Replacement Ages as a Renewal-Reward Decision
A replacement policy trades off two costs. Waiting longer can avoid preventive replacements, but it also exposes the component to failure and a potentially more expensive corrective replacement. The relevant comparison is not cost per component alone: policies generate cycles of different expected durations. Renewal-reward theory converts this decision into a cost rate per unit time, which can then be minimized over the replacement age.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \((T_k)_{k\ge1}\) be independent, identically distributed component lifetimes, with \(T_k>0\) almost surely and \(T_k<\infty\) almost surely. Write \(F(t)=\mathbb{P}(T_1\le t)\) and \(S(t)=\mathbb{P}(T_1>t)=1-F(t)\) for \(t\ge0\). A replacement at failure costs \(c_f>0\); a preventive replacement at the chosen age costs \(c_p>0\). A cycle starts with a new component and ends when that component is replaced. There is no additional downtime or operating cost in this model.
For an age threshold \(\tau>0\), replace the component at failure if its lifetime is at most \(\tau\); otherwise replace it preventively at age \(\tau\). Thus failure exactly at \(\tau\) is counted as a failure. The cycle duration and cycle cost are respectively \(L_\tau=\min(T_1,\tau)\) and \(R_\tau=c_f\mathbf{1}_{\{T_1\le\tau\}}+c_p\mathbf{1}_{\{T_1>\tau\}}\). If the lifetime distribution has no atom at \(\tau\), the tie convention has no effect. If it has an atom there, the convention must be specified because it affects the cost.
To see the denominator, for each outcome, \(\min(T_1,\tau)=\int_0^\tau \mathbf{1}_{\{T_1>t\}}\,dt\), up to an endpoint that does not affect the integral. Taking expectations gives \(\mathbb{E}[L_\tau]=\int_0^\tau S(t)\,dt\). This value is finite and strictly positive: \(L_\tau\le\tau\), and \(L_\tau>0\) almost surely. The numerator follows by splitting into the disjoint events \(T_1\le\tau\) and \(T_1>\tau\).
After each replacement, a fresh lifetime is independent and has the same distribution. The cycles are therefore independent and identically distributed. For finite \(\tau\), their durations and costs are bounded, and their mean duration is positive. The renewal-reward theorem, proved earlier in the course, then gives the long-run average cost per unit time as \(\mathbb{E}[R_\tau]/\mathbb{E}[L_\tau]\) almost surely. This theorem requires integrable cycle rewards and a positive finite mean cycle duration; both conditions hold here. If \(\tau=\infty\) is allowed as a policy, it means replacement only at failure. When \(\mathbb{E}[T_1]<\infty\), its cost rate is \(c_f/\mathbb{E}[T_1]\).
A Differentiable Interior-Optimality Condition
It is useful to have a local condition for a candidate replacement age, but the condition has specific regularity requirements. In particular, absolute continuity of \(F\) near an age does not by itself guarantee differentiability at that particular age. The proposition below assumes differentiability at the age being tested.
Proof. Differentiability of \(F\) at \(\tau\) implies continuity of \(F\), and hence of \(S=1-F\), at \(\tau\). Therefore the fundamental theorem of calculus gives \(A'(\tau)=S(\tau)\). Also \(N=c_p+(c_f-c_p)F\), so \(N'(\tau)=(c_f-c_p)f(\tau)\). Since \(A(\tau)>0\), the quotient rule applies to \(C=N/A\), giving the displayed derivative. At an interior local minimum of a differentiable real-valued function, the derivative is zero; setting this formula equal to zero gives the stated condition. \(\square\)
The condition is necessary for an interior differentiable optimum, not sufficient on its own. One must still check whether the point is a minimum and compare it with boundary policies or other candidates. At atoms of the lifetime distribution, \(C\) can jump, so a derivative condition may not apply at all. The following examples show why both the formula and its assumptions matter.
Worked Example: Exponential Lifetimes and No Finite Best Age
Suppose \(T_1\) has an exponential distribution with rate \(1\), so \(F(t)=1-e^{-t}\) and \(S(t)=e^{-t}\). Let the corrective cost be \(c_f=10\) and the preventive cost be \(c_p=2\). Find the cost rate and determine whether a finite age minimizes it.
For \(\tau>0\), the expected cycle cost is $$ N(\tau)=10(1-e^{-\tau})+2e^{-\tau}=10-8e^{-\tau}. $$ The expected cycle duration is $$ A(\tau)=\int_0^\tau e^{-t}\,dt=1-e^{-\tau}. $$ Thus \(C(\tau)=(10-8e^{-\tau})/(1-e^{-\tau})\). Subtracting \(10\) gives $$ C(\tau)-10 =\frac{10-8e^{-\tau}-10(1-e^{-\tau})}{1-e^{-\tau}} =\frac{2e^{-\tau}}{1-e^{-\tau}}>0. $$ Every finite age therefore has cost rate strictly above \(10\). Also, direct differentiation gives $$ C'(\tau)=\frac{-2e^{-\tau}}{(1-e^{-\tau})^2}<0, $$ so the rate decreases as the age increases. As \(\tau\) tends to infinity, \(e^{-\tau}\) tends to zero and \(C(\tau)\) tends to \(10\).
Check: The exponential lifetime has mean \(1\), so the failure-only policy has rate \(c_f/\mathbb{E}[T_1]=10/1=10\), matching the limit. No finite age attains this rate: for each finite \(\tau>0\), the displayed difference \(2e^{-\tau}/(1-e^{-\tau})\) is positive. The infimum is approached by increasing \(\tau\), rather than attained by a finite threshold.
Worked Example: Finding the Optimal Age for a Uniform Lifetime
Suppose \(T_1\) is uniform on \([0,2]\), with corrective cost \(c_f=8\) and preventive cost \(c_p=2\). Find the age minimizing the long-run cost rate, allowing \(0<\tau\le2\), and verify the result against the endpoints.
For \(0\le t\le2\), \(F(t)=t/2\) and \(S(t)=1-t/2\). Hence, for \(0<\tau\le2\), $$ N(\tau)=8\frac{\tau}{2}+2\left(1-\frac{\tau}{2}\right)=2+3\tau, \qquad A(\tau)=\int_0^\tau\left(1-\frac{t}{2}\right)\,dt =\tau-\frac{\tau^2}{4}. $$ The denominator is positive on this interval. Applying the quotient rule, the sign of \(C'(\tau)\) is the sign of $$ 3A(\tau)-N(\tau)\left(1-\frac{\tau}{2}\right) =3\left(\tau-\frac{\tau^2}{4}\right)-(2+3\tau)\left(1-\frac{\tau}{2}\right) =\tau+\frac{3\tau^2}{4}-2. $$ The last expression is strictly increasing for \(\tau>0\), since its derivative is \(1+3\tau/2>0\). Its unique positive zero solves \(3\tau^2+4\tau-8=0\), giving $$ \tau_*=\frac{2\sqrt{7}-2}{3}. $$ This lies in \((0,2)\): it is positive because \(\sqrt{7}>1\), and it is less than \(2\) because \(\sqrt{7}<4\). Consequently \(C'\) is negative before \(\tau_*\) and positive after it, so this point is the unique minimum in the interval.
At the minimizing age, the first-order condition also gives \(C(\tau_*)=N'(\tau_*)/A'(\tau_*)=3/(1-\tau_*/2)\). Substituting \(\tau_*/2=(\sqrt{7}-1)/3\) yields $$ C(\tau_*)=\frac{3}{1-(\sqrt{7}-1)/3} =\frac{9}{4-\sqrt{7}} =4+\sqrt{7}. $$ For comparison, at \(\tau=2\), \(N(2)=2+3(2)=8\), \(A(2)=2-2^2/4=1\), and \(C(2)=8\). As \(\tau\) decreases to zero, \(A(\tau)\) tends to zero while \(N(\tau)\) tends to \(2\), so \(C(\tau)\) diverges. These endpoint checks agree with the interior minimum.
Check: The stationary equation gives \(\tfrac34\tau_*^2+\tau_*-2=0\), so the derivative numerator vanishes there. The value at the minimum is less than the rate at \(\tau=2\), since \(4+\sqrt{7}<8\) follows from \(\sqrt{7}<4\). The derivative sign on either side and the endpoint behavior verify global minimality over \(0<\tau\le2\).
Candidate Ages and Lifetime Atoms
For a lifetime distribution with atoms, candidate-age comparisons must use the chosen tie convention and the actual probabilities \(F(\tau)\) and \(S(\tau)\). In particular, the cost rate need not vary continuously with \(\tau\). A numerical comparison among selected ages can identify the best of those candidates, but it does not prove that one is globally optimal over every possible age.
Worked Example: Comparing Ages with a Two-Point Lifetime
Suppose a component lasts one unit of time with probability \(1/2\), or three units of time with probability \(1/2\). Let \(c_f=10\), \(c_p=2\), and compare ages \(\tau=1/2\), \(2\), and \(4\). As specified above, a failure exactly at the chosen age is charged as a failure replacement.
At \(\tau=1/2\), both possible lifetimes exceed the threshold. Thus \(F(1/2)=0\), \(S(1/2)=1\), expected cycle cost is \(0\cdot10+1\cdot2=2\), and cycle duration is always \(1/2\). The rate is \(2/(1/2)=4\).
At \(\tau=2\), the one-unit lifetime is a failure and the three-unit lifetime survives to preventive replacement. Thus \(F(2)=1/2\), \(S(2)=1/2\), and the expected cost is \((1/2)10+(1/2)2=6\). The expected duration is \((1/2)1+(1/2)2=3/2\), so the rate is \(6/(3/2)=4\).
At \(\tau=4\), both lifetimes end in failure before the preventive age. The expected cost is \(10\), and the expected duration is \((1/2)1+(1/2)3=2\), giving rate \(10/2=5\). Among these three candidates, ages \(1/2\) and \(2\) tie for the lowest rate, \(4\).
Check: At age \(2\), the two possible cycle outcomes have cost-duration pairs \((10,1)\) and \((2,2)\). Their expected cost is \((10+2)/2=6\), and their expected duration is \((1+2)/2=3/2\); the ratio is \(4\), as obtained from the formula. At age \(4\), both outcomes incur cost \(10\), with durations \(1\) and \(3\), so the ratio is \(10/2=5\).
Interpreting the Objective and Avoiding a Cost Error
The numerator is the expected cost of one complete replacement cycle; the denominator is the expected time occupied by that cycle. Minimizing only the expected cycle cost can select the wrong policy, because a lower cost per cycle may come with much shorter cycles. Likewise, comparing only the failure probability omits the preventive replacement cost on survival. The cost-rate ratio accounts for both effects.
Common mistake. In the uniform-lifetime example, at age \(\tau=1/2\), the probabilities are \(F(1/2)=1/4\) and \(S(1/2)=3/4\). A wrong calculation might charge only failures, giving a numerator \(8(1/4)=2\), and then report a rate \(2/(7/16)=32/7\). This omits the preventive cost on the survival event. The correct numerator is $$ 8\left(\frac14\right)+2\left(\frac34\right)=2+\frac32=\frac72. $$ The expected duration is \(A(1/2)=1/2-(1/2)^2/4=1/2-1/16=7/16\), so the correct rate is \((7/2)/(7/16)=8\). Each outcome must be assigned its corresponding replacement cost before taking the expected cost.
For a smooth lifetime law, the first-order condition helps locate interior candidates, but it does not replace a global comparison. Check the derivative sign or other minimum criterion, inspect the limits at the ends of the age range, and include failure-only replacement if it is an allowed policy. For lifetime distributions with atoms, use the cycle formula directly at each candidate and account for jumps; differentiable calculus may not describe the behavior at an atom.
Check Your Understanding
Use the cycle-cost and cycle-duration definitions to answer the following questions.
- For an age \(\tau\), explain why the expected cycle duration is \(\int_0^\tau S(t)\,dt\), and state the identity for \(\min(T_1,\tau)\) that gives this formula.
- If \(F\) has an atom at \(\tau\), which replacement cost applies under the convention used here to a component with lifetime exactly \(\tau\)?
- In the uniform-lifetime example, calculate the cost rate at \(\tau=1\) by computing both the expected cycle cost and expected cycle duration.
- Why does the differentiable first-order condition not, by itself, establish that a candidate age is a global minimum?
- For exponential lifetimes of rate \(1\), what is the failure-only cost rate when \(c_f=10\), and why is it not attained at any finite replacement age when \(c_p=2\)?