Consider a demand -aging system with a hard drive that has an average access and

Question

Consider a demand -aging system with a hard drive that has an average access and tranfer time of 10 milliseconds. Addresses are translated through a page table in main memory, with an access time of 100 nanoseconds per memory access. By using cache/associative memoruy, 80% of memory access are completed within one main memory access. Of those remaining, through well designed paging replacement algorithms, 10% (or 2% of the total) causes page faults. What is the effective memory access time. Consider a demand -aging system with a hard drive that has an average access and tranfer time of 10 milliseconds. Addresses are translated through a page table in main memory, with an access time of 100 nanoseconds per memory access. By using cache/associative memoruy, 80% of memory access are completed within one main memory access. Of those remaining, through well designed paging replacement algorithms, 10% (or 2% of the total) causes page faults. What is the effective memory access time. By using cache/associative memoruy, 80% of memory access are completed within one main memory access. Of those remaining, through well designed paging replacement algorithms, 10% (or 2% of the total) causes page faults. What is the effective memory access time.

Explanation / Answer

Solution:

So main memory access time= 10 milliseconds

and

if the entry is found in page table then access time= 100 nanoseconds

If memory access is completed in one memory access, this means in the page table the entry was found,

So our Effective memory access time will be (EMAT)= (0.8*100*10-9)+ (0.2*(0.02*10*10-3+0.98*10-9))= 0.00004008019 Seconds= 40.08019 micro seconds

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