Path: utzoo!utgpu!news-server.csri.toronto.edu!clyde.concordia.ca!uunet!decwrl!bacchus.pa.dec.com!shlump.nac.dec.com!wjg.enet.dec.com!guineau From: guineau@wjg.enet.dec.com Newsgroups: comp.unix.questions Subject: Re: Benchmarks Message-ID: <14437@shlump.nac.dec.com> Date: 9 Aug 90 13:32:31 GMT References: <14425@shlump.nac.dec.com> <27869@netnews.upenn.edu> Sender: newsdaemon@shlump.nac.dec.com Reply-To: guineau@wjg.enet.dec.com Distribution: na Organization: Digital Equipment Corporation, Marlboro, MA Lines: 534 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 #include /* 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 ; I0.0) for(I=1 ; I0.0) for(I=1 ; I0.0) { for(I=1 ; I0) 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 ; I0.0) for(I=1 ; I0.0) for(I=1 ; I0.0) for(I=1 ; I0.0) for(I=1 ; I0.0) for(I=1 ; I