Bai Giang Quan Ly Du an - DH Bach Khoa Tp.HCM P1

8/12/2019 Bai Giang Quan Ly Du an - DH Bach Khoa Tp.HCM P1

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QUN L DNXY DNG

NCS. ThS.ng Xun Trng

E: [email protected]

[email protected]

O: HCMC University of Transport


2/65

TI LIU THAM KHO

Nguyn V Bch Uyn. Qun L D n. Trng H Bchkhoa H Ni.

Th Xun Lan. Qun L D n Xy Dng. NXB HQGTP.HCM.

Lng c Long. Cc chuyn Qun L D n. NXB

trngi hc Bch khoa TP. H Ch Minh. L Vn Kim, Ng Quang Tng. Qun L D n Bng S

Mng. Trngi hc K Thut TP.HCM, 1997.

Nguyn Vnng. Qun L D nTp 1 v 2. NXB ThngK, nm 2003

March 15, 2011 2Dang Xuan Truong, Ph.D. Candidate


3/65

NH GI KT THC HC PHN

Thi kt thc hc phn:

Hnh thc: TlunThang im : 10

Ti liu: c sdng



4/65

March 15, 2011 Dang Xuan Truong, Ph.D. Candidate 4

Qun l dn xy dngNI DUNG

Phn I. Nhng vn chung vQLDAXD

Phn II. Hnh thnh tchc QLDAPhn III. Qun l thi gian v tin dn

Phn IV. Phn phi cc ngun lc ca dn


5/65


Chng 1: Dn v qun l dn

Chng 2: Phn tch dn

Chng 3: Cc chtiu hiu qu

PHN INhng vn chung vQLDAXD


6/65March 15, 2011 6Dang Xuan Truong, Ph.D. Candidate

Khi nim dn

Cc giai on ca dn Phn loi dn

Qun l dn Cc chc nng ca qun l dn

Chng 1: Dn v QLDANI DUNG



Khi nim dn

D n l tng th cc gii php s dngcc ngun lc hu hn vn c (nh tai, nhn cng v cc ngun lc tnhin khc v tin vn) nhmem li lich cao nht cho nhu t v x hi.

Dn l cc hot ngca c nhn hay mt tchc nhm vo vic sdng c hiu qucc

ngun lc u vo tc nhng mc tiu c

th

D n l tng th nhng chnh sch,hotng v chi ph c lin quan vinhau c thit k nhm t cnhng mc tiu nhtnh

Ngn hng thgii

Khi nim khc


8/65March 15, 2011 Dang Xuan Truong, Ph.D. Candidate 8

Khi nim dn

Mt syu tca dn1. Mc tiu2. Cc ch th u tv th hng3. Nhng gii phpc thit k4. Cc ngun lcc sdng

5. Cc sn phm vt cht v dch v cc loic to ra6. Thi gian hon thnh

7. Cc hot ng din ra di cc iu kinchnh tr x hi, kinh t lut php c th cacc nc



Khi nim dn

Cc c tnh ca dnMcchNhm p ng mt nhu c u c nu ra.M c

tiu ca cc bphn thnh vin tham gia dn khxcnh.

Cc hotngCc quyt nh v cc cng vi c c quan h vinhau trong trnh t thc hin, trong s nh hngqua li , hng ti mcch

Cc rng buc

Cc rng buc v thi gian,Cc rng buc v ngun lc,

Ri ro

Mi trngu tbinng,



Cc giai on ca dnXcnh d n

Pht hin nhng lnh vc c timnng u t, hnh thnh s

b cc nhu t

Phn tch v lp d nNghin cu chi tit u t

c xut trn mi phngdin: k thut, t chc - q un l, thch x hi, thng mi, ti chnh,kinh t

Thmnh d nXc minh li ton b kt lun c a ra trong qu trnh chun

b v phn tch d n

Thc hin d n

Cc hot ng ca d n t khi btu a kinh ph vo n khi d nchm dt hotng

Nghim thu d n

Xcnh nhng thnh cng v thtbi,

Rt ra nhng kinh nghim v bihc cho qun l cc d n khctrong tng lai.



Phn loi dnTheo qui m

Dn lnDn nh

Theo mc chDn u tthay thDn u thin i hoDn u tmrngDn u tmi

Theo mi quan hgia cc dn

Dn u tc lp

Dn u tphthuc



Qun l dn

Khi nim

Qun l dn l vic huyng cc ngun lc v tchc cc cng vic thc

hin mc tiu ra



Lp khochXc nh cc cng vic cn

phi lmMc tiu v cc gii phpt c mc tiu trongphm vi gii hn vngunlc mi trng tchc

TchcHuy ng mt cch hp lcc ngun lc

Sp xp cch tin hnh cccng vic

Cc chc nng qun l dn

Lnh oHng dn v khuyn khch cc

thnh vin trong cng tc, hngvo mc tiu ca dn

Kim sotThu thp v xl thng tin vchi

ph, tin v cc tiu chun khckhi thc hin dn

iu chnh

iu chnh khach, cc hnhng, cc tiu chun cho ph hp

vi nhng thay i so vi khoch

nh



Chng 2: Phn tch dn

Cc yu cu

Ni dung

Trnh tphn tch

Yu tthi gian trong phn tch

NI DUNG



Cc yu cu

Tp hp cc chtiu xc nh mt dn l tt

hay khngLm r im mnh imyu ca dn

Gip ngi c quyn quytnh c y thng tin

cn thit



Ni dung phn tch

Thtrng Kinh t

Ti chnh

Phn bli ch Chnh tr

Php lut

Kthut



Ni dung phn tch (1)

Kinh t

Lm r hiu quca dn

theo quan im tng thquc gia

Ti chnhLm r hiu quca dn theo quanim ca cc tchc v c nhn tham

gia u tvo dn

Phn bli ch

Phn tch cc li ch v bt limang n cho cc nhm cdn khc nhau

Chnh tr

Xem xt cc nh hng v p lcc thc ca cc thlc chnh tri vi dn

Php lut

Xem xt cc thtc hnh chnh trong qu trnh lp dn c ph hpvi cc quy nh ca chnh quyn hay khng?

Xem qu trnh hot ng ca dn c vi phm cc lut php quc giahay khng?


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Thtrng

Nhu cu vsn phm hin ti, tng lai

Khch hng chyu v khch hng mi?

Khnng p ng nhu cu hin ti

Sn phm thay th

Tip thv khuyn miKhnng cnh tranh ca sn phm (Gi cv

cht lng sn phm)

Ni dung phn tch (2)


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Kthutc tnh ca sn phm v nh gi cht lng sn phm

Nghin cu kthut v phng php sn xut

Xc nh cng sut ca my mc thit bv ca dn

Nguyn vt liu u vo

Cshtng

Lao ng

a im thc hin dn

Kthut xy dng cng trnh ca dn

Ni dung phn tch (3)


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Trnh tphn tch

Phn tch tnh hnhchnh tr, kinh t,

chnh sch

Dbo sn phm v cc

ngun u vo

Xc nh chi ph

ca dn Xc nh li ch dn

nh gi hiu quu

tLi ch bng tin

Li ch khng tnh bng

tinPhn tch nhiu ch

tiu


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Yu tthi gian

Nhn dng li ch v chi ph dn

Gi trng tin theo thi gian

So snh ng tin trong cc thi kkhc nhau

Chn nm cbn

Thi on xem xt nh gi dn

Thi gian l yu tc nh hng ctnh cht quyt nhn dn

- mc tiu,- sthc hin,- chi ph,- li ch,

-


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Nhn dng chi ph v li ch Nguyn l chung

So snh

c vkhng cdn

Nm

Liich


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Nguyn l chung

Tnh tt ccc thay i vlich, vchi ph

Tt ccc thay i vli ch,vchiph gn vi mt dn phi ctnh n

Tnh nhng kt qutng thm

Li ch thm hay chi ph thm t

dn phi c tnh, m khngphi tng li ch hay tng chi ph

Loi trcc kt quchm

Cc chi ph chm v li ch chmkhng lm thay i li ch rng.

Loi trcc chi ph cnh,chi ph chung

Cc chi ph chung, chi ph c

nh khng lm bin

i l

i chrng gia cc phng n


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Yu tthi gian

Chi ph & Li ch theo thi gian

1 2 t n-1 n

0

Nm

Dng tin ca dn nm t: At = Rt - Ct

Rt Li ch ca dn nm tCt Chi ph dn nm t ( Ct = CVHT + It + Tt )

It , Tt Chi ph u t, thuthu nhp nm t


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Yu tthi gian

Ga trng tin theo thi gianGi trca 1 ng

(hm nay)

Gi trca 1 ng

(trong tng lai) u tvo sn xut sinh li

1 ng u thm nay c 2 ng trong

tng lai?

Gi tit kim ly li1 ng gi tit kim hm nay c hn 1

ng trong tng lai?

Cho vay tin ly li1 ng cho vay hm nay c 1,2 ng trong

tng lai?

>


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Yu tthi gian

Ga trng tin theo thi gian(2)

P Gi trng tin hin ti,F Gi trng tin tng lai

F = P (1 + r)n

P = F / (1 + r)n

K hiu

(1+r)n = (F/P, r, n) ~ Hstch ly

1/(1+r)n = (P/F, r, n) ~ Hschit khu

P

0

F

0 n

Tch ly

Chit khu


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Yu tthi gian

Ga trng tin theo thi gian (3)A Dng tin u hng nm

F = A (1+r)n - 1r

A

(1+r)n - 1

r

F =

(A/F, r, n)

P = A(1+r)n - 1

r(1+r)nA = P

r(1+r)n

(1+r)n-1

(F/A, r, n)

(A/P, r, n)(P/A, r, n)

0 1 2 3

t.n

A1= A2= A3=.. At=..= An=A

F

n0


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Ga trng tin theo thi gian (4)

V D:

Mt ngi mua bt ng sn theo phng thc trgp nhsau: trngay50 triu , sau 3 qu cmi qu tr5 triulin tc trong 6 qu. Nu

li sut l 8% qu th gi trhin ti ca bt ng sn ny l bao nhiu?

50 triu

5 triu

03 4 5 6 7 8

P= 50 + 5(P/A,8%,6)*(P/F, 8%,2)=69,816 triu

Hay P= 50+5(F/A,8%,6)(P/F,8%,8)


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V D:

Mt nhu tti chnh,u t20 triu u nm nay, 5 triusau 3 nm v 10triu 5 nm sau. Nu li sut l 6% nm th sau bao nhiu nm ngi c c

tng stin l 100 triu?

20 triu

5 triu

10 triu

100 triu

03 5

n

P=20 +5(P/F,6%,3)+10(P/F,6%,5)

=31,671triuF=31,671 ( )n06,01 +

n=20 nm


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Ga trng tin theo thi gian (6)V D:

Mt ngi vay 500 triu v strntheo phng thc sau: Tru n 15ln theo tng qu, ktcui qu th3. li sut theo qu l 5%. Hi mt ln trl

bao nhiu?500 triu

3 4 5 17

0

A A

?

A= 500(F/P,5%,2)(A/P,5%,15)=53,09 triu

A= 500(F/P,5%,17)(A/F,5%,15)=53,09 triu


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Mt cng ty qung co hcho vay 10 triu chyu cu khch hng trli hngthng 529.000trong 24 thng. Vy thc thcho vay vi li sut thc hng

nm l bao nhiu?

10= 0,529(P/A,r,24)

(P/A,r,24) = 10/0.529 =18,90359

Tra bng r=2%/thng Rnm= 26,82%

10 triu

0,529 triu 0,529 triu . . . . . . .. .. . . . . .

0

1 2 . . . . 23 24

V D:


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Yu tthi gian

Nm cbn

Phthuc vo chu t Nm bt u thc hin

Sau khnh thnh cng trnh

Nm c chn quy i dng tin ca dn


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Khng ng ngha vi thi gian

thc hin dn

L thi gian c xemxt nh gi trong qutrnh phn tch dn

Yu tthi gian

Thi gian dn


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Chng 3NH GI HIU QUU T

1. Cc yu cu khi so snh cc phng n

2. Chtiu NPV3. Chtiu IRR

4. Chtiu B/C5. Chtiu TP6. Mi quan hgia cc chtiu

NI DUNG


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Cng mt mi trngu t

Cng cc dliu cc dkin avo tnh tan cc phng n u

t

Cng mt hmc tiuCng cc tiu chun nh gi v cng

nguyn tc ra quyt nh

Cng mt khong thi gianthc hin

Cng vn sdngCc phng n phi a vcng qui

m vn

1. Cc yu cu khi so snh cc phng n


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2. Gi trhin ti thun NPV

Net Present Value

L gi tr quyi tt c thu nhp v chi ph ca d nv thiim hin ti (u k phn tch)

At: Dng tin ca dnMARR: Sut thu li ti thiu chp nhn cn: Thi gian thc hin dn (tnh theo n vnm)

= +=

n

t

t

t

MARR

A

NPV0 )1(


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V D= +

=n

t

t

t

MARR

ANPV

0 )1(

Cho mt dn c dng tin nhsau:

t

At

0 1

-100 110

NPV= -100+110

(1+0.1)

= 0

t

At

0 1

---100 121

NPV= -100+ 121

(1+0.1)

=10

t

At

0 1

--100 105

NPV= -100+105

(1+0.1)

=-4.5


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Gi trhin ti thun NPV (1)

NPV= 0

= +

=n

t

t

t

MARR

ANPV

0 )1(

Phng n c mc li ti thiu (=MARR)

NPV> 0

Phng n t mc li MARR v cn thu

thm mt lng bng gi trNPV

NPV< 0Phng n khng t c ti mc liMARR (ti thiu)


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Gi trhin ti thun NPV (2)

Phng n no c NPVln hn l phng ntt hn

Phng n c NPV lnnht l phng n ttnht


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V D

Cho hai phng n loi trnhau A v B c sliu nhsau:

TTCc chtiu n vtnh P A P B

1 Chi ph u tban u Triu 100 150

2 Doanh thu thun hng nm Triu 50 70

3 Chi ph vn hnh hng nm Triu 22 43

4 Gi trcn li Triu 20 0

5 MARR % 8 8

6 Thi gian thc hin Nm 5 10

Thusut thuthu nhp =0%


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V DXc nh thi gian phn tch ca dn : 10 nm v githit phng n A s

Xc nh dng tin ca cc phng n:

1 2 3 4 5 6 7 8 9 10

-150

27

0

20

-100

28

-80

1 2 3 4 5 6 7 8 9 10

0

NPVA= -100 +28(P/A,8%,10)

-80(P/F,8%,5)+20(P/F,8%,10)

NPVA=+42,69 triu

NPVB= -150 +27(P/A,8%,10)

NPVB=+31,17 triu

NPVA>NPVB Chn PAA


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Nhn xt

Net Present Value

A

B

NPV

r

r*

MARR = r* Chn A hoc BMARR = r* Chn A hoc B

MARR < r* Chn A

MARR > r* Chn B


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Tsut ni hon (1)

Internal Rate of Return - IRR

IRR l li sut m dn tora hng nmIRR cho nh u tbit chiph sdng vn cao nhtm dn c thchp nhnc


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Cng thc

Hay chnh l gi trchit khu NPV = 0

IRR biu din tlthu hi ca mi dn

Nu d n ch c t l hon vn (IRR) l bng i, th cckhon thu nhp t d n ch hon tr phn vngcu tbanu vo dn v tr li.

Mt khc, sut thu li ni ti IRR cn phn nh chi ph s dng vn ti a m nh u t c th chp nhnc.

=

=+=n

t

t

ttIRRCBNPV

0

0)1)((

Tsut ni hon (2)


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Tsut ni hon (3)

Cng thc tnh gn ng

21

1

121 )(NPVNPV

NPViiiIRR

++=

i1: L hschit khu ng vi NPV1 > 0 i2: L hschit khu ng vi NPV2 < 0

ii2

i1

IRRNPV1

NPV2


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nh gi phng n Dn c lp

IRR >MARR chp nhn phng n, phng nng gi

IRR < MARR phng n sbbc b IRR = MARR chp nhn phng n So snh cc phng n loi trnhau Nu chn phng n vi IRRmax th s c thc

li gii khc vi phng php NPV.

Tsut ni hon (3)


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Tsut ni hon (4)Nguyn tc so snh

Phng n u t ln hn ch so snh vi phng n c

u t b hn khi phng n c u t b hn l nggi theo IRR (IRR MARR)

Phng n cu t ln hnc chn khi sut thu lica gia svnu t ln hn sut thu li ti thiu chpnhn c v ngc li phng n u t b hn cchn khi sut thu li ni ti ca gia s vn u t nhhn MARR.


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Ni r mc li sut m d n c th tc.

IRRc bit hu dng khi d n vay vn u t.

Tnh ton phc tp, khi so snh ccphng n c vnu t khc nhau.

Nu c nhiu nghim, kh nh giphng n

Nhn xt


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Tsli ch v chi ph B/C

Benefit-cost (1)

B/C?

L ts gia tng gi tr

hin ti ca thu nhp vtng gi trhin ti ca chiphdn


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Cng thc

=

=

+

+== n

t

t

t

n

t

t

t

iC

iB

PVC

PVB

C

B

0

0

)1(

)1(


Benefit-cost (2)

=

=

+

+

= n

tt

t

n

tt

t

MARR

C

MARR

R

CB

0

0

)1(

)1(/


51/65



Benefit-cost (3) nh gi phng n theo tiu chun B/CCc phng n c lp:B/C >= 1 Chp nhnB/C < 1 Loi b

Cc phng n loi trnhau:nh gi nhchtiu IRR Tiu chun B/C mang tnh tngi B/C c p dng rng ri trong vic phn tch

vnh gi cc dn c qui m khc nhau.


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Benefit-cost (4)K hiu 1

B Thu nhp hin ti hng nm I Vn u t O Chi ph vn hnh M Chi ph bo dng

M)+(O+I

B=B/C

orM)]+(O+[IP

[B]P

=CostTotal

BenefitTotal

=B/C


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Benefit-cost (5)K hiu 2

B Thu nhp hin ti hng nm CRVn u t O Chi ph vn hnh M Chi ph bo dng

M)+(O+CR

B=B/C

orM)+(O+[CRP

[B]P

=CostTotal

BenefitTotal

=B/C


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Tsli ch vn u t

=

=

+

+

=n

tt

t

n

tt

vhtt

MARRI

MARRCR

CB

0

0

)1(

)1()(

/

B/C =

Gi trhin ti ca li ch rng

Gi trhin ti ca chi ph u t


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Phng n ng giB/C 1

nh giTsli ch-chi ph (B/C)


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So snh cc phng n (1)

Vn u tnhnhauPhng ntt hn

B/C cao hn

Tsli ch-chi ph (B/C)


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Vn u tnhnhau

Phng ntt nht

B/C max




58/65


Vn u tkhc nhau

B/C cao hn

Khng

chc


Phng ntt hn



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Vn u tkhc nhauPhng n

tt nhtB/C max Cha chc




60/65


u nhc im Ch r thu nhp trn mi n v vn u

t hocn v chi ph.

Khng cho chng ta bit tng li ch rngnh chtiu NPV.

B/C ph thuc vo vic la chn li sut

chit khu.


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Thi gian hon vn - Tp

Tp?

L thi gian cn thit

thu hi vn u tbanu cho dn


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Phng php thi gian hon vn

Bao gm 2 loi l thi gian hon vn ginn

v thi gian hon vn c chit khu. Thi gian hon vn Tp l khong thi gian k

vng thu hi vn u t ca d n, bng cckhon tch lu vn hng nm.

Hay khong thi gian cn thitthu hi ton

b vnu tbanu ca dn.


63/65


CH

Cho nhu t thyc lc no tin vn

thc s c thu hi. L chtiunh gi hiu qu khngy.

nh gi hiu qu chtiu ny thngikm vi cc chtiu khc.

nh gi mc ri r o ca d n.


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Mi quan hgia cc chtiu

Phng n ng gi theo NPV cng ng githeo IRR v B/C

Phng n c chn theo NPV, th cngchn theo IRR v B/C


65/65

Dang Xuan Truong, Ph.D. Candidate

KT THC PHN I

(Mi xem tip phn II)

Documents

Bai Giang Quan Ly Du an - DH Bach Khoa Tp.HCM P1