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CS241 I/O Revisited

CS241 I/O Revisited. Lawrence Angrave DMA, Polling, Disk Characteristics. Today. Networking Quiz DMA vs Polling Disk Characteristics Announcements. DMA & I/O. What is DMA? Why is useful? How does it work? Compare it Polling?. DMA. Direct Memory Access

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CS241 I/O Revisited

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  1. CS241I/O Revisited Lawrence Angrave DMA, Polling, Disk Characteristics

  2. Today • Networking Quiz • DMA vs Polling • Disk Characteristics • Announcements

  3. DMA & I/O • What is DMA? • Why is useful? • How does it work? • Compare it Polling?

  4. DMA • Direct Memory Access • DMA controls exchange of data between main memory and I/O module • Processor issues a request for transfer of block data • Process is interrupted when DMA request completes

  5. Main Advantages? Processor can perform useful work (does not need to wait for I/O device to be ready) Optimal use of memory bus: • Efficient block transfer • Use main system bus once per word (In some architectures)

  6. Cycle stealing • DMA takes control of system bus to write to main memory • CPU can not use bus at the same time • DMA has stolen cycles from the CPU!

  7. Compare to Polling • CPU continually checks I/O device to see if it is ready • Wastes CPU (and bus) cycles • So when would Polling be useful?

  8. Polling vs DMA DMA overhead: Issuing commands to DMA DMA shared resource: Requests are queued DMA efficient for block reads to memory. Interrupt overhead.

  9. Reasons for polling • Want small amount of data to be processed by CPU as soon as it is available. • Interrupt-driven I/O not available. • Not all devices, architectures support DMA transfer (or have a DMA)

  10. Speed of I/O transfer • Limitations of physical device • Limitations of I/O bus • Limitations of system bus • Limitations of read/write speed of memory

  11. Speed of I/O transfer • Limitations of physical device • Limitations of I/O bus • Limitations of system bus • Limitations of read/write speed of memory

  12. In practice • Physical device is a shared resource • So is memory, I/O bus, system bus, memory & CPU

  13. Physical device • Rotation speed • Seek time to correct track (move head to correct track)

  14. To use these • # Sectors per track • # Bytes per sector

  15. Rotation • How long before first byte is under the head? • How many milliseconds before all bytes are read from that track?

  16. HW Problem • rotational speed of 15,000 rpm, • 512 bytes per sector, • 400 sectors per track and • 1000 tracks on the disk, • average seek time is 4ms.

  17. What is the transfer time for this file? • What is the average access time for this file? • What is the rotational delay in this case? • What is the total time to read 1 sector? What is the total time to read 1 track?

  18. rpm=revolutions per minute • Transfer rate= Bytes per 360 degrees x Rotational speed (60 x rpm) • Avg Access Time for 1 sector=seek time + avg. rotational delay + transfer time for 1 sector

  19. Examples b=number of bytes to transfer, r=rotation speed, and N = number of bytes on a track Tt=transfer time = b/(rN) Ta=average access time of the whole file = Tseek+ 1/(2r)+T Average Rotational delay = 0.5/r

  20. Annoucements • HW2 due Wednesday 05/02/2007 • Newsgroup request for review lecture

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