Download Computer Systems Architecture: Lecture 17 - Storage Systems and Disks - Prof. Alan L. Suss and more Assignments Computer Science in PDF only on Docsity! CMSC 411 Computer Systems Architecture Lecture 17 Storage Systems Alan Sussman l @ d da s cs.um .e u Administrivia • Homework #4 due Tuesday • Exam #2 on Thursday April 23 • Cache simulator project questions? • Finish reading Ch. 6 (not 6.5) • Research Pizza lunch Wed., April 15, AVW 3258, if you’re interested in doing research with a CS faculty member CMSC 411 - 17 (some from Patterson, Sussman, others) 2 Storage systems • We already know about four levels of storage: – registers – cache – memory – disk b t ' b littl h th d i• u we ve een a e vague on ow ese ev ces are interconnected • In this unit, we study i t/ t t it h di k d t– npu ou pu un s suc as s s an apes – buses to connect storage devices – I/O performance issues – design of file systems (won’t talk much about this) CMSC 411 - 17 (some from Patterson, Sussman, others) 3 Disk and Tape Technologies (Hard) Disks • What it is: – a collection of 1-20 platters (like 2-sided CD's) – between 1 and 8 inches in diameter – 2.5 & 3.5 inch most common today rotating on a central spindle– – with 500-2500 tracks on each surface – divided into (maybe) 64 sectors ld di k ll t k h th b f t» o er s s: a rac s ave e same num er o sec ors » current disks: outer tracks have more sectors • larger diameter: best retrieval times • smaller diameter: cheaper and uses less power • Disk controller provides access to 1 or more disks CMSC 411 - 17 (some from Patterson, Sussman, others) 5 Disks (cont.) Used for• – file storage – slowest level of virtual memory during program execution Fig 7.1 from H&P 3ed. CMSC 411 - 17 (some from Patterson, Sussman, others) 6 Disks (cont.) • How information is retrieved by disk controller: – Wait for previous requests to be filled Time = queuing delay – A movable arm is positioned at the correct cylinder Time = seek time – The system waits for the correct sector to appear under the arm Time = rotational latency – Then a magnetic head senses » the sector number » the information recorded in the sector » an error correction code – and the information is transferred to a buffer Time = transfer time – The disk controller may impose some extra overhead Time = controller time CMSC 411 - 17 (some from Patterson, Sussman, others) 7 Disks (cont.) • Because all of this is so expensive, disk controller might also read the next sector or two, h i th t th t i f ti d d iop ng a e nex n orma on nee e s located there (prefetch or read ahead) CMSC 411 - 17 (some from Patterson, Sussman, others) 8
