memory management

Question

      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 500 nsec.  To reduce this  overhead, the computer has an
      associative  memory  which holds 32 (virtual  page,  physical  page frame)
      pairs and can do look up in 100 nsec.  What hit rate is needed  to reduce
      the mean overhead to 200 nsec?

Answer 1

Here the overhead is reading a word from the page table and we need to reduce the overhead.

So , when access is sequential memory access then 

(new mean ovehead time) = TLB hit rate (TLB access time) + TLB miss rate (TLB access time +Page table access time )

200 = h(100)+(1-h)(100+500)

h=0.8

 when access is parallel memory access then 

 (new mean ovehead time) = TLB hit rate (TLB access time) + TLB miss rate (Page table access time )

200 = h(100)+(1-h)(500)

h=0.75

Answer 2

h=75%

Look how request processed, request to access more is generated by processor. So processor generates Virtual address and it is given to TLB to give Physical address where actually data is in physical memory and parallely given to memory to search page table. Since tlb is fast enough to reply faster than memory so you get to know whether page hit  in tlb, if miss memory can reply.

access time= h(tlb access time)+(1-h)(memory access time) // as search is parallel

Comments

same doubt

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in which line of question it is mention that there is only V.A to P.A translation???can you explain??

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still facing the problem