习题答案选02_对偶理论与敏感性分析.doc

运筹学教程（胡运权主编，清华第三版）部分习题答案（第二章）2.1（1）max w = 2y1 + 3y2 + 5y3s.t. y1 + y2 + y3 23y1 + y2 + 4y3 24y1 + 3y2 + 3y3 = 4y1 0, y2 0（2）min w = 5y1 + 3y2 + 8y3s.t. y1 - y2 + 4y3 = 52y1 + 5y2 + 7y3 62y1 - y2 + 3y3 3y2 0, y3 0（3）max w = i=1,2,.,m aiui + j=1,2,.,n bjvj s.t. ui + vj cij, i = 1, 2, , m; j = 1, 2, , n（4）min w = i=1,2,.,m biyi s.t. i=1,2,.,m aijyi cj, j = 1, 2, , n1i=1,2,.,m aijyi = cj, j = n1 +1, , nyi 0, i = 1, 2, , m12.2（1）不正确；（2）不正确；（3）不正确2.3a = 2，b = 10，c = 3，d = 1/4，e = 5/4，f = -1/2，g = -3/4，h = -1/2，i = -1/4，j = -1/4，k = 0，l = 12.4原问题该写成min z = 2x1 + 3x2 + 5x3 + 6x4s.t. x1 + 2x2 + 3x3 + x4 22x1 - x2 + x3 3x4 3xj 0, j = 1, 2, 3, 4（1）max w = 2y1 + 3y2 s.t. y1 + 2y2 22y1 - y2 33y1 + y2 5y1 - 3y2 6yi 0, i=1,2（2）y1* = 8/5, y2* = 1/5, w* = 19/5（3）x1* =7/5, x2* = 0, x3* = 1/5, x4* = 0, z* = 19/52.5（1）min w = 2y1 + y2 + 2y3s.t. y1 + y2 + 2y3 1y1 - y2 + y3 2- y1 + y2 + y3 = 1y1 0, y3 0（2）对偶问题有可行解y1 =0, y2 =1, y3 = 0; w =1。z w =1。2.6该线性规划有可行解x = (0, 0, 0)，它的对偶问题min w = 2y1 + y2s.t. - y1 - 2y2 1y1 + y2 1y1 - y2 0y1 0, y2 0无可行解，故原问题目标函数无界。2.7设对偶问题为min w = 8y1 + 6y2 + 6y3 + 9y4s.t. y1 + 2y2 + y4 23y1 + y2 + y3 + y4 4 y3 + y4 1y1 + y3 1yi 0, i=1,2,3,4由x1 = 2，得y1 + 2y2 + y4 = 2由x2 = 2，得3y1 + y2 + y3 + y4 = 4由x3 = 4，得y3 + y4 = 1由x1 + x2 + x3 0.4，故购进材料扩大生产b2：从30改变为30 + Db2，单纯形表B-1b列改变为 ( 5, 3 )T + Db2 ( -1/3, 2/5 )T由5 - Db2 /3 03 + 2Db2 /5 0得-15/2 Db2 15，或45/2 b2 45应购进材料152.15（1）a11 = 0, a12 = 1, a13 = 2, a21 = 3, a22 = -1, a23 = 1; b1 = 5, b2 = 10; c1 = 6, c2 = -2, c3 = 10（2）-6 t1