Buffer, average or what?

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• Did he average someone what is this problem?

  Hello, I am a new developer here and I had this problem. I don't know where I can find references and what is the problem? If anyone knows, could you explain to me? Thank you.

  24756395 (FrontDemo) process ended the code SIGSEGV = 1 fltno = 11 ip = 001025f8 (com.example.FrontDemo.testDev_e_FrontDemob3c580db@_btext+0x980) ref = 35e5cdc0

  THX, I've solved the problem

• WAIT % - % IDLE, what is the average for a virtual machine?

  I did some research on performance of VM problems and worked with the media to see that there is maybe a bottle neck IO that occur with a virtual VMware machine I have performance problems on. The technician found that the WAITING % - % IDLE on average for this virtual machine is around 200%. The tech said it was abnormally high. I forgot to ask her what was normal but wanted to put a post to see if some people out there know what they are on average. What is an average number of see? I have the system done nothing right now and I see 579% waiting and 372% IDLE. The technology has been able to confirm that no IO queue was occurring in the VMkernel or on the HBA controller.

  Some average % positions would be useful. Thank you.

  gdewulf18480 wrote:

  Supposibly the amount of time in a State of waiting for I/O is the WAIT - IDLE % difference. I was wondering if anyone had the numbers for this feeling. Is there anyone with some of them?

  Q: How will I know the VCPU world waiting for I/O events?

  A: WAIT - % IDLE % can give you an estimate on how much time processor is spent waiting for I/O events. It is an estimate only, because the world wait perhaps for other than the i/o resources. Note that we should do it only for the worlds of the WWW, not the other kind of worlds. Because the VMM worlds represent the best guest behavior. For the disk i/o, another alternative is to read the disk latency stats that we explain in the drive section.

  WAIT % by itself cannot be used, you must develop and examine the individual vCPU % EXPECTATION values and subtract their idle time.  You can't do it on the initial view of a single line by VM CPU in esxtop.  If technology that you mentioned did not explain it, he doesn't know what it takes.

  I have lots of virtual machines that are never less than 300% when I wait - % idle and I have no problems.

  It is an example, I caught NimSoft server:

  DHSNMS1 100 100 5 192.63 193,80 0.24 0.03 291,44 1.04

  %EST USED (s) 192 - in other words, almost two processors are used 100%

  The %Idle is 1.04% WAIT 291.44 - according to the calculation of waiting idle, I have a super serious problems of e/s past  But this isn't the case, and expanding, I can see that.

  5665 vmware-vmx 100 1 0.09 0.11 100.00 0.00 0.00 0.01

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  So the changed values unfortunately because of ESXTOP bike when you develop... which is annoying, but anyway.  In this you see % waiting is 67 and 68 with another respectively being 3 to 100% (which is normal) and the idle being %, 0, 0, 0, 67 and 67.8.  Doing the calculation on the processors shows a wait - slowed virtually 0 - or very little IO wait past with processors.  But if you add up all the expectations of % and % is idling, you have 434ish, and the 134ish and do the calculation is the difference of 300 - which is the 3 who are always 100%.

  So there is no way we can answer your question, we can only help you to understand how the values should be interpreted.

  Waiting for i/o States almost always go on the disc anyway, it is unlikely, therefore, the network or the user entered, looking at the stats of disc ESXTOP is a much better way to determine if there is a problem of e/s.

• Current playback position exceeds the buffer size? VB6 with NIDAQmx 9

  I have an application written in VB6 that does HAVE and AO, with both using the Ctr0 internal as the clock to check that they are in sync.   It seems to work very well.  I use a USB-6212 and a recent version of NIDAQmx.

  I am tracking the current playback Position by calling DAQmxGetReadCurrReadPos_VB6 in the EveryNCallback routine.

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  Van

  Hi afmstm,

  The behavior you're seeing is indeed good, and I agree that the description that it is misleading.  The current description says:

  "In the samples per channel, shows the current position in the.
  buffer.

  The description should be interpreted as:

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     Total samples by acquired channel - current read Position = N

  Another distinction is that you, the user can change the current Position of playback using the properties relative to and Offset .  By default you will read from the current playback Position , which creates the behavior your described.

  I know it can be confusing, but this level of manipulation of streaming is a very advanced concept of DAQmx.  Let me know if this raises questions or concerns more!

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• Question about the keep buffer Pool and the Recycle Buffer Pool

  What will be the Pool of buffers to keep and recycle Buffer Pool contains actually in the Database Buffer Cache, especially what kind of objects? I know the definitions, but need to know the practical aspects to their topic.

  918868 wrote:
  What will be the Pool of buffers to keep and recycle Buffer Pool contains actually in the Database Buffer Cache, especially what kind of objects? I know the definitions, but need to know the practical aspects to their topic.

  When all else fails, read the Fine

  http://docs.Oracle.com/CD/E11882_01/server.112/e16638/memory.htm#PFGRF94285

• Write the list of database buffer Cache

  What is the list of write in the database buffer cache? What is he function?
  
  Thanks in advance.

  >
  What is the list of write in the database buffer cache? What is he function?
  >
  The list of writing to dirty buffers that have not been written to disk yet.

  See "Database Buffer Cache" in the Concepts of database 11g doc
  http://docs.Oracle.com/CD/B28359_01/server.111/b28318/memory.htm
  >
  Organization of the Database Buffer Cache

  The buffers in the cache are organized into two lists: list of Scripture and the list (LRU) least recently used. List of writing contains stamps Sales, but which contain data has changed, has not yet been written to disk. The LRU list holds free buffers, pinned buffers and dirty buffers that have not yet moved to the list of Scripture. Free buffers contain no useful data and can be used. Pinned buffers are being accessed.

• Copies of unexpected CR in the buffer cache

  Hello
  
  While trying to understand the mechanisms of the Oracle buffer cache, I ran a little experiment and observed an unexpected result. I believe that my expectation was wrong, and I would be so grateful, if someone could explain to me what I misunderstood.
  From what I understand, a copy of coherent reading (CR) of a buffer in the cache is created, when the old content of a buffer to be read, for example to ignore changes made by a transaction not yet committed when you query a table. I also thought, that copies of CR in the buffer cache can be reused by subsequent queries requiring a roll back image of the corresponding block.
  
  Now, I ran the following experiment on a DB of 10.2 or slowed down.
  1. I create a BC_TEST table (in a non-SAMS tablespace)
  -> V$ BH shows an A buffer with XCUR status for this table - V$ BH. COURSE number is 4, which indicates a segment according to various sources on the internet header.
  2 session 1 inserts a row into the table (and not valid)
  -> Now V$ BH shows 8 pads with XCUR status attached to the BC_TEST table. I think it's the blocks of a certain extent be allocated to the table (I expected as a block of data to load into the cache in addition to the header that was already there in step 1). There are still buffers with CLASS A # = 4 of step 1, a buffer B with XCUR and CLASS status # = 1, which indicates a block of data, according to various sources on the internet, and other 6 blocks with FREE status and CLASS # = 14 (this value is decoded differently in different sources on the internet).
  3 session 2 emits a ' select * from bc_test ".
  -> V$ BH shows 2 extra buffers with CR and the same FILE status #/ BLOCK # as buffer B in step 2. I understand that it compatible read copy needs to be done to undo the uncommitted changes in step 2 - However, I do not understand why * 2 * these copies are created.
  Note: with a slight variation of the experience, if I don't "select * from bc_test" in Session 2 between step 1 and 2, and then then I only get CR 1 copy in step 3 (I expect).
  4 session 2 issues "select * from bc_test"new"
  -> V$ BH presents yet another extra stamp with CR and the same FILE status #/ BLOCK # as buffer B step 2 (i.e. 3 these buffers in total). Here I don't understand, why the request cannot reuse the CR copy already created in step 3 (which already shows B buffer without changes to the transaction not posted in step 2).
  5 session 2 questions repeatedly ' select * from bc_test "new"
  --> The number of buffers with CR and the same FILE status #/ BLOCK # as buffer B step 2 increases by one with each additional request up to a total of 5. After that, the number of these buffers remains constant after additional queries. However various session statistics 2 ("consistent gets", "Blocks created CR", "the consistent changes", "data blocks consistent reads - undo records applied," "no work - gets consistent reading") suggests, that session 2 continues to generate copies of current reading with each "select * of bc_test" (are the buffers in the buffer cache may simply be reused from there on?).
  
  
  
  To summarize, I have the following question:
  (I) why the insertion of a single line (in step 2) load 8 blocks in the buffer cache - and what does the CLASS # = 14 indicate?
  (II) why the first select statement on the table (step 3) creates 2 copies of CR (use single) data block of the table (instead of one as I expect)?
  (III)) why other issues create copies of CR of this block of data that is unique (instead of reusing the copy created by the first select statement CR)?
  (IV) which limits the number of CR created copies to 5 (is there some parameter checking this value, it is according to some sizing of the cache or is it simply hard-coded)?
  (V) what exactly triggers the creation of a copy of CR of a buffer in the buffer cache?
  
  Thank you very much for any answer
  Kind regards
  Martin
  
  P.S. Please find below the Protocol of my experience
  
  --------------------------------------------------
  Control session
  --------------------------------------------------
  SQL > drop table bc_test;
  
  Deleted table.
  
  SQL > create table bc_test (collar number (9)) tablespace local01;
  
  Table created.
  
  SQL > SELECT bh.file #, bh.block #, bh.class #, bh.status, bh.dirty, bh.temp, bh.ping, bh.stale, bh.direct, bh.new
  2 from V$ BH bh
  3, o dba_objects
  4. WHERE bh. OBJD = o.data_object_id
  5 and o.object_name = 'BC_TEST. '
  6 order of bh.block #;
  
  FOLDER # BLOCK # CLASS # STATUS D T P S D N
  5 209 4 xcur O N N N N N
  
  
  --------------------------------------------------
  Session 1
  --------------------------------------------------
  SQL > insert into bc_test values (1);
  
  1 line of creation.
  
  --------------------------------------------------
  Control session
  --------------------------------------------------
  
  SQL > /.
  
  FOLDER # BLOCK # CLASS # STATUS D T P S D N
  5 209 4 xcur O N N N N N
  5 210 1 xcur O N N N N N
  5 211 14 free N N N N N N
  5 212 14 free N N N N N N
  5 213 14 free N N N N N N
  5 214 14 free N N N N N N
  5 215 14 free N N N N N N
  5 216 14 free N N N N N N
  
  8 selected lines.
  
  --------------------------------------------------
  Session 2
  --------------------------------------------------
  
  SQL > select * from bc_test;
  
  no selected line
  
  
  Statistics
  ----------------------------------------------------------
  28 recursive calls
  0 db block Gets
  Gets 13 coherent
  0 physical reads
  size of roll forward 172
  272 bytes sent via SQL * Net to client
  374 bytes received via SQL * Net from client
  1 SQL * Net back and forth to and from the client
  0 sorts (memory)
  0 sorts (disk)
  0 rows processed
  
  --------------------------------------------------
  Control session
  --------------------------------------------------
  SQL > /.
  
  FOLDER # BLOCK # CLASS # STATUS D T P S D N
  5 209 4 xcur N N N N N N
  5 210 1 cr N N N N N N
  5 210 1 cr N N N N N N
  5 210 1 xcur N N N N N N
  5 211 14 free N N N N N N
  5 212 14 free N N N N N N
  5 213 14 free N N N N N N
  5 214 14 free N N N N N N
  
  8 selected lines.
  
  --------------------------------------------------
  Session 2
  --------------------------------------------------
  SQL > /.
  
  no selected line
  
  
  Statistics
  ----------------------------------------------------------
  0 recursive calls
  0 db block Gets
  Gets 5 in accordance
  0 physical reads
  size of roll forward 108
  272 bytes sent via SQL * Net to client
  374 bytes received via SQL * Net from client
  1 SQL * Net back and forth to and from the client
  0 sorts (memory)
  0 sorts (disk)
  0 rows processed
  
  SQL >
  
  --------------------------------------------------
  Control session
  --------------------------------------------------
  SQL > /.
  
  FOLDER # BLOCK # CLASS # STATUS D T P S D N
  5 209 4 xcur N N N N N N
  5 210 1 cr N N N N N N
  5 210 1 cr N N N N N N
  5 210 1 cr N N N N N N
  5 210 1 xcur O N N N N N
  5 211 14 free N N N N N N
  5 213 14 free N N N N N N
  5 214 14 free N N N N N N
  
  8 selected lines.
  
  SQL >
  
  
  --------------------------------------------------
  Session 2
  --------------------------------------------------
  SQL > select * from bc_test;
  
  no selected line
  
  
  Statistics
  ----------------------------------------------------------
  0 recursive calls
  0 db block Gets
  Gets 5 in accordance
  0 physical reads
  size of roll forward 108
  272 bytes sent via SQL * Net to client
  374 bytes received via SQL * Net from client
  1 SQL * Net back and forth to and from the client
  0 sorts (memory)
  0 sorts (disk)
  0 rows processed
  
  SQL >
  
  --------------------------------------------------
  Control session
  --------------------------------------------------
  SQL > /.
  
  FOLDER # BLOCK # CLASS # STATUS D T P S D N
  5 209 4 xcur N N N N N N
  5 210 1 cr N N N N N N
  5 210 1 cr N N N N N N
  5 210 1 cr N N N N N N
  5 210 1 cr N N N N N N
  5 210 1 xcur O N N N N N
  5 211 14 free N N N N N N
  5 213 14 free N N N N N N
  
  8 selected lines.
  
  --------------------------------------------------
  Session 2
  --------------------------------------------------
  SQL > select * from bc_test;
  
  no selected line
  
  
  Statistics
  ----------------------------------------------------------
  0 recursive calls
  0 db block Gets
  Gets 5 in accordance
  0 physical reads
  size of roll forward 108
  272 bytes sent via SQL * Net to client
  374 bytes received via SQL * Net from client
  1 SQL * Net back and forth to and from the client
  0 sorts (memory)
  0 sorts (disk)
  0 rows processed
  
  --------------------------------------------------
  Control session
  --------------------------------------------------
  SQL > /.
  
  FOLDER # BLOCK # CLASS # STATUS D T P S D N
  5 209 4 xcur N N N N N N
  5 210 1 xcur O N N N N N
  5 210 1 cr N N N N N N
  5 210 1 cr N N N N N N
  5 210 1 cr N N N N N N
  5 210 1 cr N N N N N N
  5 210 1 cr N N N N N N
  5 211 14 free N N N N N N
  5 213 14 free N N N N N N
  
  9 selected lines.
  
  --------------------------------------------------
  Session 2
  --------------------------------------------------
  SQL > select * from bc_test;
  
  no selected line
  
  
  Statistics
  ----------------------------------------------------------
  0 recursive calls
  0 db block Gets
  Gets 5 in accordance
  0 physical reads
  size of roll forward 108
  272 bytes sent via SQL * Net to client
  374 bytes received via SQL * Net from client
  1 SQL * Net back and forth to and from the client
  0 sorts (memory)
  0 sorts (disk)
  0 rows processed
  
  SQL >
  
  --------------------------------------------------
  Control session
  --------------------------------------------------
  SQL > /.
  
  FOLDER # BLOCK # CLASS # STATUS D T P S D N
  5 209 4 xcur N N N N N N
  5 210 1 cr N N N N N N
  5 210 1 cr N N N N N N
  5 210 1 cr N N N N N N
  5 210 1 cr N N N N N N
  5 210 1 cr N N N N N N
  5 210 1 xcur O N N N N N
  
  7 selected lines.
  
  --------------------------------------------------
  Session 2
  --------------------------------------------------
  SQL > /.
  
  no selected line
  
  
  Statistics
  ----------------------------------------------------------
  0 recursive calls
  0 db block Gets
  Gets 5 in accordance
  0 physical reads
  size of roll forward 108
  272 bytes sent via SQL * Net to client
  374 bytes received via SQL * Net from client
  1 SQL * Net back and forth to and from the client
  0 sorts (memory)
  0 sorts (disk)
  0 rows processed
  
  SQL >
  
  --------------------------------------------------
  Control session
  --------------------------------------------------
  SQL > /.
  
  FOLDER # BLOCK # CLASS # STATUS D T P S D N
  5 209 4 xcur N N N N N N
  5 210 1 xcur O N N N N N
  5 210 1 cr N N N N N N
  5 210 1 cr N N N N N N
  5 210 1 cr N N N N N N
  5 210 1 cr N N N N N N
  5 210 1 cr N N N N N N
  
  7 selected lines.

  What version of 10.2, on what platform, and enabled to RAC? What exactly was the tablespace definition?

  (I) why the insertion of a single line (in step 2) load 8 blocks in the buffer cache - and what does the CLASS # = 14 indicate?

  It sounds like you may have formatted all of the first measure - assuming that you use 8 KB blocks and allocated system extended. But it didn't happen when I checked 10.2.0.3 on a version of the single instance of the Oracle.

  Class 14 is interpreted as "unused" in all versions of Oracle that I saw. This would be consistent with the formatting, but it bringing is not below the high water mark. You could ' alter system dump datafile N block min X block max Y' for the segment header block, the block used and unused block in the discharge.

  (II) why the first select statement on the table (step 3) creates 2 copies of CR (use single) data block of the table (instead of one as I expect)?

  Maybe because the first copy clean uncommitted transactions and clones of the second copy of the result to return to a given SNA - but that's just a guess, I had not noticed this behavior before, so I have to do some experiments to find out why it's happening.

  (III)) why other issues create copies of CR of this block of data that is unique (instead of reusing the copy created by the first select statement CR)?

  The first block of CR you create includes a moment earlier that the beginning of the second query, so your session must start from the current. If you have set the isolation level to fix SNA session (for example, "set transaction read only") before the first application, in my view, you would see that the copies of CR will not increase after the initial creation.

  (IV) which limits the number of CR created copies to 5 (is there some parameter checking this value, what is according to some sizing of the cache or is simply badly coded)?

  Hidden parameter Dbblock_max_cr_dba which, by default, is 6. The code still does not seem to obey this limit, but generally you will see only 6 copies of a block in the buffer.

  (V) what exactly triggers the creation of a copy of CR of a buffer in the buffer cache?

  There are at least two options - (a) a session needs to see a copy of a block with recent changes removed, either because they are uncommitted transactions, the session needs to see the block he was when he started to run a particular query or transaction, (b) a session changed a block when he gets hold of the current version - and in certain circumstances (possibly when the update is through a analysis), it will make a copy of the block, mark the previous CR copy, mark the new copy as xcur, and change the copy.

  Incidentally, xcur is a RAC status - it is possible to have blocks in the buffer which are xcur, but not "obtained in current mode.

  Concerning
  Jonathan Lewis

• Results of the EtreCheck report-Help

  Someone who knows how to read this report can help.  My computer is crashing randomly.  It is said that my performance is below average.  What can I do to make it better?

  Here's the report:

  EtreCheck version: 2.9.12 (265)

  Report generated 2016-06-18 07:00:41

  Download EtreCheck from https://etrecheck.com

  Time 07:05

  Performance: Below average

  Click the [Support] links to help with non-Apple products.

  Click [details] for more information on this line.

  Problem: Computer restarts

  Description:

  restart the computer on its own

  Hardware Information: ⓘ

  MacBook Pro (17-inch, mid 2010)

  [Data sheet] - [User Guide] - [warranty & Service]

  MacBook Pro - model: MacBookPro6, 1

  1 2.66 GHz Intel Core i7 CPU: 2 strands

  8 GB of RAM expandable - [Instructions]

  BANK 0/DIMM0

  OK 4 GB DDR3 1067 MHz

  BANK 1/DIMM0

  OK 4 GB DDR3 1067 MHz

  Bluetooth: Old - transfer/Airdrop2 not supported

  Wireless: en1: 802.11 a/b/g/n

  Battery: Health = Normal - Cycle count = 7

  Invalid battery 0123456789ABC serial number

  Video information: ⓘ

  Intel HD Graphics

  NVIDIA GeForce GT 330M - VRAM: 512 MB

  Color LCD 1920 x 1200

  Software: ⓘ

  OS X El Capitan 10.11.5 (15F34) - time since started: about one day

  Disc information: ⓘ

  WDC WD10JPVX-00JC3T0 disk0: (1 TB) (rotation)

  EFI (disk0s1) < not mounted >: 210 MB

  Recovery HD (disk0s3) Volumes/Recovery HD [recovery]: 650 MB (85 MB free)

  [redacted] Mac (disk1) /: 998,97 go-go (795,79 free)

  Encrypted AES - XTS unlocked

  Storage of carrots: disk0s2 999.35 GB Online

  MATSHITADVD-R UJ-868)

  USB information: ⓘ

  MediaTek WiFi

  Computer, Inc. Apple IR receiver.

  Apple Inc. BRCM2070 hub.

  Apple Inc. Bluetooth USB host controller.

  Apple Inc. Apple keyboard / Trackpad

  Guardian: ⓘ

  Mac App Store and identified developers

  Kernel extensions: ⓘ

  / Applications/Parallels Desktop.app

  [loading] com.parallels.kext.hypervisor (11.2.0). 32581 - SDK 10.9 - 2016-05-04) [Support]

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  / System/Library/Extensions

  com [loading]. Ralink.driver.RT2870USBWirelessDriver (5.0.1.25 - SDK 10.10 - 2016-06-02) [Support]

  Startup items: ⓘ

  WiFiUtilityStartUp: Path: / System/Library/StartupItems/WiFiUtilityStartUp

  Startup items are obsolete in OS X Yosemite

  Launch system officers: ⓘ

  [loaded] 8 tasks Apple

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  5% kernel_task

  Top of page process of memory: ⓘ

  3.79 GB prl_vm_app

  796 MB kernel_task

  Com.apple.WebKit.WebContent 328 MB

  Mdworker (15) 229 MB

  Safari of 106 MB

  Virtual memory information: ⓘ

  280 MB of free RAM

  used 7.72 GB RAM (384 MB Cache)

  563 MB used Swap

  Diagnostic information: ⓘ

  17 June 2016, 06:13:12 self test - passed

  Is slow when you exit Parallels too?

• Cannot add music on my iPhone via iTunes!

  I apologize if this has been asked several times, I'm afraid, I can't find a solution for myself and I have a few questions myself.

  First of all, I have an iPhone running iOS 9.2 5. Operating system my PC is Windows 8.1. My iTunes is currently 12.3. I dunno if iTunes itself set to date (although I'm sure that I have disabled the automatic updates), but I've never manually updated my iTunes since the installation. And since the installation, I was able to add music on my phone just fine. I don't have to add songs to the iTunes library and let it sync. I used to drag and drop the song directly in my phone's music library. It goes the same for movies.

  But today, my cursor turns into a 'NO' (the circle with the slash) symbol each time do I drag and drop my songs, as I've done normally. The song is not added, and my phone's music library is unchanged. I have still a few left GB which I believe is sufficient to add 2-3 songs.

  So, I tried looking for an online solution. Most of them recommended to "manually manage music and videos". This setting average/do what? I tried to check it out and apply the changes, however, I get a popup window that informs me that my phone is synchronized with another library iTunes on another computer and if I want to "erase this iPhone and sync with this iTunes library."

  He adds: "you can sync an iPhone with the library only iTunes at the same time." Erasing and syncing replaces the content of this iPhone with the contents of this iTunes library. »

  Now, as I mentioned, I used to add tracks directly on my phone, so right now, my iTunes library is completely empty. Following this and correct me if I'm wrong, but if I choose "Erase and Sync" means that all the content - not just music, but the photos and videos - my iPhone will be deleted? Because if so, I'm afraid, I can't continue to "manually manage music and videos". :/

  I also found other solutions, such as:

  1. quickly disconnect the phone after pressing "erase and Sync. However, I am a little worried about it. I do not say that it does not, but he had been warned that if you unplug a little too late, you may lose your data. And I'm not willing to take that risk.

  2 uncheck "share my library on the local network. However, mine is already disabled. Check back it will solve my problem? This setting even mean?

  It's so about it. If anyone can help or even answer some of my questions, it would be really great! Any help is appreciated. Thank you!

  In addition, if you need further details, let me know.

  See recover your iTunes from your iPod or an iOS device. You should always keep a complete copy of your media on your computer and which back up to another drive for security. You will then be able to restore or to recharge your device if need arise.

  Clear and synchronisation warning refers to music, videos, podcasts, and other media, not your data app and settings or your camera. However you would be advised to return to the top of your device and the copy on the contents of the camera roll before making any changes.

  TT2

• Games work no better in my TOSHIBA Satellite L350D?

  Guys I install 2 new games and they don't work well, I can not play my grafikkart is INTEL HD 4500GMA.

  What games do you have average and what you want to say you can t he play?

  My Satellite U400 has an Intel GMA4500 HD and I don t have any problems with it. OK, it s not the fastest graphic card games but quite fast for the older games of high quality and latest on the quality of low medium.

  I use the latest driver from Toshiba. Have you tested?
  You can download the Toshiba site:
  http://EU.computers.Toshiba-Europe.com

• Ignore the condition AVERAGEIF referenced cell is empty?

  I'm trying to run an AVERAGEIF function based on up to 3 sets of test values whose conditions are determined by the entries in cells referenced in the formula.

  Here is a picture of the table:

  

  Seminar, day and hour are the columns from the drop-down list. Registered average and average archived have AVERAGEIF formulas that use the data in the table another as test values that resolves as true when they correspond to data in the corresponding cells of the seminar, day and hour [example: IFERROR (AVERAGEIF (seminar, Schedule::Registered, $Topic, $A2, $Days, $B2, hourly seminars: $B, $C2),' ')]. This set works fine if I have all three of the cells in the columns of seminar, day and time information, but I would like to also see average which which do not take into account (i.e., what are the averages for what seminar is offered Monday, regardless of what time?). Is it possible to get these formulas to ignore the conditions that are left in white?

  Thank you!

  AVERAGEIF accept them * and? wildcard characters.  You could try to enter * in the Time column and see if your formula returns desired results.  Without seeing the data format in your time table of the seminars, however, it is difficult to know if you need to make other adjustments to make it work.

  For example, you use Date & time format of data?  If so, numbers actually stores a date and time, even if you have a cell that is formatted to display only the time. This can give results, that you are not pregnant if you do not have two cells showing that same time can be interpreted by the numbers as being different, because the part of associated of the date string date and hour is different.

  SG

• Questions about the frequency step response

  Hello

  I use the Signal Express 3.0. I'm not clear on the transfer functions in step of frequency response with different modes of calculation of the average. What I got from the help file is this H (f) = Sab (f) /Saa (f) which is cross the frequencies on the spectrum auto where is the pulse and b the signal response signal. When the mean quadratic value is used, I wonder if the transfer function becomes greatness of cross spectrum divided by the magnitude of the spectrum of the car. When an average of vector is, everthing is used in complex numbers. He averaged temporal signals, frequency domain signals, or the results of transfer functions?

  Thank you very much.

  The algorithm to calculate the spectrum is the same in both modes.  However, the method of calculation of the average can have a huge impact on the outcome.  Mean quadratic value is performed on the spectrum itself, after the calculation.  Vector averaging is done on the input signals before the calculation.  With an average of vector signals must be consistent (have the same phase) or the result will be bad due to the signal being on average by far.

• AI trigger and measure multi channels

  Hi all

  I have a simple problem (using USB-6259).

  can repeat the measurement trigger of AI and measure multi channels, but not both at the same time.

  -DAQmxCreateAIVoltageChan(hd, "DEV1/ai0",...)                    define the ai0 as the trigger channel

  -DAQmxCfgSampClkTiming

  -DAQmxCfgAnlgEdgeRefTrig

  -DAQmxStartTask

  -DAQmxReadAnalogF64 (hd, "DEV / ai0:3 ', / / I want to measure more channels"ai0:3"not just"ai0")

  Thank you

  Hassan

  Hi Hassan,.

  If you use an analog trigger with several analog channels, you will need to use the APFI0 input as source of relaxation. See this KB: Why do I get error-200264 when running analog reference trigger? All you need to do is to connect your analog signal online 0 to APFI0 (Paperback 20 in your case) and set the source of relaxation at APFI0.

  The reason is that you don't have that an NOC on Board (series E or M) and she's going to have to switch between the different lines (see this KB: modes of sampling). This parameter collides with the idea of a trigger analog reference on a specific line (constant sampling of data in a ring buffer up to what a condition is met). The APFI0 line, however, has its own CDA. Therefore, it can run simultaneously.

  However, please note that the ADC is fast but has lower resolution to HAVE it sampling ADC. See these KBs: series E and M series Analog Input Trigger resolution, be aware of a possible error between the analog trigger threshold and the value of the first sample

  Hope this clarified the issue.

  Best regards

  Peter