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

   file blending.mos
   `````````````````
   TYPE:         Blending problem
   DIFFICULTY:   1
   FEATURES:     simple LP problem, formulation of blending constraints
   DESCRIPTION:  Several ores are blended to a final product that must
                 have a certain quality (`grade'). We wish to determine 
                 the quantity of every ore to be used in the blend with 
                 the objective to maximize the total profit (calculated 
                 as sales revenues - raw material cost).     
   FURTHER INFO: `Mosel User Guide', Section 2.2 `A blending example';
                 `Applications of optimization with Xpress-MP', 
                 Section 2.7 `Blending constraints'.
                 Similar problems: Section 6.1 `Production of alloys',
                 Section 6.2 `Animal food production',  Section 6.3 `Refinery'

   (c) 2001 Dash Associates
       author: S.Heipcke
*******************************************************!)

model blending
 uses "mmxprs"                 ! Load the optimizer library

 declarations
  ROres = 1..2                 ! Range of Ores
  REV = 125                    ! Unit revenue of product
  MINGRADE = 4                 ! Min permitted grade of product
  MAXGRADE = 5                 ! Max permitted grade of product
  COST: array(ROres) of real   ! Unit cost of ores
  AVAIL: array(ROres) of real  ! Availability of ores
  GRADE: array(ROres) of real  ! Grade of ores (measured per unit of mass)

  x: array(ROres) of mpvar     ! Quantities of ores used
  
 end-declarations

 COST := [85, 93]
 AVAIL:= [60, 45]
 GRADE:= [2.1, 6.3]
                               
! Objective: maximize total profit 
 Profit:= sum(o in ROres) (REV-COST(o))* x(o)  

! Lower and upper bounds on ore quality
 LoGrade:= sum(o in ROres) (GRADE(o)-MINGRADE) * x(o) >= 0
 UpGrade:= sum(o in ROres) (MAXGRADE-GRADE(o)) * x(o) >= 0 

! Set upper bounds on variables
 forall(o in ROres) x(o) <= AVAIL(o)

! Solve the problem  
 maximize(Profit)

! Solution printing
 writeln("Solution:\n Objective: ", getobjval)
 forall(o in ROres)  writeln(" x(" + o + "): ", getsol(x(o)))

end-model