GATE Computer Science (CS) 2018 Shift 1 Solved Paper

Question : 14 of 65

Marks: +1, -0

A 32 -bit wide main memory unit with a capacity of

1 {GB}

is built using

256 {M} × 4

-bit DRAM chips. The number of rows of memory cells in the DRAM chip is

2^{14}

. The time taken to perform one refresh operation is 50 nanoseconds. The refresh period is 2 milliseconds. The percentage (rounded to the closest integer) of the time available for performing the memory read\/write operations in the main memory unit is

Your Answer:

Validate

Solution:

Total number of rows is

2^{14}

and time taken to perform one refresh operation is 50 nanoseconds. So, total time taken to perform refresh operation

=2^{14} * 50

nanoseconds = 819200 nanoseconds

=0.819200

milliseconds.
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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