1、優(yōu)化建模與LINGO計(jì)算例1 加工奶制品的生產(chǎn)計(jì)劃1桶牛奶 3kgA1 12h 8h 4kgA2 或獲利24元/kg 獲利16元/kg 50桶牛奶 時(shí)間480h 至多加工100kgA1 制訂生產(chǎn)計(jì)劃，使每天獲利最大 35元可買(mǎi)到1桶牛奶，買(mǎi)嗎？若買(mǎi)，每天最多買(mǎi)多少? 可聘用臨時(shí)工人，付出的工資最多是每小時(shí)幾元? A1的獲利增加到 30元/kg，應(yīng)否改變生產(chǎn)計(jì)劃？ 每天：?jiǎn)栴}1桶牛奶 3kgA1 12h 8h 4kgA2 或獲利24元/kg 獲利16元/kg x1桶牛奶生產(chǎn)A1 x2桶牛奶生產(chǎn)A2 獲利 243x1 獲利 164 x2 原料供應(yīng) 勞動(dòng)時(shí)間 加工能力 決策變量 目標(biāo)函數(shù) 每天獲利約

2、束條件非負(fù)約束 線性規(guī)劃模型(LP)時(shí)間480h 至多加工100kgA1 50桶牛奶 每天基本模型目標(biāo)函數(shù) 約束條件線性規(guī)劃模型(LP)model:max = 72*x1+64*x2;milk x1 + x250;time 12*x1+8*x2480;cpct 3*x1100;end 模型求解 軟件實(shí)現(xiàn) LINGO Global optimal solution found. Objective value: 3360.000 Total solver iterations: 2 Variable Value Reduced Cost X1 20.00000 0.000000 X2 30.00

3、000 0.000000 Row Slack or Surplus Dual Price 1 3360.000 1.000000 MILK 0.000000 48.00000 TIME 0.000000 2.000000 CPCT 40.00000 0.000000 結(jié)果解釋 Global optimal solution found. Objective value: 3360.000 Total solver iterations: 2 Variable Value Reduced Cost X1 20.00000 0.000000 X2 30.00000 0.000000 Row Sla

4、ck or Surplus Dual Price 1 3360.000 1.000000 MILK 0.000000 48.00000 TIME 0.000000 2.000000 CPCT 40.00000 0.000000 model:max = 72*x1+64*x2;milk x1 + x250;time 12*x1+8*x2480;cpct 3*x1100;end三種資源“資源” 剩余為零的約束為緊約束（有效約束） 原料無(wú)剩余時(shí)間無(wú)剩余加工能力剩余40結(jié)果解釋 Global optimal solution found. Objective value: 3360.000 Total

5、 solver iterations: 2 Variable Value Reduced Cost X1 20.00000 0.000000 X2 30.00000 0.000000 Row Slack or Surplus Dual Price 1 3360.000 1.000000 MILK 0.000000 48.00000 TIME 0.000000 2.000000 CPCT 40.00000 0.000000最優(yōu)解下“資源”增加1單位時(shí)“效益”的增量 影子價(jià)格 35元可買(mǎi)到1桶牛奶，要買(mǎi)嗎?35 48, 應(yīng)該買(mǎi)！ 聘用臨時(shí)工人付出的工資最多每小時(shí)幾元? 2元！原料增加1單位, 利潤(rùn)

6、增長(zhǎng)48 時(shí)間增加1單位, 利潤(rùn)增長(zhǎng)2 加工能力增長(zhǎng)不影響利潤(rùn)Ranges in which the basis is unchanged: Objective Coefficient Ranges Current Allowable AllowableVariable Coefficient Increase Decrease X1 72.00000 24.00000 8.000000 X2 64.00000 8.000000 16.00000 Righthand Side Ranges Row Current Allowable Allowable RHS Increase Decreas

7、e MILK 50.00000 10.00000 6.666667 TIME 480.0000 53.33333 80.00000 CPCT 100.0000 INFINITY 40.00000 最優(yōu)解不變時(shí)目標(biāo)函數(shù)系數(shù)允許變化范圍 敏感性分析 (“LINGO|Ranges” ) x1系數(shù)范圍(64,96) x2系數(shù)范圍(48,72) A1獲利增加到 30元/kg，應(yīng)否改變生產(chǎn)計(jì)劃? x1系數(shù)由24 3=72增加為303=90，在允許范圍內(nèi) 不變！(約束條件不變)結(jié)果解釋 Ranges in which the basis is unchanged: Objective Coefficient

8、 Ranges Current Allowable AllowableVariable Coefficient Increase Decrease X1 72.00000 24.00000 8.000000 X2 64.00000 8.000000 16.00000 Righthand Side Ranges Row Current Allowable Allowable RHS Increase Decrease MILK 50.00000 10.00000 6.666667 TIME 480.0000 53.33333 80.00000 CPCT 100.0000 INFINITY 40.

9、00000影子價(jià)格有意義時(shí)約束右端的允許變化范圍 原料最多增加10 時(shí)間最多增加53 35元可買(mǎi)到1桶牛奶, 每天最多買(mǎi)多少?最多買(mǎi)10桶!(目標(biāo)函數(shù)不變)充分條件 !IP可用LINGO直接求解整數(shù)規(guī)劃(Integer Programming,簡(jiǎn)記IP)Model:max=2*x1+3*x2+4*x3;1.5*x1+3*x2+5*x3600;280*x1+250*x2+400*x360000;gin(x1);gin(x2);gin(x3);endmax=2*x1+3*x2+4*x3;1.5*x1+3*x2+5*x3600;280*x1+250*x2+400*x30;x2*(x2-80)0;x3

10、*(x3-80)0;gin(x1);gin(x2);gin(x3);方法3：化為非線性規(guī)劃 非線性規(guī)劃（Non- Linear Programming，簡(jiǎn)記NLP） 若生產(chǎn)某類(lèi)汽車(chē)，則至少生產(chǎn)80輛，求生產(chǎn)計(jì)劃. x1=0 或 80 x2=0 或 80 x3=0 或 80最優(yōu)解同前.一般地，整數(shù)規(guī)劃和非線性規(guī)劃的求解比線性規(guī)劃困難得多，特別是問(wèn)題規(guī)模較大或者要求得到全局最優(yōu)解時(shí). 目標(biāo)函數(shù)若xij=1, 或 xij=0 0-1規(guī)劃約束條件Model：MIN=66.8*x11+75.6*x12+87*x13+58.6*x14 + +67.4*x51+71* x52+83.8*x53+62.4*x

11、54； x11+x12+x13+x14 =1； x14+x24+x34+x44+x54 =1；bin(x11); bin(x54); END 模型求解 MODEL:sets: person/1.5/; position/1.4/; link(person,position): c, x;endsetsdata: c= 66.8, 75.6, 87, 58.6, 57.2, 66, 66.4, 53, 78, 67.8, 84.6, 59.4, 70, 74.2, 69.6, 57.2, 67.4, 71, 83.8, 62.4;enddata輸入LINGO求解 min=sum(link: c*x);for(person(i): sum(position(j):x(i,j)=1;);for