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?
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Q: Consider a Disk I/O transfer, in which 1500 bytes are to be transferred, but number of bytes on a…
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Q: Example- 5.8 A hard disk with one platter rotates at 15,000 rpm and has 1024 and has 1024 tracks,…
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Q: Suppose that a disk drive rotates at 8000 RPM. It has an average Seek time of 4 milliseconds. If its…
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A: As given:- Problem: Consider a long sequence of accesses to a disk with an average seek time of 6 ms…
Q: Suppose that a disk drive rotates at 8000 RPM. It has an average seek time of 4 milliseconds. If…
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Q: Assume that 1000MB of data needs to be transferred to the hard disk at the rate of 200 MB /sec.…
A: We need to find the total context switch component time for the given scenario.
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Q: Consider a magnetic disk drive with 8 surfaces, 512 tracks per surface, and 64 sectors per track.…
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Q: Consider a disk pack with 16 surfaces, 128 tracks per surface and 256 sectors per track. 512 bytes…
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A: Here, I have to provide a solution to the above question.
Q: Let us assume a disk with rotational speed of 15,000 rpm, 512 bytes per sector, 400 sectors per…
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Q: Consider a Disk 1/O transfer, in which 1500 bytes are to be transferred, but number of bytes on a…
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Q: Q2) Consider a magnetic disk drive with 8 surfaces, 512 tracks per surface, and 64 sectors per…
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Q: What is the average time to read or write a 512-byte sector for a typical disk rotating at 15,000…
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Q: Q] Consider a disk with a sector size of 512 bytes, 2000 tracks per surface, 50 sectors Per track,…
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- Assume that 1000MB of data needs to be transferred to the hard disk at the rate of 200 MB /sec. Considering a swapout time of 2000ms, what would be the total context switch component time.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?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?
- Consider the following : 4KB sector, 7200RPM, 2ms average seek time, 80MB/s transfer rate, 0.4ms controller overhead, average waiting time in queue is 1 second. What is the estimated time to access this sector on this Hard Drive Disk? (Show and give result in ms)Suppose the user process' size is 8192Kb and is a standard hard disk where swapping has a data transfer rate of 1Mbps. Calculate for the total swap time in milliseconds (swap-in time and swap-out time) of the 8192Kb process to and from the physical memory.Q] Consider a disk with a sector size of 512 bytes, 2000 tracks per surface, 50 sectorsPer track, five double-sided platters, and average seek time of 10 msec.a) What is the capacity of a track in bytes?b) What is the capacity of each surface?c) What is the capacity of the disk?d) How many cylinders does the disk have?e) Give examples of valid block sizes. Is 256 bytes a valid block size? 2048?51200?f) If the disk platters rotate at 5400 rpm (revolutions per minute), what is theMaximum rotational delay?
- Consider a disk with an advertised average seek time of 6 ms, rotationspeed of 7,200 rpm, and 512-byte sectors with 500 sectors per track. Suppose that we wish to read a file consisting of 2500 sectors for a total of 1.28 Mbytes. Estimate the total time for the transfer when:For a magnetic disk with concentric circular tracks, the seek latency is not linearly proportional to the seek distance due to non-uniform distribution of requests arm starting and stopping inertia higher capacity of tracks on the periphery of the platter use of unfair arm scheduling policiesOn a particular brand of disk drive, the time that it takes for the disk arm to pass over a single disk track without stopping is 500ns. However, once the head reaches the track for which it has a service request, it needs 2ms to “settle” over the required track before it can start reading or writing. Based on these timings, compare the relative times required for FCFS, SSTF, and LOOK to carry out the schedule given below. You will need to compare SSTF to FCFS, LOOK to FCFS, and LOOK to SSTF As in our previous question, when the first request arrives in the disk request queue, the read/write head is at track 50, moving toward the outer (lower-numbered) tracks. The requested tracks are:35, 53, 90, 67, 79, 37, 76, 47
- 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.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.?Suppose we have a computer with the following characteristics: the magnetic disk has an accesslatency of 15 milliseconds and a streaming transfer rate of 5 megabytes per second. For the disk described,what is the effective transfer rate if an average access is followed by a streaming transfer ofa. 1024 bytes