Friday, 29 September 2017

Ugc net 2017 solved

Updated: 29-September-2017


CBSE NET 2017 PAPER II


OPERATING SYSTEMS


In a paging system, it takes 30 ns to search translation Look-a-side Buffer (TLB) and 90 ns to access the main memory. If the TLB hit ratio is 70%, the effective memory access time is


(1) 48ns
(2) 147ns
(3) 120ns
(4) 84ns


Ans :- 2


Explanation:-


A 70 % ratio means that we find the desired page number in the TLB 70 % of the time. If it takes 30 ns to search the TLB and 90 ns to access memory, then a mapped-memory access takes 30+90 ns when the page number is in the TLB. If we fail to find the page number in the TLB(30), then we must first access memory for the page table and frame number(90ns) and then access the desired byte in the memory(90ns), for a total of 210ns. To find the effective memory access time, we weight each case by its probability,
Effective access time = 0.70 X 120 + 0.30 X 210
                                 = 84 + 63 = 147 ns

Related topics

Professor Jayesh video tutorial

Please use contact page in this website if you find anything incorrect or you want to share more information about the topic discussed above.