數(shù)量模型與算法基礎(chǔ)：第六章 線性規(guī)劃

線性規(guī)劃LinearPrograming（LP)Eg1程序和結(jié)果x=26z=26%booklpch1p1ex1.1clc;clear;pkgloadoptimc=[-4-3]'A=[211101]b=[1087]Aeq=[]beq=[]lb=[0;0]ub=[]%C:\Octave\Octave-4.0.0\src\build_packages.m,%pkgloadoptim[xz]=linprog(c,A,b,Aeq,beq,lb,ub);xz=-zc=-4-3A=211101b=1087lb=00ub=[](0x0)安裝優(yōu)化包%installthepackagestry_installstruct-1.0.11.tar.gz%保留try_installoptim-1.4.1.tar.gz%保留pkgloadoptim%添加決策變量，目標(biāo)函數(shù)，約束條件MATLAB規(guī)定的linprog()標(biāo)準(zhǔn)形式其中c和x為n維列向量，A、Aeq為適當(dāng)維數(shù)的矩陣，b、beq為適當(dāng)維數(shù)的列向量。z是求出的目標(biāo)極值解：[xz]=linprog(A,b,Aeq,beq,lb,ub]一般形式如何轉(zhuǎn)換成Matlab標(biāo)準(zhǔn)型例子Ex1.2p2c=[-2-35]’A=-25-1131b=-1012Aeq=111beq=7[xz]=linprog(c,A,b,Aeq,beq,lb,ub)lb=[0;0;0]ub=[][xz]=linprog(c,A,b,Aeq,beq,lb,ub)x=6.428570.571430.00000z=14.571例子Ex1.3p3c=231a=142320b=86Aeq=[]Beq=[]lb=[0;0;0]ub=[][xz]=linprog(c,-a,-b,[],[],[0;0;0],[])例子Ex1.5書(shū)p4習(xí)題1.3（書(shū)，第9頁(yè))要求：用Octave/MATLABlinprog(c,A,b,Aeq,beq,lb,ub),并分別寫(xiě)出c,A,b,Aeq,beq,lb,ub各矩陣。習(xí)題1.3(續(xù)）習(xí)題1.3(續(xù)）

盡量獨(dú)立解題。以下提示不一定全對(duì)，僅供參考！product(1,a1)(1,a2)(1,b1)1,b21,b32,a12,a22,b13,a2,b2fee/hrequipcostavailablehrsx1x2x3x4x5x6x7x8x90.250.250.250.350.50.0530060005100.03213211000079120.06252504000680.111978370004110.05200400071.251.25222.8product(1,a1)(1,a2)(1,b1)1,b21,b32,a12,a22,b13,a2,b2fee/hravail.Hrs.peryear?x1x2x3x4x5x6x7x8x90.250.250.350.350.50.0530060000.2500000.50000.03213211000000.224700000.288900.38520.06252504000000.37500000.500.111978370000000.44742900001.2304290.05200400000000.3500001.251.25222.8c=0.750.7753-0.375-0.44743-0.351.151.3611-0.50.684371習(xí)題1.3(續(xù)）習(xí)題1.3(續(xù)）LingovsOctaveOctave一個(gè)解%z=1146.57%x=%%0.00%858.62%341.38%0.00%0.00%230.05%0.00%500.00%324.14Sourcecodesectionforex.1.3clc;clearpkgloadoptim%loadoptimpackagec(1,1)=(1.25-0.25)-300/6000*5;c(2,1)=(1.25-0.25)-321/10000*7;c(3,1)=-250/4000*6;c(4,1)=-783/7000*4;c(5,1)=-200/4000*7;c(6,1)=-300/6000*10;c(7,1)=-321/10000*9;c(8,1)=(2-0.35)-250/4000*8;c(9,1)=(2.8-.5)-321/10000*12-783/7000*11;c=-c%maxsignalA=[500001000007000090120060000800004000011000070000]b=[600010000400070004000]Aeq=[11-1-1-100000000011-10];beq=[00];lb=[000000000]’;ub=[];formatbank;[xz]=linprog(c,A,b,Aeq,beq,lb,ub);z=-zx1.2投資的效益和風(fēng)險(xiǎn)(書(shū)，p.9)

(既要。。。又要。。。)Maxz=(r0-p0)x0+(r1-p1)x1+...+(r4-p4)x4Min(max[q0x0,p1x1,...,p4x4])X0>=0,X1>=0,X2>=0,X3>=0,X4>=0

(1-p0)x0+(1-p1)x1+...+(1-p4)x4多目標(biāo)LP簡(jiǎn)化為單目標(biāo)LP：

模型1：固定風(fēng)險(xiǎn)ai=0,1,2,3,4LP1:

固定風(fēng)險(xiǎn)a=0.006，用1次linprog()LP1:

固定風(fēng)險(xiǎn)a=0.025,用1次linprog()LP模型1:plot(a,Q)

循環(huán)風(fēng)險(xiǎn)a=0.001,…,0.006,…0.025,…,0.050

用50次linprog()LP1：結(jié)果分析

-給投資咨詢顧客的報(bào)告Risk(%）

=0.000.100.200.300.400.500.600.700.800.90X=1.000.830.660.490.330.160.000.000.000.000.000.040.080.120.160.200.240.280.320.360.000.070.130.200.270.330.400.470.530.600.000.020.040.050.070.090.110.130.130.020.000.040.080.120.150.190.220.100.000.00Gain

=0.050.080.100.130.150.180.200.210.210.22LP1：結(jié)果分析

-給投資咨詢顧客的報(bào)告LP1：結(jié)果分析

-給投資咨詢顧客的報(bào)告復(fù)習(xí)題已知線性規(guī)劃問(wèn)題，復(fù)習(xí)題（1）會(huì)寫(xiě)出在MATLAB中求解所需輸入的7個(gè)參數(shù)：(！?。。┠切┦牵邢蛄浚┠切┦牵ň仃嚕┠切┦牵ㄐ邢蛄浚簺](méi)有在有Max目標(biāo)函數(shù)時(shí)，哪個(gè)參數(shù)變化，如何變化在有>=約束條件時(shí)，哪個(gè)參數(shù)變化，如何變化

復(fù)習(xí)題能寫(xiě)出在MATLAB中求解x和z的程序：(！??！)

9條基本語(yǔ)句在有Max目標(biāo)函數(shù)時(shí)，哪個(gè)參數(shù)和結(jié)果變化，如何變化在有>=約束條件時(shí)，哪個(gè)參數(shù)變化，如何變化

復(fù)習(xí)題MATLAB規(guī)定linprog()標(biāo)準(zhǔn)型是求目標(biāo)函數(shù)的（

）最小值如求最大值，如何處理？復(fù)習(xí)題搞清linprog()要求輸入的參數(shù)c,A,b,lb,ub在生產(chǎn)規(guī)劃模型中通常代表什么實(shí)際意義？產(chǎn)品價(jià)格或利潤(rùn)系統(tǒng)消耗資源約束產(chǎn)量向量上限產(chǎn)量向量下限復(fù)習(xí)題掌握怎樣在線性規(guī)劃投資組合模型中設(shè)置風(fēng)險(xiǎn)控制？（

）1約束最大風(fēng)險(xiǎn)損失率

2約束平均信用等級(jí)3