The Equipment-Replacement Problem

To illustrate these assumptions, let us consider a simple numerical example. Suppose a car is needed for three years; that is, suppose N = 3. At the beginning of the first year, we have a 2-year-old car. The annual cost of operating a car is a function of its age; and this cost function is given by: c(0) = 10, c(1) = 20, c(2) = 40, c(3) = 60, and c(4) = 70. The fact that these costs are increasing is a reflection of aging. Since the operating costs are increasing, it may become more cost effective to replace a car after it has been in operation for some periods. The price of a new car is 60, i.e., p = 60. (Our solution below can be easily adapted to reflect price variations over time.) The trade-in value of a used car is a function of its age at the time of trade in; and this function is given by: t(1) = 30, t(2) = 20, t(3) = 15, and t(4) = 10. These trade-in values are decreasing, as a reflection of the decline in desirability of a car over time. Finally, a car will no longer be needed at the end of year N ; therefore, the car in service at that time will be salvaged. The salvage value of a used car is again a function of its age; and this function is given by: s(1) = 20, s(2) = 15, s(3) = 10, s(4) = 0, and s(5) = 0. Like the trade-in values, the salvage values are also decreasing. Moreover, notice that the salvage value of a car, at any given age, is less than the trade-in value of a car with the same age.