15.3 Replacement of deteriorating equipment
A machine costs \(C\), its running cost in year \(k\) is \(r_k\), rising with age, and its resale value after \(n\) years is \(S_n\). Replacing it every \(n\) years costs, per year on average, \[ A(n) = \frac {C - S_n + \sum _{k = 1}^nr_k}n . \]
Proposition 15.3.1. If the yearly costs \(m_k = r_k + (S_{k - 1} - S_k)\) (running cost plus loss of resale value in year \(k\), with \(S_0 = C\)) increase with \(k\), then \(A(n)\) is minimised at the first \(n\) with \(m_{n + 1} \geq A(n)\): replace the machine when next year’s cost would exceed the average cost so far.
Proof. \(A(n) = \frac 1n\sum _{k \leq n}m_k\) is the running average of an increasing sequence. Then \(A(n + 1) - A(n) = \frac {m_{n + 1} - A(n)}{n + 1}\), so \(A\) decreases while the next term is below the average and increases once it is above; since the terms increase, once above they stay above, and the first such \(n\) is the minimum. □
Example 15.3.2. With \(C = 6000\), running costs \(1000, 1200, 1400, 1800, 2300, 2800, 3400, 4000\) and resale values \(3000, 1500, 750, 375, 200, 200, 200, 200\), the average costs are \(4000\), \(3350\), \(2950\), \(2756\), \(2700\), \(2717\), \(\ldots \), minimal at \(n = 5\): the sixth year would cost \(m_6 = 2800 > A(5) = 2700\). With money discounted at \(10\%\) a year, the equivalent annual cost (the level annuity with the same present value) is minimised at \(6\) years instead (Software 15.5.2): future running costs count for less, so it pays to keep the machine longer.
Items that fail suddenly, such as light bulbs or electronic components, are replaced on failure; when individual replacement is costly, group replacement of all items at fixed intervals, with failures replaced in between, can be cheaper, and the best interval is found from the failure distribution by renewal arguments.
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