AMPL编程说课讲解

AMPL编程主讲人：贾海成任意选修课IntroductionAMPL:AModelingLanguageforMathematicalProgrammingWebsite:解决最优化问题模型的计算机软件(30多种)IntroductionLINDO/LINGOExcelAMPL解决最优化问题模型的计算机软件IntroductionIntroductionIntroduction什么是最优化问题？所谓最优化问题一般是指按照给定的标准在某些约束条件下选取最优的解集。主要分支：线性规划、整数规划、二次规划、非线性规划、随机规划、动态规划、组合最优化、无限维最优化IntroductionAsimpletwo-variablelinearprogramASteelcompanytwoproducts:BandsandCoilsproductionrate(Tonsperhour):Bands200;Coils140.Profitmargin(Profitperton):Bands$25;Coils$30.

IntroductionIntroductionMathematicmodelIntroductionMaximizeSubjecttoSubjecttoSubjecttoHowtosolveit?Method1IntroductionMethod2IntroductionMethod2IntroductionMethod2IntroductionThetwo-variablelinearprograminAMPLProd0.modprod0.runIntroductionAddingonevariableASteelcompanyThreeproducts:Bands,CoilsandPlateProductionrate(Tonsperhour):Bands200;Coils140;Plate160.Profitmargin(Profitperton):Bands$25;Coils$30;Plate$29.

IntroductionIntroductionMathematicmodelIntroductionMaximizeSubjectto

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varXB;varXC;varxp;maximizeProfit:25*XB+30*XC+29*xp;subjecttoTime:(1/200)*XB+(1/140)*XC+(1/160)*xp<=40;subjecttoB_limit:0<=XB<=6000;subjecttoC_limit:0<=XC<=4000;subjecttop_limit:0<=xp<=3500;Prod1.modIntroductionmodelprod1.mod;solve;displayXB,XC,xp;quit;Prod1.runIntroductionMINOS5.5:optimalsolutionfound.2iterations,objective196400XB=6000XC=0xp=1600ToolcompletedsuccessfullyresultsIntroductionIntroductionIntroductionIntroductionmodelprod.mod;dataprod.dat;solve;displayx;quit;prod.runIntroductionIntroductionContinuepreviousexample生产这三种钢材的工艺分为2个阶段：reheat/rolling，这2个阶段最大的可利用时间为每周35hours和40hours。Introduction数学模型Sets:P–asetofproductsS–asetofprocessesinproductionParameters

=profitperunitofproductj,=unitproductiontimeinstageJforproducti,=worktimeinweekly,=upperlimitonamountofproductjsoldinperiod,数学模型Variables:Xi=unitsofproductjtobeproducedJanuary,数学模型数学模型(1)

st(2)

(3)

IntroductionSteel2.modIntroductionSteel2.datDietproblem:minimizingcostsChapter2Dietproblem:minimizingcosts每包的单价Thedinnerprovidethefollowingpercentages,perpackage,oftheminimumdailyrequirementsforvitaminsA,C,B1andB2.Dietproblem:minimizingcostsTheproblemistofindthecheapestcombinationofpackagesthatwillmeetaweek’srequirements.Dietproblem:minimizingcostsDietproblem:minimizingcostsDietproblem:minimizingcostsDietproblem:minimizingcosts数学模型Dietproblem:minimizingcostsSets:P–asetofproductsI–asetofingredientsParameters:=unitcostforproductj,=percentageoftheminimumdailyrequirementforingredientiinoneunitofproductj,

Dietproblem:minimizingcosts=minimumweek’srequirement(inpercentage)foringredientiintheblend,=maximumweek’srequirement(inpercentage)foringredientiintheblend,

=lowerlimitonamountofproductjpurchasedperweek,

=upperlimitonamountofproductjpurchasedperweek,Dietproblem:minimizingcostsVariables:Xj=numberofunitsofproductjtobepurchasedperweek,Formulation:(1)

st(2)

(3)

AMPL程序Dietproblem:minimizingcostsDietproblem:minimizingcostsParamn_mindefault700;Paramn_maxdefault10000;Dietproblem:minimizingcosts计算结果？扩展：限定每种食物一周只能吃2－10包。同时增加对2种元素的考查：钠（Na）和热量(calories)各食物中Na的含量为：beef938,chk2180,fish945,ham278,mch1182,mtl896,spg1329,tur1397.Dietproblem:minimizingcosts扩展：1、限定每种食物一周只能吃2－10包。2、增加对2种元素的考查：钠（Na）和热量(calories)各食物中calories的含量为：beef295,chk770,fish440,ham430,mch315,mtl400,spg370,tur450.Dietproblem:minimizingcosts钠（Na）的最高摄取量为40000热量(calories)的摄取量范围为16000－24000。Dietproblem:minimizingcostsDietproblem:minimizingcostsVarbuy{jinFOOD}integer>=f_min[j],<=f_max[j];Dietproblem:minimizingcostsSupposethatwehavetoproducesteelcoilsatthreemilllocation,inthefollowingamounts:TransportationModelsTheseproductsmustbeshippedinvariousamountstomeetordersatsevenlocationsofautomobilefactories.TransportationModelsTransportationModelsShippingcostspertonisgiveninbelowTransportationModelsWhatistheleastexpensiveplanforshippingthecoilsfrommillstoplans?TransportationModelsTransportationModelsTransportationModelsTransportationModelsMathematicmodelTransportationModels

supposethatweareshippingseveraldifferentproducts:bands,coils,plate.AmulticommoditytransportationmodelTheseproductsareproducedatthreemilllocationsGARYCLEVPITTBands400700800Coils80016001800plate200300300AmulticommoditytransportationmodelAmulticommoditytransportationmodelThedemandsofsevenautomobilefactoriesFRADETLANWINSTLFRELAFbands30030010075650225250Coils500750400250950850500plate100100050200100250AmulticommoditytransportationmodelTotalamountsofshipmentsfromoriginstodestinationsofallproductsmaynotexceed625tons.AmulticommoditytransportationmodelShippingcostspertonsTobandsFRADETLANWINSTLFRELAFGARY301081011716CLEV227107218213PITT19111210258315AmulticommoditytransportationmodelShippingcostspertonsTocoilsFRADETLANWINSTLFRELAFGARY3914111416828CLEV279129269517PITT24141713289920AmulticommoditytransportationmodelShippingcostspertonsToplateFRADETLANWINSTLFRELAFGARY4115121617868CLEV299139289918PITT261417133110420MathematicmodelAmulticommoditytransportationmodelAddanewparameterThefixcostoftransportationfromorigintodestinationAmulticommoditytransportationmodel线性规划模型的敏感性分析例1．某工厂在计划期内要安排I、II两种产品的生产，已知生产单位产品所需的设备台时及A、B两种原材料的消耗，以及资源的限制，如下表所示。III资源限制设备11300台时原料A21400kg原料B01250kg线性规划模型的敏感性分析该工厂每生产一单位产品I可获利50元，每生产一单位产品II可获利100元，问工厂应分别生产多少个I产品和II产品才能使工厂获利最多？线性规划模型的敏感性分析1、加载CPLEX组建，（敏感性分析和binary变量都可以用）optionsolvercplex;2、敏感性分析：optioncplex_options'sensitivity';线性规划模型的敏感性分析5、敏感性分析的各值：变量：

*.lb——指该值（变量或限制条件）的下限（lowerbound）*.ub——指该值（变量或限制条件）的上限（upperbound）*.slack——指当前值与上限或下限的差值。(thedifferenceofavariable’svaluefromitsnearerbound)线性规划模型的敏感性分析5、敏感性分析的各值：变量：*.rc——指变量的影子价格，（reducecost）。当变量值提高一个unit，则目标函数值objective将提高*.rc的值；反之亦然。(theroleofthedualvalueisplayedbythevariable’sso-calledreducedcost.)线性规划模型的敏感性分析5、敏感性分析的各值：限制条件：*.up,*.down——指限制方程的值在up和down之间波动，将遵循影子价格（*.dual）的变化规律。*.dual——影子价格，指限制条件的影子价格，即当限制条件提高一个unit，目标函数指objective将提高*.dual的值；反之亦然。

线性规划模型的敏感性分析5、敏感性分析的各值：*.current——指限制条件的右端值。*.body——指限制条件的左端值值。线性规划模型的敏感性分析setProd;setMaterials;paramprofit{iinProd};paramrate{iinProd};paramarm{iinProd,jinMaterials};paramavail_time;paramavail_res{jinMaterials};varMake{iinProd}>=0;maximizetotal_profit:sum{iinProd}profit[i]*Make[i];subjecttoman_hour:sum{iinProd}rate[i]*Make[i]<=avail_time;subjecttoresource{jinMaterials}:sum{iinProd}arm[i,j]*Make[i]<=avail_res[j];线性规划模型的敏感性分析setProd:=Prod1Prod2;setMaterials:=MAMB;paramprofit:=Prod150Prod2100;paramrate:=Prod11Prod21;paramarm:MAMB:=Prod120Prod211;paramavail_time:=300;paramavail_res:=MA400MB250;线性规划模型的敏感性分析modelsensitivity.mod;datasensitivity.dat;optionsolvercplex;optioncplex_options'sensitivity';solve;displaytotal_profit>sensitivity.sol;displayMake.lb,Make,Make.ub,Make.rc,Make.slack>sensitivity.sol;displayman_hour.up,man_hour.body,man_hour.current,man_hour.down,man_hour.dual>sensitivity.sol;displayresource.up,resource.body,resource.current,resource.down,resource.dual>sensitivity.sol;quit;线性规划模型的敏感性分析:Make.lbMakeMake.ubMake.rcMake.slack:=Prod1050Infinity050Prod2025025000线性规划模型的敏感性分析Aggregateplanningmodel某公司预测的市场需求如下表所示。每件产品需加工20小时，工人每天工作8小时。招收工人需支出广告费、考试费和培训费，折合一个工人300元，裁减一个工人需付解雇费200元。公司必须付给每名工人25元/小时的工资。假设生产中无废品和返工。为了应付需求波动，有1000件产品作为安全库存。单位维持库存费为6元/件.月。设每年的需求类型相同。因此，在计划年度开始时的工人数等于计划年度结束时的工人数。Aggregateplanningmodel预测需求量和每月工作天数月份123456789101112需求160014001200100015002000250025003000300025002000工作天212222212321212020201922参数：T—周期数；Dt—每周期的需求，t=1…T；K—每件产品需要加工的时间；h—每个工人每天工作时间；综合生产计划数学模型

参数：Ch—雇佣工人成本；Cf—解雇工人成本；I0—计划期开始时的库存；Ci—每单位产品维持库存费用；nt—每月的工作天数；W0—期初工人人数；S－工人每小时工资；综合生产计划数学模型

变量：Wt＝每一周期的在职工人数，Ht＝每一周期的雇佣工人数，Ft＝每一周期的解雇工人数，Pt＝每一周期生产产品数，It＝每一周期的库存水平，综合生产计划数学模型

综合生产计划数学模型

目标函数：约束条件：，

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综合生产计划数学模型

约束条件：，

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AMPL语音对“时间参数”的定义：mod文件paramT;paramdemand{tin1..T};data文件paramT:=12;paramdemand:=116002140031200综合生产计划数学模型

paraminvcost;paramworkdays{tin1..T};paramworkhours;paramdemand{tin1..T};paramK