Consider a demand-paging system in which the replacement of a page takes 20 mill

Question

Consider a demand-paging system in which the replacement of a page takes 20 milliseconds (access time and data transfer) Addresses are translated through a page table in main memory, with an access time of 1 microsecond per memory access. Thus, each memory reference through the page table takes two accesses. Assuming that 10% of memory access result in a page fault, compute the effective memory access time? To improve this time, a TLB is added to the system to reduces access time to one memory reference, if the page-table entry is in the associative memory. Assume that 80 percent of the accesses are in the associative memory and that, of the remaining, 20 percent (or 4 percent of the total) cause page faults. What is the effective memory access time?

Explanation / Answer

CPU -> Memory -> Disk

Demand Paging time = 20ms or 20000 us

memory access time = 1us

1) page fault = 10%

memory hit ratio = 100 - 10 = 90%

Effective Access time = ((hit ratio) * memory access time ) + (page fault ratio) * (memory access time + (2 * Demand Paging time)

Effective Access time = 0.9 * 1 + 0.1 * ( 1 + 40000)

EAT = 0.9 + 4000.1

EAT = 4001 us (micro seconds)

b) EAT = (hit ratio)*(memory access time) + (page falut ratio) * ( memory access time + ((tlb hit) * ( demand paging) + (tlb miss) *( 2* demand paging))

EAT = 0.9 + (0.1 *(1 + (0.8 * 20000 + 0.2* 40000)))

EAT = 0.9 + (0.1 *( 1 + (16000 + 8000)))

EAT = 0.9 + (0.1 * ( 24001))

EAT = 0.9 + 2400.1

EAT = 2401 us (micro sec)

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