A 50 kbps device is connected to a processor. The interrupt overhead is 50…
2021
A 50 kbps device is connected to a processor. The interrupt overhead is 50 μsec. The minimum performance achieved when interrupt is initiated and data transfer is used instead of programmed I/O is:
- A.
2.4
- B.
0.4
- C.
3
- D.
3.5
- E.
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Correct answer: B
Concept: Under polled (programmed) I/O, the CPU must continuously check device status, so its I/O cost per unit of data equals the device's own time-between-arrivals (1 / data rate). Under interrupt-driven I/O, the CPU is busy only for the fixed interrupt-handling overhead each time a unit of data arrives, and is free to do other work the rest of the time. Dividing the time-between-arrivals by the interrupt overhead gives the performance factor of switching from polling to interrupts — a factor below 1 means the overhead alone already exceeds what polling would have cost, so interrupt-driven service is then the less efficient choice for that device.
Applying it here:
The device transfers at 50 kbps = 50,000 bits per second, so one bit arrives every 1 / 50,000 second = 20 μsec — this is the time-between-arrivals under polling.
The interrupt-handling overhead is given directly as 50 μsec per interrupt.
Performance factor = time-between-arrivals ÷ interrupt overhead = 20 μsec ÷ 50 μsec = 0.4.
Cross-check: Because the overhead (50 μsec) is more than double the bit period (20 μsec), the factor comes out below 1 — confirming that interrupt-driven transfer here can keep up at only 0.4 times the rate that continuous polling could sustain, so the minimum performance is 0.4.