investmentworth investment worth. given a minimum attractive rate-of-return, be able to evaluate the...

Investment Worth

ENGM 661 Engineering

Economics for Managers

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

Investment Worth Alternatives


EUAW(MARR) > 0 Good Investment



EUAW(MARR) > 0 Good Investment

IRR > MARR Good Investment


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



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



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




Alt:FW(i*) = 0 = At(1 + i*)n - t





Alt:FW(i*) = 0 = At(1 + i*)n - t

PWrevenue(i*) = PWcosts(i*)


ExamplePW(i) = -16 + 3(P/A, i, 5) + 4(P/F, i, 5)


3,000

50

4,000

16,000



i PW(i)

12 -2.9210 -2.146 -0.374 0.645 0.12

3,000

50

4,000

16,000


i* = 5 1/4 %

i* < MARR


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

F = P(F/P,i*,16)

(F/P,i*,16) = F/P = 2.16

(1+i*)16 = 2.16

School Funding

1 2 3 16

100

216

(1+i*)16 = 2.16

16 ln(1+i*) = ln(2.16) = .7701

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

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?

School Funding

1 2 3 16

100

216


Suppose inflation = 3.5% over that same period.

School Funding

1 2 3 16

100

216


Suppose inflation = 3.5% over that same period.

di j

j

1

0493 0350

10350

. .

.

d= 1.4%

NPW > 0 Good Investment

Summary


EUAW > 0 Good Investment

Summary




Summary




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


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


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


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


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)


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


Example

SIR(t) = 3(P/A, 12%, 5) + 4(P/F, 12%, 5)16

= 3(3.6048) + 4(.5674) 16

= .818 < 1.0


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 ):

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?


How does Bill Gates changea light bulb?

He doesn’t, he declares darknessa new industry standard!!!


How many Industrial Engineers does it take to change a light bulb?


How many Industrial Engineers does it take to change a light bulb?

None, IE’s only change dark bulbs!!!!!

investmentworth investment worth. given a minimum attractive rate-of-return, be able to evaluate the...

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