investmentworth investment worth. given a minimum attractive rate-of-return, be able to evaluate the...
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Given a minimum attractive rate-of-return, be able to evaluate the investment worth of a project using Net Present Worth Equivalent Annual Worth Internal Rate of Return External Rate of Return Capitalized Cost Method
Tonight’s Learning Objectives
MARRSuppose a company can earn 12% / annum in U. S. Treasury bills
No way would they ever invest in a project earning < 12%
Def: The Investment Worth of all projects are measured at the Minimum AttractiveRate of Return (MARR) of a company.
Investment Worth
MARR is company specific utilities - MARR = 10 - 15% mutuals - MARR = 12 - 18% new venture - MARR = 20 - 30%
MARR based on firms cost of capital Price Index Treasury bills
MARR
NPW(MARR) > 0 Good Investment
EUAW(MARR) > 0 Good Investment
IRR > MARR Good Investment
Investment Worth Alternatives
Example: Suppose you buy and sell a piece of equipment.
Purchase Price $16,000 Sell Price (5 years)
$ 4,000 Annual Maintenance $ 3,000 Net Profit Contribution $ 6,000
MARR 12%Is it worth it to the company to buy the machine?
Present Worth
Present Worth
NPW = -16 + 3(P/A,12,5) + 4(P/F,12,5)
16,000
6,000
3,000
50
4,000
16,000
3,000
50
4,000
Present Worth
NPW = -16 + 3(P/A,12,5) + 4(P/F,12,5)= -16 +3(3.6048) + 4(.5674)
16,000
6,000
3,000
50
4,000
16,000
3,000
50
4,000
Present Worth
NPW = -16 + 3(P/A,12,5) + 4(P/F,12,5)= -16 +3(3.6048) + 4(.5674)= -2.916= -$2,916
16,000
6,000
3,000
50
4,000
16,000
3,000
50
4,000
Annual Worth (AW or EUAW)
AW(i) = PW(i) (A/P, i%, n) = [ At (P/F, i%, t)](A/P, i%, n)
AW(i) = Annual Worth of Investment
AW(i) > 0 **OK Investment**
Annual Worth
Annual Worth; Example
Repeating our PW example, we have
AW(12) = -16(A/P,12,5) + 3 + 4(A/F,12,5)
3,000
50
4,000
16,000
Annual Worth; Example
Repeating our PW example, we have
AW(12) = -16(A/P,12,5) + 3 + 4(A/F,12,5)
= -16(.2774) + 3 + 4(.1574)
3,000
50
4,000
16,000
Annual Worth; Example
Repeating our PW example, we have
AW(12) = -16(A/P,12,5) + 3 + 4(A/F,12,5)= -16(.2774) + 3 + 4(.1574)= -.808= -$808
3,000
50
4,000
16,000
Alternately
AW(12) = PW(12) (A/P, 12%, 5) = -2.92 (.2774) = - $810 < 0 NO
GOOD
3,000
50
4,000
16,000
Internal Rate-of-ReturnIRR - internal rate of return is that
return for which NPW(i*) = 0 i* = IRR
i* > MARR **OK Investment**
Internal Rate of Return
Internal Rate-of-ReturnIRR - internal rate of return is that
return for which NPW(i*) = 0 i* = IRR
i* > MARR **OK Investment**
Alt:FW(i*) = 0 = At(1 + i*)n - t
Internal Rate of Return
Internal Rate-of-ReturnIRR - internal rate of return is that
return for which NPW(i*) = 0 i* = IRR
i* > MARR **OK Investment**
Alt:FW(i*) = 0 = At(1 + i*)n - t
PWrevenue(i*) = PWcosts(i*)
Internal Rate of Return
ExamplePW(i) = -16 + 3(P/A, i, 5) + 4(P/F, i, 5)
Internal Rate of Return
i PW(i)
12 -2.9210 -2.146 -0.374 0.645 0.12
3,000
50
4,000
16,000
ExamplePW(i) = -16 + 3(P/A, i, 5) + 4(P/F, i, 5)
i* = 5 1/4 %
i* < MARR
Internal Rate of Return
i PW(i)
12 -2.9210 -2.146 -0.374 0.645 0.12
3,000
50
4,000
16,000
Spreadsheet Example
1
2
3
456789
10111213
A B C D E F
ExamplePeriod Cash Flow
0 (16,000)1 3,0002 3,0003 3,0004 3,0005 7,000 MARR = 12.0%
NPV = (2,916) = NPV(E9,C5:C9)+ C4PMT = (809) = -PMT(E9,5,C10)IRR = 5.2% = IRR(C4:C9,E9)
Public School Funding
0
50
100
150
200
250
1975 1980 1985 1990 1995 2000
Fu
nd
ing
Year
FSchool Funding Levels
Public School Funding
0
50
100
150
200
250
1975 1980 1985 1990 1995 2000
Fu
nd
ing
Year
FSchool Funding Levels
216%
16 yrs
School Funding
1 2 3 16
100
216
(1+i*)16 = 2.16
16 ln(1+i*) = ln(2.16) = .7701
ln(1+i*) = .0481
(1+i*) = e.0481 = 1.0493
School Funding
1 2 3 16
100
216
(1+i*)16 = 2.16
16 ln(1+i*) = ln(2.16) = .7701
ln(1+i*) = .0481
(1+i*) = e.0481 = 1.0493
i* = .0493 = 4.93%
School Funding
1 2 3 16
100
216
We know i = 4.93%, is that significant growth?
Suppose inflation = 3.5% over that same period.
School Funding
1 2 3 16
100
216
We know i = 4.93%, is that significant growth?
Suppose inflation = 3.5% over that same period.
di j
j
1
0493 0350
10350
. .
.
d= 1.4%
NPW > 0 Good Investment
EUAW > 0 Good Investment
IRR > MARR Good Investment
Note: If NPW > 0 EUAW > 0IRR > MARR
Summary
IRR Problems
1,000
4,100
5,580
2,520
n0 1 2 3Consider the following
cash flow diagram. We wish to find the InternalRate-of-Return (IRR).
IRR Problems
1,000
4,100
5,580
2,520
n0 1 2 3Consider the following
cash flow diagram. We wish to find the InternalRate-of-Return (IRR).
PWR(i*) = PWC(i*)
4,100(1+i*)-1 + 2,520(1+i*)-3 = 1,000 + 5,580(1+i*)-2
IRR Problems
Internal Rate ReturnPeriod Cash Flow
0 (1,000)1 4,1002 (5,580)3 2,520
i = NPV =5% $20.41
10% $9.0215% $2.88
20% $0.0025% ($0.96)30% ($0.91)
35% ($0.46)40% $0.0045% $0.2150% $0.0055% ($0.70)60% ($1.95)
NPV vs. Interest
($5)
$0
$5
$10
$15
$20
$25
0% 10% 20% 30% 40% 50% 60%
Interest Rate
Ne
t P
res
en
t V
alu
e
Purpose: to get around a problem of multiple roots in IRR methodNotation:
At = net cash flow of investment in period t
At , At > 0
0 , else -At , At < 0
0 , else rt = reinvestment rate (+) cash flows
(MARR) i’ = rate return (-) cash flows
External Rate of Return
Rt =
Ct =
Method
find i = ERR such that
Rt (1 + rt) n - t = Ct (1 + i’) n - t
Evaluation
If i’ = ERR > MARR Investment is Good
External Rate of Return
Example MARR = 15%
Rt (1 + .15) n - t = Ct (1 + i’) n - t
4,100(1.15)2 + 2,520 = 1,000(1 + i’)3 + 5,580(1 + i’)1
i’ = .1505
External Rate of Return
0
1
2
3
1,000
4,100
5,580
2,520
Example MARR = 15%
Rt (1 + .15) n - t = Ct (1 + i’) n - t
4,100(1.15)2 + 2,520 = 1,000(1 + i’)3 + 5,580(1 + i’)1
i’ = .1505
ERR > MARR
External Rate of Return
0
1
2
3
1,000
4,100
5,580
2,520
Example MARR = 15%
Rt (1 + .15) n - t = Ct (1 + i’) n - t
4,100(1.15)2 + 2,520 = 1,000(1 + i’)3 + 5,580(1 + i’)1
i’ = .1505
ERR > MARR Good Investment
External Rate of Return
0
1
2
3
1,000
4,100
5,580
2,520
Method 1 Let i = MARR
SIR(i) = Rt (1 + i)-t
Ct (1 + i)-t
= PW (positive flows) -
PW (negative flows)
Savings Investment Ratio
Relationships among MARR, IRR, and ERR
If IRR < MARR, then IRR < ERR < MARRIf IRR > MARR, then IRR > ERR > MARRIf IRR = MARR, then IRR = ERR = MARR
Method #2SIR(i) = At (1 + i) -t
Ct (1 + i) -t
SIR(i) = PW (all cash flows) PW (negative flows)
Savings Investment Ratio
Method #2SIR(i) = At (1 + i) -t
Ct (1 + i) -t
SIR(i) = PW (all cash flows) PW (negative flows)
Evaluation:Method 1: If SIR(t) > 1 Good Investment
Method 2: If SIR(t) > 0 Good Investment
Savings Investment Ratio
Example
SIR(t) = 3(P/A, 12%, 5) + 4(P/F, 12%, 5)16
= 3(3.6048) + 4(.5674) 16
= .818 < 1.0
Savings Investment Ratio
16
0
1 2 3 4 5
3 3 3 3
7
Example
SIR(t) = 3(P/A, 12%, 5) + 4(P/F, 12%, 5)16
= 3(3.6048) + 4(.5674) 16
= .818 < 1.0
Savings Investment Ratio
16
0
1 2 3 4 5
3 3 3 3
7
Method Find smallest value m such that
where
Co = initial investment
m = payback period investment
Payback Period
m
t = 1Rt Co
Example
Payback Period
m= 5 years
n co
1 16 3
3 16 94 16 125 16 19
2 16 6
16
0
1 2 3 4 5
3 3 3 3
7
Rtn
0
å
Perpetuity (Capitalized Cost)
Perpetuity (Capitalized Cost)• Occasionally, donors sponsor
perpetual awards or programs by a lump sum of money earning interest.
• The interest earned each period (A) equals the funds necessary to pay for the ongoing award or program.
The relationship is A = P( i )
• This concept is also called capitalized cost (where CC = P).
Perpetuity ExamplePerpetuity ExampleA donor has decided to establish a $10,000 per year scholarship. The first scholarship will be paid 5 years from today and will continue at the same time every year forever. The fund for the scholarship will be established in 8 equal payments every 6 months starting 6 months from now.
Determine the amount of each of the equal initiating payments, if funds can earn interest at the rate of 6% per year with semi-annual compounding.
Perpetuity ProblemPerpetuity ProblemGiven:
A = 10 000 per year, every year after Year 5
n = 8 payments @ 6 mo. intervals, starting @ 6 mo.
i = 6%, cpd semi-annually Find Amount of Initiating payments (Ai ):
Perpetuity ProblemPerpetuity ProblemGiven:
A = 10 000 per year, every year after Year 5
n = 8 payments @ 6 mo. intervals, starting @ 6 mo.
i = 6%, cpd semi-annually Find Amount of Initiating payments (Ai ):
A flood control project has a construction cost of $10 million, an annual maintenance cost of $100,000. If the MARR is 8%, determine the capitalized cost necessary to provide for construction and perpetual upkeep.
Class Problem
A flood control project has a construction cost of $10 million, an annual maintenance cost of $100,000. If the MARR is 8%, determine the capitalized cost necessary to provide for construction and perpetual upkeep.
Class Problem
1 2 3 4. . .
100 100 100 100
10,000
Pc = 10,000 + A/i = 10,000 + 100/.08 = 11,250
Capitalized Cost = $11.25 million
Suppose that the flood control project has major repairs of $1 million scheduled every 5 years. We now wish to re-compute the capitalized cost.
Class Problem 2
Compute an annuity for the 1,000 every 5 years:
Class Problem 2
1 2 3 4 5. . .
100 100 100 100 100
10,0001,000
Compute an annuity for the 1,000 every 5 years:
Class Problem 2
1 2 3 4 5. . .
100 100 100 100 100
10,0001,000
A = 100 + 1,000(A/F,8,5) = 100 + 1,000(.1705) = 270.5
Class Problem 2
1 2 3 4 5. . .
170.5 170.5 170.5 170.5
10,000
Pc = 10,000 + 270.5/.08
= 13,381
1 2 3 4 5. . .
100 100 100 100 100
10,0001,000
How Many to Change a Bulb?
How does Bill Gates changea light bulb?
He doesn’t, he declares darknessa new industry standard!!!