OPERATING SYSTEM

Question

Comments

Please see if the following procedure is appropriate::

step 1:- converting logical address to physical address.

0.99(tlb) + 0.01(tlb+memory)

step 2: fetching data from retrieved address.In this step there is a chance of Page fault.

(memory access)+5*10^-4(page falut service)

Emat=0.99(20)+0.01(20+100) + 100 + 5*10^-4(5000000) ns

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Plz change your comment to answer..

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@digvijay the page fault rate is 0.05%

and

Effective Access Time (EAT)

  EAT = (1 – p) x memory access

  + p (page fault overhead

             + swap page out

             + swap page in )------------ i think this should be the formula

 

and if all are dirty then u need to wirte the frame to the secondary storage also...but access time of secondary storage is not given but swap sapce is alloted in the secondary storage i think we should take the access time of swap space

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I think u are right. Since page is dirty. We have to save back the result in secondary memory

Answer 1

Case 1) Page fault doesnt occur

Effective time = TLB hit rate * ( TLB access time + Memory access time ) + (1 - TLB hit rate) * ( TLB access time + Page table access time + memory access time )

Effective time = 0.99 ( 20+100) + 0.01*(20+100+100) ---------------( I assumed lvl 1 page table)

Effective time = 121 ns

Case 2) Page fault occurs

Effective time = 121 + 5000000 + 5000000 = 10000121 ns -------------( since the page is dirty we have to write back the result in secondary storage )

So EMAT = 0.0005 * 10000121 + 0.9995 * 121 = 5121 ns

Comments

In question page table not given

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that's why I assumed lvl 1

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"In or out"  time is given not " in and out"

0.05% not 0.05

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thanks for the correction

Answer 2

step 1:- converting logical address to physical address.

0.99(tlb) + 0.01(tlb+memory)

step 2: fetching data from retrieved address.In this step there is a chance of Page fault.

(memory access)+5*10^-4(page falut service)

Emat=0.99(20)+0.01(20+100) + 100 + 5*10^-4(5000000) ns