Question

Average and minimum times for reading and writing to storage devices common measurements used to compare devices. Using techniques from Chap- ter 6, calculate values related to read and write time for disks with the following characteristics are RPM Controller Transfer Rate Average Seek Time Disk Transfer Rate 100 MB/s 90 MB/s 10 ms 7500 a. b. 200 MB/s 7 ms 40 MB/s 10,000 6.3.1 [10] <6.2, 6.3> Calculate the average time to read or write a sector for each disk listed in the table 1024-byte 6.3.2 10 6.2, 6.3> Calculate the minimum time to read or write a 2048-byte sector for each disk listed in the table. 6.3.3 10 6.2, 6.3> For each disk in the table, determine the dominant factor for performance. Specifically, if you could make an improvement to any aspect of the disk, what would you choose? If there is no dominant factor, explain why.

Answer 1

GIvemthat veading Average and minimum Times -fov Sol G.3.1fo machine a Jo vead he dda Coekime Average Yolational lat ency +- Average Time will be Controlle ove hea -inme YansferTime t C60/500) 4 ms Aveiage Yotational tcncy Transfer Ime = (024 /Co 2 20)= o.010850 69 Ms C100 2 20)= 0-D0116562 ms Comtroller oveihead Time 024 -Avevage ime 0 4o-0lD8506+ o. oo16562-- 14.0206)631 mg for machine To ead -Avevage ime = 7* 10^ -3 ) + * G0/1DC00)-t lo24/C402^20)tl024/ C2 002A 20) 6.3.2 Y oading minimum Time will be achived uhen ouf t1 ead foy lands atIhe right Ceutoi Thii way we Cam the data direty Remove the time So minimumlime for a minimum ime = lo+ o.ol 085069t 0.o091656 2 = 10.0206163) amd -for ma Chine b. |(40 2 20) lo24/20320) minimum Time e T* (O^ E 3)t (024 6.3.3 he deminant -faetor Ps Sek -time and radationml -Time for Peyf'omante ifusCem dec e4 -taht Cameho we Can fmprove oul Performanl vey efficientıy