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QUN L D NChng 2: Khi u d n

PGS. TS. Nguyn Thng

TRNG I HC BCH KHOA TP. HCM

Khoa Ky Thuat Xay Dng - BM KTTNN

Ging vin: PGS. TS. NGUYN THNG

E-mail: [email protected] or [email protected]

Web: http://www4.hcmut.edu.vn/~nguyenthong

Tl. (08) 38 640 979 - 098 99 66 719


PGS. TS. Nguyn Thng

TI LIU THAM KHO1. Qun l d n. GS. TS. Georges Hirsch, ng

Hu o, Nguyn Tin Dng, PGS. TS. Nguyn Chn. NXBGD & CFVG 1994.

2. Qun l d n. TS. Cao Ho Thi, TS.Nguyn Thy Qunh Loan. NXBHQGTp.HCM 2004.

3. Lp & thm nh d n u t xy dng. PGS. TS. Nguyn Thng. NXBXDHN 2005.

4. Nguyn l c bn qun l d n. TS. Bi ngc Ton. NXBGTVT H Ni 2008.


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NI DUNGChng 1: M uChng 2: Khi u d nChng 3: T chc d nChng 4: Hoch nh & Lp tin d nChng 4bis: S mng CPM & PERTChng 5: Kim sot d nChng 5bis: Kim sot chi phChng 6: Kt thc d nChng 7: Gii thiu phn mm MS Project.


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KHI U D NNi dung: Hnh thnh d n. nh gi & so snh la chn d

n. D n a mc tiu. Ra quyt nh a mc tiu. Phn tch ri ro d n chn.


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HNH THNH D N


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Cc giai on hnh thnh d n: Khi nim, tng & nh ngha d n; Ra quyt nh dng hay tip tc thc hin cc giai on k tip; Nghin cu tin kh thi (Lp Bo co u t);

2


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Nghin cu kh thi (Lp D n u t) Thit k chi tit (Detail design) Thc hin d n.


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CH Nhiu giai on xut pht t

iu kin ti nguyn, nhn lc,ti chnh, hn ch.

Giai on sau thc hin vi cs d liu yu cu cng chnhxc hn.


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Khi nim & nh ngha d nTr li cu hi:D n p ng nhu cu g? D n ny c ph hp vi

chuyn mn & chin lcpht trin ca Cng ty?


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Nghin cu Tin kh thi (Lp Bo co u t)

Nhm tr li cc cu hi:D n c kh thi v mt kinh t-

x hi, ti chnh? Cc bin hay ch tiu ch yu l

g? Ngun ri ro.


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Cc ni dung phn tch Th trng K thut Ngun lc Ti chnh Kinh t-X hi Nhu cu c bn


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Nghin cu Kh thi (Lp D n u t !)

Tng cng chnh xc ca cc binu vo so vi giai on trc.

Thc hin nhiu cc nghin cu cbn hn na

Cc bin php hn ch ri ro cn cnghin cu chi tit hn

3


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Nhm tr li cc cu hi: D n c kh thi v mt kinh t-x

hi, ti chnh? Mc tin cy ca cc bin ch

yu (phn tch ri ro) Ra quyt nh c hay khng tin

hnh thit k chi tit.


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NH GI & SO SNH

LA CHN D N


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GIA TR TIEN TE THEO THI GIAN

Gia tr 1$ hom nay KHAC vi1$ va ngay hom sau !!!!


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LAI SUATi (%) = 100*(Lai tc trong 1 n v thi gian; v du

la 1 nam) / von goc)LAI NT = P.N.i

T: so tien tra lai (n).P: so von cho vay (au t).N: so thi oan (nam, thang,...) trc khi thanh

toan (rut von).i: lai suat n.


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LAI KEP (GHEP)Tien lai t von ban au sau mot chu ky gi

(thang, nam..) se c gop vao von ban au &xem nh tong von cho tnh lai vao chu ky ke!!!!.

V du: Mot ngi i vay ngan hang P=2.000.000 vi lai suat kep i=1,5%/thang va se tra cavon lan lai trong N=6 thang.

Hoi anh ta phai tra bao nhieu tien ?.


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LI GIAI:Tong von va lai sau thang th 1:

P + P*i = P(1+i)Tong von va lai sau thang th 2:

P(1+i) + P(1+i)(i) = P(1+i)(1+i)= P(1+i)2

Tong quat hoa Tong von va lai sau thang th 6:P(1+i)6

Vay cuoi thang th 6, Anh ta phai tra ca goc lan lai la: 2000000(1+0.015)6 = 2186886

Nhan xet: Gia tr nay ln hn trong trng hp tnh lai n v du trc.

4


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Quan he lai suat theo thi ky khac nhauGia thiet co P (VN) tai thi iem t=0.Goi i%/nam va lai suat thang i1%/thang.Xac nh quan he i% va i1% sao cho sau 1

nam tien nhan c theo 2 cach gi labang nhau.

Giai: Ta co: P(1+i%) = P(1+i1%)12

i1% = (1+i%)1/12 1


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CAC TRNG HP KHAC Trng hp cho 3 thang (i3%):P(1+i%) = P(1+i3%)4

i3% = (1+i%)1/4 1 hay i% = (1+i3%)4 - 1

Trng hp cho 6 thang (i6%):P(1+i%) = P(1+i6%)2

i6% = (1+i%)1/2 1 hay i% = (1+i6%)2 - 1


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V du: Lai suat tien gi vao mot ngan hang lai=10%nam. Tnh lai suat tng ng 1 thang, 6thang.

Li giai: Trng hp cho 1 thang (i1%):i1%=(1+i%)1/121=(1+10%)1/121= 0.797%/1thangNhan xet: (10%/nam)/12thang > 0.797% Trng hp cho 6 thang (i6%):i6% = (1+i%)1/2 1 = (1+i%)1/2 1

= 4.88%/(6 thang)


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BIU DNG TIN THEO THI GIAN

Phc v tnh ton chtiu nh gi hiu qu vtham s ti chnh ca dn.


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BIEU O DONG TIEN TEBieu o dong tien te la o th bieu dien cac

CF theo thi gian.

0 3 4 5 N-2 N-1 N21 .

T (nam)

Ket thuc nam th 1

Mang du -

Mang du +

Hin ti


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GI TR TIN TNG NG Dng tin 2 d n so snh

thng l khc nhau theo THIGIAN. c th SO SNH Cn

phi so snh ti CNG THI IM !Nhu cu quy i tin TNG

NG.

5


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GIA TR TNG NG Quy oi gia tr tien te CUC BO

PA

F

M T(nam)0 N


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QUY OI TNG NG A VE PP(1+i%)M = A P = A / (1+i%)M

QUY OI TNG NG A VE FF = A(1+i%)N-M A = F / (1+i%)N-M

i%/nam : suat chiet khau [?? 5%14% ??]

Xem ph lc 3


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QUY I GI TR DNG TIN PHN PHI U

Ap dung nguyen tac quy oi noi tren cho tng vect giatr A ve moc thi gian tham khao quy c tai thi iemkhao sat:

P = A / (1+i%)1 + A / (1+i%)2 + A / (1+i%)3 + A / (1+i%)4+...+ A / (1+i%)N

Tong cua mot cap so nhan co N so hang vi so hang banau la A/(1+i%), cong boi la 1/(1+i%).

A

PF

TN0


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Ta co:

Hay:

N

N

1 i% 1P A

i% 1 i%

N

N

i% 1 i%A P

1 i% 1

qquS

N

1

10

Ghi chu: Tong cap so nhan N so hang co so hang au u0 va cong boi q.

Xem ph lc 5


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V du: Mot khach hang gi tien hang namvao mot ngan hang trong suot N=20 nam.Moi nam gi vao A=1 tr. . Cho biet suatchiet khau i=9%/nam. Tnh gia tr tngng hien tai P cua dong tien nay.

020

T (nam)

A =1tr.P = ?

chiet khau i=9%/nam

P=

20

20

1 9% 11 9.128

9% 1 9%

Tr.


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V du: Mot khach hang gi P=10tr. vaomot ngan hang tai thi iem ban au. Chobiet suat chiet khau i=8%/nam. Tnh giatr tng ng cua dong tien te eu trong20 nam.

0

20

T (nam)

A =?P = 10tr.

chiet khau i=8%/nam

A=

20

20

8% 1 8%10 1.018

1 8% 1

Tr.

6


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CH TIEU ANH GIA HIEU QUA D AN


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NHOM PHNG PHAP CHIET GIAM DONG TIEN

(Discount Cash Flow, DCF)


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Nhom 1: Gia tr tng ng (Equivalent Worth). Mot gia tr hien tai (Present Worth). Mot gia tr tng lai (Future Worth) hoac Mot chuoi eu gia tr hang nam (Annual

Worth).Nhom 2: Suat thu li (Rates of Return)Nhom 3: Ty so Li ch & Chi ph (Benefic Cost

Ratio). Ten goi chung la cac phng phap dong tien

te chiet giam.


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NPV(NET PRESENT VALUE)


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XT CHUI DNG TIN At vi t=0 N


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PHNG PHAP GIA TR TNG NG HIEN TAI (NPV)

At Gi tr tin ti thi im t ca chui dng tin d n (>0 Ch u t thu vo & =0 D n ng gi.NPV < 0 D n KHNG ng gi Loi

N

0tt

t

%)i1(ANPV

7


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0

20

T (nam)

NPV

chiet khau i%/nam

TNH NPV CUA CHUOI DONG TIEN

Chu y: Gia tr NPV phu thuoc vao: Ban than gia tr chuoi dong tienGia tr suat chiet khau i%/nam

Hin ti

t

At


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CH Quy i v gi tr hin ti:

P0 & At lun CNG DU.At cng XA v tng lai P0

CNG b nh hng bi i%. Tin U T T b nh hng ca

s thay i i%, trong khi LI NHUN thu v th ngc li.

tt

0 %i1AP


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CH NPV cng GIM khi i%TNG v ngc li.


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V d: Lp th dng tin & tnh NPVca hai d n A & B. Sut chit khul 10%/nm:

Hng mc D n A D n Bu t ban u (t=0) 10 15

Chi ph hng nm(t=15)

2,2 4,3

Thu nhp hng nm(t=15)

5 8,5

Tui th (nm) 5 5


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D an A

D an B

15ty

10ty

(5-2.2=2.8)ty

t

t

4.2ty

t=5

t=5

t=0

t=0


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Giai:t 0 1 2 3 4 5

D n A -10 2.8 2.8 2.8 2.8 2.8D n B -15 4.2 4.2 4.2 4.2 4.2Ta c NPV(A) = 0,61 t & NPV(B) = 0,92 t . C 2 d n ng gia v NPV > 0.

EXCEL Hm tnh gi tr hin tiNPV(sut chit khu i, i ch dng tin)

Ch : Theo quy c trong Excel, gi tr u tin cachui cui nm 1.

8


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Bi tp 1: V biu dng tin & tnh NPV vi i=10%.

Hng mc D n A D n B

u t ban u (t=0) 20 15

Chi ph hng nm (t=16) 4,7 4,3

Thu nhp hng nm (t=16) 10,1 8,5

Tui th (nm) 6 6p s: NPV(A) = 3,52 t & NPV(B) = 3,29 t .


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Bi tp 2: Tnh NPV. Sut chit khu 10%/nm.

Hng mc D n A D n Bu t ban u (t=0) 8 12

Chi ph hng nm t=14 1,2 1,5Thu nhp hng nm t=1 4 3,9 5,3

Gi tr cn li t=4 1 2Thi k phn tch (nm) 4 4


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D an A

D an B

12ty

8ty

2.7ty

T (nam)

T (nam)

3.8ty

t=4

t=4

t=0

t=0

t=4

2ty

1ty


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t0 1 2 3 4

D n A-8 2,7 2,7 2,7 3,7

D n B-12 3,8 3,8 3,8 5,8

HD: NPV(A) = 1,24 t ngNPV(B) = 1,41 t ng

9


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Bi tp 3: V th dng tin & tnh NPV. Sutchit khu 12%/nm.

Hng mc D n Au t ban u (t=0, t=1) 4 & 6 t

Chi ph hng nm t=27 1,4 tThu nhp hng nm t=2 7 4,9 t

Gi tr cn li t=7 2 tThi k phn tch 7 nm


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IRR (Internal

Rate of Return)


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nh ngha: Suat thu li noi tai i* (Internal Rate of Return IRR).

N t*

tt 0

CF 1 i 0

Gia tr i* lam cho phng trnh (1) bang 0 goi la IRRCFt: gia tr tien tai thi iem t (+ ch thu nhap, - ch

au t)N: thi gian phan tch d an.

NPV= (1)


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C TH HIU IRR l sut chit khu i lm

cho NPV ca chui dngtin xt BNG 0. Tng qut hn IRR l

sut chit khu lm cho gitr tng ng ca dngtin xt BNG 0 (ti bt kthi im no)


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TIU CH NH GI HIU QU THEO IRRMt d n gi l ng gi khi:

IRR >= MARR(Minimum Attractive Rate of Return)Gi tr MARR ca tng Cng ty s c

tnh ton ph thuc vo: C cu vn (WACC, Weight Average

Capital Cost) MARR=WACC Tnh ri ro d n, Tnh hnh hot ng Cng ty

53


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i%

NPV

0 Suat chiet khau

Gia tr hien tai thuan

Ghi ch: Trong Excel, hm IRR cho php tnh Sut thu li ni ti vi chui tin t c sn.

EXCEL IRR(a ch)

0)i1(CFNPV t*N

0tt

i*=IRR

10


PGS. TS. Nguyn ThngPGS. Dr. Nguyn Thng


PGS. TS. Nguyn ThngPGS. Dr. Nguyn Thng


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CHU Y- Gia tr IRR cua d an ch phu

thuoc vao ban than dong tien.- Gia tr IRR cua dong tien khong

thay oi khi thay oi goc toa o cuatruc thi gian dong tien.


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V d: Tnh IRR ca d n c cc sliu sau:

u t ban u: 10 Chi ph hng nm: 2,2 Thu nhp hng nm: 5,0 Tui th: 5


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ap so: IRR=12.38%/nam

Hng dan: Tnh th dan va noi suy tuyen tnh:Gia thiet i1=12% NPV1 >0 ???? (0,093)Gia thiet i2=13% NPV2

11


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Bi tp 1: Tnh IRR ca d n A vichi ph va thu nhp (ty ng) nhsau:

u t ban u (t=0): 10 Chi ph hng nm(t=1,...): 2,2 Thu nhp hng nm(t=1,...): 5,0 Gi tr cn li (t=5): 2,0 Tui th: 5


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0 3 4 521

T (nm)

P1=5t

P0=-10 tHnh: Biu dng tin t

P2=2,2t

Gi thit MARR 15%/nm

d n ng gi IRR >= MARR

Gi tr cn li 2t


PGS. TS. Nguyn Thng

Bi tp 2: Xt 2 d n A & BB. Gi thitMARR=20%/nm. Tnh IRR(A), IRR(B). Kt lun?

D n A: D n B:

3,5 t

-10 t

1 2 3 4 5

05,1 t

-15 t

1 2 3 4 5

0


PGS. TS. Nguyn Thng

Ch :Mt chui dng tin v mt ton hc c

th cho nhiu gi tr IRR (khi dng tini du nhiu ln theo t!). V mt kinh t iu ny KHNG c

ngha. Gii quyt cn 1 IRR: Quy i mt

s gi tr u t v t=0 vi sut chitkhu hp l dng tin ch cn idu 1 ln. 64


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Nm -1 0 1 2 3 4Dng tin (t) 50 -100 0 25 25 25

65

i%

NPV Dng tin c nhiu gi tr IRR !

0

V d: Cho d n c chui dng tin nh sau:


PGS. TS. Nguyn Thng

Nm Dng tin (t) i1=35% i2=63%.-1 50 50,0 50,00 -100 -74,1 -61,31 0 0,0 0,02 25 10,2 5,83 25 7,5 3,54 25 5,6 2,2

NPV -0,8 0,1

66

12


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khc phc thng thng dngMARR nh l sut chit khu quyi mt s khon u t ban u

V d trn ly MARR=10%/nm quyi 50 t t=-1 v t=0 55 t.

Kt hp ta c ti t=0 - 45 t

67


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Chui dng tin ch cn idu 1 ln ch c 1 IRR (!)

Tnh IRR tng ng (=?)68

Nm 0 1 2 3 4

Dng tin (t) -45 0 25 25 25


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Nam Dng tin (t) i=19%0 -45 -45,01 0 0,02 25 17,73 25 14,84 25 12,5

NPV 0

69

IRR ca dng tin l 19%/nm


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CH TIU MIRR(Modified Internal

Rate of Return


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Khi tnh IRR ta tha nhn ctin u t cng nh thunhp sau khi d n hotng c cng mt sutchit khu l IRR. Thc raiu ny cng khng phhp lm vi thc t.

71


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nh ngha ca IRR: Gi s c mt d nvi chui dng tin nh sau. Thi gianphn tch d n l N nm (In ch u tnm th n v Pq ch thu nhp nm thq).

0

IRR1P....

IRR1P

IRR1P

IRR1I...

IRR1I

IRR1INPV

NN

2k2k

1k1k

kk

11

00

72

13


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NN2kN2k1kN1k

kk

11

00

IRR1P....IRR1PIRR1P

IRR1I...

IRR1I

IRR1I

73

FV ca thu nhp vi sut chit

khu IRRPV ca u t vi sut chit khu IRR

Chuyn v & v 2 t tha s chung (1+IRR)-N :


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phn nh thc t hn, ngita phn bit 2 loi tin ny csut chit khu khc nhau: Tin u t Sut chit khu

u t (finance rate, r2) Tin thu hi vn Sut chit

khu cho ti u t(reinvestissement rate, r1).

74


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Ch tiu MIRR c nh ngha:

)2r,tudau(PV

)1r,nhapthu(FVMIRR1 N

75

1)2r,tudau(PV

)1r,nhapthu(FVMIRR N


PGS. TS. Nguyn Thng

V d: Cho d n sau. Tnh IRRca d n. Gi thit chit khucho tin u t l 10%/nm &chit khu cho ti u t l12%/nm. Tnh MIRR.

Year Cash flow0 -10001 -40002 50003 200076


PGS. TS. Nguyn Thng

Hng dn: IRR=25.48%/nmPV(u t,10%) = -4636.36FV(thu nhp, 12%) = 7600.

year/%91.17136.4636

7600MIRR 3

77


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Bi tp: Cho d n c dng tinnh sau. Gi thit chit khu chotin u t l r2=12%/nm & chitkhu cho ti u t lr1=14%/nm. Tnh MIRR.MARR=16%/nm. Kt lun?

78

Nm 0 1 2 3 4 5Dng tin (t) 5 -10 -30 40 35 20

14


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B/C (Benefit Cost

Ratio)


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C 2 nh ngha vtiu ch B/C:T s B/C thng T s B/C hiu chnh (sa i)


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Nhom c goi la Ty so Li ch Chi ph (Benefit Cost Ratio B/C).

Co 2 CACH TNH ve ty so B/C (xem sau): Ty so gia gia tr tng ng hien tai

(NPV) cua Li ch va Chi ph. Cac gia tr tng ng dang chuoi eu

(AW) cua Li ch & Chi ph trong thi ky phan tch.

Ca 2 cach tnh eu cho ket qua GIONG NHAU.


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CCH TNH Cch 1:NPV(B)

NPV(I+O+M)NPV


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CCH TNH Cch 2:AW(B)

AW(I+O+M) T

CHUI U TNG NG


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T S B/C THNG Trng hp s dung gia tr hien tai thuan:

Trng hp s dung chuoi gia tr eu tng ng trong thi ky phan tch:

NPV(B)B / CNPV(I O M)

AW(B)B/ CAW(I O M)

D AN ANG GIA KHI B/C >= 1

15


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B thu nhap tho cua chu au t.I chi ph au t ban au.O chi ph van hanh.M chi ph bao hanh.AW(*) ch chuoi eu tng

ng(B-O-M) THU NHP NET


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T S B/C HIU CHNHTrng hp s dung gia tr hien tai

thuan:

Hoac:

NPV(B O M)B/ CNPV(I)

D AN ANG GIA KHI B/C >= 1)I(AW

)MOB(AWC/B


PGS. TS. Nguyn Thng

CHU Y Gia tr B/C cua mot d an theo 2

nh ngha khac nhau thng laKHAC NHAU (thng s dung nhngha B/C sa oi). Hai CACH TNH B/C (thng hoac

sa oi) eu cho CNG mot ketluan (ANG GIA hoac KHONGANG GIA) cho chuoi dong tienxem xet.


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CHU Y- Gi tr B/C ca d n ph thuc

vo bn thn dng tin & sut chit khu i% (i% cng ln B/C cng b).

- Mt chui dng tin mun tnhB/C phi C gi tr C (gi tr ut)!


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V du 1: Tnh ty so B/C cua mot c hoi au t co cac d lieu ban au sau (lay suat chiet khau i=8%/nam):

Chi ph ban au (t=0) (ty) 10Chi ph van hanh bao quan

(ty)2.2

Thu nhap hang nam (ty) 5

Gia tr con lai (ty) 2Tuoi tho (nam) 5


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DNG TIN

0 1 2 3 4 5

t5 t

2,2 t

10 t

2 t

16


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B/C HIU CHNH Tnh theo chui u tng ng:

Tnh theo NPV:

Ghi chu: 0.34 l chui u tng ng ca 2 ti t=5

256.15.2

2.234.5)I(AW

)MO(AW)B(AWC/B

256.110

78.832.21)I(NPV

)MOB(NPVC/B

B/C > 1 d n ng gi

Quy i gi tr 10 chui u


PGS. TS. Nguyn Thng

CH (5.34 = 5 + 0.34) vi 0.34 l gia

tr tng ng quy v hin tica gia tr cn li 2 t ti t=5. 21.32 = 1.36 + 19.96

Gi tr NPV tng ng ca chui

u 5 t

Gi tr NPV tng ng ca gi tr 2 t

ti t=5


PGS. TS. Nguyn Thng

B/C THNG Tnh theo chui u tng ng:

Tnh theo NPV :

135.12.25.2

34,5)MO(AW)I(AW

)B(AWC/B

135.178.1832.21

)MOI(NPV)B(NPVC/B

B/C > 1 d n ng gi


PGS. TS. Nguyn Thng

Tnh B/C hieu chnh cua d an sau:So lieu ban au Aau t ban au (t=0) 15Chi ph hang nam 4.3Thu nhap hang nam 8.0Gia tr con lai 3Tuoi tho (nam) 10Suat chiet khau i (%) 12

S: B/C=1.458


PGS. TS. Nguyn Thng

CAC CH TIEU SO SANH KHAC


PGS. TS. Nguyn Thng

CAC CH TIEU SO SANH KHACIM HA VN iem hoa von (hay gia tr hoa von) la gia tr

cua mot bien so nao o, v du san lng cansan xuat, so gi van hanh mot nam cua motthiet b, so nam khai thac cua mot d an lam cho tong luy tch thu nhap bang tong

luy tch au t va chi ph (khong tnh chietkhau theo thi gian).

17


PGS. TS. Nguyn Thng


IEM HOA VONV du: au t ban au cua d an A la 10 ty

ong. Biet rang lai rong mang lai cho moisan pham cua d an la 500 ong/san pham.Xac nh iem hoa von.

Giai:iem hoa von la:

san pham.9 610.10 / 500 20.10


PGS. TS. Nguyn Thng


THI GIAN HOAN VON Tp Theo nh ngha o la thi gian can

thiet e tong thu nhap rong hang namu e hoan tra von au t ban au.

Tp la mot ch tieu anh gia hieu quakinh te n gian va dung e tham khaotng oi pho bien trong phan tch kinhte.


PGS. TS. Nguyn Thng

THI GIAN HOAN VONGia tr Tp c xac nh t phng trnh sau:

P: von au t ban au t=0.CFt: gia tr dong tien thu thi iem t >0.

Tp be d an cang c anh gia cao

p

T

1tt TCFP0

p


PGS. TS. Nguyn Thng

DNG TIN

0 1 2 3 4 5 T (nm)

2.5 t

10 t

6 7

Tp=4 nm


PGS. TS. Nguyn Thng

CAC CH TIEU SO SANH KHAC Tp trong cong thc tren s dung

vi suat chiet khau i=0% nhciem cua phng phap. Tieu ch tham khao them cung vi

IRR, B/C, NPV.


PGS. TS. Nguyn Thng

CAC CH TIEU SO SANH KHACe ke en yeu to thi gian trong cach cac nh

Tp, ngi ta a he so chiet khau vao congthc tnh Tp nh sau:

P u t, CFt thu nhp ti thi im t.Ta co the lay i%=MARR la suat chiet khau

trong cong thc tren.

p

T

1tt

t T%)i1(

CFP0p

18


PGS. TS. Nguyn Thng


Bai tap : Cho d an A va B co cac solieu nh sau. Tnh thi gian hoan vonTp cua 2 d an cho 2 trng hp :

a. Khong tnh suat chiet khau trendong tien (i=0%).

b. Tnh vi suat chiet khaui=10%/nam.


PGS. TS. Nguyn Thng

So lieu ban au D an A D an B

au t ban au (ty ong) 20 12

Thu nhap hang nam (tyong/nam)

4 2

Tuoi tho (nam) 10 10


PGS. TS. Nguyn Thng

CAC CH TIEU SO SANH KHACTHI GIAN BU VON AU T BO SUNG

ay la ch tieu s dung tng oi pho bientrong cac nc thuoc khoi Lien Xo cu. V dud an ang xem xet la A va d an thay the Bco cung so lng, chat lng va thi giannhan c san pham la nh nhau:

Chon A 0AB

BAbv TC

ICCIIT


PGS. TS. Nguyn Thng

Trong o: IA, IB: von au t ban au cua phng an A

va B (gia thiet phng an A co von au tban au ln hn).

CA, CB: chi ph eu hang nam, ke ca chi phkhau hao cua phng an A va B.

Tbv, T0: thi gian bu von au t bo sung cuaphng an so sanh va gia tr chuan cuano, tnh bang nam.


PGS. TS. Nguyn Thng

SO SANH & LA CHON D AN


PGS. TS. Nguyn Thng

DNG TIN GIA S

19


PGS. TS. Nguyn Thng

Xt 2 d n A & B c dng tin nh sau(gi thit u t B > u t A):

au t(t=0)Thu

nhap-20 6.2 6.2 6.2 6.2 6.2

-25 7.6 7.6 7.6 7.6 7.6

A

B

Gia au t Gia thu nhap-5 1.4 1.4 1.4 1.4 1.4(B-A)

Dng tin gia s


PGS. TS. Nguyn Thng

A d n c th (u t nh).

B d n tin ln (u t ln).

Ch : Nu ch xt c d n Akhng thi, ta c th xem d nc th l d n KHNG LM G (u t =0, li nhun =0).


PGS. TS. Nguyn Thng

KHI NIM V PHNG PHP

GIA S SO SNH D N


PGS. TS. Nguyn Thng

Gi thit c 2 d n A&B u ng gi v B c gi tr u t LN hn A. Xem A l d n c th Lp dng tin gia s (B-A). Tnh ch tiu nh gi (NPV,B/C,IRR)

cho dng tin gia s (B-A).* Nu ch tiu cho thy ng gi Chn d n B.* Nu ch tiu cho thy KHNG ng gi Chn d n A.


PGS. TS. Nguyn Thng

SO SNH LA CHN

D N THEO TIU CH IRR


PGS. TS. Nguyn Thng

3,5 t

10 t

1 2 3 4 5

0

5,1 t

15 t

1 2 3 4 5

0

V d: Xt 2 d n A & B. Gi thit MARR=13%/nm.Tnh IRR(A), IRR(B). Xc nh d n chn theotiu ch IRR cc i & pp. gia s.

D n A: D n B:

ap so: IRR(A)=22.11%>MARR=13%/nm d n ng giIRR(B) =20.76%>MARR=13%/nm d n ng giTiu ch IRR cc i chn A.

20


PGS. TS. Nguyn Thng

SO SNH LA CHN

D N THEO TIU CH NPV


PGS. TS. Nguyn Thng

V d: Tnh NPV ca 2 d n A & B vi sutchit khu i=10%/nm. D n chn ?

Hng mc D n A

D n B

au t ban au (t=0) 10 15

Chi ph hang nam (t=15) 2,2 4,3

Thu nhap hang nam(t=15)

5 8,5

Tuoi tho (nam) 5 5


PGS. TS. Nguyn Thng

Giai:t

0 1 2 3 4 5Phng n

A -10 2.8 2.8 2.8 2.8 2.8Phng n

B -15 4.2 4.2 4.2 4.2 4.2NPV(A) = 0,61 t ng & NPV(B) = 0,92 t ng.

Chn d n B.


PGS. TS. Nguyn Thng

Bi tp: Ly li s liu trn, dng phng php gia s la chn d n kin ngh(cho cng kt lun nh p/pso snh trc tip NPV chn phng n B ?).


PGS. TS. Nguyn Thng

SO SNH LA CHN

D N THEO TIU CH B/C


PGS. TS. Nguyn Thng

So sanh phng an theo tieu ch ty so Li ch Chi phLa chon phng an theo tieu ch B/C sa oi. Lay

suat chiet khau i=12%/nam.

So lieu A Bau t ban au (t=0) 10 15Chi ph hang nam (t=1 ..) 2.2 4.3Thu nhap hang nam (t=1 ..) 5.5 7.4Gia tr con lai 0 0Tuoi tho (nam) 5 10

21


PGS. TS. Nguyn Thng

Bi tp: Ly li s liu trn, dng phng php gia s la chn d n kin ngh(cho cng kt lun ?).


PGS. TS. Nguyn Thng

LA CHN D N VI CC TIU CH KHC

NHAU (NPV, B/C, IRR) Dng cc tiu ch KHC

NHAU c th dn n sla chn d n KHCNHAU.


PGS. TS. Nguyn Thng

Bi tp: Dng tiu ch NPV, B/C & IRR lachn 2 d n A & B sau. Ly sut chit khui=10%/nm v MARR=13%/nm.

t A B0 -10.0 -12.21 4.0 4.42 4.0 4.43 4.0 4.44 4.0 4.45 4.0 4.46 4.0 4.47 4.0 4.48 4.0 4.49 4.0 4.4

10 4.0 4.4


PGS. TS. Nguyn Thng

LA CHN D N BNG PHNG PHP SO SNH

THAY TH S dng trong trng hp

so snh & chn la 2phng n c cng KHNNG PHC V.


PGS. TS. Nguyn Thng

PHAN TCH LA CHON D AN BANG P/P THAY THE

V du: Hai d an A va B co cung nanglc san xuat. Thi gian phan tch dan la 30 nam ke t luc vao van hanh.Suat chiet khau kinh te la 10%/nam(MARR=10%/nm).

Xac nh phng an kien ngh. Dongtien chi ph au t va chi ph van hanhcua 2 d an nh sau:


PGS. TS. Nguyn Thng

t

0 1

10 ty

32

0.1 ty

D an A:

D an B:

8 tyt

32

0.6 ty

HAI D N C CNG NNG LC KINH DOANH THU NHP GING

22


PGS. TS. Nguyn Thng

032

2 ty

0.5 ty

t

Dong tien d an (A-B)

ap so : NPV=0,467ty; IRR =11,37%; B/C=1,122 Chon d an A


PGS. TS. Nguyn Thng

BI TON A MC TIU


PGS. TS. Nguyn Thng

He phng trnh toan hoc cua quy hoach a muc tieu

Rang buoc:gi(Xi) = bi------------------hi(Yi) = ci

fi ch cac ham n muc tieu.Xi , Y, Z, T bien quyet nh

)T(fMin.

)Z(fMin)Y(fMax)X(fMax

HMT

in

i3

i2

i1


PGS. TS. Nguyn Thng

CC BC GII QUYT BI TON A MC TIU

1. Xac nh cac bien quyet nh.2. Thiet lap cac ham n muc tieu.3. Thiet lap cac rang buoc cho tng ham n

muc tieu (neu co).4. Xac lap cac phng an xem xet (li giai).5. Dung phng phap phan tch a muc tieu

thch hp e la chon phng an.


PGS. TS. Nguyn Thng

CAC PHNG PHAP P/TCH A MUC TIEU Mo hnh cc. Mo hnh tnh va cho iem. Mo hnh theo hieu qua va chi ph. Phng phap ra quyet nh a yeu to. Mo hnh vi li ch chung.


PGS. TS. Nguyn Thng

CAC PHNG PHAP P/TCH A MUC TIEU Mo hnh thoa hiep. Phng phap la chon. Mo hnh AHP (Analytic Hierachy

Processus)(Sap xep theo th t t n gian en phc

tap).

23


PGS. TS. Nguyn Thng

M HNH CCay la mo hnh s dung trong giai oan au

tien e co y tng chon phng an.

NHANH

NHIU

TT

R

---: LI GII 1---: LI GII 2.:LI GII 3


PGS. TS. Nguyn Thng

Theo o th neu tren ta co 3 phng an e so sanh va a ra la chon.

LI GIAI B TROIo la li giai ma cac tieu ch cua no eu

kem li giai so sanh khac.Vi v du tren ta co li giai 3 b troi bi li giai

2 li giai 3 (b troi) se b loai trc tiep.Bai toan con lai ch con so sanh gia 1 & 2.


PGS. TS. Nguyn Thng

MO HNH TNH & CHO IMCac li giai se c anh gia theo thang iem cho

tng yeu to so sanh.Nhanh Nhieu Tot Re

iem 3 2 1 3 2 1 3 2 1 3 2 1 TongA x x x x 10B x x x x 6C x x x x 8

Li giai A c kien ngh.


PGS. TS. Nguyn Thng

MO HNH THEO HIU CH & CHI PHPhng phap c minh hoa bi o th sau:

HIU CH, R

CHI PH, I

IMAX

RMIN

li gii c th

Li giai A(IA,RA)

IA

RA


PGS. TS. Nguyn Thng

Theo phng phap nay, li giai se c xac nh t cac yeu to sau:

Hieu suat R phai ln hn gia tr Rmin yeu cau.

Chi ph I khong vut qua kha nang Imax.


PGS. TS. Nguyn Thng

Phng php raquyt nh

a yu t(MFEP)

24


PGS. TS. Nguyn Thng

Mi yu t nh hng n quyt nh gn mt h s (trng s). Cc bc phn tch nh sau:

Bc 1: Lit k tt c cc yu t & gn cho yu t th j vi gi tr trng s j vi 0 < j < 1 v:

1j

j


PGS. TS. Nguyn Thng

Bc 2: c lng gi tr theo yu t. Vi yu t j ca phng n I c nh gi FEij

Bc 3: Xc nh tng nh gi phng n i:

Chn phng n Max(TWEi)

ijj

ji FE.TWE


PGS. TS. Nguyn Thng

M HNH LI CH CHUNGPhng php c tin hnh trn bn nh gi (cho im

theo thang ging hoc khc nhau) cho cc li gii vi cctiu ch xem xt khc nhau.

Trngs

Li giiTieu ch

A1 A2 . . Am

1 F1 Z1,1 . . . .2 F2 Z2,1 Z2,2 . . .. . . . . . .n Fn . . . . Zn,m

Sum S1 S2 . . Sm


PGS. TS. Nguyn Thng

n : s tiu ch xt m : s li gii kho st (s phng n so snh) Zi,j : gi tr nh gi ca li gii j theo tiu ch

i. i trng s anh gia mc o quan trong tng

i ca cac tieu ch FiCh Cc gi tr Zij trong cng mt hng phi c

cng n v. Cc gi tr Zij trong cng mt ct c th khc

n v.


PGS. TS. Nguyn Thng

NHN XTGi tr theo hng (tng tiu

ch) phi c gi tr Min &Max KHC NHAU, nuKHNG tiu ch KHNG C NGHA phn bit phng n so snh tiu ch b LOI t u.


PGS. TS. Nguyn Thng

C 3 LOI BI TON Bi ton Max Bi ton Min Bi ton hn hp

(Max+Min)

25


PGS. TS. Nguyn Thng

NHN XT BI TONV nguyn tc ln ca cc

i lng nh gi cc tiu chl KHC NHAU (thang o). V nguyn tc n v nh

gi ca cc tiu ch l KHCNHAU.


PGS. TS. Nguyn Thng

PHI LOI B N V O CA S LIU (TNG HNG) & LOI B NH

HNG LN THANG O


PGS. TS. Nguyn Thng

N THNG KPP1: S dng php bin i gi l TRUNG TM

HO & CHUN HO loi nh hng n v &thang o:

ti bin chun ho & trung tm ho (khng cn v)

Ti bin nghin cu gi tr TB ca bin T lch chun

iT

ii

TTt

T

1;0twithiti


PGS. TS. Nguyn Thng

N THNG KPP2: Bin i Ti ti theo php

bin i:

ti KHNG N V

1,0tTT

TTt iminmax

minii


PGS. TS. Nguyn Thng

BI TON MAX


PGS. TS. Nguyn Thng

TRNG HP BI TON MAXa. Bin i Zij trong mi hng i thnh cc s bij

khng th nguyn v kh thang o:

)i(minij)i(maxij

)i(minijijij ZZ

ZZb

GI TR MAX HNG i GI TR MIN HNG i

26


PGS. TS. Nguyn Thng

CH Sau khi bin i bin bij thay i trongkhong [0 1]


PGS. TS. Nguyn Thng

b. Mi tiu ch Fi s gn vih s i ty theo mc quan trng (ch quan).


PGS. TS. Nguyn Thng

C th gn gi tr i thay ixung quanh gi tr 1: i > 1 yu t quan trng, i < 1 yu t t quan trng, i =1 yu t trung ha.C th gn dng 1

ii


PGS. TS. Nguyn Thng

M HNH LI CH CHUNGTrng

sLi gii

Tieu chA1 A2 . . Am

1 F1 b1,1 . . . .

2 F2 b2,1 b2,2 . . .

. . . . . . .

n Fn . . . . bn,m

Sum S1 S2 . . SmS1 =SUM PRODUCT


PGS. TS. Nguyn Thng

c. Tnh (Sum)j cho cc phng n j:

Li gii chn s l : (Sum)j Maxi cac tieu chj cac phng an so sanh e chon

la

n

j i iji 1

(Sum) b


PGS. TS. Nguyn Thng

TRNG HP BI TON MINa. Bin i Zij trong mi hng i thnh cc s bij

khng th nguyn v loi b nh hngthang o:

b. Mi tiu ch s gn vi h s ty theo mc quan trng (ch quan).

)i(minij)i(maxij

ij)i(maxijij ZZ

ZZb

27


PGS. TS. Nguyn Thng

c. Tnh (Sum)j cho cc phng n j:

Li gii chn s l : (Sum)j Maxi cac tieu chj cac phng an so sanh e chon la

n

j i iji 1

(Sum) b


PGS. TS. Nguyn Thng

Bi tp: Mt Ch u t mun thc hin mt d nvi 4 hm mc tiu Fi khc nhau. C 4 nh thu Ajc xem xt. Kt qu nh gi tng hp trong bngsau. Xc nh nh thu chn theo phng phpnh gi li ch chung (bi ton max).

Vi: F1: Nang lc quan ly (cao nhat c anh gia 5). F2: Von lu ong (ty VN). F3 : D an tng t a thc hien 3 nam gan nhat. F4: Cong nhan lanh nghe tren 10 nam.


PGS. TS. Nguyn Thng

Ket qua anh gia c tong hp nhsau:Nha thau

Tien ch a A1 A2 A3 A4F1 1.2 5 3 1 2F2 1.0 100 80 120 140F3 0.9 2 6 8 2F4 1.0 50 30 40 60

Kien ngh nha thau theo mo hnh li ch chung ?


PGS. TS. Nguyn Thng

Hng dan: Kien ngh A4

a A1 A2 A3 A41.20 1.00 0.50 0.00 0.251.00 0.33 0.00 0.67 1.000.90 0.00 0.67 1.00 0.001.00 0.67 0.00 0.33 1.00Sum 2.20 1.20 1.90 2.30


PGS. TS. Nguyn Thng

BI TON HN HP l bi ton m cc hm n mc tiu (tiu ch) c hm Min & MaxNG THI hin hu.


PGS. TS. Nguyn Thng

PHNG PHP GIIDng php bin i thch

hp cho tng HM MCTIU. Tnh SUM nh thng. Phng n kin ngh c

SUM Max

28


PGS. TS. Nguyn Thng

Bi tp: Mt Ch u t mun chn mt trong4 nh thu Aj da vo 4 tiu ch Fi, thchin mt d n. Kt qu nh gi tng hptrong bng sau. Xc nh nh thu chn theophng php nh gi li ch chung.

F1: Vn lu ng (tr. $) F2: S cng trnh tng t thc hin trong 5

nm gn nht (n v). F3 : S k s lnh ngh trn 15 nm (ngi). F4: T l vn vay (n ngn hng) trn tng ti

sn (%).


PGS. TS. Nguyn Thng

Ket qua anh gia c tong hp nh sau:Nha thau

Tien ch a A1 A2 A3 A4F1 1.1 6 15 10 8F2 1.0 4 6 3 8F3 0.9 20 25 35 15F4 1.2 20 32 40 35

Kien ngh nha thau theo mo hnh li ch chung ?


PGS. TS. Nguyn Thng

M HNH THA HIPM hnh tha hipn gin (Bi tonMax, Min) M hnh tha hip phc tp (tng qut).


PGS. TS. Nguyn Thng

M HNH THA HIP Mo hnh thoa hiep n gian : Trng

hp bai toan co 2 muc tieu va cac hammuc tieu co cung dang (max hoac min).

Mo hnh thoa hiep phc tap : Trnghp bai toan co nhieu hn 2 muc tieu vacac ham muc tieu co dang max hoac minbat ky.


PGS. TS. Nguyn Thng

M HNH THA HIP N GIN

BI TON MAX


PGS. TS. Nguyn Thng

MO HNH THOA HIEP N GIANMt B ni tr i ch vi s tin

c. D kin mua tht Z1 (cho Chng) & mua c Z2 (cho Con). S lng mua cho bi th. Xc nhZ1 & Z2 hp l bng phng php tha hip.

29


PGS. TS. Nguyn Thng

S (li giai ly tngkhong tng

Mop

Z1,max

Z2,max

Mien li giai khong kha thi

Z1,op

Z2,op

Z2

Z1

Mien li giai tam thng

M

LSMMin

MO HNH THOA HIEP N GIAN

0


PGS. TS. Nguyn Thng

MO HNH THOA HIEP N GIANiem S co toa o Z1,max va Z2,max la li

giai ly tng nhng KHONGTHC TE. Li giai toi u se la li giai nam

tren ng cong (Z1,max Z2,max) vaGAN vi S nhat.


PGS. TS. Nguyn Thng

vi ta im trn ng cong. Cc tr hm trn s cho cp gi

tr (Z1,Z2) chnh l li gii ca biton.

0.52 2max j max j

1 1 2 2L Z Z Z Z min j j1 2(Z , Z )


PGS. TS. Nguyn Thng

V d 1: Xc nh li gii ti u cho v d sau:

X1max=15

X2max=10

0

X2

X1

(Xem Solver_Thoa Hiep Vi du 1)


PGS. TS. Nguyn Thng

DUNG SOLVER TRONG EXCEL E GIAI


PGS. TS. Nguyn Thng

KET QUA

KET QUA

30


PGS. TS. Nguyn Thng

M HNH THA HIP N GIN

BI TON Min


PGS. TS. Nguyn Thng

MO HNH THOA HIEP N GIANMot Chu au t phai a ra la chon

Cong ty thc hien d an. Co 2 tieuch c e ra la so ty le tien nngan hang cua Cong ty tren von tco va thi gian tng oi ma cacCong ty a tre han cac d an cua 3nam gan nhat.


PGS. TS. Nguyn Thng

Tm li gii ti u cho v d sau:

Thi gian tre(Z1, %)

N (Z2, %)

O 120

100

M(Z1,Z2)

O la li giai toi u khong tng


PGS. TS. Nguyn Thng

LI GIAIPhng an tot nhat se la phng an

co Z1=0 va Z2=0. Tuy nhien trongthc te khong co phng an nay. Doo phng an tot nhat se la phngan gan vi O(0,0) nhat:

min)0()0(

)()(5,02

22

1

5,022

min2

21

min1

jj

jj

ZZ

ZZZZL


PGS. TS. Nguyn Thng

M HNH THA HIP PHC TP (TNG

QUT)


PGS. TS. Nguyn Thng

M HNH THA HIP TNG QUT Bi ton nu trn thc cht l ta c gng tm

gi tr ln nht (bi ton max) v nh nht(bi ton min) c th cho 2 hm mc tiu.

Trong thc t chng ta c th gp cc trnghp tng qut hn.

S lng hm mc tiu >= 2. Mi hm mc tiu c th l cc i hoc cc

tiu (bi ton trn c 2 hm mc tiu l cci).

31


PGS. TS. Nguyn Thng

V d: Mt d n h cha thy li bao gm cc mc tiu sau:

Cc i li nhun ti v pht in; Cc tiu s ngi ti nh c; Cc i din tch ti; Cc tiu vng rng lng h b ph b.Gi s ta c 3 phng n xem xt nh

sau:


PGS. TS. Nguyn Thng

Din tch ti (ha)

Nlm(MW)

Li nhun (106$)

S dn di di(ng)

Din tch lng h

(ha)

P/A Ti Pht in

1 20000 200 2.5 15 3000 410000

2 10000 500 1.5 65 5000 480000

3 25000 300 2.5 25 4500 390000


PGS. TS. Nguyn Thng

H s gia trng cho 4 mc tiu trn l (0.3,0.3,0.2,0.2). T s liu trn, ta c gi tr ca 4 hm mc tiu Z cho 3

phng n xem xt A nh sau:Phng n Aj A1 A2 A3

Mc tiu Zi

Max Z1 (106$) 17.5 66.5 27.5 0.3

Min Z2 (ngi) 3000 5000 4500 0.3Max Z3 (ha) ti 20000 10000 25000 0.2

Min Z4 (ha)rng

410000 480000

390000 0.2


PGS. TS. Nguyn Thng

Li gii tt nht c xc nh theo tiu ch sau:

gi tr max trong hng th i nu hm mc tiu cc b l max.

gi tr min trong hng th i nu hm mc tiu cc b l min.

Quy c so lieu theo tieu ch c trnh bay theo hang. p=2 tnh khoang cach trong khong gian Euclide.

1/ pp* jni i

j i max mini 1 i i

Z ZMin(L )

Z Z

*iZ

*iZ

S HMT


PGS. TS. Nguyn Thng

Bi gii (j=1 P/n 1,):1/pp p p p

j 166.5 17.5 3000 3000 25000 20000 39000 41000L 0.3* 0.3* 0.2* 0.2*66.5 17.5 5000 3000 25000 10000 48000 39000

1/pp p

pj 1

0.2 2L 0.3 0.2*3 9

1/ pp p

j 2L 0.3 2 0.2 1/ pp p

j 339 3L 0.3 0.3*49 4


PGS. TS. Nguyn Thng

T : Vi p=1: L1 = 0.411, L2 = 0.700, L3= 0.464L1 min Phng n A1 l tt nht Vi p=2 (khng gian Euclide): L1 = 0.310,

L2 = 0.412, L3= 0.328L1 min Phng n A1 vn l tt nht

kien ngh A1

32


PGS. TS. Nguyn Thng

Bi tp: Mt d n h cha thy li baogm cc mc tiu sau:

Cc i li nhun ti v pht in; Cc tiu s ngi ti nh c; Cc i din tch ti; Cc tiu vng rng lng h b ph b.(cac gia tr i=1)


PGS. TS. Nguyn Thng

Din tchti(ha)

Nlm(MW)

Li nhun(106$)

S dndidi

Din tchlng h(ha)

P/A Ti

in

1 15000 100 3 20 2000 380000

2 20000 400 2 60 4000 480000

3 15000 200 3 25 4500 390000Dung phng phap thoa hiep e xac nh p/a kien ngh vi p=2


PGS. TS. Nguyn Thng

PHNG PHP LA CHN


PGS. TS. Nguyn Thng

Ch cho phng n tt hn chkhng cho phng n tt nht.

Li gii c th l mt tp d n. sp hn ngi ta dng ton

t sp hng R. Ch ton t R ny v nguyn

tc KHNG c tnh bc cu.


PGS. TS. Nguyn Thng

C 2 phng n Ai & Aj. Vit Ai RAj c ngha l Ai c ATHCH hn Aj:

Ai Aj


PGS. TS. Nguyn Thng

V d c 8 phng n so snh sp hngtheo tng cp nh sau: Kernel=[1,2,5]

74 1

236

8 5

33


PGS. TS. Nguyn Thng

Ch : [2] tt hn [1] nhngkhng th kt lun gia [2] &[5] v hai phng n nykhng c cng mc tiu sosnh !


PGS. TS. Nguyn Thng

BI TP


PGS. TS. Nguyn Thng

Bi 1: Hy phn bit gia Phntch ti chnh & Phn tch kinh ttrong phn tch d n.


PGS. TS. Nguyn Thng

HT CHNG

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