john@moncol.UUCP (John Ruschmeyer) (09/25/85)
Does anyone know what heuristic is used by the Norton Utilities' 'si' program to determine the performance ratio vs. the IBM-PC? After installing a V20 in my Tandy 1000, the figure jumped from 1.0 to 1.8 (an AT&T 6300 with an 8mhz 8086 only rates 1.9). Meanwhile most other benchmarks showed only about a 10% increase in speed. Odd! Thanks in advance. -- Name: John Ruschmeyer US Mail: Monmouth College, W. Long Branch, NJ 07764 Phone: (201) 222-6600 x366 UUCP: ...!vax135!petsd!moncol!john ...!princeton!moncol!john ...!pesnta!moncol!john Disclaimer: Monmouth College is a mecca for diverse opinions. It is, therefore, highly unlikely that the above opinions are those of anyone but me. "Are we gonna be starcrossed lovers or just good friends?"
bill@hp-pcd.UUCP (bill) (09/27/85)
I don't know how much I believe that performance index. I've got a PC-AT with a 15 Mhz crystal (i.e., system running at 7.5 Mhz instead of normal 6 Mhz) for which Norton gives a performance index of 7.1 relative to a normal IBM/PC. This AT may be fast, but I have a hard time believing it's *that* fast! bill frolik hp-pcd!bill
faisal@smu (09/30/85)
Norton's "si"
x = the amount of time Norton's IBM PC took to perform the loop below;
t1 = current time;
do a bunch of times
{
mul
div
}
t2 = current time;
index = x / (t2 - t1);
I disassembled his code a few months ago; I don't remember the gritty
details, but this is pretty close.
Ralph Cramden tought Norton everything he knows.
Faisal , CS dept. SMU, Dallas, TX