A 32 – bit wide main memory unit with a capacity of 1 GB is built using 256M X 4-bit DRAM chips. The number of rows of memory cells in the DRAM chip is 2¹⁴. The time taken to perform one refresh operation is 50 nanoseconds. The refresh period is 2 milliseconds. The percentage (rounded to the closet integer) of the time available for performing the memory read/write operations in the main memory unit is _______ .

Note – This was Numerical Type question.
(A) 59
(B) 40
(C) 99
(D) None of these


Answer: (A)

Explanation: Given, total number of rows is 2¹⁴ and time taken to perform one refresh operation is 50 nanoseconds.

So, total time taken to perform refresh operation = 2¹⁴*50 nanoseconds = 819200 nanoseconds = 0.819200 milliseconds.

But refresh period is 2 milliseconds.
So, time spent in refresh period in percentage = (0.819200 milliseconds) / (2 milliseconds) = 0.4096 = 40.96%

Hence, time spent in read/write operation = 100% – 40.96% = 59.04% = 59 (in percentage and rounded to the closet integer).

So, answer is 59.


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