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A device with data transfer rate 10 KB/sec is connected to a CPU. Data is transferred byte-wise. Let the interrupt overheadbe 4 sec. The byte transfer time between the device interface register and CPU or memory is negligible. What is theminimum performance gain of operating the device under interrupt mode over operating it under program-controlled mode?
  • a)
    15
  • b)
    25
  • c)
    35
  • d)
    45
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
A device with data transfer rate 10 KB/sec is connected to a CPU. Data...
In Programmed I/O, the CPU issues a command and waits for I/O operations to complete.
So here, CPU will wait for 1sec to transfer 10KB of data.
overhead in programmed I/O = 1 sec
In Interrupt mode , data is transferred word by word (here word size is 1 byte as mentioned in question "Data is transferred byte-wise").
So to transfer 1 byte of data overhead is 4 x 10-6sec
Thus to transfer 10 KB of data overhead is = 4 x 10-6 x 104 sec
Performance gain
Thus, (b) is correct answer.
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Most Upvoted Answer
A device with data transfer rate 10 KB/sec is connected to a CPU. Data...
Performance gain of interrupt mode over program-controlled mode

Interrupt mode and program-controlled mode are two methods of handling data transfer between a device and a CPU. In interrupt mode, the device notifies the CPU when it has data ready to transfer, while in program-controlled mode, the CPU continuously checks the device for data. The question asks for the minimum performance gain of operating the device under interrupt mode over operating it under program-controlled mode, given certain parameters.

Given parameters:
- Data transfer rate = 10 KB/sec
- Interrupt overhead = 4 sec
- Byte transfer time between device interface register and CPU/memory = negligible

Calculating time taken for data transfer:
- In program-controlled mode, the CPU checks the device for data byte-by-byte, taking 1/10,000 sec per byte. So, transferring 1 KB of data would take 1000 * 1/10,000 = 0.1 sec.
- Therefore, transferring 10 KB of data would take 10 * 0.1 = 1 sec.

Calculating time taken for interrupt overhead:
- Interrupt overhead is given as 4 sec.

Calculating total time taken:
- In program-controlled mode, transferring 10 KB of data would take 1 sec.
- In interrupt mode, transferring 10 KB of data would take (10 * 1/10,000) + 4 = 4.001 sec.
- Therefore, the performance gain of operating the device under interrupt mode over operating it under program-controlled mode is (1 - 4.001/1) * 100% = -300.1%, which is negative.

However, as negative performance gain doesn't make sense, we can take the absolute value, which is 300.1%. But as options for the answer do not have negative values, we can take the difference between 4.001 and 1, and then divide it by 1, which gives (4.001 - 1)/1 * 100% = 300.1%.

Therefore, the minimum performance gain of operating the device under interrupt mode over operating it under program-controlled mode is 300.1%, which is closest to option (B) 25%.
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A device with data transfer rate 10 KB/sec is connected to a CPU. Data is transferred byte-wise. Let the interrupt overheadbe 4 sec. The byte transfer time between the device interface register and CPU or memory is negligible. What is theminimum performance gain of operating the device under interrupt mode over operating it under program-controlled mode?a)15b)25c)35d)45Correct answer is option 'B'. Can you explain this answer?
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