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  • How do you set rate limit access to your API using Iptables?

    - by Cory
    How can you set rate limit access to API using Iptables. Tried to set limit using port 80, but I don't want to set limit to the web access entirely. Is there a way to specified a subdomain rather than port. Example: set rate limit to api.example.com not example.com? If there is no way to set rate limit by subdomain, what is the suggested rate limit access to port 80 without risking blocking a legitimate web user? One connection per second would be enough?

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  • how to keep display tick rate steady when using continuous collision detection?

    - by nas Ns
    (I've just found about this forum). I hope it is ok to repost my question again here. I posted this question at stackoverflow, but it looks like I might get better help here. Here is the question: I've implemented basic particles motion simulation with continuous collision detection. But there is small issue in display. Assume simple case of circles moving inside square. All elastic collisions. no firction. All motion is constant speed. No forces are involved, no gravity. So when a particle is moving, it is always moving at constant speed (in between collisions) What I do now is this: Let the simulation time step be 1 second (for example). This is the time step simulation is advanced before displaying the new state (unless there is a collision sooner than this). At start of each time step, time for the next collision between any particles or a particle with a wall is determined. Call this the TOC time; let’s say TOC was .5 seconds in this case. Since TOC is smaller than the standard time step, then the system is moved by TOC and the new system is displayed so that the new display shows any collisions as just taking place (say 2 circles just touched each other’s, or a circle just touched a wall) Next, the collision(s) are resolved (i.e. speeds updated, changed directions etc..). A new step is started. The same thing happens. Now assume there is no collision detected within the next 1 second (those 2 circles above will not be in collision any more, even though they are still touching, due to their speeds showing they are moving apart now), Hence, simulation time is advanced now by the full one second, the standard time step, and particles are moved on the screen using 1 second simulation time and new display is shown. You see what has just happened: One frame ran for .5 seconds, but the next frame runs for 1 second, may be the 3rd frame is displayed after 2 seconds, may be the 4th frame is displayed after 2.8 seconds (because TOC was .8 seconds then) and so on. What happens is that the motion of a particle on the screen appears to speed up or slow down, even though it is moving at constant speed and was not even involved in a collision. i.e. Looking at one particle on its own, I see it suddenly speeding up or slowing down, becuase another particle had hit a wall. This is because the display tick is not uniform. i.e. the frame rate update is changing, giving the false illusion that a particle is moving at non-constant speed while in fact it is moving at constant speed. The motion on the screen is not smooth, since the screen is not updating at constant rate. I am not able to figure how to fix this. If I want to show 2 particles at the moment of the collision, I must draw the screen at different times. Drawing the screen always at the same tick interval, results in seeing 2 particles before the collision, and then after the collision, and not just when they colliding, which looked bad when I tried it. So, how do real games handle this issue? How to display things in order to show collisions when it happen, yet keep the display tick constant? These 2 requirements seem to contradict each other’s.

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  • Google Analytics says I'm Getting 50 percent bounce rate on my website ! Help!

    - by Ali
    Hi guys, I recently launched my web application and set up google analytics to monitor it. For some weird reason google analytics shows a 49.7 percent bounce rate - I'm quite suprised as to why and how is this possible. My web application works well - I've never been bounced off even once whenever I visit it from any browser. What's going on I have no idea what to look out for here?

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  • Trying to get the associated value from an Enum at runtime in Java

    - by devoured elysium
    I want to accomplish something like the following (my interest is in the toInt() method). Is there any native way to accomplish this? If not, how can I get the integer associated with an enum value (like in C#) ? enum Rate { VeryBad(1), Bad(2), Average(3), Good(4), Excellent(5); private int rate; private Rate(int rate) { this.rate = rate; } public int toInt() { return rate; } } Thanks

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  • How to prevent Google Website Optimizer from making Google Analytics spike Direct Traffic and lower Bounce Rate?

    - by Scott
    I am using Google Website Optimizer (GWO) and Google Analytics. Whenever a person (Google Website Optimizer) does a javascript redirect, Google Analytics will change the referrer. When the referrer changes, the traffic source becomes yourself and is changed to Direct Traffic. For Example: A visitor goes to google: searches for my great service. Clicks the link that goes to website page: /home/ At this point, Google Analytics tracks the referrer as Google. However, /home/ has a GWO javacript redirect to a battery of A/B tests. /home-1/ or /home-2/ or /home-3/ When the redirect from /home/ occurs to /home-1/, google analytics on the /home-1/ page now thinks the referrer is yourself and converts the referrer to Direct Traffic since the Direct Traffic bucket is the unknown. I'm really surprised that GWO and GA do this when they both come from google. Now, How does one fix this to prevent the overwrite of the referrer using GWO?

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  • "Google a raté le coche avec Chrome OS" affirment certains experts qui s'inquiètent des retards pris par le système d'exploitation

    "Google a raté le coche avec Chrome OS", affirment certains experts qui s'inquiètent des retards pris par système d'exploitation de Google Chrome OS aurait du voir le jour en 2010. Et pourtant, personne encore ne l'a vu pointer le bout de son nez. Du coup, certains experts s'inquiètent. Pour eux, le marché change vite, trop vite. Chrome OS est destiné a équiper les appareils de puissance moindre, comme les netbooks et les ordinateurs lowcost. Seulement : "il y a un an, personne n'aurait pu prédire les grands changements qui sont survenus". Ce que les analystes veulent dire, c'est qu'au moment où Google à annoncé son produit, les netbooks avaient le vent en poupe. Aujourd'hui, ils so...

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  • Why is programming sometimes viewed as a second-rate role?

    - by CaptainCodeman
    I've been a programmer for most of my life. I recently interviewed for a management job in a company and the interviewer looked at my CV asked me "How do we know you're not just a programmer". Which in my opinion is quite a rude thing to say, but it's not an isolated incident and I've heard other similar things in other settings. It does seem that for some reason being a programmer is viewed as having a lower station, especially in settings where they have a separate IT department which is viewed as a support role. Is a career in software development doomed to being a second-rate support citizen?

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  • Google a-t-il raté son entrée sur le marché des tablettes ? Android connaît une "croissance pénible" d'après DisplaySearch

    Google a-t-il raté son entrée sur le marché des tablettes ? Android connaît une "croissance pénible" d'après DisplaySearch Six mois après le lancement d'Android 3 (alias Honeycomb), force est de constater que les ventes des tablettes qui tournent dessus sont loin de défrayer les chroniques, elles ne semblent pas être en mesure en tout cas de freiner les ventes frénétiques de l'iPad d'Apple. Pour l'analyste Richard Shim de DisplaySearch, l'univers Android connaît une « croissance pénible » avec Honeycomb : « il est clair que les premières tablettes fondées sur Android ne s'écoulent pas aussi bien que ce que beaucoup attendaient, quelques marques avec lesquelles ...

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  • vim + Ruby on Rails: how do you bounce among those 4-5 files you're currently working on?

    - by glitch
    I'm just starting to get familiar with vim, and I'd like to use it as my primary Rails development tool. As a Visual Studio and RubyMine user, I find a lot of stuff to be missing from the barebones vim installation, and therefore I went ahead and attempted to soup it up with plugins such as: rails.vim tcomment ruby-vim NERDtree and a couple of others. The issue is that I still don't quite get the average work-flow of using vim as one's Rails IDE. In RubyMine (again, similarly to Visual Studio) I have a series of tabs always open, containing the main files I'm switching among, and I additionally use NERDtree to open files from the folder structure. I tried opening them as new tabs, but the tab system in vim is just a lot more awkward than that in real IDEs. (I haven't seen vim pros in action, but I imagine that they'd not be relying on tabs, but using numerous splits instead, keeping at least a couple of files per split and switching between them with CTRL + ^. Is that the case?) So, at the end of the day, how do I really squeeze the most from vim if I want to be able to quickly access several files at once? Thank you!

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  • VLOOKUP in Excel, part 2: Using VLOOKUP without a database

    - by Mark Virtue
    In a recent article, we introduced the Excel function called VLOOKUP and explained how it could be used to retrieve information from a database into a cell in a local worksheet.  In that article we mentioned that there were two uses for VLOOKUP, and only one of them dealt with querying databases.  In this article, the second and final in the VLOOKUP series, we examine this other, lesser known use for the VLOOKUP function. If you haven’t already done so, please read the first VLOOKUP article – this article will assume that many of the concepts explained in that article are already known to the reader. When working with databases, VLOOKUP is passed a “unique identifier” that serves to identify which data record we wish to find in the database (e.g. a product code or customer ID).  This unique identifier must exist in the database, otherwise VLOOKUP returns us an error.  In this article, we will examine a way of using VLOOKUP where the identifier doesn’t need to exist in the database at all.  It’s almost as if VLOOKUP can adopt a “near enough is good enough” approach to returning the data we’re looking for.  In certain circumstances, this is exactly what we need. We will illustrate this article with a real-world example – that of calculating the commissions that are generated on a set of sales figures.  We will start with a very simple scenario, and then progressively make it more complex, until the only rational solution to the problem is to use VLOOKUP.  The initial scenario in our fictitious company works like this:  If a salesperson creates more than $30,000 worth of sales in a given year, the commission they earn on those sales is 30%.  Otherwise their commission is only 20%.  So far this is a pretty simple worksheet: To use this worksheet, the salesperson enters their sales figures in cell B1, and the formula in cell B2 calculates the correct commission rate they are entitled to receive, which is used in cell B3 to calculate the total commission that the salesperson is owed (which is a simple multiplication of B1 and B2). The cell B2 contains the only interesting part of this worksheet – the formula for deciding which commission rate to use: the one below the threshold of $30,000, or the one above the threshold.  This formula makes use of the Excel function called IF.  For those readers that are not familiar with IF, it works like this: IF(condition,value if true,value if false) Where the condition is an expression that evaluates to either true or false.  In the example above, the condition is the expression B1<B5, which can be read as “Is B1 less than B5?”, or, put another way, “Are the total sales less than the threshold”.  If the answer to this question is “yes” (true), then we use the value if true parameter of the function, namely B6 in this case – the commission rate if the sales total was below the threshold.  If the answer to the question is “no” (false), then we use the value if false parameter of the function, namely B7 in this case – the commission rate if the sales total was above the threshold. As you can see, using a sales total of $20,000 gives us a commission rate of 20% in cell B2.  If we enter a value of $40,000, we get a different commission rate: So our spreadsheet is working. Let’s make it more complex.  Let’s introduce a second threshold:  If the salesperson earns more than $40,000, then their commission rate increases to 40%: Easy enough to understand in the real world, but in cell B2 our formula is getting more complex.  If you look closely at the formula, you’ll see that the third parameter of the original IF function (the value if false) is now an entire IF function in its own right.  This is called a nested function (a function within a function).  It’s perfectly valid in Excel (it even works!), but it’s harder to read and understand. We’re not going to go into the nuts and bolts of how and why this works, nor will we examine the nuances of nested functions.  This is a tutorial on VLOOKUP, not on Excel in general. Anyway, it gets worse!  What about when we decide that if they earn more than $50,000 then they’re entitled to 50% commission, and if they earn more than $60,000 then they’re entitled to 60% commission? Now the formula in cell B2, while correct, has become virtually unreadable.  No-one should have to write formulae where the functions are nested four levels deep!  Surely there must be a simpler way? There certainly is.  VLOOKUP to the rescue! Let’s redesign the worksheet a bit.  We’ll keep all the same figures, but organize it in a new way, a more tabular way: Take a moment and verify for yourself that the new Rate Table works exactly the same as the series of thresholds above. Conceptually, what we’re about to do is use VLOOKUP to look up the salesperson’s sales total (from B1) in the rate table and return to us the corresponding commission rate.  Note that the salesperson may have indeed created sales that are not one of the five values in the rate table ($0, $30,000, $40,000, $50,000 or $60,000).  They may have created sales of $34,988.  It’s important to note that $34,988 does not appear in the rate table.  Let’s see if VLOOKUP can solve our problem anyway… We select cell B2 (the location we want to put our formula), and then insert the VLOOKUP function from the Formulas tab: The Function Arguments box for VLOOKUP appears.  We fill in the arguments (parameters) one by one, starting with the Lookup_value, which is, in this case, the sales total from cell B1.  We place the cursor in the Lookup_value field and then click once on cell B1: Next we need to specify to VLOOKUP what table to lookup this data in.  In this example, it’s the rate table, of course.  We place the cursor in the Table_array field, and then highlight the entire rate table – excluding the headings: Next we must specify which column in the table contains the information we want our formula to return to us.  In this case we want the commission rate, which is found in the second column in the table, so we therefore enter a 2 into the Col_index_num field: Finally we enter a value in the Range_lookup field. Important:  It is the use of this field that differentiates the two ways of using VLOOKUP.  To use VLOOKUP with a database, this final parameter, Range_lookup, must always be set to FALSE, but with this other use of VLOOKUP, we must either leave it blank or enter a value of TRUE.  When using VLOOKUP, it is vital that you make the correct choice for this final parameter. To be explicit, we will enter a value of true in the Range_lookup field.  It would also be fine to leave it blank, as this is the default value: We have completed all the parameters.  We now click the OK button, and Excel builds our VLOOKUP formula for us: If we experiment with a few different sales total amounts, we can satisfy ourselves that the formula is working. Conclusion In the “database” version of VLOOKUP, where the Range_lookup parameter is FALSE, the value passed in the first parameter (Lookup_value) must be present in the database.  In other words, we’re looking for an exact match. But in this other use of VLOOKUP, we are not necessarily looking for an exact match.  In this case, “near enough is good enough”.  But what do we mean by “near enough”?  Let’s use an example:  When searching for a commission rate on a sales total of $34,988, our VLOOKUP formula will return us a value of 30%, which is the correct answer.  Why did it choose the row in the table containing 30% ?  What, in fact, does “near enough” mean in this case?  Let’s be precise: When Range_lookup is set to TRUE (or omitted), VLOOKUP will look in column 1 and match the highest value that is not greater than the Lookup_value parameter. It’s also important to note that for this system to work, the table must be sorted in ascending order on column 1! If you would like to practice with VLOOKUP, the sample file illustrated in this article can be downloaded from here. Similar Articles Productive Geek Tips Using VLOOKUP in ExcelImport Microsoft Access Data Into ExcelImport an Access Database into ExcelCopy a Group of Cells in Excel 2007 to the Clipboard as an ImageShare Access Data with Excel in Office 2010 TouchFreeze Alternative in AutoHotkey The Icy Undertow Desktop Windows Home Server – Backup to LAN The Clear & Clean Desktop Use This Bookmarklet to Easily Get Albums Use AutoHotkey to Assign a Hotkey to a Specific Window Latest Software Reviews Tinyhacker Random Tips DVDFab 6 Revo Uninstaller Pro Registry Mechanic 9 for Windows PC Tools Internet Security Suite 2010 Quickly Schedule Meetings With NeedtoMeet Share Flickr Photos On Facebook Automatically Are You Blocked On Gtalk? Find out Discover Latest Android Apps On AppBrain The Ultimate Guide For YouTube Lovers Will it Blend? iPad Edition

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  • What does O(log n) mean exactly?

    - by Andreas Grech
    I am currently learning about Big O Notation running times and amortized times. I understand the notion of O(n) linear time, meaning that the size of the input affects the growth of the algorithm proportionally...and the same goes for, for example, quadratic time O(n2) etc..even algorithms, such as permutation generators, with O(n!) times, that grow by factorials. For example, the following function is O(n) because the algorithm grows in proportion to its input n: f(int n) { int i; for (i = 0; i < n; ++i) printf("%d", i); } Similarly, if there was a nested loop, the time would be O(n2). But what exactly is O(log n)? For example, what does it mean to say that the height of a complete binary tree is O(log n)? I do know (maybe not in great detail) what Logarithm is, in the sense that: log10 100 = 2, but I cannot understand how to identify a function with a logarithmic time.

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  • What does the 'Burst Rate' stat mean in HDTune?

    - by UpTheCreek
    I recently upgraded my laptop's v slow hard drive to a seagate momentus 7200. Everything is working fine, but I'm a bit confused by these benchmark results: The burst rate is significantly less than the Maximim transfer rate, and not much higher than the normal minimum (if you ignore the spikes). What's going on here? On the HDtune website it defines Burst Rate as: ...the highest speed (in megabytes per second) at which data can be transferred from the drive interface (IDE or SCSI for example) to the operating system. Which begs some questions... e.g. if this is the highest, then how did the bechmarking tool record the 103MB/sec maximum? And if this really is the true maximum, then where is the bottleneck? The laptops SATA interface is on an Intel 82801GBM southbridge controller. When I check in hardware manager, I see that it's driver is iaStor.sys from 2005. Maybe that's the issue? I'll look for a newever version, but any insights would be appreciated. Thanks

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  • MacMini transmit rate stuck at 11, every other device can connect at full 54Mbit/s?

    - by chum of chance
    I have a MacMini circa 2007 that's getting very low transmit rates via wifi, 8-11. I have other devices that are getting full 54, including a MacBook Air. With everything else off, the MacMini doesn't want to seem to go any faster. Since it has been previously connected to ethernet its entire life, I was wondering if there were some settings I can change to speed up the connection. Option-clicking the network icon gives this read out: PHY Mode: 802.11g Channel: 1 (2.4 Ghz) Security: WPA2 Personal RSSI: -73 Transmit Rate: 11 My new MacBook Air has the following readout: PHY Mode: 802.11n Channel: 1 (2.4 Ghz) Security: WPA2 Personal RSSI: -66 Transmit Rate: 79 Both have full bars and the wireless router is in the same room to eliminate any obstructions from the equation. Could the MacMini be connecting at an older protocol, like 802.11b and be reporting erroneously that it is connected at 802.11g? This would explain why I haven't seen a transmit rate above 11. Any further trouble shooting I can try before buying a new USB 802.11n device? The WiFi router is a DLink DIR-615. I can see other devices, and none, even the other g connected devices, are getting below 30-40 MBit/s. What's going on here?

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  • Wireless is detected, but not connecting. Ethernet works. How to correct the wireless address?

    - by Lucas
    I am running Ubuntu 14.04 with cable internet, and my wireless is detected and connected, but I cannot connect to the internet. I know the problem is with my machine because other machines are connecting to the same router just fine. I can connect via ethernet just fine as well. Here are some notable tests: ping 192.168.0.105 works with 0% packet loss, but ping 192.168.0.1 has 100% packet loss. When I plug in my ethernet, ping 192.168.0.1 works with 0% packet loss. My wireless name is tg, and the router ip is 192.168.0.1 (where I can enter username and password). I suspect that I need to change my wireless address from 192.168.0.105 to 192.168.0.1. Any suggestions on how to proceed? extra info: [lucas@lucas-ThinkPad-W520]/home/lucas$ iwconfig eth0 no wireless extensions. lo no wireless extensions. wlan0 IEEE 802.11abgn ESSID:"tg" Mode:Managed Frequency:2.462 GHz Access Point: 00:02:6F:83:F8:F4 Bit Rate=1 Mb/s Tx-Power=15 dBm Retry long limit:7 RTS thr:off Fragment thr:off Power Management:off Link Quality=62/70 Signal level=-48 dBm Rx invalid nwid:0 Rx invalid crypt:0 Rx invalid frag:0 Tx excessive retries:52 Invalid misc:166 Missed beacon:0 [lucas@lucas-ThinkPad-W520]/home/lucas$ ifconfig eth0 Link encap:Ethernet HWaddr f0:de:f1:b2:53:53 inet addr:192.168.0.100 Bcast:192.168.0.255 Mask:255.255.255.0 inet6 addr: fe80::f2de:f1ff:feb2:5353/64 Scope:Link UP BROADCAST RUNNING MULTICAST MTU:1500 Metric:1 RX packets:980003 errors:0 dropped:0 overruns:0 frame:0 TX packets:498384 errors:0 dropped:0 overruns:0 carrier:0 collisions:0 txqueuelen:1000 RX bytes:1320506168 (1.3 GB) TX bytes:59780591 (59.7 MB) Interrupt:20 Memory:f3a00000-f3a20000 lo Link encap:Local Loopback inet addr:127.0.0.1 Mask:255.0.0.0 inet6 addr: ::1/128 Scope:Host UP LOOPBACK RUNNING MTU:65536 Metric:1 RX packets:21927 errors:0 dropped:0 overruns:0 frame:0 TX packets:21927 errors:0 dropped:0 overruns:0 carrier:0 collisions:0 txqueuelen:0 RX bytes:1781719 (1.7 MB) TX bytes:1781719 (1.7 MB) wlan0 Link encap:Ethernet HWaddr 24:77:03:29:8f:dc inet addr:192.168.0.105 Bcast:192.168.0.255 Mask:255.255.255.0 inet6 addr: fe80::2677:3ff:fe29:8fdc/64 Scope:Link UP BROADCAST RUNNING MULTICAST MTU:1500 Metric:1 RX packets:11828 errors:0 dropped:0 overruns:0 frame:0 TX packets:15444 errors:0 dropped:0 overruns:0 carrier:0 collisions:0 txqueuelen:1000 RX bytes:4855662 (4.8 MB) TX bytes:2250585 (2.2 MB) [lucas@lucas-ThinkPad-W520]/home/lucas$ lspci -nn | grep 0280 03:00.0 Network controller [0280]: Intel Corporation Centrino Ultimate-N 6300 [8086:4238] (rev 3e) [lucas@lucas-ThinkPad-W520]/home/lucas$ rfkill list 0: hci0: Bluetooth Soft blocked: no Hard blocked: no 1: tpacpi_bluetooth_sw: Bluetooth Soft blocked: no Hard blocked: no 2: phy0: Wireless LAN Soft blocked: no Hard blocked: no with ethernet unplugged: [lucas@lucas-ThinkPad-W520]/home/lucas$ route -n | grep UG 0.0.0.0 192.168.0.1 0.0.0.0 UG 0 0 0 wlan0 with ethernet plugged in: [lucas@lucas-ThinkPad-W520]/home/lucas$ route -n | grep UG 0.0.0.0 192.168.0.1 0.0.0.0 UG 0 0 0 eth0 [lucas@lucas-ThinkPad-W520]/home/lucas$ nm-tool NetworkManager Tool State: connected (global) - Device: wlan0 [tg] ---------------------------------------------------------- Type: 802.11 WiFi Driver: iwlwifi State: connected Default: no HW Address: 24:77:03:29:8F:DC Capabilities: Speed: 52 Mb/s Wireless Properties WEP Encryption: yes WPA Encryption: yes WPA2 Encryption: yes Wireless Access Points (* = current AP) tatum: Infra, 40:8B:07:D8:A5:04, Freq 2437 MHz, Rate 54 Mb/s, Strength 42 W PA WPA2 ums: Infra, 00:20:A6:72:52:BF, Freq 2437 MHz, Rate 54 Mb/s, Strength 59 Alpha 40: Infra, 28:CF:E9:86:59:5D, Freq 5260 MHz, Rate 54 Mb/s, Strength 30 W PA WPA2 thepromiselan: Infra, 58:6D:8F:51:E5:54, Freq 2452 MHz, Rate 54 Mb/s, Strength 34 $ PA WPA2 xfinitywifi: Infra, 06:1D:D5:84:27:A0, Freq 2437 MHz, Rate 54 Mb/s, Strength 52 *tg: Infra, 00:02:6F:83:F8:F4, Freq 2462 MHz, Rate 54 Mb/s, Strength 73 W PA2 ums: Infra, 00:20:A6:A1:9F:25, Freq 2452 MHz, Rate 54 Mb/s, Strength 44 BRIAN-PC_Network:Infra, 20:AA:4B:DD:93:D6, Freq 2462 MHz, Rate 54 Mb/s, Strength 35 W PA2 HOME-C0F8: Infra, 44:32:C8:D2:C0:F8, Freq 2412 MHz, Rate 54 Mb/s, Strength 40 W PA WPA2 abcsexy: Infra, 28:28:5D:27:5D:85, Freq 2412 MHz, Rate 54 Mb/s, Strength 27 W PA WPA2 IPv4 Settings: Address: 192.168.0.105 Prefix: 24 (255.255.255.0) Gateway: 192.168.0.1 DNS: 192.168.0.1 - Device: eth0 [Wired connection 1] ------------------------------------------- Type: Wired Driver: e1000e State: connected Default: yes HW Address: F0:DE:F1:B2:53:53 Capabilities: Carrier Detect: yes Speed: 100 Mb/s Wired Properties Carrier: on IPv4 Settings: Address: 192.168.0.100 Prefix: 24 (255.255.255.0) Gateway: 192.168.0.1 DNS: 192.168.0.1

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  • Form iteration function not working properly

    - by Moses
    My function is supposed to iterate all forms in the document, and bind an onclick function to each 'calculate' element int he form. The problem is, the function that executes on any of the click events executes in the context of the the last i in the loop. Here is the JavaScript that I'm using: window.onload = function(){ calculateSavings(); } function calculateSavings(){ for (i = 0; i < document.forms.length; i++) { var e = document.forms[i]; e.calculate.onclick = function() { var hours = e.hours.value; var rate = e.rate.value; alert(hours * rate); } } } And here is the HTML it is attached to: <!doctype html> <html> <body> <form> <label for="hours">Hours: </label><input type="text" id="hours" name="hours"> <label for="rate">Rate: </label><input type="text" id="rate" name="rate"> <input type="button" name="calculate" value="Calculate"> <div id="savings"></div> </form> <form> <label for="hours">Hours: </label><input type="text" id="hours" name="hours"> <label for="rate">Rate: </label><input type="text" id="rate" name="rate"> <input type="button" name="calculate" value="Calculate"> <div id="savings"></div> </form> </body> </html> I'm sure this is a really basic question but the solution is completely eluding me at this point.

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  • Why does my general frame rate slow down to 40fps randomly?

    - by Joshua
    This has been bugging me for a while. Every once in a while, I find my computer to be sort of laggy and I thought it was because it was busy or something. However, I recently noticed that it wasn't any performance issue...I thought my computer was laggy because the frame rate slowed from 75fps right down to ~40 fps and caused very visible tearing. This is not rare. It happens many, many times a day. I have no idea what is happening...I have an AMD 5670 on Windows 7 32-bit by the way, and I've heard bad things about AMD's driver support. Could this be the problem? P.S. The frame rate slowdown is not just for games (I rarely play games, and have not played games in the time since I noticed this problem), it seems it's an issue for the entirety of Windows. I first noticed the tearing when I was moving around tabs in Google Chrome.

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  • Breakout ball collision detection, bouncing against the walls [solved]

    - by Sri Harsha Chilakapati
    I'm currently trying to program a breakout game to distribute it as an example game for my own game engine. http://game-engine-for-java.googlecode.com/ But the problem here is that I can't get the bouncing condition working properly. Here's what I'm using. public void collision(GObject other){ if (other instanceof Bat || other instanceof Block){ bounce(); } else if (other instanceof Stone){ other.destroy(); bounce(); } //Breakout.HIT.play(); } And here's by bounce() method public void bounce(){ boolean left = false; boolean right = false; boolean up = false; boolean down = false; if (dx < 0) { left = true; } else if (dx > 0) { right = true; } if (dy < 0) { up = true; } else if (dy > 0) { down = true; } if (left && up) { dx = -dx; } if (left && down) { dy = -dy; } if (right && up) { dx = -dx; } if (right && down) { dy = -dy; } } The ball bounces the bat and blocks but when the block is on top of the ball, it won't bounce and moves upwards out of the game. What I'm missing? Is there anything to implement? Please help me.. Thanks EDIT: Have changed the bounce method. public void bounce(GObject other){ //System.out.println("y : " + getY() + " other.y + other.height - 2 : " + (other.getY() + other.getHeight() - 2)); if (getX()+getWidth()>other.getX()+2){ setHorizontalDirection(Direction.DIRECTION_RIGHT); } else if (getX()<(other.getX()+other.getWidth()-2)){ setHorizontalDirection(Direction.DIRECTION_LEFT); } if (getY()+getHeight()>other.getY()+2){ setVerticalDirection(Direction.DIRECTION_UP); } else if (getY()<(other.getY()+other.getHeight()-2)){ setVerticalDirection(Direction.DIRECTION_DOWN); } } EDIT: Solved now. See the changed method in my answer.

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  • Sharing Bandwidth and Prioritizing Realtime Traffic via HTB, Which Scenario Works Better?

    - by Mecki
    I would like to add some kind of traffic management to our Internet line. After reading a lot of documentation, I think HFSC is too complicated for me (I don't understand all the curves stuff, I'm afraid I will never get it right), CBQ is not recommend, and basically HTB is the way to go for most people. Our internal network has three "segments" and I'd like to share bandwidth more or less equally between those (at least in the beginning). Further I must prioritize traffic according to at least three kinds of traffic (realtime traffic, standard traffic, and bulk traffic). The bandwidth sharing is not as important as the fact that realtime traffic should always be treated as premium traffic whenever possible, but of course no other traffic class may starve either. The question is, what makes more sense and also guarantees better realtime throughput: Creating one class per segment, each having the same rate (priority doesn't matter for classes that are no leaves according to HTB developer) and each of these classes has three sub-classes (leaves) for the 3 priority levels (with different priorities and different rates). Having one class per priority level on top, each having a different rate (again priority won't matter) and each having 3 sub-classes, one per segment, whereas all 3 in the realtime class have highest prio, lowest prio in the bulk class, and so on. I'll try to make this more clear with the following ASCII art image: Case 1: root --+--> Segment A | +--> High Prio | +--> Normal Prio | +--> Low Prio | +--> Segment B | +--> High Prio | +--> Normal Prio | +--> Low Prio | +--> Segment C +--> High Prio +--> Normal Prio +--> Low Prio Case 2: root --+--> High Prio | +--> Segment A | +--> Segment B | +--> Segment C | +--> Normal Prio | +--> Segment A | +--> Segment B | +--> Segment C | +--> Low Prio +--> Segment A +--> Segment B +--> Segment C Case 1 Seems like the way most people would do it, but unless I don't read the HTB implementation details correctly, Case 2 may offer better prioritizing. The HTB manual says, that if a class has hit its rate, it may borrow from its parent and when borrowing, classes with higher priority always get bandwidth offered first. However, it also says that classes having bandwidth available on a lower tree-level are always preferred to those on a higher tree level, regardless of priority. Let's assume the following situation: Segment C is not sending any traffic. Segment A is only sending realtime traffic, as fast as it can (enough to saturate the link alone) and Segment B is only sending bulk traffic, as fast as it can (again, enough to saturate the full link alone). What will happen? Case 1: Segment A-High Prio and Segment B-Low Prio both have packets to send, since A-High Prio has the higher priority, it will always be scheduled first, till it hits its rate. Now it tries to borrow from Segment A, but since Segment A is on a higher level and Segment B-Low Prio has not yet hit its rate, this class is now served first, till it also hits the rate and wants to borrow from Segment B. Once both have hit their rates, both are on the same level again and now Segment A-High Prio is going to win again, until it hits the rate of Segment A. Now it tries to borrow from root (which has plenty of traffic spare, as Segment C is not using any of its guaranteed traffic), but again, it has to wait for Segment B-Low Prio to also reach the root level. Once that happens, priority is taken into account again and this time Segment A-High Prio will get all the bandwidth left over from Segment C. Case 2: High Prio-Segment A and Low Prio-Segment B both have packets to send, again High Prio-Segment A is going to win as it has the higher priority. Once it hits its rate, it tries to borrow from High Prio, which has bandwidth spare, but being on a higher level, it has to wait for Low Prio-Segment B again to also hit its rate. Once both have hit their rate and both have to borrow, High Prio-Segment A will win again until it hits the rate of the High Prio class. Once that happens, it tries to borrow from root, which has again plenty of bandwidth left (all bandwidth of Normal Prio is unused at the moment), but it has to wait again until Low Prio-Segment B hits the rate limit of the Low Prio class and also tries to borrow from root. Finally both classes try to borrow from root, priority is taken into account, and High Prio-Segment A gets all bandwidth root has left over. Both cases seem sub-optimal, as either way realtime traffic sometimes has to wait for bulk traffic, even though there is plenty of bandwidth left it could borrow. However, in case 2 it seems like the realtime traffic has to wait less than in case 1, since it only has to wait till the bulk traffic rate is hit, which is most likely less than the rate of a whole segment (and in case 1 that is the rate it has to wait for). Or am I totally wrong here? I thought about even simpler setups, using a priority qdisc. But priority queues have the big problem that they cause starvation if they are not somehow limited. Starvation is not acceptable. Of course one can put a TBF (Token Bucket Filter) into each priority class to limit the rate and thus avoid starvation, but when doing so, a single priority class cannot saturate the link on its own any longer, even if all other priority classes are empty, the TBF will prevent that from happening. And this is also sub-optimal, since why wouldn't a class get 100% of the line's bandwidth if no other class needs any of it at the moment? Any comments or ideas regarding this setup? It seems so hard to do using standard tc qdiscs. As a programmer it was such an easy task if I could simply write my own scheduler (which I'm not allowed to do).

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  • Does Mac OS X throttle the RATE of socket creation?

    - by pbhogan
    This may seem programming related, but this is an OS question. I'm writing a small high performance daemon that takes thousands of connections per second. It's working fine on Linux (specifically Ubuntu 9.10 on EC2). On Mac OS X if I throw a few thousand connections at it (roughly about 16350) in a benchmark that simply opens a connection, does it's thing and closes the connection, then the benchmark program hangs for several seconds waiting for a socket to become available before continuing (or timing out in the process). I used both Apache Bench as well as Siege (to make sure it wasn't the benchmark application). So why/how is Mac OS X limiting the RATE at which sockets can be used, and can I stop it from doing this? Or is there something else going on? I know there is a file descriptor limit, but I'm not hitting that. There is no error on accepting a socket, it's simply hangs for a while after the first (roughly) 16000, waiting -- I assume -- for the OS to release a socket. This shouldn't happen since all prior the sockets are closed at that point. They're supposed to come available at the rate they're closed, and do on Ubuntu, but there seems to be some kind of multi (5-10?) second delay on Mac OS X. I tried tweaking with ulimit every-which-way. Nada.

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