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.