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CAR2.SIF
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NAME CAR2
*
* A DISCRETIZED OPTIMAL CONTROL PROBLEM
*
* consider minimising the time needed for a car on a surface
* to get from x = (0,0) to (1,1) given that the initial and
* final speed is zero.
*
* This problem can be put in the following form:
*
* let (x1,x2) be the position and (x3,x4) the velocity
* let (u1,u2) be the driving force vector
* let q be the time required, then our problem becomes:
*
* MINIMISE q
*
* with the equations of motion:
*
* dx1/dt = q*x3
* dx2/dt = q*x4
* dx3/dt = q*(u1-f*x3) (1)
* dx4/dt = q*(u2-f*x4)
*
* where 't' is a rescaled time varying between 0 and 1, and
* 'f' is a coefficient of friction.
*
* the deriving force is bounded by
*
* u1^2 + u2^2 <= 1/2 (2)
*
* (the rather arbitrary no. 1/2 representing the max. power
* of the car).
* We finally have the limit conditions:
*
* x1 = x2 = x3 = x4 = 0 - at t = 0
*
* x1 = x2 = 1 , x3 = x4 = 0 - at t = 1
*
*
* This can be reduced to an NLP by discretization. Let:
*
* X(1),X(2) ... X(M) be x(t1),x(t2), ....., x(tm)
*
* and similarly
*
* U(1),U(2) ... U(M-1) be u(t1),u(t2), ..., u(tm-1)
*
* respectively, where:
*
* 0 = t1 < t2 < .... < tm = 1
*
* then, by a suitable discretization (eg. finite
* differences), (1) becomes a set of equations at each
* time nodd 'ti' of the form:
*
* K(Xi,X(i+1),U(i)) = 0 ( 1 <= i <= M-1 )
*
* while (2) and (3) simply become:
*
* mod[U(i)]^2 <= 1/2 ( 1 <= i <= M-1 )
*
* X(1) = (0,0,0,0) , X(M) = (1,1,0,0)
*
*
* The solution of the problem is obvious: the car
* will move in a straight line, initially accelerating
* at its maximum ( u1 = u2 = 1/4 ) and then decelerating
* to stop as quickly as possible ( u1 = u2 = - 1/4 ).
* In the absence of friction ( f = 0 ) the change will
* occur halfway and the minimum time will be
*
* q = 2 * sqrt(2) = 2.8284...
*
*
* SIF input: Thomas Felici, Universite de Nancy, France,
* October 1993, with modifications to compute derivatives
* by Nick Gould
* classification LOR1-RN-V-V
********************************************
* number of variables = M * ( NX + NU ) + NQ - NU
* M = Number of time nodes.
*IE M 30 $-PARAMETER n = 179 original value
IE M 300 $-PARAMETER n = 1799
IE M 1000 $-PARAMETER n = 5999
* COEFRI = Coefficient of friction.
RE COEFRI 1.0 $-PARAMETER
********************************************
IE NEQ 4
IE NCON 1
IE NLINEQ 0
IE NLINCON 0
I+ NTEQ NEQ NLINEQ
I+ NTCON NCON NLINCON
IE NX 4
IE NU 2
IE NQ 1
* DEFINE REQUIRED INTEGERS
IE 1 1
IE 2 2
IE 3 3
IE 4 4
I- M-1 M 1
* DEFINE TIME NODES
RI RM M-1
DO I 1 M
I- I-1 I 1
RI RI1 I-1
A/ T(I) RI1 RM
ND
VARIABLES
DO I 1 M
DO J 1 NX
X X(I,J)
ND
DO I 1 M-1
DO J 1 NU
X U(I,J)
ND
DO J 1 NQ
X Q(J)
OD
GROUPS
XN OBJ Q(1) 1.0
DO I 1 M-1
DO J 1 NTEQ
XE K(I,J)
ND
DO I 1 M
DO J 1 NTCON
XL G(I,J)
ND
CONSTANTS
DO I 1 M
X CAR2 G(I,1) 1.0D+0
ND
BOUNDS
XL BNDS Q(1) .10000E+01
XU BNDS Q(1) .10000E+26
XX BNDS X(1,1) .00000E+00
XX BNDS X(1,2) .00000E+00
XX BNDS X(1,3) .00000E+00
XX BNDS X(1,4) .00000E+00
XX BNDS X(M,1) .10000E+01
XX BNDS X(M,2) .10000E+01
XX BNDS X(M,3) .00000E+00
XX BNDS X(M,4) .00000E+00
DO I 2 M-1
XL BNDS X(I,1) -.10000E+26
XU BNDS X(I,1) .10000E+26
XL BNDS X(I,2) -.10000E+26
XU BNDS X(I,2) .10000E+26
XL BNDS X(I,3) -.10000E+26
XU BNDS X(I,3) .10000E+26
XL BNDS X(I,4) -.10000E+26
XU BNDS X(I,4) .10000E+26
ND
DO I 1 M-1
XL BNDS U(I,1) -.10000E+01
XU BNDS U(I,1) .10000E+01
XL BNDS U(I,2) -.10000E+01
XU BNDS U(I,2) .10000E+01
ND
START POINT
* THIS IS AN ARBITRARY, BUT JOUDICIOUS CHOICE!
X INITS Q(1) .10000E+00
DO I 1 M
X INITS X(I,1) .10000E+00
X INITS X(I,2) .10000E+00
X INITS X(I,3) .10000E+00
X INITS X(I,4) .10000E+00
ND
DO I 1 M-1
X INITS U(I,1) .00000E+00
X INITS U(I,2) .00000E+00
ND
ELEMENT TYPE
* Square
EV SQR X
* Product
EV PROD X Y
* ODE`S
EV Ktyp XI1 XF1
EV Ktyp XI2 XF2
EV Ktyp XI3 XF3
EV Ktyp XI4 XF4
EV Ktyp UV1 UV2
EV Ktyp QV1
EP Ktyp TN
EP Ktyp TN1
EP Ktyp ID
EP Ktyp COEFRI
ELEMENT USES
* ODEs
DO I 1 M-1
IA S I 1
DO J 1 NTEQ
RI ReJ J
XT Ke(I,J) Ktyp
ZP Ke(I,J) COEFRI COEFRI
ZP Ke(I,J) TN T(I)
ZP Ke(I,J) TN1 T(S)
ZP Ke(I,J) ID ReJ
ZV Ke(I,J) XI1 X(I,1)
ZV Ke(I,J) XF1 X(S,1)
ZV Ke(I,J) XI2 X(I,2)
ZV Ke(I,J) XF2 X(S,2)
ZV Ke(I,J) XI3 X(I,3)
ZV Ke(I,J) XF3 X(S,3)
ZV Ke(I,J) XI4 X(I,4)
ZV Ke(I,J) XF4 X(S,4)
ZV Ke(I,J) UV1 U(I,1)
ZV Ke(I,J) UV2 U(I,2)
ZV Ke(I,J) QV1 Q(1)
ND
* constraints
DO J 1 NTCON
DO I 1 M-1
XT Ge(I,J,1) SQR
ZV Ge(I,J,1) X U(I,1)
XT Ge(I,J,2) SQR
ZV Ge(I,J,2) X U(I,2)
ND
DO J 1 NTCON
XT Ge(M,J,1) SQR
ZV Ge(M,J,1) X U(M-1,1)
XT Ge(M,J,2) SQR
ZV Ge(M,J,2) X U(M-1,2)
ND
GROUP USES
* ODE`S
DO I 1 M-1
DO J 1 NTEQ
XE K(I,J) Ke(I,J)
ND
* constraints
DO I 1 M
DO J 1 NTCON
XE G(I,J) Ge(I,J,1) Ge(I,J,2)
ND
ENDATA
********************************************************
* DEFINITION OF THE VARIOUS NONLINEAR FUNCTIONS *
********************************************************
ELEMENTS CAR2
TEMPORARIES
R F
R GXI1
R GXI2
R GXI3
R GXI4
R GXF1
R GXF2
R GXF3
R GXF4
R GUV1
R GUV2
R GQV1
R COLLOC
F COLLOC
INDIVIDUALS
* Square group type
T SQR
F X * X
G X X + X
H X X 2.0
* Product group type
T PROD
F X * Y
G X Y
G Y X
H X Y 1.0
* ODE`S
T Ktyp
F COLLOC(
F+ XI1, XI2, XI3, XI4,
F+ XF1, XF2, XF3, XF4,
F+ UV1, UV2, QV1,
F+ GXI1, GXI2, GXI3, GXI4,
F+ GXF1, GXF2, GXF3, GXF4,
F+ GUV1, GUV2, GQV1,
F+ COEFRI, TN, TN1, ID, .FALSE. )
A F COLLOC(
A+ XI1, XI2, XI3, XI4,
A+ XF1, XF2, XF3, XF4,
A+ UV1, UV2, QV1,
A+ GXI1, GXI2, GXI3, GXI4,
A+ GXF1, GXF2, GXF3, GXF4,
A+ GUV1, GUV2, GQV1,
A+ COEFRI, TN, TN1, ID, .TRUE. )
G XI1 GXI1
G XI2 GXI2
G XI3 GXI3
G XI4 GXI4
G XF1 GXF1
G XF2 GXF2
G XF3 GXF3
G XF4 GXF4
G UV1 GUV1
G UV2 GUV2
G QV1 GQV1
ENDATA
*****************************
* EXTERNAL FUNCTIONS *
*****************************
DOUBLE PRECISION FUNCTION COLLOC(
* X1, X2, X3, X4, XP1, XP2, XP3, XP4,
* U1, U2, Q1, GX1, GX2, GX3, GX4,
* GXP1, GXP2, GXP3, GXP4, GU1, GU2, GQ1,
* COEFRI, T, T1, DIM, DERIV )
C Compute a finite-difference discretization to the system of
C ordinary differential equations
C
C dX/dT - F( X, U, Q, T ) = 0
C
C involving the values X, U, Q and XP = X + (T1 - T)
C The returned value will be the approximation function
C COLLOC( X, XP, U, Q ) and, if DERIV is true, its derivative
C with respect to each parameter.
C
INTEGER ID, I, J , NE, NX, NU, NQ, NP
DOUBLE PRECISION X1, X2, X3, X4,
* XP1, XP2, XP3, XP4, U1, U2, Q1
DOUBLE PRECISION GX1, GX2, GX3, GX4,
* GXP1, GXP2, GXP3, GXP4, GU1, GU2, GQ1
DOUBLE PRECISION COEFRI, T, T1, DIM, TOR, TC, C1, C2
C
C Parameters:
C NE = Number of differential equations
C NX = Number of variables X in the equations
C NU = Number of variables U in the equations
C NQ = Number of variables Q in the equations
C NP = Number of fixed parameters in the equations
C
PARAMETER ( NE = 4, NX = 4, NU = 2, NQ = 1, NP = 1 )
DOUBLE PRECISION X( NX ), XP( NX ), XC( NX ), U( NU ), Q( NQ )
DOUBLE PRECISION F( NX ), FP( NX ), FC( NX ), FCPRED( NX )
DOUBLE PRECISION DX( NX ), DXP( NX ), DU( NU ), DQ( NQ )
DOUBLE PRECISION DFDX ( NX, NE ), DFDU ( NU, NE )
DOUBLE PRECISION DFDQ ( NQ, NE ), DFDQP( NQ, NE )
DOUBLE PRECISION DFDXP( NX, NE ), DFDUP( NU, NE )
DOUBLE PRECISION DFDXC( NX, NE ), DFDUC( NU, NE )
DOUBLE PRECISION DFDQC( NQ, NE ), PARAM( NP )
DOUBLE PRECISION HALF, FOURTH, EIGHTH, ONEPT5
PARAMETER ( HALF = 5.0D-1 , FOURTH = 2.5D-1 )
PARAMETER ( EIGHTH = 1.25D-1, ONEPT5 = 1.5D+0 )
LOGICAL DERIV
PARAM( 1 ) = COEFRI
C
C Place scalar input in vectors, for convenience.
C
ID = DIM
X ( 1 ) = X1
X ( 2 ) = X2
X ( 3 ) = X3
X ( 4 ) = X4
XP( 1 ) = XP1
XP( 2 ) = XP2
XP( 3 ) = XP3
XP( 4 ) = XP4
U ( 1 ) = U1
U ( 2 ) = U2
Q ( 1 ) = Q1
C
C Calculate the values of F, F and FP, at X and XP, respectively.
C
TOR = T1 - T
CALL FSYS( DERIV, F , X , U, T , Q, PARAM,
* DFDX , DFDU , DFDQ )
CALL FSYS( DERIV, FP, XP, U, T1, Q, PARAM,
* DFDXP, DFDUP, DFDQP )
C
C Calculate the corrector value XC and the predicted value of F,
C FCPRED, at XC.
C
C1 = EIGHTH * TOR
C2 = ONEPT5 / TOR
DO 10 J = 1, NX
XC ( J ) = HALF * ( X( J ) + XP( J ) ) +
* C1 * ( F( J ) - FP( J ) )
FCPRED( J ) = C2 * ( XP( J ) - X( J ) ) -
* FOURTH * ( F( J ) + FP( J ) )
10 CONTINUE
C Calculate the actual value of F, FC, at XC.
TC = HALF * ( T1 + T )
CALL FSYS( DERIV, FC, XC, U, TC, Q, PARAM,
* DFDXC, DFDUC, DFDQC )
C
C The objective function value is the difference between the predicted
C and actual values of F at XC.
C
COLLOC = FCPRED( ID ) - FC( ID )
IF ( .NOT. DERIV ) RETURN
C
C Obtain the gradient values of the IDth term. First, include the
C contributions from the FC term.
C
DO 20 I = 1, NX
DX ( I ) = - FOURTH * DFDX ( I, ID )
DXP( I ) = - FOURTH * DFDXP( I, ID )
20 CONTINUE
DO 30 I = 1, NU
DU( I ) = - FOURTH * ( DFDU( I, ID ) + DFDUP( I, ID ) )
30 CONTINUE
DO 40 I = 1, NQ
DQ( I ) = - FOURTH * ( DFDQ( I, ID ) + DFDQP( I, ID ) )
40 CONTINUE
DX ( ID ) = DX ( ID ) - C2
DXP( ID ) = DXP( ID ) + C2
C
C Now include the contributions from the FCF term, using the
C chain rule for partial differentiation.
C
C Derivatives with respect to X and XP.
C
DO 80 I = 1, NX
DO 50 J = 1, NX
DX ( I ) = DX ( I ) - DFDXC( J, ID ) * C1 * DFDX ( I, J )
DXP( I ) = DXP( I ) + DFDXC( J, ID ) * C1 * DFDXP( I, J )
50 CONTINUE
DX ( I ) = DX ( I ) - DFDXC( I, ID ) * HALF
DXP( I ) = DXP( I ) - DFDXC( I, ID ) * HALF
80 CONTINUE
C
C Derivatives with respect to U.
C
DO 120 I = 1, NU
DO 90 J = 1, NX
DU ( I ) = DU ( I ) - DFDXC( J, ID ) *
* C1 * ( DFDU ( I, J ) - DFDUP( I, J ) )
90 CONTINUE
DU ( I ) = DU ( I ) - DFDUC( I, ID )
120 CONTINUE
C
C Derivatives with respect to Q.
C
DO 160 I = 1, NQ
DO 130 J = 1, NX
DQ ( I ) = DQ ( I ) - DFDXC( J, ID ) *
* C1 * ( DFDQ ( I, J ) - DFDQP( I, J ) )
130 CONTINUE
DQ ( I ) = DQ ( I ) - DFDQC( I, ID )
160 CONTINUE
C
C Assign the array components to their scalar counterparts.
C
GX1 = DX ( 1 )
GX2 = DX ( 2 )
GX3 = DX ( 3 )
GX4 = DX ( 4 )
GXP1 = DXP( 1 )
GXP2 = DXP( 2 )
GXP3 = DXP( 3 )
GXP4 = DXP( 4 )
GU1 = DU ( 1 )
GU2 = DU ( 2 )
GQ1 = DQ ( 1 )
RETURN
END
SUBROUTINE FSYS( DERIV, F, X, U, T, Q, PARAM,
* DFDX, DFDU, DFDQ )
C Subroutine defining system of ODE's for given X, U, Q and T.
C Given a set of differential equations
C
C dX/dT = F( X, U, Q, T, PARAM ),
C
C returns the values of F and, if DERIV is .TRUE., its Jacobians
C dF/dX, dF/dU and dF/dQ for given input X, U, Q, T and PARAM.
C
C DFDX( I, J ) contains the derivative of F(J) w.r.t. X(I)
C DFDU( I, J ) contains the derivative of F(J) w.r.t. U(I)
C DFDQ( I, J ) contains the derivative of F(J) w.r.t. Q(I)
INTEGER NE, NX, NU, NQ, NP
LOGICAL DERIV
PARAMETER ( NE = 4, NX = 4, NU = 2, NQ = 1, NP = 1 )
DOUBLE PRECISION T, ZERO, ONE, TWO
PARAMETER ( ZERO = 0.0D+0, ONE = 1.0D+0, TWO = 2.0D+0 )
DOUBLE PRECISION F( NE ), X( NX ), U( NU ), Q( NQ ), PARAM( NP )
DOUBLE PRECISION DFDX( 4, 4 ), DFDU( 2, 4 ), DFDQ( 1, 4 )
DOUBLE PRECISION FRIC
FRIC = PARAM( 1 )
C equation 1
F( 1 ) = Q( 1 ) * X( 3 )
IF ( DERIV ) THEN
DFDX( 1, 1 ) = ZERO
DFDX( 2, 1 ) = ZERO
DFDX( 3, 1 ) = Q( 1 )
DFDX( 4, 1 ) = ZERO
DFDU( 1, 1 ) = ZERO
DFDU( 2, 1 ) = ZERO
DFDQ( 1, 1 ) = X( 3 )
END IF
C equation 2
F( 2 ) = Q( 1 ) * X( 4 )
IF ( DERIV ) THEN
DFDX( 1, 2 ) = ZERO
DFDX( 2, 2 ) = ZERO
DFDX( 3, 2 ) = ZERO
DFDX( 4, 2 ) = Q( 1 )
DFDU( 1, 2 ) = ZERO
DFDU( 2, 2 ) = ZERO
DFDQ( 1, 2 ) = X( 4 )
END IF
C equation 3
F( 3 ) = Q( 1 ) * ( U( 1 ) - FRIC * X( 3 ) )
IF ( DERIV ) THEN
DFDX( 1, 3 ) = ZERO
DFDX( 2, 3 ) = ZERO
DFDX( 3, 3 ) = - FRIC * Q( 1 )
DFDX( 4, 3 ) = ZERO
DFDU( 1, 3 ) = Q( 1 )
DFDU( 2, 3 ) = ZERO
DFDQ( 1, 3 ) = U( 1 ) - FRIC * X( 3 )
END IF
C equation 4
F( 4 ) = Q( 1 ) * ( U( 2 ) - FRIC * X( 4 ) )
IF ( DERIV ) THEN
DFDX( 1, 4 ) = ZERO
DFDX( 2, 4 ) = ZERO
DFDX( 3, 4 ) = ZERO
DFDX( 4, 4 ) = - FRIC * Q( 1 )
DFDU( 1, 4 ) = ZERO
DFDU( 2, 4 ) = Q( 1 )
DFDQ( 1, 4 ) = U( 2 ) - FRIC * X( 4 )
END IF
RETURN
END