TM 661

1
TM 661

• Multiple Investment Alternatives

2
Summary

• NPW gt 0 Good Investment

3
Summary

• NPW gt 0 Good Investment
• EUAW gt 0 Good Investment

4
Summary

• NPW gt 0 Good Investment
• EUAW gt 0 Good Investment
• IRR gt MARR Good Investment

5
Summary (Single Investment)

• NPW gt 0 Good Investment
• EUAW gt 0 Good Investment
• IRR gt MARR Good Investment
• Note If NPW gt 0 EUAW gt 0
• IRR gt MARR

6
Multiple Investments

• NPWA gt NPWB Choose A
• Must use same planning horizon

7
Multiple Investments

• NPWA gt NPWB Choose A
• Must use same planning horizon
• EUAWA gt EUAWB Choose A
• Same Planning Horizon implicit in computation

8
Multiple Investments

• NPWA gt NPWB Choose A
• Must use same planning horizon
• EUAWA gt EUAWB Choose A
• Same Planning Horizon implicit in computation
• IRRA gt IRRB Choose A

9
Multiple Investments

• NPWA gt NPWB Choose A
• Must use same planning horizon
• EUAWA gt EUAWB Choose A
• Same Planning Horizon implicit in computation
• IRRA gt IRRB Choose A
• Must use Incremental Rate-of-Return
• IRRB-A lt MARR Choose A

10
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

11
Present Worth A

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

12
Present Worth A

• A
• NPW(10) -50 9(P/A,10,10) 10(P/F,10,10)
• -50 9(6.1446) 10(.3855)

13
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

14
Present Worth B

• B

80
NPW(10) -80 12(P/A,10,10) 20(P/F,10,10)
15
Present Worth B

• B

80
NPW(10) -80 12(P/A,10,10)
20(P/F,10,10) -80 12(6.1446) 20(.3855)
16
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
17
Conclusion

• NPWA gt NPWB
• Choose A

18
Equivalent Worth
A
EUAW(10) -50(A/P,10,10) 9 10(A/F,10,10)
19
Equivalent Worth
A
EUAW(10) -50(A/P,10,10) 9
10(A/F,10,10) -50 (.1627) 9 10(.0627)
20
Equivalent Worth
A
EUAW(10) -50(A/P,10,10) 9
10(A/F,10,10) -50 (.1627) 9 10(.0627)
1,492
21
Equivalent Worth
B
EUAW(10) -80(A/P,10,10) 12
20(A/F,10,10)
22
Equivalent Worth
B
EUAW(10) -80(A/P,10,10) 12
20(A/F,10,10) -80(.1627) 12 20(.0627)
23
Equivalent Worth
B
EUAW(10) -80(A/P,10,10) 12
20(A/F,10,10) -80(.1627) 12 20(.0627)
238
24
Conclusion
EUAWA gt EUAWB Choose A
25
Future Worth Analysis

• Is an extension of present value.
• FW PW (F/P, i, n)
• In the previous example,
• FWA 9,156 (F/P, 10,10)
• 9,156 (2.5937)
• 23747.92
• FWB (F/P, 10,10)
• 1,445 (2.5937)
• 3747.90
• A is selected.

26
Comparing alternatives with Different Planning
Horizon

• The Present Worth of the Alternatives must be
  compared over the same number of years (planning
  horizon).
• If projects have been defined over different
  planning periods, compare them over a period of
  time equal to the least common multiple (LCM) of
  their lives.

27
Example NPW
A
NPW -4 3.5(P/A, 10,3)
4.5(P/F,10,3)
28
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)
29
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
30
Example NPW
5,000
3,000
B
3
6
NPW -5 3(P/A,10,6) 5(P/F,10,6)
5,000
31
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
32
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
33
Planning Horizons

• Least Common Multiple
• Shortest Life
• Longest Life
• Standard Planning Horizon

34
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)
35
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)
36
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)
37
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
38
Example NPW
5,000
3,000
B
3
6
NPW -5 3(P/A,10,6) 5(P/F,10,6)
5,000
39
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
40
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
41
Conclusion

• NPWA gt NPWB
• Choose A

42
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
43
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
44
EUAW

• Equivalent Uniform Annual Worth
• method implicitly assumes that you are
• comparing alternatives on a least
• common multiple planning horizon
• You do not need to repeat the projects to
• find the LCM if you are Annual Worth
• analysis.

45
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.

46
Class Problem
EUAWA -11(A/P,10,5) 5 -
1(A/G,10,5) -11(.2638) 5 - 1(1.8101)
.2881 288
47
Class Problem
EUAWB -5(A/P,10,3) 2 1(A/G,10,3)
-5(.4021) 2 1(.9366) .9261 926
48
Class Problem
EUAWB gt EUAWA Choose B
49
Critical Thinking
A
Use Net Present Worth and least common multiple
of lives to compare alternatives A B.
B
50
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
51
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
52
Incremental Analysis

• Suppose we have two investment alternatives

53
Incremental Analysis

• Suppose we have two investment alternatives

IRRB gt IRRA Choose B
54
Incremental Analysis

• Suppose we have two investment alternatives

IRRB gt IRRA Choose B
55
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,
56
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,
57
Correction
IRRA gt IRRB Choose A
58
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.

59
Example 2
IRRA gt IRRB Choose A
60
Example 2
IRRA gt IRRB Choose A
61
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 gt NPWA Choose B
62
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

63
Example 2
if we invest in A, we have an extra 50,000 which
can be invested at MARR (15).
64
Example 2
If we invest in B, we have an extra 10,000 which
can be invested at MARR (15).
65
Example 2
IRRcB gt IRRcA Choose B
66
Alternative Comparison with ROR

• Define B - A , which is a new cash flow.
• If ? iB A lt MARR, select alternative A.
• If ? iB A MARR, select B.

67
Incremental Analysis
68
ROR with Different Planning Horizon

• If alternatives have different planning horizon,
  you need to repeat them (find the least common
  multiplier).

69
Differing Planning Horizons
70
Differing Planning Horizons
71
Example

• A company that manufactures magnetic membrane
  switches is investigating two production options
  that have the estimated cash flows below. (a)
  Determine which option is preferable at an
  interest rate of 20 per year. (b) If the options
  are independent, determine which are economically
  acceptable (All dollar values are in millions).
• Alternative In-house
  License
• First Cost, 40
  2
• Annual cost,/year 5
  0.2
• Annual income,/year 14 6
• Salvage value, 7
  __
• Life, years 10
  8

72
Solution

• AWIn-house -40(A/P,20,10) 5 14
  7(A/F,20,10)
• -40(0.23852) 9 7(0.03852)
• -0. 27116
• AWLicense - 2(0.2) -0.2 6
• -.04 5.8
• 5.4
• License is selected.
• B) Only the license option can be selected.

73
Example

• The manager of a canned food processing plant is
  trying
• to decide between two labeling machine. Determine
• which should be selected on the basis of rate of
  return
• with a MARR of 20 per year.
• Alternative Machine
  A Machine B
• First Cost,
  -15,000 -25,000
• Annual operating cost, -1,600
  -400
• Salvage value, 3,000
  6,000
• Life, years
  3 6

74
Solution

• Form cash flow for B-A
• Year A
  B B-A
• 0 -15,000
  -25,000 -15,000
• 1 -1,600
  -400 -1,200
• 2 -1,600
  -400 -1,200
• 3 -1600 (3000 15,000) -400
  -1,200 -12,000
• 4 -1600
  -400 -1,200
• 5 -1600
  -400 -1,200
• 6 -1600 3000 -400 6000
  -1,200 3000
• NPW -15,000 -1,200 (P/A, i, 6) -12,000(P/F,i,
  3) 3000 (P/F,i,6)0
• i26.8