Super User is a question and answer site for computer enthusiasts and power users. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm using chrome/firefox on windows 7, on fiber, at the office. When browsing the web and opening multiple concurrent pages some of them will 404, timeout, or hang. This doesn't seem to happen to anyone else on my switch. I asked IT and they thought it may be a tcpip-conncurrent-connection issue but that isn't supposed to occur on Windows 7.

Any idea what would cause this to occur?

share|improve this question
Technically, it is concurrent- incomplete -connection. Basically, you can only make 10 incomplete connections at one time. If they haven't connected yet, then others are put into a queue. But the number of complete or connected connections is a different number. . . – surfasb Jul 29 '11 at 16:15
That's crazy considering the number of http requests a single page can make. Especially one with advertisements. – Zach Shallbetter Jul 29 '11 at 17:08
I don't think you understand what an incomplete connection is or how TCPIP works. An http request != TCP connection. Theoretically, you can make as many requests over a TCP connection as you want. . . – surfasb Jul 29 '11 at 23:37
So how do I increase the number of incomplete-connections limit. – Zach Shallbetter Aug 1 '11 at 20:38… The limit was removed in Windows 7. – surfasb Aug 1 '11 at 20:53

multiple? 1000, 100, 10, 2?

windows used to have a low limit for concurrent connections. If you do a search in google you'll find many tips about this issue; for example:

share|improve this answer
Excellent have you seen any similar tweaks to Chrome? – Zach Shallbetter Jul 29 '11 at 15:47
Multiple meaning 2-5 – Zach Shallbetter Jul 29 '11 at 17:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.