turc@stl-08sima.army.mil (Andrea K. Turczynski) (11/29/89)
Hello, I am looking for Unix benchmark products, commercial and/or public domain. If anyone knows of any or has used any, please let me know. Thanks ahead of time. A. K. Turczynski
frech@mwraaa.army.mil (Norman R. Frech CPLS) (07/17/90)
I just finished running the byte benchmarks on our 5000/80. Does anyone
have any information on how to apply the results? Some of the output is
as follows:
Start Benchmark Run (BYTE Version 2.6)
Mon Jul 16 10:32:33 CDT 1990
8 interactive users.
=============================================================================
Dhrystone 2 without register variables
-----------------------------------------------------------------------------
Arithmetric Geometric Variance
Mean Mean (6 tests)
Microseconds/loop: 374 374 6.40
Dhrystones/sec: 2674 2674 329.48
=============================================================================
Dhrystone 2 using register variables
-----------------------------------------------------------------------------
Arithmetric Geometric Variance
Mean Mean (6 tests)
Microseconds/loop: 347 347 7.60
Dhrystones/sec: 2882 2882 525.56
End Benchmark Run (Mon Jul 16 10:32:33 CDT 1990) ....bradley@grip.cis.upenn.edu (John Bradley) (08/08/90)
Can anybody point me towards some publicly available floating point benchmarks, written in C. I've got some hardware I'd like to compare. The code should be non-trivial (ie, more than just a couple of lines), but it shouldn't use large amounts of memory. Please EMAIL any responses. John Bradley - University of Pennsylvania - bradley@cis.upenn.edu
guineau@wjg.enet.dec.com (08/09/90)
In article <27869@netnews.upenn.edu>, bradley@grip.cis.upenn.edu (John Bradley) writes: |> From: bradley@grip.cis.upenn.edu (John Bradley) |> Newsgroups: comp.unix.questions,sci.math,sci.math.num-analysis,comp.lang.c |> Subject: Benchmarks |> Reply-To: bradley@grip.cis.upenn.edu (John Bradley) |> |> Can anybody point me towards some publicly available floating point benchmarks, |> written in C. I've got some hardware I'd like to compare. The code should be |> non-trivial (ie, more than just a couple of lines), but it shouldn't use |> large amounts of memory. |> |> Please EMAIL any responses. |> |> John Bradley - University of Pennsylvania - bradley@cis.upenn.edu |> (I've mailed them to him, but for others...) Following is WHETS.FOR and following that is WHETS.C The fortran one is the whetstone FP benchmark. The C version is one I haved from the fortran code. It's meant as a comparison only. -- W. John Guineau guineau@wjg.enet.dec.com Digital Equipment Corporation Marlboro MA. 01752
guineau@wjg.enet.dec.com (08/09/90)
In article <14425@shlump.nac.dec.com>, I write: > |> From: guineau@wjg.enet.dec.com > |> Newsgroups: comp.unix.questions,sci.math,sci.math.num-analysis,comp.lang.c > |> Subject: Re: Benchmarks > |> Reply-To: guineau@wjg.enet.dec.com > > |> (I've mailed them to him, but for others...) |> > > |> Following is WHETS.FOR and following that is WHETS.C Well, I kinda forgot to post the actual source. Here it is: WHETS.FOR C----------------------------------------------------------------------------- C C Copy this if you like. I don't think it's the "Official" VAX C CPU Whetstone benchmark, but is close. John C C WHETS.FOR 09/27/77 TDR C ...WHICH IS AN IMPROVED VERSION OF C WHET1A.FTN 01/22/75 RBG C SINGLE-PRECISION VARIANT OF PROGRAM C C THIS PROGRAM IS THE C "WHETSTONE INSTRUCTIONS PER SECONDS" MEASURE OF FORTRAN C AND CPU PERFORMANCE. C C IT WAS DEVELOPED BY THE BRITISH CENTRAL COMPUTER AGENCY AND C OBTAINED BY A ROUNDABOUT MEANS FROM A CUSTOMER WHO RECEIVED C A LISTING OF THE SOURCE PROGRAM FROM DG MARKETING. C DIMENSION TIMES(3) C C COMMON WHICH REFERENCES LOGICAL UNIT ASSIGNMENTS C COMMON /LUNS/ ICRD,ILPT,IKBD,ITTY C COMMON T,T1,T2,E1(4),J,K,L T=0.499975E00 T1=0.50025E00 T2=2.0E00 TYPE 1 ACCEPT 2,I 1 FORMAT(' TYPE LOOP COUNT (I4 FORMAT)'/) 2 FORMAT(I4) C C C C ***** BEGININNING OF TIMED INTERVAL ***** DO 200 ILOOP = 1,3 I = ILOOP*100 TIMES(ILOOP) = SECNDS(0.) C ***** ***** C ISAVE=I N1=0 N2=12*I N3=14*I N4=345*I N5=0 N6=210*I N7=32*I N8=899*I N9=616*I N10=0 N11=93*I N12=0 X1=1.0E0 X2=-1.0E0 X3=-1.0E0 X4=-1.0E0 IF(N1)19,19,11 11 DO 18 I=1,N1,1 X1=(X1+X2+X3-X4)*T X2=(X1+X2-X3+X4)*T X4=(-X1+X2+X3+X4)*T X3=(X1-X2+X3+X4)*T 18 CONTINUE 19 CONTINUE CALL POUT(N1,N1,N1,X1,X2,X3,X4) E1(1)=1.0E0 E1(2)=-1.0E0 E1(3)=-1.0E0 E1(4)=-1.0E0 IF(N2)29,29,21 21 DO 28 I=1,N2,1 E1(1)=(E1(1)+E1(2)+E1(3)-E1(4))*T E1(2)=(E1(1)+E1(2)-E1(3)+E1(4))*T E1(3)=(E1(1)-E1(2)+E1(3)+E1(4))*T E1(4)=(-E1(1)+E1(2)+E1(3)+E1(4))*T 28 CONTINUE 29 CONTINUE CALL POUT(N2,N3,N2,E1(1),E1(2),E1(3),E1(4)) IF(N3)39,39,31 31 DO 38 I=1,N3,1 38 CALL PA(E1) 39 CONTINUE CALL POUT(N3,N2,N2,E1(1),E1(2),E1(3),E1(4)) J=1 IF(N4)49,49,41 41 DO 48 I=1,N4,1 IF(J-1)43,42,43 42 J=2 GOTO44 43 J=3 44 IF(J-2)46,46,45 45 J=0 GOTO47 46 J=1 47 IF(J-1)411,412,412 411 J=1 GOTO48 412 J=0 48 CONTINUE 49 CONTINUE CALL POUT(N4,J,J,X1,X2,X3,X4) J=1 K=2 L=3 IF(N6)69,69,61 61 DO 68 I=1,N6,1 J=J*(K-J)*(L-K) K=L*K-(L-J)*K L=(L-K)*(K+J) E1(L-1)=J+K+L E1(K-1)=J*K*L 68 CONTINUE 69 CONTINUE CALL POUT(N6,J,K,E1(1),E1(2),E1(3),E1(4)) X=0.5E0 Y=0.5E0 IF(N7)79,79,71 71 DO 78 I=1,N7,1 X=T*ATAN(T2*SIN(X)*COS(X)/(COS(X+Y)+COS(X-Y)-1.0E0)) Y=T*ATAN(T2*SIN(Y)*COS(Y)/(COS(X+Y)+COS(X-Y)-1.0E0)) 78 CONTINUE 79 CONTINUE CALL POUT(N7,J,K,X,X,Y,Y) X=1.0E0 Y=1.0E0 Z=1.0E0 IF(N8)89,89,81 81 DO 88 I=1,N8,1 88 CALL P3(X,Y,Z) 89 CONTINUE CALL POUT(N8,J,K,X,Y,Z,Z) J=1 K=2 L=3 E1(1)=1.0E0 E1(2)=2.0E0 E1(3)=3.0E0 IF(N9)99,99,91 91 DO 98 I=1,N9,1 98 CALL P0 99 CONTINUE CALL POUT(N9,J,K,E1(1),E1(2),E1(3),E1(4)) J=2 K=3 IF(N10)109,109,101 101 DO 108 I=1,N10,1 J=J+K K=J+K J=J-K K=K-J-J 108 CONTINUE 109 CONTINUE CALL POUT(N10,J,K,X1,X2,X3,X4) X=0.75E0 IF(N11)119,119,111 111 DO 118 I=1,N11,1 118 X=SQRT(EXP(ALOG(X)/T1)) 119 CONTINUE CALL POUT(N11,J,K,X,X,X,X) C C ***** END OF TIMED INTERVAL ***** 200 TIMES(ILOOP)=SECNDS(TIMES(ILOOP)) C C WHET. IPS = 1000/(TIME FOR 10 ITERATIONS OF PROGRAM) WHETS = 10000./(TIMES(3)-TIMES(2)) TYPE 201,WHETS 201 FORMAT(' SPEED IS: ',F8.0,' THOUSAND WHETSTONE', 2 ' SINGLE PRECISION INSTRUCTIONS PER SECOND') C ***** ***** C STOP END SUBROUTINE PA(E) COMMON T,T1,T2 DIMENSION E(4) J=0 1 E(1)=(E(1)+E(2)+E(3)-E(4))*T E(2)=(E(1)+E(2)-E(3)+E(4))*T E(3)=(E(1)-E(2)+E(3)+E(4))*T E(4)=(-E(1)+E(2)+E(3)+E(4))/T2 J=J+1 IF(J-6)1,2,2 2 CONTINUE RETURN END SUBROUTINE P0 COMMON T,T1,T2,E1(4),J,K,L E1(J)=E1(K) E1(K)=E1(L) E1(L)=E1(J) RETURN END SUBROUTINE P3(X,Y,Z) COMMON T,T1,T2 X1=X Y1=Y X1=T*(X1+Y1) Y1=T*(X1+Y1) Z=(X1+Y1)/T2 RETURN END SUBROUTINE POUT(N,J,K,X1,X2,X3,X4) C C WRITE STATEMENT COMMENTED OUT TO IMPROVE REPEATABILITY OF TIMINGS C C WRITE(2,1)N,J,K,X1,X2,X3,X4 1 FORMAT(1H,3I7,4E12.4) RETURN END WHETS.C #include <stdio.h> #include <time.h> /* C <<< SQM::DISK$TSDPERF:[NOTES$LIBRARY]VMSTUNING.NOTE;1 >>> C -< ** VMS Tuning Tricks ** >- C ======================================================================= C Note 36.4 Whetstone Benchmark C AURORA::HALLYB 210 lines 19-SEP-1985 C----------------------------------------------------------------------------- C C Copy this if you like. I don't think it's the "Official" VAX C CPU Whetstone benchmark, but is close. John C C WHETS.FOR 09/27/77 TDR C ...WHICH IS AN IMPROVED VERSION OF C WHET1A.FTN 01/22/75 RBG C SINGLE-PRECISION VARIANT OF PROGRAM C C THIS PROGRAM IS THE C "WHETSTONE INSTRUCTIONS PER SECONDS" MEASURE OF FORTRAN C AND CPU PERFORMANCE. C C IT WAS DEVELOPED BY THE BRITISH CENTRAL COMPUTER AGENCY AND C OBTAINED BY A ROUNDABOUT MEANS FROM A CUSTOMER WHO RECEIVED C A LISTING OF THE SOURCE PROGRAM FROM DG MARKETING. C DIMENSION TIMES(3) C C COMMON WHICH REFERENCES LOGICAL UNIT ASSIGNMENTS C */ int ICRD,ILPT,IKBD,ITTYC; int J,K,L; double T,T1,T2,E1[5]; int N1,N2,N3,N4,N5,N6,N7,N8,N9,N10,N11,N12; double X1,X2,X3,X4; double X,Y,Z; long TIMES[4],start,end; main() { char buf[10]; int n,I,ILOOP,ISAVE; T=0.499975E00; T1=0.50025E00; T2=2.0E00; n=0; while((n<1)||(n>3)) { printf("Loop Count (1-3)?"); gets(buf); n = atoi(buf); }; /* ***** BEGININNING OF TIMED INTERVAL ***** */ for(ILOOP=1 ; ILOOP <= n ; ILOOP++) { printf("LOOP: %d\n",ILOOP); I = ILOOP*100; time(&start); ISAVE = I; N1 = 10*I; N2 = 12*I; N3 = 14*I; N4 = 345*I; N5 = 0; N6 = 210*I; N7 = 32*I; N8 = 899*I; N9 = 616*I; N10 = 0; N11 = 93*I; N12 = 0; X1 = 1.0E0; X2 = -1.0E0; X3 = -1.0E0; X4 = -1.0E0; printf(" N1...\n"); if(N1>0.0) for(I=1 ; I<N1 ; I++) { X1=(X1+X2+X3-X4)*T; X2=(X1+X2-X3+X4)*T; X4=(-X1+X2+X3+X4)*T; X3=(X1-X2+X3+X4)*T; }; E1[1] = 1.0E0; E1[2] = -1.0E0; E1[3] = -1.0E0; E1[4] = -1.0E0; printf(" N2...\n"); if(N2>0.0) for(I=1 ; I<N2 ; I++) { E1[1]=(E1[1]+E1[2]+E1[3]-E1[4])*T; E1[2]=(E1[1]+E1[2]-E1[3]+E1[4])*T; E1[3]=(E1[1]-E1[2]+E1[3]+E1[4])*T; E1[4]=(-E1[1]+E1[2]+E1[3]+E1[4])*T; }; printf(" N3...\n"); if(N3>0.0) for(I=1 ; I<N3 ; I++) PA(); J=1; printf(" N4...\n"); if(N4>0.0) { for(I=1 ; I<N4 ; I++) { if((J-1) == 0) J=2; else J=3; if((J-2)>0) J=0; else J=1; if((J-1)<0) J=1; else J=0; }; }; printf(" N5...\n"); J=1; K=2; L=3; printf(" N6...\n"); if(N6>0.0) for(I=1 ; I<N6 ; I++) { J=J*(K-J)*(L-K); K=L*K-(L-J)*K; L=(L-K)*(K+J); E1[L-1]=J+K+L; E1[K-1]=J*K*L; }; X = 0.5E0; Y = 0.5E0; printf(" N7...\n"); if(N7>0.0) for(I=1 ; I<N7 ; I++) { X=T * atan(T2 * sin(X) * cos(X) / (cos(X+Y)+cos(X-Y)-1.0E0)); Y=T * atan(T2 * sin(Y) * cos(Y) / (cos(X+Y)+cos(X-Y)-1.0E0)); }; X = 1.0E0; Y = 1.0E0; Z = 1.0E0; printf(" N8...\n"); if(N8>0.0) for(I=1 ; I<N8 ; I++) P3(X,Y,Z); J = 1; K = 2; L = 3; E1[1] = 1.0E0; E1[2] = 2.0E0; E1[3] = 3.0E0; printf(" N9...\n"); if(N9>0.0) for(I=1 ; I<N9 ; I++) P0(); J = 2; K = 3; printf(" N10...\n"); if(N10>0.0) for(I=1 ; I<N10 ; I++) { J=J+K; K=J+K; J=J-K; K=K-J-J; }; X=0.75E0; printf(" N11...\n"); if(N11>0.0) for(I=1 ; I<N11 ; I++) X=sqrt(exp(log(X)/T1)); /* ***** END OF TIMED INTERVAL ***** */ time(&end); TIMES[ILOOP-1] = end-start; printf("Time = %d seconds\n",TIMES[ILOOP-1]); }; /* WHET. IPS = 1000/(TIME FOR 10 ITERATIONS OF PROGRAM) */ /* WHETS = 10000./(TIMES(3)-TIMES(2)) */ } PA() { int J; for(J=1 ; J<7 ; J++) { E1[1]=(E1[1]+E1[2]+E1[3]-E1[4])*T; E1[2]=(E1[1]+E1[2]-E1[3]+E1[4])*T; E1[3]=(E1[1]-E1[2]+E1[3]+E1[4])*T; E1[4]=(-E1[1]+E1[2]+E1[3]+E1[4])/T2; }; return(0); } P0() { E1[J]=E1[K]; E1[K]=E1[L]; E1[L]=E1[J]; return(0); } P3(X,Y,Z) double X,Y,Z; { double X1,Y1; X1=X; Y1=Y; X1=T*(X1+Y1); Y1=T*(X1+Y1); Z=(X1+Y1)/T2; return(0); } -- W. John Guineau guineau@wjg.enet.dec.com Digital Equipment Corporation Marlboro MA. 01752