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A $32\text{-bit}$ wide main memory unit with a capacity of $1\;\textsf{GB}$ is built using $256\textsf{M} \times 4\text{-bit}$ 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 $50\;\text{nanoseconds}$. The refresh period is $2\;\text{milliseconds.}$ The percentage (rounded to the closest integer) of the time available for performing the memory read/write operations in the main memory unit is _________.
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116 votes

One refresh operation takes $50\;\text{ns}$.

Total number of rows $= 2^{14}$

Total time to refresh all Rows $= 2^{14}\times 50\; \text{ns} = 819200 \;\text{ns} =  0.819200\;\text{ms}$
The Refresh Period is $2\;\text{ms}.$

$\%$ Time spent in refresh  $= \frac{\text{Total time to Refresh all Rows}}{\text{Refresh period}}\ast 100
$
                                             $= \frac{0.8192\;\text{ms}}{2.0\;\text{ms}}\ast 100 = 40.96\%$

$\%$ Time spent in Read/Write $= 100 - 40.96 = 59.04\%$

$= 59\%$ (Rounded to the closest Integer)

Reference: https://en.wikipedia.org/wiki/Memory_refresh

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27 votes
27 votes

On a 32bit machine, the maximum amount of memory is around 4GB and it's ok to have a capacity of less than 4GB. Because it depends on OS as well and might be less due to parts of address space being reserved, So the usable capacity of the main memory is 1GB, and this main memory may have been designed using more than 1 DRAM chips, 
One DRAM chip size is 256M x 4bits=2^28 x 2^2 =2^30bits.
No. of DRAM chips needed to design main memory of size 1GB= 2^30*8 / 2^30=8. so 8 DRAM chips are needed. Now there are 2^14 rows in each DRAM chips. then how many cells(column) are there in one row or in one DRAM chip=2^30bits(Size of one DRAM) / 2^14=2^16.
The time taken to perform one refresh operation is for only 1 row of 1 DRAM chip and it is 50nsec. The refresh period is 2 msec means after every 2msec the DRAM chips need to be refreshed otherwise data stored in it will get lost. Now they are asking what is the percentage of time for meaningful operations(R/W) other than refresh overhead. so if we calculate refresh overhead itself, we can also derive percentage of time for meaningful operations(100%- refresh overhead%).
 Given, the 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 on 1DRAM chip = 2^14*50 nanoseconds = 819200 nanoseconds = 0.819200 milliseconds. But the refresh period is 2 milliseconds.

Note:- All DRAM chips are refreshed simultaneously.

So, time spent in refresh period in percentage or REFRESH OVERHEAD= TIME TAKEN FOR 1 REFRESH OPERATION TO 1 DRAM CHIP / REFRESH INTERVAL 

= (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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2 votes

Time taken to refresh one row = 50 ns
There are 2^14 rows, so time taken to refresh all the rows = 2^14 * 50ns = 0.82 milliseconds
It is given that total refresh period is 2ms. The refresh period contains the time to refresh all the rows and also the time to perform read/write operation.
So % time spent in refresh = (Time taken to refresh all rows / refresh period)*100
= (0.82 ms / 2ms)*100
= 41%
So the % of time for read/write operation = 100 - 41 = 59%

Answer:

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