CIS 415 Operating Systems - Spring 2003

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Discussion: Week 8 - Disk Systems - Solutions

Disk Transfer Rate

Suppose a hard disk drive has the following characteristics:

2 platters (4 heads = 4 tracks/cylinder)
16383 cylinders
64 sectors/track
512 bytes/sector
Rotational speed = 7200 rpm = 120 rps
Average seek time: 10ms
Adjacent cylinder seek time: 1ms
Head switch time: 0.1ms

How much data can this disk hold?
b = 16383 * 4 * 64 * 512 = 2,147,352,576 bytes = 1.9999GB

Calculate the sustained transfer rate (no initial seek, sequential access across entire disk).
For every track, the disk must spin twice - once to line up the data and once to seek to the next track. The disk spins once every 8.3ms, and sectors go by in 0.13ms. That means that even with only a head switch, the first sector on the next track will already be past after the head switch and we need the rest of the rotational delay anyway.
Transfer time T = (16383*(4+4)*T_r) = 1092.2 sec
Transfer rate R_S = b/T = 1,966,080 bytes/sec = 1.875MB/sec

Now suppose that the last sector on each track is left blank, so in essence we now have 63 sectors per track with a one-sector gap at the end of the track. How much data can this new disk hold?

Calculate the sustained transfer rate for this 63-sector/track disk.
Now a head switch can occur during the time it takes to traverse the gap, eliminating the rotational delay due to head switching. A full rotational delay is still necessary for adjacent cylinder seeks.
Transfer time T = 16384*(4+1)*T_r = 682.625 sec.
Transfer rate R_S = b/T = 3,145,728 bytes/sec = 3MB/sec

RAID

Here is a summary of the different RAID levels for your convenience:

Data striped across n disks, no redundancy
RAID 0 + mirrors (n extra disks)
RAID 0 + error-correcting codes (log n extra disks)
RAID 0 + parity (1 extra disk)
RAID 3 with large strips (block-level parity, 1 extra disk)
RAID 4 with parity strips distributed among all disks (1 extra disk)
RAID 5 with dual parity (2 extra disks)

What RAID level would you use for the following applications and why?
There are many answers to this question. These are what I would use.

Streaming media internet server
RAID 4 - Fast for multiple connections, good availability, and we don't care about the write cost.
Bank record database server
RAID 6 - Fast for multiple transactions, best failure tolerance, doesn't have RAID 1's complete synchronizion requirement for writing.
Small business video editing box
RAID 0 - Low cost, high performance for one application, data loss is inconvenient but not catastrophic, as the video exists elsewhere, such as on DV cameras.

Suppose disk 2 goes down in a 5-disk RAID 3 array. The contents of the remaining disks are:

Reconstruct X2(i).

    11001001
    10101010
    11111111  (remember XOR is associative and commutative, like +)
XOR 00000000
------------
    10011100

Disk Scheduling

Schedule the following sequence of requests using FCFS, SSTF and SCAN:
(Assume the starting position is 100 in the "up" direction)

	FCFS	SSTF	SCAN
110	110	110	110
5	5	57	176
32	32	48	205
205	205	32	57
176	176	23	48
3	3	5	32
1	1	3	23
57	57	1	5
23	23	176	3
48	48	205	1
total seek length:	634	323	309