Applied Practice test

Question

Suppose: TLB lookup time = 20 ns

TLB hit ratio = 80%

Memory access time = 75 ns

PFST = 500,000 ns

50% of the pages are dirty

OS uses a single level page table

What is the approximated effective access time (EAT) if we assume the page fault rate is 10%? Assume the cost to update the TLB, the page table, and the frame table (if needed) is negligible.

Comments

@darshak_devani

But I am not being able to analyze as to how they solved this question...

EAT

= 0.8 (TLB + MEM) + 0.2 (0.9(TLB + MEM + MEM) + 0.1 (TLB + MEM + 0.5 (DISK)+ 0.5 (2 DISK) + MEM)) ...

EAT

= 0.8 (20 + 75) + 0.2 (0.9 (20 + 75 + 75) + 0.1 (20 + 75 + 0.5 (500000) + 0.5 (1000000) +75)) EAT

= 76 + 0.2 (153 + .1 (750170))

= 76 + 15034

= 15110 ns..

 

1. https://gateoverflow.in/238144/test-series 

 

2. https://stackoverflow.com/questions/14192649/demand-paging-calculating-effective-memory-access-time 

 

3. https://csedoubts.gateoverflow.in/28686/gate-applied-course

 

4. https://gateoverflow.in/288644/effective-memory-access-time

 

 

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ans 15110 nano sec

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Yes, I too didn’t get how they are solving the question.

Answer 1

we have a simple formulae for this but before that few thing to say:-

dirty means that the page is modified
tlb hit=tlb access time+main memory access time
tlb miss=tlb access time+k(k is number of levels of page table)*main memory access time+(1-p)*main memory access time{if no page fault}+p(PFST).
if PFST is not given then d(m+2D)+(1-d)(m+D) d is probability of dirty pages and D is Secondary memory access time
emat=tlb hit ratio(2)+tlb miss ratio(3)

calculate and find answer