1 / 4

Answers to Quiz4, Question 1

Answers to Quiz4, Question 1. The four components of a disk access are: The seek-time, the time for the read/write heads to move to the right track. The average rotational-latency, which is halfway around the disk, the time it takes the disk to perform half a revolution

irisa
Download Presentation

Answers to Quiz4, Question 1

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Answers to Quiz4, Question 1 • The four components of a disk access are: • The seek-time, the time for the read/write heads to move to the right track. • The average rotational-latency, which is halfway around the disk, the time it takes the disk to perform half a revolution • The transfer-time, the time it takes to transfer a block of data, which the data size/bandwidth. • The controller-time, the overhead of the disk controller. • The disk access time to read 2Kbyes of data is:8ms + (0.5/7000)RPM + 2K/8MB/sec + 2ms = 8ms + 7.14e-5 min + 0.00025 seconds + 2ms = 8ms + 4.28ms + 0.25ms + 2ms = 14.53ms

  2. Question 2 • Each DMA transfer takes 1400 CPU cycles. 1000 to setup the DMA and 400 to notify (by interrupt) the processor that the transfer has completed. • Each data transfer takes 1K/6MB/sec = 0.167ms, so there are 6000 transfers per second. But the disk is being accessed only 50% of the time so there are only 3000 accesses. 3000*1400 = 4,200,000 processor cycles per second.

  3. Question 3 • ETA = 120M*3.3/450M = 0.88secETB = 105M*3.9/500M = 0.819secSo computer B is 7% (0.88/0.819) faster than computer A. • For computer A, using Amdahl's we have a FE of 60%, and a SE of 1.22 (3.3/2.7). So the new execution time of A is:(FE/SE + 1 -FE)ETold = (0.6/1.22 + 0.4)0.88 = 0.784 • For computer B, using Amdahl's we have a FE of 40%, and a SE of 1.34 (3.9/2.9). So the new execution time of A is:(FE/SE + 1 -FE)ETold = (0.4/1.34 + 0.6)0.819 = 0.735 • B is faster than A by 6.6% (0.784/0.735)

  4. Question 4 • The wall-clock time or elapsed time = CPU time + I/O time. The CPU time was improved by 50% so the SE is 1.5, FE is 0.9 so the new elapsed time is:0.9/1.5 + 0.1 = 0.7. So the improvement is 1/0.7 = 43%. • In order to improve elapsed time by 1.15 (15%) the new elapsed time must be 1/x = 1.15 . x = 0.87. But the CPU time itself is 0.90, so it is impossible to gain an improvement by only improving the I/O time.

More Related