Mining and process industries
Description
| Problem name and type, features | Difficulty | Related examples |
| A‑1 | Production of alloys: Blending problem
formulation of blending constraints; data with numerical indices, solution printout, if-then, getsol | * | blending_graph.mos |
|---|
| A‑2 | Animal food production: Blending problem
formulation of blending constraints; data with string indices, as, formatted solution printout, use of getsol with linear expressions, strfmt | * | a1alloy.mos |
|---|
| A‑3 | Refinery : Blending problem
formulation of blending constraints; sparse data with string indices, dynamic initialization, union of sets | ** | a2food.mos |
|---|
| A‑4 | Cane sugar production : Minimum cost flow (in a bipartite graph)
mo
ceil, is_binary, formattext | * | e2minflow.mos, mincostflow_graph.mos |
|---|
| A‑5 | Opencast mining: Minimum cost flow
encoding of arcs, solving LP-relaxation only, array of set | ** | a4sugar.mos |
|---|
| A‑6 | Production of electricity: Dispatch problem
inline if, is_integer, looping over optimization problem solving | ** | |
|---|
Further explanation of this example:
'Applications of optimization with Xpress-MP', Chapter 6: Mining and process industries (blending problems)
Source Files
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Data Files
a1alloy.mos
(!******************************************************
Mosel Example Problems
======================
file a1alloy.mos
````````````````
Production of alloys
An order of steel must be met with a range of carbon,
copper, and manganese grades. Given the raw materials in
stock, the objective is to determine the composition of
the steel that minimizes the production cost.
Simple linear programming blending problem formulation includes
data with numerical indices and introduction of if-then statement
for solution printout.
(c) 2008 Fair Isaac Corporation
author: S. Heipcke, Feb. 2002
*******************************************************!)
model "A-1 Production of Alloys"
uses "mmxprs"
declarations
COMP = 1..3 ! Components (chemical elements)
RAW = 1..7 ! Raw materials (alloys)
P: array(RAW,COMP) of real ! Composition of raw materials (in percent)
PMIN,PMAX: array(COMP) of real ! Min. & max. requirements for components
AVAIL: array(RAW) of real ! Raw material availabilities
COST: array(RAW) of real ! Raw material costs per tonne
DEM: real ! Amount of steel to produce
use: array(RAW) of mpvar ! Quantity of raw mat. used
produce: mpvar ! Quantity of steel produced
end-declarations
initializations from 'a1alloy.dat'
P PMIN PMAX AVAIL COST DEM
end-initializations
! Objective function
Cost:= sum(r in RAW) COST(r)*use(r)
! Quantity of steel produced = sum of raw material used
produce = sum(r in RAW) use(r)
! Guarantee min. and max. percentages of every chemical element
forall(c in COMP) do
sum(r in RAW) P(r,c)*use(r) >= PMIN(c)*produce
sum(r in RAW) P(r,c)*use(r) <= PMAX(c)*produce
end-do
! Use raw materials within their limit of availability
forall(r in RAW) use(r) <= AVAIL(r)
! Satisfy the demand
produce >= DEM
! Solve the problem
minimize(Cost)
! Solution printing
declarations
NAMES: array(RAW) of string
end-declarations
initializations from 'a1alloy.dat' ! Get the names of the alloys
NAMES
end-initializations
writeln("Total cost: ", getobjval)
writeln("Amount of steel produced: ", getsol(produce))
writeln("Alloys used:")
forall(r in RAW)
if(getsol(use(r))>0) then
write(NAMES(r), ": ", getsol(use(r))," ")
end-if
write("\nPercentages (C, Cu, Mn): ")
forall(c in COMP)
write( getsol(sum(r in RAW) P(r,c)*use(r))/getsol(produce), "% ")
writeln
end-model
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