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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
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
A 32 - bit wide main memory unit with a capacity of 1 GB is built usin...
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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Most Upvoted Answer
A 32 - bit wide main memory unit with a capacity of 1 GB is built usin...
Given:
- Main memory unit has a capacity of 1 GB
- Main memory unit is built using 256M X 4-bit DRAM chips
- Number of rows of memory cells in the DRAM chip is 214
- Time taken to perform one refresh operation is 50 nanoseconds
- Refresh period is 2 milliseconds

We need to calculate the percentage of time available for performing memory read/write operations in the main memory unit.

Let's calculate the number of DRAM chips required to build the main memory unit:
1 GB = 2^30 bytes
Each DRAM chip has a capacity of 256M X 4 bits = 256M/8 bytes = 32M bytes
Number of DRAM chips required = (2^30 bytes) / (32M bytes) = 2^5 * 2^20 / 2^25 = 2^5 = 32

Let's calculate the number of rows in the main memory unit:
Number of rows = Number of DRAM chips * Number of rows in each DRAM chip = 32 * 2^14 = 32 * 16384 = 524,288

Let's calculate the total time taken for all refresh operations during the refresh period:
Number of refresh operations = Number of rows = 524,288
Time taken for one refresh operation = 50 nanoseconds
Total time taken for all refresh operations = Number of refresh operations * Time taken for one refresh operation = 524,288 * 50 nanoseconds = 26,214,400 nanoseconds = 26.2144 milliseconds

Let's calculate the time available for performing memory read/write operations:
Refresh period = 2 milliseconds
Time taken for all refresh operations = 26.2144 milliseconds
Time available for performing memory read/write operations = Refresh period - Time taken for all refresh operations = 2 milliseconds - 26.2144 milliseconds = -24.2144 milliseconds

Since the time available for performing memory read/write operations is negative, it means that the refresh operations take longer than the refresh period. This is not possible.

Therefore, the given information is not correct or there is an error in the calculation.
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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)59b)40c)99d)None of theseCorrect answer is option 'A'. Can you explain this answer?
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