6.4 Server Failures

In Section 6.2 and Section 6.3 we assumed that servers are never interrupted while serving a job. However, this assumption is not always satisfied; for example, a machine may fail in the middle of processing a job. In this section, we develop a model¹¹There is not much point in combining the models of Section 6.2--Section 6.4 into one large model. For example, when a machine can be set up while a part is being repaired, there may not be time wasted on setup. Everything depends on specific strategies to combine outages. to calculate the influence on the mean waiting time of such preemptive outages, again based on Sakasegawa’s formula for the \(G/G/1\) queue.

As in Section 6.3, we only have to derive expressions for the expectation and variance of the effective service time \(S\). The other components of Sakasegawa’s formula are not affected.

We assume that a failure can occur at any time during the service time of a job. The repair times \(R_i \sim R\) have common mean \(\E R\) and finite second moment \(\E{R^{2}}\). Supposing that \(N\) interruptions occur during the net job service time \(S_0\), the effective service time becomes²²In an insurance context, if we interpret \(\{R_i\}\) as a set of claims, then \(\sum_{i=1}^N R_i\) is the total claim size of \(N\) claims. Likewise, in an inventory system, this sum is the total demand of \(N\) customers.

\begin{equation*} S= S_0 + S_N = S_0 + \sum_{i=1}^N R_i. \tag{6.4.1} \end{equation*}

In general, \(N\) is a random number, so we need to adjust for this in the computation of the mean and variance of \(S\).

It is common to assume that the time between two interruptions is memoryless with mean \(1/\lambda_f\), where \(\lambda_f\) is the failure rate. Defining the MTTF³³MTTF := mean time to fail. as \(m_{f} = 1/\lambda_f\), then the availability is given by

\begin{equation*} A:=\frac{m_f}{m_f + \E R}, \end{equation*}

from which follows easily that ⁴⁴Exercise 7.

\begin{equation*} A=(1+\lambda_f \E R)^{-1}. \tag{6.4.2} \end{equation*}

Next, because by assumption the time between two failures is \(\sim \Exp{\lambda_{f}}\) so that \(N\sim \Pois{\lambda_{f} S_{0}}\) if \(S_{0}\) is given,

\begin{equation*} \E N = \E{\E{N|S_{0}}} = \E{\lambda_{f}S_{0}} = \lambda_{f} \E{S_{0}}. \tag{6.4.3} \end{equation*}

Therefore,

\begin{align*} \E S &= \E{S_0} + \E N \E R = \E{S_0} + \lambda_f \E{S_0} \E R \\ &= \E{S_0}(1+\lambda_f \E R) = \E{S_{0}}/A. \end{align*}

With this expression for the expected effective service time, the load becomes

\begin{equation*} \rho = \lambda \E S = \lambda \frac{\E{S_0}}A. \end{equation*}

Clearly, \(A\in (0,1]\), so the load increases due to failures.

With some more work, we obtain⁵⁵Exercise 8. from (6.4.1),

\begin{align*} \V{S} &= \frac{\V{S_0}}{A^2} + \lambda_f \E{R^2} \E{S_0}. \tag{6.4.4} \end{align*}

If the repair times are exponentially distributed, it is possible to find a neat expression for the scv of the effective service time:⁶⁶Exercise 10.

\begin{equation*} C_s^2 = C_0^2 + 2 A(1-A) \frac{\E{R}}{\E{S_0}}, \tag{6.4.5} \end{equation*}

where \(C_0^2\) is the scv of \(S_0\).