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quy mo dau tu
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Prepared by Luu TRuong Van, M.Eng., adapted from C.H.Thi's presentation 1
QUI MO VAQUI MO VATHTHI I IEIEM M AAU TU T
CHO DCHO D AAN N
Prepared by Luu TRuong Van, M.Eng., adapted from C.H.Thi's presentation 2
QUI MO DQUI MO D AANN
Chuyen g xay ra cho d an neu quy mola qua ln hoac qua nho?
Qui mo qua nho hoac qua ln co the lamhong mot d an tot
Prepared by Luu TRuong Van, M.Eng., adapted from C.H.Thi's presentation 3
QUI MO DQUI MO D AANN
NPV r %NPV Max
Stoi u Qui mo
MNPV
MARR
MIRR
Prepared by Luu TRuong Van, M.Eng., adapted from C.H.Thi's presentation 4
QUI MO DQUI MO D AANN
TAI QUI MO TOI U: NPV Max NPV(gia so) = 0 IRR (gia so) = MARR
Prepared by Luu TRuong Van, M.Eng., adapted from C.H.Thi's presentation 5
QUI MO DQUI MO D AANN
NamQui mo
0 1 2 . . . . n NPV
NCF(S1)
NCF(S2)
NCF(Stoiu ) NPV Max
NCF(Sm)
Prepared by Luu TRuong Van, M.Eng., adapted from C.H.Thi's presentation 6
QUI MO DQUI MO D AANNNam
Qui mo0 1 2 . . . . n NPV
Gia soIRRGia so
NCF(S2 - S1) + > MARR
NCF(S3 S2) + > MARR
+ > MARR
NCF(Si SI-1) + > MARR
NCF(Stoiu SI) 0 = MARR
- < MARR
NCF(Sm) - < MARR
Prepared by Luu TRuong Van, M.Eng., adapted from C.H.Thi's presentation 7
C hart s o f N PV ,IR R ,M N PV ,M IR R
( 3 0 . 0 0 )
( 2 5 . 0 0 )
( 2 0 . 0 0 )
( 15 . 0 0 )
( 10 . 0 0 )
( 5 . 0 0 )
0 . 0 0
5 . 0 0
10 . 0 0
15 . 0 0
2 0 . 0 0
2 5 . 0 0
3 0 . 0 0
3 0 0 0 0 0 3 5 0 0 0 0 4 0 0 0 0 0 4 5 0 0 0 0 5 0 0 0 0 0 5 5 0 0 0 0 6 0 0 0 0 0 6 5 0 0 0 0 7 0 0 0 0 0 7 5 0 0 0 0 8 0 0 0 0 0 8 5 0 0 0 0 9 0 0 0 0 0
Quy mo
5 . 0 0 %
6 . 0 0 %
7 . 0 0 %
8 . 0 0 %
9 . 0 0 %
10 . 0 0 %
11. 0 0 %
12 . 0 0 %
13 . 0 0 %
14 . 0 0 %
15 . 0 0 %
NPV
MNPV
IRR
MIRR
MARR
Prepared by Luu TRuong Van, M.Eng., adapted from C.H.Thi's presentation 8
THTHI I IEIEM M AAU TU T
Luc nao la thi iem thch hp e batau d an
Luc nao la thi iem thch hp e ketthuc d an
Prepared by Luu TRuong Van, M.Eng., adapted from C.H.Thi's presentation 9
THTHI I IEIEM M AAU TU TCACAC TRC TRNG HNG HP TP TNH TOANH TOANN
Li ch rong tang lien tuc theo thi gianlch. Chi ph au t oc lap vi thi gianlch
Li ch rong tang lien tuc theo thi gianlch. Chi ph au t thay oi theo thigian lch
Chi ph va li ch khong thay oi motcach co he thong vi thi gian lch
Prepared by Luu TRuong Van, M.Eng., adapted from C.H.Thi's presentation 10
THTHI I IEIEM M AAU TU TB(tB(t) ) tangtang theotheo t, K = Constt, K = Const
B(t)
t
K=Const
Bt+1
Ktt t+1
Prepared by Luu TRuong Van, M.Eng., adapted from C.H.Thi's presentation 11
THTHI I IEIEM M AAU TU TB(tB(t) ) tangtang theotheo t, K = Constt, K = Const Neu au t thi iem t (cuoi nam t) --> Li ch thu c: Bt+1
Neu hoan au t sang thi iem t+1 (cuoi nam t+1) --> Li ch thu c: r* K t = r* K
au t thi iem t: Bt+1 > r* K t
Prepared by Luu TRuong Van, M.Eng., adapted from C.H.Thi's presentation 12
THTHI I IEIEM M AAU TU TB(tB(t) ) tangtang theotheo t, t, K(tK(t) ) tangtang theotheo tt
B(t)
t
K(t)
Bt+1
Kt
t t+1
Kt+1
Prepared by Luu TRuong Van, M.Eng., adapted from C.H.Thi's presentation 13
THTHI I IEIEM M AAU TU TB(tB(t) ) tangtang theotheo t, t, K(tK(t) ) tangtang theotheo tt Neu au t thi iem t (cuoi nam t) --> Li ch thu c: Bt+1+ (Kt+1- Kt )
Neu hoan au t sang thi iem t+1 (cuoi nam t+1) --> Li ch thu c: r* Kt
au t thi iem t: Bt+1+ (Kt+1- Kt ) > r* Kt
Prepared by Luu TRuong Van, M.Eng., adapted from C.H.Thi's presentation 14
THTHI I IEIEM KEM KET THUT THUC DC D AANN
t
SV(t)
B(t)
t t+1
Bt+1
SVt
SVt+1
Prepared by Luu TRuong Van, M.Eng., adapted from C.H.Thi's presentation 15
THTHI I IEIEM KEM KET THUT THUC DC D AANN
Neu ket thuc thi iem t (cuoi nam t) --> Li ch b mat i: Bt+1--> Li ch thu c: (SVt - SVt+1) + r*SVt
Ket thuc thi iem t: (SVt - SVt+1) + r*SVt > Bt+1