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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

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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.

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Question 3• ETA = 120M*3.3/450M = 0.88sec

ETB = 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)

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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.