Question

In a computer system, memory access time is 10 nsec when there is a hit in cache. In case of cache miss data is accessed from either main memory or hard disk. It takes 500 nsec time to access memory when miss in cache but hit in main memory and 5000 nsec time when miss in both cache and main memory. It is observed that 90% time cache hit occurs. 80% time cache miss, hit in main memory. Average memory access time in nsec is _____________.

Answer 1

option (c)

$T_{avg}=\left(0.9{}^{\ast }10\right)+0.1\left[\right(0.8{}^{\ast }500)+(0.2{}^{\ast }5000\left)\right]$
$=9+0.1(400+1000)$
$=9+140$
$=149nsec$