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PsL Monthly 1995 April
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pslmonthlyvol3-4utilitiesapril1995.iso
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diskperf.txt
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1986-12-27
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HARD DISK PERFORMANCE vs. INTERLEAVE FACTOR
System Interleave factors 2 3 4 5 6
Theoretical
Maximum 261K 174K 131K 104K 87K
80286 measured (+) 243K 162K
8086 measured 26K 28K 100K 80K
-----------------------------------------------------------------------
Compaq Portable 27K 140K 110K 98K 77K
(An interesting exercise in technical logic is to determine why it is
that an 80286 running at 6Mhz outperforms the exact same machine running
at 8Mhz in terms of effective data transfer rates.)
The Compaq Portable is an obvious exception to the above in that it is
easily able to fully utilize an interleave factor of three.
CALCULATION OF DATA TRANSFER RATES
The above chart shows the significance of the interleave factor as it
relates to effective data transfer rates for all 5-1/4" hard disk
systems. The top curve represents the theoretical data transfer rate
available from hard disks at each interleave factor (2 to 6). The plus
sign (+) shows the actual measured data transfer rate using interleave
factors of 2 and 3 on a Compaq Deskpro 286 system. The lower line shows
actual measured data transfer rates on a Compaq Deskpro (not a 286) at
interleave factors of 3 to 6.
What is obvious from the chart is that the Deskpro 286 system should be
configured with an interleave factor of 2.
Most disturbing from the chart is the effect of setting up a Deskpro
(non-286) with an interleave factor of 3 as is the standard proceedure
utilized by many retailers. In essence, the resulting system is left
with a hard disk subsystem that is performing at 1/4th of it's ability.
By simply setting the interleave factor to 5 (NOT 6!!!) the effective
data transfer rate will increase from 25,500 bytes per second to
100,400.
Following is the method used in determining theoretical data transfer
rates (does not include control information transfer).
Assumptions:
Disk spin rate = 3600 RPM
Rated speed = 5000000 bits per second
Time to rotate once = 16.66666666 msecs (60/3600)
Maximum bits per track = 83333 (.016666666*5000000)
Number of sectors per track = 17
Maximum bytes per track = 10416 (83333/8)
Maximum bytes per sector = 612 (10416/17)
Therefore, bits per sector = 4901 (612*8)
Time to transmit 1 sector = 980.4 usecs (4901/5000)
Data bytes per sector = 512
Data transfer rate = 522240 Data bytes per second
(Assuming interleave factor of 1)
However, we have seen that between data bytes there are only 100 bytes
(800 bits) which takes nearly 160 usecs to pass under the heads at 3600
RPM. The 80286, using programmed I/O moves a word (2 bytes) in six
machine cycles. At 6Mhz, it takes 256 usecs to obtain the 512 data
bytes just read, thus, there is not enough time to do so and be ready
to read the next sector before that sector reaches the head. Therefore,
an interleave factor of at least 2 must be used. This provides nearly a
full millisecond (980.4 usecs) between the end of a sector and the
beginning of the next logical one.
That means that the Data transfer rate which is theoretically possible
at an interleave factor of 1 is cut in half when the factor becomes 2.
Therefore, the maximum theoretical data trasfer rate possible on such a
system is 261,120 bytes per second. Similarly, with an interleave
factor of three we must divide the maximum posible by 3 to yield 174,080
bytes per second, and so forth.