I have looked at the classic method by Maged Michael and Michael Scott (1996) but their version uses linked lists. I would like an implementation that makes use of bounded circular buffer. Something that uses atomic variables? On a side note, I am not sure why these classic methods are designed for linked lists that require a lot of dynamic memory management. In a multi-threaded program, all memory management routines are serialized. Aren't we defeating the benefits of lock-free methods by using them in conjunction with dynamic data structures? I am trying to code this in C/C++ using pthread library on a Intel 64-bit architecture. Thank you, Shirish

    Read the article

  • jQuery .queue() not working as expected

    - by Andrew
    The page in question is right here: http://s289116086.onlinehome.us/lawjournaltv/index.php I'm 90% of the way there, I'm just assuming there's an error with my syntax. Focus on the blue callout area with "Workers' Compensation" as the title: Basically, I've created a little slideshow there with the controls at the top and when you click on any of them both the callout and the background are to slide. As you can see it's working now, but the problem is that I want the new background to slide in over the old one. I've tried several things including the queue() method, delay(), and adding everything else to the animate() callback with no success (right now the background goes black and then the new image slides in). You guys have never let me down before, so I'm hoping this is an easy fix. Thanks in advance!

    Read the article

  • How to create a delayed queue in RabbitMQ?

    - by eandersson
    What is the easiest way to create a delay (or parking) queue with Python, Pika and RabbitMQ? I have seen an similar questions, but none for Python. I find this an useful idea when designing applications, as it allows us to throttle messages that needs to be re-queued again. There are always the possibility that you will receive more messages than you can handle, maybe the HTTP server is slow, or the database is under too much stress. I also found it very useful when something went wrong in scenarios where there is a zero tolerance to losing messages, and while re-queuing messages that could not be handled may solve that. It can also cause problems where the message will be queued over and over again. Potentially causing performance issues, and log spam.

    Read the article

  • C++: A variation of priority queue

    - by Helltone
    I need some kind of priority queue to store pairs <key, value>. Values are unique, but keys aren't. I will be performing the following operations (most common first): random insertion; retrieving (and removing) all elements with the least key. random removal (by value); I can't use std::priority_queue because it only supports removing the head. For now, I'm using an unsorted std::list. Insertion is performed by just pushing new elements to the back (O(1)). Operation 2 sorts the list with std::sort (O(N*logN)), before performing the actual retrieval. Removal, however, is O(n), which is a bit expensive. Any idea of a better data structure?

    Read the article

  • How to Create a Queue

    - by regex
    Hello All, I have an application built that hits a third party company's web service in order to create an email account after a customer clicks a button. However, sometimes the web service takes longer than 1 minute to respond, which is way to long for my customers to be sitting there waiting for a response. I need to devise a way to set up some sort of queuing service external from the web site. This way I can add the web service action to the queue and advise the customer it may take up to 2 minutes to create the account. I'm curious of the best way to achieve this. My initial thought is to request the actions via a database table which will be checked on a regular basis by a Console app which is run via Windows Scheduled tasks. Any issues with that method? Is there a better method you can think of?

    Read the article

  • Which data structure(s) to back a Final Fantasy ATB-style queue? (a delay queue)

    - by ZoFreX
    Situation: There are several entities in a simulated environment, which has an artificial notion of time called "ticks", which has no link to real time. Each entity takes it in turns to move, but some are faster than others. This is expressed by a delay, in ticks. So entity A might have a delay of 10, and B 25. In this case the turn order would go: A A B A A I'm wondering what data structure to use. At first I automatically thought "priority queue" but the delays are relative to "current time" which complicates matters. Also, there will be entities with larger delays and it's not unforseeable that the program will run through millions of ticks. It seems silly for an internal counter to be building higher and higher when the delays themselves stay relatively small and don't increase. So how would you solve this?

    Read the article

  • Using MySQL as a job queue

    - by user237815
    I'd like to use MySQL as a job queue. Multiple machines will be producing and consuming jobs. Jobs need to be scheduled; some may run every hour, some every day, etc. It seems fairly straightforward: for each job, have a "nextFireTime" column, and have worker machines search for the job with the nextFireTime, change the status of the record to "inProcess", and then update the nextFireTime when the job ends. The problem comes in when a worker dies silently. It won't be able to update the nextFireTime or set the status back to "idle". Unfortunately, jobs can be long-running, so a reaper thread that looks for jobs that have been inProcess too long isn't an option. There's no timeout value that would work. Can anyone suggest a design pattern that would properly handle unreliable worker machines?

    Read the article

  • Tomcat 6 thread safe email queue (javax.mail.*)

    - by Eric V
    Hi I have design/architecture question. I would like to send emails from one of my jsp pages. I have one particular issue that has been a little bit of a problem. there is an instance where one of the pages will need to send around 50 emails at near the same time. I would like the messages sent to a queue where a background thread will actually do the email sending. What is the appropriate way to solve this problem? If you know of a tutorial, example code or tomcat configuration is needed please let me know. Thanks,

    Read the article

  • Bad disk performance on HP DL360 with Smarty Array P400i RAID controller

    - by sarge
    I have a HP DL360 server with 4x 146GB SAS disks and a Smart Array P400i RAID controller with 256MB cache. The disks are in RAID 5 (3 disks + 1 hot spare). The server is running VMware ESX 3i. The disk write performance is really bad. Here are some numbers: ns1:~# hdparm -tT /dev/sda /dev/sda: Timing cached reads: 3364 MB in 2.00 seconds = 1685.69 MB/sec Timing buffered disk reads: 18 MB in 3.79 seconds = 4.75 MB/sec ns1:~# time sh -c "dd if=/dev/zero of=ddfile bs=8k count=125000 && sync" 125000+0 records in 125000+0 records out 1024000000 bytes (1.0 GB) copied, 282.307 s, 3.6 MB/s real 4m52.003s user 0m2.160s sys 3m10.796s Compared to another server those number are terrible: Dell R200, 2x 500GB SATA disks, PERC raid controller (disks are mirrored). web4:~# hdparm -tT /dev/sda /dev/sda: Timing cached reads: 6584 MB in 2.00 seconds = 3297.79 MB/sec Timing buffered disk reads: 316 MB in 3.02 seconds = 104.79 MB/sec web4:~# time sh -c "dd if=/dev/zero of=ddfile bs=8k count=125000 && sync" 125000+0 records in 125000+0 records out 1024000000 bytes (1.0 GB) copied, 35.2919 s, 29.0 MB/s real 0m36.570s user 0m0.476s sys 0m32.298s The server isn't very loaded and the VMware Infrastructure Client performance monitor is showing 550KBps average read and 1208KBps average write for the last 30 minutes (highest write rate: 6.6MBps). This has been a problem from the start. Any ideas?

    Read the article

  • Windows 7 keeps insisting that it needs to check disk for consistency, but never does

    - by Mike
    Lately Windows 7 has been telling me that I need to check disk D: for consistency. This happens more than 50% of the time when booting up. The first time, I didn't touch anything so that it would go ahead and do its scan. It didn't seem to do anything - just booted straight into Windows. The second time I tried to skip it by pressing any key. It ignored all of my keystrokes and still counted down to 0 (then skipped the disk check). Sometimes, it gets down to 0 but then just hangs... no indication that anything is going on. This is happening on a < 3 month old laptop. C: and D: are on the same physical disk - just two partitions. I never get any notification that C: needs to be checked for consistency. It's a ~300GB HD. C: has 60gb (32gb free) and D: has ~240GB (122gb free). What could be causing this to keep coming up? What can I do to fix it? Thanks!

    Read the article

facebook twitter digg google delicious technorati stumbleupon myspace wordpress linkedin gmail igoogle windows live tumblr viadeo yahoo buzz yahoo mail yahoo bookmarks favorites email print

< Previous Page | 7 8 9 10 11 12 13 14 15 16 17 18  | Next Page >