UGC NET CSE | October 2022 | Part 1 | Question: 21

Question

Consider a memory system having address spaced at a distance of $m, T=$ Bank cycle time and $n$ number of banks, then the average data access time per word access in synchronous organization is

• A. $ t=\left\{\begin{array}{l} m, \frac{T}{n} \text { for } m\ll n \\ T \text { for } m \gg n \end{array}\right.$   
• B. $ t=\left\{\begin{array}{l} T / n \text { for } m\ll n \\ T \text { for } m\gg n \end{array}\right. $ 
• C. $ t=\left\{\begin{array}{l} m . T \text { for } m \ll n \\ T \text { for } m \gg n \end{array}\right. $
• D. $ t=\left\{\begin{array}{l} m \cdot T \text { for } m \ll n \\ m / T \text { for } m\gg n \end{array}\right. $

Answer 1

• When the distance at which address is spaced is very less than the number of banks, that is, m << n, then the average data access time per word access is - m ⋅ T n

• • When the distance at which address is spaced is much more than the number of banks, that is, m >> n, then the average data access time per word access is - T

• • Combining above two points, we get the average data access time per word access in m ⋅ Tn for m << n synchronous organisation as Option 1