In our explanation of the physical elements of disk performance, we said that switching from 7,200rpm to 10,000rpm disks may increase performance by 10% to 50%. What would be the reason for a 10% increase? Is it possible that the situation will remain the same? Explain. (Hint: Disk performance is not only determined by rotational delay.)
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In our explanation of the physical elements of disk performance, we said that switching from 7,200rpm to 10,000rpm disks may increase performance by 10% to 50%. What would be the reason for a 10% increase? Is it possible that the situation will remain the same? Explain. (Hint: Disk performance is not only determined by rotational delay.)
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- If a microprocessor has a cycle time of 0.5 nanoseconds, what’s the processor clock rate? If the fetch cycle is 40% of the processor cycle time, what memory access speed is required to implement load operations with zero wait states and load operations with two wait states?Consider a Disk I/O transfer, in which 1500 bytes are to be transferred, but number of bytes on a track is 1000, and rotation speed of disk is 1500 rps but the average time required to move the disk arm to the required track is 15 ms, then what will be total access time?Suppose that a disk drive rotates at 8000 RPM. It has an average seek time of 4 milliseconds. If its transfer rate is 15 Mbps, determine the average time it takes for a 15K byte request to be transferred from the time the disk starts the seek? From this, determine its throughput.
- In our discussion of the physical aspects of disk performance, we stated that replacing 7,200rpm disks with 10,000rpm disks can bring a 10% to 50% performance improvement. Why would an improvement of only 10% occur? Could it be that no improvement at all would occur? Explain. (Hint: Rotational latency is not the only determining factor of disk performance.)Suppose that a magnetic disk has an average seek time of 4 ms, a rotation rate of 5400 RPM, a transfer rate of 50 MB/second, a sector size of 512 bytes, and controller overhead of 2 ms. What is the average amount of time to read a single sector?In Feynman's lecture he wondered "why cannot we write the entire 24 volumes of the Encyclopedia Brittanica on the head of a pin". For a modern hard disk drive the area of a single bit of information on the surface of the disk is roughly 10 x 40 nm² in area. In such a hard drive would the area required to store the data for 24 volumes of the Encyclopedia Brittanica be greater or less than the head of a pin? In Feynman's lecture he estimates the 24 million books as big as the Encyclopedia Brittanica has 10¹5 bits of information.
- Consider a disk pack with 16 surfaces, 128 tracks per surface and 256 sectors per track. 512 bytes of data are stored in a bit serial manner in a sector. The capacity of the disk pack and the number of bits required to specify a particular sector in the disk.?Consider a disk pack with 16 surfaces, 128 tracks per surface and 256 sectors per track. 512 bytes of data are stored in a bit serial manner in a sector. Calculate the capacity of the disk pack and the number of bits required to specify a particular sector in the disk.Consider a disk drive specifications. with the following 16 surfaces, 512 tracks/surface, 512 sectors/ track, 1 KB/sector, rotation speed 3000 rpm. The disk is operated in cycle stealing mode whereby whenever one byte word is ready it is sent to memory; similarly, for writing, the disk interface reads a 4 byte word from the memory in each DMA cycle. Memory cycle time is 40 nsec. The maximum percentage of time that the CPU gets blocked during DMA operation?
- Consider a single-platter disk with the following parameters: rotation speed: 7200 rpm; number of tracks on one side of platter: 30,000; number of sectors per track: 600; seek time: one ms for every hundred tracks traversed. Let the disk receive a request to access a random sector on a random track and assume the disk head starts at track 0. a. What is the average seek time? b. What is the rotational latency? c. What is the transfer time for a sector? d. What is the total average time to satisfy a request?Transfer rate of a disk drive can be no faster than the bit density (bits/track) times the rotational speed of the disk. Figure 7.15 gives a data transfer rate of 112GB/sec. Assume that the average track length of the disk is 5.5 inches. What is the average bit density of the disk?Consider an audio CD that contains exactly one quarter of an hour of stereo sound. Ignoring any additional requirements for format information and other data to ensure the integrity of the sound samples, how many bytes of storage does the CD need to contain for both of the stereo channels? Assume the sample rate is 44500 samples per second and each sample requires two bytes of storage.