题目:
Consider a demand-paging system with a paging disk that has an average access
and transfer time of 10 milliseconds ( you can ignore extra time to swap a
page out to make a free frame). Addresses are translated through a page table
in real memory, with an access time of 1 micrsecond per memory access. Thus,
each memory reference through the page table take two accesses. To improve
this time, we have added a TLB that reduces access time to one memory referenc
if the page-table entry is in the TLB. The TLB lookup time is 10 nanoseconds.
Assume that 99 percent of the accesses are in the TLB and that, of those
remaining, 20 percent (or 0.2 percent of th total) cause page faults.
What is the effiective memory access time in microseconds?
我的答案是：
我算出了0.99*1us + 0.01*(1us+(1-0.2)*(1us)+(0.1)*(10ms)
= 0.99 + 0.01*(1+0.8*1+1000)
= 0.99 + 10.018 = 11.008us
不太确定对不对，有人可以帮忙看一下吗Orz