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| Teaching Since: | May 2017 |
| Last Sign in: | 408 Weeks Ago |
| Questions Answered: | 66690 |
| Tutorials Posted: | 66688 |
MCS,PHD
Argosy University/ Phoniex University/
Nov-2005 - Oct-2011
Professor
Phoniex University
Oct-2001 - Nov-2016
1. A(P) at L, V(P) at M, A(P) at N, V(P) at O, P initially 1
2. A(P) at L, V(P) at M, A(Q) at N, V(Q) at O, P and Q initially 1
3. A(P) at L, V(Q) at M, A(Q) at N, V(P) at O, P and Q initially 1
4. V(P) at L, V(Q) at M, A(P) at N, A(Q) at O, P and Q initially 1
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For questions 7 and 8 refer to the data given below:
For translation from 32 bit virtual to 32 bit physical address the processor uses 2-level page tables and for both levels the page tables are stored in the main memory. The memory is byte addressable. For translation, 10 MSB’s of the virtual address are used as index into the first level page table and the next 10 bits are used as index into the second level page table. The 12 LSB’s of the virtual address are used as offset within the page. Assume that the translation look aside buffer of the processor has the hit rate of 96% and it caches recently used virtual page numbers and the corresponding physical page numbers. Also assume that the page table entries in both levels of page tables are 4 bytes wide. The physically addressed cache of the processor has a hit rate of 90%. Cache access time = TLB access time = 1ns and main memory access time = 10ns
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