FIN 403 Lecture 6 Homework
Annual Equivalent Costs
Solutions Chapter 7
7.1 A company has the choice between two different types of dies. One cost less, but it
also has a shorter life expectancy. The expected cash flows after taxes for the two
different dies are as follows:
Die
0
1
2
3
4
A
(10,000) 8,000 8,000
B
(12,000) 5,000 5,000 5,000 5,000
The cost of money of the firm is 10%. Assume that the selected die will be used for
many years to come.
One approach is to reinvest in A so that it has the same life as B, call this A*. We
can then compare the NPV’s.
A* (10,000) 8,000
8,000 8,000 8,000
(10,000)
4
10,000
8,000

 7,094.50 , IRR=38%

2
t
1.1
t 1 1.10
4
5,000
NPVB  12,000  
 3,849.30 , IRR=24.1%
t
t 1 1.10
Under this assumption A is desirable: higher NPV and also higher IRR.
CF0 = -10,000
CF1 = 8,000
CF2 = -2,000
CF3 = 8,000
CF4 = 8,000
I/YR=10
Press NPV & IRR to comput
NPVA*  10,000 
The second approach is to compute the equivalent annual cost/benefit for each
alternative—compute an annuity corresponding to the life of the alternative based on its
NPV. Since we already know the NPV for project B, the corresponding equivalent
annual annuity is computed as follows:
4
XB
 3,849.30

X B  1,214.34

t
t 1 1.10
For investment A, we must first compute the NPV of the payments, then the corresponding equivalent annual annuity:
2
NPVA*  10,000  
t 1
2
X
t 1
A
 3,844.30

8,000
 3,884.30
1.10t
X A  2,238.10
Again, A is the superior investment.
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FIN 403 Lecture 6 Homework
7.2 Assume that there are two mutually exclusive investments that have the following
cash flows:
Investment
0
1
IRR
A
(10,000) 12,000 20
B
(5,000) 6,100 22
Assume that either investment will require modification to the basic building structure,
which will cost $1,000 and that this amount is not included in the preceding
computations. The cost of money is 10%.
a) Compute the actual internal rates of return of the investments.
Solution:
12,000
 90.90 , IRR=9.1%
1.1
6,100
NPVB  5,000  1,000 
 454.50 , IRR=1.7%
1.1
The reason the NPV’s are negative is that both investments are returning less than
the firm’s cost of money—10%.
NPVA  10,000  1,000 
b) Does the additional $1,000 alter the ranking of the two investments? Explain.
Solution:
12,000
 909.10 , IRR=20%
1.1
6,100
NPVA  5,000 
 545.50 , IRR=22%
1.1
The IRR ranking changes, not NPV. Note: the reason for the reversal in ranking is
that B is a smaller investment than A, and therefore, more sensitive to the additional
$1,000 cost.
NPVA  10,000 
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FIN 403 Lecture 6 Homework
7.12 State Electric wants to decide whether to repair or replace electric meters when they
break down. A new meter cost $30 and, on the average, will operate for 12 years without
repair. It costs $18 to repair a meter, and a repaired meter will, on average, operate for 8
years before it again needs a repair. Repairs can be made repeatedly to meters because
they are essentially rebuilt each time they are repaired. It costs $6 to take out and reinstall
a meter. The time value of money is 0.05. Should the company repair old meters or buy
new meters?
New (A)
Cost
($30)
Take out and reinstall cost ($ 6)
Horizon
12 years
Old (B)
($18)
($ 6)
8 years
Solution:
First note that you have to pay the cost of removal and installation/reinstallation under
both options. But, we cannot eliminate this cost from the analysis because, under the first
option, we incur it every twelve years, and under the second, every eight years.
The simplest way to approach this problem is to compute that annual equivalent
cost of each. The first step is to compute the NPV for each option, the second, the
corresponding annuity.
12
XA
 $36

X A  $4.06

t
t 1 1.05
8
XB
 $24

X B  $3.71

t
t 1 1.05
Option B is superior because it has lower annual costs.
A more messy way is to repeat both options—reinvest—until the lives are equal, then compare the NPV’s. We must repeat A twice,
and B three times.
$36
 $56.05
1.0512
($24) ($24)
B  $24 

 51.24
1.058 1.0516
Same result. The lifecycle cost of A for 24 years is greater than B.
A  $36 
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FIN 403 Lecture 6 Homework
7.20 A new piece of equipment being considered cost $75,816 and has a useful life of
five years. It will cost $10,000 in out-of-pocket expenses per year to operate. The cost of
money is 10%. An alternative to the equipment is to use a labor-intensive process that
would cost $38,000 per year. This $38,000 includes $15,000 of depreciation expense for
the currently used machine (the machine has no other use but does have a replacement
cost of $25,000). The tax rate is zero.
A
B
Initial cost
($75,816) ($25,000)
Expense/year
($10,000) ($38,000)
Depreciation (add-back)
$15,000
Horizon
5 years
5 years
If we have no tax, the effect of depreciation has no significance and should be
excluded in our analysis. Remember, depreciation is not a real expense; the sole
purpose of depreciation expense is to reduce the tax burden.
a) Prepare an analysis of the alternatives. What should the firm do?
5
$10,000
 $113,723.9
1.10t
t 1
5
($38,000  $15,000)
PVOld  $25,000  
 $112,188.1
1.10t
t 1
The old machine should be retained because it yields a lower cost.
PVNew  $75,816  
b) If the equipment will make 1,000 units of product per year, what will be the cost per
unit if the new equipment is purchased?
This is a tricky question and somewhat unclear. You could simply divide the $10,000 per
year marginal cost of production by 1000. But that would fail to acknowledge the cost of
the machine. You could also divide the cost of the machine by 1000 and add it to the
marginal costs of production, but that would fail to address the time value of money. The
proper way to compute this is to take the PV that you computed above, compute a
corresponding annual annuity, and divide that by 1000:
1  1.15 
X New  $30,000
X New  
  $113,723.9 
 0.1 
Thus, the cost of production is $30 per unit.
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FIN 403 Lecture 6 Homework
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FIN 403 Lecture 6 Homework
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