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- HARD DISK PERFORMANCE vs. INTERLEAVE FACTOR
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- System Interleave factors 2 3 4 5 6
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- Theoretical
- Maximum 261K 174K 131K 104K 87K
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- 80286 measured (+) 243K 162K
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- 8086 measured 26K 28K 100K 80K
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- -----------------------------------------------------------------------
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- 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.