type *.for

 
FACULTY:[IJL.COMPUTER.CS323.FORTRAN]ARNOLD.FOR;4
 
C	Stephanie Arnold
C	Project 1
C	This is a program that will calculate possible molecular combinations
C 	of Carbon, Oxygen, and Hydrogen, given the total weight.

	PROGRAM PROJ1A

C	WC = weight of carbon, WH = weight of hydrogen, WO = weight of oxygen,
C	CW = calculated total weight, OW = observed total weight,
C	ERROR = error tolerance,
C	NC = # carbon items, NH = # hydrogen atoms, NO = # oxygen items,
C	SOLN = the number of solutions that we have found
	DOUBLE PRECISION WC, WH, WO, CW, OW
	INTEGER NC, NH, NO, SOLN

C	Enter the known weights for C, H, and O
	DATA WC, WH, WO /12.001, 1.0079, 15.9994/
C	Set initial values for the everything else
	DATA CW, OW, ERROR, NC, NH, NO, SOLN /0.,0.,0.,0,1,0,0/

C	Read in the observed weight and error tolerance
 10	FORMAT (' Observed weight:     Error Tolerance:')
	WRITE (*,10)
 15	FORMAT (2D20.6)
	READ (*,15) OW,ERROR

C	Increment the number of oxygen atoms to include
C	This, in effect, the outer loop.
 30	NC = 0
	NO = NO + 1
C	If the molecule is too big already, then kick out of loop
	IF ((NO * WO + WC + WH - OW) .GT. OW) GOTO 115

C	Increment the number of carbon atoms - inner loop
 40	NC = NC + 1
C	If the molecule is too big, then kick out of loop
	IF ((NO * WO + NC * WC + WH - OW) .GT. OW) GOTO 115
	IF (NO .GT. (NC * 2)) GOTO 40

C	Calculate the number of hydrogen atoms
C	Until I find out how to round, I will just truncate
	NH = (OW - NO * WO - NC * WC) / WH
C	If there isn't at least one atom of H then go to outer loop
	IF (NH .LT. 1) GOTO 30
C	Find the calculated total weight
	CW = NH * WH + NO * WO + NC * WC

	IF (NH .GT. (NC * 4)) GOTO 40
	IF (ABS(OW - CW) .LE. ABS(ERROR)) THEN
	  GOTO 110 
	ELSE
	  GOTO 40
	ENDIF
 
C	We have a solution
 110	SOLN = SOLN + 1
	IF (SOLN .EQ. 1) CALL PRINT1(OW,ERROR)
	CALL PRINT2(CW,OW,NC,NH,NO)
	GOTO 40
C	We have no more solutions
 120  	FORMAT (' No solutions')
 115    IF (SOLN .EQ. 0) WRITE(*,120)

C	The End of the program
	END
C	The End of the program
        	
CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC
C 								    C
C		Subroutine that prints the solution		    C
C								    C
CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC

	SUBROUTINE PRINT1(OBSW,TLRNC)
 400	FORMAT (' Observed Molecular Weight:    ',D20.6)
	WRITE (*,400) OBSW
 500	FORMAT (' Experimental Error (+or-):    ',D20.6)
	WRITE (*,500) TLRNC
 600	FORMAT (' Possible Formula(s)    Calculated Weight   Error')
	WRITE (*,600)
 700	FORMAT (' ------------------     -----------------   -----')
	WRITE (*,700)
	RETURN
        END
C	End of subroutine print1

	SUBROUTINE PRINT2(CALCW,OBSW,C,H,O)
        INTEGER C, H, O
	DOUBLE PRECISION OE
	OE = OBSW - CALCW
 420	FORMAT (' C   H   O    ',2D20.6)
	WRITE (*,420), CALCW, OE
	N = C + H + O
 430	FORMAT (I3,2I4)
	WRITE (*,430), C, H, O
	RETURN
	END
C	End of the subroutine print2
 
FACULTY:[IJL.COMPUTER.CS323.FORTRAN]CONVEY.FOR;1
 
       PROGRAM CHEMISTRY
C
C      -- CURRENT VALUE FOR NUMBER OF EACH ATOM BEING CONSIDERED
       INTEGER CCOUNT, HCOUNT, OCOUNT
C
C      -- INDICATED WHETHER OR NOT ANY FORMULAE WERE FOUND
       LOGICAL EVENONE
C      
C      -- USER-ENTERED SAMPLE AND TOLLERANCE MASSES
       DOUBLE PRECISION SAMPLE, TOL
C
C      -- MIN AND MAX POSSIBLE OCCURANCES OF EACH ATOM IN THE SAMPLE
       INTEGER OMAX, HMAX, CMAX
C
C      -- INTERMEDIATE VALUES USED WHEN COMPUTING ATOMIC MASSES
       DOUBLE PRECISION DTEMP1, DTEMP2
C
C      *********************     PROGRAM CODE      **********************
C
C      -- CONSTANT DATA FOR MASSES OF SINGLE ATOMS
       DOUBLE PRECISION CMASS, HMASS, OMASS
       DATA CMASS, HMASS, OMASS / 12.011, 1.0079, 15.9994 /
C
C      -- GET PARAMETERS FROM USER, ECHO THEM BACK
10     CALL PARAMS(SAMPLE, TOL)
C
C      -- FIND OUT WHETHER OR NOT USER CHOSE THE 'QUIT' OPTION
       IF ((SAMPLE.EQ.0.0).OR.(TOL.EQ.0.0)) GOTO 500

C      -- PUT AN UPPER LIMIT ON THE LOOPING DONE
       CALL BOUNDS(CMAX, HMAX, OMAX, CMASS, HMASS, OMASS, SAMPLE)

C      -- SOLVE THE PROBLEM
       EVENONE = .FALSE.
       DO 200 CCOUNT = 1,CMAX,1
          DO 100 OCOUNT = 1,OMAX,1
C            -- TOTAL MASS USED BY CURRENT C, O GUESSES
             DTEMP1 = (CCOUNT * CMASS) + (OCOUNT * OMASS)
C            -- REMAINING MASS TO BE TAKEN UP BY HYDROGEN
             DTEMP2 = (SAMPLE - DTEMP1)
C            -- CLOSEST NUMBER OF H TO ACCOUNT FOR REMAINING MASS
             HCOUNT = IROUND(DTEMP2 / HMASS)
C            -- CHECK FOR ACCURACY OF GUESS
             DTEMP1 = (CCOUNT * CMASS) + (OCOUNT * OMASS) 
     2                + (HCOUNT * HMASS)
C            -- SKIP OUTPUT OF FORMULA IF IT ISN'T VALID
             IF ( (ABS(DTEMP1-SAMPLE).GT.TOL).OR.(HCOUNT.LT.0)
     2           .OR.((4*HCOUNT).GT.CCOUNT).OR.((2*OCOUNT).GT.CCOUNT))
     3          GOTO 50
             EVENONE = .TRUE.
             CALL WORKS(CCOUNT, HCOUNT, OCOUNT)
50     CONTINUE
100    CONTINUE
200    CONTINUE
C
C      -- IF NOT VALID FORMULAE, SAY SO
       IF (.NOT.EVENONE)
     2    WRITE(*,300)
300    FORMAT (/' SORRY, NO VALID COMBINATIONS WITHIN TOLERANCE'//)

C      -- LOOP BACK TO BEGINNING FOR MORE INPUT
       GOTO 10
500    CONTINUE
       END
C
C      ********************* FUNCTIONS/SUBROUTINES **********************
C
       SUBROUTINE PARAMS(S, T)
C      -- GETS THE INPUT PARAMETERS FROM THE USER
       REAL S, T
       WRITE(*, 600)
600    FORMAT (//' PLEASE ENTER SAMPLE ATOMIC MASS AND TOLLERANCE'/ )
       READ(*,*) S, T
       RETURN
       END       
C
C      --------------------------------------------------------------------
C
       SUBROUTINE BOUNDS(CMAX, HMAX, OMAX, CMASS, HMASS, OMASS, MOLMAS)
C      -- COMPUTES UPPER BOUNDS OF POSSIBLE PRESENCE OF EACH ELEMENT IN MOL.
       INTEGER CMAX, HMAX, OMAX
       REAL CMASS, HMASS, OMASS, MOLMAS
       CMAX = INT(MOLMAS / CMASS + 2.)
       HMAX = INT(MOLMAS / HMASS + 2.)
       OMAX = INT(MOLMAS / OMASS + 2.)
       RETURN
       END
C
C      --------------------------------------------------------------------
C
       INTEGER FUNCTION IROUND(X)
C      -- GIVEN A REAL, RETURNS NEAREST INTEGER
       DOUBLE PRECISION X
       IROUND = INT(X + .5)
       RETURN
       END
C
C      --------------------------------------------------------------------
C
       SUBROUTINE WORKS(C,H,O)
       INTEGER C, H, O
       WRITE (*,1000) C, H, O
1000   FORMAT (/' C = 'I4', H = 'I4', O = 'I4)
       RETURN
       END
 
FACULTY:[IJL.COMPUTER.CS323.FORTRAN]DEMO.FOR;5
 
 1	WRITE(6,*) '1234567.8901'
	READ(5,10,END=1000,ERR=100) FLOATV
 10	FORMAT(F12.4)
	WRITE(6,*) FLOATV
	WRITE(6,*) 'X1234567890'
	READ(5,20,END=1000,ERR=100) INTV
 20	FORMAT(1X,I10)
	WRITE(6,*) INTV
	GO TO 1
 100    WRITE(6,*) 'INPUT IS NOT VALID'
	GO TO 1
 1000	STOP 'END-OF-FILE SIGNALED'
	END
 
FACULTY:[IJL.COMPUTER.CS323.FORTRAN]DEMO2.FOR;2
 
	DO 100 I = 3.100
	PRINT *, DO100I
	END
 
FACULTY:[IJL.COMPUTER.CS323.FORTRAN]EQUIV.FOR;16
 
	REAL X
	INTEGER I
	LOGICAL L(4)
	CHARACTER * 1 C(4)

	EQUIVALENCE (X,I,L,C)

	X=42.0
 	WRITE(6,*) X
	WRITE(6,*) I
 	WRITE(6,*) L
 	WRITE(6,*) C

	I=42
 	WRITE(6,*) X
	WRITE(6,*) I
 	WRITE(6,*) L
 	WRITE(6,*) C

	L(1)=.TRUE.
	L(2)=.FALSE.
	L(3)=.FALSE.
	L(4)=.TRUE.
 	WRITE(6,*) X
	WRITE(6,*) I
 	WRITE(6,*) L
 	WRITE(6,*) C

	C(1) = 'J'
	C(2) = 'O'
	C(3) = 'H'
	C(4) = 'N'
 	WRITE(6,*) X
	WRITE(6,*) I
 	WRITE(6,*) L
 	WRITE(6,*) C

	END
	
 
FACULTY:[IJL.COMPUTER.CS323.FORTRAN]FORTRANQ.FOR;1
 
	PROGRAM Quiz


C DECLARATIONS OF VARIABLES (ALL VARIABLES USED BELOW MUST BE DECLARED)


	INTEGER C, H


	DOUBLE PRECISION CWeight, HWeight, maxWt


C INITIALIZATION OF CWeight, HWeight = ATOMIC WEIGHTS OF C AND H


	DATA CWeight, HWeight / 12.011, 1.0079/


C READ THE SPECIFIED MAXIMUM WEIGHT, maxWt


	READ (*,101) maxWt


 101	FORMAT(F10.4)


C LOOP THROUGH ALL POSSIBLE NUMBERS OF C'S AND H'S


	DO 105, C = 1, INT((maxWt - HWeight) / CWeight)


	    DO 104, H = 1, INT((maxWt - C*CWeight) / HWeight)


		WRITE (6, 102) C*CWeight + H*HWeight


 102		FORMAT(1X, 'C   H     has molecular weight', F8.4)


		WRITE (6, 103) C, H


 103		FORMAT(1X, 2I4)


 104	    CONTINUE


 105 	CONTINUE


	END
 
FACULTY:[IJL.COMPUTER.CS323.FORTRAN]OUT.FOR;1
 
C Just a couple of words about your evaluation of FORTRAN ... first, remember 
C that it really matters!  The value of the written evaluation is equivalent to  

C 1/3 the value of the project, itself ... so this isn't something that you 
C should just do really quickly on the way to class because we'll be looking 
C for a thoughtful write-up, prepared with good style.

C Also remember that the evaluation is to be done individually even though you 
C may have worked with a partner.

C When writing the evaluation try not to think of it in terms of, "Was this 
C mini-project easy?  In that case this language was a good one to use..."  
C because it's to be expected that your first program of substance in any new 
C language will be a bit difficult to produce.  Instead, think of the various 
C LANGUAGE EVALUATION CRITERIA which were presented in class and in the
C textbook.

C You may find it to be useful to organize your paper in the following way...

C PROS

C    - List the language criteria for which the language excels.  Be specific 
C	in explaining HOW the language excels

C CONS

C    - List the language criteria for which the language seems to be lacking.
C	Again be specific in your criticism by giving actual examples

C Personal evaluation

C    - Would you ever willfully use the language again for some other project?
C	If not why not (specifically) ... if so, what would the reason be
C	for choosing FORTRAN ... what kind of specific requirement would
C	prompt you to choose the language rather than, say, Pascal, C or C++?

C --------------------------------------

C The project is due in class on Wednesday ... it will also be accompanied by 
C the project quiz ... In order to ENCOURAGE EXCELLENCE in the writing of the 
C evaluation, I will accept the evalaution any time on Wednesday ... just slip 
C it under the door if the door is closed, Don!  ;)

C By the way, go ahead and extract this and execute it! 

 10	FORMAT(1X18HHAPPY PROGRAMMING!)
	WRITE(6,10)
	STOP 'REMEMBER -- QUALITY COUNTS!'
	END

 
FACULTY:[IJL.COMPUTER.CS323.FORTRAN]PLOT.FOR;19
 
	PROGRAM PLOT 
C
C	BASED ON 'PROGRAMMING WITH FORTRAN' SCHAUM'S OUTLINE SERIES
C	EXAMPLE ON PAGE 258
C
	INTEGER BLANK, DOT, CROSS, LINE(61), LINEY(61)
	DATA BLANK, DOT, CROSS, LINE, LINEY /' ', '.', 'X', 61*' ', 61*'.'/
	WRITE(6,10)
  10	FORMAT('1',10X,'GRAPH OF A FUNCTION'//3X,'X',6X,'Y'//)
	X = -4.
  50	Y = F(X)
C
C	THIS NEXT STEP IS VERY EMPIRICAL.  NOT AT ALL GENERAL.
C
	J = INT((Y + 39.5)/1.5)
	IF (J.GT.61.OR.J.LT.1) STOP 'RESULT OUT OF RANGE'
C
C	TEST TO SEE IF WE'RE NEAR THE AXIS
C
	IF (ABS(X).LT.0.001) GO TO 100
	LINE(30) = DOT
	LINE(J) = CROSS
	WRITE(6,20) X, Y, LINE
  20	FORMAT(1X,2(F6.2,1X),61A1)
	LINE(J) = BLANK
	GO TO 200
 100	LINEY(J) = CROSS
	WRITE(6,20) X, Y, LINEY
 200	X = X + 0.1
	IF (X.LE.4.0) GO TO 50
	STOP 'EXECUTION COMPLETE'
	END
C
	REAL FUNCTION F(VAL)
	REAL VAL
	F = ((2.0*VAL-1.0)*VAL-22.0)*VAL+21.
	RETURN
	END
 
FACULTY:[IJL.COMPUTER.CS323.FORTRAN]PROJ1.FOR;1
 
C
C CS323 FORTRAN PROJECT - PROFESSOR'S VERSION
C
C THIS PROGRAM READS AN OBSERVED MOLECULAR WEIGHT AND EXPERIMENTAL ERROR FOR
C A CHEMICAL UNKNOWN COMPOSED OF C, H, AND O.  IT THEN PRINTS ALL CHEMICALLY
C FEASIBLE FORMULAS WHOSE CALCULATED MOLECULAR WEIGHT IS SUFFICIENTLY CLOSE.
C ANY NUMBER OF DATA SETS MAY BE READ, ONE TO A LINE OF INPUT
C
	DOUBLE PRECISION OBSERVED, EXPERROR, CWEIGHT, HWEIGHT, OWEIGHT, 
	1                CALCULATED
	DATA  CWEIGHT, HWEIGHT, OWEIGHT /12.011, 1.0079, 15.9994/
	INTEGER C, H, O
	LOGICAL FOUND
C
1	READ (*,101,END=999), OBSERVED, EXPERROR
101	FORMAT(D19.0,1X,D19.0)
C
	WRITE (*,102) OBSERVED, EXPERROR
102	FORMAT(1X, 'OBSERVED MOLECULAR WEIGHT: ', F13.6 / 
	1      1X, 'EXPERIMENTAL ERROR (+OR-): ',F13.6 // 
	2      1X, 'POSSIBLE FORMULA(S)    CALCULATED WEIGHT',
	3          '        ERROR' /
	4      1X, '-------------------    -----------------',
	5          '        -----' /)
C
	FOUND = .FALSE.
C
	DO 2, O = 1, INT((OBSERVED+EXPERROR - CWEIGHT - HWEIGHT)
	1                     / OWEIGHT)
	    DO 2, C = 1, INT((OBSERVED+EXPERROR - O*OWEIGHT - HWEIGHT)
	1	              / CWEIGHT)
		H = NINT((OBSERVED - O*OWEIGHT - C*CWEIGHT) / HWEIGHT)
		CALCULATED = O*OWEIGHT + C*CWEIGHT + H*HWEIGHT
		IF (ABS(OBSERVED - CALCULATED) .LE. EXPERROR .AND.
	1	    H .LE. 4*C .AND. O .LE. 2*C)
	2	    CALL PRINTIT(C, H, O, OBSERVED, CALCULATED, FOUND)
2	CONTINUE
C
	IF (.NOT. FOUND) WRITE(*,103)
103	FORMAT(1X,'NO FORMULAS POSSIBLE')
	WRITE(*,104)
104	FORMAT(1H0)
C
	GOTO 1
C
999	STOP
	END
C
C SUBROUTINE TO ACTUALLY PRINT A FORMULA
C
	SUBROUTINE PRINTIT(C, H, O, OBSERVED, CALCULATED, FOUND)
	INTEGER C, H, O
	DOUBLE PRECISION OBSERVED, CALCULATED
	LOGICAL FOUND
C
	WRITE (*,101), CALCULATED, CALCULATED - OBSERVED, C, H, O
101	FORMAT(1X, 'C      H      O            ', F13.6, 4X, F9.6 /
	1      2X, I4, 2X, I4, 2X, I4 / )
C
	FOUND = .TRUE.
C
	RETURN
	END
 
FACULTY:[IJL.COMPUTER.CS323.FORTRAN]QUIZ.FOR;2
 
	I = 0
	J = 1
  10	IF (J.GT.10) GO TO 100
	K = 0
  20	IF (K.GT.J) GO TO 30
	I = I + J * K
	K = K + 1
	GO TO 20
  30	J = J + 1
	GO TO 10
  40	FORMAT(1X,I6)
 100	WRITE(6,40) I
	END
 
FACULTY:[IJL.COMPUTER.CS323.FORTRAN]QUIZB.FOR;8
 
	INTEGER VAL
	REAL PERCENT, TOLERANCE
	VAL=42		
	PERCENT=0.1
	TOLERANCE=1.0E-07
 10	FORMAT('VAL =',I2//F3.1/F5.3)
	WRITE (6,10) VAL,PERCENT,TOLERANCE
 20	FORMAT(' VAL =',1X,I2//1X,F3.1/1X,F5.3)
	WRITE (6,20) VAL,PERCENT,TOLERANCE
	END
 
FACULTY:[IJL.COMPUTER.CS323.FORTRAN]QUIZC.FOR;18
 
	DOUBLE PRECISION D
	D=12.34D+01
	I=IROUND(D)
 10	FORMAT(1X,D12.4,5X,I5,5X,I5)
	WRITE(6,10)D,I,IROUND(D)
	END
C
	FUNCTION IROUND(VALUE)
	INTEGER RESULT
	DOUBLE PRECISION VALUE
	RESULT = IDINT(VALUE+0.5)
	RETURN RESULT
	END

 
FACULTY:[IJL.COMPUTER.CS323.FORTRAN]STYLE_ZERO.FOR;1
 



C	PROGRAM TO CALCULATE THE ZERO OF A FUNCTION IN A GIVEN INTERVAL
C	WRITTEN BY IRVIN J. LEVY, GORDON COLLEGE
C
C	P.S. YOUR COMMENTS HAD BETTER BE CONSIDERABLY BETTER THAN THIS!
C
	PROGRAM ZERO
C
	DOUBLE PRECISION MINI,MAXI,STEP,OLDVAL,PARM,XIN,COMP,
     2	                 THRESH,VALUE,TEST1,TEST2,CALC
 10	FORMAT (' PRECISION DESIRED IN RESULT: '$)
	WRITE(6,10)
	READ (5,*) THRESH
	IF (THRESH.LT.1E-6) WRITE(6,15)
	IF (THRESH.LT.1E-6) THRESH=1E-6
 15	FORMAT('+>> THRESHOLD TOO SMALL.  AUTO-SET TO 1 PPM <<')
 20	FORMAT(/' FIND FIRST ZERO IN RANGE (0,0 = DONE): ',$)
 23	WRITE(6,20)
	READ (5,*) MINI,MAXI
 25	FORMAT(1X)
	WRITE(6,25)
	IF (MINI-MAXI) 31,220,31
 31	STEP=(MAXI-MINI)/100.
 50	OLDVAL=CALC(MINI+0.1*STEP)
	DO 100 PARM=MINI,MAXI,STEP
		XIN=PARM
		VALUE=CALC(XIN)
		TEST1=OLDVAL/ABS(OLDVAL)
		TEST2=VALUE/ABS(VALUE)
		IF (TEST1.NE.TEST2) GOTO 150
 100	CONTINUE
 150	MINI=PARM-STEP
	STEP=STEP/10.
 175	FORMAT (' SIGN CHANGE.  X = ',G,' VALUE = ',G)
	WRITE(6,175) PARM,VALUE
	COMP=VALUE/PARM
	IF (ABS(COMP).LE.THRESH) GOTO 200
	GOTO 50
 200	PREC=VALUE/PARM*1E6
	WRITE(6,210) PARM,VALUE,PREC
 210	FORMAT (////' RESULT!!!   X = ',G,' VALUE = ',G,
     2             /' PRECISION = ',G,' PPM')
	GOTO 23
 220	STOP 'PROGRAM TERMINATED - RANGE OF ZERO'
	END
C
C * * * * * * * * * * *  FUNCTION SECTION * * * * * * * * * * *
C
	FUNCTION CALC(X)
  	DOUBLE PRECISION X
	CALC=((2.7182818**(-2.*X))*(2.*X**2+2.*X+1.))-0.100
	RETURN
	END
 
FACULTY:[IJL.COMPUTER.CS323.FORTRAN]TEST.FOR;1
 
	i = 10
	goto 10
	write(6,5)
  5     format(' hello')
 10	stop 'did i just say hello?'
	end
 
FACULTY:[IJL.COMPUTER.CS323.FORTRAN]VARIABLE.FOR;7
 
	I = 10
	J = 4
	K = 12
	L = 5
	M = ((I-J) * ((K-L) / ((K-I) * ((K-I) * ((L-J) / ((J-K) * ((L-J))))))))
	END
 
FACULTY:[IJL.COMPUTER.CS323.FORTRAN]YIKES.FOR;2
 
	DO 50 I = 1. 10
 50	CONTINUE
	WRITE(6,*)DO50I
	END
 
FACULTY:[IJL.COMPUTER.CS323.FORTRAN]ZERO.FOR;1
 



C	Program to calculate the zero of a function in a given interval
C	Written by Irvin J. Levy, Gordon College
C
C	P.S. Your comments had better be considerably better than this!
C
	PROGRAM ZERO
C
	DOUBLE PRECISION MINI,MAXI,STEP,OLDVAL,PARM,XIN,COMP,
     2	                 THRESH,VALUE,TEST1,TEST2,CALC
 10	FORMAT (' Precision desired in result: '$)
	WRITE(6,10)
	READ (5,*) THRESH
	IF (THRESH.LT.1E-6) WRITE(6,15)
	IF (THRESH.LT.1E-6) THRESH=1E-6
 15	FORMAT('+>> Threshold too small.  Auto-set to 1 ppm <<')
 20	FORMAT(/' Find first zero in range (0,0 = done): ',$)
 23	WRITE(6,20)
	READ (5,*) MINI,MAXI
 25	FORMAT(1X)
	WRITE(6,25)
	IF (MINI-MAXI) 31,220,31
 31	STEP=(MAXI-MINI)/100.
 50	OLDVAL=CALC(MINI+0.1*STEP)
	DO 100 PARM=MINI,MAXI,STEP
		XIN=PARM
		VALUE=CALC(XIN)
		TEST1=OLDVAL/ABS(OLDVAL)
		TEST2=VALUE/ABS(VALUE)
		IF (TEST1.NE.TEST2) GOTO 150
 100	CONTINUE
 150	MINI=PARM-STEP
	STEP=STEP/10.
 175	FORMAT (' Sign change.  X = ',G,' Value = ',G)
	WRITE(6,175) PARM,VALUE
	COMP=VALUE/PARM
	IF (ABS(COMP).LE.THRESH) GOTO 200
	GOTO 50
 200	PREC=VALUE/PARM*1E6
	WRITE(6,210) PARM,VALUE,PREC
 210	FORMAT (////' RESULT!!!   X = ',G,' Value = ',G,
     2             /' Precision = ',G,' ppm')
	GOTO 23
 220	STOP 'PROGRAM TERMINATED - RANGE OF ZERO'
	END
C
C * * * * * * * * * * *  FUNCTION SECTION * * * * * * * * * * *
C
	FUNCTION CALC(X)
  	DOUBLE PRECISION X
	CALC=((2.7182818**(-2.*X))*(2.*X**2+2.*X+1.))-0.100
	RETURN
	END

F::IJL.COMPUTER.CS323.FORTRAN>