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1、吉林建筑工程学院城建学院人文素质课线性规划单纯形法例题8511.4(1)分别用图解法和单纯形去求解线性规划问题。max z = 2xl + x23X + 5x2 15I6x + 2x2 0在上述线性规划问题中 分别加入松驰变量5些，得到该线性规划问题的标准型max乙=2“+x2 + 0x3 + 0x43X + 5x2 + x3= 15|6%j + 2x2+xA = 24xl,x2,x3,x4 0选择心，兀为初始基变量,Cj21000CBXrb兀2心兀40Xy15351015-5- 丁 70兀4246201生640100碎斗43 6 J6 = 2-(0x3 + 0x6) = 2(r2 = 1 一
2、 (0 x 5 + 0 x 2) = 16 = 0-(0xl + 0x0) = 0cr4 = 0-(0x0 + 0xl) = 0所以选择山为进基变量，丸为出基变量。Cj210005Xrb兀2七尢40兀33041-1/232411 /301/6=12 1/3Cj-J01/30-1/3ninP, 1-3/4 b 1/3Jb = 2-(OxO + 2xl) = O6 =l (0x4 + 2xl/3) = l/3 tr3 = 0-(0xl + 2x0) = 0cr4 =0 (0x1/2 + 2xl/6) = 1/3所以选择x2为进基变量，勺为出基变量。Cj2i0005Xbb兀2兀3兀41x23/401
3、1/4-1/82x15/410-1/125/24cj zJ001/12-7/24b = 2-(1 x 0 +2x 1) = 06 = 1 - (1 x 1 + 2 x 0) = 06 =0-(lx 1/4 +2x-l/12) =-1/126 = 0-(1 X-1/8 +2x5/24) = -7/24所以,最优解 M=(x2,x,)r=(- , 4)T.4415 333故有 : max z 2x. + 工= 2x i=4 44【8 页 1.4(2)分别用图解法和单纯形去求解线性规划问题。max z = 2X1 + 5xXj 12I2x. 123X + 2x2 0在上述线性规划问题电分别加入松驰变
4、量得到该线性规划问题的际准型max z = 2X +x2 + 0x3 + 0x4 + 0x51= 4二24xs = 18 01+X32X2+X43X + 2x2 +Cj25000CbXbb兀1兀2勺兀4兀50兀341010140兀41202010虫虫60x5183200018八=925000 J12 1一 a nje ，5 = 2-(0xl + 0x0 + 0x3) = 2(r2 = 5-(0x0 + 0x2 + 0x2) = 5cr. = 0-(0xl + 0x0 + 0x0) = 06 = 0 (0x0 + 0xl + 0x0) = 06 = 0-(0xl + 0x0 + 0x0) = 0
5、所以选择七为进基变量，兀为出基变量。Cj250000cBXbb兀2兀3兀4兀50x341010045X260101/20肚0x56300-113ci zJ2005/20吨仆一寸=2b = 2-(0xl + 5x0 + 0x3) = 2 ( r2 = 5-(OxO + 5xl + OxO) = O6 = O-(Oxl + 5xO + OxO) = Obq =0 (0x0 + 5xl/2 + 0xl) = 5/26 = 0-(0xl + 5x0 + 0xl) = 0所以山为进基变量，x5为111基变量52500005Xbb兀2七兀4兀50兀320011/31/35X260101/202兀121001/31/3CjZj00011/6-2/36 =2-(OxO + 5xO + 2xl) = Ob? = 5-(OxO + 5xl + 2xO) = O6 = O-(Oxl + 5xO + 2xO) = O6 =0-(0xl/3 + 5xl/2 + 2x-l/3) = -ll/6 =O-(Ox-l/3 + 5xO+2xl/3) = -2/3单纯形表得计算结果潮：X” =(2,6,2,0,()T为最优解。max z * = 2x2 + 5x6 = 34
