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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【8页1.4】分别用图解法和单纯形法求解线性规划问题。max z = 2 x + x3x + 5x 15 1 2(s.t) 6x + 2x 0在上述线性规划问题中，分别加入松驰变量色,x 4，得到该线性规划问题的标准型 max z = 2 x + x + 0 x + 0 x3 x + 5 x + x = 15(st) 0V 1234选择,气为初始基变量，cj21009cXbxxxx0x153510 = 5；0x24620124 = 4c - z,2100. 15 241 / min Its |=4b = 2 - (0 x 3 + 0 x 6
2、) = 2b = 1 - (0 x 5 + 0 x 2) = 12b = 0 - (0 x 1 + 0 x 0) = 03b = 0 - (0 x 0 + 0 x 1) = 04所以选择气为进基变量，x4为出基变量。cj21000.cXbbxxxx0x3041-1/232x411/301/6=12 1/3c - z,01/30-1/3min 4,1/3 = 3/4b = 2 - (0 x 0 + 2 x 1) = 0b = 1 - (0 x 4 + 2 x 1/3) = 1/32b = 0 - (0 x 1 + 2 x 0) = 03b = 0 - (0 x-1/2 + 2 x 1/6) =
3、-1/34所以选择x2为进基变量，x3为出基变量。cj21000.cXbbxxxx1x3/4011/4-1/82x15/410-1/125/24c - z00-1/12-7/24b = 2 -(1 x 0 + 2 x 1) = 0b = 1 -(1 x 1 + 2 x 0) = 02b = 0 -(1 x 1/4 + 2 x-1/12) = -1/123b = 0 -(1 x-1/8 + 2 x 5/24) = -7/24415 3所以，最优解为x = E，x1)=(4)，故有 :max z = 2 x + x关415 3=2 x + =4 4【8页1.4(2)】分别用图解法和单纯形法求解线性
4、规划问题。max z = 2 x + 5 x12(s.t )2 x 1223 气+ 2 x 2 0在上述线性规划问题中，分别加入松驰变量尤,x ,x，得到该线性规划问题的标准型 345max z = 2 x + x + 0 x + 0 x + 0 xx + x = 4(st ) 012345c j250009cXbxxxxx0x4101014 =0x1202010?=60x1832000与=9c - z250001 12 181 右 min fl 卜 6b = 2 -(0 X1 + 0 X 0 + 0 X 3)= 2b = 5 -(0 x 0 + 0 x 2 + 0 x 2)= 5b = 0
5、-(0 x 1 + 0 x 0 + 0 x 0) = 03b = 0 -(0 x 0 + 0 x 1 + 0 x 0) = 04b = 0 - (0 x 1 + 0 x 0 + 0 x 0) = 0 5所以选择x2为进基变量，x4为出基变量。cj250000.cXbbxxxxx0x4101004 = 45x60101/206 =_0x6300-113 = 2c - z,200-5/20min卜牛2b = 2 -(0 x 1 + 5 x 0 + 0 x 3) = 2b = 5 -(0 x 0 + 5 x 1 + 0 x 0) = 0b = 0 -(0 x 1 + 5 x 0 + 0 x 0) =
6、 03b = 0 -(0 x 0 + 5 x 1/2 + 0 x-1) = -5/24b = 0 - (0 x 1 + 5 x 0 + 0 x 1 ) = 05所以气为进基变量，气为出基变量cj250000.cXbbxxxxx0x20011/3-1/35x60101/202x2100-1/31/3c - z,000-11/6-2/3b = 2 - (0 x 0 + 5 x 0 + 2 x 1 )= 0b = 5 -(0 x 0 + 5 x 1 + 2 x 0) = 0b = 0 - ( 0 x 1 + 5 x 0 + 2 x 0 ) = 03b = 0 -(0 x 1/3 + 5 x 1/2 + 2 x-1/3) = -11/64b = 0 -(0 x-1/3 + 5 x 0 + 2 x 1/3) = -2/35单纯形表得计算结果表明：X* = (2,6,2,0,0)t为最优解。max z * = 2 x 2 + 5 x 6 = 34
