A device with data transfer rate 10 KB/sec is connected to a CPU. Data is transferred byte-wise. Let the interrupt overhead be 4 microsec. The byte transfer time between the device interface register and CPU or memory is negligible. What is the minimum 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?

Question

Answer

Given:
Data transfer rate = 10 KB/sec
Interrupt overhead = 4 microseconds
Byte transfer time = negligible

To find: Minimum performance gain of operating the device under interrupt mode over operating it under program controlled mode

Approach:
In program-controlled mode, the CPU continuously polls the device to check if any data is available for transfer. This results in wastage of CPU time and hence, slows down the system. On the other hand, in interrupt mode, the device signals the CPU when data is available for transfer. This saves CPU time and increases system performance. The performance gain can be calculated as follows:

Performance gain = (Time taken in program-controlled mode - Time taken in interrupt mode) / Time taken in program-controlled mode * 100%

We know that data transfer rate = 10 KB/sec = 10,000 bytes/sec

Time taken to transfer one byte = 1 / 10,000 sec = 0.1 milliseconds = 100 microseconds

Therefore, time taken to transfer n bytes = n * 100 microseconds

In program-controlled mode:
CPU time wasted in polling = 100 microseconds (for each byte)
Total CPU time wasted in polling for n bytes = n * 100 microseconds
Total time taken in program-controlled mode = n * 100 microseconds + n * byte transfer time + interrupt overhead
= n * 100 microseconds + n * 0 + 4 microseconds (negligible byte transfer time)
= n * 100 + 4 microseconds

In interrupt mode:
CPU time saved = 4 microseconds (for each byte)
Total CPU time saved in interrupt mode for n bytes = n * 4 microseconds
Total time taken in interrupt mode = n * byte transfer time + interrupt overhead + time taken to handle interrupts
= n * 0 + 4 microseconds + n * interrupt handling time
= 4 microseconds + n * interrupt handling time

Now, we need to find the minimum value of n for which the performance gain is maximum.

Let's assume that interrupt handling time per byte is t microseconds.

Then, the performance gain can be calculated as follows:

Performance gain = (n * 100 + 4 - 4 - n * t) / (n * 100 + 4) * 100%
= (100 - n * t / (n * 100 + 4)) * 100%

To maximize the performance gain, we need to minimize n * t / (n * 100 + 4)

Differentiating with respect to n, we get:

t / (n * 100 + 4)^2 - t / (n * 100 + 4) = 0
=> t / (n * 100 + 4) * (1 - t / (n * 100 + 4)) = 0

Since t and (n * 100 + 4) are positive, we get:

1 - t / (n * 100 + 4) = 0
=> t = n * 100 + 4
=> n = (t - 4) / 100

Substituting this value of n in the expression for performance gain, we get:

Performance gain = (100 - t / 100) * 100%

To maximize the performance gain, we need to minimize t, which means interrupt handling time should be as small as possible.

Therefore, the minimum performance gain of operating the device under

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