4.3-4 解：Max. Z?4 S.t. 3x?3x?6x
123x?x?3x?x1234?30 (1)
2x1?2x2?3x3?x5?40 (2)
And Initialization:
x?0, for i=1,2,3,4,5
i2x?0，x1?0,x3?0; so,x4?30,x5?40
The basic solution (0,0,030,40) is not a optimal solution.
Interation 1 : Step1: Z?4 Step2:
334x?3x?6x, 6>4>3, so, the entering basic variable is x
12322x?0，x?0
3x?x?30,x?30?3x?x?10?min
403x?x?40,x?40?3x?x?314335533 so, the leaving basic variable is
x
4Z?2x1?x2?2x4?60(0)Step3:
x?13x?x?13x?60(1) ?x?x?x?x?10(2)12341245 The BF solution is (0,0,10,0,10), z=60 Interation 2 : Step1: Z??2x?x?2x?60, and xis the entering basic variable.
1242x,x14?0
??10(x3?10?1x2?x2?30)3x2x33 ?x2?x5?10(x5?10?x2?x2?10?min) So, the leaving basic variable is Step2:
1x
5Z?x1?x4?x5?70(0)4??2x4?1x5?20(1) 5x1x3333x1?13x2?x3?13x4?10(2) So, the BF solution is (0,10,20/3,0,0), Z=70
4.4-3 解：
(a)图略；Z?2x?x;
1212 CPF: (0,0),(25,0),(0,40),(20,20); Optimal solution is (20,20)
x?x?0 x?x?x?40
S.t.
4x?x?x?100(c) Max. Z?2123124Interation 1 :Z?2133x?x, and xis the entering basic variable, xx?x?40(x?40?x?x?40)
4x?x?100(x?100?4x?x?25?min)12111144112?0
So, the leaving basic variable is
x
4Z?1?502x43??1x4?15 4x2x34x1?14x2?14x4?25 The BF solution is (25,0,15,0), z=50 Interation 2:
2x2?1xis the entering basic variable,x24?0
3x4x2?11??20?min)4x2x2 1?25,(?25???100)x14x24x2x2?x3?15，(x3?15?3 So, the leaving basic variable is
x
3Z?22?1x?60x33x?43x?13x?20 x?13x?13x?203434134 The BF solution is (20,20,0,0), Z=60.
(e) Interation 0 Basic Eq Variable Z 0 1 2 Coefficient of Z 1 0 0 x1 -2 1 4 x2 -1 1 1 x3 0 1 0 x4 0 0 1 Right Side 0 40 100 x3 x4 1 Z 0 1 2 0 1 2 1 0 0 1 0 0 0 0 1 0 0 1 -1/2 3/4 1/4 0 1 0 0 1 0 2/3 4/3 -1/3 1/2 -1/4 1/4 1/3 -1/3 1/3 50 15 25 60 20 20 x3 x1 2 Z x2 x1 The BF solution is (20,20,0,0), Z=60. 4.4-5解:(a)
Z?2x1?4x2?3x3 Max.
3x1?4x2?2x3?x4?602x1?x2?2x3?x5?40
x?3x12?2x3?x6?80Interation 1 :
xis the entering basic variable, x,x2125523?0
4x2?x4?60(x4?60?4x2?x2?15?min)x?x?40(x?40?x?x?40)3x?x?80(x?80?3x?x?80)3226622
So, the leaving basic variable is
x
4Z?x1?x3?x4?60?x?1x?1x?15x424
531????254x2x4xx?5x?1x?3x?x?354241123434513463Interation 2 :
xis the entering basic variable, x,x314?0
x2x3?2xx1?2xx233?1??30)2x3x3?25(x5?25?3x3?x3?50?min) 523?35(x6?35?1x3?x2?70)623?15(x2?15?1 So, the leaving basic variable is
x
5?5x?2x?230x66331?x?1x?1x?20x3333 5?x?1x?2x?50x6633?5x?2x?1x?x?803333Z?11114524513451456 The BF solution is (0,20/3,50/3,0,0,80/3), Z=230/3.
(b) Interation 0 Basic Eq Coefficient of Variable Z x1 Z 0 1 2 3 0 1 2 3 0 1 2 3 1 0 0 0 1 0 0 0 1 0 0 0 -2 3 2 1 1 3/4 5/4 -5/4 11/6 1/3 5/6 =5/3 Right Side 0 60 40 80 60 15 25 35 230/3 20/3 50/3 80/3 x2 -4 4 1 3 0 1 0 0 0 1 0 0 x3 -3 2 2 2 -1 1/2 3/2 1/2 0 0 1 0 x4 0 1 0 0 1 1/4 -1/4 -3/4 5/6 1/3 -1/6 -2/3 x5 0 0 1 0 0 0 1 0 2/3 -1/3 2/3 -1/3 x6 0 0 0 1 0 0 0 1 0 0 0 1 x4 x5 x6 1 Z x2 x5 x6 2 Z x2 x3 x6 The BF solution is (0,20/3,50/3,0,0,80/3), Z=230/3. 4.6-1 解： (a)图略；
x?2,x12?1；Z=7；
(b) Big M method:
Z?2x1?3x2?Mx4?0x?2x?x?4x?x?x?3112324
Interation Basic Eq Variable Z 0 1 2 Coefficient of Z 1 0 0 x1 -M-2 1 1 x2 -3-M 2 1 x3 0 1 0 x4 0 0 1 Right Side -3M 4 3 0 x3 x4 Initial BF solution is (0,0.4.3); the entering basic variable is x2,the leaving basic variable is x3
(c) Interation Basic Eq Variable Z 0 1 2 0 1 2 Coefficient of Z 1 0 0 1 0 0 x1 x2 x3 (3+M)/2 1/2 -1/2 1 1 -1 x4 0 0 1 1+M =1 2 Right Side 6-M 2 1 7 1 2 1 (-1-M)/2 0 1/2 1/2 0 0 1 2 0 0 1 0 x2 x4 2 Z x2 x1 The BF solution is (2,1,0,0), Z=7. 4.6-7 解：
(a) Big M method:
Z?2x1?5x2?3x3?Mx5?Mx6?0x?2x?x?x?x?202x?4x?x?x?50123451236
Interation Basic Eq Coefficient of Variable Z x1 Z 0 1 2 0 1 2 0 1 2 0 1 2 1 0 0 1 0 0 1 0 0 1 0 0 -3M-2 1 2 0 1 0 0 1 0 x2 -2M-5 -2 4 -8M-9 -2 8 0 0 1 x3 -2M-3 1 1 M-1 1 -1 -2.125 0.75 -0.125 0 1 0 x4 M -1 0 -2M-2 -1 2 0.25 -0.5 0.25 -1.167 -0.67 x5 0 1 0 x6 0 0 1 Right Side -70M 20 50 -10M+40 20 10 51.25 22.5 0 x5 x6 1 Z 3M+2 0 1 -2 M-0.25 0.5 -0.25 0 1 M+1.125 0.25 x1 x6 2 Z x1 x2 3 Z 0.125 1.25 2.833 0 1.333 0 0.167 1 1.167 1.833 115 0.67 0.33 30 x3 x2 0.167 -0.16 0.167 5
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