运筹学(第五版) 习题答案

(1)min z=P1(d1?+d1?)+P2(2d2?+d3?) s.t. x1-10x2+d1?-d1?=50 3x1+5x2+d2?-d2?=20 8x1+6x2+d3?-d3?=100

x1,x2,d1?,d1?,d2?,d2?,d3?,d3??0

(2)min z=P1(d3?+d4?)+P2d1?+P3d2?+P4(d3?+1.5d4?) s.t.

x1+x2+d1?-d1?=40

x1+x2+d2?-d2?=100 x1+d3?-d3?=30 x2+d4?-d4?=15

x1,x2,d1?,d1?,d2?,d2?,d3?,d3?,d4?,d4??0

(3) min z=P1(d1?+d1?)+P2 d2?+P3d3? s.t. x1+x2+d1?-d1?=10 3x1+4x2+d2?-d2?=50 8x1+10x2+d3?-d3?=300

x1,x2,d1?,d1?,d2?,d2?,d3?,d3??0

(1)满意解是:(50,0) (2)满意解是:(25,15) (3)满意解是:(10,0)

4.3使用单纯形法求解下列目标规划问题。

(1)min z=P1 d1?+P2 d2?+P3(5d3?+3 d4?)+P4 d1? s.t.

x1+x2+d1?-d1?=80

x1+x2+d2?- d2?=90

x1+d3?-d3?=70 x2+d4?-d4?=45

x1,x2,d1?,d1?,d2?,d2?,d3?,d3?,d4?,d4??0

(2)min z=P1 d2?+P1 d2?+P2 d1? s.t. x1+2x2+d1?-d1?=10 10x1+12x2+d2?-d2?=62.4

x1+2x2?8

x1,x2,d1?,d1?,d2?,d2? ?0

(3)min z=P1(d1?+ d2?)+P2 d3? s.t. x1+x2+d1?-d1?=1 2x1+2x2+d2?-d2?=4 6x1-4x2+d3?-d3?=50

x1,x2,d1?,d1?,d2?,d2?,d3?,d3??0

解:

(1)把原问题转化为: Min z=P1d2?+P1d2?+P2d1? S.T.

x1+2x2+d1?-d1?=10

10x1+12x2+d2?-d2?=62.4 2x1+x2+x3=8

x1,x2,x3,d1?,d1?,d2?,d2??0 x3是松弛变量

单纯形法计算得: cj 0 0 0 P2 0 P1 P2 ?i

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