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