1. Department of Computer and IT Engineering University of Kurdistan Operating systems Disk Management By: Dr. Alireza Abdollahpouri

2. Hard Disk Drives Western Digital Drive http://www.storagereview.com/guide/ Read/Write Head Side View

3. Mechanics of Disks Platters circular platters covered with magnetic material to provide nonvolatile storage of bits Tracks concentric circles on a single platter Sectors segment of the track circle – usually each contains 512 bytes – separated by non-magnetic gaps. The gaps are often used to identify beginning of a sector Cylinders corresponding tracks on the different platters are said to form a cylinder Spindle of which the platters rotate around Disk heads read or alter the magnetism (bits) passing under it. The heads are attached to an arm enabling it to move across the platter surface

4. Hard Disk Drives

5. + Rotational delay + Transfer time Seek time Disk access time = + Other delays Disk platter Disk arm Disk head Disk Access Time

6. Measure Disk Performance Transfer rate is rate at which data flows between drive and computer Positioning time (random-access time) is time –The time to move disk arm to desired cylinder (seek time) –and the time for desired sector to rotate under the disk head (rotational latency)

7. Magnetic Disks  Platters range from.85” to 14” (historically)  Commonly 3.5”, 2.5”, and 1.8”  Range from 30GB to 3TB per drive  Performance  Transfer Rate – theoretical – 6 Gb/sec  Effective Transfer Rate – real – 1Gb/sec  Seek time from 3ms to 12ms – 9ms common for desktop drives  Average seek time measured or calculated based on 1/3 of tracks  Latency based on spindle speed  1/(RPM * 60)  Average latency = ½ latency (From Wikipedia)

8. Magnetic Disk Performance  Access Latency = Average access time = average seek time + average latency  For fastest disk 3ms + 2ms = 5ms  For slow disk 9ms + 5.56ms = 14.56ms  Average I/O time = average access time + (amount to transfer / transfer rate) + controller overhead  For example to transfer a 4KB block on a 7200 RPM disk with a 5ms average seek time, 1Gb/sec transfer rate with a.1ms controller overhead =  5ms + 4.17ms + 4KB / 1Gb/sec + 0.1ms =  9.27ms + 4 / 131072 sec =  9.27ms +.12ms = 9.39ms

9. Disk Structure  Disk drives are addressed as large 1-dimensional arrays of logical blocks, where the logical block is the smallest unit of transfer  The 1-dimensional array of logical blocks is mapped into the sectors of the disk sequentially  Sector 0 is the first sector of the first track on the outermost cylinder  Mapping proceeds in order through that track, then the rest of the tracks in that cylinder, and then through the rest of the cylinders from outermost to innermost  Logical to physical address should be easy  Except for bad sectors  Non-constant # of sectors per track via constant angular velocity

10. Disk Scheduling  Disk can do only one request at a time; What order do you choose to do queued requests?  Seek time  seek distance  Several algorithms exist to schedule the servicing of disk I/O requests 32 10 752 Head User Requests cylinder # of requested block

11. Disk Scheduling (Cont.)  There are many sources of disk I/O request  OS  System processes  Users processes  I/O request includes input or output mode, disk address, memory address, number of sectors to transfer  OS maintains queue of requests, per disk or device  Idle disk can immediately work on I/O request, busy disk means work must queue  Optimization algorithms only make sense when a queue exists  Note that drive controllers have small buffers and can manage a queue of I/O requests (of varying “depth”)  Several algorithms exist to schedule the servicing of disk I/O requests  The analysis is true for one or many platters  We illustrate scheduling algorithms with a request queue (0-199) 98, 183, 37, 122, 14, 124, 65, 67 Head pointer 53

12. Disk Scheduling: FCFS  Fair among requesters, but order of arrival may be to random spots on the disk  Very long seeks  Example در اين روش سيلندرهاي متقاضي به ترتيب درخواست‌شان سرويس‌دهي مي‌شوند. بعبارتي هر درخواست در صف اجرا قرار مي‌گيرد. ساده‌ترين روش است اما كارآيي چنداني ندارد.

13. Disk Scheduling: SSTF  در اين روش، هر لحظه سيلندر متقاضي كه به محل هد در آن لحظه نزديكتر باشد مورد پردازش قرار مي‌گيرد.

14. Disk Scheduling: SCAN The disk arm starts at one end of the disk, and moves toward the other end, servicing requests until it gets to the other end of the disk, where the head movement is reversed and servicing continues. Sometimes called the elevator algorithm

15. Disk Scheduling: C-SCAN  The head moves from one end of the disk to the other, servicing requests as it goes. When it reaches the other end, however, it immediately returns to the beginning of the disk, without servicing any requests on the return trip  Treats the cylinders as a circular list that wraps around from the last cylinder to the first one

16. C-LOOK  Arm only goes as far as the last request in each direction, then reverses direction immediately, without first going all the way to the end of the disk  Total number of cylinders?

17. مثال فرض کنید یک دیسک 200 شیار داشته و صف درخواست دیسک درخواستهای Random را در خود دارد. شیار (track)های در خواست شده به ترتیب دریافت عبارتند از : 55, 58, 39,18,90,160,38,184 در هر یک از حالات زیر: اگر زمان حرکت از شیار به شیار دیگر 4 میلی ثانیه طول بکشد، و بازوی دیسک در ابتدا در روی شیار 100 قرار داشته باشد :  ترتیب سرویس دهی به در خواستها و طول متوسط Seek چقدر است ؟  کل زمان جستجو چقدر خواهد بود ؟  الف ) از روش FIFO استفاده کنید.  ب ) از الگوریتم SSTF استفاده شود.  ج ) از الگوریتم آسانسور استفاده شود.  د ) از الگوریتم C-SCAN استفاده شود.

18. فرض کنید در روش scan و c-scan جهت اولیه حرکت به سمت افزایش شماره شیار می باشد.

19. request in queue 100 55 58 39 18 90 160 150 38 184 order of service 0 1 2 3 4 5 6 7 8 9 FIFO تعداد track های پیموده شده 0 45 3 19 21 72 70 10 112 146 55.3 Average Seek Length =تعداد track های پیموده498

20. زمان جستجو کل = =تعداد track های پیموده498 498 * 4 (ms) =1992 msec FIFO

21. order of service request serviced تعداد track های پیموده شده 0100 19010 25832 3553 43916 5381 61820 7150132 816010 918424 زمان جستجو کل = =تعداد track های پیموده248 248 * 4 (ms) = 299 msec 27.5 Average Seek Length SSTF

22. order of service request serviced تعداد track های پیموده شده 0100 19010 25832 3553 43916 5381 61820 7150132 816010 918424

23. request in queue 100 55 58 39 18 90 160 150 38 184 order of service 0 1 2 3 4 5 6 7 8 9 request serviced 27.8 Average Seek Length 100 150 160 184 90 58 55 39 38 18 SCAN or ELEVATOR آسانسور)) زمان جستجو کل = =تعداد track های پیموده250 250 * 4 (ms) = 1000 msec تعداد track های پیموده شده 50 10 24 94 32 3 16 1 20

24. order of service request serviced 0100 1150 2160 3184 490 558 655 739 838 918 SCAN or ELEVATOR آسانسور))

25. order of service request serviced تعداد track های پیموده شده 0100 115050 216010 318424 418166 53820 6391 75516 8583 99032 35.8 Average Seek Length زمان جستجو کل = =تعداد track های پیموده322 322 * 4 (ms) = 1288 msec C-SCAN

26. order of service request serviced تعداد track های پیموده شده 0100 115050 216010 318424 418166 53820 6391 75516 8583 99032

28. 28 QuestionsQuestions