運(yùn)籌學(xué)基礎(chǔ)及應(yīng)用習(xí)題解答習(xí)題一P461.1(a)x244x2x4123210123x14x6x612該問(wèn)題有無(wú)窮多最優(yōu)解，即滿足且1的所有，此時(shí)目標(biāo)函數(shù)值4x6x60xx,x121222z3。(b)x232014x1用圖解法找不到滿足所有約束條件的公共范圍，所以該問(wèn)題無(wú)可行解。1.3(a)(1)圖解法天天快樂(lè)x243210123x134x9x335的解1,，最大值xz22最優(yōu)解即為125x12x28(2)單純形法首先在各約束條件上添加松弛變量，將問(wèn)題轉(zhuǎn)化為標(biāo)準(zhǔn)形式maxz10x15x20x30x43x4xx9s.t.1235x12x2x48則,組成一個(gè)基。令x20PP4得基可行解0,0,9,8，由此列出初始單純形表xx13cj10500基bx1x2x3x4cB341000x3x498[5]20110500czjj898。min,53512cj10500x1x2x3x4基bcB211430110x3x1555852511005天天快樂(lè)cz0102jj21830，min1422,2新的單純形表為cj10500基bx1x2x3x4cB35314501010x2x121141012775142514cz0jj32352,，表明已找到問(wèn)題最優(yōu)解x1,x21,x30,x0。最大值z(mì)*4012(b)(1)圖解法6x12x224x212963x1x250369x1\\62x24x73,，最大值z(mì)17最優(yōu)解即為12的解xx1x25222(2)單純形法首先在各約束條件上添加松弛變量，將問(wèn)題轉(zhuǎn)化為標(biāo)準(zhǔn)形式天天快樂(lè)maxz2xx0x0x0x512345xx15236x2xx24124xxx5s.t.125則P，P，P組成一個(gè)基。令x03451x2得基可行解x0,0,15,24,5，由此列出初始單純形表cj21000\基bxxx2x4x5cB13x0510000153x4[6]2010024x551211000010czjj245min,,4。6112cj21000基bxxx2x4x5cB13051000x01531311000x46204216x51131313cz000jj15533min,24,，0222新的單純形表為天天快樂(lè)cj21000基bxxx2x4x5cB131551521x001027413x410020234231x500102414212czj00jx1，x7，x，x0，x0。最15,，表明已找到問(wèn)題最優(yōu)解大022121234517值z(mì)*21.8表1-23x12x2x3x4x5x4x56143-2210011cz31200jj表1-24x1x2x3x4x5x13112511121201x500czj75320j1.10354000x1x2x3x4x5x62310130100015x2835430232300x5x61432935304cz13045300jj天天快樂(lè)x1x21x30x4x5x60231305x2834154115012152151545040x3x614158915001cz1115001715450jjx10x2x3x4x5x610154164184154110415x2x3x15041001441436241894110024112411541cz000454124411141jj最后一個(gè)表為所求。習(xí)題二P762.2(a)錯(cuò)誤。原問(wèn)題存在可行解，對(duì)偶問(wèn)題可能存在可行解，也可能無(wú)可行解。(b)錯(cuò)誤。線性規(guī)劃的對(duì)偶問(wèn)題無(wú)可行解，則原問(wèn)題可能無(wú)可行解，也可能為無(wú)界解。(c)錯(cuò)誤。(d)正確。2.8將該問(wèn)題化為標(biāo)準(zhǔn)形式：maxz2x1x2x30x40x5x6xxx4123s.t.x12xx54i201,5xi用單純形表求解c211x310x410x50jx1x21基cbB[1]10x4x56402001cz21100jj6天天快樂(lè)基bx1x21x31x41x50cB120x1x5610003111cz3-120jj由于0，所以已找到最優(yōu)解*6,0,0,0,10，目標(biāo)函數(shù)值z(mì)*12Xj(a)令目標(biāo)函數(shù)3maxz（2）x（-1+）x（1+）x11223(1)令0，將反映到最終單純形表中123cj211100基bx1x2x3xx45cB1111021x64031110x1050-3--1-2-0cz1jj1111為最優(yōu)的條件：-3-0-1-02-0，從而表中解，，1110(2)令，將反映到最終單純形表中1322112cj00基bx1xxxx5cB2341111020x6103111x1050-3-120czjj2為最優(yōu)的條件：2-30，從而3表中解20(3)令，將反映到最終單純形表中123cj211300基bx1xxxx5cB234天天快樂(lè)1111020x6103111x105-123cz0-30jj13-10表中解為最優(yōu)的條件：，從而3(b)令線性規(guī)劃問(wèn)題為maxz2xxx1236xxx1234s.t.x2x451201,3xii(1)先分析的變化1bB1b1011101106bb0，從而6使問(wèn)題最優(yōu)基不變的條件是1116102(2)同理有0，從而102(c)由于x(6,0,0,0,10)代入x2x62，所以將約束條件減去剩余變量后的方13程x2xx2直接反映到最終單純形表中136cj2-11000基xbxx23x4xx56cB12x610-2111110010x010301100050x-2016cz-3-1-200jj對(duì)表中系數(shù)矩陣進(jìn)行初等變換，得cj2-1100基xbxx23x4xx56cB1天天快樂(lè)2x61111010000101110x10000350x-8-1