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Sludge production planning solved by recursion

The two model versions show how to solve a production planning problem via iterative LP solves (sludge.mos) or by using a nonlinear formulation (sludge2.mos).

Further explanation of this example: This model is discussed in Section 11.2.1 of the book 'J. Kallrath: Business Optimization Using Mathematical Programming - An Introduction with Case Studies and Solutions in Various Algebraic Modeling Languages' (2nd edition, Springer, Cham, 2021, DOI 10.1007/978-3-030-73237-0).

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   Mosel Example Problems

   file sludge.mos
   Sludge problem illustrating recursion
     Example discussed in section 11.2.1 of
     J. Kallrath: Business Optimization Using Mathematical Programming -
     An Introduction with Case Studies and Solutions in Various Algebraic 
     Modeling Languages. 2nd edition, Springer Nature, Cham, 2021 

   author: S. Heipcke, June 2018

   (c) Copyright 2020 Fair Isaac Corporation
    Licensed under the Apache License, Version 2.0 (the "License");
    you may not use this file except in compliance with the License.
    You may obtain a copy of the License at
    Unless required by applicable law or agreed to in writing, software
    distributed under the License is distributed on an "AS IS" BASIS,
    WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
    See the License for the specific language governing permissions and
    limitations under the License.


model 'sludge'
  uses "mmxprs"

    I= 2                        ! Number of sources
    J= 2                        ! Number of processing areas
    K= 3                        ! Number of destinations (end points)
    C= 2                        ! Number of components

    LAM: array(RC,RJ) of real   ! Assumed fraction of c at j (recursed)
    COST: array(RJ,RK) of real  ! Per tonne disposal cost from j to k
    MAXD: array(RC,RK) of real  ! Maximum tonnes disposal of c at k
    FRACT: array(RC,RI) of real ! Fraction of c in material from i
    AVAIL: array(RI) of real    ! Availability at i

  AVAIL::[100, 150]
  FRACT::[0.1, 0.2,
          0.2, 0.15]
  COST::[2, 3, 1,
         1, 5, 0.5]
  MAXD::[20.0, 20.0, 10.0,
         15.0, 15.0, 15.0]
  LAM::[0.20, 0.15,             ! Starting guesses for LAM(c,j)
        0.12, 0.18]

    x: array(RI,RJ) of mpvar    ! Amount (tonnes) sent from i to j
    y: array(RJ,RK) of mpvar    ! Amount (tonnes) sent from j to k
    t: array(RJ) of mpvar       ! Throughput at j (tonnes)
    q: array(RC,RJ) of mpvar    ! Quantity of component c passing through j
    LastSol: real

  ! Disposal cost
  Cost:=sum(j in RJ,k in RK) COST(j,k)*y(j,k)

  ! Get rid of all sludge at i
  forall(i in RI)
    Arid(i):= sum(j in RJ) x(i,j) = AVAIL(i)

  ! Mass balance into j
  forall(j in RJ)
    Tpx(j):= sum(i in RI) x(i,j) = t(j)

  ! Mass balance out of j
  forall(j in RJ)
    Tpy(j):= sum(k in RK) y(j,k) = t(j)

  ! Define quantities
  forall(c in RC,j in RJ)
    Q(c,j):= q(c,j) = sum(i in RI) FRACT(c,i)*x(i,j)

  writeln("***LAM initial values: ", LAM)
  LastSol:= -INFINITY; itercnt:=1
    ! Limits on disposal: not too much c at k
    forall(c in RC,k in RK)
      Mx(c,k):= sum(j in RJ) LAM(c,j)*y(j,k)<= MAXD(c,k)

    ! Solve the problem
    writeln("Iteration ", itercnt, ". Solution: ", getobjval)
    forall(i in RI,j in RJ) write(" x(", i, ",", j, "):", x(i,j).sol)
    forall(j in RJ,k in RK) write(" y(", j, ",", k, "):", y(j,k).sol)

    ! Calculate new LAM values or stop if converged
    forall(c in RC,j in RJ) do
      if t(j).sol<>0 then
        writeln("***Division by 0")
    gap:=(sum(c in RC,j in RJ) abs(LAM(c,j)-NEWL(c,j))) /
         (sum(c in RC,j in RJ) LAM(c,j))
    ifperturb:=LastSol>getobjval     ! Perturbation to prevent cycling
    if gap<TOL then
      writeln("***Minimum deviation reached: ", gap, "%")
    elif itercnt>=10000 then
      writeln("***Iteration limit reached. Difference: ", gap, "%")
      forall(c in RC,j in RJ) LAM(c,j):=NEWL(c,j)*if(ifperturb,0.9+0.2*random,1)
      writeln("***LAM updated values: ",LAM, " difference: ", gap, "%")
    LastSol:=getobjval; itercnt+=1
  until false


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