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IIS 6.0请求减速(IIS 6.0 requests slowdown)

我在IIS 6.0上有一个webapplicaton。 它不断处理大量的短时请求(15-30 ms处理时间)。 当有一些(1-10)长时间请求时,所有短时间请求都会减慢(最多2000-6000 ms的处理时间,其中一些时间超过100000)。

IIS中的请求之间是否应该隔离? 不应该假设一个请求不应该中断另一个请求吗?

在IIS日志中,它看起来像:

[Normal work]
cs-host sc-status sc-substatus sc-win32-status sc-bytes cs-bytes time-taken
192.168.1.7 200 0 0 2394 524 734
192.168.1.7 200 0 0 2394 524 0   
192.168.1.7 200 0 0 2394 524 0
192.168.1.7 200 0 0 2394 524 15
192.168.1.7 200 0 0 2394 524 15
192.168.1.7 200 0 0 2394 524 0
192.168.1.7 200 0 0 2394 524 0
192.168.1.7 200 0 0 2394 524 15
192.168.1.7 200 0 0 2394 524 46

[Slowdown]
cs-host sc-status sc-substatus sc-win32-status sc-bytes cs-bytes time-taken
192.168.1.7 200 0 64 0 522 508251
192.168.1.7 200 0 64 0 522 91827
192.168.1.7 200 0 64 0 522 386438
192.168.1.7 200 0 64 0 522 445947
192.168.1.7 200 0 0 178 522 35545
192.168.1.7 200 0 64 0 522 274130

sc-win32-status 64表示“指定的网络不再可用”,但没有断开连接。 我尝试使用像IISTuner( http://iistuner.codeplex.com/ )这样的工具调整IIS,但它没有任何效果。

  • 为什么会出现这种情况
  • 如何排除故障?

I have a webapplicaton on IIS 6.0. It constantly processes huge amount of short-time requests (15-30 ms process time). When there comes some (1-10) long-time requests all short-time requests slow down (up to 2000-6000 ms process time and more than 100000 for some of them).

Should there be like an isolation between requests in IIS? It isn't supposed that one requests should not interrupt another?

In IIS logs it is look like:

[Normal work]
cs-host sc-status sc-substatus sc-win32-status sc-bytes cs-bytes time-taken
192.168.1.7 200 0 0 2394 524 734
192.168.1.7 200 0 0 2394 524 0   
192.168.1.7 200 0 0 2394 524 0
192.168.1.7 200 0 0 2394 524 15
192.168.1.7 200 0 0 2394 524 15
192.168.1.7 200 0 0 2394 524 0
192.168.1.7 200 0 0 2394 524 0
192.168.1.7 200 0 0 2394 524 15
192.168.1.7 200 0 0 2394 524 46

[Slowdown]
cs-host sc-status sc-substatus sc-win32-status sc-bytes cs-bytes time-taken
192.168.1.7 200 0 64 0 522 508251
192.168.1.7 200 0 64 0 522 91827
192.168.1.7 200 0 64 0 522 386438
192.168.1.7 200 0 64 0 522 445947
192.168.1.7 200 0 0 178 522 35545
192.168.1.7 200 0 64 0 522 274130

sc-win32-status 64 means "The specified network is no longer available" but there was no disconnections. I tried to tune IIS up with tools like IISTuner (http://iistuner.codeplex.com/) it causes no effect.

  • Why such situation happens?
  • How to troubleshoot that?

原文:https://stackoverflow.com/questions/18442022
更新时间:2022-06-10 06:06

最满意答案

为什么你需要停下等待的未来,而不是反复轮询? 答案很明显,恕我直言。 因为在一天结束时它更快更有效!

为了反复轮询未来(可能被称为“ 忙碌等待 ”),图书馆必须决定是经常做还是很少做,而且回答都不令人满意。 经常这样做,你浪费CPU周期,很少做,代码反应迟钝。

所以是的,你需要在等待某事时停放任务,然后在等待时取消停放任务。 喜欢这个:

#![allow(deprecated)]

extern crate futures;

use std::sync::{Arc, Mutex};
use futures::*;
use futures::task::{park, Task};

struct Status {
    ready: bool,
    task: Option<Task>,
}

#[allow(dead_code)]
struct F(Arc<Mutex<Status>>);

impl Future for F {
    type Item = ();
    type Error = ();

    fn poll(&mut self) -> Result<Async<Self::Item>, Self::Error> {
        println!("Check if flag is set");
        let mut status = self.0.lock().expect("!lock");
        if status.ready {
            Ok(Async::Ready(()))
        } else {
            status.task = Some(park());
            Ok(Async::NotReady)
        }
    }
}

#[test]
fn test() {
    let flag = Arc::new(Mutex::new(Status {
                                       ready: false,
                                       task: None,
                                   }));
    let future = F(flag.clone());
    ::std::thread::spawn(move || {
        ::std::thread::sleep_ms(10);
        println!("set flag");
        let mut status = flag.lock().expect("!lock");
        status.ready = true;
        if let Some(ref task) = status.task {
            task.unpark()
        }
    });
    let result = future.wait();
    println!("result: {:?}", result);
}

请注意, Future::poll在这里做了几件事:它正在检查一个外部条件并且它正在停止任务,因此可以进行比赛 ,例如:

  1. poll检查变量并发现它是false ;
  2. 外部代码将变量设置为true ;
  3. 外部代码检查任务是否停放并发现它不是;
  4. poll了任务,但繁荣! 现在为时已晚,没有人会再停车了。

为了避免任何比赛,我使用Mutex来同步这些交互。

PS如果您只需要将一个线程结果包装到Future那么请考虑使用oneshot通道:它具有已实现Future接口的Receiver


Why would you need to park a waiting future instead of polling it repeatedly? The answer is rather obvious, IMHO. Because at the end of the day it's faster and more efficient!

To repeatedly poll a future (which might be dubbed "busy-waiting") the library would have to decide whether to do it often or seldom and neither answer is satisfactory. Do it often and you're wasting the CPU cycles, do it seldom and the code is slow to react.

So yeah, you need to park the task when you're waiting for something and then unpark it when you've done waiting. Like this:

#![allow(deprecated)]

extern crate futures;

use std::sync::{Arc, Mutex};
use futures::*;
use futures::task::{park, Task};

struct Status {
    ready: bool,
    task: Option<Task>,
}

#[allow(dead_code)]
struct F(Arc<Mutex<Status>>);

impl Future for F {
    type Item = ();
    type Error = ();

    fn poll(&mut self) -> Result<Async<Self::Item>, Self::Error> {
        println!("Check if flag is set");
        let mut status = self.0.lock().expect("!lock");
        if status.ready {
            Ok(Async::Ready(()))
        } else {
            status.task = Some(park());
            Ok(Async::NotReady)
        }
    }
}

#[test]
fn test() {
    let flag = Arc::new(Mutex::new(Status {
                                       ready: false,
                                       task: None,
                                   }));
    let future = F(flag.clone());
    ::std::thread::spawn(move || {
        ::std::thread::sleep_ms(10);
        println!("set flag");
        let mut status = flag.lock().expect("!lock");
        status.ready = true;
        if let Some(ref task) = status.task {
            task.unpark()
        }
    });
    let result = future.wait();
    println!("result: {:?}", result);
}

Note that Future::poll is doing several things here: it's checking for an external condition and it's parking the task, so it's possible to have a race, like when:

  1. the poll checks the variable and finds it to be false;
  2. the outer code sets the variable to true;
  3. the outer code checks if the task is parked and finds that it's not;
  4. the poll parks the task, but boom! it is too late, nobody is going to unpark it any longer.

In order to avoid any races, I've used a Mutex to synchronize these interactions.

P.S. If all you need is to wrap a thread result into a Future then consider using the oneshot channel: it has the Receiver that implements the Future interface already.

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