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Assume we have a demand-paged memory. The page table is held in registers. It takes $8$ milliseconds to service a page fault if an empty page is available or the replaced page is not modified, and $20$ milliseconds if the replaced page is modified. Memory access time is $100$ ns. Assume that the page to be replaced is modified $70$ percent of the time. The maximum acceptable page-fault rate for an effective access time of no more than $200$ ns is __________ $\%$
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Answered 0.0006 and got wrong :(
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me too bro....

range should be there 0.0006-0.0007
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Let p =  the page fault rate ( the probability that a memory access results in a page fault )

Effective access time
= (1-p) memory access time  +  page fault rate  ×   ( probability of empty or unmodified page
× time to access empty or unmodified page   + probability of page modification
× time to access modified page )

We have been given that,

Effective access time = 0.20 microsecond
Memory access time = 0.1 microsecond
Probability of empty or un-modified page   = 0.3
Time to access empty or unmodified page = 8000 microseconds
probability of modified page = 0.7

Time to access modified page = 20000 microseconds

Substituting these values in the formula, we get,

0.2 = ( ( 1 - P) 0.1 + P( 0.3 (8000) + 0.7(20000) )  ) 

=> P = 6.1 * 10(-6)

         = 0.00061 %

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exact answer should be 0.000609
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@bikram sir what is the meaning of "PAGE TABLE IS IN REGISTER"??
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@ As P.T is in register, there is no memory access required for accessing the page table from memory.(Registers are supremely fast and require negligible access time)

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Answer:

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