Copy of Mô hình toan kinh tế -1

8/2/2019 Copy of M hnh toan kinh t -1

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Phn 2:M Hnh Ton Kinh TChng 1:Gii Thiu M Hnh Ton Kinh T


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I. Khi nim m hnh kinh t v m hnh ton kinh t

1. M hnh kinh t:

- M hnh ca mt i tng l s phn nh hinthc khch quan ca mt i tng; s hnhdung, tng tng i tng bng ngh ca

ngi nghin cu v vic trnh by, th hin, dint ngh bng li vn, ch vit, s , hnhv, hoc mt ngn ng chuyn ngnh.- M hnh bao gm ni dung ca m hnh v hnh

thc th hin ni dung.


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2. M hnh ton kinh t:

L m hnh kinh t c trnh by bng ngn ng

ton hc.V d:

Gi s chng ta mun nghin cu, phn tch

qu trnh hnh thnh gi c mt loi hng ho Atrn th trng vi gi nh cc yu t khc nh

iu kin sn xut hng ho A, thu nhp, s thch

ngi tiu dng cho trc v khng thayi.


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M hnh bng li:Ti th trng hng ho A, ni ngi bn, ngimua gp nhau v xut hin mc gi ban u. Vimc gi lng hng ho ngi bn mun bngi l mc cung, lng hng ho ngi mua munmua gi l mc cu. Nu cung ln cu th ngi b

phi gim gi do hnh thnh mc gi mi thphn. Nu cu ln hn cung th ngi mua sn sngtr gi cao hn mua c hng do mc gi

mi cao hn c hnh thnh. Vi mc gi mi xuthin mc cung, mc cu mi. Qu trnh tip din chn khi cung bng cu.


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M hnh ton kinh t:

- Gi S, D l ng cung, ng cu tng ng.

- ng vi mc gi p ta c: S = S(p); D = D(p)Ta c m hnh cn bng th trng k hiu MHIA

di y:

S = S(p)

D = D(p)

S = D

0D

)(' dp

dpD

0)(' dp

dS

pS


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Khi mun cp n tc ng ca thu nhp (M)

v thu (T) ti qu trnh hnh thnh gi ta c m

hnh MHIB di y:S = S(p, T)

D = D(p, M, T)

S = D

0

D

p

0

p

S


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II. Cu trc m hnh ton kinh t:

- M hnh ton kinh t l mt tp hp gm cc bin

s v cc h thc ton hc lin h gia chngnhm din t i tng lin quan n s kin,hin tng kinh t.M hnh ton kinh t gm: cc bin, cc phngtrnh, cc bt phng trnh.


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1. Cc bin s ca m hnh:- Bin ni sinh (bin c gii thch):+ L cc bin m v bn cht chng phn nh, th

hin trc tip s kin, hin tng kinh t v gitr ca chng ph thuc vo gi tr ca cc binkhc trong m hnh.

+ Nu bit gi tr ca cc bin khc trong m hnhta c th xc nh gi tr c th ca bin ni sinh

bng cch gii cc h thc.V d:Trong m hnh MHIA cc bin S, D, p l ccbin ni sinh.- Bin ngoi sinh (bin gii thch)

L cc bin c lp vi cc bin khc trong mhnh, gi tr ca chng tn ti bn ngoi m hnh.V d:Trong m hnh MHIB cc bin M, T l ccbin ngoi sinh.


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- Tham s(thng s):l cc bin s m trong phmvi nghin cu chng th hin cc c trngtng i n nh, t bin ng.

Cc tham s ca m hnh phn nh xu hng,mc nh hng ca cc bin ti cc bin nisinh.

V d:Nu trong m hnh MHIB cS = p.T

th , , l cc tham s ca m hnhLu :Cng mt bin s, trong cc m hnh khcnhau c th ng vai tr khc nhau


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2. Mi lin h gia cc bin s- Cc phng trnh

ca m hnh:

a. Phng trnh nh ngha:phng trnh thhin quan h nh ngha gia cc bin s hochai biu thc hai v ca phng trnh.

V d:

+ Li nhun (LN) c nh ngha l hiu s catng doanh thu (TR) v tng chi ph (TC):

LN = TR TC

+ trong m hnh MHIA, cc phng trnh

l cc phng trnh nh ngha.

0D

)(' dp

dpD0)('

dp

dSpS


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b. Phng trnh hnh vi:

l phng trnh m t quan h gia cc bin do

tc ng ca cc quy lut hoc do gi nh.- T phng trnh hnh vi ta c th bit s bin

ng ca bin ni sinh- hnh vi ca bin nykhi cc bin s khc thay i.

V d:

Trong m hnh MHIA c S = S(p), D = D(p) chng

phn nh phn ng ca ngi sn xut v ngi

tiu dng trc s thay i ca gi c.


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c. Phng trnh iu kin:

L phng trnh m t quan h gia cc bin s

trong cc tnh hung c iu kin m m hnh cp.

V d:

Trong m hnh MHIA, phng trnh S = D l phngtrnh iu kin v n th hin iu kin cn bng th

trng.


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III. Phn loi m hnh ton kinh t:1. Phn loi m hnh theo c im cu trc v cngc ton hc s dng:

- M hnh ti u:l m hnh phn nh s la chn cch thc hotng nhm ti u ho mt hoc mt s ch tiu nhtrc.

- M hnh cn bng:l lp m hnh xc nh s tn ti ca trng thi bnnu c v phn tch s bin ng ca trng thi nykhi cc bin ngoi sinh hay cc tham s thay i.- M hnh tt nh, m hnh ngu nhin: M hnh vicc bin l tt nh (phi ngu nhin) gi l m hnhtt nh, nu c cha bin ngu nhin gi l m hnhngu nhin.


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- M hnh tnh, m hnh ng:

M hnh c cc bin m t hin tng kinh t

tn ti mt thi im hay mt khong thigian xc nh gi l m hnh tnh. M hnhm t hin tng kinh t trong cc bin sph thuc vo thi gian gi l m hnh ng.


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2. Phn loi m hnh theo quy m, phm vi, thi gian:

- M hnh v m:M hnh m t cc hin tng kinh

t lin quan n mt nn kinh t, mt khu vc kinht gm mt s nc.

- M hnh vi m:M hnh m t mt thc th kinh tnh hoc nhng hin tng kinh t vi cc yu t

nh hng trong phm vi hp v mc chi tit.- Theo thi hn m m hnh cp ta c: M hnh

ngn hn, m hnh di hn.


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Chng 2:M Hnh Ti u TuynTnh


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I. Mt s v d thc t dn n Bi ton quy hoch tuyntnh (QHTT):

VD 1: u t ti chnh:Mt cng ty u t nh dng khon qu u t 500triu ng mua mt s c phiu trn th trng chngkhon. Cng ty a ra cc gii hn trn ca s tin mua

tng loi chng khon.Bng di y cho cc s liu v cc gii hn nycng nh li sut ca cc chng khon .


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Loi chngkhon

Li sut (trungbnh)

Gii hn mua

A

B

C

D

7%

8,5%

7,8%

8,2%

100 triu

300 triu

250 triu

320 triu

ngn nga ri ro trong u t, cng ty cn quy nhkhon u t vo loi c phiu A v C phi chim t nht l55%,loi c phiu B phi chim t nht 15%trong tng sdanh mc u t thc hin.

Hy xc nh s tin cng ty mua tng loi c phiu saokhng vt qu khon d kin ban u, m bo i hiv a dng ho ng thi t mc li (trung bnh) cao nht.


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VD 2: Bi ton vn ti

Mt cng ty kinh doanh xng du hng tun cung ng

xng du cho 3 trm bn l A, B, C. Cng ty c th axng t kho I v II. D tr cung ng xng ca kho I l 20

tn, kho II l 40 tn.

Chi ph cho vic cung ng xng t kho n cc trm

c cho trong bng di y:

n v: Nghn ng/tn


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Trm

kho A B C

I 500 400 700

II 600 500 500

Nhu cu tiu th xng hng tun ca 3 trm ln lt l20, 15, 15 (tn).Cng ty cn lp k hoch cung ng xng t d tr ca cckho n cc trm m bo nhu cu ca cc trmvi

tng chi ph l nh nht.


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M hnh ho:- Gi lng xng chuyn t kho I, kho II n cc trm ln

lt l x1A, x1B, x1C v x2A, x2B, x2C(tn).

- Tng lng xng chuyn t kho I n cc trm: x1A+x1B+x1C- Tng lng xng chuyn t kho II n cc trm:x2A+x2B+x2C- Tng lng xng trm A nhn c t 2 kho: x1A + x2A- Tng lng xng trm B nhn c t 2 kho: x1B + x2B- Tng lng xng trm C nhn c t 2 kho: x

1C

+ x2C

- Tng chi ph tng ng l:

500x1A+400x1B+700x1C+600x2A+500x2B+500x2C


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Ta c bi ton sau:

Xc nh vect x = ( x1A, x1B, x1C, x2A, x2B, x2C )sao cho:

f(x) = 500x1A+400x1B+700x1C+600x2A+500x2B+500x2C minVi iu kin:x1A + x2A = 20

x1B + x2B = 15

x1C + x2C = 15x1A + x1B + x1C 20

x2A + x2B + x2C 40

x1A 0, x1B 0, x1C 0, x2A 0, x2B 0, x2C 0


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II. Bi ton QHTT tng qut v cc dng c bit:

1. Dng tng qut: Tm x = (x1, x2, , xn) sao cho

1)

2)

max)min(xcf(x)n

1j

jj

n

1j1ijji )I(ibxa

n

1j

2ijji )I(ibxa

n

1j3ijji )I(ibxa

Nu k hiu D l tp tt c cc vect x tho mn h iukin 2)th y chnh l bi ton tm min (max) ca hm f(x)trn D.


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VD 1:Cho bi ton QHTT

Tm x = (x1, x2, x3, x4) sao cho

1)

2)

minxx-2xxf(x) 4321

x1 + x2 x3 = 2 (1)

x2 0 (4)2x1 + x2 3 (2)

x2 + x3 + x4 4 (3)x3 0 (5)x4 0 (6)


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2. Mt s khi nim v nh ngha:f(x): gi l hm mc tiuMi phng trnh hoc bt phng trnh trong h iukin 2)gi l mt rng buc. Nhng rng buc dng cbit: xj 0 hay xj 0, gi l cc rng buc du i vi bin xjng vi rng buc th i ta k hiu A*i l vect dng c ccthnh phn l cc h s ca bin xj

Mt nhm rng buc c h vect A*i tng ng c lptuyn tnh c gi l cc rng buc c lp tuyn tnh.

Xt cc rng buc khng phi rng buc du, h vect A*itng ng vi cc rng buc ny to thnh mt ma trn, khiu A. Ma trn A c n ct, vect ct th j k hiu l Aj.


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V d 2:

Xc nh x = (x1, x2, x3, x4, x5) sao cho

f(x) = 3x1 +x2 -5x3 + 2x4 + x5 minx1 +x2 +x3 + x4 + x5 = 21

2x1 +6x2 -3x3 + 2x4 - 2x5 2-3x1 +x2 +2x3 -3x4 + 3x5 = 28

xj 0 (j = 1, 2, , 5)A*1 = (1, 1, 1, 1, 1);

A*2 = (2, 6, -3, 2, -2);

A*3 = (-3, 1, 2, -3, 3)


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Phng n:Mt vect x tha mn mi rng buc ca bi ton gi l

mt phng n (PA).

+Nu i vi PA x m rng buc i tho mn vi du ng

thc th ta ni PA x tho mn cht rng buc i hay rng

buc i l cht i vi PA x.

+Nu i vi PA x m rng buc i tha mn vi du bt

ng thc thc s th ta ni PA x tho mn lng rng buc

i hay rng buc l lng i vi PA x.


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Phng n cc bin (PACB):Mt phng n tho mn cht n rng buc c lp

tuyn tnh gi l phng n cc bin (PACB).

PACB tho mn cht ng n rng buc gi l PACBkhng suy bin, tho mn cht hn n rng buc gi lPACB suy bin.Phng n ti u (PAT):

Mt phng n m ti tr s hm mc tiu t cctiu (hoc cc i) gi l PAT.

Mt bi ton c t nht mt PAT gi l bi ton giic, bi ton khng c PAT gi l bi ton khng giic.


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VD 3: Xt bi ton

f(x) = x1

+ 6x2 max

x1 + 5x2 20

x1 5x2 4

x2 0Bi ton c cc PACB: xA = (5, 3), xB = (5, 0), xC=(20, 0)


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Dng th biu din tp phng n:

x2

4 A

3B C

0 5 20 x1


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3. Cc dng c bit:

a. Bi tondng chnh tc:Tm vt x = (x1, x2, , xn) sao cho

max)min(xcf(x)1)n

1jjj

)n1,(j0x

)m1,(ibxa2)

j

n

1jijji

Mnh :Mi bi ton quy hoch tuyn tnh u c th a v biton dng chnh tc tng ng theo ngha tr ti uca hm mc tiu trong hai bi ton l trng nhau v tPA, PAT ca bi ton ny suy ra PA, PAT ca bi tonkia.


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Cch a mt bi ton v dng chnh tc:Nu xj0 th t tj = -xj tj 0. Nu bin s xj khngc rng buc du th ta t xj= vi

n

1jijij bxa

,,j

,j xx 0x,x

,,j

,j

Nu mt rng buc c dng: th thay bng

n

1ji

pijij bxxa 0x

pi vi

pixv h s ca trong f(x) bng 0.

Tng t nu rng buc c dng

n

1jijij bxa

n

1ji

pijij bxxa 0x

pi

th thay bng

vi


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b. Bi ton dng chun:l bi ton c dng

x1 + a1,m+1xm+1 + + a1nxn = b1x2 + a2,m+1xm+1 + + a2nxn = b2

xm+ am,m+1xm+1 + + amnxn = bm

Bton c mt PACB l x0 = (b1, b2, , bm, 0, , 0)

max)min(xcf(x)n

1j

jj

m,1i,0b),n,1j(0x ij

III C h h h bi QHTT


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III. Cc tnh cht chung ca bi ton QHTT:

Tnh cht 1:S tn ti PACB

Nu bi ton c PA v hng ca ma trn h rng buc

bng n th bi ton c PACB.Tnh cht 2:S tn ti PAT

Nu bi ton c phng n v tr s hm mc tiu b chn

di (trn) trn tp phng n th bi ton c PAT (gii

c).

Nu bton c PACB v gii c th bton c PACB ti

u.

Tnh cht 3:Tnh hu hn ca s PACBS PACB ca mi bi ton quy hoch tuyn tnh u hu

hn.


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IV. Phng php n hnh gii bi ton QHTT:1. Ni dung ca phng php:

Xut pht t mt PACB tm cch nh gi PACB y, nu

n cha ti u th tm cch chuyn sang mt PACB mitt hn, qu trnh c lp li, v s PACB l hu hn nnsau mt s hu hn bc hoc s kt lun bi ton khnggii c hoc s tm c PACB ti u.

Ta s xt bi ton dng chnh tc trong qu trnh giithiu phng php n hnh.


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2. c im ca PACB ca bi ton dng chnh tc:

nh l:

PA x = (x1, x2, , xn) ca bi ton dng chnh tc l cc

bin khi v ch khi h cc vect Aj / xj>0l .lp tuyn tnh.Nhn xt:

Khng lm mt tnh tng qut ta lun c th gi thit h

phng trnh rng buc ca bi ton dng chnh tc gm m

phng trnh c lp tuyn tnh vi m < n, tc r(A) = m.

Khi mt PACB s c khng qu m thnh phn dng.

PACB khng suy bin c ng m thnh phn dng, PACB

suy bin c t hn m thnh phn dng.


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3. C s ca PACB ca bi ton dng chnh tc:

N:Ta gi mt h m vect Ajc lp tuyn tnh bao hmh cc vect Aj/ xj > 0l c s ca PACB x.

K hiu mt cch quy c l J, trong

J = {j: Ajnm trong c s}

Ch : PACB x c c s l J, cn phi hiu:

- S phn t ca J l m- {Aj, jJ} c lp tuyn tnh- {Aj, j J} {Aj, xj>0}

i vi PACB x =(x1, x2, , xn) c s J ta gi cc thnh

phn xj (jJ) l thnh phn c s, xk (kJ) l thnh phnphi c s. D thy xk = 0 (kJ).


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PACB x, c s J ta c:- Cc vect Ak (kJ) cng biu din c qua c s J.Gi cc h s phn tch ca Ak l xjk tc l:Ak =

- Ta xc nh i lng k (kJ) bng cng thc sau- k cgi l c lng ca bin xktheo c s J.- Ni ring th

Jj

jjkAx

Jj kjkjk cxc

J)(j0j


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4. Quan h gia PACB v PA ca bi ton dng ctc:

i vi PACB x0c s J0, vi mi ch s kJ0 xc nh mtvect zkgi l phng zkc cc thnh phn nh sau:

k)(j1

k)j,J(j0

)J(jx

z 0

0jk

jk

Xt s di chuyn theo phng zk, tc l xt cc vect cdng x() = x0 + .zkvi 0.

Thay vt x() = x0 + .zk vo cc phng trnh rng buclun tho mn.

x() l PA th ch cn x() 0.

T


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Ta c:

Vy tm li x() l phng n th ch cn x0j - .xjk 0(jJ0). C 2 trng hp xy ra:TH 1:Nu xjk 0 (jJ0) th x() l PA 0. Khi ta gi

zkl phng v hn.TH 2:Nu xjk > 0, t x0j - .xjk0 ng vi xjk > 0, suy ra x0j /xjk. Gi

Nh vy x() l PA khi 0 0.Trng hp ny zkgi l phng hu hn v x(0) l

PACB mi.

k)(j

k)j,J(j0

)J(j.xx

)(x 0

0jkj0

j

jkj

jkx

/xx0

00 min


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Trong c 2 trng hp ta lun c:

f(x()) = f(x0) - .kVi > 0 ta c Nu k > 0 th f(x()) < f(x0), khi zkgi l phng gim.Nu k < 0 th zkgi l phng tng.Nu k = 0 th zkl phng khng i.

C


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5. Ccnh l c bn:

Di y ta xt bi ton dng chnh tc vi hm mc

tiu f(x) min.nh l 1:(Du hiu ti u ca PACB)

Nu i vi PACB x0, c s J0ca bi ton dng chnh

tc m k 0 (kJ0) th x0 l PAT.Ch :

+)Nu k < 0, kJ0 th x0l PAT duy nht.+) Nu x0l PACB khng suy bin th x0l PAT khi v ch

khi k 0 (kJ0).


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nh l 2:(Du hiu bi ton khng gii c)

Nu i vi PACB x0c s J0ca bi ton dng chnh

tc tn ti mt k > 0 m xjk 0 (jJ0) th bi ton khnggii c.

nh l 3:(Du hiu iu chnh PACB)

Nu i vi PACB x0c s J0ca bi ton dng chnh

tc m mi k > 0 u tn ti xjk > 0 th c th chuynsang mt PACB mi tt hn trong trng hp bi tonkhng suy bin (ngha l bi ton m mi PACB u

khng suy bin ).


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6. Thut ton ca phng php n hnh:Gi thit bi ton dng chnh tc c hm mc tiu f(x)

min, bit mt PACB x0c s J0, khng lm mt tnh tng

qut c th gi thit J0 = 1, 2, , mtc l c s gm ccvect {A1, A2, , Am}.

Thut ton gm cc bc sau:Bc 1:Lp bng n hnh ng vi PACB x0


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Hs

Cs

Phngn

c1 c2 cr cm cm+1 cs cnx1 x2 xr xm xm+1 xs xn

c1c2cr

cm

x1x2xr

xm

x01x02x0r

x0m

1 0 0 0 x1,m+1 x1s x1n0 1 0 0 x2,m+1 x2s x2n 0 0 1 0 xr,m+1 [xrs] xrn

0 0 0 1 xm,m+1 xms xmn

f(x) f(x0) 0 0 0 0 m+1 s n


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Bc 2:Kim tra du hiu ti u:

Nu k 0, kJ0 th x0l PAT, nu tn ti mt k > 0th chuyn sang bc 3.Bc 3:Kim tra tnh khng gii c ca bi ton:

Nu tn ti mt k > 0 m xjk 0, jJ0 th bi tonkhng gii c.

Nu vi mi k> 0 u c xjk> 0 th chuyn sang bc 4.

4 C


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Bc 4:Chn vect a vo c s v xc nh vect loikhi c s.+ Chn vect a vo:

Gi s maxk = s (k>0). Vect Asc a vo c s.+ Chn vect loi ra:

Tnh , gi s,

vect Arb loi khi c s, phn t trc ca php bin i

l xrs, trong bng ng khung phn t ny.

Thnh lp mt mu bng n hnh mi, v tr xr ghi xs

v ghi cs thay cho cr. Chuyn sang bc 5

js

j

jsxJj x

x0

0,00 min

)0,( 0

0

0 rsrs

r xJrx

x


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Bc 5:Bin i bng:Tnh cc dng ca bng mi (bt

u t ct th 3 tr i) theo quy tc sau

- tnh dng vect a vo (xs) trong bng mi ta lydng vect loi ra (xr) trong bng c chia cho phn ttrc. Dng ny c gi l dng chun.

- tnh dng (xj) trong bng mi, ta ly dng (xj) trongbng c tr i tch dng chun vi xjs

- tnh dng cui trong bng mi ta ly dng cui cabng c tr i tch dng chun vi s

VD 1: Gii bng phng php n hnh bi ton


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VD 1:Gii bng phng php n hnh bi tonf(x) = -4x1 + 3x3 - x4 -5x5 min

4x1 + x2 +4x4 + 2x5 = 6 (1)2x1 + x3 + 3x4 -3x5 = 8 (2)

3x1 + 5x4 + 5x5 10 (3)xj 0 (j = 1, 2, 3, 4, 5)

Gii:a bi ton v dng chnh tc.

f(x) = -4x1

+ 3x3

- x4

-5x5

+ 0x6 min

4x1 + x2 +4x4 + 2x5 = 6 (1)2x1 + x3 + 3x4 -3x5 = 8 (2)3x1 + 5x4 + 5x5 +x6= 10 (3)

xj0 (j = 1, 2,, 6)Bi ton trn dng chun, c PACB x0=(0, 6, 8, 0, 0, 10)c s J0 = {2, 3, 6} l c s n v. Lp bng n hnh:


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HS CS PA

f(x) 24 0010 -4 010

-430

x1x3x6 1

0

3/2 1 1/4 0 1 1/2 0 dc

f(x)

n bng n hnh th 2 ta c k 0 (kJ) nn bi tonc PAT x* = (3/2, 0, 5, 0, 0) vi fmin = f(x*) = 9.

5 0 -1/2 1 1 -4

9 0 -5/2 0 0 -9 0

11/2 0 -3/4 0 2 7/2

030

x2x3x6

6810

- 4 0 3 -1 -5 0

x1 x2 x3 x4 x5 x6

[4] 1 0 4 2 02 0 1 3 -3 0

3 0 0 5 5 1

C


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Cc ch khi p dng thut ton

1)i vi bi ton c hm f(x) max th c th chuynv gii bi ton vi hm g(x) = - f(x) min, vi fmax = - gminhoc cng c th gii trc tip vi du hiu ti u l k 0(kJ0).2)Trng hp bi ton suy bin th 0c th bng 0, khi 0= 0 vn thc hin thut ton mt cch bnh thng, nghal vect ng vi 0 vn b loi khi c s.3)Nu khi chn vect a vo c s hoc a ra khi c

s c nhiu vect thuc din la chn th ta tu chn mt

trong s .

4) Khi d th t t 2 t h


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4)Khi p dng thut ton s c 2 trng hp xy ra:

TH 1:Bi ton dng chun, n cho ngay mt PACB x0, c

s J0l c s n v, ta a ton b cc h s v tri ca

cc phng trnh rng buc vo bng n hnh v lpc ngay bng n hnh i vi PACB ny.

TH 2: Khi PACB x0, c s J0cha phi l c s n v, ta

phi bin i ma trn h s m rngA bng cc php bini s cp trn dng ca ma trn, a cc vect c sthnh cc vect n v khc nhau. Sau a ton b cc

phn t trong ma trn m rng cui cng vo trong bng

n hnh v thc hin tip thut ton.

VD 2 Ch bi t


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VD 2: Cho bi ton

f(x) = -2x1 - 6x2 + 8x3 5x4 minx1 + 2x2 3x3 + x4 = 8 (1)

-2x1 + x2 + x3 5x4 2 (2)

4x1 + 7x2 -8x3 +2x4 20 (3)xj 0 (j =14 )

v vect x0

= (8, 0, 0, 0).a.Chng t x0l phng n cc bin, li dng x0gii

bi ton bng phng php n hnh.

b.Tm mt phng n x c tr s f(x) = - 50.

Gii:

a Vect x0 tho mn mi rng buc ca bi ton tho mn


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a. Vect x0tho mn mi rng buc ca bi ton, tho mn

cht rng buc (1) v 3 rng buc du, cc rng buc ny

ltt nn x0l PACB ca bi ton.

a bi ton v dng chnh tc:f(x) = -2x1 - 6x2 + 8x3 5x4 min

x1 + 2x2 3x3 + x4 = 8

-2x1

+ x2

+ x3

5x4

+ x5

= 2

4x1 + 7x2 - 8x3 + 2x4 - x6 = 20

xj 0 (j = 1 6)PACB tng ng x0 = (8, 0, 0, 0, 18, 12) c s {A1, A5, A6}.y khng phi l c s n v. lp bng n hnh taphi bin i ma trn m rngA

8001321


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Bng n hnh:

2010287420151128001321

A

3d14d3d12d2d2d

12102410180135508001321


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HS CS PA -2 -6 8 -5 0 0x

1x

2x

3x

4x

5x

6

-2

0

0

x1

x5

x6

8

18

12

1 2 -3 1 0 0

0 5 -5 -3 1 0

0 1 -4 2 0 1

f(x) -16 -220 0 03

-20-5

x1x5x4 6 0 1/2 -2 1 0 1/2

3/2

2 1 3/2 -1 0 0 -1/2

dc

f(x)

n bng n hnh th 2 ta c 3 = 4 > 0 m xj3 < 0 (jJ)nn bi ton khng gii c.

[ ]

36 0 13/2 -11 0 1

-34 0 1/2 4 0 0 -3/2

b T t h t f( ) 50


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b.Tm mt phng n c tr s f(x) = - 50.

Gi PA tng ng bng n hnh th 2 l x*, t x* di

chuyn theo phng z3 l phng gim v hn ta c

cc PA c dng:

x() = x* + .z3, 0Do : f(x()) = f(x*) - .3Nh vy -50 = -34 4. = 4Ta c: x* = (2,0,0,6,36,0)

z3 = (1,0,1,2,11,0)

Vy PA cn tm l: x = (6,0,4,14,80,0)

V. Tm phng n cc bin:


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Khi bi ton dng chnh tc nhng khng phi dngchun ng thi khng bit PACB, nh vy mun pdng thut ton cn tm mt PACB ca bi ton.

Xt bi ton dng chnh tc:

bi 0 (i = 1m)

n

1jjj minxcf(x)

n

1j ijji)m1,(ibxa

)n1,0(jxj


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T bi ton cho xy dng mt bi ton ph, k hiu

l P bng cch cng vo v tri phng trnh rng buc i

mt bin gi xgi (i = 1 m) vi hm mc tiu l tng ccbin gi thm vo v hm mc tiu ny phi t cc

tiu.

K hiu xg = (xg1, xg2, , xgm) l vect cc bin gi vhm mc tiu ca bi ton ph l P(x, xg).

Khi bi ton ph c dng:

minx)xP(xm g

ig


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Nhn xt: Vto xl PA ca bi ton xut pht khi v ch khi (x,xg= 0) lPA ca bi ton ph P. Do x l PACB ca bi ton xutpht khi v ch khi (x, xg= 0) l PACB ca bi ton P.Vic tm PACB ca bi ton xut pht (nu c) s dn titm PACB ca bi ton P c dng (x, xg= 0).Bi ton P c dng chun v P(x, xg) 0, PA (x, xg) nnbi ton P lun gii c. Do P(x, xg = 0) = 0 nn (x, xg = 0) lPAT ca bi ton P.

minx)xP(x,1i

i

n

1ji

gijji

)m1,(ibxxa

),n1,0(jxj )m1,0(ixgi


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Nh vy:Vic tm PACB ca bi ton xut pht dn ti vicgii bi ton P.

Dng thut ton n hnh gii bi ton P tm c PAT

vC 2 trng hp xy ra:

TH1: Pmin > 0

Khi bi ton xut pht khng c phng n.

)x,x(

g

min

g

P)x,xP(

TH 2: Pmin = 0


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min

Khi = 0 (i = 1m) hay = 0, PAT ca P c dngdo x l PACB ca bi ton xut pht.

Hai kh nng c th xy ra:a.Trong c s ca PACB ti u khng c cc

vect tng ng vi cc bin gi. Ta loi cc ct xgi , tnh

li hng c lng ktheo hm f v tip tc thut ton.b.Trong c s ca PACB T c t nht mt vectbin gi.Ta loi cc ct ng vi k(P) < 0 v cc ct xgi phic s, sau tnh li cc c lng k theo theo hm f vtip tc thut ton.

0)x,x(g

i

gx

gx

0)x,x(g

0)x,x( g

VD 1: Gii bi ton sau bng phng php n hnh:


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VD 1:Gii bi ton sau bng phng php n hnh:f(x) = 3x1 + 4x2 + 2x3 + 2x4 min

2x1 + 2x2 + x4 = 28

x1 + 5x2 + 3x3 2x4 312x1 2x2 + 2x3 + x4 = 16xj 0 (j = 1 4)

Gii:a bi ton v dng chnh tc:

f(x) = 3x1 + 4x2 + 2x3 + 2x4 min2x1 + 2x2 + x4 = 28

x1 + 5x2 + 3x3 2x4 + x5 = 312x1 2x2 + 2x3 + x4 = 16

xj 0 (j = 15)

Ta thy bi ton khng phi dng chun nn thnh lp bi


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Ta thy bi ton khng phi dng chun nn thnh lp biton ph:

P(x, xg) = xg1 + xg

3 min2x1 + 2x2 + x4 + xg1 = 28x1 + 5x2 + 3x3 2x4 + x5 = 312x1 2x2 + 2x3 + x4 + xg3 = 16

xj 0 (j = 15), xg 0

HS CS PA 1 13 4 2 2 0


66/147

44 4 0 2 2 0 0 0

x1 x2 x3 x4 x5 xg

1 xg

3

1

01

xg1

x5xg3

28

3116

2 2 0 1 0 1 0

P

1 5 3 -2 1 0 02 - 2 2 1 0 0 1[ ]

xg1

x5x1

1

00 8 1 -1 1 1/2 0 0

12 0 4 -2 0 0 1

00 6 2 -5/2 123

0P 12 -20 4 0 0

[ ]

3 0 1 -1/2 0 0 d

10 0 5 -5/2511 1 0 1/2 1/2 0

dc

f(x) 45 -5/2 -1/20 0 0

x2

x5x1

4

03

Bi ton c PAT duy nht x* = (11,3,0,0) v fmin = 45


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Khi gii bi ton P cn ch mt s c im sau: Khi xy dng bi ton ph ch cng thm bin gi vo

nhng phng trnh cn thit. Mt bin gi b loi khi c s th ct tng ngkhng cn tnh cc bc tip theo.

Ch c p dng cng thc i c s cho hng c

lng khi hai bng k tip c cng tn hm mc tiu. Khi tt c cc bin gi b loi khi c s th kt thc vicgii bi ton P, tnh li dng c lng k theo hm f vtip tc thut ton.

VD 2: Gii bi ton sau bng phng php n hnh:


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VD 2:Gii bi ton sau bng phng php n hnh:f(x) = - 4x1 - 2x2 + 4x3 - x4 min

x1 + x2 + 2x3 + x4 = 15

2x2 + x3 2x4 = 62x1 + 5x2 x4 = 45

xj 0 (j = 1 4)Gii:

Lp bi ton ph:P(x, xg) = xg1 + xg2+ xg3 minx1 + x2 + 2x3 + x4 +x

g1 = 15

2x2 + x3 2x4 + xg2 = 6

2x1 + 5x2 x4 +xg

3 = 45xj 0 (j = 1 5), xg 0

HS CS PA 1 1 1


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66 3 8 3 -2 0 0 0

x1 x2 x3 x4 xg

1 xg

2 xg

3

1

11

xg1

x

g

2xg3

15

645

1 1 2 1 1 0 0

P

0 2 1 -2 0 1 02 5 0 -1 0 0 1[ ]

xg1

x2xg3

1

01 30 2 0 -5/2 4 0 1

12 1 0 3/2 2 1 0

00 1 1/2 -1 03

0P 42 -13 0 6 0

[ ]

6 1/2 0 3/4 1 0 d

01/2 1 5/4 096 0 0 -11/2 0 1

dc

P 6 -11/2 00 0 0

x4

x2xg3

0

01

Do Pmin = 6 >0 nn bi ton k0

c PA. Vy bi ton k0

gii

VD 3:Gii bi ton sau bng phng php n hnh:


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g p g p pf(x) = 6x1 + 2 x2 + x3 min

2x1 + 5x2 + 3x3 104x1 - 3x2 + 2x3 = 162x1 + 4x2 + x3 = 8

xj (j = 1, 2, 3)Gii:

a bi ton v dng chnh tc v lp bi ton P.P(x, xg) = xg2 + xg3 min

2x1 + 5x2 + 3x3 + x4 = 104x1 - 3x2 + 2x3 + x

g2 =16

2x1 + 4x2 + x3 + xg

3 =8

xj 0(j = 1 4); xg2, xg3 0

HS CS PA 1 16 2 1 0


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24 6 1 3 0 0 0

x1 x2 x3 x4 xg

2 xg

3

0

11

x4

xg

2xg3

10

168

2 5 3 1 0 0

P

4 -3 2 0 1 02 4 1 0 0 1[ ]

x4

x

g

2x1

0

10 4 1 2 1/2 0 0

2 0 1 2 1 0

0 -11 0 0 10

0P 0 00 -11 0

[ ]

7/2 1 0 -1/4 0

0 1 1/210 0 0 1

0

0

dc

22f(x) -10 0

0 dx3xg2x1

106

0

06

f(x) 24 20 0 0

Xt bi ton QHTT:


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f(x) = x1 + 3x2 + 3x3 min5x1 + 3x2 + 6x3 8 (1)-x1 - 2x2 -2 (2)

x3 1/4 (3)3x1 + x2 + x3 4 (4)-x1 - x3 -1 (5)

xj 0 (j = 1, 2, 3)Bi ton ny nu gii trc tip bng ph ng php nhnh s rt di, v khi bi ton ph c 5 n gi. Chnh v

vy mi BT QHTT ta u thnh lp mt bton QHTT khctheo mt quy tc nht nh gi l Bi ton i ngu caBton cho v ta s i nghin cu mi q.h gia cp biton i ngu.

VI. Bi ton i ngu


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1- Cch thnh lp:a. Cp bi ton i ngu khng i xng:Xt bi ton dng chnh tc (I):

Bi toni ngu ca bi ton (I), k/h c dng sau:

n

1jjj minxcf(x)

n

1jijij m1ibxa

n1j0xj )I

~(

m

1iii

~

maxyb(y)f

m

1ijiij n1jcya

Nhn xt: ~


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Nu f(x) min th max v h rng buc cabi ton i ngu c dng .

Nu f(x) max th min v h rng buc cabi ton i ngu c dng . S rng buc (khng k rng buc du ca bi tonny) bng s bin s ca bi ton kia.

H s trong hm mc tiu ca bi ton ny l v phica h rng buc trong bi ton kia. Ma trn iu kin trong hai bi ton l chuyn v canhau.

(y)f

~

(y)f~

Cp rng buc i ngu:


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p g gTa gi 2 rng buc bt ng thc (k c rng buc du)

trong hai bi ton cng t ng ng vi mt ch sl mt

cp rng buc i ngu.Trong bi ton (I) v c n cp rng buc i ngu:

0xj m

1ijiij n1jcya

)I~(

V d 1: Vit bi ton i ngu ca bi ton sau v ch


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r cc cp rng buc i ngu:f(x)= -4x1 + 3x2 - 5x4 + x5 min

2x2

- x3

+ 4x4

- 2x5

+ 7x6= -12

-2x1+ 2x2 +5x3 + 3x5 = 33x1 - 3x2+ 5x3- x4 + x5 + 2x6 = 7

xj 0 (j = 1 6)Bi ton i ngu:

-2y2 + 3y3 -4 (1)2y1 + 2y2 - 3y3 3 (2)-y1 + 5y2 + 5y3 0 (3)

4y1 -y3 -5 (4)

-2y1 + 3y2 + y3 1 (5)7y1 + 2y3 0 (6)

max7y3y12y(y)f321

~

V d 2: Vit bi ton i ngu ca bi ton sau:


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f(x)= 5x2 + 4x3 - 2x4 + x5 max- 2x1 - 3x2 - x3 + 6x4 - 2x5 = -14

- x1 + 2x2 + 5x3 + 3x5 = 86x1 - 3x2 + 2x3 - x4 + x5 = 12xj 0 (j = 1 5)

Bi ton i ngu:

- 2y1 - y2 + 6y3 0 (1)- 3y1 + 2 y2 - 3y3 5 (2)- y1 + 5 y2 + 2y3 4 (3)6y1 - y3 -2 (4)

- 2y1 + 3 y2 + y3 1 (5)

min12y8y14y(y)f 321

~

Ch :


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Ch :i vi bi ton bt k th a v bi ton dng chnh

tc, xy dng bi ton i ngu ca bi ton ny v gin l bi ton i ngu ca bi ton cho.

b. Cp bi ton i ngu i xng- Xt bi ton sau gi lbi ton (II) n


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bi ton (II)

-a bi ton (II) v dng chnh tc, k hiu l (II)

Bi ton i ngu ca (II) v cng l i ngu ca (II)k hiu c dng:

n

1jjj minxcf(x)

n

1jijij m1ibxa n1j0xj

n

1jjj minxcf(x)

n

1jiinjij m1ibxxa

mn1j0xj

~

II

m

ii

~maxyb(y)f


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Hai bi ton (II) v c n+m cp rng buc i ngusau:

1i

m

1ijiij n1jcya

m1i0yi ~

II

n1jcya0xm

1ijiijj

m1i0ybxan

1jiijij

V d 3: Vit bi ton i ngu ca bi ton sau:


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V d 3: Vit bi ton i ngu ca bi ton sau:f(x)= -5x2 + 4x3 min- x1 - 3x2 - x3 -14- 2x1 + 5x2 + 5x3 86x1 - 3x2 + 2x3 12

xj 0 (j = 1 3)

c. Cp bi ton i ngu tng qut:


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p g g q

Ta c th s dng cc q.tc nu trongl- c tng qut

d- i y vit trc tip bi ton i ngu m khng cn- a bi ton v dng chnh tc.

Lc Tng QutBi ton gc Bi ton i ngu


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Bi ton gc Bi ton i ngu

n

1jjj minxcf(x)

n

1j1ijij Iibxa

n

1j2ijij Iibxa

2j Jj0x

xj khng c rng buc du jJ1

m

1iii

~

maxyb(y)f

2i Ii0y

m

1i1jiij Jjcya

m

1i2jiij Jjcya

yi khng c rng buc du iI1

3j Jj0x

m

1i3jiij Jjcya

n

1j3ijij Iibxa 3i Ii0y

Nhn xt:+ Nu mt bin s khng c rng buc du trong bi ton


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ny th rng buc t- ng ng trong bi ton kia c dubng v ng- c li.

+ Nu mt bin s c rng buc du trong bi ton ny thrng buc t- ng ng trong bi ton kia c du bt ngthc v ng- c li.

+ Chiu ca cc du bt ng thc ca bi ton i ngu- c quyt nh bi hm mc tiu phi t cc i hay cctiu.+ Nu max v bin s yic rng buc du th yiv rng buc tng ng cng chiu bt ng thc.

(y)f~

V d 4 : Vit bi ton i ngu ca bi ton sau v ch


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V d 4 : Vit bi ton i ngu ca bi ton sau v chr cc cp rng buc i ngu:f(x) = -4x1 + x2 +5x3 +3x5 min

3x1 -6 x2 - x3 +2x4 +4x5 -15 (1)-2x1 +3 x2 +4x3 -5x4 +x5 8 (2)

-6 x2 +3x3 +8x4 -4x5 = 9 (3)

3x1 +2 x2 -3x4 +x5 24 (4)x1 0 , x3 0, x5 0

V d 5 : Vit bi ton i ngu ca bi ton sau v ch


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gr cc cp rng buc i ngu:f(x) = -2x1 + x2 +8x4 max

3x1 -6 x2 - x3 +2x4 = -5 (1)x1 +5 x2 -5x4 3 (2)

-3 x2 +3x3 +8x4 = 2 (3)3x1 +2 x2 -3x4 24 (4)

x2 0 , x3 0, x4 0

2- Cc tnh cht v nh l i ngu


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Xt cp bi ton i ngu tng qut vi hm mc tiuf(x) min (max) v max(min).Tnh cht 1:Vi mi cp ph ng n x v y ca hai bi ton i nguta lun c:

f(x)

()Tnh cht 2:

Nu i vi hai ph ng n x* v y* ca 1 cp bi toni ngu m: th x* v y* t ng ng l 2 PAT.

(y)f~

(y*)ff(x*)~

(y)f~

nh l 1(i ngu):


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Nu mt trong hai bi ton i ngu gii c th biton kia cng gii c v khi mi cp PAT x* v y*

ta lun c:H qu 1:

iu kin cn v hai bi ton i ngu gii cl mi bi ton phi c t nht mt PA.

H qu 2:iu kin cn v mt bi ton c PA cn mt

bi ton khng c ph ng n l tr s hm mc tiu ca

bi ton c ph ng n khng b chn trn tp ph ngn ca n.

(y*)f*xf~

VD: Cho bi ton:


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f(x) = x1 + 8x2 + 10x3 minx1 + x2 + 4x3 = 2 (1)

x1 - x2 + 2x3 = 0 (2)(xj 0, j = 1 3)

Khng gii hy chng minh bi ton c pacb t .

Ch : T t/c2 v l 1 ta suy ra iu kin cn v ~


90/147

hai PA x v y ca mt cp BTN ti u l

nh l 2 (i ngu):

iu kin cn v hai PA x v y ca mt cp BTNti u l trong cc cp rng buc i ngu nu mt rngbuc tho mn lng th rng buc kia phi tho mn cht.H qu:

Nu mt rng buc l lng i vi mt PAT ca biton ny th rng buc i ngu ca n phi l cht ivi mi PAT ca bi ton kia.

(y)ff(x)

3- ng dng:a) Phn tch tnh cht ti u ca mt PA: Xt xem 1 PA xo


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a) Phn tch tnh cht ti u ca mt PA: Xt xem 1 PA xoca bton gc c phi PAT hay khng:Gi s xo l PAT, theo nh l 2 i ngu, mi PAT y caBTN phi tho mn cht cc rng buc i ngu vi ccrng buc m xo tho mn lng. Tp hp cc rng buc nyto thnh h p.trnh i vi y.Gii hpt ny Nu h VN th xo khng phi l PAT. Nu h c nghim th phi th nghim vo cc rng

buc cn li ca BTN.- Nu mi nghim u khng tho mn th xo khng phi

PAT.

- Nu c 1 nghim yo tho mn th xo l PAT, ng thi yocng l PAT ca BTN.

b) Xc nh tp PAT:


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- Nu xo l PAT ca bi ton gc, theo cch phn tchtrn ta xc nh c tp PAT ca BTN.

- T 1 PAT no ca BTN ta xc nh c tp PATca bi ton gc.

VD: Cho bi ton:


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f(x) = 3x1 + 9x2 - 2x3 + x4 - 4x5 min-x1 + 5x2 - 3x3 + x4 - 2x5 = - 6 (1)

3x1 - 4x2 +2x3 - x4 + x5 = 4 (2)4x1 - x3 +2x4 - 3x5 2 (3)(xj 0, j = 1 5)

v vc t xo = (2, 0, 0, 8, 6)

a) Vit bi ton i ngu.b) Phn tch cc tnh cht ca xo i vi bi ton cho.c) Xc nh tp ph ng n ti u v cc PACB ti u ca

hai bi ton.

Gii:

a) Bi ton i ngu:y + 3y + 4y 3 (1)

max2y4y6y(y)f 321~


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- y1 + 3y2 + 4y33 (1 )5y1 - 4y2 9 (2)

- 3y1 + 2 y2 - y3 -2 (3)y1 - y2 + 2y3 1 (4)- 2y1 + y2 - 3y3 - 4 (5)

y3 0

b) - Vt xo= (2, 0, 0, 8, 6) tho mn mi rng buc ca

bi l PA PA th ht b (1) (2)


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bi ton nn n l PA. PA xotho mn cht rb (1), (2)

v 2 rb du nn xo khng l PACB.

- Gi s xol PAT, theo nh l 2 (i ngu), mi PATy ca bi ton i ngu phi t/m:

- H pt trn c nghim duy nht yo

= (3, 2, 0). Th yo

vocc rb cn li ca BTN u t/m.

- Vy yol PAT ca BTN. Do xo, yol 2 PAT.

432

12 343

0

321

321

321

3

yyy

yyy yyy

y

42

133

0

21

21

21

3

yy

yy yy

y

0

2

3

3

2

1

y

y

y

Chng II: Phng Php Bng Cn i Lin Ngnh


96/147

Chng II:Phng Php Bng Cn i Lin NgnhI.M hnh bng cn i lin ngnh:1. Dng hin vt:

Khi nghin cu qu trnh ti sn xut x hi, phng phpbng cn i lin ngnh xem ton b nn KTQD l mt ththng nht bao gm n ngnh sn xut vt cht thun tu

khc nhau.Gia cc ngnh c mi quan h qua li mt thit thng qua

mt m hnh ton hc phn nh cc mt ca qu trnh ti snxut.

GiQ l tng sn lng ca ngnh i trong nm (i = 1 n)


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Qil tng sn lng ca ngnh i trong nm (i = 1n).- Mt phn phn phi cho cc ngnh khc di dng nguyn,nhin vt liu, t liu sn xut trong nm, k hiu qij (i, j=1n) Qi qij 0- Phn cn li l sn phm cui cng ca ngnh i, k hiu qi(i = 1n) dng tch lu, tiu dng cho nm sau (qi 0).Ta c cc phng trnh phn phi sn phm dng hin vt:

(1))n1,(iqqQ in

1jiji

Gi:Q l t l t b ki h t


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Qol tng n v lao ng sngca ton b nn kinh tquc dn s dng trong nm.

qoj l tng s n v lao ng s dng ngnh jqol tng s n v lao ng s dng trong cc ngnh phisn xut.

Ta c: Qo > 0, qoj > 0, qo > 0

Phng trnh phn phi lao ng dng hin vt l:

)2(qqQn

1j

00j0

Ghp (1) v (2) c bng cn i lin ngnhdng hin vt.


99/147

Ngnhsx

TngSL

n vtnh

Phn phi s dng cc ngnh

1 2 . . . n

SPcuicng

12...n

Q1Q2...Qn

KW/h1000T

...

1000m3

q11 q12 . . . q1nq21 q22 . . . q2n. . . . . .

. . . . . .

. . . . . .qn1 qn2 . . . qnn

q1q2.

.

.qn

Lao

ng Q0q0q01 q02 . . . q0n

Ngy

(ngi)

2. Dng gi tr:


100/147

Gi:

pi:gi tr mt n v sn phm ngnh i (tnh theo n v

quy c).p0:gi tr mt n v lao ng x hi.

Ta c:

-Tng gi tr sn lng trong nm ca ngnh i l:Xi = piQi (i = 1 n)


101/147

Xi piQi (i 1 n)- Gi trphn sn phm ca ngnh i cung cp cho ngnh j l

Xij = pi.qij (i, j = 1 n)- Gi tr sn phm cui cng ca ngnh i l:

xi = pi.qi (i = 1 n)- Tng gi tr lao ng sng ca ton x hi l:

X0 = p0.Q0

- Gi tr khi lng lao ng s dng trong ngnh sn xutth j l:X0j = q0j.p0

- Gi tr ca lao ng x hi ca cc ngnh phi sn xut vtcht l: x0 = p0.q0

Ta c cc phng trnh:n


102/147

(3)l cc phng trnh phn phi sn phm dng gi tr.

(4)l phng trnh phn phi lao ng dng gi tr.

Sau mi nm, mi ngnh sn xut vt cht u sngto thm mt phn gi tr mi cho x hi (gi l gi tr

ng gp cho x hi, k hiu mj).

Ta c bng cn i lin ngnh dng gi tr:

)3()n1,(ixXX1j

iiji

)4(xXX

n

1j

00j0

Ngnh

SXGtr Tng Phn phi s dng cc ngnh

Snphm


103/147

SX sn lng1 2 . . . npCuicng

12..

.n

X1X2..

.Xn

X11 X12 . . . X1nX21 X22 . . . X2n

Xn1 Xn2 . . . Xnn

x1x2..

.xn

Laong

X0 X01 X02 . . . X0nx0

Gi tr ng gpcho x hi

m1 m2 . . . mn Nm

II. Cc h s chi ph trc tip v ton b:

T n


104/147

Ta c:

t:

aij:gi l h s chi ph trc tip dng gi tr

Khi ta c:

hay

)n1,(ixXX i

n

1jiji

)n1,(ixX

X.XX in

1j j

jij

i

)n1,(jX

Xa

j

ij

ij )n1,j(i,1a0 ji

(*))n1,(i.XaXx n

1jjijii

)n1,(ixX.aX i

n

1jjiji

(*))n1,(i.XaXxn

1jjijii


105/147

Gi:

E:ma trn n v cp nA=[aij]nxn l ma trn h s chi ph trc tip

X v xl cc ma trn ct c cc thnh phn l Xi v xi(*) x = X AX

= (E-A)XX = (E-A)-1.xt: (E-A)-1 = B X = BxB = [b

ij]nxn

c gi l ma trn h s chi ph ton b.

1j

III. ng dng lp k hoch nm sau (dng A):

Mt t h h th t t k h h l


106/147

Mt tnh hung c th xy ra trong cng tc k hoch l

ngi ta d kin trc nhng sn phm cui cng ca

nm k hoch (chnh l nm sau, t+1).T mc xi(t) (t: nm trc), pht trin, m rng n mc

xi(t+1) nm d kin k hoch (xi(t+1)>xi(t))

Vic xy dng d n k hoch trong tnh hung ny gi

l lp k hoch cn i dng A.

lp d n k hoch dng A cho nm t +1 ta lm nh sau:B1:Tm A(t+1)


107/147

( )

T c thc hin k hoch nm t ta tnh aij(t) = Xij(t)/Xj(t)suy ra aij(t+1) v A(t+1).

B2:Tm B(t+1)Tnh E A(t+1) tnh [E A(t+1)] -1 = B(t+1)

B3:Tm X(t+1) = B(t+1).x(t+1)

s tnh c cc Xj(t+1)

B4:Tm Xij(t+1)Xij(t+1) = aij(t+1).Xj(t+1)

B5:Tm Xoj(t+1)

+ Tnh a0j(t) = X0j(t)/Xj(t) ri suy ra a0j(t+1)

+ Tnh Xoj(t+1) = aoj(t+1).Xj(t+1)+ Tnh

n

i 10jijjj XXXm

VD1: Cho

) th hi k h h t l


108/147

a) c thc hin k hoch nm t l:

b) aij(t+1) aij(t); a0j(t+1) a0j(t);c) x1(t+1) = 40; x2(t+1) = 60

Hy tm d n k hoch nm t+1 cn i dng A

Ngnh Xi Xij xi

1

2

80

100

16 32

15 40

32

45

X0 10 20 Nm

tmj 39 8

Gii:


109/147

B1: Tm A(t+1):

Tm aij(t) = X

ij(t)/X

j(t)

a11 (t) = X11/X1 = 16/80

a12 (t) = X12/X2 = 32/100

a21 (t) = X21/X1 = 15/80

a22 (t) = X22/X2 = 40/100

A(t+1) A(t) =

4,01875,0

32,02,0

= 0,2

= 0,1875

= 0,32

= 0,4

B2: Tm B(t+1)

[E A( )] C 01 32020

32,08,0


110/147

[E A(t+1)] = C=

[C]-1 = B(t+1) =

B3: Tm X(t+1) = B(t+1).x(t+1)

10

01

4,01875,0

32,02,0

6,01875,0

,,

8,01875,0

32,06,0.42,0

1

9048,14464,0

7619,04286,1

6040.

9048,14464,07619,04286,1

144,132858,102

B4: Tm Xij(t+1) = aij(t+1).Xj(t+1)

X (t+1) = a (t+1) X (t+1)


111/147

X11(t+1) = a11(t+1).X1(t+1)

= 0,2 x 102,858

= 20,5716X12(t+1) = a12(t+1).X2(t+1)

= 0,32 x 132,144

= 42,2861

X21(t+1) = a21(t+1).X1(t+1)

= 0,1875 x 102,858

= 19,2859

X22(t+1) = a22(t+1).X2(t+1)= 0,4 x 132,144

= 52,8576

B5: Tm X0j(t+1)

a (t) = X /X = 10/80 = 0 125


112/147

a01(t) = X01/X1 = 10/80 = 0,125

a02(t) = X02/X2 = 20/100 = 0,2

a01(t+1) a01(t) ; a02(t+1) a02(t)X0j(t+1) = a0j(t+1). Xj(t+1)

X01(t+1) = a01(t+1). X1(t+1)

= 0,125 x 102,858 = 12,8573

X02(t+1) = a02(t+1). X2(t+1)

= 0,2 x 132,144 = 26,4288

Bng cn i lin ngnh dng gi tr nm t+1

N h X X


113/147

Ngnh Xi Xij xi1

2

102.858

132,144

20,5716 42,2861

19,2859 52,8576

40

60

X0 12,8573 26,4288 Nmt+1

mj

m1 = X1 (X11+X21) X01= 102,858 20,5716 - 19,2859- 12,8573= 50,1432

m2 = X2 (X12+X22) X02= 132,144 42,2861 - 52,8576 - 26,4288= 10,5715

50,1432 10,5715

VD 2: Cho bita) c thc hin k hoch nm t th hin bng cn


114/147

i lin ngnh dng gi tr sau:(n v tnh: 1000 triu ng)

b) aij(t+1) aij(t); a0j(t+1) a0j(t); (i,j = 1, 2, 3)c) x(t+1) = (500, 300, 400)Hy hon thin bng v lp d n k hoch nm t+1 cn idng A

Ngnh Xi Xij xi

123

397,8197,9300,5

159,7 40,1 87,816,5

80,1 38,7

110,2118,9151,3

X0 73,7 39,6 Nm t

mj 58,6 130,2

H h thi b l d k h h t+1


115/147

Hy hon thin bng v lp d n k hoch nm t+1 cn

i dng A

Gii:Hon thin bng

Ngnh Xi Xij xi1 397 8 159 7 40 1 87 8 110 2


116/147

1

2

3

397,8

197,9

300,5

159,7 40,1 87,8

16,5

80,1 38,7

110,2

118,9

151,3

X0 73,7 39,6 Nm t

mj 58,6 130,2

9,206,41

4.30

6,35

7,42

B1: Tm A(t+1):

(t) X (t)/X (t)


117/147

aij(t) = Xij(t)/Xj(t)

a11

= X11

/X1

= 159,7/397,8

a12 = X12/X2= 40,1/197,9

a13 = X13/X3= 87,8/300,5

a21 = X21/X1= 41,6/397,8

a22 = X22/X2= 20,9/197,9

a23 = X23/X3= 16,5/300,5

a31 = X31/X1= 80,1/397,8

a32 = X32/X2= 38,7/197,9

a33 = X33/X3= 30,4/300,5

4015,0

0549,0

2922,0

1956,01012,0

1056,0

1046,0

2014,0

2026,0

A(t+1) A(t)=

054901056010460

2922,02026,04015,0


118/147

A(t+1) A(t)=B2: Tm B(t+1)

[E A(t+1)] = C =

C = 0,5985.0,8944.0,8988+(-0,2026).(-0,0549).(-0,2014)++(-0,1046)(-0,1956)(-0,2922)-(-0,2014).0,8944.(-0,2922)

-(-0,1046).(-0,2026).0,8988-(-0,1956).(-0,0549).0,5985

= 0,3948

1012,01956,02014,0

0549,01056,01046,0

8988,01956,02014,0

0549,08944,01046,0

2922,02026,05985,0

0 89880 1956

0,05490,8944C11

= 0,7931


119/147

0,89880,1956

0,89880,2014

0,05490,1046-C12

0,89880,2014

0,29220,5985C22

0,89880,1956

0,29220,2026-C21

0,1956-0,2014

0,89440,1046-C13

= 0,1051

= 0,2006

= 0,2393

= 0,4791

0,20260,5985C23

= 0 1579


120/147

0,0549-0,8944

0,29220,2026-C31

0,1956-0,2014C23

0,89440,1046

0,20260,5985C33

0,0549-0,1046

0,29220,5985

C32

= 0,1579

= 0,2725

= 0,0634

= 0,5141

C11 = 0,7931; C21 = 0,2393 ; C31 = 0,2725C 0 1051 C 0 4791 C 0 0634


121/147

C -1 =B(t+1) =

5141,01579,02006,00634,04791,01051,02725,02393,07931,0

3948,0

1

3022,13999,05081,01606,02135,12662,06902,06061,00089,2

C12 = 0,1051; C22 = 0,4791 ; C32 = 0,0634C13 = 0,2006; C23 = 0,1579 ; C33 = 0,5141


122/147

9,89439,561

36,1462

B3: X(t+1) = B(t+1). x(t+1)

400300

500

.3022,13999,05081,01606,02135,12662,0

6902,06061,00089,2

B4: Tm Xij(t+1)

X (t+1) 0 4015 x 1462 36 587 1375


123/147

X11(t+1) = 0,4015 x 1462,36 =

X12

(t+1) = 0,2026 x 561,39 =

X13(t+1) = 0,2922 x 894,9 =

X21(t+1) = 0,1046 x 1462,36 =

X22(t+1) = 0,1056 x 561,39 =

X23(t+1) = 0,0549 x 894,9 =

X31(t+1) = 0,2014 x 1462,36 =

X32(t+1) = 0,1956 x 561,39 =

X33(t+1) = 0,1012 x 894,9 =

587,1375

113,7376

261,4898

152,9629

59,2828

49,13

294,5193

109,8079

90,5639

B5: Tm X0j(t+1)

a01(t+1) a01(t) =X01/X1= 73 7/397 8 = 0 1853


124/147

a01(t+1) a01(t) =X01/X1= 73,7/397,8 = 0,1853a02(t+1) a02(t) =X02/X2= 39,6/197,9 = 0,2001a03(t+1) a03(t) =X03/X3= 35,6/300,5 = 0,1185X01(t+1) = a01(t+1). X1(t+1)

= 0,1853 x 1462,36 = 270,9753X02(t+1) = a02(t+1). X2(t+1)

= 0,2001 x 561,39 = 112,3341

X03(t+1) = a03(t+1). X3(t+1)

= 0,1185 x 894,9 = 106,0457

Ngnh Xi Xij xi

1 1462 36 587 1375 113 7376 261 4898 500


125/147

1

2

3

1462,36

561,39

894,9

587,1375 113,7376 261,4898

152,9629 59,2828 49,13

294,5193 109,8079 90,5639

500

300

400

X0 270,9753 112,3341 106,0457 Nm

t+1mj 156,765 166,2276 387,6706

m1 = X1 (X11+X21+X31) X01= 1462,36 - 587,1375 - 152,9629 - 294,5193 - 270,9753

= 156,765

m2 = 166,2276m3 = 387,6706

Bi 1:Cho bita) c thc hin k hoch nm t th hin bng cni lin ngnh dng gi tr sau:


126/147


b) aij(t+1) aij(t); a0j(t+1) a0j(t); (i,j = 1, 2, 3)c) x(t+1) = (120, 150, 100)Hy lp d n k hoch nm t+1 cn i dng A

Ngnh Xi Xij xi

123

750,15890,2

350,18 268,47 300,89210,5 148,34

215,51 210,1

100,81130,9674,14

X0 50,37 75,63 Nm t

mj 18,85 26,25 43,23


127/147

Ngnh Xi Xij xi123

1020,35750,15890,2

350,18 268,47 300,89210,5 148,34 260,35390,45 215,51 210,1

100,81130,9674,14

X0 50,37 91,58 75,63 Nm t

mj 18,85 26,25 43,23

B1: Tm A(t+1):


128/147

B1: Tm A(t+1):

a11 = 350,18 / 1020,35

a12 = 268,47 / 750,15

a13= 300,89 / 890,2

a21= 210,5 / 1020,35

a22= 148,34 / 750,15

a23= 260,35 / 890,2

a31= 390,45 / 1020,35

a32= 215,51 / 750,15

a33= 210,1 / 890,2

3432,0

2925,0

338,0

2873,0236,0

1977,0

2063,0

3827,0

3579,0

A(t+1) A(t)=

2925,01977,02063,0338,03579,03432,0


129/147

( ) ( )

B2: Tm B(t+1)

[E A(t+1)] = C =

C = 0,6568.0,8023.0,764 + (-0,3579).(-0,2925).(-0,3827) ++ (-0,2063)(-0,2873)(-0,338) - (-0,3827).0,8023.(-0,338)

- (-0,2063).(-0,3579).0,764 - (-0,2873).(-0,2925).0,6568

= 0,1271

236,02873,03827,0

2925,01977,02063,0

764,02873,03827,02925,08023,02063,0338,03579,06568,0

0,7640,28730,29250,8023

C11

= 0,5289


130/147

,,

0,7640,3827 0,29250,2063-C12

0,7640,38270,3380,6568

C22

0,7640,28730,3380,3579-

C21

0,2873-0,38270,80230,2063-

C13

= 0,2696

= 0,3663

= 0,3705

= 0,3724

0 28730 38270,35790,6568

C23

= 0,3257


131/147

0,2925-0,80230,3380,3579-

C31

0,2873-0,382723

0,80230,20630,35790,6568

C33

0,2925-0,2063

0,3380,6568

C32

,

= 0,3759

= 0,2618

= 0,4531

375903705052890


132/147

B(t+1) = C-1

=

4531,03257,03663,0 2618,03724,02696,0

3759,03705,05289,0

1271,0

1

5649,35625,2882,20598,293,21212,2

9575,2915,21613,4


133/147

705,1086024,900356,1232

B3: X(t+1) = B(t+1). x(t+1)

100150120

.5649,35625,2882,20598,293,21212,29575,2915,21613,4


X11(t+1) = 0 3432 x 1232 356 = 422,9446


134/147

X11(t+1) = 0,3432 x 1232,356 =

X12(t+1) = 0,3579 x 900,024 =

X13(t+1) = 0,338 x 1086,705 =

X21(t+1) = 0,2063 x 1232,356 =

X22(t+1) = 0,1977 x 900,024 =

X23(t+1) = 0,2925 x 1086,705 =

X31(t+1) = 0,3827 x 1232,356 =

X32(t+1) = 0,2873 x 900,024 =

X33(t+1) = 0,236 x 1086,705 =

422,9446

322,1186

367,3063

254,235

177,9347

317,8612

471,6226

258,5769

256,4624

B5: Tm X0j(t+1)

a01(t+1) a01(t) = 50,37/1020,35 = 0,0494


135/147

01( ) 01( ) , , ,

a02(t+1) a02(t) = 91,58/750,15 = 0,1221a03(t+1) a03(t) = 75,63/890,2 = 0,085X01(t+1) = 0,0494 x 1232,356 = 60,8784

X02(t+1) = 0,1221 x 900,024 = 109,8929X03(t+1) = 0,085 x 1086,705 = 92,3699

Bng CLN nm t+1

Ngnh Xi Xij xi

1 1232,356 422,9446 322,1186 367,3063 120


136/147

1

2

3

1232,356

900,024

1086,705

422,9446 322,1186 367,3063

254,235 177,9347 317,8612

471,6226 258,5769 256,4624

120

150

100

X0 60,8784 109,8929 92,3699 Nm

t+1mj 22,6754 31,5009 52,7052

m1 = 22,6754

m2 = 31,5009

m3 = 52,7052

Bi 2:Cho bita) c thc hin k hoch nm t th hin bng cni li h d i t


137/147


b) aij(t+1) aij(t); a0j(t+1) a0j(t); (i,j = 1, 2, 3)c) x(t+1) = (96,14; 165,37; 195,85)

Hy lp d n k hoch nm t+1 cn i dng A

Ngnh Xi Xij xi

123

648,36820,57973,15

150,24 246,54 203,12230,17 193,82 293,18206,03 346,19 297,84

48,46103,4

123,09

X0 40,26 21,07 113,21 Nm t

mj 21,66 12,95 65,8

B1: Tm A(t+1):

a11 = 150,24/648,36 23170


138/147

a11 = 150,24/648,36

a12 = 246,54/820,57

a13= 203,12/973,15

a21= 230,17 /648,36

a22= 193,82 /820,57

a23= 293,18 /973,15

a31= 206,03 /648,36

a32= 346,19 /820,57

a33= 297,84 /973,15

2317,0

3013,0

2087,0

4219,0

3061,0

2362,0

355,0

3178,0

3004,0

A(t+1) A(t)=

3013,02362,0355,0

2087,03004,02317,0


139/147

B2: Tm B(t+1)

[E A(t+1)] = C =

C= 0,1249

3061,04219,03178,0

6939,04219,03178,0

3013,07638,0355,0

2087,03004,07683,0

0,69390,4219

0,30130,7638C11

= 0,4029


140/147

0,69390,3178

0,30130,355-

C12

0,69390,3178

0,20870,7683

C22

0,69390,4219

0,20870,3004-C21

0,4219-0,3178

0,76380,355-C13

= 0,3421

= 0,3925

= 0,2965

= 0,4668

0 4219-0 3178

0,30040,7683C23

= 0,4196


141/147

0,3013-0,76380,20870,3004-C31

0,4219-0,3178

0,76380,355

0,30040,7683C33

0,3013-0,355

0,20870,7683C

32

= 0,2499

= 0,3056

= 0,4802

2499,02965,04029,0

1


142/147

B(t+1) = C-1 =

4802,04196,03925,0

3056,04668,03421,01249,0

1

8447,33595,31425,3

4468,27374,3739,2

0008,23739,22258,3

B3: X(t+1) = B(t+1). x(t+1)


143/147

665,1610

5871,1360

5569,1094

85,195

37,16514,96

.

8447,33595,31425,3

4468,27374,3739,20008,23739,22258,3


X11(t+1) = 0,2317 x 1094,5569 = 253,6088


144/147

11( ) , ,

X12(t+1) = 0,3004 x 1360,5871 =

X13(t+1) = 0,2087 x 1610,665 =

X21(t+1) = 0,355 x 1094,5569 =

X22(t+1) = 0,2362 x 1360,5871 =

X23(t+1) = 0,3013 x 1610,665 =

X31(t+1) = 0,3178 x 1094,5569 =

X32(t+1) = 0,4219 x 1360,5871 =

X33(t+1) = 0,3061 x 1610,665 =

408,7204

336,1458

388,5677

321,3707

485,2934

347,8502

574,0317

493,0246

B5: Tm X0j(t+1)

a01(t+1) a01(t) = 40,26/648,36 = 0,0621


145/147

a02(t+1) a02(t) = 21,07/820,57 = 0,0257a03(t+1) a03(t) = 113,21/973,15 = 0,1163X01(t+1) = 0,0621 x 1094,5569 = 67,972

X02(t+1) = 0,0257 x 1360,5871 = 34,9671X03(t+1) = 0,1163 x 1610,665 = 187,3203

Bng CLN nm t+1

Ng Xi Xij xi

1 1094,5569 253,6088 408,7204 336,1458 96,14


146/147

2

3

1360,5871

1610,665

388,5677 321,3707 485,2934

347,8502 574,0317 493,0246

165,3

195,85

X0 67,972 34,9671 187,3203 Nm

t+1mj 36,5582 21,4972 108,8809

m1 = 36,5582

m2 = 21,4972

m3 = 108,8809

Hs

Cs

Phngn x1 x2 xr xm xm+1 xs xn

c1 c2 cr cm cm+1 cs cn


147/147

f(x0)

s s

x01x01x0rx0m

x1x2xrx

m

c1c2crc

m

1 0 0 0 x1,m+1 x1s x1n0 1 0 0 x2,m+1 x2s x2n 0 0 1 0 xr,m+1 [xrs] xrn 0 0 0 1 xm,m+1 xms xmn

f(x) 0 0 0 0 m+1 s n

Copy of Mô hình toan kinh tế -1

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