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Chapter
feasible alternatives
single proj.
Loan Payment
Pay loan ti at the end of sur
ur interest Total owed End-of-yr remaining
of ur at ar pasment unpaid.
interest+
pay 20% of principal back urly.
ur interest Total owed End-of-yr remaining
of ur at ar pasment unpaid.
pay equally every yr.
ur interest Total owed End-of-r remaining
of ur at ar payment unpaid
multiple proj
mutually exclusive
-Pick A only
independent
- do several proj
interest rate
ex. borrow 500,000 & repay
530,000 interest = 30,000
interest rate = 30,000 = 6%
Chapter 2
→ Fm - P (1+1)
P=Fn (1+i)
+ P = A [(1+i)" -1
i(1+i)"
A = pi(+)
(1+i)-1
+ A = Fi
(1+1)"-1
+F=A[(1+i)"-1"
+ P/G & A/G
P=G(+)-in-1
12(1+i)"
A-G-n
A-61
use n!
Chapter 5
(i+1)-1
501,000
0
Fn
ใน
0122
1) PV (present value)
- MARR = opportunity cost
PV transform future cash
flow into equivalent NOW
A+2G
·A+G
10,000
0
10,000
0
10,000
11000
11000
2 1100
12100
..
0
11000
1000
11000
2000+1000
-30 00 POGO
11,000 2637.97 $312.03
12100
2
800
8800
2800
6000
2 131-2 148.23 2637.46560-25
:
:
0
S
C
200 2200
2200
0
16105-1
To
tal
13000
S239.12 2637.97 2637.47
13189-37
5 1466.1 16105.1 16105.1
MIRR % per yr.
Min rate that firm
require to meet/
Chapter 3
inflation decrease in value.
- simple interest = คิดแค่เงินต้น
I - Plijn period
ex. How long will money triple
its value if i=8%
10,000
A = 10,00
-1500
G=-500
1212 13
Sol" F = (1+i)" → aP = P(1+0.08)" →
Chapter 4
-
no compounding
• nominal rate (r) + r% per time period
ex. 1.5% per m. for 24 m.compounded
= 1.5%. (24) = 36% per 24 m. m times a
effective interest rate cis- year
=true, periodic interest rate
A+ (n-1)
aredient (6)
base
n annuity (A)-
find equivalent
present value
| 2) single proj
ex. invest 1 mill
PV>0√
PV cox
generate 200,000/yr., salvage
value 300,000 at yrs.
MARR -12%.
L salvage value value of PV = รวมหมด!
) independent proj. decision = 21 proj. can be selected
w/o budget, decide proj. independently like single
4) mutually exclusive w/
equal life = choose 1
+ Rule of 72 (1≤25%, astro0)
n-92 approx. time of investment
to double in value
P10,000 (P/A, 1, 4)
P92-500 (P/G, i, 4)
P9 = 891-892
/ex. 12% per yr., compounded
quarterly, eff quarterly rate?
127.
n"
1093
109 1.08
= 14.27 = 15 yr
Tame
1) eff rate per compounding period I
-
= 37
2) EAIR - (1+1)=-1 → ex interest rate 12%, compounded
monthly, find true annual interest rate
•EAIR = (1+ 1)^-1 = (1+ 1/2/1) - 1 = 12.68%
eff monthly rate = 1% per month
eff annual
rate
•Compounded frequently = whole amount
ex. eff rate per quarter if r-12% cm.
3) eff rate per
payment period
cp=compounding period
Lapp = payment period *
EAIR Effective
Annual Interest Rate
used in real world
4) interest rate comparison
same period
2011-16% -4%
ex. payments done every
6 m. 16% per ur. cq.
15% per ur. em. - Ims
ime 12% 1% per month
m
12
(1+-12.03%
leff per pp (1+leff per cp)"-1
15%
1.25%iamo (1+0.0125) ° -1
lam (1+0.04)2-15) Apply eff rate to single cash flows (CP & PP)
6) Applying eff rate to
cash flow series
*find eff rate per PP
ex. pay 1000 every 6 m.
Lzapproach:
ex. i-12% cq.
adjust time scale to match CP
find EAIR, count time in ur.
$1000 is deposited now, how much a urs later?
FPCF/P, 3., 12) EAIA-(+)-1-12.55%.
for 10 yrs, 6%. /yr cq, find P? 7) Cont. compounding F-P(F/P, 12.55%,3)
#CP-1&r= nominal interest rate that is compounded
Sol PP-6 mo.
ex. -15%/yr, cc. sol" a) eff annual rate: i=15-1
b) eff montly rate: r = 1.25% i
imo=5% = 1.5%
ismo (1+0.015)-1
PA(P/A, isme, 2(10))
ex. i -20%, cc. find nominal rate?
j-207.-e-1
5) mutually exclusive w/
ex. at MARR = 12% | A
first cost
diff liver find LCM
6) Future Value Analysis
r = In 1.218.2327
B
Sol LCM-12 PV-10,000-2000 (P/A, 12%., 12)
calculate F from calculated PV
10,000 19,000
Annual operating-200
-1000
Salvage value
life
4
+1000 +2000
6
2000
10.400
10,000
10,000
$10,000
- 9000 ((P/F, 127., 4) + (P/F, 12%, 8) + (P/F, 127, 12)] 7) capitalized cost-PV of proj. that last forever: cc
PV=CC= ex. 10,000 can be withdrawn from fund each yr.
to forever having return of 10%, size of fund = ?
P=A=10,000 = 100,00
10%

ページ2:

end of cumulative F
-10,000
1-2000 -12,000
12+4000 -8000
07.
7) Mo recurring (period & repeat)? | 8) Payback Period
-non-recurring cost = PV
- recurring cost clast forever)
I convert to Acc
ex. initial cost 500,000, i = 6%
annual maintainence 30,000
reconditioning 100,000/3yrs.
park is expected forever
Sol" as CC of proj
3000
A₁-3000CC,-0.06 500,000
Az 100,000 (A/F, %,3)-31411
+ Shorter PP is √
how lone to recover
initial investment
Extra: 10, 12%, syrs.
each payment=10'CA/P, 12%,5) = 277,410
pay towards interest = 12%. *10° 120,000
pay towards principal = 277, 410-120,000
= 157,410
(it PPT)
+0--P+ & NCF, (P/F, 1%,+)|
→ 2 Forms
+=
21-12%
End of
r. (CF)
(P/F, 127, +) CF, (/F, 127, +) culm
-10,000
-10,000
1
-11,000
-10000
1-2000
0-89286
-1785.72 -1178572
2 4000 0.79719
3188.76 -8591-96
A = 9
+7000 +10,000
57000 0.56763 3992.00 2747.07
F = 100,000
31411
CC₂ = 0.06
523,516.67
Ct 500,000 500,000 + 523,516.67
A
diff from PV bec.
if alternatives have
unequal life just AV!
3) CR (capital recovery)
Alternative selection
using annual value
ex. MAAR 12% A
first cost
000'01-
Annual operating -2000
2) analysis using AV
b) CC=A= CC (0.06)
Chapter
1) AV Cannual Value)
+ annual cost of owing assets Salvage Val.
1000
→ Convert proj CF into ACR = PCA/P,i,n) - (A/F, i,n)
Life
4
ex. invest 1 mill, income
Sestimated fut. salvage val.
it is sold in yr. n
Sol"
CR₁ = -10000 (A/P, 12%, 4)
+1000 (A/F, 127, 4)
200,000 per ur for syrs
salvage val 300,000 at yrs
MARR 12%, find Annual Val.
Sol" A300,000 (A/F, 12%, 5)
A₁₂ = 10° 1 A/P, 12%, 5)
AV-200,000-A,+Az
AVA 10,000 CA/P, 12%, 4)-2000
CR (P-S) (A/P, i, n) + S(1)
5) AV of perpetual invest
last forever + p = 4 A = P(i) ≈ capitalized cost
ex. 2 locations buy 1 mill, MARR = 10% AV = Pri) = 10' (0.1) = 10
→ lower AV = lower PV, also sign =
@rent 500,000 every 30 yrs o
1 Cycle 30
AV of cycle AV of n cycles + annual cost 4000
4000
300,000
=57040/yr.
A=500,000 (A/P, 107., 30)-53640/yr.
AV-4000+53040
renting is cheaper (salvage val, renting prices)