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7 votes
7 votes

Consider a 32-bit machine where four-level paging scheme is used. If the hit ratio to TLB is 98%, and it takes 20 nanosecond to search the TLB and 100 nanoseconds to access the main memory what is effective memory access time in nanoseconds?

  1. 126
  2. 128
  3. 122
  4. 120
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4 Answers

Best answer
13 votes
13 votes

Let  $C$ is  the $TLB$ access time  and $M$ is main memory access time.

So Effective memory access time ($EMAT$) in multilevel cache,

$EMAT\ =h(C+M)+(1-h)(C+(n+1)M)$

 Here no of level $n$=4 So,

$EMAT\ =h(C+M)+(1-h)(C+5M)$

Given here $h=.98$ , Main memory access time ($M$)=100 ns and

$TLB$ access time ($C$)=20ns.

$EMAT\ =.98(20+100)+.02(20+5\times 100)$

             $=.98(120)+.02(520)$

             $=128ns$

Reference:Multilevel paging

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9 votes
9 votes

Suppose page tables be in main memory .

Effective access time = 0.98 * (TLB + MM) + 0.02 * (TLB + 4 level page tables + MM)

Effective access time = 0.98 * 120 + 0.02 * 520 = 128ns

5 votes
5 votes

Total number of level used in paging = 4. So if TLB miss occurs then to get a page from main memory we need to go through 4 memory access to get the exact location of the page in main memory.

TLB hit ratio = 98%.

So effective memory access = time to get the address of page + memory access time.

                                                    =  ((0.98 )* 20 + (0.02)*(20+4*100 ) + 100ns.

                                                    = 19.6 + 8.4 + 100 ns

                                                    = 128ns   (Ans.)

3 votes
3 votes
effective memory access time =  access TLB ( if page number is present in TLB, get its frame number goto main memory directly and access page + if page is not present in TLB, access 4level page tables and then access main memory )

emat = 20 ns  + ( 0.98 * 100 + 0.02 * 500 ) = 128ns

option B
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