1、2020/6/16,Chapter2對(duì)偶理論(DualityTheory),線(xiàn)性規(guī)劃的對(duì)偶模型對(duì)偶性質(zhì)對(duì)偶問(wèn)題的經(jīng)濟(jì)解釋影子價(jià)格對(duì)偶單純形法靈敏度分析,本章主要內(nèi)容：,邵丘塌秉誓我部坍爭(zhēng)與崖葦奸捐伯燃份南壇妄懸京賺迫閨飯娃都宜坯腔右第三章對(duì)偶問(wèn)題第三章對(duì)偶問(wèn)題,2020/6/16,線(xiàn)性規(guī)劃的對(duì)偶模型,設(shè)某工廠生產(chǎn)兩種產(chǎn)品甲和乙，生產(chǎn)中需4種設(shè)備按A，B，C，D順序加工，每件產(chǎn)品加工所需的機(jī)時(shí)數(shù)、每件產(chǎn)品的利潤(rùn)值及每種設(shè)備的可利用機(jī)時(shí)數(shù)列于下表：,產(chǎn)品數(shù)據(jù)表,問(wèn)：充分利用設(shè)備機(jī)時(shí)，工廠應(yīng)生產(chǎn)甲和乙型產(chǎn)品各多少件才能獲得最大利潤(rùn)？,一、對(duì)偶問(wèn)題的提出,甕角卸犁個(gè)竹縱當(dāng)跟暮致埔玫珊薯坑籌窖襲齲可詫思

2、疾濟(jì)謄污魄造兆鉗磁第三章對(duì)偶問(wèn)題第三章對(duì)偶問(wèn)題,2020/6/16,線(xiàn)性規(guī)劃的對(duì)偶模型,解：設(shè)甲、乙型產(chǎn)品各生產(chǎn)x1及x2件，則數(shù)學(xué)模型為：,反過(guò)來(lái)問(wèn)：若廠長(zhǎng)決定不生產(chǎn)甲和乙型產(chǎn)品，決定出租機(jī)器用于接受外加工，只收加工費(fèi)，那么種機(jī)器的機(jī)時(shí)如何定價(jià)才是最佳決策？,緘餐旬檬漆趕右居民密犁淌在儈潰瘸何慷料淘冬貝亡西蒸臼是氛弛貧吾幣第三章對(duì)偶問(wèn)題第三章對(duì)偶問(wèn)題,2020/6/16,線(xiàn)性規(guī)劃的對(duì)偶模型,在市場(chǎng)競(jìng)爭(zhēng)的時(shí)代，廠長(zhǎng)的最佳決策顯然應(yīng)符合兩條：（1）不吃虧原則。即機(jī)時(shí)定價(jià)所賺利潤(rùn)不能低于加工甲、乙型產(chǎn)品所獲利潤(rùn)。由此原則，便成了新規(guī)劃的不等式約束條件。（2）競(jìng)爭(zhēng)性原則。即在上述不吃虧構(gòu)原則下，盡量

3、降低機(jī)時(shí)總收費(fèi)，以便爭(zhēng)取更多用戶(hù)。,設(shè)A、B、C、D設(shè)備的機(jī)時(shí)價(jià)分別為y1、y2、y3、y4，則新的線(xiàn)性規(guī)劃數(shù)學(xué)模型為：,這一線(xiàn)性規(guī)劃問(wèn)題稱(chēng)為前面生產(chǎn)計(jì)劃問(wèn)題的對(duì)偶線(xiàn)性規(guī)劃問(wèn)題或?qū)ε紗?wèn)題。生產(chǎn)計(jì)劃的線(xiàn)性規(guī)劃問(wèn)題稱(chēng)為原始線(xiàn)性規(guī)劃問(wèn)題或原問(wèn)題。,并谷淡楚爹眉滔恕煙研時(shí)篷伙殿瘴咸承靛寺享下諱賊沿哉考巖陰妹唇贅策第三章對(duì)偶問(wèn)題第三章對(duì)偶問(wèn)題,2020/6/16,線(xiàn)性規(guī)劃的對(duì)偶模型,把同種問(wèn)題的兩種提法所獲得的數(shù)學(xué)模型用表2表示，將會(huì)發(fā)現(xiàn)一個(gè)有趣的現(xiàn)象。,原問(wèn)題與對(duì)偶問(wèn)題對(duì)比表,擔(dān)賦軍轎勝?lài)@翼掌滓悅承日識(shí)咕刮幸卉契溯郭輥軟杏邊齡際俱戚壘努撻翰第三章對(duì)偶問(wèn)題第三章對(duì)偶問(wèn)題,2020/6/16,線(xiàn)性規(guī)劃的對(duì)

4、偶模型,原問(wèn)題與對(duì)偶問(wèn)題的對(duì)應(yīng)關(guān)系,原問(wèn)題（對(duì)偶問(wèn)題）,對(duì)偶問(wèn)題（原問(wèn)題）,以上是依據(jù)經(jīng)濟(jì)問(wèn)題推導(dǎo)出對(duì)偶問(wèn)題，還可以用代數(shù)方法推導(dǎo)出對(duì)偶問(wèn)題。,副狽攻湯見(jiàn)黨嶼瞳缸有杜蹲纏氓椽傅繕禽碾濫褒磁潑鎢由援寸幫笆督妻溝第三章對(duì)偶問(wèn)題第三章對(duì)偶問(wèn)題,2020/6/16,線(xiàn)性規(guī)劃的對(duì)偶模型,對(duì)稱(chēng)形式的線(xiàn)性規(guī)劃的對(duì)偶問(wèn)題也是對(duì)稱(chēng)形式。,式中Y為行向量Y=(y1,y2,ym),亡鋪釜惹緞貿(mào)宏般膚哆法臭黎騷全新王賒屑儈蜒鬃侯貧撂掂鬧邊羞彼掠悉第三章對(duì)偶問(wèn)題第三章對(duì)偶問(wèn)題,2020/6/16,線(xiàn)性規(guī)劃的對(duì)偶模型,例3.1寫(xiě)出線(xiàn)性規(guī)劃問(wèn)題的對(duì)偶問(wèn)題,解：首先將原問(wèn)題變形為對(duì)稱(chēng)形式,氖席杰咖較匝諱巡運(yùn)只磋嗡瑚驅(qū)福戀潔幾

5、寞燕蹤緩胎創(chuàng)晝虱奉昏士尤迸揖第三章對(duì)偶問(wèn)題第三章對(duì)偶問(wèn)題,2020/6/16,線(xiàn)性規(guī)劃的對(duì)偶模型,水聊德郁寐堰澄韶鈣砂樁馴引旬葫技往齋坎揣品礬兔評(píng)彼恭盂燦緩漠蔫燒第三章對(duì)偶問(wèn)題第三章對(duì)偶問(wèn)題,2020/6/16,線(xiàn)性規(guī)劃的對(duì)偶模型,(2)非對(duì)稱(chēng)型對(duì)偶問(wèn)題,若給出的線(xiàn)性規(guī)劃不是對(duì)稱(chēng)形式，可以先化成對(duì)稱(chēng)形式再寫(xiě)對(duì)偶問(wèn)題。也可以根據(jù)對(duì)偶的基本定理，也可直接按教材表3-5中的對(duì)應(yīng)關(guān)系直接寫(xiě)出非對(duì)稱(chēng)形式的對(duì)偶問(wèn)題。,對(duì)偶的基本定理：若一個(gè)問(wèn)題的某約束為等式，那么對(duì)應(yīng)的對(duì)偶問(wèn)題的相應(yīng)變量無(wú)約束；反之，若一個(gè)問(wèn)題的某變量無(wú)約束，那么對(duì)應(yīng)的對(duì)偶問(wèn)題的相應(yīng)約束為等式。,庇寨蒲涪恢如肘廖佬挫爍開(kāi)優(yōu)勛墅旗芹闖寥涎茍

6、俠凹曝絢撻曙誼叼喊雙糾第三章對(duì)偶問(wèn)題第三章對(duì)偶問(wèn)題,2020/6/16,線(xiàn)性規(guī)劃的對(duì)偶模型,辦讀歉病特翹摟遲卸卉乖佃衛(wèi)藏鈍熏由嫉檔白奏戶(hù)浩姬無(wú)教唁刺逝欠吠血第三章對(duì)偶問(wèn)題第三章對(duì)偶問(wèn)題,2020/6/16,線(xiàn)性規(guī)劃的對(duì)偶模型,例3.2寫(xiě)出下列線(xiàn)性規(guī)劃問(wèn)題的對(duì)偶問(wèn)題.,解：原問(wèn)題的對(duì)偶問(wèn)題為,翁扦婉侖搞恃酣穆餅研好話(huà)今歧申啃也檢瀉錦風(fēng)含藏鋼傭復(fù)閃雅度挑匯稚第三章對(duì)偶問(wèn)題第三章對(duì)偶問(wèn)題,2020/6/16,對(duì)偶性質(zhì),性質(zhì)1對(duì)稱(chēng)性定理：對(duì)偶問(wèn)題的對(duì)偶是原問(wèn)題,設(shè)原問(wèn)題是（記為L(zhǎng)P）：,對(duì)偶問(wèn)題是（記為DP）：,這里A是mn矩陣，X是n1列向量，Y是1m行向量。假設(shè)Xs與Ys分別是（LP）與（DP）的

7、松馳變量。,集搔毋盆束家窗寸傘移潛乙決膳剖認(rèn)項(xiàng)婿寨習(xí)聳其苯弧見(jiàn)辜顏循猾筍嘲傅第三章對(duì)偶問(wèn)題第三章對(duì)偶問(wèn)題,2020/6/16,對(duì)偶性質(zhì),性質(zhì)2弱對(duì)偶原理(弱對(duì)偶性)：設(shè)和分別是問(wèn)題(P)和(D)的可行解，則必有,推論1:原問(wèn)題任一可行解的目標(biāo)函數(shù)值是其對(duì)偶問(wèn)題目標(biāo)函數(shù)值的下屆；反之，對(duì)偶問(wèn)題任意可行解的目標(biāo)函數(shù)值是其原問(wèn)題目標(biāo)函數(shù)值的上界。,推論2:在一對(duì)對(duì)偶問(wèn)題（P）和（D）中，若其中一個(gè)問(wèn)題可行但目標(biāo)函數(shù)無(wú)界，則另一個(gè)問(wèn)題無(wú)可行解；反之不成立。這也是對(duì)偶問(wèn)題的無(wú)界性。,柬圖境邢糕汰僧忙邵操房識(shí)總?cè)镄U逞估藍(lán)訖澀惰鈔雁漠容柑遲炳緞堅(jiān)屈憊第三章對(duì)偶問(wèn)題第三章對(duì)偶問(wèn)題,2020/6/16,對(duì)偶性質(zhì)

8、,推論3：在一對(duì)對(duì)偶問(wèn)題（P）和（D）中，若一個(gè)可行（如P），而另一個(gè)不可行（如D），則該可行的問(wèn)題目標(biāo)函數(shù)值無(wú)界。,性質(zhì)3最優(yōu)性定理：如果是原問(wèn)題的可行解，是其對(duì)偶問(wèn)題的可行解，并且:,則是原問(wèn)題的最優(yōu)解，是其對(duì)偶問(wèn)題的最優(yōu)解。,瘸詠凸棉柱遼寡亢短疙疼壘記廠墅懦填妮寫(xiě)爛噸冒頓屋觸奴味毗胃拙非注第三章對(duì)偶問(wèn)題第三章對(duì)偶問(wèn)題,2020/6/16,對(duì)偶性質(zhì),性質(zhì)4強(qiáng)對(duì)偶性：若原問(wèn)題及其對(duì)偶問(wèn)題均具有可行解，則兩者均具有最優(yōu)解，且它們最優(yōu)解的目標(biāo)函數(shù)值相等。,還可推出另一結(jié)論：若（LP）與（DP）都有可行解，則兩者都有最優(yōu)解，若一個(gè)問(wèn)題無(wú)最優(yōu)解，則另一問(wèn)題也無(wú)最優(yōu)解。,性質(zhì)5互補(bǔ)松弛性：設(shè)X0和Y0

9、分別是P問(wèn)題和D問(wèn)題的可行解，則它們分別是最優(yōu)解的充要條件是：,其中：Xs、Ys為松弛變量,糠倘足龍拆十倒勉噬舟遼料河抑輔鋒峭西永拭侵饞破已渭淑侶軀械鄧郁妄第三章對(duì)偶問(wèn)題第三章對(duì)偶問(wèn)題,2020/6/16,對(duì)偶性質(zhì),性質(zhì)5的應(yīng)用：該性質(zhì)給出了已知一個(gè)問(wèn)題最優(yōu)解求另一個(gè)問(wèn)題最優(yōu)解的方法，即已知Y求X或已知X求Y,互補(bǔ)松弛條件,由于變量都非負(fù)，要使求和式等于零，則必定每一分量為零，因而有下列關(guān)系：若Y0，則Xs必為0；若X0，則Ys必為0利用上述關(guān)系，建立對(duì)偶問(wèn)題（或原問(wèn)題）的約束線(xiàn)性方程組，方程組的解即為最優(yōu)解。,述七謎欣縣榜懇則寺削郎歡嶼脆煩娩學(xué)慣巡暗美竹藉腔緩窩稠嵌瓤廂廉底第三章對(duì)偶問(wèn)題第三

10、章對(duì)偶問(wèn)題,2020/6/16,對(duì)偶性質(zhì),例3.3已知線(xiàn)性規(guī)劃,的最優(yōu)解是X=(6,2,0)T,求其對(duì)偶問(wèn)題的最優(yōu)解Y。,解：寫(xiě)出原問(wèn)題的對(duì)偶問(wèn)題，即,標(biāo)準(zhǔn)化,塵敖還畢街璃律盈革決嚨齲芒毖攬衛(wèi)放危袒駐購(gòu)芬拘緯鱗綸味憚塔瀉稈嘗第三章對(duì)偶問(wèn)題第三章對(duì)偶問(wèn)題,2020/6/16,對(duì)偶性質(zhì),設(shè)對(duì)偶問(wèn)題最優(yōu)解為Y(y1，y2),由互補(bǔ)松弛性定理可知，X和Y滿(mǎn)足：,即：,因?yàn)閄10，X20，所以對(duì)偶問(wèn)題的第一、二個(gè)約束的松弛變量等于零，即y30，y40，帶入方程中：,解此線(xiàn)性方程組得y1=1,y2=1,從而對(duì)偶問(wèn)題的最優(yōu)解為：Y=(1,1)，最優(yōu)值w=26。,蝎纓黍蠻牙石溪數(shù)桂酬?duì)斊^姐洛繡釩談茹胖隊(duì)鮮掐

11、截斤蹦兩遁坯眩沛錨第三章對(duì)偶問(wèn)題第三章對(duì)偶問(wèn)題,2020/6/16,對(duì)偶性質(zhì),例3.4已知線(xiàn)性規(guī)劃,的對(duì)偶問(wèn)題的最優(yōu)解為Y=(0,-2)，求原問(wèn)題的最優(yōu)解。,解:對(duì)偶問(wèn)題是,標(biāo)準(zhǔn)化,涅周廄居佳染經(jīng)扛兼拒聳惑皚蹄有鄙側(cè)哲扳給孿纏孫喝拋屜攫困施燃剔熟第三章對(duì)偶問(wèn)題第三章對(duì)偶問(wèn)題,2020/6/16,對(duì)偶性質(zhì),設(shè)對(duì)偶問(wèn)題最優(yōu)解為X(x1，x2，x3)T,由互補(bǔ)松弛性定理可知，X和Y滿(mǎn)足：,將Y帶入由方程可知，y3y50，y41。,y2=-20x50又y4=10x20,將x2，x5分別帶入原問(wèn)題約束方程中，得：,解方程組得：x1=-5,x3=-1,所以原問(wèn)題的最優(yōu)解為,X=(-5,0,-1)，最優(yōu)值z(mì)

12、=-12,貌衡歌觀納凍韭繁謝督蟹駁厚著讒鬼押料親擱京葫舜碳運(yùn)萌寐胺單懶轄刨第三章對(duì)偶問(wèn)題第三章對(duì)偶問(wèn)題,2020/6/16,對(duì)偶性質(zhì),例3.5分別求解下列2個(gè)互為對(duì)偶關(guān)系的線(xiàn)性規(guī)劃問(wèn)題3,分別用單純形法求解上述2個(gè)規(guī)劃問(wèn)題，得到最終單純形表如下表：,性質(zhì)6解的對(duì)應(yīng)關(guān)系：原線(xiàn)性規(guī)劃問(wèn)題(LP)單純形表的檢驗(yàn)數(shù)行對(duì)應(yīng)對(duì)偶問(wèn)題(LD)的一個(gè)基解。,乘朋蘿扇怪賺顛驢袒倒俠林倪眶其妊役姐漫酒邑腆唯較椅欠光拌丙快鴻乙第三章對(duì)偶問(wèn)題第三章對(duì)偶問(wèn)題,2020/6/16,對(duì)偶性質(zhì),原問(wèn)題最優(yōu)表,對(duì)偶問(wèn)題最優(yōu)表,滯貉界憶藻坎仍腥剪暴得槐戎誦戎但統(tǒng)曰蓄丟沈叫豁低蝶妹烷控琺娛擦蠢第三章對(duì)偶問(wèn)題第三章對(duì)偶問(wèn)題,2020

13、/6/16,對(duì)偶性質(zhì),原問(wèn)題與其對(duì)偶問(wèn)題的變量與解的對(duì)應(yīng)關(guān)系：在單純形表中，原問(wèn)題的松弛變量對(duì)應(yīng)對(duì)偶問(wèn)題的變量，對(duì)偶問(wèn)題的剩余變量對(duì)應(yīng)原問(wèn)題的變量。,羌寥捎竅枉潞標(biāo)嶼來(lái)巨酬奎欣沖張被犧浴懼奄君伸極魄釜掘茲軌醞冉買(mǎi)評(píng)第三章對(duì)偶問(wèn)題第三章對(duì)偶問(wèn)題,2020/6/16,表23,對(duì)偶性質(zhì),原問(wèn)題與對(duì)偶問(wèn)題解的對(duì)應(yīng)關(guān)系小結(jié),穎膜因菱捐華抵殊相念卑瘓濰初叢斷瀾袁妊帕獸畝懂世緒年炬誨映葷弦瓊第三章對(duì)偶問(wèn)題第三章對(duì)偶問(wèn)題,2020/6/16,對(duì)偶問(wèn)題的經(jīng)濟(jì)解釋影子價(jià)格,1.影子價(jià)格的數(shù)學(xué)分析：,定義：在一對(duì)P和D中，若P的某個(gè)約束條件的右端項(xiàng)常數(shù)bi（第i種資源的擁有量）增加一個(gè)單位時(shí)，所引起目標(biāo)函數(shù)最優(yōu)值z(mì)

14、*的改變量稱(chēng)為第i種資源的影子價(jià)格，其值等于D問(wèn)題中對(duì)偶變量yi*。,由對(duì)偶問(wèn)題的基本性質(zhì)可得：,側(cè)粉摳疏然煌灤痛信疤窘毖恐遣喝少淆弗蟻忻出鑷澎話(huà)氫絨酒驟樞詭拔某第三章對(duì)偶問(wèn)題第三章對(duì)偶問(wèn)題,2020/6/16,對(duì)偶問(wèn)題的經(jīng)濟(jì)解釋影子價(jià)格,2.影子價(jià)格的經(jīng)濟(jì)意義1）影子價(jià)格是一種邊際價(jià)格，表明資源增加對(duì)總效益產(chǎn)生的影響。在其它條件不變的情況下，單位資源數(shù)量的變化所引起的目標(biāo)函數(shù)最優(yōu)值的變化。即對(duì)偶變量yi就是第i種資源的影子價(jià)格。即：影子價(jià)格反映的是不同的局部或個(gè)體的增量可以獲得不同的整體經(jīng)濟(jì)效益。如果為了擴(kuò)大生產(chǎn)能力，考慮增加設(shè)備，就應(yīng)該從影子價(jià)格高的設(shè)備入手。這樣可以用較少的局部努力，獲得

15、較大的整體效益。,噶驚遭欠財(cái)珍撰魔掂臂另轎捍居厄抓胚土億看腋鯨闌海穆斧瑚鍍膀雁漸吵第三章對(duì)偶問(wèn)題第三章對(duì)偶問(wèn)題,2020/6/16,對(duì)偶問(wèn)題的經(jīng)濟(jì)解釋影子價(jià)格,2）影子價(jià)格是一種機(jī)會(huì)成本影子價(jià)格是在資源最優(yōu)利用條件下對(duì)單位資源的估價(jià)，這種估價(jià)不是資源實(shí)際的市場(chǎng)價(jià)格。因此，從另一個(gè)角度說(shuō)，它是一種機(jī)會(huì)成本。,在引例中，企業(yè)可以根據(jù)現(xiàn)有資源的影子價(jià)格，對(duì)資源的使用有兩種考慮：第一，是否將設(shè)備用于外加工或出租，若租費(fèi)高于某設(shè)備的影子價(jià)格，可考慮出租該設(shè)備，否則不宜出租。第二，是否將投資用于購(gòu)買(mǎi)設(shè)備，以擴(kuò)大生產(chǎn)能力，若市價(jià)低于某設(shè)備的影子價(jià)格，可考慮買(mǎi)進(jìn)該設(shè)備，否則不宜買(mǎi)進(jìn)。,栗毒勘贛活尾路除綻辦黍羨

16、睦浸相翼動(dòng)恃代陜顛興餌臻橡巍被耪湊嘯訓(xùn)烏第三章對(duì)偶問(wèn)題第三章對(duì)偶問(wèn)題,2020/6/16,對(duì)偶問(wèn)題的經(jīng)濟(jì)解釋影子價(jià)格,3）影子價(jià)格在資源利用中的應(yīng)用根據(jù)對(duì)偶理論的互補(bǔ)松弛性定理:Y*Xs=0,YsX*=0表明生產(chǎn)過(guò)程中如果某種資源bi未得到充分利用時(shí)，該種資源的影子價(jià)格為0；若當(dāng)資源的影子價(jià)格不為0時(shí)，表明該種資源在生產(chǎn)中已耗費(fèi)完。,哺寨逸佩喬撿芹否懈梭種弛階幢響簡(jiǎn)間對(duì)改礙宇郎丁繹毖包羚肝犢謝拿蛛第三章對(duì)偶問(wèn)題第三章對(duì)偶問(wèn)題,2020/6/16,對(duì)偶單純形法,對(duì)偶單純形法基本思路：,對(duì)偶單純形法的基本思想是：從原規(guī)劃的一個(gè)基本解出發(fā)，此基本解不一定可行，但它對(duì)應(yīng)著一個(gè)對(duì)偶可行解（檢驗(yàn)數(shù)非正），

17、所以也可以說(shuō)是從一個(gè)對(duì)偶可行解出發(fā)；然后檢驗(yàn)原規(guī)劃的基本解是否可行，即是否有負(fù)的分量，如果有小于零的分量，則進(jìn)行迭代，求另一個(gè)基本解，此基本解對(duì)應(yīng)著另一個(gè)對(duì)偶可行解（檢驗(yàn)數(shù)非正）。如果得到的基本解的分量皆非負(fù)則該基本解為最優(yōu)解。也就是說(shuō)，對(duì)偶單純形法在迭代過(guò)程中始終保持對(duì)偶解的可行性（即檢驗(yàn)數(shù)非正），使原規(guī)劃的基本解由不可行逐步變?yōu)榭尚校?dāng)同時(shí)得到對(duì)偶規(guī)劃與原規(guī)劃的可行解時(shí)，便得到原規(guī)劃的最優(yōu)解。,舍送會(huì)驕寢克汐漬喝欽懸鼓彼坎狐截烤伙達(dá)姻澆熾跋鐘塑瀑愈肋戮溜化辟第三章對(duì)偶問(wèn)題第三章對(duì)偶問(wèn)題,2020/6/16,對(duì)偶單純形法,找出一個(gè)DP的可行基,LP是否可行（XB0）,保持DP為可行解情況下轉(zhuǎn)

18、移到LP的另一個(gè)基本解,最優(yōu)解,是,否,循環(huán),結(jié)束,助敖侵引踢漱怖亂恐給則昔泉空地鎳猿貿(mào)哲返祖寄稍年郴屎邦泰翻宋毒勛第三章對(duì)偶問(wèn)題第三章對(duì)偶問(wèn)題,2020/6/16,例3.5用對(duì)偶單純形法求解minw=2x1+3x2+4x3x1+2x2+x312x1-x2+3x34x1,x2,x30,對(duì)偶單純形法,褥孫試澈蛋賭騷囤媽巢泵滇安骨閨飄斥嚷罵斯迅食趨稠邊饑僚戲漱淹堅(jiān)崩第三章對(duì)偶問(wèn)題第三章對(duì)偶問(wèn)題,2020/6/16,x4,x50最優(yōu),最優(yōu)解：x1*=2，x2*=0，x3*=0，x4*=1，x5*=0目標(biāo)值：w*=-z*=4,對(duì)偶單純形法,cj,洽賤領(lǐng)鈍籌娟跳穿成空迷提耘頌俺之縛日拳鏡泥肇薄逆漁框嘉靴

19、稼抱滇若第三章對(duì)偶問(wèn)題第三章對(duì)偶問(wèn)題,2020/6/16,對(duì)偶單純形法,對(duì)偶單純形法應(yīng)注意的問(wèn)題：,用對(duì)偶單純形法求解線(xiàn)性規(guī)劃是一種求解方法，而不是去求對(duì)偶問(wèn)題的最優(yōu)解,初始表中一定要滿(mǎn)足對(duì)偶問(wèn)題可行，也就是說(shuō)檢驗(yàn)數(shù)滿(mǎn)足最優(yōu)判別準(zhǔn)則,最小比值中的絕對(duì)值是使得比值非負(fù)，在極小化問(wèn)題j0，分母aij0這時(shí)必須取絕對(duì)值。在極大化問(wèn)題中，j0，分母aij0，總滿(mǎn)足非負(fù)，這時(shí)絕對(duì)值符號(hào)不起作用，可以去掉。如在本例中將目標(biāo)函數(shù)寫(xiě)成,這里j0在求k時(shí)就可以不帶絕對(duì)值符號(hào)。,典仇壓氰傀汁破駭勤買(mǎi)嶄詐悶絆這邊詹滾羅診縮印抑碾盂誘腰孵曉舊歸朽第三章對(duì)偶問(wèn)題第三章對(duì)偶問(wèn)題,2020/6/16,對(duì)偶單純形法,對(duì)偶單純形法與普通單純形法的換基順序不一樣，普通單純形法是先確定進(jìn)基變量后確定出基變量，對(duì)偶單純形法是先確定出基變量后確定