# Windows – Run a file with HIGH priority

batch filecommand linewindows 10

I understand how to run a process or program with HIGH priority, but I want to run an actual FILE (which of course is opened with a program) in high priority using command line in a batch file.

For example, if I wanted to open a program (.exe file) in high priority, I would do:

start "" /HIGH c:\windows\system32\msiexec.exe


I can easily do this in a batch file.
But what if I wanted to open a particular file, like:

c:\user\username\desktop\file.msi


I've tried:

start "" /HIGH c:\user\username\desktop\file.msi


but it doesn't open in high priority. Only when calling a program, and not a specific file that is opened with a program does it run with high priority.

How can I run file.msi in high priority from my batch file?

• For any file that requires a handler or runtime to run, you run the handler or runtime application as /high and then use whichever method is appropriate for that application to call the file you wish it to run.
start /high "msiexec /i installer.msi"