UGC NET CSE | December 2014 | Part 3 | Question: 01

Question

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 ? 

• A. $157.5$ n.sec. 
• B. $110$ n.sec.
• C. $75$ n.sec.
• D. $82.5$ n.sec. 

Comments

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

Answer 1

Option A). 

T_Read = 0.8 * 50 + 0.2 * (300 + 50) = 110 ns

T_Write = 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

Comments

can you please explain why max(300,50) not 350???

Answer 2

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