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  • How are Reads Distributed in a Workload

    - by Bill Graziano
    People have uploaded nearly one millions rows of trace data to TraceTune.  That’s enough data to start to look at the results in aggregate.  The first thing I want to look at is logical reads.  This is the easiest metric to identify and fix. When you upload a trace, I rank each statement based on the total number of logical reads.  I also calculate each statement’s percentage of the total logical reads.  I do the same thing for CPU, duration and logical writes.  When you view a statement you can see all the details like this: This single statement consumed 61.4% of the total logical reads on the system while we were tracing it.  I also wanted to see the distribution of reads across statements.  That graph looks like this: On average, the highest ranked statement consumed just under 50% of the reads on the system.  When I tune a system, I’m usually starting in one of two modes: this “piece” is slow or the whole system is slow.  If a given piece (screen, report, query, etc.) is slow you can usually find the specific statements behind it and tune it.  You can make that individual piece faster but you may not affect the whole system. When you’re trying to speed up an entire server you need to identity those queries that are using the most disk resources in aggregate.  Fixing those will make them faster and it will leave more disk throughput for the rest of the queries. Here are some of the things I’ve learned querying this data: The highest ranked query averages just under 50% of the total reads on the system. The top 3 ranked queries average 73% of the total reads on the system. The top 10 ranked queries average 91% of the total reads on the system. Remember these are averages across all the traces that have been uploaded.  And I’m guessing that people mainly upload traces where there are performance problems so your mileage may vary. I also learned that slow queries aren’t the problem.  Before I wrote ClearTrace I used to identify queries by filtering on high logical reads using Profiler.  That picked out individual queries but those rarely ran often enough to put a large load on the system. If you look at the execution count by rank you’d see that the highest ranked queries also have the highest execution counts.  The graph would look very similar to the one above but flatter.  These queries don’t look that bad individually but run so often that they hog the disk capacity. The take away from all this is that you really should be tuning the top 10 queries if you want to make your system faster.  Tuning individually slow queries will help those specific queries but won’t have much impact on the system as a whole.

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  • Webcast - June 27th: Announcing Oracle SuperCluster T5-8: Our Fastest Engineered System

    - by Javier Puerta
    Join us for a live webcast with Oracle Executive Vice President, John Fowler, as he announces the new Oracle SuperCluster T5-8: Our Fastest Engineered System. Learn how the new Oracle SuperCluster T5-8 delivers: Extreme performance through Oracle Exadata, Oracle Exalogic, Oracle’s virtualization solutions, and the world’s fastest servers Highest availability with no single point of failure and 99.999% uptime Highest efficiency with unmatched price/performance and the lowest operating costs A complete engineered system ideal for database and application consolidation and private cloud Register here

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  • MySQL: operation of summing and division ?

    - by Nick
    Alright, so I have a user table and would like to get the max value for the user with the highest amount of points divided by a score. Below is a rough idea of what I'm looking for: SELECT MAX(SUM(points)/SUM(score)) FROM users I'm not interested in adding up both columns and dividing, rather I'm interested in dividing the points and score for each user and retrieve the highest value out of the lot.

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  • How to sum and divide in MySql [migrated]

    - by Nick
    Alright, so I have a user table and would like to get the max value for the user with the highest amount of points divided by a score. Below is a rough idea of what I'm looking for: SELECT MAX(SUM(points)/SUM(score)) FROM users I'm not interested in adding up both columns and dividing, rather I'm interested in dividing the points and score for each user and retrieve the highest value out of the lot.

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  • Felix Baumgartner Skydives from the Edge of Space [Video]

    - by Jason Fitzpatrick
    Yesterday Felix Baumgartner broke the record for highest skydive by leaping out of a capsule 128,100 feet above the Earth. Check out his jump in the following videos. After flying to an altitude of 39,045 meters (128,100 feet) in a helium-filled balloon, Felix Baumgartner completed a record breaking jump for the ages from the edge of space, exactly 65 years after Chuck Yeager first broke the sound barrier flying in an experimental rocket-powered airplane. Felix reached a maximum of speed of 1,342.8 km/h (833mph) through the near vacuum of the stratosphere before being slowed by the atmosphere later during his 4:20 minute long freefall. The 43-year-old Austrian skydiving expert also broke two other world records (highest freefall, highest manned balloon flight), leaving the one for the longest freefall to project mentor Col. Joe Kittinger. The above video is a 2 minute highlight reel of the ascent and jump; check out the full 15 minute descent video here. For an in-depth look at the technology used to keep Baumgartner safe during his record setting journey, hit up the link below. The Tech Behind Felix Baumgartner’s Stratospheric Skydive [ExtremeTech] HTG Explains: What is the Windows Page File and Should You Disable It? How To Get a Better Wireless Signal and Reduce Wireless Network Interference How To Troubleshoot Internet Connection Problems

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  • Does it matter the direction of a Huffman's tree child node?

    - by Omega
    So, I'm on my quest about creating a Java implementation of Huffman's algorithm for compressing/decompressing files (as you might know, ever since Why create a Huffman tree per character instead of a Node?) for a school assignment. I now have a better understanding of how is this thing supposed to work. Wikipedia has a great-looking algorithm here that seemed to make my life way easier. Taken from http://en.wikipedia.org/wiki/Huffman_coding: Create a leaf node for each symbol and add it to the priority queue. While there is more than one node in the queue: Remove the two nodes of highest priority (lowest probability) from the queue Create a new internal node with these two nodes as children and with probability equal to the sum of the two nodes' probabilities. Add the new node to the queue. The remaining node is the root node and the tree is complete. It looks simple and great. However, it left me wondering: when I "merge" two nodes (make them children of a new internal node), does it even matter what direction (left or right) will each node be afterwards? I still don't fully understand Huffman coding, and I'm not very sure if there is a criteria used to tell whether a node should go to the right or to the left. I assumed that, perhaps the highest-frequency node would go to the right, but I've seen some Huffman trees in the web that don't seem to follow such criteria. For instance, Wikipedia's example image http://upload.wikimedia.org/wikipedia/commons/thumb/8/82/Huffman_tree_2.svg/625px-Huffman_tree_2.svg.png seems to put the highest ones to the right. But other images like this one http://thalia.spec.gmu.edu/~pparis/classes/notes_101/img25.gif has them all to the left. However, they're never mixed up in the same image (some to the right and others to the left). So, does it matter? Why?

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  • STOP PRESS: FY15 Q1 Oracle ZS3 Contest for Partners

    - by Cinzia Mascanzoni
    04 JUNE 2014 Oracle EMEA Partners Stop Press Stay Connected Oracle Media Network   OPN on PartnerCast   STOP PRESS: FY15 Q1 Oracle ZS3 Contest for PartnersShare an unforgettable experience at the Teatro Alla Scala in Milan Dear valued Partner, We are pleased to launch a partner contest exclusive to our partners dedicated to promoting and selling Oracle Systems! You are essential to the success of Oracle and we want to recognize your contribution and effort in driving Oracle Storage to the market. To show our appreciation we are delighted to announce a contest, giving the winners the opportunity to attend a roundtable chaired by Senior Oracle Executives and spend an unforgettable evening at the magnificent Teatro Alla Scala in Milan, followed by a stay at the Grand Hotel et de Milan, courtesy of Oracle. Recognition will be given to 12 partner companies (10 VARs & 2 VADs) who will be recognized for their ZFS storage booking achievement in the broad market between June 1st and July 18th 2014. Criteria of Eligibility A minimum deal value of $30k is required for qualification Partners who are wholly or partially owned by a public sector organization are not eligible for participation Winners The winning VARs will be: The highest ZS3 or ZBA bookings achievers by COB on July 18th, 2014 in each Oracle EMEA region (1) The highest Oracle on Oracle (2) ZS3 or ZBA bookings achievers by COB on July 18th, 2014 in each Oracle EMEA region The winning VADs (3) will be: The highest ZS3 or ZBA bookings achiever by COB on July 18th 2014 in EMEA The highest Oracle on Oracle (2) ZS3 or ZBA bookings achiever by COB on July 18th 2014 in EMEA (1) Two VAR winners for each EMEA region – Eastern Europe & CIS, Middle East & Africa, South Europe, North Europe, UK/Ireland & Israel - as per the criteria outlined above(2) Oracle on Oracle, in this instance, means ZS3 or ZBA storage attached to DB or DB options, Engineered Systems or Sparc servers sold to the same customer by the same partner within the contest timelines.(3) Two VAD winners, one for each of the criteria outlined above, will be selected from across EMEA. Oracle shall be the final arbiter in selecting the winners. All winners will be notified via their Oracle account manager. Full details about the contest, expenses covered by Oracle and timetable of events can be found on the Oracle EMEA Hardware (Servers & Storage) Partner Community workspace (FY15 Q1 ZFS Partner Contest). Access to the community workspace requires membership. If you are not a member please register here. The Prize Winners will be invited to participate to a roundtable chaired by Oracle on Monday September 8th 2014 in Milan and to be guests of Oracle in the evening of September 8th, 2014 at the Teatro Alla Scala. The evening will comprise of a private tour of the Scala museum, cocktail reception at the elegant museum rooms and attending the performance by the renowned Soprano, Maria Agresta. Our guests will then retire for the evening to the Grand Hotel et de Milan, courtesy of Oracle. Good Luck!! For more information, please contact Sasan Moaveni. Regards, Olivier TordoSenior Director - Systems Business DevelopmentOracle EMEA Alliances & Channels Resources EMEA Hardware Partner Community EMEA Oracle Partner Days Find Partner Events EMEA Partner News Blog EMEA Partner Enablement Blog Oracle PartnerNetwork Copyright © 2014, Oracle and/or its affiliates.All rights reserved. Contact Us | Legal Notices and Terms of Use | Privacy Statement

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  • Oracle ZS3 Contest for Partners: Share an unforgettable experience at the Teatro Alla Scala in Milan

    - by Claudia Caramelli-Oracle
    12.00 Dear valued Partner, We are pleased to launch a partner contest exclusive to our partners dedicated to promoting and selling Oracle Systems! You are essential to the success of Oracle and we want to recognize your contribution and effort in driving Oracle Storage to the market. To show our appreciation we are delighted to announce a contest, giving the winners the opportunity to attend a roundtable chaired by Senior Oracle Executives and spend an unforgettable evening at the magnificent Teatro Alla Scala in Milan, followed by a stay at the Grand Hotel et de Milan, courtesy of Oracle. Recognition will be given to 12 partner companies (10 VARs & 2 VADs) who will be recognized for their ZFS storage booking achievement in the broad market between June 1st and July 18th 2014. Criteria of Eligibility A minimum deal value of $30k is required for qualification Partners who are wholly or partially owned by a public sector organization are not eligible for participation  Winners The winning VARs will be: The highest ZS3 or ZBA bookings achievers by COB on July 18th, 2014 in each Oracle EMEA region (1) The highest Oracle on Oracle (2) ZS3 or ZBA bookings achievers by COB on July 18th, 2014 in each Oracle EMEA region The winning VADs (3) will be: The highest ZS3 or ZBA bookings achiever by COB on July 18th 2014 in EMEA The highest Oracle on Oracle (2) ZS3 or ZBA bookings achiever by COB on July 18th 2014 in EMEA  The Prize Winners will be invited to participate to a roundtable chaired by Oracle on Monday September 8th 2014 in Milan and to be guests of Oracle in the evening of September 8th, 2014 at the Teatro Alla Scala. The evening will comprise of a private tour of the Scala museum, cocktail reception at the elegant museum rooms and attending the performance by the renowned Soprano, Maria Agresta. Our guests will then retire for the evening to the Grand Hotel et de Milan, courtesy of Oracle. Oracle shall be the final arbiter in selecting the winners and all winners will be notified via their Oracle account manager.Full details about the contest, expenses covered by Oracle and timetable of events can be found on the Oracle EMEA Hardware (Servers & Storage) Partner Community workspace (FY15 Q1 ZFS Partner Contest). Remember: access to the community workspace requires membership. If you are not a member please register here. Good Luck!! For more information, please contact Sasan Moaveni. (1) Two VAR winners for each EMEA region – Eastern Europe & CIS, Middle East & Africa, South Europe, North Europe, UK/Ireland & Israel - as per the criteria outlined above (2) Oracle on Oracle, in this instance, means ZS3 or ZBA storage attached to DB or DB options, Engineered Systems or Sparc servers sold to the same customer by the same partner within the contest timelines.(3) Two VAD winners, one for each of the criteria outlined above, will be selected from across EMEA. Normal 0 14 false false false IT X-NONE X-NONE MicrosoftInternetExplorer4 /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-priority:99; mso-style-qformat:yes; mso-style-parent:""; mso-padding-alt:0cm 5.4pt 0cm 5.4pt; mso-para-margin-top:0cm; mso-para-margin-right:0cm; mso-para-margin-bottom:10.0pt; mso-para-margin-left:0cm; line-height:115%; mso-pagination:widow-orphan; font-size:11.0pt; font-family:"Calibri","sans-serif"; mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin; mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin; mso-bidi-font-family:"Times New Roman"; mso-bidi-theme-font:minor-bidi; mso-fareast-language:EN-US;}

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  • min() and max() give error: TypeError: 'float' object is not iterable

    - by PythonUser3.3
    markList=[] Lmark=0 Hmark=0 while True: mark=float(input("Enter your marks here(Click -1 to exit)")) if mark == -1: break markList.append(mark) markList.sort() mid = len(markList)//2 if len(markList)%2==0: median=(markList[mid]+ markList[mid-1])/2 print("Median:", median) else: print("Median:" , markList[mid]) Lmark==(min(mark)) print("The lowest mark is", Lmark) Hmark==(max(mark)) print("The highest mark is", Hmark) My program is a basic grade calculator using lists. My program asks the user to input their grades into a list in which it then calculates your average and finds your lowest and highest mark. I have found the average but I can't seem to figure out how to find the lowest and highest grade. Can you please show me pr tell me what to do?

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  • Prevent initial array from sorting

    - by George
    I have an array where the order of the objects is important for when I finally output the array in a document. However, I'm also sorting the array in a function to find the highest value. The problem is that I after I run the function to find the highest value, I can't get the original sort order of the array back. // html document var data = [75,300,150,500,200]; createGraph(data); // js document function createGraph(data) { var maxRange = getDataRange(data); // simpleEncode() = google encoding function for graph var dataSet = simpleEncode(data,maxRange); } function getDataRange(dataArray) { var num = dataArray.sort(sortNumber); return num[0]; } I've also tried setting data to dataA and dataB and using dataB in the getDataRange function and dataA in the simpleEncode function. Either way, data always end up being sorted from highest to lowest.

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  • Setting Higher Z-Index for Sprite

    - by Siddharth
    For my game, I have to set highest z index for my sprite. At present, I wrote following code but didn't work for me. Sprite houseSprite = new Sprite(pX, pY, textureManager.houseBgRegion.deepCopy(), mVertexBufferObjectManager); attachChild(houseSprite); houseSprite.setZIndex(500); sortChildren(); My requirement did not satisfied with setting sprite in the HUD. So any how I have to apply highest z index. Also in my game sprites are dynamically generated as per game play. So members please share your thoughts.

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  • Auto DOP and Concurrency

    - by jean-pierre.dijcks
    After spending some time in the cloud, I figured it is time to come down to earth and start discussing some of the new Auto DOP features some more. As Database Machines (the v2 machine runs Oracle Database 11.2) are effectively selling like hotcakes, it makes some sense to talk about the new parallel features in more detail. For basic understanding make sure you have read the initial post. The focus there is on Auto DOP and queuing, which is to some extend the focus here. But now I want to discuss the concurrency a little and explain some of the relevant parameters and their impact, specifically in a situation with concurrency on the system. The goal of Auto DOP The idea behind calculating the Automatic Degree of Parallelism is to find the highest possible DOP (ideal DOP) that still scales. In other words, if we were to increase the DOP even more  above a certain DOP we would see a tailing off of the performance curve and the resource cost / performance would become less optimal. Therefore the ideal DOP is the best resource/performance point for that statement. The goal of Queuing On a normal production system we should see statements running concurrently. On a Database Machine we typically see high concurrency rates, so we need to find a way to deal with both high DOP’s and high concurrency. Queuing is intended to make sure we Don’t throttle down a DOP because other statements are running on the system Stay within the physical limits of a system’s processing power Instead of making statements go at a lower DOP we queue them to make sure they will get all the resources they want to run efficiently without trashing the system. The theory – and hopefully – practice is that by giving a statement the optimal DOP the sum of all statements runs faster with queuing than without queuing. Increasing the Number of Potential Parallel Statements To determine how many statements we will consider running in parallel a single parameter should be looked at. That parameter is called PARALLEL_MIN_TIME_THRESHOLD. The default value is set to 10 seconds. So far there is nothing new here…, but do realize that anything serial (e.g. that stays under the threshold) goes straight into processing as is not considered in the rest of this post. Now, if you have a system where you have two groups of queries, serial short running and potentially parallel long running ones, you may want to worry only about the long running ones with this parallel statement threshold. As an example, lets assume the short running stuff runs on average between 1 and 15 seconds in serial (and the business is quite happy with that). The long running stuff is in the realm of 1 – 5 minutes. It might be a good choice to set the threshold to somewhere north of 30 seconds. That way the short running queries all run serial as they do today (if it ain’t broken, don’t fix it) and allows the long running ones to be evaluated for (higher degrees of) parallelism. This makes sense because the longer running ones are (at least in theory) more interesting to unleash a parallel processing model on and the benefits of running these in parallel are much more significant (again, that is mostly the case). Setting a Maximum DOP for a Statement Now that you know how to control how many of your statements are considered to run in parallel, lets talk about the specific degree of any given statement that will be evaluated. As the initial post describes this is controlled by PARALLEL_DEGREE_LIMIT. This parameter controls the degree on the entire cluster and by default it is CPU (meaning it equals Default DOP). For the sake of an example, let’s say our Default DOP is 32. Looking at our 5 minute queries from the previous paragraph, the limit to 32 means that none of the statements that are evaluated for Auto DOP ever runs at more than DOP of 32. Concurrently Running a High DOP A basic assumption about running high DOP statements at high concurrency is that you at some point in time (and this is true on any parallel processing platform!) will run into a resource limitation. And yes, you can then buy more hardware (e.g. expand the Database Machine in Oracle’s case), but that is not the point of this post… The goal is to find a balance between the highest possible DOP for each statement and the number of statements running concurrently, but with an emphasis on running each statement at that highest efficiency DOP. The PARALLEL_SERVER_TARGET parameter is the all important concurrency slider here. Setting this parameter to a higher number means more statements get to run at their maximum parallel degree before queuing kicks in.  PARALLEL_SERVER_TARGET is set per instance (so needs to be set to the same value on all 8 nodes in a full rack Database Machine). Just as a side note, this parameter is set in processes, not in DOP, which equates to 4* Default DOP (2 processes for a DOP, default value is 2 * Default DOP, hence a default of 4 * Default DOP). Let’s say we have PARALLEL_SERVER_TARGET set to 128. With our limit set to 32 (the default) we are able to run 4 statements concurrently at the highest DOP possible on this system before we start queuing. If these 4 statements are running, any next statement will be queued. To run a system at high concurrency the PARALLEL_SERVER_TARGET should be raised from its default to be much closer (start with 60% or so) to PARALLEL_MAX_SERVERS. By using both PARALLEL_SERVER_TARGET and PARALLEL_DEGREE_LIMIT you can control easily how many statements run concurrently at good DOPs without excessive queuing. Because each workload is a little different, it makes sense to plan ahead and look at these parameters and set these based on your requirements.

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  • The First Annual Crappy Code Games

    - by Testas
    SQLBits announced some super-exciting news! A tie-up with our platinum sponsor, Fusion-io. Together we'll be running a series of events called "The Crappy Code Games" where SQL Server developers will compete to write the worst-performing code and win some very cool prizes including:   •        Gold: A hands-on, high performance flying day for two at Ultimate High plus Fusion-io flight jackets•        Silver: One day racing experience at Palmer Sports where you will drive seven different high performance cars•        Bronze: Pure Tech Racing 10 person package at PTR’s F1 racing facility includes FI tees, food and drinks. …plus iPods, Windows Mobile phones, X-box 360s, t-shirts and much more. There will be two qualifying events in Manchester on March 17th and London on March 31st, and the third qualifier as well as the grand finale will be held in the evening of Thursday April 7th at SQLBits. And if that isn’t cool enough, Fusion-io's Chief Scientist Steve Wozniak (yes, that Steve Wozniak, tech industry legend and co-founder of Apple) will be on hand in Brighton to hand out the prizes! If you'd like to take part you'll need to register, and since places are limited we recommend you do so right away. For more details and to register, go to http://www.crappycodegames.com/ The Games: In conjunction with SQL Bits, dbA-thletes (that’s you) will compete  head-to-head in one of three separate qualifying events to be held in Manchester, London and Brighton.  Four separate SQL  rounds make up the evening’s Games, and will challenge you to write code that pushes the boundaries of SQL performance.  The four events are: ?  The High Jump: Generate the highest I/O per second ?  The 100 m dash: Cumulative highest number of I/O’s in 60 seconds ?  The SSIS-athon: Load one billion row fact table in the shortest time ?  The Marathon: Generate the highest MB per second in 60 seconds

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  • New Science and Technology Centers

    NSF supports integrative partnerships that require large-scale, long-term funding to produce research and education of the highest quality National Science Foundation - Education - Science in Society - Educational Resources - United States

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  • How to discriminate from two nodes with identical frequencies in a Huffman's tree?

    - by Omega
    Still on my quest to compress/decompress files with a Java implementation of Huffman's coding (http://en.wikipedia.org/wiki/Huffman_coding) for a school assignment. From the Wikipedia page, I quote: Create a leaf node for each symbol and add it to the priority queue. While there is more than one node in the queue: Remove the two nodes of highest priority (lowest probability) from the queue Create a new internal node with these two nodes as children and with probability equal to the sum of the two nodes' probabilities. Add the new node to the queue. The remaining node is the root node and the tree is complete. Now, emphasis: Remove the two nodes of highest priority (lowest probability) from the queue Create a new internal node with these two nodes as children and with probability equal to the sum of the two nodes' probabilities. So I have to take two nodes with the lowest frequency. What if there are multiple nodes with the same low frequency? How do I discriminate which one to use? The reason I ask this is because Wikipedia has this image: And I wanted to see if my Huffman's tree was the same. I created a file with the following content: aaaaeeee nnttmmiihhssfffouxprl And this was the result: Doesn't look so bad. But there clearly are some differences when multiple nodes have the same frequency. My questions are the following: What is Wikipedia's image doing to discriminate the nodes with the same frequency? Is my tree wrong? (Is Wikipedia's image method the one and only answer?) I guess there is one specific and strict way to do this, because for our school assignment, files that have been compressed by my program should be able to be decompressed by other classmate's programs - so there must be a "standard" or "unique" way to do it. But I'm a bit lost with that. My code is rather straightforward. It literally just follows Wikipedia's listed steps. The way my code extracts the two nodes with the lowest frequency from the queue is to iterate all nodes and if the current node has a lower frequency than any of the two "smallest" known nodes so far, then it replaces the highest one. Just like that.

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  • Catalyst forgets monitor settings after reboot or log out in 12.10

    - by Mate Weisz
    I have a Samsung monitor connected to my ATI Radeon graphic card via HDMI. By default the screen has a black border around, but I can disable it in the Catalyst menu to get full screen. (There is a scalebar that I have to set to the highest value.) My problem is that every time I turn off and on my computer it changes back and I have to set it up again. It is really annoying. Is there any way to make this setting permanent? Notes: 1. I open Catalyst with admin rights. 2. When I open the Catalyst settings menu, it looks like that it keeps my settings, because the scalebar stays at the highest point, but still it doesn't scale up my screen until I move back and forth the scale bar to the same point

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  • Which creative framework can create these games? [closed]

    - by Rahil627
    I've used a few game frameworks in the past and have run into limitations. This lead me to "creative frameworks". I've looked into many, but I cannot determine the limitations of some of them. Selected frameworks ordered from highest to lowest level: Flash, Unity, MonoGame, OpenFrameworks (and Cinder), SFML. I want to be able to: create a game that handles drawing on an iPad create a game that uses computer vision from a webcam create a multi-device iOS game create a game that uses input from Kinect Can all of the frameworks handle this? What is the highest level framework that can handle all of them?

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  • Memory leak in Google Chrome

    - by jasondavis
    As a developer it is very common for me to have 2-3 different IDE's open, 10-15 google chrome windows which can hold up to 200 open tabs (I know I get out of hand some times), Photoshop, couple twitter bots for promo, and a few other programs but my system still runs fast and smooth. I have an i7 processor with 12gb ram. Now with all my usual stuff running my Physical memory is usually running around 50-60% however over the course of the day or much less even, I will gradually grow to 98% The highest Memory usage processes will be from Google Chrome, if I sort in the task manager by highest memory usage and end the 1 highest process which will be a google chrome one, my memory usage will jump back down to about 60%. Also by ending that 1 process, all my Chrome windows will remain open and in use, so it doesn't affect me at all by ending that process. Based on this research I am assuming that that 1 runaway process is likely the Adobe Flash as I also can say that it gets up to the 98% much faster when I am using flash items like video or music player. But even without using any of them it will still climb up to that high number eventually. Has anyone else experienced similar results?

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  • Beginning Java (Working with Arrays; Class Assignment)

    - by Jason
    I am to the point where I feel as if I correctly wrote the code for this homework assignment. We were given a skeleton and 2 classes that we had to import (FileIOHelper and Student). /* * Created: *** put the date here *** * * Author: *** put your name here *** * * The program will read information about students and their * scores from a file, and output the name of each student with * all his/her scores and the total score, plus the average score * of the class, and the name and total score of the students with * the highest and lowest total score. */ // import java.util.Scanner; import java.io.*; // C:\Users\Adam\info.txt public class Lab6 { public static void main(String[] args) throws IOException { // Fill in the body according to the following comments Scanner key boardFile = new Scanner(System.in); // Input file name String filename = getFileName(keyboardFile); //Open the file // Input number of students int numStudents = FileIOHelper.getNumberOfStudents(filename); Student students[] = new Student[numStudents]; // Input all student records and create Student array and // integer array for total scores int totalScore[] = new int[students.length]; for (int i = 0; i < students.length; i++){ for(int j = 1; j < 4; j++){ totalScore[i] = totalScore[i] + students[i].getScore(j); } } // Compute total scores and find students with lowest and // highest total score int maxScore = 0; int minScore = 0; for(int i = 0; i < students.length; i++){ if(totalScore[i] >= totalScore[maxScore]){ maxScore = i; } else if(totalScore[i] <= totalScore[minScore]){ minScore = i; } } // Compute average total score int allScores = 0; int average = 0; for (int i = 0; i < totalScore.length; i++){ allScores = allScores + totalScore[i]; } average = allScores / totalScore.length; // Output results outputResults(students, totalScore, maxScore, minScore, average); } // Given a Scanner in, this method prompts the user to enter // a file name, inputs it, and returns it. private static String getFileName(Scanner in) { // Fill in the body System.out.print("Enter the name of a file: "); String filename = in.next(); return filename; // Do not declare the Scanner variable in this method. // You must use the value this method receives in the // argument (in). } // Given the number of students records n to input, this // method creates an array of Student of the appropriate size, // reads n student records using the FileIOHelper, and stores // them in the array, and finally returns the Student array. private static Student[] getStudents(int n) { Student[] myStudents = new Student[n]; for(int i = 0; i <= n; i++){ myStudents[i] = FileIOHelper.getNextStudent(); } return myStudents; } // Given an array of Student records, an array with the total scores, // the indices in the arrays of the students with the highest and // lowest total scores, and the average total score for the class, // this method outputs a table of all the students appropriately // formatted, plus the total number of students, the average score // of the class, and the name and total score of the students with // the highest and lowest total score. private static void outputResults( Student[] students, int[] totalScores, int maxIndex, int minIndex, int average ) { // Fill in the body System.out.println("\nName \t\tScore1 \tScore2 \tScore3 \tTotal"); System.out.println("--------------------------------------------------------"); for(int i = 0; i < students.length; i++){ outputStudent(students[i], totalScores[i], average); System.out.println(); } System.out.println("--------------------------------------------------------"); outputNumberOfStudents(students.length); outputAverage(average); outputMaxStudent(students[maxIndex], totalScores[maxIndex]); outputMinStudent(students[minIndex], totalScores[minIndex]); System.out.println("--------------------------------------------------------"); } // Given a Student record, the total score for the student, // and the average total score for all the students, this method // outputs one line in the result table appropriately formatted. private static void outputStudent(Student s, int total, int avg) { System.out.print(s.getName() + "\t"); for(int i = 1; i < 4; i++){ System.out.print(s.getScore(i) + "\t"); } System.out.print(total + "\t"); if(total < avg){ System.out.print("-"); }else if(total > avg){ System.out.print("+"); }else{ System.out.print("="); } } // Given the number of students, this method outputs a message // stating what the total number of students in the class is. private static void outputNumberOfStudents(int n) { System.out.println("The total number of students in this class is: \t" + n); } // Given the average total score of all students, this method // outputs a message stating what the average total score of // the class is. private static void outputAverage(int average) { System.out.println("The average total score of the class is: \t" + average); } // Given the Student with highest total score and the student's // total score, this method outputs a message stating the name // of the student and the highest score. private static void outputMaxStudent( Student student, int score ) { System.out.println(student.getName() + " got the maximum total score of: \t" + score); } // Given the Student with lowest total score and the student's // total score, this method outputs a message stating the name // of the student and the lowest score. private static void outputMinStudent( Student student, int score ) { System.out.println(student.getName() + " got the minimum total score of: \t" + score); } } But now I get an error at the line totalScore[i] = totalScore[i] + students[i].getScore(j); Exception in thread "main" java.lang.NullPointerException at Lab6.main(Lab6.java:42)

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  • Simple question on database query.

    - by GK
    I have been asked in an interview, To write a sql query which fetches the first three records with highest value on some column from a table. i had written a query which fetched all the records with highest value, but didnt get how exactly i can get only first three records of those. could you help me in this. thanks.

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  • Updating a sql server table with data from another table

    - by David G
    I have two basic SQL Server tables: Customer (ID [pk], AddressLine1, AddressLine2, AddressCity, AddressDistrict, AddressPostalCode) CustomerAddress(ID [pk], CustomerID [fk], Line1, Line2, City, District, PostalCode) CustomerAddress contains multiple addresses for the Customer record. For each Customer record I want to merge the most recent CustomerAddress record where most recent is determined by the highest CustomerAddress ID value. I've currently got the following: UPDATE Customer SET AddressLine1 = CustomerAddress.Line1, AddressPostalCode = CustomerAddress.PostalCode FROM Customer, CustomerAddress WHERE Customer.ID = CustomerAddress.CustomerID which works but how can I ensure that the most recent (highest ID) CustomerAddress record is selected to update the Customer table?

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  • Organization &amp; Architecture UNISA Studies &ndash; Chap 6

    - by MarkPearl
    Learning Outcomes Discuss the physical characteristics of magnetic disks Describe how data is organized and accessed on a magnetic disk Discuss the parameters that play a role in the performance of magnetic disks Describe different optical memory devices Magnetic Disk The way data is stored on and retried from magnetic disks Data is recorded on and later retrieved form the disk via a conducting coil named the head (in many systems there are two heads) The writ mechanism exploits the fact that electricity flowing through a coil produces a magnetic field. Electric pulses are sent to the write head, and the resulting magnetic patterns are recorded on the surface below with different patterns for positive and negative currents The physical characteristics of a magnetic disk   Summarize from book   The factors that play a role in the performance of a disk Seek time – the time it takes to position the head at the track Rotational delay / latency – the time it takes for the beginning of the sector to reach the head Access time – the sum of the seek time and rotational delay Transfer time – the time it takes to transfer data RAID The rate of improvement in secondary storage performance has been considerably less than the rate for processors and main memory. Thus secondary storage has become a bit of a bottleneck. RAID works on the concept that if one disk can be pushed so far, additional gains in performance are to be had by using multiple parallel components. Points to note about RAID… RAID is a set of physical disk drives viewed by the operating system as a single logical drive Data is distributed across the physical drives of an array in a scheme known as striping Redundant disk capacity is used to store parity information, which guarantees data recoverability in case of a disk failure (not supported by RAID 0 or RAID 1) Interesting to note that the increase in the number of drives, increases the probability of failure. To compensate for this decreased reliability RAID makes use of stored parity information that enables the recovery of data lost due to a disk failure.   The RAID scheme consists of 7 levels…   Category Level Description Disks Required Data Availability Large I/O Data Transfer Capacity Small I/O Request Rate Striping 0 Non Redundant N Lower than single disk Very high Very high for both read and write Mirroring 1 Mirrored 2N Higher than RAID 2 – 5 but lower than RAID 6 Higher than single disk Up to twice that of a signle disk for read Parallel Access 2 Redundant via Hamming Code N + m Much higher than single disk Highest of all listed alternatives Approximately twice that of a single disk Parallel Access 3 Bit interleaved parity N + 1 Much higher than single disk Highest of all listed alternatives Approximately twice that of a single disk Independent Access 4 Block interleaved parity N + 1 Much higher than single disk Similar to RAID 0 for read, significantly lower than single disk for write Similar to RAID 0 for read, significantly lower than single disk for write Independent Access 5 Block interleaved parity N + 1 Much higher than single disk Similar to RAID 0 for read, lower than single disk for write Similar to RAID 0 for read, generally  lower than single disk for write Independent Access 6 Block interleaved parity N + 2 Highest of all listed alternatives Similar to RAID 0 for read; lower than RAID 5 for write Similar to RAID 0 for read, significantly lower than RAID 5  for write   Read page 215 – 221 for detailed explanation on RAID levels Optical Memory There are a variety of optical-disk systems available. Read through the table on page 222 – 223 Some of the devices include… CD CD-ROM CD-R CD-RW DVD DVD-R DVD-RW Blue-Ray DVD Magnetic Tape Most modern systems use serial recording – data is lade out as a sequence of bits along each track. The typical recording used in serial is referred to as serpentine recording. In this technique when data is being recorded, the first set of bits is recorded along the whole length of the tape. When the end of the tape is reached the heads are repostioned to record a new track, and the tape is again recorded on its whole length, this time in the opposite direction. That process continued back and forth until the tape is full. To increase speed, the read-write head is capable of reading and writing a number of adjacent tracks simultaneously. Data is still recorded serially along individual tracks, but blocks in sequence are stored on adjacent tracks as suggested. A tape drive is a sequential access device. Magnetic tape was the first kind of secondary memory. It is still widely used as the lowest-cost, slowest speed member of the memory hierarchy.

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