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  • How to generate unique serial number of machine in Delphi?

    - by noxwow
    Hi, I have question how to generate unique serial number of machine in Delphi? I tried to do this using the ID the motherboard or processor, but unfortunately it's unfortunately supported. Partition serial numbers, etc. fall off, because it is changing after the formatted. I'm looking for something that doesn't change after the formatted. Has anyone any idea?

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  • How to prove worst-case number of inversions in a heap is O(nlogn)?

    - by Jacques
    I am busy preparing for exams, just doing some old exam papers. The question below is the only one I can't seem to do (I don't really know where to start). Any help would be appreciated greatly. Use the O(nlogn) comparison sort bound, the theta(n) bound for bottom-up heap construction, and the order complexity if insertion sort to show that the worst-case number of inversions in a heap is O(nlogn).

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  • How to get number of entries in a Lua table?

    - by romkyns
    Sounds like a "let me google it for you" question, but somehow I can't find an answer. The Lua # operator only counts entries with integer keys, and so does table.getn: tbl = {} tbl["test"] = 47 tbl[1] = 48 print(#tbl, table.getn(tbl)) -- prints "1 1" count = 0 for _ in pairs(tbl) do count = count + 1 end print count -- prints "2" How do I get the number of all entries?

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  • what is the probability that the next random number will equal current one?

    - by I__
    if i do this in a worksheet: =RAND() i also specified that i want only 2 decimal places and for example let's say that rand() = 0.07 what is the probability that when i call this function again i will get 0.07 ?? i know that ideally if we assume 100% randomness, the answer would be 1/ (10 * 10) because there are only 100 possible combinations, but what would it be according to the way excel generates a random number?

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  • Mysql: Order Results by number of matching rows in Second Table.

    - by KyleT
    I'm not sure the best way to word this question so bear with me. Table A has following columns: id name description Table B has the following columns: id a_id(foreign key to Table A) ip_address date Basically Table B contains a row for each time a user views a row from Table A. My question is how do I sort Table A results, based on the number of matching rows in Table B. i.e SELECT * FROM TableA ORDER BY (SELECT COUNT(*) FROM TableB where TableB.a_id = TableA.id) Thank you!

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  • Number of 0/1 colorings of a m X n rectangle which have no monochromatic subrectangles with both dimension greater than 1.

    - by acbruptenda
    A m x n rectangular matrix is give, and each cell is to be filled with 0/1 colour. I have to find number of colorings possible so that there is no monochromatic coloured (same colour) sub-rectangle whose both dimension is greater than 1 (eg - 2x2, 2x3,4x3) I have found a slightly different version of it here But they have said nothing about the algorithm. So, I am looking for an algorithm here !!

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  • How to read large number of rows efficiently using Zend_Db?

    - by Alex N.
    Is there a simple :) and efficient way or reading very large number of rows sequentially using Zend_Db? Basically I need to process entire table, row by row. Table is large, primary key sequence is not guaranteed(i.e. not an autoincrement, but is UNSIGNED INT). What's the best way to approach this? Environment: PHP 5.2, Zend Framework 1.10, MySQL 5.1

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  • How to map a long integer number to a N-dimensional vector of smaller integers (and fast inverse)?

    - by psihodelia
    Given a N-dimensional vector of small integers is there any simple way to map it with one-to-one correspondence to a large integer number? Say, we have N=3 vector space. Can we represent a vector X=[(int32)x1,(int32)x2,(int32)x3] using an integer (int48)y? The obvious answer is "Yes, we can". But the question is: "What is the fastest way to do this and its inverse operation?"

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  • How can I limit the number of registrants to an event?

    - by user356900
    I've set up a basic html/php submission form where people can register for our event, but need a way to replace the submission form webpage with one that reads something like "We have reached our registration limit" when we reach a certain number of submitted forms. Our database is MySQL (if that makes a difference) I've looked around on the web but people either say to count the entries by hand, or the ones that do have an automated system use CMS like drupal or joomla. Is it possible to setup an automated script that will do this?

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  • Bladecenter-E Power Module fault

    - by Lihnjo
    We have problem on IBM Bladecenter-E Critical Events Power module 2 is off. DC fault. Power module 4 is off. DC fault. Warnings and System Events Insufficient chassis power to support redundancy What is the best solution for this problem? Thanks AMM Service Data Help SPAPP Capture Available 10/13/2010 17:03:47 1090347 bytes Time: 11/19/2012 11:02:31 UUID: 42E1 5D2F D7BF 41A6 A4A2 48D1 3FB7 0540 MAC Address xx:xx:xx:xx:xx:xx MM Information Name: nnnnn Contact: aaa, bbb, ccc, England Location: [email protected] IP address: 111.222.333.444 Date Time Information GMT offset: +1:00 - Central Europe Time (Western Europe, Algeria, Nigeria, Angola) Adjust for DST: Yes NTP: Enabled NTP Hostname/IP: 111.222.333.444 System Health: Critical System Status Summary One or more monitored parameters are abnormal. Critical Events Power module 2 is off. DC fault. Power module 4 is off. DC fault. Warnings and System Events Insufficient chassis power to support redundancy CHASSIS (BladeCenter-E) in CHASSIS slot: 01 TopoPath is "CHASSIS[1]". Description : BladeCenter-E Width : 1 Sub Type : BladeCenter (BC) Power Mode : 220 v KVM Owner : CHASSIS[1]/BLADE[9] MT Owner : CHASSIS[1]/MGMT_MOD[1] Component Type : CHASSIS Inventory: VPD ID: 336 (decimal) POS ID EXT: 0 (decimal) POS ID: 8 (decimal) Machine Type/Model: 86773RG Machine Serial Number: 99ZL816 Part Number: 39R8561 FRU Number: 39R8563 FRU Serial Number: YK109174W1HV Manufacturer ID: IBM Hardware Revision: 3 (decimal) Manufacture Date: 18 (wk), 07 (yr) UUID: 42E1 5D2F D7BF 41A6 A4A2 48D1 3FB7 0540 (hex) Type Code: 97 (decimal) Sub-type Code: 0 (decimal) IANA Num: 336 (decimal) Product ID: 8 (decimal) Manufacturer Sub ID: FOXC Enviroment data: -------------- Type: : POWER_USAGE Unit: : WATTS Reading: : 0xa Sensor Label: : Midplane Sensor ID: : 0x0 MGMT MOD (Advanced Management Module) in MGMT_MOD slot: 01 TopoPath is "CHASSIS[1]/MGMT_MOD[1]". Description : Advanced Management Module Name : kant Width : 1 Component Role : Primary Component Type : MGMT MOD Insert Time : 28050132 Inventory: VPD ID: 288 (decimal) POS ID EXT: 0 (decimal) POS ID: 4 (decimal) Part Number: 39Y9659 FRU Number: 39Y9661 FRU Serial Number: YK11836CE2RC Manufacturer ID: IBM Hardware Revision: 4 (decimal) Manufacture Date: 50 (wk), 06 (yr) UUID: 1D95 9937 8CA5 11DB 9499 0014 5EDF 1C98 (hex) Type Code: 81 (decimal) Sub-type Code: 1 (decimal) IANA Num: 20301 (decimal) Product ID: 65 (decimal) Manufacturer Sub ID: ASUS Firmware data: Type : AMM firmware Build ID : BPET50P File Name : CNETCMUS.PKT Release Date : 03/26/2010 Release Level : 50 Revision - Major: 80 Port info: ======================================================== Topology Path ID : 1 Label : External Phy Orientation : EXTERNAL Port Number : 1 Type : MGT Physical Meidum : Copper Number of Link Intferfaces : 1 ------------------------------------ Link Ifc ID Number : 1 Link Ifc Transport Protocol : ENET Link Ifc Addr Type : MAC Link Ifc Burned-in Addr : xx:xx:xx:xx:xx:xx Link Ifc Admin Addr : 00:00:00:00:00:00 Link Ifc Addr in use : xx:xx:xx:xx:xx:xx ---------------------------------------------------------- Configuration behaviors: Save Only Enviroment data: -------------- Type: : TEMPERATURE Unit: : DEGREES_C Reading: : 38.00 Sensor Label: : MM Ambient Sensor ID: : 0x0 -------------- Type: : VOLTAGE Unit: : VOLTS Reading: : +4.81 Sensor Label: : +5V Sensor ID: : 0x1b -------------- Type: : VOLTAGE Unit: : VOLTS Reading: : +3.26 Sensor Label: : +3.3V Sensor ID: : 0x19 -------------- Type: : VOLTAGE Unit: : VOLTS Reading: : +11.97 Sensor Label: : +12V Sensor ID: : 0x16 -------------- Type: : VOLTAGE Unit: : VOLTS Reading: : -4.88 Sensor Label: : -5V Sensor ID: : 0x1e -------------- Type: : VOLTAGE Unit: : VOLTS Reading: : +2.47 Sensor Label: : +2.5V Sensor ID: : 0x18 -------------- Type: : VOLTAGE Unit: : VOLTS Reading: : +1.76 Sensor Label: : +1.8V Sensor ID: : 0x15 -------------- Type: : POWER_USAGE Unit: : WATTS Reading: : 0x19 Sensor Label: : kant Sensor ID: : 0x0

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  • Quantifying the effects of partition mis-alignment

    - by Matt
    I'm experiencing some significant performance issues on an NFS server. I've been reading up a bit on partition alignment, and I think I have my partitions mis-aligned. I can't find anything that tells me how to actually quantify the effects of mis-aligned partitions. Some of the general information I found suggests the performance penalty can be quite high (upwards of 60%) and others say it's negligible. What I want to do is determine if partition alignment is a factor in this server's performance problems or not; and if so, to what degree? So I'll put my info out here, and hopefully the community can confirm if my partitions are indeed mis-aligned, and if so, help me put a number to what the performance cost is. Server is a Dell R510 with dual E5620 CPUs and 8 GB RAM. There are eight 15k 2.5” 600 GB drives (Seagate ST3600057SS) configured in hardware RAID-6 with a single hot spare. RAID controller is a Dell PERC H700 w/512MB cache (Linux sees this as a LSI MegaSAS 9260). OS is CentOS 5.6, home directory partition is ext3, with options “rw,data=journal,usrquota”. I have the HW RAID configured to present two virtual disks to the OS: /dev/sda for the OS (boot, root and swap partitions), and /dev/sdb for a big NFS share: [root@lnxutil1 ~]# parted -s /dev/sda unit s print Model: DELL PERC H700 (scsi) Disk /dev/sda: 134217599s Sector size (logical/physical): 512B/512B Partition Table: msdos Number Start End Size Type File system Flags 1 63s 465884s 465822s primary ext2 boot 2 465885s 134207009s 133741125s primary lvm [root@lnxutil1 ~]# parted -s /dev/sdb unit s print Model: DELL PERC H700 (scsi) Disk /dev/sdb: 5720768639s Sector size (logical/physical): 512B/512B Partition Table: gpt Number Start End Size File system Name Flags 1 34s 5720768606s 5720768573s lvm Edit 1 Using the cfq IO scheduler (default for CentOS 5.6): # cat /sys/block/sd{a,b}/queue/scheduler noop anticipatory deadline [cfq] noop anticipatory deadline [cfq] Chunk size is the same as strip size, right? If so, then 64kB: # /opt/MegaCli -LDInfo -Lall -aALL -NoLog Adapter #0 Number of Virtual Disks: 2 Virtual Disk: 0 (target id: 0) Name:os RAID Level: Primary-6, Secondary-0, RAID Level Qualifier-3 Size:65535MB State: Optimal Stripe Size: 64kB Number Of Drives:7 Span Depth:1 Default Cache Policy: WriteBack, ReadAdaptive, Direct, No Write Cache if Bad BBU Current Cache Policy: WriteThrough, ReadAdaptive, Direct, No Write Cache if Bad BBU Access Policy: Read/Write Disk Cache Policy: Disk's Default Number of Spans: 1 Span: 0 - Number of PDs: 7 ... physical disk info removed for brevity ... Virtual Disk: 1 (target id: 1) Name:share RAID Level: Primary-6, Secondary-0, RAID Level Qualifier-3 Size:2793344MB State: Optimal Stripe Size: 64kB Number Of Drives:7 Span Depth:1 Default Cache Policy: WriteBack, ReadAdaptive, Direct, No Write Cache if Bad BBU Current Cache Policy: WriteThrough, ReadAdaptive, Direct, No Write Cache if Bad BBU Access Policy: Read/Write Disk Cache Policy: Disk's Default Number of Spans: 1 Span: 0 - Number of PDs: 7 If it's not obvious, virtual disk 0 corresponds to /dev/sda, for the OS; virtual disk 1 is /dev/sdb (the exported home directory tree).

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  • July, the 31 Days of SQL Server DMO’s – Day 24 (sys.dm_db_index_operational_stats)

    - by Tamarick Hill
    The sys.dm_db_index_operational_stats Dynamic Management Function returns information about the IO, locking, and access methods for the indexes that you currently have on your SQL Server Instance. This function takes four input parameters which are (1) database_id, (2) object_id, (3) index_id, and (4) partition_number. Let’s have a look at the results from this function against our AdventureWorks2012 database. This function returns a ton of columns, so not only will I not attempt to describe each of the columns, I wont even attempt to display all of them here. My query below will give you a subset of the columns returned from this function. SELECT database_id, object_id, index_id, partition_number, leaf_insert_count, leaf_delete_count, leaf_update_count, leaf_ghost_count, nonleaf_insert_count, nonleaf_delete_count, nonleaf_update_count, range_scan_count, forwarded_fetch_count, row_lock_count, row_lock_wait_count, page_lock_count, page_lock_wait_count, Index_lock_promotion_attempt_count, index_lock_promotion_count, page_compression_attempt_count, page_compression_success_count FROM sys.dm_db_index_operational_stats(db_id('AdventureWorks2012'), NULL, NULL, NULL) The first four columns in the result set represent the values that we passed in as our input parameters. If you use NULL’s as I did, then you will see results for every index on your system. I specified a database_id so my result set only shows those records pertaining to my AdventureWorks2012 database. The next columns in the result set provide you with information on how may inserts, deletes, or updates that have taken place on your leaf and nonleaf index levels. The nonleaf levels would refer to the intermediate and root index levels. In the middle of these you see a leaf_ghost_count column, which represents the number of records that have been logically deleted and marked as “ghosted”  and are waiting on the background ghost cleanup process to physically remove them. The range_scan_count column represents the number of range or table scans that have been performed against an index. The forwarded_fetch_count column represents the number of rows that were returned from a forwarding row pointer. The row_lock_count and row_lock_wait_count represent the number of row locks that have been requested for an index and the number of times SQL has had to wait on a row lock respectively. The page_lock_count and page_lock_wait_count represent the number of page locks that have been requested for an index and the number of times SQL has had to wait on a page lock respectively. The index_lock_promotion_attempt_count represents the number of times the database engine has attempted to promote a lock to the index level. The index_lock_promotion_count column displays how many times that index lock promotion was successful. Lastly the page_compression_attempt_count and page_compression_success_count represents how many times a page was attempted to be compressed and how many times the attempt was successful. As you can see there is a ton of information returned from this DMV. The DMV we reviewed on yesterday (sys.dm_db_index_usage_stats) provided you with good information on when and how indexes have been used, but this DMF takes an even deeper dive into these statistics. If you are interested in performing a very detailed analysis on the operational stats of your indexes, this is not only a good place to start, but more than likely the best place. For more information on this Dynamic Management Function, please see the below Books Online link: http://msdn.microsoft.com/en-us/library/ms174281.aspx Follow me on Twitter @PrimeTimeDBA

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  • How to get better at solving Dynamic programming problems

    - by newbie
    I recently came across this question: "You are given a boolean expression consisting of a string of the symbols 'true', 'false', 'and', 'or', and 'xor'. Count the number of ways to parenthesize the expression such that it will evaluate to true. For example, there is only 1 way to parenthesize 'true and false xor true' such that it evaluates to true." I knew it is a dynamic programming problem so i tried to come up with a solution on my own which is as follows. Suppose we have a expression as A.B.C.....D where '.' represents any of the operations and, or, xor and the capital letters represent true or false. Lets say the number of ways for this expression of size K to produce a true is N. when a new boolean value E is added to this expression there are 2 ways to parenthesize this new expression 1. ((A.B.C.....D).E) ie. with all possible parenthesizations of A.B.C.....D we add E at the end. 2. (A.B.C.(D.E)) ie. evaluate D.E first and then find the number of ways this expression of size K can produce true. suppose T[K] is the number of ways the expression with size K produces true then T[k]=val1+val2+val3 where val1,val2,val3 are calculated as follows. 1)when E is grouped with D. i)It does not change the value of D ii)it inverses the value of D in the first case val1=T[K]=N.( As this reduces to the initial A.B.C....D expression ). In the second case re-evaluate dp[K] with value of D reversed and that is val1. 2)when E is grouped with the whole expression. //val2 contains the number of 'true' E will produce with expressions which gave 'true' among all parenthesized instances of A.B.C.......D i) if true.E = true then val2 = N ii) if true.E = false then val2 = 0 //val3 contains the number of 'true' E will produce with expressions which gave 'false' among all parenthesized instances of A.B.C.......D iii) if false.E=true then val3=( 2^(K-2) - N ) = M ie. number of ways the expression with size K produces a false [ 2^(K-2) is the number of ways to parenthesize an expression of size K ]. iv) if false.E=false then val3 = 0 This is the basic idea i had in mind but when i checked for its solution http://people.csail.mit.edu/bdean/6.046/dp/dp_9.swf the approach there was completely different. Can someone tell me what am I doing wrong and how can i get better at solving DP so that I can come up with solutions like the one given above myself. Thanks in advance.

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  • How to show button ‘Done’ on number pad on iPhone OS 4?

    - by Will Harris
    I'd like to add a Done button to the iPhone number pad keyboard. There's even a handy space at the bottom left for just such a button. Previously, I was using a similar trick to those described in Question 584538 and Luzian Scherrer's excellent blog post, but that stopped working in iOS 4. I can do it by creating a custom inputView, but I'd prefer to extend Apple's keyboard instead of writing my own. Is there a new way to add a view to the standard keyboard? Has someone published an OSS inputView for this? Is there another way?

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  • How do i specify wcf behaviorExtension class type without the assembly version number?

    - by Mr Bell
    I have a web app that uses a WCF service that utilizes a behaviorExtension like so: <behaviorExtensions> <add name="clientCredentialsExtension" type="Simon.Web.Giftcard.WCFSecurity.ClientCredentialsExtensionElement, Simon.Web.Giftcard, Version=1.0.3736.20411, Culture=neutral, PublicKeyToken=null"/> </behaviorExtensions> The problem is this web app's version changes with every compile (i think) and thus invalidating this entry. How can I avoid having to change the version number every time I compile this? Can I specify the extension in code somewhere?

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