Trong cong fhUc (3.15) ta th~y r~ng th\lc ra tlli thai diem t =0 khong co khoan thu nao, chi co v6n dAu tu ban dAuYo. Ngoai ra tlli thai diem n, trj s6 SV phat sinh khong pMi tir hOllt dQng san xu~t kinh doanh Cllad\l an rna tir hOllt dQng b~t thuemg (thanh If tai san). Neu ta tach cac gia trj Vo va SV ra thi dong ti~n con Illi Ia dong thu - chi tir hOllt dQng san xu~t kinh doanh cua d\l an. Trj s6 NPV luc nay co the xac djnh . theo cong thuc sau:
NPV=-V +:t(B,-C.)+ SV (3.16a)
o ,., (l+i)' (l+i)"
Neu tlli m6i thai diem co phat sinh thu chi, ta I~y thu trir chi thi dong ti~n con Il;Iila dong cac khoan thu nh?P hoan v6n N (vi C Iii chi phi v?n hanh khong ke kh~u hao nen thu nh?P hoan van N ciing chinh b~ng Igi nhu?n L cQng khllu hao - KH):
Cac dong ti~n clladg an11k nay co d\Ulgoou hinh (3.2) sau: Bt B2 Bn-I Bn BI SV • CI Ct Cn t C2 Cn_1 Vo HiOO3.2
Cong thuc (3.16a) co th~ vi6t l\1ithitoo:
NPV =-v +:i:~+ SV (3.16c)
o ,=1 (1+i)' (1+i)"
N6u trj s6 Nt Ia d~u d(in hang niim (va bfuIgN) thi cong thuc (3.16c) co th~ vi6t I\li thaOO:
NPV=-Vo+N(1+i)" -1 + SV (3.17)
i(1+i)" (1+i)"
Sau khi ki~m tra sg dang gia clla cac phuang an, phuang an nao co tri s6 NPV IOnOO~tIa phuang an t6t ook Khi hai phuang.an co s6 v6n d~u tll khac nhau, nguai ta phai gia diOOco mQt phuang an d~u tll tai chinh ph\! them vao phuang an co v6n dAu tll 006 han (trong thi truang v6n hoan hao NPV clla phu<Jllg an d~u tll tai chiOO ph\! them bfuIg 0). Khi cac phuang an co tuAi thQkhac OOauthi thai kY tinh toan d~ so slinh dugc I~y bAng bQi s6 chung 006 ooAt clla cac tuAi thQ clla cac phu<Jllg an. N6u thai ky t6n t\li clla dg an dugc xac diOOro rang, thi thai ky philn tich dugc I~y bfuIg thai ky t6n t\li do.
b. Chi tieu hi~u s6 thu chi quy v~ thai di~m tllang Iai (NFV)
Qui dAi hi~u s6 thu chi v~ thai di~m tuang Iai la qui dAi t~t ca cac khoan thu chi clla dg an v~ mQt m6c thai gian nao do trong tllang Iai (thong thuang la eu6i thai kY philn tich). Tieu chuk dlinh gia clla phuang an Ia:
va:
" "
NFV = IB"Cl+ir-' - IC.(1+i)"-';:: 0
t=O 1",,0
Tuang tg, quy dAi cac cong thuc khac v~ tllang Iai,ta co:
"
NFV"; -Vo(1 +i)" + ICB, -C,)(1 +i)"-' +SV
t=1 " (l+i)"-1 NFV=-Vo(l+i) +(B+C) . +SV I (3.18) (3.19) (3.20)
Sau khi ki~m tra S\f dang gill, phuong an nllOco tri s6 NFV IOn nhdt Iii phuong an t6t hhdt. Con vS thm kY tinh toan cling xac dinh gi6ng nhu wimg hqp quy dili vS thm di~m hi~tlili NPV.
c. Chi tieu h~ sa thu chi quy san dSu hang nam (NAV)
f)~tinh chi tieu hi~u sa thu chi san dSu hang nam, neu cac tri sa thu chi hang nam khong dSu d(in thi trucrc het phai tinh gia trj hi~n tliliCllahi~u sa thu chi (NPV) sau do theo cong thuc tinh dili tinh trj sa NA V
NAV=NPV i(l+i)" (3.21)
(I+i)"-I
Phuong an dang gia Iii phuong an co NAV > O. Con phuong an tat nhdt Iii phuong an co tri sa NA V IOnnhilt.
Neu cac tri sa thu chi pMt sinh dSu d(in hang nam thi S\f dang gia clla cac phuong an se Ill:
NAV=Bid-C'd -CR ~o (3.22)
Trong do: (adsbygoogle = window.adsbygoogle || []).push({});
B'd: Tri sa thu dSu d(in hang nam; C,: Tri s6 chi dSu d(in hang nam;
CR: Tilng sa khilu hao co ban hang nam (10' hi~u IIIA) vii liii trung binh hang nam pMi tra cho sa gia trj tili san c6 djnh chua khdu hao het & d~u meli nam (10' hi~u la R) vm gia thiet la chi phi cho d~u ill ban d~u Iii nher ngu6n v6n di vay va khdu hao co ban hang nam se duQ'ctra nQ'het.
CR=A+R (3.23) A= (V-SV) (3.24) n i(l +i)" n iV . R = (I+1')"- IL'.1(I+1~)" (3.25) Trong do:
Ve,: Gia trj tili san ca djnh con chua duQ'c tinh khilu hao co ban & nam t clla phuong an;
t: Khoang cach tinh til cu6i nam t den thm di~m b~t d~u sil d\lng phuong an. N; SV: Nhutren
Phuong an co trj s6 NA V IOn nhdt Iii phuong an t6t nhilt. Khi quy dili hi~u sa thu chi san dSu hfutg nam khong pMi quy dili thm gian tinh toan clla cac phuong an thanh bAng nhau (bQi sa chung nh6 nhilt) vll nhu v\iy pMi gia djnh rfutg cac trj s6 thu chi III)<hongthay dili qua cac thm kyo
ru~, - ~ • ~ " chit" hi' <th chi ... " "d
u" •.•.•••".ro-c ••';u.V\Ulu\'-"5 cae ..leU yuSO U - ta gIarcac VI \I sau: . d. Vi d •••sa 1: DUng chi tieu gili trj hi~ 4ri ho~c gili trj tlrang lai de chQn phuang an diu tlr thea s6 li~u nhu bang sau
Chi tieu Niim thu PAl PAl
LVon diu tlr ban db V
0 100 100 2.Thu nh~p hoan v6n N I 30 40 " 40 30 2 " 30 20 3 " 20 30 4 " 20 20
3.Gili trj con I?i 5 2 2
4.SuAt chiet khAu 5 0,1 0,1
Gilii:
+ Bucrc 1- Viet cong thuc: Vi thu nh~P hoan van N Iii khong deu d~ hang nam nen ta lip d\mg cong thuc (3 .l6c):
NPV=-V + :t~+ SV
o .=1 (l+i)' (l+i)" Holic cong thuc (3.19):
"
NFV= -Vo(l+i)" +L(B, -C,)(l +i)"-I +SV
t",1
+ Bucrc 2 - Thay sa:
- Dai vm NPV: NPV}=-100 + (30/1,1 + 40/1,12 + 30/1,13 + 20/1,14 + 20/1,15) + 2/1,15 = 10,19 NPV2=-100 + (40/1,1 + 30/1,12 + 20/1,13 + 30/1,14 + 20/1,15) + 2/1,15 = 10,33 - Dai vm NFW: NFV}=-100xl,15 + 30xl,14 + 40xl,13 + 30xl,12+ 2OxI,1 + 20 + 2 = 16,41 NFV2=-100xl,15 + 40xl,14 + 30xl,13 + 20xl,12+ 30xl,1 + 20 + 2 = 16,64 -+ Bucrc 3: So sanh, h,rachQn
NPV} <NPV2 hay dai vcriNFV
NFV} < NFV 2 .-Vrjy fa ch(Jn phucmg an 2
d. Vi d\l sA 2: Dung chi tieu hi~u sa thu chi qui ve thOi diem hi~n t?i ho~c chi tieu hi~u sa thu chi qui ve thai diem tuang lai de chQn phuang an dAu tlr theo tili li~u nhu bang sau:
Chi tieu PAl PA2
LV on dau tIr ban dau 100 150
2.Thu nh~p hoan van hang nam 30 40
3.Gia tIi con l(li 2 0
4.Tu6i thQ d\I an 5 10 (adsbygoogle = window.adsbygoogle || []).push({});
5.Suat chi~t khau 0,12 0,12
Giili:
+ Phuong phap 1:
- Buac 1 - Nh~ xet chung: Tu6i thQ cac phuong an kh6ng bfutg nhau, d~ so
. sanh chUng ta phili dung phuong phap dl1utIr b6 sung. Tnrac h~t, ta quy cac phuong an v€ cilng mQt thai h\lIl tinh toano Thai h\lIl tinh toan b~g bQi s6 chung nhi>nhat cua cac tu6i thj cac phuong an dem ra so sanh. Trong tnrang hqp C\l th~ thai h\lIl tinh toan Iii: BQi s6 chung nhi> nhat(5,10) =10. V~y phuong an I co dl1utIr b6 sung (mQt IAn).
- Buac 2 - L~p dong ti€n t~:
Chi tieu Nam PA2 PA I
VO 0 150 100 NvilSV I 40 30 2 40 30 3 40 30 4 40 30 5 40 30-100+2 6 40 30 7 40 30 8 40 30 9 40 30 10 40 30+2
- Buac 3 - Vi~t c6ng thirc: Dung c6ng thfrc (3.17)
NPV=-V +N (1+i)"-I+ SV
o i(1+ i)" (1+ I)" . Hoac cong thirc (3.20):
." (I+i)"-I
NFV=-Vo(l+l) +(B+C) . +SV
I
3030 30 30 30 30
NPV clla yang doi 1
30t 3030 30 30 30 30 30 *1)6i vai NPV: NPV1= -100 + 30x(l,1210 -1)xO,12-1x1,12-1O - lOOx1,12-5+ 2x1,12-10 = 14,54 NPV2 =-150 + 40x(1,121o - 1)xO,12-1x1,12-1O=76,00 *1)6i vai NFV: NFV1 =-100x1,1210 + 30x(1,1210 - 1)xO,12-1- 100x1,125 + 2x1,125 +2 =45,17 NFV2 = -150x1,1210 + 40x(l,1210 - 1)xO,12-1= 236,07
- Buae 5 - So sanh, 19a ehQn: NPV1 <NPV2
NFVI <NFV2
Vgyfa eh9n phUO'Ylg an 2.
+ Phuong phap 2:
Tilbuae 2 ta l~p dong tiBn t~ theo phuang ngang, sau do vai phuang an co d~u tu b6 sung thi truae hSt tinh NPV ho(ie NFV d6i vai m6i vong dol. Dong tiBn t~ hIe nay duge thay bfmg dong cae NPV ho(ie NFV co gili trj bfmg nhau va phat sinh d€u d(in t~i eacthoi diSm 1athm diSm d~u (d6i vai NPV) va thm diSm eu6i (d6i vai NFV) eua tirng vong dm. NPV va NFV ella phuang an 13.NPV va NFV duge tinh d6i vai dong tiBn t~mai l~p duge (hinh 3.3)
NPV clla yang doi 1
o 1 2 3 4 6 7 8 9 10 t
100 100
Hinh 3.3. Dong tiBn t~ ban d~u eua phuong an 1dugc thay bAng cae NPV eua m6i vong doi (thS hi~n bAng net doi)
e. Vi d•••sA 3: Dung chi tieu NPV ho(ie NFV dS so sanh Iga ehQn phuong an d~u tu theo s6li~u nhu bang 3.3 sau:
Phuangan 1 2 3 Vo ISO 200 300 N 100 105 110 SV 50 100 150 n 3 4 6 1 0,1 0,1 0,1
Cae kth qua tinh toan nhu bang sau:
Phucmg an 1 2 3 (adsbygoogle = window.adsbygoogle || []).push({});
NPV 373 432 412
NFV 1172 1357 1295
Qua ket qua tinh toan nhu bang tren, ta so chQn phucmg an:
+D6i vai NPV: NPV1 < NPV3< NPV2 +D6ivmNFV:NFV1 < NFV3<NFV2
V(iyfaeh9n phuang an 2.
f.Uu, nhu'lc di~m ella phucmg phap dUng chi tieu hi~u s6 thu chi + Uudi~m:
- Co tinh d~nsg bi~n dQng ella cae chi tieu theo thai gian; - Co tinh den gia tri ella ti~n t~ theo thm giari;
- Co th~ tinh d~n wgt gia va 1\Iffi pMt thOng qua vi~c di~u chinh cae chi tieu:
doanh thu, chi phi va trj s6 ella suAtchi~t khAu; .
- Co th~ tinh d~n nhan t6 rili ro thOng qua muc dQ tang tri s6 ella suAt chi~t
- Co th~ so sanh vai phucmg an co v6n dAutu khac nhau vm dieu ki~n liii suAt di vay va liii suAt cho vay bkg nhau mQt each gAn dung. (BB sung cho phucmg an co v6n dAu tu nh6mQt phucmg an dAu tu tai chinh b6 sung, vi d\l di vay v6n d~ ho~t dQng tren thi wang v6n. Nhrmg vi liii suAt di vay va liii suAt cho vay bkg nhau mQt each gAn dung nen hi~u s6 thu chi ella phucmg an dAu tu tai chinh b6 sung nay bkg
khong. Do do trong truang hgp thi wang v6n hoan Mo khong cAn eM y d~n di~u
ki~n v6n dAutu ban dAub~ng nhau). + Nlm'lc di~m:
~ Chi dam bao tinh chinh xac trong wang hgp thi truang v6n hoan Mo, mQt
dieu ki~n kho dam bao trong thgc t~;
- Kho dg bao chinh xac cae chi tieu cho ca dm dg an;
- K~t qua Iga chQn phucmg an ph\! thuQc rAt nhi~u vao dQ Ian ella suAt chiSt khAu i, vi~c xac dinh no rAtkho khan;
- Thuang nang do cae phucmg an co v6n dAutu it va ngful h~;
- Hi~u qua khong du'lc bi~u dien duai d~g tY s6, chua du'lc so slinh vai mQt nguang hi~u qua co trj s6 ducmg khlic 0;
- Khi xet phucmg an bB sung khi 2 phucmg an co v6n dAutu ban dAukhac nhau dii coi liii suAt vay va liii suAt di vay va suAt thu I'li nhu~n t6i thi~u (suAt chi~t khAu) la nhu nhau, mQt dieu khong phil hgp vai thgc t~.
3. Chi tieu suat thu l{1inpi tv.i (IRR) (Internal rate o/return method):
a. Y nghia cua chi tieu IRR
Sudt thu lqi nQi qU (IRR) lil muc liii sudt mit n~u dUng no lam sudt chi~t khdu d~ quy d6i cae dong tib t~ ella phucrng an thi gia tri hi~n t~i cua thu nh~p PVa se can b~g vo; gia tri hi~n t~i cua chi phi PVc, nghia lit NPV =O. Hay noi khac di IRR lit nghi~m cua phucrng trinh
(3.28) (3.26)
hay:
NPV = i: B, i: C,
,=0 (I+IRR)' ,=0 (I+IRR)'
V~ thgc chdt chi sAIRR lit sudt thu lqi tinh theo cae k~t sA con I~i cua vAn dfru tu i:J dfru cae thai do~ vit khi sir d\lIlg chi tieu IRR nhu lit muc sinh lqi do nQi bQ cua
dg an sinh ra ngum ta dii ngArocong nh~ r~ng nhUng hi~u sA thu chi ducrng thu du'lc trong qua trinh ho~t dQng cua dg an d~u du'lc dem dAu tu I~i ngay I~p mc cho dg an vai.sudt thu I'li b~g chinh tri s5 IRR, vit ngu'lc I~i nhUng hi~u sAthu chi am se du'lc bi! dlip ngay b~ng ngu6n vAn chiu liii sudt b~g chinh trisA IRR.
V~ ban chdt IRR lit kha nang cho liii cua dg an, lit liii sudt Ian nhdt co th~ vay vAn d~ dAutu mit khong bi 16. (adsbygoogle = window.adsbygoogle || []).push({});
Sg dang gia cua cae phucrng an lit khi thOa miin di~uki~n sau:
IRR ~ MAAR (3.27)
Trong do: MARR lit sudt thu I'li tAi thi~u chdp nh~ du'lc (Minimal Attractive Rate of Return).
b. Phucrng phlip xac dinh IRR
D~ tim IRR, ta dUng phucrng phlip nQi suy gAn dung.
Tren hinh 3.4 d6 thi bi~u di~n cae mAi qua h~ giiia NPV vit sudt chi~t khdu i clit tI1lChoitnh t~i di~m I. Hoitnh dQ cua I chinh lit IRR.
Dfru tien, ta dn pMi xac dinh mQt tri sA NPWI ducrng (cling nha cling tAt) tucrng frng vai gia tri sudt chi~t khdu lit il sau do ta I~i xac dinh mQt tri sANPW2 <0 (cling gAn 0 cling tAt) tucrng frng vai iz. Tren d6 thi ta thdy 2 tam giac Ale vit BID lit d6ng d~ng. V~y: AC CI -=- BD ID NPV1 _ IRR-il INPV,I i, - IRR
Tri sAIRR cAntim du'lc xac dinh theo cong thuc sau:
IRR . (' ') NPV1
NPV D IRR I A c o NPV2 ---B"
Hlnh 3.4:Dbthj bl~u diBnmlll quan h~ glua NPV vii su~t chl8t kh~u I
Phuong pMp chi sir d\lI1gchi tieu IRR d~ so saM, Iga chQn phuong an t6i llU
+ Tnrimg hgp co 2 phuang an
- N8u v6n dAutu ban dAucho tili san c6 diOOIii bfuIg OOauthi chQn phuong an nilo co IRR 16n han Iiiphuang an t6t ooAt;
- N8u v6n dAutu ban dAukMc OOauthi pMi so saM Iga chQn theo nguyen t~c "gia 56 dAutu". Nghia Iii phuang an co v6n dAu tu ban dAu 16n han duQ'c chQn Iii phuong an t6t ooAt n8u gia s6 dAutucua no dang gili (tirc IRR(~) >MARR). N8u gia s6 dAutu cua no IRR(~) <MARR thi chQn phuang an co v6n dAutu 006 hon.
+ Tnrimg hgp co OOi~uphuang an
Tnrimg hgp co OOi~uphuang an thi cac buac ti8n hanh so saOO Iga chQn phuong an duQ'cti8n hanh OOusau:
- X8p h(Ulg cac phuang an theo thir tg tang dAn cua v6n dAu tu ban dAu. LAy phuong an s60 lam phuang an C(J so tiOOtoan (phuang an s6 0co v6n dAutu=0);
- TiOOsuAtthu IQ'inQi t~i cua gia s6 dAu tu cua phuang an I so vai phuang an O. Tirc hl tiOOIRRI, n8u IRRI <MARR thi lo~i b6 phuang an I, va lAyphuang an 2
so saOOvai phuang an O. N8u IRR2 <MARR thi lo~i b6 phuang an 2, cir OOu v~y
cho d8n khi tim duQ'c phuang an th(r n nao do co IRRn > MARR hk nay se chQn
phuong an co so la phuang an thir n;
- Ti8p t\lC so saOOphuang an thir n vm phuang an n+ I bfuIg cach xae diOO chuai ti~n t~ clla gia s6 dAutu bfuIg dong ti~n t~ cua phuang an co v6n dAutu 16n trir di dong ti~n t~ clla phuang an co v6n diiu tu 006) sau do cling xac diOOtri s6 IRR clla gia s6 diiu tu n8u IRR(~) >MARR thi se 10~ibo phuang an n va phuang an (n+l) se duQ'cdimg lam phuang an co so d~ ti8p t\lc so saOO.N8u IRR(~) <MARR thi phuong an (n+ 1) bi lo~i b6 va phuang an n se ti8p t\lc so saOOval phuong an (n+2).
Vi~e so sanh tlmg e~p hai phu<Jllg an nhu v~y I~p I~i eho den khi con I~i mQt phu<Jllgan, do la phlI<Jllgan duge ehQn.
e. Vi dl} sa4: Hiiy tinh suAt thu lOi nQi t~i IRR va danh gia dg an theo s8 li~u nhu bang sau:
Chi tieu So li~u
Von dau ttl ban dau 100
Thu nh~p hang nam 55
Chi phi hang nam 25
Lai suAttien vay hang nam 12% (adsbygoogle = window.adsbygoogle || []).push({});
Tu6i thi dg an 5
Oiai:
+Buae I - L~p lu~ chung:
IRR Ia kha nang eho liii ella dg an, la liii suAt 160 nhAt co the yay v6n de dAuttl rna khong bi 16. Vi v~y de danh gia dg an, truae het ta tinh IRR ella no rBi so sanh vai liii suAttien vay hang nam. Neu IRR 160 h<JllliiisuAttien vay thi dg an la dang gia.
De tinh IRR ta ehQn i1 160 nhAt th6a man dieu ki~n NPV1>0 va i2 nh6 nhAt th6a miin dieu ki~n NPV2<0, luuy i21uon luon phili 160 h<Jllil•
Ta ehQn i2= 0,2 va il = 0,1 +Buae 2 - Tinh NPV: Ap d\lng eong thue (3.17) NPV=-V +N (l+i)"-I+ SV o i(l+i)" (l+i)" Thay s6: NPVI = - 100+(55 -25) x (1,15 -I) x o,rl x 1,1-5= 13,73 NPV2 = -100+(55 - 25) x (1,25 -I) X 0,2"1x 1,2-5= 10,27
+Buae 3 - Tinh IRR: Ap d\lng eong thue (3.28) IRR . (. .) NPV, =11+ 12 -11 NPV1 + NPV,I I Thay s6: IRR=0,1+(0,2-0,1) 13,73 -0,1572 (13,73+ 10, 27) . Hay IRR = 15,72%
+Buae 4 - Danh gia:
IRR (kha nang eho liii ella dg an) = 15,72% 160 h<Jllliii suAttien vay =12%. V~y dg an la dang gia.
d. Vi dy s6 5: Dimg chi tieu IRR de chQn phuang an dAu tu theo tili li~u OOu bangsau
Chi tieu PAl PA2 PA3
1.Von dau tu ban dau 100 190 100
2.TIm oo~ hoan van hang nam 33 35 37
3.Gia tIi con l\li .0 20 20
4.Tu6i thj dg an 5 10 10
5.Ui suAt ti~n di vay 0,1 0,1 0,1
Giai