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GATE | GATE-CS-2003 | Question 78
  • Difficulty Level : Hard
  • Last Updated : 24 Nov, 2014

A processor uses 2-level page tables for virtual to physical address translation. Page tables for both levels are stored in the main memory. Virtual and physical addresses are both 32 bits wide. The memory is byte addressable. For virtual to physical address translation, the 10 most significant bits of the virtual address are used as index into the first level page table while the next 10 bits are used as index into the second level page table. The 12 least significant bits of the virtual address are used as offset within the page. Assume that the page table entries in both levels of page tables are 4 bytes wide. Further, the processor has a translation look-aside buffer (TLB), with a hit rate of 96%. The TLB caches recently used virtual page numbers and the corresponding physical page numbers. The processor also has a physically addressed cache with a hit rate of 90%. Main memory access time is 10 ns, cache access time is 1 ns, and TLB access time is also 1 ns.

Assuming that no page faults occur, the average time taken to access a virtual address is approximately (to the nearest 0.5 ns)
(A) 1.5 ns
(B) 2 ns
(C) 3 ns
(D) 4 ns

Answer: (D)


The possibilities are
 TLB Hit*Cache Hit +
 TLB Hit*Cache Miss + 
 TLB Miss*Cache Hit +
 TLB Miss*Cache Miss
= 0.96*0.9*2 + 0.96*0.1*12 + 0.04*0.9*22 + 0,04*0.1*32
= 3.8
≈ 4 

Why 22 and 32?
22 is because when TLB miss occurs it takes 1ns and the for the physical address it has to go through two level page tables which are in main memory and takes 2 memory access and the that page is found in cache taking 1 ns which gives a total of 22

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