Re: [理工] [ OS ]-中山95
※ 引述《lineageorc (who I am)》之銘言:
: ※ 引述《KarmaPolice (Karma Police)》之銘言:
: : 4. Assume we have a demand-paged memory. The page table is held in registers.
: : It takes 8 milliseconds to service a page fault if an empty page is available
: : or the replaced page is not modified, and 20 milliseconds if the replaced page
: : is modified. Memory access time is 100 nanoseconds. Assume that the page to be
: : replaced is modified 70 percent of the time. What is the maximum acceptable
: : page-fault rate for an effective access time of no more than 200 nanoseconds?
: : 這題想請教一下該如何解?
: : 我解出來的答案 讓我覺得有點離譜 想問問大家都是算多少?
: EAT=(1-p)*100ns + p * (0.3*8ms+0.7*20ms)
: =0.1ms-0.1ms*p+p*16.4ms
: =0.1ms+(16.3ms)*p
: and EAT<200ns
: 故
: 16.3ms*p<200ns-100ns=0.1ms
: => p<0.1ms/16.3ms =0.6%
EAT=(1-p)*100ns + p * (0.3*8ms+0.7*20ms)
1.
請問上色部份 為何要(1-p) ???
我有上洪兔的課 他也是教要1-p
當初我沒有多想 也認為是理所當然
但是現在想想跟張凡教的計組不同
我覺得計組的說法好像比較正確耶!
100ns + p * (0.3*8ms+0.7*20ms)
因為miss之後 還是會存取memory吧?
而後面加的是miss penalty
2.
p * 0.3 * 8ms 這個0.8是什麼?
我英文沒有很好 0.3*8ms 這個部份我不太了解是哪部份耗掉的時間
題目看不懂@@"
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