Take the 2-minute tour ×
Super User is a question and answer site for computer enthusiasts and power users. It's 100% free, no registration required.

I'm running a windows 2003 guest in Linux xiamx-pc 3.2.0-24-generic #39-Ubuntu SMP Mon May 21 16:52:17 UTC 2012 x86_64 x86_64 x86_64 GNU/Linux

I notice that virtualbox process constantly takes around 50% of cpu time but the process manager in 2003 shows only 5% of cpu usage. What could be the cause? Is there anyway that I can lower down the cpu usage of virtualbox process?

share|improve this question
    
check out this answer, seems related: superuser.com/a/156655/65379 –  Baarn Jun 19 '12 at 22:25
    
That's a good tip. Keep in mind it will limit Windows to a single virtual core though. Another good link for improving Windows VM performance: blog.jdpfu.com/2010/06/22/virtualbox-performance-improved The network tip has helped me a lot for a network-heavy VM, and the SATA tip is a VERY noticeable improvement. –  Vickash Jun 21 '12 at 3:56
add comment

1 Answer

While the SATA and NIC statements in the linked article above are still true, the best performance comes from:

  • fully allocating storage (don't use sparse files unless you are on SSDs)
  • properly allocating RAM and CPU (too little for the host or the guests is bad)
  • using VirtIO drivers for storage and networking (these drivers are available for both Linux and Windows)

There are a few other settings like disabling any video acceleration, 2D and 3D. I've written another article after seeing a Core i7 brought to 30 minute Ubuntu desktop logins with default VirtualBox settings.

If you are doing server virtualization, not desktop OSes, please do not use VirtualBox. Use KVM, LXC or ESXi instead. If you can get Spice working, even remote desktop performance over the LAN and WAN can be impressive with KVM as the host.

VirtualBox is best for desktop virtualization. Only VMware Workstation is better, IMHO.

share|improve this answer
add comment

Your Answer

 
discard

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.