bre44380_ch06_132-161.indd 148 09/30/15 12:46 PM
148 Part One Value
Because the two machines produce exactly the same product, the only way to choose
between them is on the basis of cost. The present value of each machine’s cost is as follows:Costs ($ thousands)Year: 0 1 2 3PV at 6%
($ thousands)
Machine A 15 5 5 5 $28.37
Machine B 10 6 6 — 21.00Should we take machine B, the one with the lower present value of costs? Not necessarily. All
we have shown is that machine B offers two years of service for a lower total cost than three
years of service from machine A. But is the annual cost of using B lower than that of A?
Suppose the financial manager agrees to buy machine A and pay for its operating costs out
of her budget. She then charges the plant manager an annual amount for use of the machine.
There will be three equal payments starting in year 1. The financial manager has to make sure
that the present value of these payments equals the present value of the costs of each machine.
When the discount rate is 6%, the payment stream with such a present value turns out to be
$10,610 a year. In other words, the cost of buying and operating machine A over its three-year
life is equivalent to an annual charge of $10,610 a year for three years.BEYOND THE PAGE
mhhe.com/brealey12eEquivalent annual
costsCosts ($ thousands)Year: 0 1 2 3PV at 6%
($ thousands)
Machine A 15 5 5 5 28.37
Equivalent annual cost 10.61 10.61 10.61 28.37We calculated the equivalent annual cost by finding the three-year annuity with the same
present value as A’s lifetime costs.PV of annuity = PV of A’s costs = 28.37
= annuity payment × 3-year annuity factorAt a 6% cost of capital, the annuity factor is 2.673 for three years, soAnnuity payment = _____28.37
2.673= 10.61A similar calculation for machine B gives an equivalent annual cost of $11,450:Costs ($ thousands)Year: 0 1 2PV at 6%
($ thousands)Machine B 10 6 6 21.00
Equivalent annual cost 11.45 11.45 21.00Machine A is better, because its equivalent annual cost is less ($10,610 versus $11,450 for
machine B).