Path: utzoo!attcan!uunet!cs.utexas.edu!tut.cis.ohio-state.edu!ucbvax!inmos.co.uk!des
From: des@inmos.co.uk (David Shepherd)
Newsgroups: comp.sys.transputer
Subject: Re: TDS Questions, and the Speed of Light on a T800
Message-ID: <13814.9002061032@elberton.inmos.co.uk>
Date: 6 Feb 90 10:32:54 GMT
References: <268:vorbrueggen@mpih-frankfurt.mpg.dbp.de>
Sender: daemon@ucbvax.BERKELEY.EDU
Organization: INMOS Limited, Bristol, UK.
Lines: 111

In reply to article <268:vorbrueggen@mpih-frankfurt.mpg.dbp.de>
>Hello Transputer wizards,

hi!

>3. Now comes the strange part. In Parallelogram 17/89, Jerry Sanders
>calls peak theoretical performance "only that the manufacturer
>guarantees that programmes will not exceed these rates, a speed of
>light for a computer". Just to play around (and to see whether I could
>beat Topexpress's Veclib), I wrote an assembler routine for the T800 to
>compute the magnitude of an n-dimensional vector. The core of the 
>instructions per vector element are:
>
>         fpldnlsn      ( 3)
>         fpdup         ( 1)
>         fpmul         (11)
>         fpadd         ( 7)
>
>The numbers are the processor cycles required, according to the
>Transputer Instruction Set. If you add them up, you get 22 cycles
>per element - assuming all code and operands on-chip, and *not*
>accounting for the fact that they are long instructions, i.e.,
>require a prefix to build the operation code. This should be overlapped
>for the load (with the preceding add) and the add (with the preceding
>multiply), but not for the duplicate and the multiply. Thus we expect
>two cycles in addition. 

I think the prefix for the fpdup will get overlapped (I can't remember
exactly how that overlappping works - that is if I ever understood it !)

> ..... This leaves us with 160 cycles per 8 elements or 20 cycles per
>element. We've beaten the speed of light of the T800! 
>
>BTW, the times given for the fpadd and fpmul instructions are quite
>consistent with other documentation. 

glad to hear this :-)

However if you look carefully at the Transputer Instruction Set book
you will find the following paragraph

	The time taken to execute many floating point instructions
	is extremely data dependent. The timings given here are
	typical times for such instructions, such as arithmetic on
	normalised data. Certain values in the registers may cause
	execution to take more or less time.

basically your problem is due to the fact that the time for fpadd its
*totally* data dependent. 7 cycles is on the pessimistic side and 5 or 6
is probably more likely for data that doesn't involve denorms.

>Communicating Process Architecture, the T800 multiplies three bits per
>cycle and normalises in two cycles. This gives: 
>
>fpadd: 1 (get inst) + 2 (denormalize) + 1 (add) + 2 (normalize result) = 6
                          ^^^^^^^^^^^                 ^^^^^^^^^^^^^^^^
these ought to read       align fractions             round result
        ^^^^^^^^^^^
        get inst shouldn't be here - we hope this is overlapped.

>fpmul: 1 (get inst) + 8 (24 bits/3 multiply) + 2 (normalize result) = 11
        ^^^^^^^^^^^^   ^^^^^^^^^^^^^^^^^^^^^^     ^^^^^^^^^^^^^^^^^
        shouldn't be counted here                 round result
                       should be 9 as you need to generate 26+ bits for rounding

>Now the question goes to somebody who knows about the internal workings
>of the T800: Why do I get these results?

see above - fpu timings very data dependant -- eg arithmetic involving
0s. infs or NaNs typically takes 3 cycles for all instructions as the
result is immediate. Other timings vary depending on whether denorms
need to be normalized (mul and div) or fractions need to be aligned (add
and sub) etc.

>4. Lastly, I'd like to know why some instructions have been done the
>way they are. I can understand that cj (conditional jump) behaves as a
>`jump zero' or `jump false', that's a matter of taste - a RISC
>processor just doesn't have the orthogonal instruction set of a CISC.
>But why, for whoever's sake, does it remove the operand if not jumping
>and leave the zero when jumping?

a very good question! i believe someone was being over zealous in tuning
the instructions to implement short circuit boolean evalution in an
optimal manner.!

> Most of the time, I find myself
>preceding it with a dup so I have a copy of my laboriously computed
>loop count after I've checked whether it's zero! Well, of course the
>zero is easier to remove (with a diff) than a non-zero, but in that
>case, we could invert the result of the comparison.

well, at least you have a dup instruction now :-)

>In general, I find the selection of direct and short instructions very
>reasonable. Exceptions are the startp and endp instructions: Why do
>they, taking 10 and 11 cycles, get the privilege of a one-byte
>instruction, while the much more useful dup now takes two cycles just
>because it has to be a two-byte instruction? (Why dup wasn't there
>from the start, but was added as an afterthought to the T800, would
>probably make an amusing historical anecdote. Does anybody know the
>reason?)

again i believe someone said that it wouldn't be needed for an occam
compiler.

david shepherd
INMOS ltd

disclaimer: all the above is probably based on personal opinion and bias
and should not be taken as representing anything approaching official
INMOS support, opinion or whatever .... etc.