Quy Mo Dau Tu 14Oct2004
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description
quy mo dau tu
Transcript of Quy Mo Dau Tu 14Oct2004
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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
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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
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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
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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
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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)
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
