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
a6electrg.mos
(!******************************************************
Mosel Example Problems
======================
file a6electrg.mos
``````````````````
Production of electricity.
Plotting parametrics in the reserve capacity.
Four types of power generators are available to meet daily
electricity demands and up to 20% above. Each type of
generator has a set maximum capacity and minimum power output.
A generator can only be started or stopped at the beginning
of a time period. The objective is to determine which
generators should be used in each period so that total daily
cost is minimized.
Similar model formulation as a6electr.mos however defines data
directly in code and is set up to solve for varying reserve
demand (0-20%), redefining the reserve constraints for every loop
iteration and reloading the basis saved after the initial LP solve.
Displays results with a graph.
(c) 2008 Fair Isaac Corporation
author: S. Heipcke, Oct. 2004, rev. June 2017
*******************************************************!)
model "A-6 Electricity production (with graph drawing)"
uses "mmxprs", "mmsvg"
declarations
NT = 7
TIME = 1..NT ! Time periods
TYPES = 1..4 ! Power generator types
LEN, DEM: array(TIME) of integer ! Length and demand of time periods
PMIN,PMAX: array(TYPES) of integer ! Min. & max output of a generator type
CSTART: array(TYPES) of integer ! Start-up cost of a generator
CMIN: array(TYPES) of integer ! Hourly cost of gen. at min. output
CADD: array(TYPES) of real ! Cost/hour/MW of prod. above min. level
AVAIL: array(TYPES) of integer ! Number of generators per type
bas: basis ! LP basis
start: array(TYPES,TIME) of mpvar ! No. of gen.s started in a period
work: array(TYPES,TIME) of mpvar ! No. of gen.s working during a period
padd: array(TYPES,TIME) of mpvar ! Production above min. output level
end-declarations
! Time period data
LEN:: [ 6, 3, 3, 2, 4, 4, 2]
DEM:: [12000, 32000, 25000, 36000, 25000, 30000, 18000]
! Power plant data
PMIN:: [ 750, 1000, 1200, 1800]
PMAX:: [1750, 1500, 2000, 3500]
CSTART:: [5000, 1600, 2400, 1200]
CMIN:: [2250, 1800, 3750, 4800]
CADD:: [ 2.7, 2.2, 1.8, 3.8]
AVAIL:: [ 10, 4, 8, 3]
! Objective function: total daily cost
Cost:= sum(p in TYPES, t in TIME) (CSTART(p)*start(p,t) +
LEN(t)*(CMIN(p)*work(p,t) + CADD(p)*padd(p,t)))
! Number of generators started per period and per type
forall(p in TYPES, t in TIME)
NumStart(p,t):= start(p,t) >= work(p,t) - if(t>1, work(p,t-1), work(p,NT))
! Limit on power production above minimum level
forall(p in TYPES, t in TIME)
Limit(p,t):= padd(p,t) <= (PMAX(p)-PMIN(p))*work(p,t)
! Satisfy demands
forall(t in TIME)
Demand(t):= sum(p in TYPES) (PMIN(p)*work(p,t) + padd(p,t)) >= DEM(t)
! Security reserve (reserve initially at 0%)
forall(t in TIME)
Reserve(t):= sum(p in TYPES) PMAX(p)*work(p,t) >= DEM(t)
! Limit number of available generators; numbers of generators are integer
forall(p in TYPES, t in TIME) do
work(p,t) <= AVAIL(p)
work(p,t) is_integer
end-do
! Uncomment the following line to see the Optimizer log
! setparam("XPRS_verbose", true)
! Solve as an LP problem and save the basis
setparam("XPRS_PRESOLVE", 0)
minimize(XPRS_LPSTOP, Cost)
savebasis(bas)
setparam("XPRS_PRESOLVE", 1)
! Define a user graph
svgaddgroup("Graph", "Reserve %")
! Solve the problem with different values of reserve (from 0% to 20%)
forall(r in 0..20) do
forall(t in TIME) ! Redefine the reserve constraints
Reserve(t):= sum(p in TYPES) PMAX(p)*work(p,t) >= (1+r/100)*DEM(t)
loadprob(Cost) ! Load problem into the Optimizer
loadbasis(bas) ! Load the saved basis
minimize(Cost) ! Solve the modified problem
writeln(r, "%: ", getobjval) ! Print solution value
svgaddpoint("Graph", r, getobjval/1000) ! Display solution in user graph
end-do
! Scale the size of the displayed graph
svgsetgraphscale(10)
svgsetgraphpointsize(3)
svgsave("a6electr.svg")
svgrefresh
svgwaitclose
end-model
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