PHƯƠNG PHÁP ĐỊNH LƯỢNG 1

7/29/2019 PHNG PHP NH LNG 1

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PHNG PHP NH LNG

Lp C09

Hc k 1 Nm hc 2010-2011


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Ging vin

Name: TRN PHC HUY

Email: [email protected]: 0903 715335

Add: Vn phng Khoa QTKD
mailto:[email protected]:[email protected]


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Gii thiu mn hc

Tn: Phng php nh lng

Tn ch: 3(2,1,6)

Mn hc trc:+ Xc sut thng k+ Qun tr hc


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Mc tiu

Kin thc:Cc phng php tm li gii cho biton trong qun l

K nng:T duy, phn tch, ra quyt nh, phthin & gii quyt vn , tm kim & la

chn kin thc vo mc ch ring bit Yu thch ngnh hc


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Tm tt

Cung cp 1 s phng php c bn quyhoch tuyn tnh, bi ton ra quyt nh,lp lch trnh d n

Trang b cc m hnh ton c ngdng vo nhng tnh hung kinh doanh,sn xut, ti chnh, tip th

Vn dng my tnh gii quyt vn


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Ni dung chi tit

Chng 1: Gii thiu v PPL

Chng 2: C s ca l thuyt ra quyt nh

Chng 3:Cy quyt nh v l thuyt li ch

Chng 4: Quy hoch tuyn tnh

Chng 5:Cc bi ton c bit lin quan n QHTT

Chng 6:Bi ton s mng trong qun l d n


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Ti liu

Bt buc:Hunh Trung Lng & Trng Tn Hin cPP nh lng trong qun l & vn hnh

NXB Khoa hc & K thut 2003 Tham kho:

ng Hn

Quy hoch tuyn tnhTrng HKT TP H Ch Minh - 1995


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T chc dy hc

Lch trnh chung

Ni dungHnh thc t chc

TngLT BT TL TH T hc

Chng 1 3 0.5 07 10.5Chng 2 6 2.0 0.5 0.5 18 27.0

Chng 3 6 1.5 0.5 1.0 19 27.0

Chng 4 6 2.0 1.0 0.5 19 28.5

Chng 5 3 1.0 0.5 09 13.5

Chng 6 6 2.0 1.0 0.5 19 28.5

Tng 9.0 3.0 3.0 90 135.0


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T chc dy hc

Lch trnh c th Yu cu t chc

+ Hc l thuyt tp trung phng c my chiu, micro

+ Bi tp, tho lun, thc hnh t chc thnh nhmkhng qu 30 sinh vin

Yu cu sinh vin+ Tham d y theo quy nh ca Trng

+c trc ti liu+ Tch cc tham gia hc tp+ Lm bi tp & np bi ng hn


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nh gi

Ni dungT lim

Thi gianThi lng

(pht)

Hot ng trn lp 15% Ton b thigian

Bi tp nh 15% Theo yu cuca GV

Kim tra gia k 20% Tun 8 60-90

Thi cui k 50% Theo lch caTrng

90 pht


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PP nh lng Cc khi nim

Cuc cch mng qun tr c tnh khoa hc u nm 1900, khi xng bi FredericW.Taylor. Nhng nghin cu khoa hc qun tr

hin i bt u trong thi k th chin th 2 Nhng thnh tu nh hng n phng php

nh lng:+ Phng php n hnh gii cc bi ton Quy hoch

tuyn tnh ca George Dantzig - nm 1947+ S bng n ca my tnh


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PP nh lng Cc khi nim

Cch tip cn nh lng ra quyt nhc nhiu tn gi: Khoa hc qun tr,Vntr hc v Khoa hc quyt nh

Ti liu c thm:+ Khi nim chung & ngun gc - Gio trnh (tr.2)

+ S lc v cc PPL trong nghin cu th trng -

www.viet-studies.info/kinhte/quantitative_%20methods%20_ttphuong.pdf
http://www.viet-studies.info/kinhte/%0Bquantitative_%20methods%20_ttphuong.pdfhttp://localhost/var/www/apps/conversion/tmp/scratch_1/S%C6%A1%20l%C6%B0%E1%BB%A3c%20v%E1%BB%81%20c%C3%A1c%20PP%C4%90L%20trong%20nghi%C3%AAn%20c%E1%BB%A9u%20th%E1%BB%8B%20tr%C6%B0%E1%BB%9Dng_TTPhuong.pdfhttp://localhost/var/www/apps/conversion/tmp/scratch_1/S%C6%A1%20l%C6%B0%E1%BB%A3c%20v%E1%BB%81%20c%C3%A1c%20PP%C4%90L%20trong%20nghi%C3%AAn%20c%E1%BB%A9u%20th%E1%BB%8B%20tr%C6%B0%E1%BB%9Dng_TTPhuong.pdfhttp://www.viet-studies.info/kinhte/%0Bquantitative_%20methods%20_ttphuong.pdfhttp://www.viet-studies.info/kinhte/%0Bquantitative_%20methods%20_ttphuong.pdfhttp://www.viet-studies.info/kinhte/%0Bquantitative_%20methods%20_ttphuong.pdfhttp://www.viet-studies.info/kinhte/%0Bquantitative_%20methods%20_ttphuong.pdf


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PP nh lngCc bc chnh trong phn tch nh lng

Gii quyt vn l qu trnhnhn dngskhc nhau gia trng thi thc t vi trng thimong mun ca cc cng vic v thc hingii quyts khc nhau

Cc bc chnh+ Xc nh vn + Pht trin m hnh+ Thu thp d liu u vo

+ Tm kim gii php (Gii m hnh)+ Kim nghim gii php+ Phn tch kt qu & phn tch nhy+ ng dng kt qu nhn c


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PP nh lng

Tin ch ca vic m hnh ha

Phn loi m hnh+ Xc nh

+ Xc sut

Cc kh khn thng gp trong vic pdng phng php phn tch nh lngGio trnh (tr.6-7)


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L thuyt ra Q Cc khi nim

L thuyt nh lng trong qun l datrn quan im:Qun tr l quyt nh v mun vic qun

tr c hiu qu th cc quyt nh phing n"

L thuyt v ra quyt nh l phng

php tip cn gii tchv c h thng


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L thuyt ra Q Cc bc

Xc nh vn cn phi ra quyt nh Tm tt c cc phng n Xc nh cc tnh hung kt qu xy ra Tnh ton li ch/thit hi ng vi tng

tnh hung kt qu (nh gi cc PA) Xc nh mi trng ra quyt nh & la

chn 1 m hnh ra quyt nhph hpp dng m hnh & ra quyt nh sau

cng


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L thuyt ra QV d 1Bi ton Thompson

B1. Vn : c nn sn xut mt snphm mi kinh doanh

B2. C tt c 3 phng n thc hin:+ Xy nh my ln sn xut+ Xy nh my nh sn xut+ Khng sn xut

B3. Cc tnh hung c th c:+ Th trng thun li (ha hn)+ Th trng bt li (khng ha hn)


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L thuyt ra QV d 1Bi ton Thompson

B4. c tnh cc li ch (chi ph) pht sinh khi chn ccphng n ng vi mi tnh hung

Bng ra quyt nh:

Phng n Cc trang thi t nhinTh trng tt Th trng xu

Nh my ln 200,000 -180,000

Nh my nh 100,000 -20,000

Khng SX 0 0

B5 v B6: Xc nh m hnh ton gii, tm li gii, RQ


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L thuyt ra QV d 2Bi ton v c hi vic lm

B1. Gi s c ngi ang tht nghip vmong mun c vic lm va

B2. C 4 phng n v ni lm vic+ H Ni

+ Nng

+ TP H Ch Minh

+ Cn Th


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B3. Cc tnh hung c th c:+ Ch c mt tiu chun: tin lng, th phng n la

chn tt nht s l lng khi im cao nht.+ C 3 tiu chun: lng khi im, tim nng thng

tin v v th ngh nghip.

Nhng vn m vic tm li gii tt nht ch lu n mttiu chun gi l nhng vn ra quyt nh mt tiu

chun (single-criterion decision problems).Nhng vn gm nhiu hn mt tiu chun la chngi l vn ra quyt nh nhiu tiu chun(multicriteria decision problem).


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B4. nh gi cc phng n

Phng nCc trng thi t nhin

Lng khi im(1000ng) Tim nngthng tin V th nghnghip

H Ni 1200 Kh Trung bnh

Nng 800 Rt tt Tt

Si Gn 1000 Trung bnh Tt

Cn Th 700 Rt tt Tt


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B5. Chn phng n:+ Kh khn trong la chn chnh l tm quan

trng ca cc phng n khng nh nhau v

khng c phng n l no tt nht vi mitiu chun.

+nh gi cn thn d liu Bng trn,

chng ta quyt nh chn phng n 3.Phng n 3 c gi l mt quyt nh


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L thuyt ra Q Cc bc

Xc nh vn

Tm tt c cc phng n

Xc nh cc tnh hung kt qu

nh gi cc phng n

Chn phng n (Quyt nh)

Thc hin phng n

& nh gi kt qu

Gii

quytvn

Raquyt

nh


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L thuyt ra Q Cc giai on

Cu trc vn v (bc 1, 2, 3)+ Xc nh vn + Tm tt c cc phng n

+ Xc nh cc tnh hung kt qu Phn tch vn (bc 4, 5)

+nh gi cc phng n Phn tch nh tnh

Phn tch nh lng Tm lt

+ Chn phng n


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L thuyt ra QNghin cu nh lng v nh tnh

Ra quyt nh l nhim v quan trng ca nh quntr. Kinh nghim, kh nng xt on, c sng tocha th m bo c c nhng quyt nh phhp & ti u nu thiu kh nng nh lng.


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L thuyt ra QNghin cu nh lng & nh tnh

NCT khng s dng cng c o lng, tnhton. NCT tm bit nhng c im, tnh chtca i tng nghin cu (TNC) cng nhnhng yu t nh hng n suy ngh, hnh vica TNC.

NCL s dng nhng cng c o lng, tnhton vi nhng con s c th.

NCL i tm tr li cho cu hi bao nhiu, mcno (how many, how much)NCT i tm tr li cho cu hi Ci g (what),nh th no (how), ti sao (why).


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L thuyt ra QSo snh gia NCL & NCT

Nghin cu nh tnh Nghin cu nh lngDng m t, khm ph, thm d Dng khng nh,suy rng& d bo

Ch tiu, i tng NC, mc nghincu c th cha r rng

Ch tiu, i tng NC, mc nghincu r rng

Linh ng trong hng nghin cu,khm ph hng nghin cu cha bit

Yu cu phi o lng

Ngi nghin cu l cng c thu thpthng tin

Ngi nghin cu s dng cc cngc (bn cu hi) thu thp thng tin

Ngi nghin cu bit s b nhngiu m h mun nghin cu

Ngi nghin cu bit r rng nhngiu m h mun nghin cu

Ch quan: kin ca c nhn l quantrng, vd: quan st, phng vn su

Khch quan: o lng v phn tchqua iu tra

Quy np gi thuyt Kim tra gi thuyt

Kh khi qut ha Khi qut ha

T ng, hnh nh Con s, thng k


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L thuyt ra QTi sao phi phn tch nh lng?

Vn phc tp

Vn quan trng c bit m nh qun trmun phn tch trc khi a ra quyt nh;

Vn mi m nh qun tr khng c kinhnghim

Nhng vn c c trng lp i lp li, nh

qun tr mun tit kim thi gian v ngun lcbng vic da vo th tc nh lng quytnh hng ngy


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L thuyt ra QMc tiu ca nghin cu nh lng

Khng nh, suy rng v d bo nhn dng vn Kim nh mt l thuyt hay mt gi thit,o lng cc con s, v phn tch bng

cc k thut thng k Lp k hoch sn xut

tnh ton la chn phng n ti u(Quyt nh u t, la chn cc phngn qui hoch)


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L thuyt ra QMi trng ra Quyt nh

Xc nh+ Bit chc chn kt qu tng ng vi tng PA+ Thng tin thu thp c xem l hon ho

Ri ro+ Khng bit chc chn kt qu, bit xc sut xy raca tnh hung

+ Thng tin thu thp c xem l khng y

Khng xc nh+ Khng bit bt c thng tin no+ Thng tin thu thp c xem l khng hon ho


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L thuyt ra QM hnh ra Quyt nh

+ i -phng n hng i trong bng RQ

+j - trng thi ct j trong bng RQ

+ PO(i,j) - li ch c c nu chn

phng n i v trng thi j xy ra


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RQ trong mi trng ri ro+ M hnh Max EMV:

Tiu chun ra quyt nh da trn gi tr k vng

ti a EMV(i) =

EMV(i)gi tr k vng ca phng n th i

ns lng cc trng thi PO(i,j)li ch c c phng n i & trng thi th j

P(j)xc sut xy ra ca trng thi t nhin j

Phng n c chn tng ng phng n cEMV(i) = maxi {EMV(i)}

n

j

jPjiPO1

)(*),(


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V d1: Bi ton Thompson

Gi s mi trng thi c xc sut xy ra 0.5{th trng tt, th trng xu}

Gi tr k vng

> EMV(1) = 0.5*200000 + 0.5*(-180000) = 10000

> EMV(2) = 0.5*100000 + 0.5*(- 20000) = 40000

> EMV(3) = 0.5*0 + 0.5*0 = 0

Phng n 2 l phng n tt nht c chn


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V d 2: Bi ton bo tr my

Cc my ging nhau cn phi phc hi sau khong thigian (nm) vn hnh. Trong khong thi gian , my cth c sa cha nu h hng. Sau bao nhiu nm thphc hi my chi ph bo tr l thp nht. Bit rng xcsut h hng mi nm ca tng my:

Nm (t)Xc sut

1 2 3 4 5 6

P(t) 5% 7% 10% 13% 18%


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RQ trong mi trngri ro+ M hnh Max EMV:


Gi s> C 50 my (n)

> Thi gian tt nht phc hi my l T

> Chi ph sa cha trung bnh ca mi my $100 (c1)

> Chi ph phc hi trung bnh ca mi my $200 (c2) Gi tr k vng ca chi ph bo tr trung bnh hng nm

EC(t)


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Chi ph phc hi sau T nm: c2* n

Chi ph sa cha .cho ton b n my: c1* n

Chi ph sa cha.cho n my nm th t: c1* n* P(t)Chi ph sa cha.trong T nm:

Gi tr k vng ca chi ph bo trtrung bnh trong T nm (EC):

T

nctPncT

t

*2))(**1(1

1

1

1

)(**1T

t

tPnc


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Gi tr k vng ca chi ph bo tr trong T nm (EC):t P(t) EC(t)

1 5% 500

2 7% 375

3 10% 3674 13% 400

5 18% 450

Phng n phc hi theo chu k 3 nm l tt nht


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RQ trong mi trng ri ro+ Gi tr ca thng tin hon ho:

Bit chnh xc v s p ng ca th trng

Gi tr k vng ca thng tin hon ho c nghaquyt nh

Gi tr k vng vi thng tin hon ho: EVWPI =

BO(j) = maxi {PO(i,j)}gi tr kt qu tt nht ca trng thi j P(j)xc sut xy ra ca trng thi j

Gi tr k vng ca thng tin hon ho: EVPI = EVWPI maxi {EMV(i)}

J

j

jPjBO1

)(*)(


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RQ trong mi trngri ro+ Gi tr ca thng tin hon ho:

V d bi ton Thompson

Gi tr kt qu tt nht ca tng trang thi:> Th trng tt: BO(1) = 200000> Th trng xu: BO(2) = 0

EVWPI = 200000*0.5 + 0*0.5 = 100000 EVPI = 100000 40000 = 60000

y cng chnh l gi ti a chp nhn cho chi ph thngtin hon ho


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RQ trong mi trngri ro+ M hnh EOL: Gi tr ca s mt mt c hi

Tiu chun ra quyt nh da trn gi tr k vngti thiu

Cc bc: Lp bng mt mt c hi: PO(i,j) = BO(j) PO(i,j) Xc nh gi tr k vng (EOL) Phng n c chn tng ng phng n c

EOL(i)= mini {EOL(i)}


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RQ trong mi trngri ro+ M hnh EOL: Gi tr ca s mt mt c hi

V d bi ton Thompson Bng mt mt c hi: PO(i,j) = BO(j) PO(i,j)

BO(1) = maxi {PO(i,j)} = 200000BO(2) = maxi {PO(i,j)} = 0

Phng nCc trang thi t nhin

Th trng tt Th trng xu

Nh my ln 0 180,000

Nh my nh 100,000 20,000

Khng SX 200,000 0


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RQ trong mitrngri ro+ M hnh EOL: Gi tr ca s mt mt c hi

V d bi ton Thompson Gi tr k vng

> EOL(1) = 0*0.5 + 180000*0.5 = 90000

> EOL(2) = 100000*0.5 + 20000*0.5 = 60000

> EOL(3) = 200000*0.5 + 0*0.5 = 100000

Phng n 2 c chn

Nhn xt: EVWPI = EMV(i) + EOL(i) EVPI = Mini {EOL(i)} i vi bi ton chi ph, cc ch Max s thnh Min


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RQ trong mitrngri ro+ Phn tch nhy

Mt yu t thay i trong 1 phm vi kh rng mkhng lm thay i phng n la chn th biton c xem l khng nhy vi yu t .

V d bi ton Thompson Gi s xc sut ca mi trng tt l Pth mi trng

xu la 1-P

Ta c:EMV(1) = 380000*P - 180000EMV(2) = 120000*P - 20000EMV(3) = 0


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RQ trong mitrngri ro+ Phn tch nhy V d bi ton Thompson

th phn tch nhy

-200,000

-100,000

0

100,000

200,000

300,000

0 0.2 0.4 0.6 0.8 1

P

E

MV

EMV(1)

EMV(2)

EMV(3)


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RQ trong mitrngri ro+ Phn tch nhy V d bi ton Thompson

Giao im ca th

EMV(1) v EMV(2) ti P = 0.6154EMV(2) v EMV(3) ti P = 0.1667

EMV(1) v EMV(3) ti P = 0.4737

Nu P

[0,0.1667]: Phng n 3Nu P [0.1667,0.6154]: Phng n 1

Nu P [0.6154,1]: Phng n 2


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RQtrongmitrngkhngxcnh+ Tiu chun Maximax (lc quan)

Phng n c gi tr kt qu tt nht tng ngvi trng thi t nhin tt nht: Maxi {Maxj PO(i,j)}

V d bi ton Thompson: Chn phng n xy nh my ln


Th trng tt Th trng xuNh my ln 200,000 -180,000

Nh my nh 100,000 -20,000

Khng SX 0 0

L h Q


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RQtrongmitrngkhngxcnh+ Tiu chun Maximin (bi quan)

Phng n c gi tr kt qu xu nht tng ngvi trng thi t nhin tt nht: Maxi {Minj {PO(i,j)}

V d bi ton Thompson: Chn phng n khng sn xut


Th trng tt Th trng xuNh my ln 200,000 -180,000

Nh my nh 100,000 -20,000

Khng SX 0 0

L th t Q


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RQtrongmitrngkhngxcnh+ Tiu chun Laplace (c hi xy ra nh nhau)

Phng n c gi tr li nhun trung bnh cao nht(chi ph trung binh thp nht):

V d bi ton Thompson: Phng n xy nh my nh

Phng nCc trang thi t nhin Gi tr li nhun

trung binh

Th trng tt

Th trng xu

Nh my ln 200,000 -180,000 10,000

Nh my nh 100,000 -20,000 40,000

Khng SX 0 0 0

N

jiPOMax

N

ji1 ),(

L th t Q


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RQtrongmitrngkhngxcnh+ Tiu chun Hurwicz (hin thc/trung bnh c

trng s)

Kt hp gia tiu chun lc quan & bi quan phthuc vo h s [0,1]:Maxi{*Maxj{PO(i,j)} + (1)*Minj{PO(i,j)}}

V d bi ton Thompson vi = 80% CH(1) = 0.8*200000 + (1-0.8)*(-180000) = 124000

CH(2) = 0.8*100000 + (1-0.8)*(-20000) = 76000CH(3) = 0

Chn phng n xy nh my ln

L th t Q


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RQ trong mi trng khng xc nh+ Tiu chun Minimax

Ti tiu ha tn tht vn c d tnh c th l tia: Minj {Maxi {BO(i,j)}}

V d bi ton Thompson Chn phng n xy nh my nh

Bng ra mt mt c hi:


Th trng tt Th trng xuNh my ln 0 180,000

Nh my nh 100,000 20,000

Khng SX 200,000 0

L th t Q


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RQ trong mi trng khng xc nh+ Tiu chun Minimax

V d 2.9 tr.56

Bng ra quyt nh (dng chi ph)


MaximinS1 S2

1 11,000 902 10,000 10,000


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Cy quyt nhKhi nim

Bi ton ra quyt nh c th cnhiu cp. Cc quyt nh sau phthuc vo cc quyt nh trc.

Vic thit lp bng ra quyt nh rtphc tp cho nn vic thit lp cy

quyt nh thng c s dng


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Cy quyt nhKhi nim

Cy Q c thit lp theo trnh tthi gian v logic ca cc cp.

K hiu trn biu ca cy Q:+vung - nt quyt nh

+ trn - nt trng thix


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Cy quyt nh Cc bc

Xc nh vn

V cy Q

Xc nh xc sut xy ra ca tngtrng thi

Xc nh gi tr kt qu

Gii bi ton da vo EMV cho tngnt trng thi (t phi sang tri)

C t h


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Cy quyt nhV d bi ton Thompson

1

2

3

Th trng ha hn (50%)

Th trng khng ha hn (50%)


Th trng ha hn (50%)

$200,000

($180,000)

$100,000

$0

($20,000)

EMV(3)=$0

EMV(2)=$40,000

EMV(1)=$10,000

Xy nhmy ln

Xy nhmy nh

Khngsn xut

$40,000

Payoff

C t h


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Cy quyt nhV d bi ton Thompson (m rng)

Gi s c s xem xt t vic thm d kho stth trng.+ Chi ph c tnh l $10,000

+ Xc sut ca kt qu thm d th trngTh trngha hn

45%

Th trngkhng ha hn55%

Thc s Khng thc s Thc s Khng thc s78% 22% 27% 73%

C t h


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1

2

3

Thc s ha hn (78%)

Khng thc s ha hn (22%)


Thc s ha hn (78%)

$190,000

($190,000)

$90,000

($10,000)

($30,000)

EMV(3)=($10,000)

EMV(2)=$63,600

EMV(1)=$106,400

Xy nh

my ln

Xy nhmy nh

Khngsn xut

$106,400

PayoffKho st th trng ha hn

C q t nh


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1

2

3

Thc s ha hn (27%)



Thc s ha hn (27%)

$190,000

($190,000)

$90,000

($10,000)

($30,000)

EMV(6)=($10,000)

EMV(5)=$2,400

EMV(4)=($87,400)

Xy nh

my ln

Xy nhmy nh

Khngsn xut

$2,400

PayoffKho st th trng khng ha hn

Cy quyt nh


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1

2

3

Th trng ha hn (50%)



Th trng ha hn (50%)

$200,000

($180,000)

$100,000

$0

($20,000)

EMV(9)=$0

EMV(8)=$40,000

EMV(7)=$10,000

Xy nhmy ln

Xy nhmy nh

Khngsn xut

$40,000

PayoffKhng kho st th trng

Cy quyt nh


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Cy quyt nhV d bi ton Thompson (m rng)

0

Thm d ha hn (45%)

Thm d khng ha hn (55%)

$106,400

$2,400

$40,000

EMV(0)=$49,200

Thm d

Khng

thm d

$49,200

C t h


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Cy quyt nhGi tr ca thng tin ly mu

Gi tr k vng ca thng tin ly mu (EVSI) lChi ph ti a cho vic kho st th trng(trc khi ra Q)

EVSI =Gi tr k vng ca Q tt nht khi c thng tin(khng tnh chi ph thu thp thng tin)

Gi tr k ca Q tt nht khi khng c thng tin

C q t nh


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Cy quyt nhGi tr ca thng tin ly mu V d

Theo v d k trn:

+ Gi tr k vng ca Q tt nht khi c thngtin: $49,200 + $10,000

+ Gi tr k vng ca Q tt nht khi khng cthng tin: $40,000

EVSI = $59,200 - $40,000 = $19,200

Cy quyt nh


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Cy quyt nhXc nh XS trong cy Q bng cng thc Bayes

Cy quyt nh


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Cy quyt nhL thuyt v v li M u

V d:

1

Kh nng 1 (50%)

Kh nng 2 (50%)

$5,000,000

$0

$2,000,000

EMV(1)=$2,500,000

PA 1

PA 2

$2,500,000

Payoff

2

EMV(2)=$2,000,000

Dng EMV trong ra Q khng xt nmc chpnhn ri ro

Kt hp mc chp nhn ri ro vo bi ton l vics dng khi nim V LI

Cy quyt nh


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Cy quyt nhXc nh v li

Gn v li+ 0kh nng xu nht+ 1kh nngtt nht

Xc nh v li ca cc trng hp trung gianGi s:+ (p) Xc sut xy ra kt qu tt nht

(1-p) Xc sut xy ra kt quxu nht+ ng vi mi gi tr trung gian, thay i dn gi tr ca

psao cho khng th Q nn chn PA no+ Gi tr pxc nh c chnh l VL ca gi tr kt

qu trung gian ang xt

Cy quyt nh


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Cy quyt nhXc nh v li

Thc hin xc nh VL vi nhiu gi trkhc nhau, ta v c ng biu dinca VL

1

(p)

(1-p)

Kt qu tt nht

Kt qu trung gian

PA 1

PA 2

2

Kt qu xu nht

Cy quyt nh


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Cy quyt nhL thuyt v v li

C 2 PA:+u t vn vo a c: li nhun $10,000 vi xc

sut phoc khng c g

+ Gi tin vo ngn hng Vi 1 gi tr li nhun trung gian, cn tm p sao

cho c 2 PA l nh nhau Gi s ti gi tr $5,000, 2 PA l nh nhau trong

80% trng hp; K hiu U($5,000) = 80% Tip tc nh trn, ta c VL ca mi gi tr

trung gian

V d: Biu din VL cho 1 c hi kinh doanh

Cy quyt nh


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Biu din v li

U

$

Trnh ri ro

a thch ri ro

Cn bng ri ro

C 3 dng ng cong biu din v li

Cy quyt nh


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Lu :Ngi ra Q c th c nhng ngcong biu din VL khc nhau cho nhng

tnh hung khc nhau Vic ra Q c th c thc hin da

trn tiu chun v k vng ca VL

Cy quyt nh


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Cy quyt nhL thuyt v v li V d bi ton tung ng xu

1

c (45%)

Mt (55%)

$10,000

($10,000)

$0

EMV(1)=($1,000)

Chp nhncc

Khng chpnhn cc

$0

2

EMV(2)=0

Theo tiu chun EMV

Cy quyt nh


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Cy quyt nhL thuyt v v li V d bi ton tung ng xu

1

c (45%)

Mt (55%)

$10,000U=0.30

($10,000)U=0.05

$0U=0.15

U(1)=$0.1625

Chp nhn

cc

Khng chpnhn cc

$0,1625

2

U(2)=0.15

Phn tch VL


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L thuyt ra Q


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RQ trong iu kin khng xc nh+ M hnh Maximax: Maxi (Max Pij)

+ M hnh Maximin: Maxi (Min Pij)

+ M hnh ng u ngu nhin: Maxi Pij+ M hnh Hurwieztrung bnh c trng s:

l mt trng s (0


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PP nh lng


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PP nh lngCc phng php ton ng dng

Thng k k ton:L mt b phn ca ton hc ng dng dnh cho ccphng php x l v phn tch s liu thng k. Ccng dng ch yu ca n trong qun l l cc phngphp x l kim tra v d on (d on, iu tra chn

mu,) M hnh ton:L s phn nh nhng thuc tnh c bn nht nh cacc i tng nghin cu kinh t, l cng c trng chovic tru tng ho mt cch khoa hc cc qu trnhv hin tng kinh t.Khoa hc kinh t t lu bit s dng cc m hnhkinh t lng nh m hnh hm sn sut Cobb -Douglas, m hnh cung cu, gi c v.v...

PP nh lng


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PP nh lngCc phng php ton ng dng

Vn tr hc:L khoa hc c mc ch nghin cu cc phng phpphn tch nhm chun b cn c chnh xc cho ccquyt nh, i tng ca n l h thng, tc l tp hpcc phn t v h thng cn c tc ng qua li vinhau nhm t ti mt mc tiu nht nh.Vn tr hc bao gm nhiu nhnh khoa hc ng dnggp li: (1) L thuyt ti u (bao gm: quy hoch tuyntnh, quy hoch ng, quy hoch ngu nhin, quyhoch nguyn, quy hoch 0 - 1, quy hoch a mc tiu,l thuyt tr chi...); (2) L thuyt th v s mngli; (3) L thuyt d tr bo qun; (4) L thuyt tmkim;...

PP nh lng


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PP nh lngCc phng php & m hnh c bn

Thng k m t Phng php Phn tch hi qui

Cc phng php D bo

M hnh ton (qui hoch tuyn tnh, quihoch nguyn, qui hoch phi tuyn)

M hnh mng

Phn tch Markov,

PHƯƠNG PHÁP ĐỊNH LƯỢNG 1

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