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A computer whose processes have $1024$ pages in their address spaces keeps its page tables in memory. The overhead required for reading a word from the page table is $5\: nsec.$ To reduce this overhead, the computer has a $TLB,$ which holds $32$ (virtual page, physical page frame) pairs, and can do a lookup in $1\: nsec.$ What hit rate is needed to reduce the mean overhead to $2\: nsec?$
in Operating System by Veteran (58.8k points) | 32 views

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Answer:

The Effective instruction time $=1 \mathrm{ \times h + (1+5) \times(1-\mathrm h)},$ where $\mathbf h$ is the $\textbf{hit rate}$.

$\therefore 1\times \mathrm h + 6  -6\times \mathrm h = 2 \\ \Rightarrow \mathrm h = 0.8$

$\therefore$ The hit rate needed is $\mathbf{0.8}$

 
by Boss (18.9k points)
edited by
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effective instruction time = 1*h + (1+5)*(1-h)

please correct me if i was wrong
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Why $1+5$?
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when we have an tlb miss we have to look up the TLB so it takes 1 nsec to do that
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What if you are simultaneously looking at both the places?
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If we look at TLB then we can know that whther it was a miss or hit. If it was a miss then we will read from the page table
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Yeah right!

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