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:
Answer: B. 0.4 — Concept: For a rate R bits/s, an interrupt that services b bits has a transfer interval T = b / R. If the interrupt overhead is Ti, the relative performance…
- A.
2.4
- B.
0.4
- C.
3
- D.
3.5
- E.
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Correct answer: B
Concept: For a rate R bits/s, an interrupt that services b bits has a transfer interval T = b / R. If the interrupt overhead is Ti, the relative performance factor is S = T / Ti = b / (R × Ti). Because b is a positive whole number of bits, the minimum possible factor occurs at the smallest transfer granularity, b = 1 bit.
Application:
Set R = 50 kbps = 50,000 bits per second and Ti = 50 μsec.
For the minimum, use b = 1 bit, so T = b / R = 1 / 50,000 second = 20 μsec.
Substitute into the factor: Smin = T / Ti = 20 μsec / 50 μsec = 0.4.
Cross-check: Equivalently, R × Ti = 50,000 × 50 × 10−6 = 2.5, so Smin = 1 / 2.5 = 0.4. Thus the saved value is the theoretical minimum under the question’s stated bit-rate convention; larger values of b give larger factors.