0 votes 0 votes Consider a k-level paging system along with a TLB. A TLB takes 10ns, and a memory takes 100ns on average. The hit ratio of TLB is equal to 0.8. If it is known that the average memory access time is 70ns, then the value of k is? Operating System operating-system paging translation-lookaside-buffer memory-management made-easy-booklet + – Gaurav Padole asked Dec 29, 2022 • retagged Dec 30, 2022 by Lakshman Bhaiya Gaurav Padole 760 views answer comment Share Follow See all 4 Comments See all 4 4 Comments reply Kabir5454 commented Dec 29, 2022 reply Follow Share K will be 2 . $10+0.2*(k+1)*100=70$ So ,$k=2$ 0 votes 0 votes Gaurav Padole commented Dec 30, 2022 reply Follow Share why will it be (k+1) and not k? the question is asking for k-level no? 0 votes 0 votes Kabir5454 commented Dec 30, 2022 reply Follow Share (k+1) because say you have k level page table so , you will look up in each k page table which reside in memory. So k*memory access time it will take. Now from the last page table (k'th) you will get the address of the memory location where actual data reside . For which you need one more extra memory access. So total memory access needed =K* memory access time+memory access time =(K+1) memory access time 0 votes 0 votes Gaurav Padole commented Dec 31, 2022 reply Follow Share ok i got it thanks 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes TLB = t = 10ns Memory time = m = 100ns Hit ratio (tlb) = p = 0.8 Miss ratio = 1 – p = 0.2 Let assume there are k level for checking : then formula for the Effective time calculation : T effective = p( t + m ) + (1-p)(t + km + m ) T effective = pt + pm + t + km – pt – pkm + m – mp → t +m(k+1) ( 1-p) Now , we only put the value in te formula : T effective = t + (1-p) m (k+1) 70 = 10 + ( 1 – 0.8) * (k+1)* 100 60 = 0.2 * (k+1) * 100 60 = 20 *(k+1) k = 3 – 1 = 2 ANSWER IS 2 LEVEL WILL REQUIRE :) Shikhar. answered Mar 4, 2023 Shikhar. comment Share Follow See all 0 reply Please log in or register to add a comment.