GATE | GATE-CS-2014-(Set-3) | Question 65
Consider a paging hardware with a TLB. Assume that the entire page table and all the pages are in the physical memory. It takes 10 milliseconds to search the TLB and 80 milliseconds to access the physical memory. If the TLB hit ratio is 0.6, the effective memory access time (in milliseconds) is _________.
Explanation: TLB stands for Translation Lookaside Buffer. In Virtual memory systems, the cpu generates virtual memory addresses. But, the data is stored in actual physical memory i.e. we need to place a physical memory address on the memory bus to fetch the data from the memory circuitry. So, a special table is maintained by the operating system called the Page table. This table contains a mapping between the virtual addresses and physical addresses. So, every time a cpu generates a virtual address, the operating system page table has to be looked up to find the corresponding physical address. To speed this up, there is hardware support called the TLB. The TLB is a high speed cache of the page table i.e. contains recently accessed virtual to physical translations. TLB hit ratio- A TLB hit is the no of times a virtual-to-physical address translation was already found in the TLB, instead of going all the way to the page table which is located in slower physical memory. TLB hit ratio is nothing but the ratio of TLB hits/Total no of queries into TLB. In the case that the page is found in the TLB (TLB hit) the total time would be the time of search in the TLB plus the time to access memory, so
TLB_hit_time := TLB_search_time + memory_access_time
In the case that the page is not found in the TLB (TLB miss) the total time would be the time to search the TLB (you don’t find anything, but searched nontheless) plus the time to access memory to get the page table and frame, plus the time to access memory to get the data, so
TLB_miss_time := TLB_search_time + memory_access_time + memory_access_time
But this is in individual cases, when you want to know an average measure of the TLB performance, you use the Effective Access Time, that is the weighted average of the previous measures
EAT := TLB_miss_time * (1- hit_ratio) + TLB_hit_time * hit_ratio.
EAT := (TLB_search_time + 2*memory_access_time) * (1- hit_ratio) + (TLB_search_time + memory_access_time)* hit_ratio.
As both page table and page are in physical memory
T(eff) = hit ratio * (TLB access time + Main memory access time) +
(1 – hit ratio) * (TLB access time + 2 * main memory time)
= 0.6*(10+80) + (1-0.6)*(10+2*80)
= 0.6 * (90) + 0.4 * (170)
This solution is contributed Nitika Bansal
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