这是我的作业问题。我只想确认我的方法，从而确认答案。
具有2级分页方案的计算机系统，其中常规内存访问需要300纳秒并且服务于页面故障需要500ns。平均指令需要200 ns的CPU时间和一次内存访问。 TLB命中率为80％，页面错误率为20％。平均指令执行时间是？
我的方法=＆gt;
**Average time To Execute Instruction = CPU Time + Memory Access Time**
It is given that CPU Time = 200 ns
Probability of having a page fault for an instruction = 20% = 1/5
Hence, probability of not having a page fault = 4/5
If TLB hit occurs,
then memory Access time = 0 + 300 = 300 ns ( Here, TLB is taken negligible, so, 0 )
and
if TLB miss occurs,
then Memory Access Time = TLB access time + Access time Page Table 1 +
Access time Page table 2 + one memory Access = 0+ 300 + 300 + 300 = 900 ns
( Assume all the page tables in main memory )
Hit ratio of TLB = 80 %
Hence, Memory Access Time = prob. of no page fault (...Memory
access...........) + prob. of page fault (.......Page fault service
time..........)
Memory Access Time = 4/5 ( 0.80 * 300 + 0.20 * 900 )
+ 1/5 ( (0.80 * 300 + 0.20 * 900) + 500 ns for servicing a page fault )
= 336 + 184
= 520 ns
Average time To Execute Instruction = CPU Time + Memory Access Time
Average time To Execute Instruction = 200 ns + 520 ns = 720 ns .
如果我犯了任何错误，请纠正我。