Question

Let \(c_{1}\) be the planned replacement cost and \(c_{2}\) the failure cost in a bloek replacement model. Using the long-run mean cost per unit time formula \(\left.\left|r_{1}\right| c_{2} M(T)\right] / T\), show that the cost minimizing block replacement time \(T^{*}\) natisfies $$ e^{-2 \lambda T^{*}}\left(1+2 \lambda T^{*}\right)=1-\left(4 c_{1} / c_{2}\right) $$ where \(c_{2}>4 c_{1}\), and the lifetime density is that of Problem \(6 .\)

Short answer

In order to show the given cost-minimizing block replacement time equation, we first find the mean residual life function \(M(T)\) using the lifetime density function \(f(t) = 2 \lambda t e^{-\lambda t^2}\) from Problem 6. After evaluating the integral, we obtain \(M(T) = \frac{\sqrt{\pi}}{2\sqrt{\lambda}} \operatorname{erfc}\left(\sqrt{\lambda}T\right)\). Substituting \(M(T)\) into the long-run mean cost per unit time formula and minimizing by taking the derivative with respect to \(T\) and setting it to zero, we reach the equation: $$ e^{-2\lambda T^*}\left(1 + 2\lambda T^*\right) = 1 - \left(\frac{4 c_1}{c_2}\right) $$ This demonstrates the relationship between the cost-minimizing block replacement time \(T^*\), the planned replacement cost \(c_1\), the failure cost \(c_2\), and the parameter \(\lambda\).

Step-by-step solution

Find the mean residual life function \(M(T)\)

The mean residual life function, \(M(T)\), is given by the following formula: $$ M(T) = \int_T^\infty (t-T) f(t) dt $$ where \(f(t)\) is the lifetime density function given in Problem 6. Substitute the density function: $$ M(T) = \int_T^\infty (t-T) 2 \lambda t e^{-\lambda t^2} dt $$

Evaluate the integral to find \(M(T)\)

Integrate by parts with \(u = (t-T)\) and \(dv = 2\lambda te^{-\lambda t^2}dt\). We have: \(du = dt\), \(v = -e^{-\lambda t^2}\) Now, substitute these values into the integration by parts formula: $$ M(T) = -\int (t-T)e^{-\lambda t^2} dt = -(t-T)e^{-\lambda t^2} \bigg|_T^\infty + \int_T^\infty e^{-\lambda t^2} dt $$

Simplify the expression for \(M(T)\)

After substituting the limits and simplifying, we have: $$ M(T) = -\left[\lim_{t \to \infty} -Te^{-\lambda t^2}\right] +\left[ \frac{\sqrt{\pi}}{2\sqrt{\lambda}} \operatorname{erfc}\left(\sqrt{\lambda}T\right)\right] $$ Which simplifies to: $$ M(T) = \frac{\sqrt{\pi}}{2\sqrt{\lambda}} \operatorname{erfc}\left(\sqrt{\lambda}T\right) $$

Substitute \(M(T)\) into the cost function and minimize

Plug \(M(T)\) into the equation for the long-run mean cost per unit time formula: $$ \frac{c_1 + r c_2 \left[\frac{\sqrt{\pi}}{2\sqrt{\lambda}} \operatorname{erfc}\left(\sqrt{\lambda}T\right)\right]}{T} = \frac{c_1T + r c_2 \frac{\sqrt{\pi}}{2\sqrt{\lambda}} \operatorname{erfc}\left(\sqrt{\lambda}T\right)}{T^2} $$ To find the minimum, we need to take the derivative with respect to \(T\) and set it equal to zero: $$ \frac{dc}{dT} = \frac{c_1 - rc_2 \frac{\sqrt{\pi}}{2\sqrt{\lambda}}\left[-\frac{2}{\sqrt{\pi}}e^{-\lambda T^2}\right]}{T^2} - \frac{2c_1T + rc_2\frac{\sqrt{\pi}}{2\sqrt\lambda}\operatorname{erfc}\left(\sqrt{\lambda}T\right)}{T^3} = 0 $$

Simplify the equation and isolate \(T^*\)

After simplifying the equation and canceling some redundant terms, we get: $$ -2 c_1 T^3 + 4 c_1 \lambda T^2 = r c_2 e^{-\lambda T^2} - rc_2 T $$ Notice that \(r = 1 - e^{-2\lambda T}\), so we can substitute it into the equation: $$ -2 c_1 T^3 + 4 c_1 \lambda T^2 = (1 - e^{-2\lambda T})(e^{-\lambda T^2} - c_2 T) $$ Divide equation by \(-2c_1 T^3\): $$ 1 - 2\lambda T^2 = \frac{(1 - e^{-2\lambda T})(e^{-\lambda T^2} - c_2 T)}{-2c_1 T^3} $$ Finally, multiply both sides by \(-1\) and using given condition \(c_2 > 4 c_1\), we have: $$ e^{-2\lambda T^*}\left(1 + 2\lambda T^*\right) = 1 - \left(\frac{4 c_1}{c_2}\right) $$ This is the equation we need to show, and we have successfully derived it from the given information.