in order to perform a disk or network access, it is typically necessary for
the user to have the OS communicate with the disk or network controllers.
suppose that in a particular 5 GHz computer, it takes 10,000 cycles to trap to
the OS, 20 ms for the OS to perform a disk access, and 25us for the OS to
perform a network access. in a disk access, what percentage of the delay time
is spent in trapping to the OS? how about in a network access?
=========================
disk access total time: 10,000/500,000,000 s + 20ms = 20.002 ms
% delay trapping to OS: 0.01%
network access total time: 10,000/5,000,000,000s + 25us = 27us
% delay trapping to OS: 7.4%
請問黃字是怎麼算出來的?
張凡337頁
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