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\$350 \$350 \$350 \$350 \$350
3 5
4
1 2 8
6 10
0 7 9

-\$1500
The net present value of investing in the 5-year project twice is \$693, while the net
present value of the 10-year project remains at \$478. These net present values now can be
compared since they correspond to two investment choices that have the same life.
This approach has its limitations. On a practical level, it can become tedious to
use when the number of projects increases and the lives do not fit neatly into multiples of
each other. For example, an analyst using this approach to compare a 7-year, a 9-year and
a 13-year project would have to replicate these projects to 819 years to arrive at an
equivalent life for all three. Theoretically, it is also difficult to argue that a firmâ€™s project
choice will essentially remain unchanged over time, especially if the projects being
compared are very attractive in terms of net present value.

Illustration 6.3: Project Replication to compare projects with different lives
Suppose you are deciding whether to buy a used car, which is inexpensive but
does not give very good mileage, or a new car, which costs more but gets better mileage.
The two options are listed in Table 6.1.
Table 6.1: Expected Cash Flows on New versus Used Car
9

Used Car New Car
Initial cost \$ 3000 \$ 8000
Maintenance costs/year \$ 1500 \$ 1000
Fuel costs/mile \$ 0.20/ mile \$ 0.05/mile
Assume that you drive 5000 miles a year and that you have an opportunity cost of 15%.
This choice can be analyzed with replication:
Step 1: Replicate the projects until they have the same lifetime; in this case, that would
mean buying used cars five consecutive times and new cars four consecutive times.
A. Buy a used car every 4 years for 20 years.
|____________|___________|____________|____________|__________|
Year: 0 4 8 12 16 20
Investment -\$3000 -\$3000 -\$3000 -\$3000 -\$3000
Maintenance costs: \$ 1500 every year for 20 years
Fuel costs: \$ 1000 every year for 20 years ( 5000 miles at 20 cents a mile).
B. Buy a new car every 5 years for 20 years
|_______________|_______________|_______________|_____________|
Year: 0 5 10 15 20
Investment: -\$8000 -\$8000 -\$8000 -\$8000
Maintenance costs: \$1000 every year for 20 years
Fuel costs: \$ 250 every year for 20 years (5000 miles at 5 cents a mile)
Step 2: Compute the NPV of each stream.
NPV of replicating used cars for 20 years = -22225.61
NPV of replicating new cars for 20 years = -22762.21
The net present value of the costs incurred by buying a used car every 4 years is less
negative than the net present value of the costs incurred by buying a new car every 5
years, given that the cars will be driven 5000 miles every year. As the mileage driven
increases, however, the relative benefits of owning and driving the more efficient new car
will also increase.

Equivalent Annuities
10

We can compare projects with different lives by converting their net present
values into equivalent annuities. In this method, we convert the net present values into
annuities. Since the NPV is annualized, it can be compared legitimately across projects
with different lives. The net present value of any project can be converted into an annuity
using the following calculation.
Equivalent Annuity = Net Present Value * [A(PV,r,n)]
where
r = Project discount rate,
A(PV,r,n) = annuity factor, with a discount rate of r and an annuity of n years
Note that the net present value of each project is converted into an annuity using that
projectâ€™s life and discount rate. Thus, this approach is flexible enough to use on projects
with different discount rates and life times. Consider again the example of the 5-year and
10-year projects given in the previous section. The net present values of these projects
can be converted into annuities as follows:
Equivalent Annuity for 5-year project = \$442 * PV(A,12%,5 years) = \$ 122.62
Equivalent Annuity for 10-year project = \$478 * PV(A,12%,10 years) = \$ 84.60
The net present value of the 5-year project is lower than the net present value of the 10-
year project, but using equivalent annuities, the 5-year project yields \$37.98 more per
year than the 10-year project.
While this approach does not explicitly make an assumption of project replication,
it does so implicitly. Consequently, it will always lead to the same decision rules as the
replication method. The advantage is that the equivalent annuity method is less tedious
and will continue to work even in the presence of projects with infinite lives.

eqann.xls: This spreadsheet allows you to compare projects with different lives,
using the equivalent annuity approach.

Illustration 6.4: Equivalent Annuities To Choose Between Projects With Different Lives
Consider again the choice between a new car and a used car described in
illustration 12.4. The equivalent annuities can be estimated for the two options as
follows:
11

Step 1: Compute the net present value of each project individually (without replication)
Net present value of buying a used car = - \$3,000 - \$ 2,500 * PV(A,15%,4 years)
= - \$10,137
Net present value of buying a new car = - \$8,000 - \$ 1,250 * PV(A,15%,5 years)
= - \$12,190
Step 2: Convert the net present values into equivalent annuities
Equivalent annuity of buying a used car = -\$10,137 * (A(PV,15%, 4 years))
= -\$3,551
Equivalent annuity of buying a new car = -12,190 * (A(PV,15%, 5 years))
= -\$3,637
Based on the equivalent annuities of the two options, buying a used car is more
economical than buying a new car.

Calculating Break-even
When an investment that costs more initially but is more efficient and economical
on an annual basis is compared with a less expensive and less efficient investment, the
choice between the two will depend on how much the investments get used. For instance,
in illustration 6.4, the less expensive used car is the more economical choice if the
mileage driven is less than 5000 miles. The more efficient new car will be the better
choice if the car is driven more than 5000 miles. The break-even is the number of miles at
which the two alternatives provide the same equivalent annual cost, as is illustrated in
Figure 6.6.
12

Figure 6.6: Equivalent Annual Costs as a function of Miles Driven

\$-

\$(500)

\$(1,000)

\$(1,500)
Equivalent Annual Cost

\$(2,000)

\$(2,500)

\$(3,000)
Break Even Point
\$(3,500)

\$(4,000)

\$(4,500)

\$(5,000)
1000 2000 3000 4000 5000 6000 7000 8000 9000 10000
Miles Driven

Equivalent Annuity - Used Car Equivalent Annuity - New Car

The break-even occurs at roughly 5500 miles; if there is a reasonable chance that the
mileage driven will exceed this break-even, the new car becomes the better option.

Illustration 6.5: Using Equivalent Annuities as a General Approach for Multiple Projects
The equivalent annuity approach can be used to compare multiple projects with
different lifetimes. For instance, assume that Disney is considering three storage
alternatives for its retailing division:
Alternative Initial Investment Annual Cost Project Life
Build own storage system \$ 10 million \$ 0.5 million Infinite
Rent storage system \$ 2 million \$ 1.5 million 12 years
Use third-party storage ----- \$ 2.0 million 1 year
These projects have different lives; the equivalent annual costs have to be computed for
the comparison. If the correct cost of capital for the retail business is 12.5%, the
equivalent annual costs can be computed as follows:
Alternative Net Present Value Equivalent Annual Cost
Build own storage system \$ 14.00 million \$ 1.75 million
Rent storage system \$ 11.08 million \$ 1.83 million
13

Use third-party storage \$ 2.00 million \$ 2.00 million
Based on the equivalent annual costs, Disney should build its own storage system, even
though the initial costs are the highest for this option.

6.2. â˜ž: Mutually exclusive projects with different risk levels
Assume that the cost of the third-party storage option will increase 2.5% a year forever.
What would the equivalent annuity for this option be?
a. \$2.05 million
b. \$2.50 million
c. \$ 2 million
d. None of the above

Project Comparison Generalized
To compare projects with different lives, we can make specific assumptions about
the types of projects that will be available when the shorter term projects end. To
illustrate, we can assume that the firm will have no positive net present value projects
when its current projects end; this will lead to a decision rule whereby the net present
values of projects can be compared, even if they have different lives. Alternatively, we
can make specific assumptions about the availability and the attractiveness of projects in
the future, leading to cash flow estimates and present value computations. Again, going
back to the 5-year and 10-year projects described in figure 6.4, assume that future
projects will not be as attractive as current projects. More specifically, assume that the
annual cash flows on the second 5-year project that will be taken when the first 5-year
project ends will be \$320 instead of \$400. The net present values of these two investment
streams can be computed as shown in Figure 6.7.
14

Figure 6.7: Cash Flows on Projects with Unequal Lives: Replicated with poorer project
Five-year Project: Replicated
\$400 \$400 \$400 \$400 \$400 \$320 \$320 \$320 \$320 \$320
3 5
4
1 2 8
6 10
0 7 9

-\$1000 -\$1000 (Replication)
Longer Life Project
\$350 \$350 \$350 \$350 \$350
\$350 \$350 \$350 \$350 \$350
3 5
4
1 2 8
6 10
0 7 9

-\$1500
The net present value of the first project, replicated to have a life of 10 years, is \$ 529.
This is still higher than the net present value of \$478 of the longer life project. The firm
will still pick the shorter-life project, though the margin in terms of net present value has
shrunk.
This problem is not avoided by using internal rates of return. When the internal
rate of return of a short-term project is compared to the internal rate of return of a long
term project, there is an implicit assumption that future projects will continue to have
similar internal rates of return.

The Replacement Decision: A Special Case of Mutually Exclusive Projects
In a replacement decision, we evaluate the replacement of an existing investment
with a new one, generally because the existing investment has aged and become less
efficient. In a typical replacement decision,
the replacement of old equipment with new equipment will require an initial cash
â€¢
outflow, because the money spent on the new equipment will exceed any proceeds
obtained from the sale of the old equipment.
there will be cash inflows during the life of the new machine as a consequence of
â€¢
either the lower costs of operation arising from the newer equipment or the higher
revenues flowing from the investment. These cash inflows will be augmented by the
tax benefits accruing from the greater depreciation that will arise from the new
investment.
15

the salvage value at the end of the life of the new equipment will be the differential
â€¢
salvage value â€“â€“ i.e., the excess of the salvage value on the new equipment over the
salvage value that would have been obtained if the old equipment had been kept for
the entire period and had not been replaced.
This approach has to be modified if the old equipment has a remaining life that is much
shorter than the life of the new equipment replacing it.

replace.xls: This spreadsheet allows you to analyze a replacement decision.

Illustration 6.6: Analyzing a Replacement Decision
Bookscape would like to replace an antiquated packaging system with a new one.
The old system has a book value of \$50,000 and a remaining life of 10 years and could be
sold for \$15,000, net of capital gains taxes, right now. It would be replaced with a new
machine that costs \$150,000 and has a depreciable life of 10 years, and annual operating
costs are \$40,000 lower than with the old machine. Assuming straight line depreciation
for both the old and the new system, a 40% tax rate, and no salvage value on either
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