習(xí)題答案選02_對偶理論與敏感性分析

1、運籌學(xué)教程（胡運權(quán)主編，清華第三版）部分習(xí)題答案（第二章）2.1（1）max w = 2y1 + 3y2 + 5y3s.t. y1 + y2 + y3 £ 23y1 + y2 + 4y3 £ 24y1 + 3y2 + 3y3 = 4y1 ³ 0, y2 £ 0（2）min w = 5y1 + 3y2 + 8y3s.t. y1 - y2 + 4y3 = 52y1 + 5y2 + 7y3 ³ 62y1 - y2 + 3y3 £ 3y2 £ 0, y3 ³ 0（3）max w = åi=1,2,.,m aiui

2、 + åj=1,2,.,n bjvj s.t. ui + vj £ cij, i = 1, 2, , m; j = 1, 2, , n（4）min w = åi=1,2,.,m biyi s.t. åi=1,2,.,m aijyi ³ cj, j = 1, 2, , n1åi=1,2,.,m aijyi = cj, j = n1 +1, , nyi ³ 0, i = 1, 2, , m12.2（1）不正確；（2）不正確；（3）不正確2.3a = 2，b = 10，c = 3，d = 1/4，e = 5/4，f = -1/2，

3、g = -3/4，h = -1/2，i = -1/4，j = -1/4，k = 0，l = 12.4原問題該寫成min z = 2x1 + 3x2 + 5x3 + 6x4s.t. x1 + 2x2 + 3x3 + x4 ³ 22x1 - x2 + x3 3x4 ³ 3xj ³ 0, j = 1, 2, 3, 4（1）max w = 2y1 + 3y2 s.t. y1 + 2y2 £ 22y1 - y2 £ 33y1 + y2 £ 5y1 - 3y2 £ 6yi ³ 0, i=1,2（2）y1* = 8/5, y2*

4、 = 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 ³ 1y1 - y2 + y3 £ 2- y1 + y2 + y3 = 1y1 ³ 0, y3 £ 0（2）對偶問題有可行解y1 =0, y2 =1, y3 = 0; w =1。z £ w =1。2.6該線性規(guī)劃有可行解x = (0, 0, 0)，它的對偶問題min w = 2y1 + y2s.t. - y1 - 2y

5、2 ³ 1y1 + y2 ³ 1y1 - y2 ³ 0y1 ³ 0, y2 ³ 0無可行解，故原問題目標(biāo)函數(shù)無界。2.7設(shè)對偶問題為min w = 8y1 + 6y2 + 6y3 + 9y4s.t. y1 + 2y2 + y4 ³ 23y1 + y2 + y3 + y4 ³ 4 y3 + y4 ³ 1y1 + y3 ³ 1yi ³ 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 +

6、y4 = 1由x1 + x2 + x3 < 9，得y4 = 0得到y(tǒng)1 = 4/5，y2 = 3/5，y3 = 1，y4 = 02.8y1 = y1 /5 - 3y3/5，y2 = 5yi，y3 = y3 2.9（1）x* = (0, 1.5, 1), z* = 36；（2）x* = (2/3, 2, 0), z* = 22/32.10x1* = 5/6，x2* = 2/3，x3* = 0；z* = 230/3 2.11原問題最優(yōu)解：x* = (6, 0, 0), z* = 12（1）x* = (8/3, 10/3, 0), z* = 46/3（2）x* = (3, 0, 0), z*

7、= 6（3）x* = (10/3, 0, 8/3), z* = 28/32.12原問題最優(yōu)解：x* = (1, 2, 0, 0, 0), z* = 8（1）x* = (2, 0, 1, 0, 0), z* = 10（2）3/4 £ c1 £ 3, 2 £ c2 £ 8（3）x* = (5, 1, 0, 0, 0), z* = 13（4）x* = (0, 2, 0, 0, 0, 1/3), z* = 25/3（5）x* = (2, 1, 0, 0, 1), z* = 72.14（1）max z = 3x1 + x2 + 4x3s.t. 6x1 + 3x2

8、+ 5x3 + s1 = 453x1 + 4x2 + 5x3 + s2 = 30xj ³ 0, j=1,2,3si ³ 0, i=1,2最優(yōu)解單純形表：cBxBB-1bx1x2x3s1s23x151-1/301/3-1/34x33011-1/52/50-20-1/5-3/5最優(yōu)解：x*1 = 5，x*2 = 0，x*3 = 3；z* = 27（2）c1：從3改變?yōu)? + Dc1，單純形表行改變?yōu)椋?，-2 + Dc1/3，0，-1/5 - Dc1/3，-3/5 + Dc1/3）由-2 + Dc1/3 £ 0-1/5 - Dc1/3 £ 0-3/5 + D

9、c1/3 £ 0得-3/5 £ Dc1 £ 9/5，或12/5 £ c1 £ 24/5（3）在（1）的單純形表中加入x4的列：B-1P4 = ( 2, -4/5 )T4 = 1/5求得最優(yōu)解：x1 = 0，x2 = 0，x3 = 3，x4 = 5/2；z=27.5（4）由（1）中的單純形表，材料的影子價格 = 3/5 = 0.6 > 0.4，故購進(jìn)材料擴(kuò)大生產(chǎn)b2：從30改變?yōu)?0 + Db2，單純形表B-1b列改變?yōu)?( 5, 3 )T + Db2 ( -1/3, 2/5 )T由5 - Db2 /3 ³ 03 + 2Db2 /5 ³ 0得-15/2 £ Db2 £ 15，或45/2 £ b2 £ 45應(yīng)購進(jìn)材料152.15（1）a11 = 0, a12 = 1, a13 = 2, a21 = 3, a22 = -1, a23 = 1; b1 = 5