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Overview of Mosel examples for 'Business Optimization' book

Description

List of FICO Xpress Mosel implementations of examples discussed in 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).

List of provided model files

(Examples marked with * are newly introduced in the 2nd edition, all other models have been converted from the mp-model versions that were provided with the 1st edition of the book in 1997.)
FilenameDescriptionSection
absval.mos Modeling absolute value terms (linearization) 6.5
*absval2.mos, absval2a.mos Modeling absolute value as general constraints 6.5
bench101.mos Parkbench production planning problem (solution to Exercise 10.1) 10.9
bench102.mos Parkbench production planning problem, MIP problem (solution to Exercise 10.2) 10.9
blend, blend_graph Ore blending problem 2.7.1
*boat.mos Boat renting problem (solution for Exercise 3.3) 1.4.1, 3.7
*boat2.mos Multi-period Boat renting problem 1.6
boatdual.mos Dual of the Boat renting problem (solution for Exercise 3.3) 3.7
brewery.mos Brewery production planning (data files: brewery.xlsx, brewdata.dat) 8.3
burgap.mos Generalized assignment problem (solution for Exercise 7.3) 7.10
burglar Small knapsack problem 7.1.1
buscrew.mos Bus crew scheduling 7.8.4
calves.mos Calves and pigs problem 3.3.1
carton.mos Carton production scheduling problem (data file: carton.dat) 10.3
*ch-2tri.mos Minimal perimeter convex hull for two triangles 10.3
*contract.mos, contract_graph Contract allocation problem with semi-continuous variables 10.2.1
couples.mos Feasibility puzzle problem (solution for Exercise 6.5) 6.11
dea.mos Data envelopment analysis (solution for Exercise 5.2) 5.3
dual.mos Dual problem for a small LP (solution for Exercise 3.1b) 3.5.1
*dynbigm.mos Dynamic computation of big-M coefficients for production planning 14.1.2.1
*dynbigm2.mos Production planning problem formulation using indicator constraints 14.1.2.1
*ea_smpld.mos Evolutionary algorithm for supply management 14.1.3.3
euro.mos Choosing investment projects (solution for Exercise 7.5) 7.10
flowshop.mos, flowshop_graph Flowshop scheduling problem (solution for Exercise 7.7) 7.10
*fracprog.mos Fractional programming example 11.1
gap.mos Generalized assignment problem (solution for Exercise 7.2) 7.10
*goalprog.mos Lexicographic Goal Programming 5.4.3
*lagrel.mos Lagrange relaxation applied to the GAP 14.1.3.3
lim1.mos, coco Multi-period, multi-site production planning, LP model (solution for Exercise 5.1) 5.3
lim2.mos, coco_fixbv Multi-period, multi-site production planning, MIP model (solution for Exercise 6.7) 6.11
*manufact.mos Production scheduling problem with SOS formulation (solution for Exercise 6.9) 6.11
multk.mos Multi-knapsack problem (solution for Exercise 7.4) 7.10
network.mos Network flow problem (solution for Exercise 4.3) 4.7
network2.mos Generic formulation of network flow problem (Exercise 4.3) 4.7
*newsvendor.mos Newsvendor problem: 2-stage stochastic programming 11.3.2.1
npv.mos Net present value problem (solution for Exercise 6.8) 6.11
*optgrid.mos Optimal breakpoints for piecewise linear approximation 14.2.3
*portfolio.mos Multi-stage stochastic portfolio investment model 11.3.2.5
primal.mos Primal problem for a small LP (solution for Exercise 3.1a) 3.7
prodx.mos Simple production planning example 2.5.2
projschd.mos, projplan_graph Project scheduling case study 10.2.3
quadrat.mos Linearized quadratic programming example 11.4
*quadrat2.mos Quadratic programming example solved as NLP 11.4
set.mos Set covering problem (solution for Exercise 7.6) 7.10
simple1.mos Simple LP problem (solution for Exercise 2.2) 2.13
simple2.mos Simple LP problem (solution for Exercise 2.3) 2.13
slab.mos Lifting slabs (solution for Exercise 6.6) 6.11
sludge.mos Sludge production planning example illustrating recursion 11.2.1
*sludge2.mos Recursion example solved as NLP 11.2.1
*teams.mos (solution for Exercise 6.4) 6.11
trim1.mos Trimloss problem LP formulation (solution for Exercise 4.1) 4.1.1
trim2.mos Trimloss problem MIP formulation (solution for Exercise 4.2) 4.1.2
*trimminlp.mos Trimloss problem formulated as a MINLP problem 13.3
*trimminlp2.mos Alternative NLP solver choice for trimloss problem 13.3
tsp.mos Traveling salesman problem (solution for Exercise 7.1) 7.10
*vrp, vrp_graph Vehicle routing - heating oil delivery problem 7.2.3
*woods.mos (solution for Exercise 2.1) 2.13
yldmgmt.mos Yield management, financial modeling 8.4.2

bsnsoptbook_ovw.zip[download all files]

Source Files

Data Files





carton.mos

(!*********************************************************************
   Mosel Example Problems
   ======================

   file carton.mos
   ```````````````
   Carton scheduling problem 
   
     Example discussed in section 10.3 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
 
       http://www.apache.org/licenses/LICENSE-2.0
 
    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 'carton'
  uses 'mmsystem', 'mmxprs'

  declarations
    M= 5                    ! Number of machines (m = 1,...,M)
    CX= 15                  ! Maximum number of configurations on each
                            !   machine (c = 0,1,... c(m)), where c = 0 means
                            !   setting up
    G= 16                   ! Groups of orders (g = 1, ... ,G)
    T= 20                   ! Shifts, where t = 0 means the last shift of
                            !   the preceeding week.
    RM=1..M
    RT=1..T
    RG=1..G
    RCX=1..CX

    C: array(RM) of integer     ! Number of configurations of machine m
    S: array(RT) of real        ! Setting up cost (per machine, per shift)
    P: array(RG) of real        ! Penalty cost of not meeting an order
                                !   (per carton)
    D: array(RG) of real        ! Demand for cartons in group g
    O: array(RT) of real        ! Overtime cost (per machine, per shift)
    H: array(RM,RCX,RCX) of real! Number of shifts required to change
                                !   from configuration c to k on machine m
    R: array(RM,RCX,RG) of real ! Production rate (number of cartons of
                                !   group g produced on machine m in
                                !   configuration c per shift)
    I: array(RM) of integer     ! Initial configuration of machine m
    OT: array(RT) of real       ! Set of overtime shifts
    SU: array(RT) of real       ! Set of shifts in which setting up is
                                !   permitted
    REQ: array(RM,RCX) of real  ! 0 if and only if configuration is not required

  end-declarations

  initializations from "carton.dat"
    C  OT  SU  D  H  R  I
  end-initializations

  forall(t in  2..T) O(t):= 400.00
  forall(t in  2..T) S(t):= 300.00 + OT(t)*O(t)
  forall(g in  RG) P(g):= 1000.00
  forall(m in RM,c in 1..C(m)) REQ(m,c):= sum(g in RG) R(m,c,g)*D(g)


 ! Configuration: equal to unity if machine m is in configuration c
 ! in shift t and zero otherwise
  declarations
    x: dynamic array(RM, RCX, RT) of mpvar
  end-declarations
  forall(m in  RM,c in  RCX,t in  RT| (c=1 and SU(t)<>0) or
        (c<>1 and REQ(m,c)<>0) or (I(m)=c))
    create(x(m,c,t))

 ! Overtime production: equal to unity if machine m is producing
 ! in configuration c in overtime shift t
  declarations
    y: dynamic array(RM, RCX, RT) of mpvar
  end-declarations
  forall(m in  RM,c in  RCX,t in  RT| OT(t)<>0 and ((c=1 and SU(t)<>0) or
         (c<>1 and REQ(m,c)<>0) or (I(m)=c)))
    create(y(m,c,t))

 ! Shortfall of order group g (in cartons)
  declarations
    z: array(RG) of mpvar
  end-declarations

 ! Objective: minimise the sum of setting up, shortfall and overtime costs
  Cost:= sum(m in  RM,t in  2..T) S(t)*x(m,1,t) + 
    sum(g in  RG) P(g)*z(g) +
    sum(m in  RM,c in  2..C(m),t in 2..T| OT(t)<>0) O(T)*y(m,c,t)

  ! Subject to the following constraints.
  ! The overtime production can only be in the current
  ! configuration for each machine
  forall(m in  RM,c in  1..C(m),t in  RT)
    OTime(m,c,t):= y(m,c,t)<= x(m,c,t)

  ! No machine, m, can be in two distinct producing configurations,
  ! say k and c, u shifts apart if it takes u or more shifts
  ! to change from one configuration to the other
  forall(m in  RM,c in  2..C(m),u in  1..4,t in 1..(T-u))
    Chgvr(m,c,u,t):=
      sum(k in  2..C(m)| k<>c and H(m,k,c)>=u) x(m,k,t) + x(m,c,t+u)<= 1

  ! The weekly demand, less the shortfall in production cannot exceed
  ! the number of cartons produced, so
  forall(g in RG)
    Demnd(g):=
      sum(m in  RM,c in  2..C(m),t in RT| OT(t)=0) R(m,c,g)*x(m,c,t) +
      sum(m in  RM,c in 2..C(m),t in  RT) R(m,c,g)*y(m,c,t) +
      z(g)>= D(g)

  ! Each machine must be in exactly one configuration (including setting up)
  ! in each shift
  forall(m in  RM,t in  2..T)
    Xcnvx(m,t):= sum(c in  1..C(m)) x(m,c,t) = 1
  forall(m in  RM,t in  2..T | OT(t)<>0)
    Ycnvx(m,t):= sum(c in  1..C(m)) y(m,c,t) = 1

  ! The model must begin in the configuration of each machine
  ! at the end of the previous week
  forall(m in  RM,c= I(m),t= 1) x(m,c,t) = 1
  forall(m in  RM,c in  1..C(m),t= 1| c<>I(m)) x(m,c,t) = 0

  ! The configuration variables are binary
  forall(m in  RM,c in  1..C(m),t in  2..T | exists(x(m,c,t)))
    x(m,c,t) is_binary
  forall(m in  RM,c in  1..C(m),t in  2..T | exists(y(m,c,t)))
    y(m,c,t) is_binary
     
  ! Display the solver logging output
  setparam("Xprs_verbose",true)

  ! Solve the problem
  minimise(Cost)
  writeln("Solution: Cost=", getobjval)
  writeln("Quantities:")
  forall(g in RG)
    writeln(formattext("Group %2d: %4.1f (%4.1f) shortfall: %4.1f", g, 
            Demnd(g).act-z(g).sol, D(g), z(g).sol))
  writeln("Configuration of machines:")
  write("Period:  I  ")
  forall(t in RT) write(strfmt(t,2),if(OT(t)>0,"* ","  "))
  writeln
  forall(m in RM) do
    write("Mach ", m, ":  ", I(m), " ")
    forall(t in RT)
      write(formattext("%3.0f ",round(getsol(sum(c in  1..C(m)) c*x(m,c,t)))))
    writeln
  end-do
end-model

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