Consider a system with a two-level paging scheme in which a regular memory…
2004
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: D. 1260 nanoseconds — Step 1 — average memory access time including TLB misses: On a TLB hit (90%) a memory access costs 150 ns. On a TLB miss (10%) a two-level page table requires…
- A.
645 nanoseconds
- B.
1050 nanoseconds
- C.
1215 nanoseconds
- D.
1260 nanoseconds
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Correct answer: D
Step 1 — average memory access time including TLB misses: On a TLB hit (90%) a memory access costs 150 ns. On a TLB miss (10%) a two-level page table requires two extra memory reads plus the actual access (3 accesses total) = 3 * 150 = 450 ns. Therefore average memory access = 0.9 * 150 + 0.1 * 450 = 180 ns.
Two memory accesses per instruction: 2 * 180 ns = 360 ns.
Add CPU time per instruction: 100 ns. Subtotal without page faults = 360 + 100 = 460 ns.
Page-fault overhead: 8 ms = 8,000,000 ns. Page-fault rate = 1/10,000 = 0.0001 per instruction. Expected page-fault cost per instruction = 0.0001 * 8,000,000 ns = 800 ns.
Total effective instruction time = 460 ns + 800 ns = 1260 ns.
Answer: 1260 nanoseconds.