[問題] OS-memory計算

[nana0130 (小那)]

Consider a system with following properties:

●A 64K byte logical address space.
●A physical memory of 16K bytes
●A simple (one-level)page table address translation implementation using
  page size of 4096 byte

●The page tables are stored in physical memory.
●The access time of physical memory is 100μs.

(a) How many bits wide is a logical address ?

    我算是 14bits
    2^16 byte / 2^12 byte = 4=2^2
    so p+d = 2+12 = 14
    這樣對嗎 ?

(b)What is the min number of bits a pointer in a C program compiled for
   this machine would need to occupy?
   這題怎麼算啊 ？？

(c)Assuming it also needs to hold a vaild bit, a reference bit
   and a modified bit , what is the min number of bits wide a
   page table entry need to be?
   我算出來是 5bit耶，好奇怪喔！很懷疑我有沒有算錯？！
   physical page number 我算是2^14 / 2^12 = 2^2
   所以 3+2 = 5 !!
   如果我有算錯，請您指正！謝謝你。



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◆ From: 220.139.157.70
推 s987692:a.16 b.14 c.17
推 joe1986112:A.16 B.16 C.9(含Page Number)
推 joe1986112:a.64K=2^16=>16bit
→ nana0130:喔，對，我真是熬夜熬昏了>"<
推 ieaan:c也是算5
推 s987692:C是五沒錯,我也昏頭了...
推 decimal:那B呢？（我的C也是5）
→ nana0130:thank you all!