NIELIT Scientific Assistant A 2020 November: 70

Question

A computer has a single cache (off-chip) with a $3$ ns hit time and a $95\%$ hit rate. Main memory has a $50$ ns access time. If we add an on-chip cache with a $0.6$ ns hit time and a $98\%$ hit rate, the computer’s effective access time:

• A. $\text{2.8 ns}$
• B. $\text{5.5 ns}$
• C. $\text{0.7 ns}$
• D. None of the options

Answer 1

Average access time = [H1*T1]+[(1-H1)*H2*T2]+[(1-H1)(1-H2)*Hm*Tm]

 

Simultaneous access-

0.98(0.6) + 0.02(0.95)(3) + 0.02(0.05)(50) = 0.695 ns

 

 

Hierarchal access- 

0.98(0.6) + 0.02(0.95)(0.6+3) + 0.02(0.05)(0.6+3+50) = 0.69974 ns

 

in both case ~ 0.7 ns

so, option C