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C s ca phng php n hnh

ci xi bi x1 x2 x3 x4 x5-1 x3 3 0 -1 1 -1 0

2 x1 2 1 1 0 -1 0

4 x5 0 0 -2 0 0 1f(x) 1 0 -2 0 0 0

Phng n ti u ca bi ton l: (2,0,3,0,0).Gi tr hm mc tiu t c l : f(x) = 1

T6.(1.5)

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C s phng php n hnh

Xy dng PA mi tt hn. (4 = 2 > 0 ln nhtnn x4 l n a vo, ly bi chia cho ai4 > 0 gi trnh nht t ti hng i = 1 nn x1 l n a ra.a14 = 1 l phn t ch yu.

ci xi bi1 -1 0 -2 2 3

x1 x2 x3 x4 x5 x61 x1 2 1 0 0 [1] 1 -1

-1 x2 12 0 1 0 1 0 1

0 x3 9 0 0 1 2 4 3

f(x) -10 0 0 0 2 -1 1

T7(2.5)

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C s phng php n hnh

ci xi bi

1 -1 0 -2 2 3

x1 x2 x3 x4 x5 x6-2 x4 2 1 0 0 1 1 -1

-1 x2 10 -1 1 0 0 -1 2

0 x3 5 -2 0 1 0 2 [5]f(x) -14 -2 0 0 0 -3 3


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C s phng php n hnh

Phng n ti u ca bi ton l : (0,8,0,3,0,1)Gi tr hm mc tiu t c l : f(x) = -17

ci xi bi1 -1 0 -2 2 3

x1 x2 x3 x4 x5 x6-2 x4 3 3/5 0 1/5 1 7/5 0

-1 x2

8 -1/5 1 -2/5 0 -9/5 0

-3 x6 1 -2/5 0 1/5 0 2/5 1

f(x) -17 -4/5 0 -3/5 0 -21/5 0

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C s phng php n hnhV d: Gii bi ton QHTT:

f(x) = -7x1 + x2 + 3x3 + x4 + x5 + 6x6 p Min

Gii:

1 3 5 6

1 4 6

1 2 3 6

j

x x x x 15

2x x 2x 9

3x x 2x 4x 2

x 0 j 1,6

! !

! u !

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C s phng php n hnh

ci xi bi-7 1 3 1 1 6

x1 x2 x3 x4 x5 x61 x5 15 [1] 0 -1 0 1 -1

1 x4 9 -2 0 0 1 0 -2

1 x2 2 -3 1 2 0 0 4

f(x) 26 3 0 -2 0 0 -5


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C s ca phng php n hnh

Bi ton khng c phng n ti u

ci xi bi-7 1 3 1 1 6

x1 x2 x3 x4 x5 x6-7 x1 15 1 0 -1 0 1 -1

1 x4

39 0 0 -2 1 2 -4

1 x2 47 0 1 -1 0 3 1

f(x) -19 0 0 1 0 -3 -2

V tn ti (3 = 1 v ai3 < 0 vi mi i = 1,2,3.

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Phng n ti uV d: Cho bi ton QHTT:

f(x) = 5x1+ 4x2 + 5x3 + 2x4 + x5 + 3x6 p Min

a) Gii bi ton trn

b) Bi ton c PAT khc hay khng? Nu chy tm mt PAT khc?

c) Hy tm PAT x* ca bi ton c x*4 = 36

1 2 3 4

1 2 3 5

1 3 6

2x 4x 3x x 152

4x 2x 3x x 60

3x x x 36x 0 1, 6

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Phng n ti ua) Gii bi ton:

5 4 5 2 1 3ci xi bi x1 x2 x3 x4 x5 x62 x4 152 2 4 3 1 0 0

1 x5 60 4 2 3 0 1 03 x6 36 [3] 0 1 0 0 1

f(x) 472 12 6 7 0 0 0

Xy dng PA mi tt hn. (1 = 12 > 0 lnnht nn x1 l n a vo, ly bi chia cho ai1 > 0 gitr nh nht t ti hng i = 3 nn x6 l n a ra.

a31 = 1 l phn t ch yu.

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Phng n ti u

ci xi bi5 4 5 2 1 3

x1 x2 x3 x4 x5 x62 x4 128 0 4 7/3 1 0 -2/3

1 x5 12 0 2 5/3 0 1 -4/3

5 x1 12 1 0 1/3 0 0 1/3f(x) 328 0 6 3 0 0 -4


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Phng n ti u

Phng n ti u ca bi ton l : (12,6,0,104,0,0)Gi tr hm mc tiu t c l : f(x) = 292

ci xi bi5 4 5 2 1 3

x1 x2 x3 x4 x5 x62 x4 104 0 0 -1 1 -2 2

4 x2 6 0 1 5/6 0 1/2 -2/3

5 x1 12 1 0 1/3 0 0 1/3f(x) 292 0 0 -2 0 -3 0

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Phng n ti ub) Lp bng n hnh mi vi x6 l n a vo vx

1l n a ra:

PATU khc ca bi ton l: (0, 30,0, 32,0,36)Gi tr hm mc tiu t c l : f(x) = 292

ci xi bi5 4 5 2 1 3

x1 x2 x3 x4 x5 x6

2 x4 32 -6 0 -3 1 -2 04 x2 30 2 1 9/6 0 1/2 0

3 x6 36 3 0 1 0 0 1

f(x) 292 0 0 -2 0 -3 0

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Phng n ti uc) T cui bng n hnh trong cu a) ta c: PATx0 = (12, 6, 0,104, 0, 0) v x

6khng l n c s

nhng (6= 0 nn ta c b s:

Khi tp PAT ca bi ton c dng:

z6 = (1/3, -2/3, 0, 2, 0, -1)

xP = x0 - P.z6 = (12 - P/3; 6 + 2 P/3, 0, 104 - 2 P, 0, P)

PAT x* c x*4 = 32 th: P = 36. Ta cPAT cn tm l:

Vi 0 e P e 36

x* = (0, 30, 0, 32, 0, 36)

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Phng php gii bi ton MaxCch 1: Nu bi ton Max c hm mc tiu:

f(x)p

MaxKhi ta chuyn v bi ton Min c hm mc tiu:

g(x) = - f(x) p MinNu bi ton Min khng c PAT th bi ton

Max cng khng c PATNu bi ton Min c PAT l x0 th cng l

PAT ca bi ton Max v: fmax = - g(x0)

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Phng php gii bi ton MaxCch 2: Thut ton n hnh i vi bi ton Max,ta lp bng n hnh i vi PACB x0 :

i) (j u 0 vi j th PAT

ii) (k < 0 v aik e 0 vi i th bi ton

khng c PATiii) (k < 0 m (k < 0 u aik > 0. Ta xy

dng PA mi tt hn. n c a vo ng vi (k

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