[理工] [計組]-台大97-資工
http://www.lib.ntu.edu.tw/exam/graduate/97/97419.pdf
第二題的(d)小題
張凡的解答寫
(100-x)/(0.5*8) = x
請問這個式子要怎麼解釋??
爬文之後發現問題似乎是出在這一句敘述
for an 8-processor run, 50% of time a processor has to stall
because the processor is waiting to access the disk
我自己的解釋是這樣
(run_CPU_time + I/O_time) = 100
I/O_time = x
total_CPU_time = run_CPU_time_8pro + stall_CPU_time
stall_CPU_time = 50%*total_CPU_time = run_CPU_time_8pro
total_CPU_time = run_CPU_time_8pro/(0.5)
run_CPU_time = 100-x
run_CPU_time_8pro = (100-x)/8
total_CPU_time = (100-x)/(0.5*8)
如果是這樣
那題目中有哪一句提到total_CPU_time = I/O_time 呢?
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