ace test series

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 tlb which holds 32  entries and can do look up in 100 nsec.  What hit rate is needed  to reduce the mean overhead to 200 nsec?

 

can anyone solve this

Comments

200=h(tlb+m)+(1-h)(tlb+m.a+m.a)

why is wrong ???

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I think the search here is parallel here.

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@vijju532,

Please add question source in title.

Answer 1

i’m agree with diksha as question is not talking about page access time it is only about overhead in finding page address in memory 

so Average overhead time = TLB hit( TLB acess time) + TLB Miss(TLB Access time + page table access time)

let TLB hit ratio h then

200 = h x 100 + (1-h)(100+500)

h = 0.8