fa10-sl-s2 - ECE 3035 4 problems 7 pages Computing Mechanisms Exam Two Solutions Fall 2010 27 October 2010 Disk Organization Problem 1(3 parts 34 points

fa10-sl-s2 - ECE 3035 4 problems 7 pages Computing...

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ECE 3035 Computing Mechanisms Fall 2010 4 problems, 7 pages Exam Two Solutions 27 October 2010 Problem 1 (3 parts, 34 points) Disk Organization A hard disk has the following specifications. recording density 1M bits/cm track density 128K tracks/cm average track circumference 16 cm usable radial disk size 4 cm maximum track circumference 32 cm rotation speed 128 revs/sec sector size 512 bytes number of platters 2 Part A (18 points) Complete the following disk summary. Use power of two approximation; but express the final answer in decimal (e.g., 16 instead of 2 4 ). Show work. areal density 1 Mbits/cm * 1 byte/8bits * 128K tk/cm = 2^(20-3+17) = 16Gbytes/cm 2 avg. track capacity 16cm/tk * 1Mbits/cm * 1byte/8bits = 2^(4+20-3) = 2 Mbytes avg. cylinder capacity 2M bytes * 4 tk/cyl = 8 Mbytes surface capacity (1 head) 2M bytes * 128K tk/cm * 4cm = 2^(21+17+2) = 1 Tbytes total disk capacity 1 Tbytes * 4 surfaces = 4 Tbytes max data transfer rate 32 cm/rev * 128 rev/sec * 1M bits/cm = 2^(5+7+20) = 4Gbits/sec Part B (12 points) The disk in Part A is a 3.74” diameter disk used in desktops. A laptop disk must satisfy a smaller form factor (2.5” diameter). This reduces the maximum track circumference to 16 cm, the average track circumference to 8 cm, and the usable radial disk size to 2 cm. All other physical parameters remain unchanged. Recompute the following parameters. total disk capacity 8cm/tk * 1Mbits/cm * 1byte/8bits * 128K tk/cm * 2cm* 4 surfaces = 2^(3+20-3+17+1+2) = 1 Tbytes max data transfer rate 16 cm/rev * 128 rev/sec * 1M bits/cm = 2^(4+7+20) = 2Gbits/sec Part C (4 points) A desktop 3.5” form factor 7200 RPM disk platter is typically 3.74” in diameter. A 10,000 RPM platter in the same form factor is typically 3.0” in diameter. A 15,000 RPM platter is typically 2.5” in diameter. If form factor is not responsible for this reduced platter diameter, why would disk designers adopt a smaller disk? Concisely explain the reasons.
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