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In a paging system, it takes $30$ ns to search translation Look-a-side Buffer (TLB) and $90$ ns to access the main memory. If the TLB hit ratio is $70\%$, the effective memory access time is :

  1. $48$ ns
  2. $147$ns
  3. $120$ns
  4. $84$ns
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let c be the time required to check whether corresponding frame no  to the page is   in TLB .

If a page is present in TLB(hit) then Effective Memory Access time(EMAT)= c+m ( c to visit TLB and m to visit the memory using the frame no(that we got using TLB) and offset)

If a page is not present in TLB then EMAT=c+2m  (  here additional m because if the page is not in TLB then you have to visit page table to get  frame no so additional m )

Assume hit ratio is x ,

so EMAT=    x(c+m) + (1-x)(c+2m)

given x=70% and c=30ns and m=90ns

so, EMAT= (0.7)(30+90) + (0.3)(30+2*90)

EMAT=147ns
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2 votes
2 votes
70% tlb hit ratio 30 ns to search tlb

(0.7*30) in hit case: to find the page

0.3*(90+30) in miss case: to find the page

90 ns to access the data from main memory

so total time:

(0.7*30)+ (0.3*(90+30)) + 90 = 147

 

ans: option 2
2 votes
2 votes
Effective access time = Hit ratio*Time during hit + Miss Ratio * Time During Miss
=0.7*(30+90) + 0.3 (30+90+90) 
=84 + 63
=147 ns

OPTION B

Answer:

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