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bài giảng vận trù học phần quy hoạch tuyến tính
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Chng 1 Quy hoch tuyn tnh
1
1.1. Bi ton quy hoch tuyn tnh (QHTT)
1.2. Phng php th vi bi ton QHTT hai bin
1.3. Phng php n hnh (simplex method)
1.4. Cc phng n xut pht
1.5. i ngu
Chng 1 Quy hoch tuyn tnh
Khi ta gp bi ton QHTT 2 bin th c phng php th gii. Nu bi ton QHTT c 3 bin tr ln th sao??
Phng php g?
Phng php n hnh
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Chng 1 Quy hoch tuyn tnh 1.3 Phng php n hnh (Simplex method)
Dng 1: Dng tng qut Cc i (cc tiu) hm mc tiu n bin (n 2) tha cc h rng buc c du (, , = ), du ca cc bin ( 0, 0).
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Chng 1 Quy hoch tuyn tnh 1.3 Phng php n hnh (Simplex method)
Dng 2: Dng chun (tc) (Standard Form)
(2.1) Cc tiu, du h s khng m, cc bin khng m.
(2.2) Cc i, du h s khng m, cc bin khng m.
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Chun min Chun max
Chng 1 Quy hoch tuyn tnh 1.3 Phng php n hnh (Simplex method)
Chun max
V d 1A ( xt)
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Chng 1 Quy hoch tuyn tnh 1.3 Phng php n hnh (Simplex method)
Chun min
p dng 2( xt)
V d 1B ( xt)
Gi x1, x2, x3 l s kg thc n A, B, C
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Chng 1 Quy hoch tuyn tnh 1.3 Phng php n hnh (Simplex method)
V d 2: Bi ton vn ti
Dng tng qut
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Chng 1 Quy hoch tuyn tnh 1.3 Phng php n hnh (Simplex method)
Dng tng qut
V d 1
V d 2
min
max
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Chng 1 Quy hoch tuyn tnh 1.3 Phng php n hnh (Simplex method)
H pttt
H Pt,
bpt
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Chng 1 Quy hoch tuyn tnh 1.3 Phng php n hnh (Simplex method)
(basic variables)
(nonbasic variables) 10
(n/bin khng c s)
Trong STT, c s l g?
Chng 1 Quy hoch tuyn tnh 1.3 Phng php n hnh (Simplex method)
V sao??
Nghim c s X = (0, 0, 40, 12, 40, 0)
Nghim c s X = (x1, x2, s1, s2, s3, Z)
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Chng 1 Quy hoch tuyn tnh 1.3 Phng php n hnh (Simplex method)
Phn tch v thit lp phng php n hnh
Z B
2 4 1 0 0 0 40
1 1 0 1 0 0 12
5 1 0 0 1 0 40
-20 -30 0 0 0 1 0 12
Chng 1 Quy hoch tuyn tnh 1.3 Phng php n hnh (Simplex method)
Z B
2 4 1 0 0 0 40
1 1 0 1 0 0 12
5 1 0 0 1 0 40
-20 -30 0 0 0 1 0
Ct then cht: Ct 2 Hng then cht:
hng 1
Phn t then cht: 4 13
Chng 1 Quy hoch tuyn tnh 1.3 Phng php n hnh (Simplex method)
Z B
2 4 1 0 0 0 40
1 1 0 1 0 0 12
5 1 0 0 1 0 40
-20 -30 0 0 0 1 0
1/2 1 1/4 0 0 0 10
1/2 0 -1/4 1 0 0 2
9/2 0 -1/4 0 1 0 30
-5 0 15/2 0 0 1 300
1/2 1 1/4 0 0 0 10
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Chng 1 Quy hoch tuyn tnh 1.3 Phng php n hnh (Simplex method)
Z B
1/2 1 1/4 0 0 0 10
1/2 0 -1/4 1 0 0 2
9/2 0 -1/4 0 1 0 30
-5 0 15/2 0 0 1 300
CTC: ct 1
HTC: hng 2
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Chng 1 Quy hoch tuyn tnh 1.3 Phng php n hnh (Simplex method)
Z B
1/2 1 1/4 0 0 0 10
1/2 0 -1/4 1 0 0 2
9/2 0 -1/4 0 1 0 30
-5 0 15/2 0 0 1 300
1 0 -1/2 2 0 0 4
1 0 -1/2 2 0 0 4
0 1 1/2 -1 0 0 8
0 0 2 -9 1 0 12
0 0 5 10 0 1 320
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Chng 1 Quy hoch tuyn tnh 1.3 Phng php n hnh (Simplex method)
Z B
0 1 1/2 -1 0 0 8
1 0 -1/2 2 0 0 4
0 0 2 -9 1 0 12
0 0 5 10 0 1 320
Dng, v sao?
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Chng 1 Quy hoch tuyn tnh 1.3 Phng php n hnh (Simplex method)
Trnh by tm tt theo dng bng
Z B
2 4 1 0 0 0 40
1 1 0 1 0 0 12
5 1 0 0 1 0 40
-20 -30 0 0 0 1 0
Z B
1/2 1 1/4 0 0 0 10
1 1 0 1 0 0 12
5 1 0 0 1 0 40
-20 -30 0 0 0 1 0 18
= A1
Nghim c s X1 = (0, 0, 40, 12, 40, 0) CTC: 2, HTC: 1, PTTC: a12 = 4
Chng 1 Quy hoch tuyn tnh 1.3 Phng php n hnh (Simplex method)
Z B
1/2 1 1/4 0 0 0 10
1/2 0 -1/4 1 0 0 2
9/2 0 -1/4 0 1 0 30
-5 0 15/2 0 0 1 300
Z B
1/2 1 1/4 0 0 0 10
1 0 -1/2 2 0 0 4
9/2 0 -1/4 0 1 0 30
-5 0 15/2 0 0 1 300 19
= A2
Nghim c s X2 = (0, 10, 0, 2, 30, 300) CTC: 1, HTC: 2, PTTC: a21 = 1/2
Chng 1 Quy hoch tuyn tnh 1.3 Phng php n hnh (Simplex method)
Z B
0 1 1/2 -1 0 0 8
1 0 -1/2 2 0 0 4
0 0 2 -9 1 0 12
0 0 5 10 0 1 320
Dng, v tt c cc h s ca hng cui (hm mc tiu) u khng m
20
= A3
Nhn xt: PP n hnh di chuyn t nh ny qua nh kia ca min rng buc (nh O(0, 0) qua D(0, 10) qua C(4, 8))
Chng 1 Quy hoch tuyn tnh 1.3 Phng php n hnh (Simplex method)
Bng 1: nh O Bng 2: nh D Bng 3: nh C
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Chng 1 Quy hoch tuyn tnh 1.3 Phng php n hnh (Simplex method)
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Chng 1 Quy hoch tuyn tnh 1.3 Phng php n hnh (Simplex method)
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Chng 1 Quy hoch tuyn tnh 1.3 Phng php n hnh (Simplex method)
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Chng 1 Quy hoch tuyn tnh 1.3 Phng php n hnh (Simplex method)
IV. Cc bc gii BT QHTT dng chun tc max
25 4) Trnh by dng bng nh Slide 18 Slide 20
Chng 1 Quy hoch tuyn tnh 1.3 Phng php n hnh (Simplex method)
V. Lu : Khi gii bi ton chun max bng pp n hnh
1) Ct then cht: * Trong hng cui xc nh s m b nht, * Ct cha s m b nht ny l ct then cht. Nu c nhiu s m b nht v bng nhau th ta chn bt k trong chng.
2) Hng then cht: * Ly cc s ct cui (B) chia cho cc phn t dng ca ct then cht, * T s nh nht tng ng vi hng then cht. Nu c nhiu t s nh nht ging nhau, ta chn bt k trong chng.
3) Phn t then cht: nm trn giao im ca ct then cht v hng then cht.
4) Nghim c s: nghim Xi ma trn n hnh th i, c c bng cch cho cc bin c s bng bi , cc bin cn li u bng 0.
Nu cc phn t < 0 hoc = 0 th sao?
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1. BT c dng chun tc max khng?
2. Bin i c v dng chun tc max khng?
3. Dng, xem pp n hnh vi dng bt khc (4.6/page 124) or mc 1.4
4. Bin i v dng chun.
Chng 1 Quy hoch tuyn tnh 1.3 Phng php n hnh (Simplex method)
VI. Lu ca phng php n hnh
1 3 2
4 5
6
8
10
7
9
No No
No
No
Yes Yes
Yes
Yes
5. a thm bin ph, vit ma trn n hnh xut pht.
6. C h s m hng cui khng?
7. Dng, thu c ma trn n hnh ch.
8. C phn t dng trn ct then cht khng?
9. Dng, bi ton v nghim
10. Chn phn t then cht v thc hin php tnh then cht (bin i ma trn)
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Chng 1 Quy hoch tuyn tnh 1.3 Phng php n hnh (Simplex method)
p dng
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Chng 1 Quy hoch tuyn tnh 1.3 Phng php n hnh (Simplex method)
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Bi 2. Cho bi ton QHTT chun max (P) c ma trn n
hnh th nht nh sau:
x1 x2 s1 s2 s3 s4 Z B
3 -1 1 0 0 0 0 5
2 0 0 1 0 0 0 7
1 3 0 0 1 0 0 30
2 1 0 0 0 1 0 12
-2 -5 0 0 0 0 1 0
A1 =
a) Tm phn t then cht b) Vit bi ton (P)
c) Gii (P) d) Phng php n hnh s dng
li ma trn th my?
Chng 1 Quy hoch tuyn tnh 1.3 Phng php n hnh (Simplex method)
SV lm tt c
bi tp dng
chun max
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