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TM 661. Multiple Investment Alternatives. Summary. NPW > 0 Good Investment. Summary. NPW > 0 Good Investment EUAW > 0 Good Investment. Summary. NPW > 0 Good Investment EUAW > 0 Good Investment IRR > MARR Good Investment. Summary.

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Tm 661

TM 661

Multiple Investment Alternatives


Summary
Summary

  • NPW > 0 Good Investment


Summary1
Summary

  • NPW > 0 Good Investment

  • EUAW > 0 Good Investment


Summary2
Summary

  • NPW > 0 Good Investment

  • EUAW > 0 Good Investment

  • IRR > MARR Good Investment


Summary3
Summary

  • NPW > 0 Good Investment

  • EUAW > 0 Good Investment

  • IRR > MARR Good Investment

    Note: If NPW > 0 EUAW > 0

    IRR > MARR


Multiple investments
Multiple Investments

  • NPWA > NPWB Choose A

    • Must use same planning horizon


Multiple investments1
Multiple Investments

  • NPWA > NPWB Choose A

    • Must use same planning horizon

  • EUAWA > EUAWB Choose A

    • Same Planning Horizon implicit in computation


Multiple investments2
Multiple Investments

  • NPWA > NPWB Choose A

    • Must use same planning horizon

  • EUAWA > EUAWB Choose A

    • Same Planning Horizon implicit in computation

  • IRRA > IRRB Choose A

    • Must use Incremental Rate-of-Return

      IRRB-A < MARR Choose A


Example
Example

Suppose we have two projects, A & B

A B Initial cost $50,000 $80,000

Annual maintenance 1,000 3,000

Increased productivity 10,000 15,000

Life 10 10

Salvage 10,000 20,000


Present worth a

10

9

9

. . .

0

1

2

3

10

50

Present Worth A

A

NPW(10) = -50 + 9(P/A,10,10) + 10(P/F,10,10)


Present worth a1

10

9

9

. . .

0

1

2

3

10

50

Present Worth A

A

NPW(10) = -50 + 9(P/A,10,10) + 10(P/F,10,10)

= -50 + 9(6.1446) + 10(.3855)


Present worth a2

10

9

9

. . .

0

1

2

3

10

50

Present Worth A

A

NPW(10) = -50 + 9(P/A,10,10) + 10(P/F,10,10)

= -50 + 9(6.1446) + 10(.3855)

= $9,156


Present worth b

20

12

12

. . .

0

1

2

3

10

Present Worth B

B

80

NPW(10) = -80 + 12(P/A,10,10) + 20(P/F,10,10)


Present worth b1

20

12

12

. . .

0

1

2

3

10

Present Worth B

B

80

NPW(10) = -80 + 12(P/A,10,10) + 20(P/F,10,10)

= -80 + 12(6.1446) + 20(.3855)


Present worth b2

20

12

12

. . .

0

1

2

3

10

Present Worth B

B

80

NPW(10) = -80 + 12(P/A,10,10) + 20(P/F,10,10)

= -80 + 12(6.1446) + 20(.3855)

= $1,445


Conclusion
Conclusion

NPWA > NPWB

Choose A


Equivalent worth

10

9

9

. . .

0

1

2

3

10

50

Equivalent Worth

A

EUAW(10) = -50(A/P,10,10) + 9 + 10(A/F,10,10)


Equivalent worth1

10

9

9

. . .

0

1

2

3

10

50

Equivalent Worth

A

EUAW(10) = -50(A/P,10,10) + 9 + 10(A/F,10,10)

= -50 (.1627) + 9 + 10(.0627)


Equivalent worth2

10

9

9

. . .

0

1

2

3

10

50

Equivalent Worth

A

EUAW(10) = -50(A/P,10,10) + 9 + 10(A/F,10,10)

= -50 (.1627) + 9 + 10(.0627)

= $1,492


Equivalent worth3

20

12

12

. . .

0

1

2

3

10

Equivalent Worth

B

EUAW(10) = -80(A/P,10,10) + 12 + 20(A/F,10,10)


Equivalent worth4

20

12

12

. . .

0

1

2

3

10

Equivalent Worth

B

EUAW(10) = -80(A/P,10,10) + 12 + 20(A/F,10,10)

= -80(.1627) + 12 + 20(.0627)


Equivalent worth5

20

12

12

. . .

0

1

2

3

10

Equivalent Worth

B

EUAW(10) = -80(A/P,10,10) + 12 + 20(A/F,10,10)

= -80(.1627) + 12 + 20(.0627)

= $238


Conclusion1
Conclusion

EUAWA > EUAWB

Choose A


Different planning horizons

30

115

1

2

3

4

5

100

100

Different Planning Horizons

Example: Suppose MARR is 10%. Suppose also that we can invest in T-bill @15% or we can invest in a 5 year automation plan.

A

B

NPW = 30(P/A,10,5) - 100

= $13,724

NPW = 115(1.1)-1 - 100

= $4,545

B


Problem
Problem

But this ignores reinvestment of T-bills for full

5-year period.

201,135

0

5

100

NPW = 201.135(P/F,10,5) - 100

= $24,889

A


Conclusion2
Conclusion

Projects must be

compared using same

Planning Horizon


Example npw

4,500

3,500

3

4,000

Example; NPW

A

NPW = -4 + 3.5(P/A, 10,3)

+ 4.5(P/F,10,3)


Example npw1

4,500

3,500

3

4,000

Example; NPW

A

NPW = -4 + 3.5(P/A, 10,3)

+ 4.5(P/F,10,3)

= -4 + 3.5(2.4869) + 4.5(.7513)

= 8.085

= $8,085


Example npw2
Example: NPW

5,000

3,000

B

3

6

NPW = -5 + 3(P/A,10,6)

+ 5(P/F,10,6)

5,000


Example npw3
Example: NPW

5,000

3,000

B

3

6

NPW = -5 + 3(P/A,10,6)

+ 5(P/F,10,6)

= -5 + 3(4.3553) + 5(.5645)

5,000


Example npw4
Example: NPW

5,000

3,000

B

3

6

NPW = -5 + 3(P/A,10,6)

+ 5(P/F,10,6)

= -5 + 3(4.3553) + 5(.5645)

= 10.888

= $10,888

5,000


Planning horizons
Planning Horizons

  • Least Common Multiple

  • Shortest Life

  • Longest Life

  • Standard Planning Horizon


Example npw5

4,500

4,500

3,500

3

6

4,000

4,000

Example; NPW

A

NPW = -4 -4(P/F,10,3) + 3.5(P/A,10,6) + 4.5(P/F,10,3)

+ 4.5(P/F,10,6)


Example npw6

4,500

4,500

3,500

3

6

4,000

4,000

Example; NPW

A

NPW = -4 -4(P/F,10,3) + 3.5(P/A,10,6) + 4.5(P/F,10,3)

+ 4.5(P/F,10,6)

= -4 + .5(P/F,10,3) + 3.5(P/A,10,6) + 4.5(P/F,10,6)


Example npw7

4,500

4,500

3,500

3

6

4,000

4,000

Example; NPW

A

NPW = -4 -4(P/F,10,3) + 3.5(P/A,10,6) + 4.5(P/F,10,3)

+ 4.5(P/F,10,6)

= -4 + .5(P/F,10,3) + 3.5(P/A,10,6) + 4.5(P/F,10,6)

= -4 + .5(.7513) + 3.5(4.3553) + 4.5(.5645)


Example npw8

4,500

4,500

3,500

3

6

4,000

4,000

Example; NPW

A

NPW = -4 -4(P/F,10,3) + 3.5(P/A,10,6) + 4.5(P/F,10,3) + 4.5(P/F,10,6)

= -4 + .5(P/F,10,3) + 3.5(P/A,10,6) + 4.5(P/F,10,6)

= -4 + .5(.7513) + 3.5(4.3553) + 4.5(.5645)

= 14.159 = $14,159


Example npw9
Example: NPW

5,000

3,000

B

3

6

NPW = -5 + 3(P/A,10,6)

+ 5(P/F,10,6)

5,000


Example npw10
Example: NPW

5,000

3,000

B

3

6

NPW = -5 + 3(P/A,10,6)

+ 5(P/F,10,6)

= -5 + 3(4.3553) + 5(.5645)

5,000


Example npw11
Example: NPW

5,000

3,000

B

3

6

NPW = -5 + 3(P/A,10,6)

+ 5(P/F,10,6)

= -5 + 3(4.3553) + 5(.5645)

= 10.888

= $10,888

5,000


Conclusion3
Conclusion

NPWA > NPWB

Choose A


4,500

3,500

3

4,000

EUAW

A

EUAW = -4(A/P,10,3) + 3.5

+ 4.5(A/F,10,3)

= -4(.4021) + 3.5 + 4.5(.3021)

= 3.251

= $3,251

Note: NPW = 3,251(P/A,10,6) = 3,251(4.3553) = $14,159


EUAW

5,000

3,000

B

3

6

EUAW = -5(A/P,10,6) + 3

+ 5(A/F,10,6)

= -5(.2296) + 3 + 5(.1296)

= 2.500

= $2,500

Note: NPW = 2,500(P/A,10,6) = $10,888

5,000


EUAW

Equivalent Uniform Annual Worth method implicitly assumes that you are comparing alternatives on a least common multiple planning horizon


Class problem
Class Problem

Two alternatives for a recreational facility are being considered. Their cash flow profiles are as follows. Using a MARR of 10%, select the preferred alternative.


Class problem1

5

4

3

2

1

1 2 3 4 5

11

Class Problem

EUAWA = -11(A/P,10,5) + 5 - 1(A/G,10,5)

= -11(.2638) + 5 - 1(1.8101)

= .2881 = $288


Class problem2

4

3

2

1 2 3

5

Class Problem

EUAWB = -5(A/P,10,3) + 2 + 1(A/G,10,3)

= -5(.4021) + 2 + 1(.9366)

= .9261 = $926


Class problem3
Class Problem

EUAWB > EUAWA

Choose B


Critical thinking

5

4

3

2

1

1 2 3 4 5

11

4

3

2

1 2 3

5

Critical Thinking

A

Use Net Present Worth and

least common multiple of lives

to compare alternatives A & B.

B


Critical thinking1

5

4

3

2

1

1 2 3 4 5

11

4

3

2

1 2 3

5

Critical Thinking

A

Use Net Present Worth and

least common multiple of lives

to compare alternatives A & B.

B

NPWA = 288(P/A,10,15)

= 288(7.6061)

= $2,191


Critical thinking2

5

4

3

2

1

1 2 3 4 5

11

4

3

2

1 2 3

5

Critical Thinking

A

Use Net Present Worth and

least common multiple of lives

to compare alternatives A & B.

NPWA = 288(P/A,10,15)

= 288(7.6061)

= $2,191

NPWB = 926(P/A,10,15)

= 926(7.6061)

= $7,043

B



Incremental analysis

226

A

B

110

1

1

100

200

IRRA = 10%

IRRB = 13%

Incremental Analysis

Suppose we have two investment alternatives


Incremental analysis1

226

A

B

110

1

1

100

200

IRRA = 10%

IRRB = 13%

Incremental Analysis

Suppose we have two investment alternatives

IRRB > IRRA Choose B


Incremental analysis2

226

A

B

110

1

1

100

200

IRRA = 10%

IRRB = 13%

Incremental Analysis

Suppose we have two investment alternatives

IRRB > IRRA Choose B


Correction
Correction

Investment alternative B costs $200. If we forego

B for $100 invested in A, we have an extra $100

which can be invested at MARR. If MARR = 20%,


Correction1

230

A

110

120

+

=

1

1

1

100

100

200

IRRA = 15%

Correction

Investment alternative B costs $200. If we forego

B for $100 invested in A, we have an extra $100

which can be invested at MARR. If MARR = 20%,


Correction2

230

226

A

B

1

1

200

200

IRRA = 15%

IRRB = 13%

Correction

IRRA > IRRB Choose A


Example 2

106,200

A

B

60,000

1

1

50,000

90,000

IRRA = 20%

IRRB = 18%

Example 2

  • Suppose we have $100,000 to spend and we have two mutually exclusive investment alternatives both of which yield returns greater than MARR = 15%.


Example 21

106,200

A

B

60,000

1

1

50,000

90,000

IRRA = 20%

IRRB = 18%

Example 2

IRRA > IRRB Choose A


Example 22

106,200

A

B

60,000

1

1

50,000

90,000

IRRA = 20%

IRRB = 18%

Example 2

IRRA > IRRB Choose A


Example 23
Example 2

106,200

A

B

60,000

1

1

50,000

90,000

NPWA = -50 + 60(1.15)-1

= $2,170

NPWB = -90 + 106.2(1.15)-1

= $2,350

NPWB > NPWA Choose B


Example 24
Example 2

Remember, we have $100,000 available in funds so we could spend an additional $50,000 above alternative A or an additional $10,000 above alternative B. If we assume we can make MARR or 15% return on our money, then


Example 25

A

60,000

57,500

117,500

+

=

1

1

1

50,000

50,000

100,000

i = 20%

i = 15%

ic = 17.5%

Example 2

if we invest in A, we have an extra $50,000 which

can be invested at MARR (15%).


Example 26

B

106,200

11,500

117,700

+

=

1

1

1

90,000

10,000

100,000

i = 18%

i = 15%

ic = 17.7%

Example 2

If we invest in B, we have an extra $10,000 which

can be invested at MARR (15%).


Example 27

117,500

117,700

A

B

1

1

100,000

100,000

IRRA = 17.5%

IRRB = 17.7%

Example 2

IRRcB > IRRcA Choose B





Other analysis
Other Analysis

  • Choose max AWj

    • max incremental Awj - i >0

  • Choose max FWj

    • max incremental FWj - i > 0

  • Choose max IRRj

    • max incremental IRRj - i > MARR

  • Choose max ERRj

    • max incremental ERRj - i > MARR


Other analysis1
Other Analysis

  • Choose max SIRj

    • max incremental SIRj - i

  • Minimum payback

  • Max capitalized worth

    • valid only if cash flows continue indefinately


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