 FICO Xpress Optimization Examples Repository
 FICO Optimization Community FICO Xpress Optimization Home   Successive linear programming (SLP) model for a financial planning problem

Description
• modifying constraint coefficients
• while statement
• setting/accessing optimiser parameters
• procedure
Further explanation of this example: 'Mosel User Guide', Section 12.1 Recursion

Source Files

recurse.mos

```(!*******************************************************
* Mosel Example Problems                              *
* ======================                              *
*                                                     *
* file recurse.mos                                    *
* ````````````````                                    *
* Example for the use of the Mosel language           *
* (Financial application of Recursion, that is        *
*  Non-linear programming)                            *
*                                                     *
*  The problem is to solve                            *
*     net(t) = Payments(t)  - interest(t)             *
*     balance(t) = balance(t-1) - net(t)              *
*     interest(t) = (92/365) * balance(t) *           *
*                                   interest_rate     *
*  where                                              *
*     balance(0) = 0                                  *
*     balance(T) = 0                                  *
*     for interest_rate                               *
*                                                     *
*  The problem is that we have the T products:        *
*     balance(t) * interest_rate                      *
*  which cannot be modelled just by LP.               *
*                                                     *
* (c) 2008 Fair Isaac Corporation                     *
*     author: S. Heipcke, 2001                        *
*******************************************************!)

model Fin_nlp                       ! Start a new model

uses "mmxprs"                       ! Load the optimizer library

forward procedure solverec          ! Declare a procedure defined later

declarations
T=6                                ! Time horizon
RT=1..T                            ! Range of time periods
P,R,V: array(RT) of real           ! Payments
B: array(RT) of real               ! An INITIAL GUESS as to balances b(t)
X: real                            ! An INITIAL GUESS as to interest rate x

i: array(RT) of mpvar              ! Interest
n: array(RT) of mpvar              ! Net
b: array(RT) of mpvar              ! Balance
x: mpvar                           ! Interest rate
dx: mpvar                          ! Change to x
epl, emn: array(RT) of mpvar       ! + and - deviations
end-declarations

X:= 0.00
B:: [1, 1, 1, 1, 1, 1]
P:: [-1000, 0, 0, 0, 0, 0]
R:: [206.6, 206.6, 206.6, 206.6, 206.6, 0]
V:: [-2.95, 0, 0, 0, 0, 0]

! net = payments - interest
forall(t in RT) net(t):= n(t) = (P(t)+R(t)+V(t)) - i(t)

! Money balance across periods
forall(t in RT) bal(t):= b(t) = if(t>1, b(t-1), 0) - n(t)

!       interest = (92/365)*( balance * interest_rate)
! i.e.  i(t)     = (92/365)*( b(t-1)  * x )
!       i(t)     = (92/365)*( b(t-1)  * ( X + dx ) )
!       i(t)     = (92/365)*( b(t-1)*X + (B(t-1)+db(t-1))*dx )
!       i(t)     = (92/365)*( b(t-1)*X + B(t-1)*dx ) (approx)
!       Use penalty variables (epl and emn) to ensure feasibility:
forall(t in 2..T)
interest(t):= -(365/92)*i(t) + X*b(t-1) + B(t-1)*dx + epl(t) - emn(t) = 0

def:= X + dx = x                    ! Define the interest rate: x = X + dx

feas:= sum(t in RT) (epl(t)+emn(t)) ! Objective: get feasible

i(1) = 0                            ! Initial interest is zero
forall (t in RT) n(t) is_free
forall (t in 1..T-1) b(t) is_free
b(T) = 0                            ! Final balance is zero
dx is_free

minimize(feas)                      ! Solve the LP-problem

solverec                            ! Recursion loop

! Print the solution
writeln("\nThe interest rate is ", x.sol)
write(strfmt("t",5), strfmt(" ",4))
forall(t in RT) write(strfmt(t,5), strfmt(" ",3))
write("\nBalances ")
forall(t in RT)  write(strfmt(b(t).sol,8,2))
write("\nInterest ")
forall(t in RT)  write(strfmt(i(t).sol,8,2))
writeln

!************************************************************************
!  Recursion loop: we recurse on X and the b(t)'s.
!    The 'B(t-1)' in rows interest(t) get the prior value of b(t-1)
!    The 'X' in rows interest(t) get the prior value of x
!    The 'X' in row def gets the prior value of x
!    We say we have converged when the change in dx is less than 1.0E-6
!************************************************************************
procedure solverec

declarations
TOLERANCE=0.000001                  ! Convergence tolerance
variation: real                     ! Variation of x
BC: array(RT) of real
bas: basis
end-declarations

setparam("zerotol", TOLERANCE)      ! Set Mosel comparison tolerance
variation:=1.0
ct:=0

while(variation>0) do
savebasis(bas)                    ! Save the current basis
ct+=1
forall(t in 2..T)
BC(t-1):= b(t-1).sol            ! Get the solution values for b(t)'s
XC:= x.sol                        ! and x
write("Round ", ct, " x:", x.sol, " (variation:", variation,"), ")
writeln("Simplex iterations: ", getparam("XPRS_SIMPLEXITER"))

forall(t in 2..T) do              ! Update coefficients
interest(t)+= (BC(t-1)-B(t-1))*dx
B(t-1):=BC(t-1)
interest(t)+= (XC-X)*b(t-1)
end-do
def+= XC-X
X:=XC
oldxval:=XC                       ! Store solution value of x   