MadeEasy Test Series: CO & Architecture - Memory Interfacing

Question

Consider a DRAM that must be given a refresh cycle $64$ times per msec. Each refresh operation requires $100 \ nsec$ and a memory cycle requires $200 \ nsec$. The percentage of the memory’s total operating time must be given to refresh is _______ (upto $2$ decimal places)

Comments

$(\frac{6400}{200})/(\frac{10^{6}}{200})\times 100=0.64$%

https://gateoverflow.in/3754/gate2005-it-9

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Thanks @srestha for the link.

Answer 1

Refreshments done in $1 \text{ msec } = 64$

$\implies$ refreshments done in $10^6 \text{ nsec } = 64$,  $(\because 1 \text{ msec } = 10^6 \text{ nsec})$

Now, refreshments done in $200 \text{ nsec } = \frac{64 \times 200}{10^6} = \frac{128}{10^4}$

$\implies$ in $1$ memory cycle, $\frac{128}{10^4}$ refreshes could be done.

Time spent in doing $\frac{128}{10^4}$ refreshes $ = 100 \text{ nsec } \times \frac{128}{10^4} = 1.28 \text{ nsec }$

$\implies$ Out of $200 \text{ nsec}, 1.28 \text{ nsec } $ is spent in doing refreshments.

$\implies$ Percentage of CPU time spent in refreshments $ = \frac{1.28 \text{ nsec }}{200 \text{ nsec }} \times 100 = 0.64$ %