Chuong2-Ra Quyết Định Trong Quản Lí

7/25/2019 Chuong2-Ra Quyt nh Trong Qun L

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Q U N T R K I N H D O A N H C H O K SCHNG 2

RA QUYTNH (RQ)TRONG QUN L

Nguyn Ngc Bnh [email protected]


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Ni dung1. Gii thiu vRQtrong qun l

2. RQtrongiu kin ri ro

3. RQtrongiu kin khng chc chn

4. RQbng quy hoch tuyn tnh

5. RQa yu t6. RQtheo l thuyt hu ch

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1. Gii thiu vRQtrong qun l

RQl mt qu trnh la chn c thc gia hai

hoc nhiu phng n (PA) (gii php gii quytvn ) chn ra mt PA v PA ny sto ra

c mt kt qumong mun trong cc iu kinrng buc bit.

nh ngha RQ

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4

Cc quyt nh c a ra l kt quca mt sla chn c lptrng v vi mc tiu l ti u (cci hay cc tiu) mt gi trno trong nhng iu kin rng buc cth. Ngi RQc mc tiu cth. Tt ccc PA c thc u c xc nh y . Sa thch ca ngi RQcn phi r rng, cn lng ha cc tiu

chun ca cc PA v xp hng cc tiu chun theo thta thchca ngi RQ.

Sa thch ca ngi RQl khng thay i trong qu trnh RQ,ngha l cc tiu chun v trng sca cc tiu chun l khng i. Khng c shn chvthi gian v chi ph, ngha l c iu

kin thu thp y thng tin trc khi RQ. Sla chn cui cng sl tiu mc tiu mong mun.

Githit vshp l


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Vnc cu trc tt: Mc tiu r rng, thng tin y ,

dng bi ton quen thucRQtheo chng trnh (quytrnh, lut, chnh sch)

Vnc cu trc km: Dng bi ton mi m, thng tinkhng y , khng r rngRQkhng theo chngtrnh

RQtheo cu trc ca vn

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RQtrongiu kin khng chc chn

(uncertainty): Khi RQ, khng bit c xcsut xy ra ca mi trng thi hoc khng bitc cc dliu lin quan n cc vn cn

gii quyt. RQtrongiu kin ri ro (risk): Khi RQ,

bit c xc xut xy ra ca mi trng thi.

RQtrongiu kin chc chn (certainty):Khi RQ, bit chc chn trng thi no sxy ra, do sddng v nhanh chng ra

quyt nh.

RQtheo tnh cht ca vn

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Phng phpnh lng (PPL) l phng php khoa hc

dng cc php tnh ton gii quyt cc vn RQ.

Cc thut ngchPPL:Phng phpnh lng (Quantitative Methods)

Phn tchnh lng (Quantitative Analysis) Tip cnnh lng (Quantitative Approaches)Nghin cu tc v/Vn tr hc (Operations Research)Khoa hc qun l (Management Science)

Khi nim chung vPPL trong qun l

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Cc cng cnh lng

Cc m hnhnh lng

Phn mm: Excels Solver, ABQM, QM for Windows,...

Khi nim chung vPPL trong qun l

L thuyt RQ (Xc sut c iukin, Bng/cy quyt nh, Lthuyt tr chi,...)

Bi ton phn cng

Bi ton dng chy tia

Quy hoch tuyn tnh S mng (CPM/PERT)

Quy hoch nguyn Kim sot tn kho

Quy hochng L thuyt xp hng

Bi ton vn ti Phn tch Markov ...

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B1: Xc nh r vn/chi cn gii quyt.

B2: Lit k tt ccc PA c thc.

B3: Nhn ra cc tnh hung/trng thi.

B4: c lng tt cli ch/chi phcho mi PA ng vimi trng thi.

B5: La chn mt m hnh ton trong phng phpnh lng tm li gii ti u.

B6: p dng m hnh tm li gii v da vo RQ.

Cc bc ca qu trnh RQ

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Bc 1: C nn sn xut mt sn phm mi tham gia thtrng hay khng? Bc 2: C 3 phng n sn xut l:

Lp 1 nh my c qui m lnsn xut sn phm Lp 1 nh my c qui m nhsn xut sn phm Khng lm g c

Bc 3: C 2 tnh hung ca thtrng sxy ra l: thtrng tt, thtrng xu.

Bc 4: c lng li nhun ca cc phng n ng vi cc tnh hung:

Bc 5 v 6: Chn mt m hnh ton trong phng php nh lng p dng vo

bi ton ny. Vic chn la m hnh c da vo shiu bit, vo thng tin t hay nhiuvkhnng xut hin cc trng thi ca hthng.

V d: Bi ton ca Cng ty xgThompson

Phng nTrng thi

Thtrng tt Thtrng xu

Nh my ln 200.000 -180.000

Nh my nh 100.000 -20.000

Khng sn xut 0 0

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2. RQtrong iu kin ri ro

Khi RQtrong iu kin ri ro, bit xc sut xy ra ca

mi trng thi. Khi RQtrong iu kin ri ro, thng sdng hai tiu

chun/m hnh (tngng nhau): Cci gi trkvngc tnh bng tin (ExpectedMoneytary Value) (Maxi EMVi) Cc tiu thit hi kvng (Expected Opportunity Loss)

(MiniEOL

i)

Hai phng php thng dng: Phng php bng quytnh (decision table)

Phng php cy quytnh (decision tree) 11


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2. RQtrong iu kin ri roPhng php bng quytnh

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i i


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EMV1 = 0,5 x 200.000 + 0,5 x (-180.000) = 10.000

EMV2

= 0,5 x 100.000 + 0,5 x (-20.000) = 40.000

EMV3 = 0,5 x 0 + 0,5 x 0 = 0

Ra quytnh EMVi > 0 Phng n c li

Maxi EMVi = 40.000Chn phng n xy nh my nhGi trkvng vi thng tin hon ho

(EVWPI, Expected Value With Perfect Information)

EVWPI = 0,5 x (200.000) + 0,5 x (0) = 100.000

Gi trkvng ca thng tin hon ho(EVPI, Expected Value Of Perfect Information)

EVPI = 100.000 - 40.000 = 60.000

Phng php bng quytnh

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i i


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2. RQtrong iu kin ri roPhng php bng quytnh

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OL11 = 200.000-200.000 = 0

OL12 = 0 - (-180.000) = 180.000

OL21 = 200.000 -100.000 = 100.000

OL22 = 0 - (-20.000) = 20.000OL31 = 200.000 0 = 200.000

OL32 = 0 - 0 = 0

Phng php bng quytnh

T.Trng tt T.Trng xu

Nh my ln 0 180.000

Nh my nh 100.000 20.000

Khng sn xut 200.000 0

P(Sj) 0,5 0,5

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Bng thit hi chi OLij


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EOL1 = 0,5 x 0 + 0,5 x 180.000 = 90.000

EOL2 = 0,5 x 100.000 + 0,5 x 20.000 = 60.000EOL3 = 0,5 x 200.000 + 0,5 x 0 = 100.000

RQtheo tiu chun Mini EOLi

Mini EOLi= Min(90.000, 60.000, 100.000) = 60.000Chn PA xy nh my nh

Ghi ch: EVWPI = Maxi EMVi + Mini EOLi

EVPI = Mini EOLi

Phng php bng quytnh

T.Trng tt T.Trng xu

Nh my ln 0 180.000

Nh my nh 100.000 20.000

Khng sn xut 200.000 0

P(Sj) 0,5 0,5


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Phn tchnhy:

Xc sut p thay i thno th quyt nh sthay i?380.000 p 180.000 = 120.000 p 20.000

p = 16/26 = 0,62

Phng php bng quytnh

-20.000

-180.000

0,167 0,62

EMV Khng lm g

Gi trEMV

Xc sut ca thtrng a thch


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Cc k hiu & quyc

Phng php cy quytnh

TTtt

Nh my ln

TT xuNh my

nh

Khng lm g

: Nt quyt nh

: Nt trng thi/tnh hung

VD: Xt li Cng ty gThompson


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Cc bc ca vic phn tch bi ton cy quytnh

B1: Xc nh vn /chi cn gii quyt

B2: Vcy quyt nh

B3: Gn xc sut cho cc trng thi B4: c tnh li nhun hay chi ph cho mt skt hp gia

mt phng n v mt trng thi

B5: Gii bi ton bng phng php Maxi EMVi (tphiqua tri).

Phng php cy quytnh

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VD: Gii v dtrc bng cy quytnh

B1: C nn sn xut sn phm mi kinh doanh? B2: Vcy quyt nh

B3: Gn xc sut 0,5 cho cc loi thtrng

B4: Dng gi trbng sliu ghi vo

B5: Tnh cc gi trEMVi ti cc nt

- Ti nt 1 : EMV1 = 0,5 x 200.000 + 0,5 x (-180.000) = 10.000- Ti nt 2 : EMV2 = 0,5 x 100.000 + 0,5 x (-20.0000) = 40.000

- Ti nt 3 : EMV3 = 0

Phng php cy quytnh

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=

=

m

1j

ijji )xPP(SEMV


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2. RQtrong iu kin ri ro (tt)Phng php cy quytnh

MaxiEMV

i= 40.000 Ch

n PA nh my nh

21

200.00010.000TT tt (0.5)

- 180.000

100.00040.000

-20.0000

Nh my ln

TT xu (0.5)

Nh my nh

Khng lm g

TT xu (0.5)

TT tt (0.5)

==

m

1j ijji

)xPP(SEMV


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Cp Q1 Cp Q2


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3. RQtrong iu kin khng chc chn

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M hnh Maximax (lc quan) M hnh Maximin (bi quan) M hnh Laplace (ng u ngu nhin)

M hnh Hurwicz (trung bnh c trng s, hin thc) M hnh Minimax


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i: PA hng i trong bng RQj: Trng thi ct j trong bng RQ

Pij: Li ch c c nu chn PA i v trng thi j xy ra.

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j

j

Strngthi

j


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M hnh Maximax (tiu chun lc quan)

Maxi (Maxj Pij) = Maxi(200.000, 100.000, 0) = 200.000RQ: Chn PA xy nh my ln (i = 1)

M hnh Maximin (tiu chun bi quan)

Maxi (Minj Pij) = Maxi (-180.000, -20.000, 0) = 0

RQ: Khng lm g c(i = 3).

M hnh Laplace (ngu ngu nhin)

= Maxi (10.000, 40.000, 0) = 40.000

RQ: Chn PA xy nh my nh(i = 2)

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M hnh Hurwicz (m hnh trung bnh c trng s)

Chn = 0,8Maxi [0,8 x 200.000 + 0,2 x (-180.000),

0,8 x 100.000 + 0,2 x (-20.000), 0,8 x 0 + 0,2 x 0]

= Maxi (124.000, 76.000, 0) = 124.000

RQ: Chn PA xy nh my ln (i = 1).

M hnh Minimax

Mini [Maxj OLij] = Mini [180.000, 100.000, 200.000] = 100.000

RQ: Chn PA xy nh my nh(i = 2).

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Quy hoch tuyn tnh (QHTT Linear Programming)

l mt kthut ton hc nhm xc nh gi trca ccbinx1,x2,...,xn (c gi l cc bin quyt nh -decision variable) sao cho: Lm cc i hay cc tiu gi trca hm mc tiu

(objective function):Z=z(x1,x2,...,xn) Tha mn cc rng buc (constraint): R

j= r

j(x

1,x

2,...,

xn)

Trong , hm mc tiuzv cc rng buc rjl nhng

biu thc tuyn tnh (bc nht) i vi cc binx1,x2,...,xn

Gii thiu bi ton QHTT


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Bi ton lp khoch sn xut (bi ton cci)

Mt nh qun l dn nng nghip c cc sliu sau:

Hy xc nh lng la go v lc m cn c sn xut

lm cc i li nhun ca dn.


Sliu u vo i vimt n vsn phm

Loi sn phm Khnng ln nht cacc ngun ti nguyn

sn c

La go La m

Din tch [Ha/tn] 2 3 50 HaLng nc [103m3/tn] 6 4 90 x 103m3

Nhn lc [cng/tn] 20 5 250 cng

Li nhun [USD/tn] 18 21


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Bi ton lp khoch sn xut (bi ton cci)

Bc 1: Xc nh cc bin quyt nh Gix1,x2 ln lt l stn la go v la m cn c sn xut

Bc 2: Xc nh hm mc tiuHm mc tiu l li nhunZ Cc i li nhun:Z= 18x1 + 21x2Max

Bc 3: Xc nh cc rng buc Rng buc vdin tch: 2x1 + 3x2 < 50 Rng buc vlng nc: 6x1 + 4x2 < 90 Rng buc vnhn lc: 20x1 + 5x2 0



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Bi ton trn khu phn (bi ton cc tiu)

Mt nh qun l tri g dnh mua 2 loi thc n trn ra khu phn tt v gi r.

Mi n vthc n loi 1 gi 2ng c cha: 5g thnh phn A

4g thnh phn B

0,5g thnh phn C Mi n vthc n loi 2 gi 3ng c cha:

10g thnh phn A

3g thnh phn B

khng c cha thnh phn C Trong 1 thng, 1 con g cn ti thiu 90g thnh phn A, 48g thnh phn B v 1,5g thnh

phn C.

Hy tm slng mi loi thc n cn mua c m bonhu cu ti thiu vdinhdng cho 1 con g vi gi rnht.


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Bi ton trn khu phn (bi ton cc tiu)

Bc 1: Xc nh cc bin quyt nh Gix1,x2 ln lt l slng n vthc phm loi 1 v loi 2cn cho 1 con g trong 1 thng

Bc 2 : Xc nh hm mc tiuHm mc tiu l gi muaZ Cc tiu gi mua:Z= 2x1 + 3x2Min

Bc 3 : Xc nh cc rng buc Thnh phn A : 5x1 + 10x2 > 90 Thnh phn B : 4x1 + 3x2 > 48 Thnh phn C : 0,5x1 >1,5

Cc bin khng m: x1,x2 > 0


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Bi ton cci chun Hm mc tiu:Z = c1x1 + c2x2 + .... + cnxnMax

Rng buc:a11x1 + a12x2 + .... + a1nxn < b1a21x1 + a22x2 + .... + a2nxn < b2am1x1 + am2x2 + .... + amnxn < bm xj > 0 , j = 1,n

M hnh tng qut ca bi ton QHTT

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ngha (c n sn phm v m

ti nguyn) xj l sn vsn phm

thj c

jl li nhun do 1n v

sn phm thjem li aij l lng ti nguyn thi

cn cho 1n vsn phm

thj bi l tng lng ti nguyn

thi sn c


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Bi ton cc tiu chun Hm mc tiu:Z = c1x1 + c2x2 + .... + cnxnMin

Rng buc:a11x1 + a12x2 + .... + a1nxn b1a21x1 + a22x2 + .... + a2nxn b2am1x1 + am2x2 + .... + amnxn bm xj > 0 , j = 1,n

M hnh tng qut ca bi ton QHTT

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B1:Nhn dng ccbin quytnhvhm mc tiu.

B2:Din thm mc tiuvcc rng buc theo cc bin quyt nh.

B3:Kim tra xem c phi tt ccc quan htrong hm mc tiu v trong cc rng bucc tuyntnhkhng? Nu khng, phi tm m hnh phi tuyn khcgii.

B4:Kim travng khng gian li gii,xem xtiu kin nghim ca bi ton. Cc khnngc thxy ra l:

Khng cvng khthi (v nghim)Ni lng cc rng buc.

Vng khthiv hn v khng cim cc trCu trc li m hnh, c tha thmrng buc vo m hnh

Vng khthiv hn v c

im c

c tr

Sang B5. Vng khthic gii hnSang B5.

B5:Tm racc li gii tiu c thc, c thdng: Phng phpth (Graphical method)

Phng phpn hnh (Simplex method)

Qu trnh gii bi ton QHTT

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b h h h


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Gii bi ton v dcciHm mc tiu: 18x

1+ 21x

2Max

Rng buc: 2x1 + 3x2 50 (1)6x1 + 4x2 90 (2)20x1 + 5x2 250 (3)

x1 0 (4)x2 0 (5)

Gii Trong mt phng ta 0x1x2, vcc ng thng:(D1) 2x1 + 3x2 = 50

(D2) 6x1 + 4x2 = 90(D3) 20x1 + 5x2 = 250(D4) x1 = 0

(D5) x2 = 0

PPngng li hayngng ph (iso profit/cost line)

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10

50

5025100

x2

x1

22.5

16.67

15

C

(D3)

(D1)

(D2)

Z0 = 0

Z = 18x1+21x2 Max

2x1+3y


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10

50

5025100

(x2)

(x1)

22.5

16.67

A15

C

(D3)

(D1)

(D2)

7

12

Nghim ti u:x1=7; x2=12 hay C(7; 12)

Rng buc tch cc: D1 v D2(SRBTC = Sbin Q)

Ta ca im C l nghim ca hphng trnh:2x1 + 3x2 = 50 x1* = 7

6x1 + 4x2 = 90 x2* = 12Gi trca hm mc tiu:Z = Zmax = 18x1* + 21x2*

= 18 x7 + 21x12

Zmax = 378

ng ng li

Vng nghim khd(min khng gian li gii,khng gian sch lc)

Z0 = 0 B

D

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4 RQ b h h t t h


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Gii bi ton v dcc tiuHm MT: Min Z = 2x

1

+ 3x2Rng buc: 5x1 + 10x2 90 (1)

4x1 + 3x2 48 (2)0,5x1 1,5 (3)

x1 0 (4)x2 0 (5)

Gii: Trong mt phng ta 0x1x2 , ta vcc ng thng:(D1): 5x1 + 10x2 = 90

(D2): 4x1 + 3x2 = 48(D3): 0,5x1 = 1,5(D4): x1 = 0(D5): x

2= 0

PPngng li hayngng ph (iso profit/cost line)

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322824201612

32

28

24

20

16

12

8

4

0

Min

D2

Min khng gian li gii v hn,

nhng c cc tiu.

D3

D1

CB

A

3 8.4

4.8

x2

x1

ng ng ph

Z0 = 0

Ta im B l nghim ca hpt:5x1 + 10x2 = 90 x1* = 8,44x1 + 3x2 = 48 x2* = 4,8

Gi trca hm mc tiu:Z = Zmin = 2x1 + 3x2= 2 x 8,4 + 3 x 4,8 = 31,2

Nghim ti u:

x1* = 8,4x2* = 4,8

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4 RQ b h h t t h


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Ccimnh (corner point, extreme point) l giao im

ca cc rng buc nm trong khng gian li gii gi l ccnh ca khng gian li gii.

Nu bi ton QHTT c li gii tiu, th li gii snmtrn cc nh ca khng gian li gii.

p dng kt quny, tm gi trca hm mc tiu bngcch so snh gi trca cc nh ca khng gian li gii.

Phng php dngimnh

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4 RQ bng quy hoch tuyn tnh


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Gii bi ton v dcci

So snh gi trti 5nh O, A, B, C, D nh O (0, 0) ==> ZO =0

nh A (12,5) ==> ZA = 18 x 15,5 + 21 x 0 = 225

nh B (11, 6) ==> ZB = 8 x 11 + 21 x 6 = 324 nh C (7, 12) ==> ZC = 18 x 7 + 21 x 12 = 378

nh D (0, 16,67) ==> ZD = 18 x 0 + 21 x 16,67 = 350,07

==>Zmax = ZC = 378 ==>x1 = 7vx2 = 12

Phng php dngimnh

Hm MT: Z = 18x1 + 21x2Max

Max

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4 RQ bng quy hoch tuyn tnh


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Gii bi ton v dcc tiu

So snh gi trti 3 nh A, B, C:

nh A (18, 0) ==> ZA = 2 x 18 + 3 x 0 = 36

nh B (8.4 , 4.8) ==> ZB = 2 x 8.4 + 3 x 4.8 = 31.2 nh C (3, 12) ==> ZC = 2 x 3 + 3 x 12 = 42

==>Zmin = ZB = 31.2 ==> x1 = 8.4 v x2 = 4.8

Phng php dngimnh

Min

Hm MT: Z= 2x1 + 3x2Min

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5 RQa yu t


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5. RQa yu t

V d: Mt sinh vin tt nghip mun tm vic lm th c nhiu yu tsnh hng n quyt nh chn vic ca anh ta: Lng khi im, Chithng tin, Vtr ca ni lm vic, Nhng ngi m mnh slm vic vih, Loi cng vic bn cn phi lm, Nhng li nhun khc ngoi lng...

Cch gii quyt bi ton RQa yu t:

Nhiu ngi xem xt cc yu tkhc nhau ny mt cch chquan vtrc gic

Dng phng phpnh gi yu tMFEP (Multi Factor EvaluationProcess): Mi yu tquan trng nh hng n quyt nh scgn mt hsni ln tm quan trng tng i gia cc yu tvi

nhau

nh gi PA theo cc hsny

Gii thiu bi ton RQa yu t

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5 RQa yu t


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5. RQa yu t

B1: Lit k tt ccc yu tv gn cho yu tthi mt trng sFWi:Tm quan trng ca mi yu tmt cch tng i, 0 < FW

i< 1,

FWi = 1

B2: Lng gi theo yu t. Vi mi yu ti, ta nh gi PA j bng cchgn mt hsFE

ij: Lng gi ca PA j i vi yu ti

B3: Tnh tng lng gi trng sca tng PA j

Chn PA ng vi Max TWEj

Cc bc thc hin phng php MEFP

i: Yu t

j: Phng n

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FW: Factor weightFE: Factor EvaluationTWE: Total Weighted evaluation

5 RQa yu t


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5. RQa yu t

B1: Xc nh FWi

Sau khi nghin cu, bn bc vi thy, bn b, gia nh..., sinh vin S nhnthy 3 yu tquan trng nht i vi vic chn slm l:

Lng Chi thng tin Vtr ni lm vic

Sinh vin S gn cho cc yu tcc trng ssau:Yu t i Trng sFWiLng 0.3

Chi thng tin 0.6Vtr ni lm vic 0.1

VD: Bi ton tm vic lm ca sinh vin

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5 RQa yu t


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5. RQa yu t

B2: Xc nh FEij

Sinh vin S nghrng c 3 cng ty A, B, C snhn mnh vo lm vic.i vi mi cng ty, sinh vin S nh gi theo 3 yu ttrn v c bnglng gi nhsau:

Lm sao xc nh gi trtrong bng ny?


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5 RQa yu t


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5. RQa yu t

i vi Lng, Anh S mong rng lng sl 1.000.000. Nhng thc tcng ty A tr700.000, cng ty B tr800.000, cng ty C tr900.000.

B3: Tnh cc tng lng gi trng sTWEjTWE(A) = 0.3 x 0.7 + 0.6 x 0.9 + 0.1 x 0.6 = 0.81

TWE(B) = 0.3 x 0.8 + 0.6 x 0.7 + 0.1 x 0.8 = 0.74

TWE(C) = 0.3 x 0.9 + 0.6 x 0.6 + 0.1 x 0.9 = 0.72

Kt lun: Sinh vin S chn cng ty A


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6 RQ theo l thuyt hu ch


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6. RQtheo l thuyt hu ch

hu ch/vli (Utility) l o mc u tin ca ngi

RQi vi li nhun. L thuythu ch l l thuyt nghin cu cch kt hp mc

u tin vmay ri ca ngi RQi vi cc yu tkhc

trong qu trnh RQ. VD: Gisbn c mt tm v sc bit m khi thy ng xu

ln nu mt nga xut hin th bn trng thng 5.000.000,nu mt sp xut hin th bn khng c g ht. Vn t ra:Nu c ngi no nghmua li tm v sca bn trc

khi tung ng xu vi gi 2.000.000th bn c bn hay khng?

Khi nim vhu ch

48



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VD:Gisbn c mt tm v sc bit m khi thy ng xu ln nu mt nga xut hinth bn trng thng 5.000.000, nu mt sp xut hin th bn khng c g ht.

Vn t ra: Nu c ngi no nghmua li tm v sca bn trc khi tung ngxu vi gi 2.000.000th bn c bn hay khng?

Gii:

Khi nim vhu ch

Nu xt theo tiu chun EMV:EMV(khng bn) = EMV1 =

= 5.000.000 x 0,5 + 0 x 0,5 = 2.500.000EMV(bn) = EMV2 = 2.000.000

EMV (khng bn) > EMV (bn)Kt lun: Khng bn tm v s

Nu xt trn quan im thc t, th a smi ngi sbn

v t ai thch may ri trnhng ngi tph thch may ri. 49



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hu ch c c tnh nhsau: Kt qutt nht sc hu ch l 1 U(tt nht) = 1 Kt quxu nht sc hu ch l 0 U(xu nht) = 0 Kt qukhc sc hu ch (0,1) 0 < U(khc) < 1

Cch tnhhu ch ca kt qukhc:hu ch ca kt qukhc c tnh da trn sxem xt mt

tr chi chun gm 2 kt qu:

+ Kt qutt nht c xc sut l p+ Kt quxu nht c xc sut l (1 - p)

Cch tnhhu ch

50

6. RQ theo l thuyt hu ch


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Ta c 2 PA:+ PA 1: Chp nhn tr chi, ta sc kt qutt nht hayc kt quxu nht.

+ PA 2: Khng chp nhn tr chi,c mt kt quchc chn trnhc ri ro.Vn:Xc nh xc sut p (tt nht),2 PA nyc xem l tng ng i vi ngi RQi vi ngi RQ,2 PAc xem l tngng nhaunu kvnghu ch ca 2 PAbng nhau:

GiEU lkvng cahu ch (Expected Utility)EU(khng chi) = EU(chi)EU(chi) = p * U(T) + (1 -p) * U(X) = p * 1 + (1 -p) * 0 = pU(kt qukhc) = EU(khng chi) = p

Kt lun: p chnh l kvng cahu chlm cho 2 PA tng ng nhaui vi ngi RQ. Nhvy,hu ch hon ton chquan, ty thuc vo mccm nhn vri ro ca ngi

RQ.

Cch tnhhu ch

51



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VD: CXmun vng hu ch i vi tin, t0 n 10.000

vi U(10.000) = 1 v U(0) = 0CXcmt s tin, c c th mua btng sn hoc b vo qu titkim ca ngn hng. NucX u tvo bt ng snthsau3nm cthuc 10.000 hoc l b mt trng. Nu c X gi tit kim th sau 3nm s chc chn thu c 5.000.V mt ch quan, c X cho rng nu 80% c c may thuc 10.000sau3nmthcXmi u tvo bt ng sn nu khng c X s gi

tit kim.Nhvy vi xc sutp=0 ,8 mua btng sn thnh cng th 2 PAmuabt ng snvgi tin tit kimlnhnhau.

Cch tnhhu ch

52



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i vi c X: U(5.000) = p = 0,8

Gisi vi c X: U(7.000) = 0,9; U(3.000) = 0,5

ng cong hu ch i vi tin ca c X

Cch tnhhu ch

53



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Cch tnhhu ch

54

Cc dng ng cong hu ch

UU

Ngi RQtrnh ri ro Ngi RQthch ri ro Ngi RQkhng csthin lch vri ro

U

$ $ $



55/56

Q y

VD: ng Bang xem xt c nnu tvo mt dn hay khng. Nu d

n thnh cng ng B thu

c 10.000 tri li m

t 10.000. Theo ng B d

n c 45% cmay thnh cng. Ngoi ra, nghu ch ca ng B c

dng sau:

nh gi PA bnghu ch

55



56/56

Q y

nh gi PA bnghu ch

Nu dng EU:EU(tham gia) = 0,45*0,3 + 0,55* 0,05 = 0,1625EU(khng tham gia) = 0,15Kt lun: ng B tham gia u tvo dn.

Nu dng EMV:EMV(tham gia) = 0,45*10.000 + 0,55*(-10.000) = -1.000EMV(khng tham gia) = 0Kt lun: ng B khng tham gia vo dn. 56

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