Question

Consider a system with a two-level paging scheme in which a regular memory access takes 150 nanoseconds, and servicing a page fault takes 8 milliseconds. An average instruction takes 100 nanoseconds of cpu time, and two memory accesses. The tlb hit ratio is 90%, and the page fault rate is one in every 10,000 instructions. What is the effective average instruction execution time?

Answer 1

The effective average instruction execution time is 1260 nanoseconds.

Explanation:

• The formula for Average Instruction execution time is

• Average Instruction execution time = Average CPU execution time + Average time for getting data

Average CPU execution time + Average address translation time for each instruction + Average memory fetch time for each instruction + Average page fault time for each instruction

= 100 + 2×(0.9×(0)+0.1×(2×150)) + 2×150 + 1 /10000 × 8 × 106

= 100 + 60 + 300 + 800

= 1260 nanoseconds.

Hence the effective average instruction execution time is 1260 nanoseconds.

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Assume that for a certain processor, a read request takes 50 nanoseconds on a cache miss and 5  nanoseconds on a cache hit. Suppose while running a program, it was observed that 80% of the  processor's read requests result in a cache hit. The average read access time in nanoseconds is?

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