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Blending ores

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).
• simple LP problem
• data in model or input from file
• bounds on variables
Further explanation of this example: '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'

Source Files

Data Files

blending_graph.mos

(!******************************************************
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) 2008 Fair Isaac Corporation
author: S.Heipcke, Jan. 2001, rev. Sep. 2017
*******************************************************!)

model blending
uses "mmxprs", "mmsvg"

declarations
ROres = 1..2                 ! Range of Ores
REV = 125                    ! Unit revenue 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]

! Objective: maximize total profit
Profit:= sum(o in ROres) (REV-COST(o))* x(o)

! Lower and upper bounds on ore quality

! 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)))

! Solution drawing
FACT:=20
ttl:=sum(o in ROres) x(o).sol
cum:=0.0
forall(o in ROres) do
cum+=x(o).sol/ttl
end-do