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  • Developing a Cost Model for Cloud Applications

    - by BuckWoody
    Note - please pay attention to the date of this post. As much as I attempt to make the information below accurate, the nature of distributed computing means that components, units and pricing will change over time. The definitive costs for Microsoft Windows Azure and SQL Azure are located here, and are more accurate than anything you will see in this post: http://www.microsoft.com/windowsazure/offers/  When writing software that is run on a Platform-as-a-Service (PaaS) offering like Windows Azure / SQL Azure, one of the questions you must answer is how much the system will cost. I will not discuss the comparisons between on-premise costs (which are nigh impossible to calculate accurately) versus cloud costs, but instead focus on creating a general model for estimating costs for a given application. You should be aware that there are (at this writing) two billing mechanisms for Windows and SQL Azure: “Pay-as-you-go” or consumption, and “Subscription” or commitment. Conceptually, you can consider the former a pay-as-you-go cell phone plan, where you pay by the unit used (at a slightly higher rate) and the latter as a standard cell phone plan where you commit to a contract and thus pay lower rates. In this post I’ll stick with the pay-as-you-go mechanism for simplicity, which should be the maximum cost you would pay. From there you may be able to get a lower cost if you use the other mechanism. In any case, the model you create should hold. Developing a good cost model is essential. As a developer or architect, you’ll most certainly be asked how much something will cost, and you need to have a reliable way to estimate that. Businesses and Organizations have been used to paying for servers, software licenses, and other infrastructure as an up-front cost, and power, people to the systems and so on as an ongoing (and sometimes not factored) cost. When presented with a new paradigm like distributed computing, they may not understand the true cost/value proposition, and that’s where the architect and developer can guide the conversation to make a choice based on features of the application versus the true costs. The two big buckets of use-types for these applications are customer-based and steady-state. In the customer-based use type, each successful use of the program results in a sale or income for your organization. Perhaps you’ve written an application that provides the spot-price of foo, and your customer pays for the use of that application. In that case, once you’ve estimated your cost for a successful traversal of the application, you can build that into the price you charge the user. It’s a standard restaurant model, where the price of the meal is determined by the cost of making it, plus any profit you can make. In the second use-type, the application will be used by a more-or-less constant number of processes or users and no direct revenue is attached to the system. A typical example is a customer-tracking system used by the employees within your company. In this case, the cost model is often created “in reverse” - meaning that you pilot the application, monitor the use (and costs) and that cost is held steady. This is where the comparison with an on-premise system becomes necessary, even though it is more difficult to estimate those on-premise true costs. For instance, do you know exactly how much cost the air conditioning is because you have a team of system administrators? This may sound trivial, but that, along with the insurance for the building, the wiring, and every other part of the system is in fact a cost to the business. There are three primary methods that I’ve been successful with in estimating the cost. None are perfect, all are demand-driven. The general process is to lay out a matrix of: components units cost per unit and then multiply that times the usage of the system, based on which components you use in the program. That sounds a bit simplistic, but using those metrics in a calculation becomes more detailed. In all of the methods that follow, you need to know your application. The components for a PaaS include computing instances, storage, transactions, bandwidth and in the case of SQL Azure, database size. In most cases, architects start with the first model and progress through the other methods to gain accuracy. Simple Estimation The simplest way to calculate costs is to architect the application (even UML or on-paper, no coding involved) and then estimate which of the components you’ll use, and how much of each will be used. Microsoft provides two tools to do this - one is a simple slider-application located here: http://www.microsoft.com/windowsazure/pricing-calculator/  The other is a tool you download to create an “Return on Investment” (ROI) spreadsheet, which has the advantage of leading you through various questions to estimate what you plan to use, located here: https://roianalyst.alinean.com/msft/AutoLogin.do?d=176318219048082115  You can also just create a spreadsheet yourself with a structure like this: Program Element Azure Component Unit of Measure Cost Per Unit Estimated Use of Component Total Cost Per Component Cumulative Cost               Of course, the consideration with this model is that it is difficult to predict a system that is not running or hasn’t even been developed. Which brings us to the next model type. Measure and Project A more accurate model is to actually write the code for the application, using the Software Development Kit (SDK) which can run entirely disconnected from Azure. The code should be instrumented to estimate the use of the application components, logging to a local file on the development system. A series of unit and integration tests should be run, which will create load on the test system. You can use standard development concepts to track this usage, and even use Windows Performance Monitor counters. The best place to start with this method is to use the Windows Azure Diagnostics subsystem in your code, which you can read more about here: http://blogs.msdn.com/b/sumitm/archive/2009/11/18/introducing-windows-azure-diagnostics.aspx This set of API’s greatly simplifies tracking the application, and in fact you can use this information for more than just a cost model. After you have the tracking logs, you can plug the numbers into ay of the tools above, which should give a representative cost or in some cases a unit cost. The consideration with this model is that the SDK fabric is not a one-to-one comparison with performance on the actual Windows Azure fabric. Those differences are usually smaller, but they do need to be considered. Also, you may not be able to accurately predict the load on the system, which might lead to an architectural change, which changes the model. This leads us to the next, most accurate method for a cost model. Sample and Estimate Using standard statistical and other predictive math, once the application is deployed you will get a bill each month from Microsoft for your Azure usage. The bill is quite detailed, and you can export the data from it to do analysis, and using methods like regression and so on project out into the future what the costs will be. I normally advise that the architect also extrapolate a unit cost from those metrics as well. This is the information that should be reported back to the executives that pay the bills: the past cost, future projected costs, and unit cost “per click” or “per transaction”, as your case warrants. The challenge here is in the model itself - statistical methods are not foolproof, and the larger the sample (in this case I recommend the entire population, not a smaller sample) is key. References and Tools Articles: http://blogs.msdn.com/b/patrick_butler_monterde/archive/2010/02/10/windows-azure-billing-overview.aspx http://technet.microsoft.com/en-us/magazine/gg213848.aspx http://blog.codingoutloud.com/2011/06/05/azure-faq-how-much-will-it-cost-me-to-run-my-application-on-windows-azure/ http://blogs.msdn.com/b/johnalioto/archive/2010/08/25/10054193.aspx http://geekswithblogs.net/iupdateable/archive/2010/02/08/qampa-how-can-i-calculate-the-tco-and-roi-when.aspx   Other Tools: http://cloud-assessment.com/ http://communities.quest.com/community/cloud_tools

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  • CodePlex Daily Summary for Monday, June 02, 2014

    CodePlex Daily Summary for Monday, June 02, 2014Popular ReleasesPortable Class Library for SQLite: Portable Class Library for SQLite - 3.8.4.4: This pull request from mattleibow addresses an issue with custom function creation (define functions in C# code and invoke them from SQLite as id they where regular SQL functions). Impact: Xamarin iOSTweetinvi a friendly Twitter C# API: Tweetinvi 0.9.3.x: Timelines- Added all the parameters available from the Timeline Endpoints in Tweetinvi. - This is available for HomeTimeline, UserTimeline, MentionsTimeline // Simple query var tweets = Timeline.GetHomeTimeline(); // Create a parameter for queries with specific parameters var timelineParameter = Timeline.GenerateHomeTimelineRequestParameter(); timelineParameter.ExcludeReplies = true; timelineParameter.TrimUser = true; var tweets = Timeline.GetHomeTimeline(timelineParameter); Tweets- Add mis...Sandcastle Help File Builder: Help File Builder and Tools v2014.5.31.0: General InformationIMPORTANT: On some systems, the content of the ZIP file is blocked and the installer may fail to run. Before extracting it, right click on the ZIP file, select Properties, and click on the Unblock button if it is present in the lower right corner of the General tab in the properties dialog. This release completes removal of the branding transformations and implements the new VS2013 presentation style that utilizes the new lightweight website format. Several breaking cha...Tooltip Web Preview: ToolTip Web Preview: Version 1.0Database Helper: Release 1.0.0.0: First Release of Database HelperCoMaSy: CoMaSy1.0.2: !Contact Management SystemImage View Slider: Image View Slider: This is a .NET component. We create this using VB.NET. Here you can use an Image Viewer with several properties to your application form. We wish somebody to improve freely. Try this out! Author : Steven Renaldo Antony Yustinus Arjuna Purnama Putra Andre Wijaya P Martin Lidau PBK GENAP 2014 - TI UKDWAspose for Apache POI: Missing Features of Apache POI WP - v 1.1: Release contain the Missing Features in Apache POI WP SDK in Comparison with Aspose.Words for dealing with Microsoft Word. What's New ?Following Examples: Insert Picture in Word Document Insert Comments Set Page Borders Mail Merge from XML Data Source Moving the Cursor Feedback and Suggestions Many more examples are yet to come here. Keep visiting us. Raise your queries and suggest more examples via Aspose Forums or via this social coding site.SEToolbox: 01.032.014 Release 1: Added fix when loading game Textures for icons causing 'Unable to read beyond the end of the stream'. Added new Resource Report, that displays all in game resources in a concise report. Added in temp directory cleaner, to keep excess files from building up. Fixed use of colors on the windows, to work better with desktop schemes. Adding base support for multilingual resources. This will allow loading of the Space Engineers resources to show localized names, and display localized date a...ClosedXML - The easy way to OpenXML: ClosedXML 0.71.2: More memory and performance improvements. Fixed an issue with pivot table field order.Vi-AIO SearchBar: Vi – AIO Search Bar: Version 1.0Top Verses ( Ayat Emas ): Binary Top Verses: This one is the bin folder of the component. the .dll component is inside.Traditional Calendar Component: Traditional Calender Converter: Duta Wacana Christian University This file containing Traditional Calendar Component and Demo Aplication that using Traditional Calendar Component. This component made with .NET Framework 4 and the programming language is C# .SQLSetupHelper: 1.0.0.0: First Stable Version of SQL SetupComposite Iconote: Composite Iconote: This is a composite has been made by Microsoft Visual Studio 2013. Requirement: To develop this composite or use this component in your application, your computer must have .NET framework 4.5 or newer.Magick.NET: Magick.NET 6.8.9.101: Magick.NET linked with ImageMagick 6.8.9.1. Breaking changes: - Int/short Set methods of WritablePixelCollection are now unsigned. - The Q16 build no longer uses HDRI, switch to the new Q16-HDRI build if you need HDRI.fnr.exe - Find And Replace Tool: 1.7: Bug fixes Refactored logic for encoding text values to command line to handle common edge cases where find/replace operation works in GUI but not in command line Fix for bug where selection in Encoding drop down was different when generating command line in some cases. It was reported in: https://findandreplace.codeplex.com/workitem/34 Fix for "Backslash inserted before dot in replacement text" reported here: https://findandreplace.codeplex.com/discussions/541024 Fix for finding replacing...VG-Ripper & PG-Ripper: VG-Ripper 2.9.59: changes NEW: Added Support for 'GokoImage.com' links NEW: Added Support for 'ViperII.com' links NEW: Added Support for 'PixxxView.com' links NEW: Added Support for 'ImgRex.com' links NEW: Added Support for 'PixLiv.com' links NEW: Added Support for 'imgsee.me' links NEW: Added Support for 'ImgS.it' linksToolbox for Dynamics CRM 2011/2013: XrmToolBox (v1.2014.5.28): XrmToolbox improvement XrmToolBox updates (v1.2014.5.28)Fix connecting to a connection with custom authentication without saved password Tools improvement New tool!Solution Components Mover (v1.2014.5.22) Transfer solution components from one solution to another one Import/Export NN relationships (v1.2014.3.7) Allows you to import and export many to many relationships Tools updatesAttribute Bulk Updater (v1.2014.5.28) Audit Center (v1.2014.5.28) View Layout Replicator (v1.2014.5.28) Scrip...Microsoft Ajax Minifier: Microsoft Ajax Minifier 5.10: Fix for Issue #20875 - echo switch doesn't work for CSS CSS should honor the SASS source-file comments JS should allow multi-line comment directivesNew Projects[ISEN] Rendu de projet Naughty3Dogs - Pong3D: Pong3D est un jeu qui reprend le principe classique du Pong en le portant dans un environnement 3D à l'aide du langage c# et du moteur Unity3DBootstrap for MVC: Bootstrap for MVC.F. A. Q. - Najczesciej zadawane pytania: FAQForuMvc: Technifutur short projecthomework456: no.iStoody: Studies organize solution, available through app for Windows and Windows Phone.liaoliao: ???????????Price Tracker: Allows a user to track prices based on parsed emailsRoslynEval: RoslynRx Hub: Rx Hub provides server side computation which initiate by subscriber requestSharepoint Online AppCache Reset: We are an IT resource company providing Virtual IT services and custom and opensource programs for everyday needs. UnitConversionLib : Smart Unit Conversion Library in C#: Conversion of units, arithmetic operation and parsing quantities with their units on run time. Smart unit converter and conversion lib for physical quantities,

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  • When is a Seek not a Seek?

    - by Paul White
    The following script creates a single-column clustered table containing the integers from 1 to 1,000 inclusive. IF OBJECT_ID(N'tempdb..#Test', N'U') IS NOT NULL DROP TABLE #Test ; GO CREATE TABLE #Test ( id INTEGER PRIMARY KEY CLUSTERED ); ; INSERT #Test (id) SELECT V.number FROM master.dbo.spt_values AS V WHERE V.[type] = N'P' AND V.number BETWEEN 1 AND 1000 ; Let’s say we need to find the rows with values from 100 to 170, excluding any values that divide exactly by 10.  One way to write that query would be: SELECT T.id FROM #Test AS T WHERE T.id IN ( 101,102,103,104,105,106,107,108,109, 111,112,113,114,115,116,117,118,119, 121,122,123,124,125,126,127,128,129, 131,132,133,134,135,136,137,138,139, 141,142,143,144,145,146,147,148,149, 151,152,153,154,155,156,157,158,159, 161,162,163,164,165,166,167,168,169 ) ; That query produces a pretty efficient-looking query plan: Knowing that the source column is defined as an INTEGER, we could also express the query this way: SELECT T.id FROM #Test AS T WHERE T.id >= 101 AND T.id <= 169 AND T.id % 10 > 0 ; We get a similar-looking plan: If you look closely, you might notice that the line connecting the two icons is a little thinner than before.  The first query is estimated to produce 61.9167 rows – very close to the 63 rows we know the query will return.  The second query presents a tougher challenge for SQL Server because it doesn’t know how to predict the selectivity of the modulo expression (T.id % 10 > 0).  Without that last line, the second query is estimated to produce 68.1667 rows – a slight overestimate.  Adding the opaque modulo expression results in SQL Server guessing at the selectivity.  As you may know, the selectivity guess for a greater-than operation is 30%, so the final estimate is 30% of 68.1667, which comes to 20.45 rows. The second difference is that the Clustered Index Seek is costed at 99% of the estimated total for the statement.  For some reason, the final SELECT operator is assigned a small cost of 0.0000484 units; I have absolutely no idea why this is so, or what it models.  Nevertheless, we can compare the total cost for both queries: the first one comes in at 0.0033501 units, and the second at 0.0034054.  The important point is that the second query is costed very slightly higher than the first, even though it is expected to produce many fewer rows (20.45 versus 61.9167). If you run the two queries, they produce exactly the same results, and both complete so quickly that it is impossible to measure CPU usage for a single execution.  We can, however, compare the I/O statistics for a single run by running the queries with STATISTICS IO ON: Table '#Test'. Scan count 63, logical reads 126, physical reads 0. Table '#Test'. Scan count 01, logical reads 002, physical reads 0. The query with the IN list uses 126 logical reads (and has a ‘scan count’ of 63), while the second query form completes with just 2 logical reads (and a ‘scan count’ of 1).  It is no coincidence that 126 = 63 * 2, by the way.  It is almost as if the first query is doing 63 seeks, compared to one for the second query. In fact, that is exactly what it is doing.  There is no indication of this in the graphical plan, or the tool-tip that appears when you hover your mouse over the Clustered Index Seek icon.  To see the 63 seek operations, you have click on the Seek icon and look in the Properties window (press F4, or right-click and choose from the menu): The Seek Predicates list shows a total of 63 seek operations – one for each of the values from the IN list contained in the first query.  I have expanded the first seek node to show the details; it is seeking down the clustered index to find the entry with the value 101.  Each of the other 62 nodes expands similarly, and the same information is contained (even more verbosely) in the XML form of the plan. Each of the 63 seek operations starts at the root of the clustered index B-tree and navigates down to the leaf page that contains the sought key value.  Our table is just large enough to need a separate root page, so each seek incurs 2 logical reads (one for the root, and one for the leaf).  We can see the index depth using the INDEXPROPERTY function, or by using the a DMV: SELECT S.index_type_desc, S.index_depth FROM sys.dm_db_index_physical_stats ( DB_ID(N'tempdb'), OBJECT_ID(N'tempdb..#Test', N'U'), 1, 1, DEFAULT ) AS S ; Let’s look now at the Properties window when the Clustered Index Seek from the second query is selected: There is just one seek operation, which starts at the root of the index and navigates the B-tree looking for the first key that matches the Start range condition (id >= 101).  It then continues to read records at the leaf level of the index (following links between leaf-level pages if necessary) until it finds a row that does not meet the End range condition (id <= 169).  Every row that meets the seek range condition is also tested against the Residual Predicate highlighted above (id % 10 > 0), and is only returned if it matches that as well. You will not be surprised that the single seek (with a range scan and residual predicate) is much more efficient than 63 singleton seeks.  It is not 63 times more efficient (as the logical reads comparison would suggest), but it is around three times faster.  Let’s run both query forms 10,000 times and measure the elapsed time: DECLARE @i INTEGER, @n INTEGER = 10000, @s DATETIME = GETDATE() ; SET NOCOUNT ON; SET STATISTICS XML OFF; ; WHILE @n > 0 BEGIN SELECT @i = T.id FROM #Test AS T WHERE T.id IN ( 101,102,103,104,105,106,107,108,109, 111,112,113,114,115,116,117,118,119, 121,122,123,124,125,126,127,128,129, 131,132,133,134,135,136,137,138,139, 141,142,143,144,145,146,147,148,149, 151,152,153,154,155,156,157,158,159, 161,162,163,164,165,166,167,168,169 ) ; SET @n -= 1; END ; PRINT DATEDIFF(MILLISECOND, @s, GETDATE()) ; GO DECLARE @i INTEGER, @n INTEGER = 10000, @s DATETIME = GETDATE() ; SET NOCOUNT ON ; WHILE @n > 0 BEGIN SELECT @i = T.id FROM #Test AS T WHERE T.id >= 101 AND T.id <= 169 AND T.id % 10 > 0 ; SET @n -= 1; END ; PRINT DATEDIFF(MILLISECOND, @s, GETDATE()) ; On my laptop, running SQL Server 2008 build 4272 (SP2 CU2), the IN form of the query takes around 830ms and the range query about 300ms.  The main point of this post is not performance, however – it is meant as an introduction to the next few parts in this mini-series that will continue to explore scans and seeks in detail. When is a seek not a seek?  When it is 63 seeks © Paul White 2011 email: [email protected] twitter: @SQL_kiwi

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  • Organization &amp; Architecture UNISA Studies &ndash; Chap 4

    - by MarkPearl
    Learning Outcomes Explain the characteristics of memory systems Describe the memory hierarchy Discuss cache memory principles Discuss issues relevant to cache design Describe the cache organization of the Pentium Computer Memory Systems There are key characteristics of memory… Location – internal or external Capacity – expressed in terms of bytes Unit of Transfer – the number of bits read out of or written into memory at a time Access Method – sequential, direct, random or associative From a users perspective the two most important characteristics of memory are… Capacity Performance – access time, memory cycle time, transfer rate The trade off for memory happens along three axis… Faster access time, greater cost per bit Greater capacity, smaller cost per bit Greater capacity, slower access time This leads to people using a tiered approach in their use of memory   As one goes down the hierarchy, the following occurs… Decreasing cost per bit Increasing capacity Increasing access time Decreasing frequency of access of the memory by the processor The use of two levels of memory to reduce average access time works in principle, but only if conditions 1 to 4 apply. A variety of technologies exist that allow us to accomplish this. Thus it is possible to organize data across the hierarchy such that the percentage of accesses to each successively lower level is substantially less than that of the level above. A portion of main memory can be used as a buffer to hold data temporarily that is to be read out to disk. This is sometimes referred to as a disk cache and improves performance in two ways… Disk writes are clustered. Instead of many small transfers of data, we have a few large transfers of data. This improves disk performance and minimizes processor involvement. Some data designed for write-out may be referenced by a program before the next dump to disk. In that case the data is retrieved rapidly from the software cache rather than slowly from disk. Cache Memory Principles Cache memory is substantially faster than main memory. A caching system works as follows.. When a processor attempts to read a word of memory, a check is made to see if this in in cache memory… If it is, the data is supplied, If it is not in the cache, a block of main memory, consisting of a fixed number of words is loaded to the cache. Because of the phenomenon of locality of references, when a block of data is fetched into the cache, it is likely that there will be future references to that same memory location or to other words in the block. Elements of Cache Design While there are a large number of cache implementations, there are a few basic design elements that serve to classify and differentiate cache architectures… Cache Addresses Cache Size Mapping Function Replacement Algorithm Write Policy Line Size Number of Caches Cache Addresses Almost all non-embedded processors support virtual memory. Virtual memory in essence allows a program to address memory from a logical point of view without needing to worry about the amount of physical memory available. When virtual addresses are used the designer may choose to place the cache between the MMU (memory management unit) and the processor or between the MMU and main memory. The disadvantage of virtual memory is that most virtual memory systems supply each application with the same virtual memory address space (each application sees virtual memory starting at memory address 0), which means the cache memory must be completely flushed with each application context switch or extra bits must be added to each line of the cache to identify which virtual address space the address refers to. Cache Size We would like the size of the cache to be small enough so that the overall average cost per bit is close to that of main memory alone and large enough so that the overall average access time is close to that of the cache alone. Also, larger caches are slightly slower than smaller ones. Mapping Function Because there are fewer cache lines than main memory blocks, an algorithm is needed for mapping main memory blocks into cache lines. The choice of mapping function dictates how the cache is organized. Three techniques can be used… Direct – simplest technique, maps each block of main memory into only one possible cache line Associative – Each main memory block to be loaded into any line of the cache Set Associative – exhibits the strengths of both the direct and associative approaches while reducing their disadvantages For detailed explanations of each approach – read the text book (page 148 – 154) Replacement Algorithm For associative and set associating mapping a replacement algorithm is needed to determine which of the existing blocks in the cache must be replaced by a new block. There are four common approaches… LRU (Least recently used) FIFO (First in first out) LFU (Least frequently used) Random selection Write Policy When a block resident in the cache is to be replaced, there are two cases to consider If no writes to that block have happened in the cache – discard it If a write has occurred, a process needs to be initiated where the changes in the cache are propagated back to the main memory. There are several approaches to achieve this including… Write Through – all writes to the cache are done to the main memory as well at the point of the change Write Back – when a block is replaced, all dirty bits are written back to main memory The problem is complicated when we have multiple caches, there are techniques to accommodate for this but I have not summarized them. Line Size When a block of data is retrieved and placed in the cache, not only the desired word but also some number of adjacent words are retrieved. As the block size increases from very small to larger sizes, the hit ratio will at first increase because of the principle of locality, which states that the data in the vicinity of a referenced word are likely to be referenced in the near future. As the block size increases, more useful data are brought into cache. The hit ratio will begin to decrease as the block becomes even bigger and the probability of using the newly fetched information becomes less than the probability of using the newly fetched information that has to be replaced. Two specific effects come into play… Larger blocks reduce the number of blocks that fit into a cache. Because each block fetch overwrites older cache contents, a small number of blocks results in data being overwritten shortly after they are fetched. As a block becomes larger, each additional word is farther from the requested word and therefore less likely to be needed in the near future. The relationship between block size and hit ratio is complex, and no set approach is judged to be the best in all circumstances.   Pentium 4 and ARM cache organizations The processor core consists of four major components: Fetch/decode unit – fetches program instruction in order from the L2 cache, decodes these into a series of micro-operations, and stores the results in the L2 instruction cache Out-of-order execution logic – Schedules execution of the micro-operations subject to data dependencies and resource availability – thus micro-operations may be scheduled for execution in a different order than they were fetched from the instruction stream. As time permits, this unit schedules speculative execution of micro-operations that may be required in the future Execution units – These units execute micro-operations, fetching the required data from the L1 data cache and temporarily storing results in registers Memory subsystem – This unit includes the L2 and L3 caches and the system bus, which is used to access main memory when the L1 and L2 caches have a cache miss and to access the system I/O resources

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  • Un-failing over a Cisco PIX 515e

    - by ABrown
    We had a power outage at our data center last week and when our dual PIX 515E running IOS 7.0(8) (configured with a failover cable) came back, they were in a failed over state where the Secondary unit is active and the Primary unit is standby I have tried 'failover reset', 'failover active', and 'failover reload-standby' as well as executing reloads on both units in a variety of orders, and they don't come back Primary/Active Secondary/Standby. The only thing in my arsenal that I haven't tried is driving to the data center and performing a hard reboot, which I hate to do. I have read How Failover Works on the Cisco Secure Firewall and it seems like this should be wicked straight forward. output of show failover on Primary: Failover On Cable status: Normal Failover unit Primary Failover LAN Interface: N/A - Serial-based failover enabled Unit Poll frequency 15 seconds, holdtime 45 seconds Interface Poll frequency 15 seconds Interface Policy 1 Monitored Interfaces 2 of 250 maximum Version: Ours 7.0(8), Mate 7.0(8) Last Failover at: 02:52:05 UTC Mar 10 2010 This host: Primary - Standby Ready Active time: 0 (sec) Interface outside (x.x.x.165): Normal Interface inside (y.y.y.3): Normal Other host: Secondary - Active Active time: 897045 (sec) Interface outside (x.x.x.164): Normal Interface inside (y.y.y.4): Normal Stateful Failover Logical Update Statistics Link : Unconfigured. output of show failover on Secondary: Failover On Cable status: Normal Failover unit Secondary Failover LAN Interface: N/A - Serial-based failover enabled Unit Poll frequency 15 seconds, holdtime 45 seconds Interface Poll frequency 15 seconds Interface Policy 1 Monitored Interfaces 2 of 250 maximum Version: Ours 7.0(8), Mate 7.0(8) Last Failover at: 02:03:04 UTC Feb 28 2010 This host: Secondary - Active Active time: 896925 (sec) Interface outside (x.x.x.164): Normal Interface inside (y.y.y.4): Normal Other host: Primary - Standby Ready Active time: 0 (sec) Interface outside (x.x.x.165): Normal Interface inside (y.y.y.3): Normal Stateful Failover Logical Update Statistics Link : Unconfigured. I'm seeing the following in my syslog: Mar 10 03:05:00 fw1 %PIX-5-111008: User 'enable_15' executed the 'failover reset' command. Mar 10 03:05:09 fw1 %PIX-5-111008: User 'enable_15' executed the 'failover reload-standby' command. Mar 10 03:05:12 fw1 %PIX-6-720032: (VPN-Secondary) HA status callback: id=3,seq=200,grp=0,event=406,op=20,my=Active,peer=Failed. Mar 10 03:05:12 fw1 %PIX-6-720028: (VPN-Secondary) HA status callback: Peer state Failed. Mar 10 03:06:09 fw1 %PIX-6-720032: (VPN-Secondary) HA status callback: id=3,seq=200,grp=0,event=401,op=0,my=Active,peer=Failed. Mar 10 03:06:09 fw1 %PIX-6-720024: (VPN-Secondary) HA status callback: Control channel is down. Mar 10 03:06:09 fw1 %PIX-6-720032: (VPN-Secondary) HA status callback: id=3,seq=200,grp=0,event=401,op=1,my=Active,peer=Failed. Mar 10 03:06:10 fw1 %PIX-6-720024: (VPN-Secondary) HA status callback: Control channel is up. Mar 10 03:06:10 fw1 %PIX-6-720032: (VPN-Secondary) HA status callback: id=3,seq=200,grp=0,event=411,op=2,my=Active,peer=Failed. Mar 10 03:06:23 fw1 %PIX-6-720032: (VPN-Secondary) HA status callback: id=3,seq=200,grp=0,event=406,op=80,my=Active,peer=Standby Ready. Mar 10 03:06:23 fw1 %PIX-6-720028: (VPN-Secondary) HA status callback: Peer state Standby Ready. Mar 10 03:06:24 fw2 %PIX-6-720027: (VPN-Primary) HA status callback: My state Standby Ready. Mar 10 03:07:05 fw1 %PIX-5-111008: User 'enable_15' executed the 'failover reset' command. Mar 10 03:07:31 fw1 %PIX-5-111008: User 'enable_15' executed the 'failover active' command. Mar 10 03:08:04 fw1 %PIX-5-611103: User logged out: Uname: enable_1 Mar 10 03:08:04 fw1 %PIX-6-315011: SSH session from admin1_int on interface inside for user "pix" terminated normally Mar 10 03:08:39 fw1 %PIX-6-720032: (VPN-Secondary) HA status callback: id=3,seq=200,grp=0,event=406,op=20,my=Active,peer=Failed. Mar 10 03:08:39 fw1 %PIX-6-720028: (VPN-Secondary) HA status callback: Peer state Failed. Mar 10 03:09:10 fw1 %PIX-6-605005: Login permitted from admin1_int/36891 to inside:192.168.4.4/ssh for user "pix" Mar 10 03:09:23 fw1 %PIX-5-111008: User 'enable_15' executed the 'failover reset' command. Mar 10 03:09:38 fw1 %PIX-6-720032: (VPN-Secondary) HA status callback: id=3,seq=200,grp=0,event=401,op=0,my=Active,peer=Failed. Mar 10 03:09:39 fw1 %PIX-6-720024: (VPN-Secondary) HA status callback: Control channel is down. Mar 10 03:09:39 fw1 %PIX-6-720032: (VPN-Secondary) HA status callback: id=3,seq=200,grp=0,event=401,op=1,my=Active,peer=Failed. Mar 10 03:09:39 fw1 %PIX-6-720024: (VPN-Secondary) HA status callback: Control channel is up. Mar 10 03:09:39 fw1 %PIX-6-720032: (VPN-Secondary) HA status callback: id=3,seq=200,grp=0,event=411,op=2,my=Active,peer=Failed. Mar 10 03:09:52 fw1 %PIX-6-720032: (VPN-Secondary) HA status callback: id=3,seq=200,grp=0,event=406,op=80,my=Active,peer=Standby Ready. Mar 10 03:09:52 fw1 %PIX-6-720028: (VPN-Secondary) HA status callback: Peer state Standby Ready. Mar 10 03:09:53 fw2 %PIX-6-720027: (VPN-Primary) HA status callback: My state Standby Ready. I'm not exactly sure how to interpret that syslog data. Primary doesn't seem to even try to become Active. When I reload the individual units separately, my connections are retained, so it doesn't seem like I have a real hardware failure. Is there something I can query (IOS or SNMP) to check for hardware issues? Any thoughts? My IOS-fu is weak. Thanks for any help you might provide, Aaron

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  • Is there a good dual monitor arm solution for iMac 27" i7s?

    - by Darren Newton
    I currently have an iMac 27" and am considering purchasing another to run in target display mode. My desk space is a little limited. Is there a dual monitor arm solution that can support the weight of two iMac 27" units (30.5 pounds (13.8 kg)) as well as their width (25.6 inches (65.0 cm)) in a side-by-side landscape configuration? I looked at the Ergotron LX Dual Side by Side but the iMacs appear to exceed the width and weight limit this device is rated for. I'm open to alternate solutions to arms, such as a multi-unit desk stand/mount, but a wall mount is not possible for me at this time. Thanks!

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  • Rails3 environment running very slow on Windows XP, Ubuntu 9.04, Ubuntu 9.10

    - by bergyman
    I've tried all three (granted the Ubuntu versions were via VirtualBox with XP as a host, but I gave the images all the available RAM my system has). Loading the rails environment is taking 30-60 seconds. rails console, rake test:units - anything that requires rails to load up. And not just on the first go - every time. I've even used autotest to see if it helps with execution time for unit tests, but it doesn't. Any time I change one test, it still takes 30 seconds to load them, and then about 4 seconds to execute. Has anyone else come across this issue? Has anyone figured out any way to fix this?

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  • Linux partitioning problem

    - by Claudiu
    I am using cfdisk to repartition my hdd as from OS install I only got 1 big partition a swap. I wanted to resize the big partition to 1 GB /boot and use the rest of the space for an extended partition. After I do cfdisk, I recheck the partitions with fdisk -l and I get these: Disk /dev/sda: 320 GB, 320070320640 bytes 255 heads, 63 sectors/track, 38913 cylinders Units = cylinders of 16065 * 512 = 8225280 bytes Device Boot Start End Blocks Id System /dev/sda3 1 38455 308881755 f Extended LBA Warning: Partition 3 does not end on cylinder boundary. /dev/sda2 38455 38698 1951897 82 Linux swap /dev/sda1 * 38699 38913 311349654 83 Linux My problem is the Warning message, I think I know the cause, I think its because of sda1 Blocks size. How could that be soo big if Start and End interval is small?

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  • Summing / Running totals

    - by John
    I have 4 columns: Name, Week, Batch and Units Produced (Cols, A,b,c,d). In column e, I need to keep running totals based on name and week. When the week changes for the same person, restart the total. Fred, 12, 4001, 129.0 Answer in e: 129.0 Fred, 12, 4012, 234.0 Answer in e: 363.0 Fred, 13, 4023, 12.0 Answer in e: 12.0 John, 12, 4003, 420.0 Answer in e: 420.0 John, 13, 4021, 1200.0 Answer in e: 1200.0 John, 13, 4029, 120.0 Answer in e: 1320.0 I need to be able to copy the formula to over 1000 rows.

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  • All Xen domU LVM volumes corrupt after reboot

    - by zcs
    I'm running a Debian Squeeze dom0, and after rebooting it all 7 of my domUs have data corruption. Each is setup as ext3 partition directly on a separate lvm2 volume. None of the lvm volumes will mount; all have bad superblocks. I've tried e2fsck with each superblock to no avail. What else can I try? Each domU has two LVM volumes connected to it, one for the disk and one for swap. The disk is mounted at root, formatted as a normal ext3 partition as a xen-blk device. The volumes are never mounted outside of the guest OS. I'm running Ubuntu 11.04 using the instructions here. I'm not sure that they didn't shutdown properly, all I know is they were corrupt after I issues a clean 'reboot' on the dom0. Here's a sample Xen config file; the rest are the same except for name, vcpus, memory, vif and disk. name = 'load1' vcpus = 2 memory = 512 vif = ['bridge=prbr0', 'bridge=eth0'] disk = ['phy:/dev/VolGroup00/load1-disk,xvda,w','phy:/dev/VolGroup00/load1-swap,xvdb,w'] #============================================================================ # Debian Installer specific variables def check_bool(name, value): value = str(value).lower() if value in ('t', 'tr', 'tru', 'true'): return True return False global var_check_with_default def var_check_with_default(default, var, val): if val: return val return default xm_vars.var('install', use='Install Debian, default: false', check=check_bool) xm_vars.var("install-method", use='Installation method to use "cdrom" or "network" (default: network)', check=lambda var, val: var_check_with_default('network', var, val)) # install-method == "network" xm_vars.var("install-mirror", use='Debian mirror to install from (default: http://archive.ubuntu.com/ubuntu)', check=lambda var, val: var_check_with_default('http://archive.ubuntu.com/ubuntu', var, val)) xm_vars.var("install-suite", use='Debian suite to install (default: natty)', check=lambda var, val: var_check_with_default('natty', var, val)) # install-method == "cdrom" xm_vars.var("install-media", use='Installation media to use (default: None)', check=lambda var, val: var_check_with_default(None, var, val)) xm_vars.var("install-cdrom-device", use='Installation media to use (default: xvdd)', check=lambda var, val: var_check_with_default('xvdd', var, val)) # Common options xm_vars.var("install-arch", use='Debian mirror to install from (default: amd64)', check=lambda var, val: var_check_with_default('amd64', var, val)) xm_vars.var("install-extra", use='Extra command line options (default: None)', check=lambda var, val: var_check_with_default(None, var, val)) xm_vars.var("install-installer", use='Debian installer to use (default: network uses install-mirror; cdrom uses /install.ARCH)', check=lambda var, val: var_check_with_default(None, var, val)) xm_vars.var("install-kernel", use='Debian installer kernel to use (default: uses install-installer)', check=lambda var, val: var_check_with_default(None, var, val)) xm_vars.var("install-ramdisk", use='Debian installer ramdisk to use (default: uses install-installer)', check=lambda var, val: var_check_with_default(None, var, val)) xm_vars.check() if not xm_vars.env.get('install'): bootloader="/usr/sbin/pygrub" elif xm_vars.env['install-method'] == "network": import os.path print "Install Mirror: %s" % xm_vars.env['install-mirror'] print "Install Suite: %s" % xm_vars.env['install-suite'] if xm_vars.env['install-installer']: installer = xm_vars.env['install-installer'] else: installer = xm_vars.env['install-mirror']+"/dists/"+xm_vars.env['install-suite'] + \ "/main/installer-"+xm_vars.env['install-arch']+"/current/images" print "Installer: %s" % installer print print "WARNING: Installer kernel and ramdisk are not authenticated." print if xm_vars.env.get('install-kernel'): kernelurl = xm_vars.env['install-kernel'] else: kernelurl = installer + "/netboot/xen/vmlinuz" if xm_vars.env.get('install-ramdisk'): ramdiskurl = xm_vars.env['install-ramdisk'] else: ramdiskurl = installer + "/netboot/xen/initrd.gz" import urllib class MyUrlOpener(urllib.FancyURLopener): def http_error_default(self, req, fp, code, msg, hdrs): raise IOError("%s %s" % (code, msg)) urlopener = MyUrlOpener() try: print "Fetching %s" % kernelurl kernel, _ = urlopener.retrieve(kernelurl) print "Fetching %s" % ramdiskurl ramdisk, _ = urlopener.retrieve(ramdiskurl) except IOError, _: raise elif xm_vars.env['install-method'] == "cdrom": arch_path = { 'i386': "/install.386", 'amd64': "/install.amd" } if xm_vars.env['install-media']: print "Install Media: %s" % xm_vars.env['install-media'] else: raise OptionError("No installation media given.") if xm_vars.env['install-installer']: installer = xm_vars.env['install-installer'] else: installer = arch_path[xm_vars.env['install-arch']] print "Installer: %s" % installer if xm_vars.env.get('install-kernel'): kernelpath = xm_vars.env['install-kernel'] else: kernelpath = installer + "/xen/vmlinuz" if xm_vars.env.get('install-ramdisk'): ramdiskpath = xm_vars.env['install-ramdisk'] else: ramdiskpath = installer + "/xen/initrd.gz" disk.insert(0, 'file:%s,%s:cdrom,r' % (xm_vars.env['install-media'], xm_vars.env['install-cdrom-device'])) bootloader="/usr/sbin/pygrub" bootargs="--kernel=%s --ramdisk=%s" % (kernelpath, ramdiskpath) print "From CD" else: print "WARNING: Unknown install-method: %s." % xm_vars.env['install-method'] if xm_vars.env.get('install'): # Figure out command line if xm_vars.env['install-extra']: extras=[xm_vars.env['install-extra']] else: extras=[] # Reboot will just restart the installer since this file is not # reparsed, so halt and restart that way. extras.append("debian-installer/exit/always_halt=true") extras.append("--") extras.append("quiet") console="hvc0" try: if len(vfb) >= 1: console="tty0" except NameError, e: pass extras.append("console="+ console) extra = str.join(" ", extras) print "command line is \"%s\"" % extra root There are two LVM logical volumes connected to each VM. Here's the fdisk -l output for the disk volume: Disk /dev/VolGroup00/VMNAME-disk: 8589 MB, 8589934592 bytes 255 heads, 63 sectors/track, 1044 cylinders Units = cylinders of 16065 * 512 = 8225280 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x00029c01 Device Boot Start End Blocks Id System /dev/VolGroup00/VMNAME-disk1 1 1045 8386560 83 Linux And the swap volume: Disk /dev/VolGroup00/VMNAME-swap: 536 MB, 536870912 bytes 37 heads, 35 sectors/track, 809 cylinders Units = cylinders of 1295 * 512 = 663040 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x0004faae Device Boot Start End Blocks Id System /dev/VolGroup00/VMNAME-swap1 2 809 522240 82 Linux swap / Solaris Partition 1 has different physical/logical beginnings (non-Linux?): phys=(0, 32, 33) logical=(1, 21, 19) Partition 1 has different physical/logical endings: phys=(65, 36, 35) logical=(808, 4, 28)

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  • Barracuda spam filter alternative - virtualization/appliance friendly?

    - by ewwhite
    I've sold and deployed Barracuda spam and web filters for years. I've always thought that the functionality was good (Barracuda Central, easy interface, effective filtering), but the hardware on the entry to midrange units is a weak point. They have single power supplies, no RAID and limited monitoring support. Personally, I think Barracuda would make a killing selling their software as a VMWare appliance, but I'm looking for something similar that I can deploy as a consultant, but will be easy for customers to manage. It should have support for server-grade hardware or the ability to be deployed as a virtual machine. Is there anything out there that's close?

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  • Rails3 environment running very slow on Windows XP, Ubuntu 9.04, Ubuntu 9.10

    - by bergyman
    I've tried all three (granted the Ubuntu versions were via VirtualBox with XP as a host, but I gave the images all the available RAM my system has). Loading the rails environment is taking 30-60 seconds. rails console, rake test:units - anything that requires rails to load up. And not just on the first go - every time. I've even used autotest to see if it helps with execution time for unit tests, but it doesn't. Any time I change one test, it still takes 30 seconds to load them, and then about 4 seconds to execute. Has anyone else come across this issue? Has anyone figured out any way to fix this?

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  • Create new partition on live production CentOS server

    - by Kimmel
    I have a production server that is running on CentOS. I'd like to create a partition on the server without having to reinstall everything. I have CLI and VNC access to the remote server. Is there anyway that I can create a partition safely? Here's my output from fdisk -l Disk /dev/sda: 85.9 GB, 85899345920 bytes 255 heads, 63 sectors/track, 10443 cylinders Units = cylinders of 16065 * 512 = 8225280 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x00033d5e Device Boot Start End Blocks Id System /dev/sda1 * 1 10444 83885056 83 Linux Thanks.

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  • Improving terminal server performance for a specfic app

    - by Matt
    We have a windows 2003 terminal server running 2X application load balancign that is hosting a client's application that is accessed by around 50 users. Each user has there own database. The database is a file based database. The application is developed under Delphi so I think the database may be BDE based. As you can imagine, there is probably quite a lot of disk i/o. Here are some of the perfmon settings. Logged in users (average) 20 - 25 CPU Utilization (average) 80 - 100% Disk Queue Length (average) 1.6 % Disk time (average) 111 Page faults/sec (average) 1400 The application takes on average about a minute to load up. As usual, the budget is tight. Is there basic windows performance tuning tips that people can recommend to improve things before we fork out on more RAM etc. Server is a 2.8GHz Xeon with 3GB of RAM.

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  • How do I recover from a Linux CentOS 4.6 Operating System Crash

    - by Greg Omebije
    Our x86 Linux server running CentOS4.6 has crashed. The machine boots only to the Grub prompt. We have tried using the "rescue mode" to recover the System, but it hasn't worked. How can we fix this problem, so that the machine boots normally? How can we fix this problem to the point were we can recover our files from the server Our Linux Server Configuration: Dell PowerEdge 1950 Intel Xeon 2 HDD (146GB each) 4GB RAM Hardware and Software raid setup CentOS 4.6 We used Sysrecord to boot the computer: the following are the output of fdisk -l Disk /dev/sda: 293.3 GB, 292326211584 255 heads, 63 sectors/track, 35539 Cylinders Units = cylinders of 16065 * 512 = 8225280 bytes Disk identifier: 0x00000080 Device Boot Start End Blocks Id System /dev/sda1 * 1 13 104391 83 Linux /dev/sda2 14 17769 142625070 8e Linux LVM

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  • How to stop a group of systemd custom services

    - by tsingyue
    I wrote three service units, say a.service b.service and c.service. C requires and runs after b, b requires and runs after a, so when I execute "systemctl start c.service", all three of them will be launched one by one. But when I want to stop all of them, I have to execute "systemctl stop a.service b.service c.service". Is there any other way to stop all of them with less typing? I know with "Bindto=" I can use "systemctl stop a.service" to stop all of them, but what if I got c Bindto a and b, while a and b have no required relationship to each other?

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  • 1Tb disk formatted on Linux won't mount on windows nor mac

    - by Pedro MC
    I have an external HD (western digital) with 1Tb. I use Linux but I wanted to reserve a cross platform partition on the disk. I decided to create two partitions and used the "disks" application to do it. I created one partition with the LUKS (version 1) encryption and the other one, cross platform, in NTFS filesystem. Things work fine on my OS but when I try to use the disk (the cross platform partition) on both windows and mac the device is not recognized. What could it be? Next, output of "sfdisk -l /dev/sdb": Disk /dev/sdb: 121600 cylinders, 255 heads, 63 sectors/track Units = cylinders of 8225280 bytes, blocks of 1024 bytes, counting from 0 Device Boot Start End #cyls #blocks Id System /dev/sdb1 0+ 36473- 36473- 292968750 83 Linux /dev/sdb2 36473+ 121600- 85128- 683789062+ 83 Linux /dev/sdb3 0 - 0 0 0 Empty /dev/sdb4 0 - 0 0 0 Empty

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  • program for drawing diagrams

    - by ldigas
    I need to draw a few diagrams that describe the working of some smaller programs (nothing real, but some exercise to illustrate some points). They're supposed to be 'old kind' diagrams, like you remember from the "old school of programming" - I gather some users here will know what I mean. And it's be good if they have ready blocks for program units, IF conditions, FOR/DO loops and such ... So, anyone knows of an application for such ? (I've tried Visio which I have on my work machine (for whatever reason ... I sure didn't put it there) ... but it is "too visual" for this kinda thing ... I just want to draw a few simple black on white diagrams ... nothing special or in colours).

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  • fdisk -l shows a partition is not in /dev directory

    - by zitronic
    I am having troubles on mounting my ntfs hard drive on linux. I am running ophcrack live CD. fdisk shows me the hard drive I have installed on my computer fdisk -l Disk /dev/hdc: 30.0 GB, 30005821440 bytes 255 heads, 63 sectors/track, 3648 cylinders Units = cylinders of 16065 * 512 = 8225280 bytes Device Boot Start End Blocks Id System /dev/hdc1 * 1 3647 29294496 7 HPFS/NTFS but I dont have that device on my /dev directory so I can not mount it ls /dev/hd* /dev/hda /dev/hda4 /dev/hda8 /dev/hdb2 /dev/hdb6 /dev/hdc /dev/hda1 /dev/hda5 /dev/hda9 /dev/hdb3 /dev/hdb7 /dev/hdd /dev/hda2 /dev/hda6 /dev/hdb /dev/hdb4 /dev/hdb8 /dev/hda3 /dev/hda7 /dev/hdb1 /dev/hdb5 /dev/hdb9 What could be going on_

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  • Restoring a Ubuntu Linux LVM drive

    - by user848106
    Scenario: I have a hard disk with a bootable sda1 partition and an Linux LVM. I seem to have corrupted the MBR? or Grub? not sure. I know it no longer boots and even with boot-repair and a Ubuntu live usb I was able to restore it. Following a "fix" with boot-repair I get "missing operating system" Can't i just reinstall grub or a new menu.lst? Disk /dev/sda: 320.1 GB, 320072933376 bytes 255 heads, 63 sectors/track, 38913 cylinders, total 625142448 sectors Units = sectors of 1 * 512 = 512 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x0007b68f Device Boot Start End Blocks Id System /dev/sda1 * 2048 499711 248832 83 Linux /dev/sda2 501758 625141759 312320001 5 Extended /dev/sda5 501760 625141759 312320000 8e Linux LVM

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  • Can't deploy an ejb bean on jboss

    - by leo
    I am try to deploy a jar to an jboss server. It works on my environment. But when I deployed the same jar on another server, i kept getting an error saying that the persistence unit is already registered. There is no other bean using the same name and the same persistence unit name. I tried to restart the server and remove the tmp, work, data directory but still get the same error. here is my error: ObjectName: jboss.j2ee:service=EJB3,module=wess_jpa.jar State: FAILED Reason: java.lang.RuntimeException: javax.management.InstanceAlreadyExistsException: persistence.units:unitName=dses_wess already registered. This is almost identical to this issue in jboss forum but there is no solution: http://www.jboss.org/index.html?module=bb&op=viewtopic&p=4211687#4211687

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  • Performance required to improve Windows Experience Index?

    - by Ian Boyd
    Is there a guide on the metrics required to obtain a certain Windows Experience Index? A Microsoft guy said in January 2009: On the matter of transparency, it is indeed our plan to disclose in great detail how the scores are calculated, what the tests attempt to measure, why, and how they map to realistic scenarios and usage patterns. Has that amount of transparency happened? Is there a technet article somewhere? If my score was limited by my Memory subscore of 5.9. A nieve person would suggest: Buy a faster RAM Which is wrong of course. From the Windows help: If your computer has a 64-bit central processing unit (CPU) and 4 gigabytes (GB) or less random access memory (RAM), then the Memory (RAM) subscore for your computer will have a maximum of 5.9. You can buy the fastest, overclocked, liquid-cooled, DDR5 RAM on the planet; you'll still have a maximum Memory subscore of 5.9. So in general the knee-jerk advice "buy better stuff" is not helpful. What i am looking for is attributes required to achieve a certain score, or move beyond a current limitation. The information i've been able to compile so far, chiefly from 3 Windows blog entries, and an article: Memory subscore Score Conditions ======= ================================ 1.0 < 256 MB 2.0 < 500 MB 2.9 <= 512 MB 3.5 < 704 MB 3.9 < 944 MB 4.5 <= 1.5 GB 5.9 < 4.0GB-64MB on a 64-bit OS Windows Vista highest score 7.9 Windows 7 highest score Graphics Subscore Score Conditions ======= ====================== 1.0 doesn't support DX9 1.9 doesn't support WDDM 4.9 does not support Pixel Shader 3.0 5.9 doesn't support DX10 or WDDM1.1 Windows Vista highest score 7.9 Windows 7 highest score Gaming graphics subscore Score Result ======= ============================= 1.0 doesn't support D3D 2.0 supports D3D9, DX9 and WDDM 5.9 doesn't support DX10 or WDDM1.1 Windows Vista highest score 6.0-6.9 good framerates (e.g. 40-50fps) at normal resoltuions (e.g. 1280x1024) 7.0-7.9 even higher framerates at even higher resolutions 7.9 Windows 7 highest score Processor subscore Score Conditions ======= ========================================================================== 5.9 Windows Vista highest score 6.0-6.9 many quad core processors will be able to score in the high 6 low 7 ranges 7.0+ many quad core processors will be able to score in the high 6 low 7 ranges 7.9 8-core systems will be able to approach 8.9 Windows 7 highest score Primary hard disk subscore (note) Score Conditions ======= ======================================== 1.9 Limit for pathological drives that stop responding when pending writes 2.0 Limit for pathological drives that stop responding when pending writes 2.9 Limit for pathological drives that stop responding when pending writes 3.0 Limit for pathological drives that stop responding when pending writes 5.9 highest you're likely to see without SSD Windows Vista highest score 7.9 Windows 7 highest score Bonus Chatter You can find your WEI detailed test results in: C:\Windows\Performance\WinSAT\DataStore e.g. 2011-11-06 01.00.19.482 Disk.Assessment (Recent).WinSAT.xml <WinSAT> <WinSPR> <DiskScore>5.9</DiskScore> </WinSPR> <Metrics> <DiskMetrics> <AvgThroughput units="MB/s" score="6.4" ioSize="65536" kind="Sequential Read">89.95188</AvgThroughput> <AvgThroughput units="MB/s" score="4.0" ioSize="16384" kind="Random Read">1.58000</AvgThroughput> <Responsiveness Reason="UnableToAssess" Kind="Cap">TRUE</Responsiveness> </DiskMetrics> </Metrics> </WinSAT> Pre-emptive snarky comment: "WEI is useless, it has no relation to reality" Fine, how do i increase my hard-drive's random I/O throughput? Update - Amount of memory limits rating Some people don't believe Microsoft's statement that having less than 4GB of RAM on a 64-bit edition of Windows doesn't limit the rating to 5.9: And from xxx.Formal.Assessment (Recent).WinSAT.xml: <WinSPR> <LimitsApplied> <MemoryScore> <LimitApplied Friendly="Physical memory available to the OS is less than 4.0GB-64MB on a 64-bit OS : limit mem score to 5.9" Relation="LT">4227858432</LimitApplied> </MemoryScore> </LimitsApplied> </WinSPR> References Windows Vista Team Blog: Windows Experience Index: An In-Depth Look Understand and improve your computer's performance in Windows Vista Engineering Windows 7 Blog: Engineering the Windows 7 “Windows Experience Index”

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  • Anyone love/hate the PowerConnect line of switches from Dell?

    - by Rob Bergin
    I am looking at replacing some unmanaged 16 port store bought GB switches and wanted to go with Cisco but it may be cost prohibitive. Instead I am looking at ProCurve or Dell's PowerConnect line up. I am looking for SNMP, Management, VLANs, and SFLOW would icing on the switch cupcake. I would get the 6224 or the 6248 and then maybe add the RPS-600 to it for redundant power. I think the RPS-600 supports multiple switches. Rackspace is also a little challenge so I am trying to do it with as little Rack Units as possible. Ideally I would go with two 6224's or a single 6248 and then do two VLANs. Thanks for any feedback. Rob

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  • CentOS Insufficient space in download directory /var/cache/yum/base/packages

    - by Joao Heleno
    Hello! I was trying to yum install libpcap when I got Error Downloading Packages: 14:libpcap-0.9.4-15.el5.i386: Insufficient space in download directory /var/cache/yum/base/packages * free 0 * needed 108 k Here's output from df -h: Filesystem Size Used Avail Use% Mounted on /dev/sda1 20G 19G 0 100% / /dev/sda3 202G 38G 154G 20% /home tmpfs 1.5G 0 1.5G 0% /dev/shm And fdisk -l: Disk /dev/sda: 250.0 GB, 250000000000 bytes 255 heads, 63 sectors/track, 30394 cylinders Units = cylinders of 16065 * 512 = 8225280 bytes Device Boot Start End Blocks Id System /dev/sda1 * 1 2611 20972826 83 Linux /dev/sda2 2612 3251 5140800 82 Linux swap / Solaris /dev/sda3 3252 30394 218026147+ 83 Linux I have launched yum clean all with no success clearing up space. Please advise. Thanks.

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  • Lexmark C532 error code won't go away

    - by Paul Phillips
    I can't get rid of error code 84 - Yellow PC unit missing. I've just replaced all 4 photoconductors with genuine Lexmark parts and now the error code keeps appearing. I've swapped the PC units around so know it's not a faulty unit. I've followed the suggestions in the online Lexmark manuals and even called the Lexmark helpline but to no avail. any suggestions? Printer was fine until it gave me a PC unit life warning and I renewed them. Thanks.

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