4 votes 4 votes A hierarchical memory system that uses cache memory has cache access time of $50$ nano seconds, main memory access time of $300$ nano seconds, $75$% of memory requests are for read, hit ratio of $0.8$ for read access and the write-through scheme is used. What will be the average access time of the system both for read and write requests ? $157.5$ n.sec. $110$ n.sec. $75$ n.sec. $82.5$ n.sec. CO and Architecture ugcnetcse-dec2014-paper3 co-and-architecture cache-memory + – makhdoom ghaya asked Jul 24, 2016 retagged Nov 13, 2017 by Arjun makhdoom ghaya 6.4k views answer comment Share Follow See 1 comment See all 1 1 comment reply Sanjay Sharma commented Aug 21, 2018 i edited by Sanjay Sharma Aug 21, 2018 reply Follow Share read operation will use the cache memory but write operation will simply write on memory so no need of cache here so let 100 operation are there 75 will be read and 25 will be write (75(0.8x50+0.2x350)+25(300))/100=157.5ns 1 votes 1 votes Please log in or register to add a comment.
Best answer 9 votes 9 votes Option A). TRead = 0.8 * 50 + 0.2 * (300 + 50) = 110 ns TWrite = 1 * max(300,50) = 1* 300 = 300 ns Hence, Average access time for both read and write requests = 0.75 * 110 + 0.25 * 300 = 157.5 ns Kapil answered Jul 24, 2016 selected Jul 24, 2016 by srestha Kapil comment Share Follow See 1 comment See all 1 1 comment reply baijnath559 commented Jan 21, 2023 reply Follow Share can you please explain why max(300,50) not 350??? 0 votes 0 votes Please log in or register to add a comment.
2 votes 2 votes The average access time of the system for memory read cycle is 0.8x50+0.2x(300+50)=110ns The average access time of the system for both read and writes request is 0.75x110+0.25x300=82.5+75=157.5 Pranabesh Ghosh 1 answered Jun 1, 2016 Pranabesh Ghosh 1 comment Share Follow See all 0 reply Please log in or register to add a comment.