Flashcards on Computer I/O and Storage Systems
Computer I/O & Storage Systems: A Student's Guide
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Storage and I/O
86 cards
Card 1
Question: What are the three ways to characterize an I/O device mentioned in the content?
Answer: Behaviour (input, output, storage), partner (human or machine), and data rate (bytes/sec, transfers/sec) including I/O bus connections.
Card 2
Question: What dependability concepts are defined for I/O and storage systems?
Answer: Fault (component failure), service interruption (deviation from specified service), failure (service not delivered as specified), and restoration (ret
Card 3
Question: How are reliability, mean time to repair (MTTR), and mean time to failure (MTTF) related to availability?
Answer: MTBF = MTTF + MTTR; Availability = MTTF / (MTTF + MTTR).
Card 4
Question: Name two strategies to improve availability and briefly how they help.
Answer: Increase MTTF via fault avoidance, fault tolerance, and fault forecasting; reduce MTTR via improved diagnosis and repair tools/processes.
Card 5
Question: List the main performance measures for I/O systems and how they differ in importance between desktops/embedded systems and servers.
Answer: Latency (response time) and throughput (bandwidth). Desktops/embedded systems prioritize response time and device diversity; servers prioritize throug
Card 6
Question: What are the typical components of a disk sector?
Answer: Sector ID, data (typically 512 bytes, with 4096 bytes proposed), error-correcting code (ECC), synchronization fields, and gaps.
Card 7
Question: What delays contribute to access time for a disk sector?
Answer: Queuing delay, seek time (moving the heads), rotational latency, data transfer time, and controller overhead.
Card 8
Question: Using the given example parameters (512B sector, 15,000 rpm, 4 ms avg seek, 100 MB/s transfer, 0.2 ms controller overhead), what is the average read t
Answer: Average read time = 4 ms seek + 2 ms rotational latency (½ rotation at 15,000 rpm) + 0.005 ms transfer + 0.2 ms controller = 6.2 ms.
Card 9
Question: If the actual average seek time reduces to 1 ms in the previous example, what is the new average read time?
Answer: Average read time = 1 ms seek + 2 ms rotational latency + 0.005 ms transfer + 0.2 ms controller = 3.2 ms.
Card 10
Question: Why do actual average seek times often differ from manufacturer-quoted average seek times?
Answer: Manufacturers quote averages over all possible seeks; locality and OS scheduling usually result in smaller actual average seek times in practice.