Bai Tap cao học Phan Tich Dinh Luong

Bi1.1Ngi ch chnh ca Cng ty l Kenneth ra quyt nh c tnh lc quan v th la chn theo tiu chun cc iKh nng la chnTh trng tt (USD)Th trng xu(USD)S cc i ca dng

SUB100300.000-200.000300.000

OILERJ250.000-100.000250.000

TEXAN75.00018.00075.000

Nh vy Kenneth s quyt nh chn mua h thng SUB100Bi 1.2 Bob ra quyt nh c tnh bi quan v th la chn theo tiu chun cc i s cc tiu ca dngKh nng la chnTh trng tt (USD)Th trng xu(USD)S cc tiu ca dng

SUB100300.000-200.000-200.000

OILERJ250.000-100.000-100.000

TEXAN75.00018.00018.000

Nh vy Kenneth s quyt nh chn mua h thng TEXANBi 1.3Kenneth s ra quyt nh da trn tiu chun hin thc vi = 0,7Cng thc tnh: Th trng tt + th trng xu (1-)Kh nng la chnTh trng tt (USD)Th trng xu(USD)Tiu chun hin thc = 0,7

SUB100300.000-200.000150.000

OILERJ250.000-100.000145.000

TEXAN75.00018.00047.100

Kenneth s ra quyt nh chn h thng SUB100 v = 0,7 gn 1 th quy nh nghing v pha lc quanBi 1.4a) Bng 1.4.1: La chn quyt nh theo tiu chun MaximaxTm c ca trmTh trng tt (USD)Th trng trung bnh (USD)Th trng xu (USD)S cc i ca dng

Nh50.00020.000-10.00050.000

Trung bnh80.00030.000-20.00080.000

Ln100.00030.000-40.000100.000

Rt ln300.00025.000-160.000300.000

Quyt nh maximax l trm xng c quy m rt ln

b) Bng 1.4.2: La chn quyt nh theo tiu chun MaximinTm c ca trmTh trng tt (USD)Th trng trung bnh (USD)Th trng xu (USD)S cc tiu ca dng

Nh50.00020.000-10.000-10.000

Trung bnh80.00030.000-20.000-20.000

Ln100.00030.000-40.000-40.000

Rt ln300.00025.000-160.000-160.000

Quyt nh maximin l trm xng c quy m nhc) Bng 1.4.3: La chn quyt nh theo tiu chun cc i gi tr trung bnh cc dngCng thc tnh: Th trng tt + th trng xu

s th trng

Bi tp c 3 loi th trngTm c ca trmTh trng tt (USD)Th trng trung bnh (USD)Th trng xu (USD)Gi tr trung bnh ca dng

Nh50.00020.000-10.00020.000

Trung bnh80.00030.000-20.00030.000

Ln100.00030.000-40.00030.000

Rt ln300.00025.000-160.00055.000

Quyt nh theo cc i gi tr trung bnh ca cc kh nng th chn phng n quy m rt lnd)Bng 1.4.4: La chn quyt nh theo tiu chun hin thcCng thc tnh: Th trng tt + th trng trung bnh + th trng xu Tm c ca trmTh trng tt (USD)Th trng trung bnh (USD)Th trng xu (USD)Tiu chun hin thc = 0,8; =0,1;=0,1

Nh50.00020.000-10.00041.000

Trung bnh80.00030.000-20.00065.000

Ln100.00030.000-40.00079.000

Rt ln300.00025.000-160.000226.500

Quyt nh theo tiu chun hin thc th chn phng n rt ln

e) Bng 1.4.5: L khi b l c hi u tTm c ca trmTh trng tt (USD)Th trng trung bnh (USD)Th trng xu (USD)

Nh300.000-50.000=250.00030.000-20.000=10.0000-(-10.000)=10.000

Trung bnh300.000-80.000=220.00030.000-30.000=00-(-20.000)=20.000

Ln300.000-100.000=200.00030.000-30.000=00-(-40.000)=40.000

Rt ln300.000-300.000=030.000-25.000=5.0000-(-160.000)=160.000

Bi 2.1Gi S1 l chin lc c iu tra trc khi quyt nh cho vay, S2 l chin lc khng iu tra. S3 l chin lc quyt nh cho vay, S4 l chin lc t chi khng cho vay.Cc bin c l T1,T2 v E1 v E2Ghi ch: - Li nhun t vic cho vay: 80.000 12%=9.600 - Li sut mua cng tri: 80.000 5%=4.000

Ta c: P(E1/T1) = ; khi P(E2/T1) = . Vy E(S3) = nhnh th 2:

P(E1/T2) = ; khi P(E2/T2) = . Vy E(S3) = nhnh cui cng P(E1) = 0,95; P(E2) = 0,05Vy E(S3) = 9.600 0,95 + (-80.000) 0,05 = 5.120

V P(T1) = v P(T2) =

Nn E(S1) = Nu tr i 400 chi ph iu tra th ta c:E(S1) = 7.080 - 400 = 6.680 ln hn 5.120vy chin lc ca ngn hng l tin hnh iu tra trc khi quyt nh cho vaya) Nu kt qu iu tra l T1 th quyt nh cho vayb) Nu kt qu iu tra l T2 th t chi khng cho vay v s tin 80.000$ dng mua cng triKhi li nhun trung bnh ca mi khon tin 80.000$ m ngn hng t c mt nm l 7.080

Chn S1

S2 S1

7.080 P(T2) P(T1) T2 T1 S3 S4 S3 S3 S4 S4 5.120 - -5.848 8.338 P(E2) E1 P(E2/T2) E1 P(E2/T1) E1 E2 P(E1) E2 P(E1/T2) E2 P(E1/T1) 4.000 -80.000 9.600 4.000 -80.000 9.600 4.000 -80.000 9.600

S cy biu din quan h gia cc chin lc cho vay ca ngn hng v cc bin c c lin quan

8.3384.0005.120

Bi 2.2 (VT: 1000)Gi S1 l chin lc c iu tra trc khi quyt nh cho vay, S2 l chin lc khng iu tra. S3 l chin lc quyt nh cho vay, S4 l chin lc t chi khng cho vay.Cc bin c l T1,T2 v E1 v E2Ghi ch: - Li nhun t vic cho vay: 500.000 15%=75.000 - Li sut chuyn tit kim: 500.000 6%=30.000

Ta c: P(E1/T1) = ; khi P(E2/T1) = . Vy E(S3) = nhnh th 2:

P(E1/T2) = ; khi P(E2/T2) = . Vy E(S3) = nhnh cui cng P(E1) = 0,95; P(E2) = 0,05Vy E(S3) = 75.000 0,95 + (-500.000) 0,05 = 46.250

V P(T1) = v P(T2) =

Nn E(S1) = E(S1) =61.825 > 46.250Vy chin lc ca ngn hng l tin hnh iu tra trc khi quyt nh cho vaya) Nu kt qu iu tra l T1 th quyt nh cho vayb) Nu kt qu iu tra l T2 th t chi khng cho vay v s tin 500 triu ng chuyn vo tit kimKhi li nhun trung bnh ca mi khon tin 500 triu ng m ngn hng t c mt nm l 61.825 nghn ng

Chn S1

S2 S1

61.825 P(T2) P(T1) T2 T1 S3

S3 S3 S4 46.250 -46.969 68.114 S4 S4 P(E2) E1 P(E2/T2) E1 P(E2/T1) E1 E2 P(E1) E2 P(E1/T2) E2 P(E1/T1)

30.000 500.000 75.000 30.000 500.000 75.000 30.000 500.000 75.000

46.25068.11430.000

S biu din quan h gia cc chin lc cho vay ca ngn hng v cc bin c lin quan

Bi 3.1a) Phng php loi tr bng cch lp biu tiu chun chp nhn c ca tng thuc tnhBng 3.1: Tiu chun chp nhn c ca tng thuc tnhCc thuc tnhCc kh nng la chnTiu chun chp nhn c

H thng AH thng BH thng CTi thiuTi a

Chi ph ban u140.000$180.000$100.000$180.000$

bn vngTtTuyt viTrung bnhTt

an tonTtTtTtTt

Kiu dngTtTuyt viTrung bnhTrung bnh

Cht lng sn phmTtTtTrung bnhTt

Cn c vo biu tiu chun chp nhn c ca tng thuc tnh ta nhn thy h thng A v B l p ng c yu cu, tuy nhin h thng B c tiu chun bn vng v kiu dng li vt hn ln chn phng n B l hp l nht.b) Theo phng php sp xp theo li t inBng 3.2: Sp xp th t ca cc thuc tnh theo tm quan trngThuc tnhSo snhThuc tnhThuc tnhSo snhThuc tnh

an ton> bn vng bn vng>Chi ph ban u

an ton>Cht lng sn phm bn vng>Kiu dng

an ton>Chi ph ban uCht lng sn phm>Chi ph ban u

an ton>Kiu dngCht lng sn phm>Kiu dng

bn vng>Cht lng sn phmChi ph ban u>Kiu dng

Cn c vo bng 3.1 ta thy an ton c 4 im, bn vng c 3 im, cht lng sn phm c 2 im, chi ph ban u c 1 im, kiu dng c 0 im.Bng 3.3: Th t cc kh nng (phng n) la chnSTTCc thuc tnhim sSp xp th t cc phng n la chn

1 an ton4A=B=C

2 bn vng3B>A>C

3Cht lng sn phm2A=B>C

4Chi ph ban u1B>A>C

5Kiu dng0B>A>C

Nh vy bng phng php t in th phng n B l phng n c la chnc) Theo phng php trng sB1: Cho im tng thuc tnh- Chi ph ban u: 180.000-100.000=80.000Cng thc: 180.000-CPBi

80.000

Bng tnh im v chi ph ban u ca cc phng nTinim

Phng n A140.000

Phng n B180.000

Phng n C100.000

- bn vng: ta xp hng bn vng trung bnh l 1, tt l 2, tuyt vi l 3. Ly 3 - 1 = 2Cng thc: HangcuaPAi - 1

2

Bng tnh im v bn vng ca cc phng n bn vngXp hngim

Phng n ATt2

Phng n BTuyt vi3

Phng n CTrung bnh1

- Kiu dng: ta xp hng kiu dng trung bnh l 1, tt l 2, tuyt vi l 3. Ly 3 - 1 = 2Cng thc: HangcuaPAi - 1

2

Bng tnh im v kiu dng ca cc phng nKiu dngXp hngim

Phng n ATt2

Phng n BTuyt vi3

Phng n CTrung bnh1

- Cht lng sn phm: ta xp hng cht lng sn phm trung bnh l 1, tt l 2. Ly 2 - 1 = 1Cng thc: HangcuaPAi - 1

1

Bng tnh im v kiu dng ca cc phng nKiu dngXp hngim

Phng n ATt2

Phng n BTt2

Phng n CTrung bnh1

Bng 3.4: Bng tnh im cho cc thuc tnh ca cc phng n u tSTTCc thuc tnhim cho cc phng n u t

H thng AH thng BH thng C

1Chi ph ban u0,501

2 bn vng0,510

3Kiu dng0,510

4Cht lng sn phm110

B2: Gn cc trng s cho cc thuc tnhSp xp theo tnh quan trng ca cc thuc tnh: an ton> bn vng > cht lng sn phm > chi ph ban u > kiu dng. Ta gn chng theo th t cc s 5>4>3>2>1. Khi : 1+2+3+4+5=15

Nh vy trng s ca an ton l ; bn vng ; cht lng sn phm ;chi ph ban u;

kiu dng B3: Tnh im trung bnh tng th theo trng s ca tng phng n la chnBng 3.5: Bng tnh im trung bnh theo trng s ca cc phng n u tSTTCc thuc tnhTrng sim cho cc phng n u t

H thng AH thng BH thng C

1Chi ph ban u

2 bn vng

3 an ton

4Kiu dng

5Cht lng sn phm

Gi tr trung bnh theo trng s0,46660,69980,3333

Nh vy im trung bnh theo trng s ca phng n B l cao nht. C ngha l nn trang b h thng my mi cho nh my theo phng n B.Bi 4.1 = [0,25; 0,25; 0,25; 0,25]

P =

a) Cc phn phn chia th trng ca 4 cng ty nm sau:

1 = P = [0,25; 0,25; 0,25; 0,25] = [0,250,6+0,250,1+0,250,05; 0,250,2+0,250,7+0,250,1+0,250,05; 0,250,1+0,250,2+0,250,8+0,250,1; 0,250,1+0,250,1+0,250,8] = [0,1875; 0,2625; 0,3; 0,25]b) Cc phn phn chia th trng ca 4 cng ty trong nm th 4:

2 = 1 P = [0,1875; 0,2625; 0,3; 0,25] = [0,18750,6+0,30,1+0,250,05; 0,18750,2+0,26250,7+0,30,1+0,250,05; 0,18750,1+0,26250,2+0,30,8+0,250,1; 0,18750,1+0,26250,1+0,250,8] = [0,155; 0,26375; 0,33625; 0,245]

3 = 2 P = [0,155; 0,26375; 0,33625; 0,245] = [0,1550,6+0,336250,1+0,2450,05; 0,1550,2+0,263750,7+0,336250,1+0,2450,05; 0,1550,1+0,263750,2+0,336250,8+0,2450,1; 0,1550,1+0,263750,1+0,2450,8] = [0,138875; 0,2615; 0,36175; 0,237875]Vy phn phn chia th trng ca 4 cng ty trong nm th 4 biu din bi vct 3 = [0,138875; 0,2615; 0,36175; 0,237875]Bi 4.2a) t c 2 trng thi hot ng:Trng thi 1: t n my cTrng thi 2: t khng n my cNu hm trc t n my c th kh nng sng hm sau cng n my c l 0,9 (P11 = 0,9), do xc sut hm sau khng n my c l 0,1 (P12 = 0,1).Tng t ta c: P21 = 0,3; P22 = 0,7

Vy ma trn cc xc sut chuyn i trng thi l: P = b) - Nu ngy hm nay (ngy th 2) t n my ( trng thi 1) th xc sut trng thi 1 l 1, trng thi 2 l 0. Vy tnh trng hot ng ngy th 2 ca t biu din bi vct 1 = [1;0].Vi ma trn P bit ta d on c kh nng n my ca t ngy th 3 l:

2 = 1 . P = [1;0] . 2 = [10,9; 10,1] = [0,9;0,1].Ngy th 4 l:

3 = 2 . P = [0,9;0,1] . 3 = [0,90,9+ 0,10,3 ; 0,90,1+0,10,7] = [0,84;0,16].Ngy th 5 l:

4 = 3 . P = [0,84;0,16] . 4 = [0,840,9+ 0,160,3 ; 0,840,1+0,160,7] = [0,81;0,19].Ngy th 6 l:

5 = 4 . P = [0,81;0,19] . 5 = [0,810,9+ 0,190,3 ; 0,810,1+0,190,7] = [0,79;0,21].Vy ngy th 6 xc sut ng A c cuc (xe t n my) l 0,79, khng c cuc (xe t khng n my) l 0,21. - Nu ngy hm nay (ngy th 2) t khng n my ( trng thi 2) th xc sut trng thi 1 l 0, trng thi 2 l 1. Vy tnh trng hot ng ngy th 2 ca t biu din bi vct 1 = [0;1].Vi ma trn P bit ta d on c kh nng n my ca t ngy th 3 l:

2 = 1 . P = [0;1] . 2 = [10,3; 10,7] = [0,3 ; 0,7].Ngy th 4 l:

3 = 2 . P = [0,3;0,7] . 3 = [0,30,9+ 0,70,3 ; 0,30,1+0,70,7] = [0,48;0,52].Ngy th 5 l:

4 = 3 . P = [0,48;0,52] . 4 = [0,480,9+ 0,520,3 ; 0,480,1+0,520,7] = [0,59;0,41].Ngy th 6 l:

5 = 4 . P = [0,59;0,41] . 5 = [0,590,9+ 0,410,3 ; 0,590,1+0,410,7] = [0,65;0,35].Vy ngy th 6 xc sut ng A c cuc (xe t n my) l 0,65, khng c cuc (xe t khng n my) l 0,35. Bi 5.2b)Y1Y2

X121116

X2893

S nh nht ca dng 1 l 21, dng 2 l 3, theo tiu chun maximin ta chn c s 21S ln nht ca ct 1 l s 89, s ln nht ca ct 2 l s 116, theo tiu chun minimax ta chn c s 89.Hai s c chn khng trng nhau nn khng tn ti im yn nga. Ta lp bng Xs Payoff matrixBng Xs Payoff matrixY1:PY2:1-P

X1:Q21116

X2:1-Q893

Ta c phng trnh vi X : 21Q + 89(1-Q) = 116Q + 3(1-Q)Gii phng trnh: 21Q + 89 - 89Q = 116Q + 3 - 3Q

181Q = 86 ; do Tng t ta c phng trnh vi Y: 21P + 116(1-P) = 89P + 3(1-P)Gii phng trnh: 21P + 116 116P = 89P + 3 3P

181P = 113 do Ta c kt qu bng Xs Payoff matrix mi nh sau:

Y1:Y2:

X1:21116

X2:893

Bng phn phi xc sut gi tr ca tr chiZ21116893

P(Z)

=

=

Vy ta c gi tr trung bnh ca tr chi l:

E(Z) = Ni cch khc nu chi nhiu ln th s im thng trung bnh bn X l 56,69 cn im thua trung bnh bn Y l 56,69.c)Y1Y2

X1-5-10

X2128

X3412

X4-40-5

Ta thy ngay l X s khng bao gi chi 2 chin lc X1 v X4 v i vi X th 2 chin lc X2 v X3 c th tri hn hn. Do ta c th rt gn ma trn v dng ma trn cp 2x2 sau y:Y1Y2

X2128

X3412

S nh nht ca dng 1 l 8, dng 2 l 4, theo tiu chun maximin ta chn c s 8S ln nht ca ct 1 l s 12, s ln nht ca ct 2 l s 12, theo tiu chun minimax ta chn c s 12.Hai s c chn khng trng nhau nn khng tn ti im yn nga. Ta lp bng Xs Payoff matrixBng Xs Payoff matrixY1:PY2:1-P

X1:Q128

X2:1-Q412

Ta c phng trnh vi X : 12Q + 4(1-Q) = 8Q + 12(1-Q)Gii phng trnh: 12Q + 4 - 4Q = 8Q + 12 - 12Q

12Q =8 ; do Tng t ta c phng trnh vi Y: 12P + 8(1-P) = 4P + 12(1-P)

Gii phng trnh: 21P + 8 8P = 4P + 12 12P 21P = 4 do Ta c kt qu bng Xs Payoff matrix mi nh sau:

Y1:Y2:

X1:128

X2:412

Bng phn phi xc sut gi tr ca tr chiZ128412

P(Z)

Vy ta c gi tr trung bnh ca tr chi l: E(Z) = Ni cch khc nu chi nhiu ln th s im thng trung bnh bn X l 9,33 cn im thua trung bnh bn Y l 9,33.Bi 6.1a) Lp bng phn phi xc sut, bng phn phi xc sut tch lu v khong cc s ngu nhin cho X:XTn s (s gi)niP(X=x) = ni/nXc sut tch luKhong cc s ngu nhin

X30000

4200,10,1T 1 n 10

5300,150,25T 11 n 25

6500,250,5T 26 n 50

7600,30,8T 51 n 80

8400,21T 81 n 00

X90010

Cng200

b) Bng kt qu thLn thS ngu nhinS xe ra c trong ngyLn thS ngu nhinS xe ra c trong ngyLn thS ngu nhinS xe ra c trong ngy

1828811515094

2577927616255

36871079717366

42861190818777

50541287819697

69481392820024

703414416

Nh vy sau 20 ln th tng s xe c ra l 125 vy trung bnh mt ngy ra c 6,25 chic xeBi 6.2a) Bng phn phi xc sut tch lu v phn khong cc s ngu nhin ca bin X (s cc chuyn tu n mi ngy) XXc sutXc sut tch luKhong cc s ngu nhin

00,130,13T 01 n 13

10,170,3T 14 n 30

20,150,45T 31 n 45

30,250,7T 46 n 70

40,20,9T 71 n 90

50,11T 91 n 00

Bng phn phi xc sut tch lu v phn khong cc s ngu nhin ca bin Y( s tu c bc than trong mt ngy) YXc sutXc sut tch luKhong cc s ngu nhin

10,030,03T 01 n 03

20,120,15T 04 n 15

30,40,55T 16 n 55

40,280,83T 56 n 83

50,120,95T 84 n 95

60,051T 96 n 00

Bng kt qu 15 ln th ca bin X v YCch tnh:- S tu phi ch bc than ngy hm sau = s tu phi nm ch bc than + X (s tu n mi ngy)- S tu nm ch bc than = s tu phi bc than ngy hm sau Y (s tu c bc than trong mt ngy)Ghi ch: Y = X khi Y>X, Y = Y khi Y