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Machines in a factory break down at an exponential rate of six per hour. There is a single repairman who fixes machines at an exponential rate of eight per hour. The cost incurred in lost production when machines are out of service is \(\$ 10\) per hour per machine. What is the average cost rate incurred due to failed machines?

Short Answer

Expert verified
The average cost rate incurred due to failed machines is \(\$30\) per hour.

Step by step solution

01

Identify the Failure Rate and Repair Rate

In the problem, we are given the failure rate and repair rate for machines. The failure rate \(\lambda\) is six per hour, and the repair rate \(\mu\) is eight per hour.
02

Calculate Average Number of Failed Machines

Using the fact that the average number of failed machines can be found using the formula \(\frac{\lambda}{\mu-\lambda}\), we can plug in our failure and repair rates: \[\text{Average Number of Failed Machines} = \frac{6}{8-6} = \frac{6}{2} = 3\]
03

Calculate the Average Cost Rate

Now that we have the average number of failed machines, we can calculate the average cost rate. Given that the cost incurred per hour per failed machine is \(\$10\), the average cost rate is: \[ \text{Average Cost Rate} = 3 \cdot 10 = \$ 30 \text{ per hour} \] Thus, the average cost rate incurred due to failed machines is \(\$30\) per hour.

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Most popular questions from this chapter

In an \(\mathrm{M} / \mathrm{G} / 1\) queue, (a) what proportion of departures leave behind 0 work? (b) what is the average work in the system as seen by a departure?

For the \(M / G / 1\) queue, let \(X_{n}\) denote the number in the system left behind by the nth departure. (a) If $$ X_{n+1}=\left\\{\begin{array}{ll} X_{n}-1+Y_{n}, & \text { if } X_{n} \geqslant 1 \\ Y_{n}, & \text { if } X_{n}=0 \end{array}\right. $$ what does \(Y_{n}\) represent? (b) Rewrite the preceding as $$ X_{n+1}=X_{n}-1+Y_{n}+\delta_{n} $$ where $$ \delta_{n}=\left\\{\begin{array}{ll} 1, & \text { if } X_{n}=0 \\ 0, & \text { if } X_{n} \geqslant 1 \end{array}\right. $$ Take expectations and let \(n \rightarrow \infty\) in Equation ( \(8.64\) ) to obtain $$ E\left[\delta_{\infty}\right]=1-\lambda E[S] $$ (c) Square both sides of Equation \((8.64)\), take expectations, and then let \(n \rightarrow \infty\) to obtain $$ E\left[X_{\infty}\right]=\frac{\lambda^{2} E\left[S^{2}\right]}{2(1-\lambda E[S])}+\lambda E[S] $$ (d) Argue that \(E\left[X_{\infty}\right]\), the average number as seen by a departure, is equal to \(L\).

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Consider a \(M / G / 1\) system with \(\lambda E[S]<1\). (a) Suppose that service is about to begin at a moment when there are \(n\) customers in the system. (i) Argue that the additional time until there are only \(n-1\) customers in the system has the same distribution as a busy period. (ii) What is the expected additional time until the system is empty? (b) Suppose that the work in the system at some moment is \(A\). We are interested in the expected additional time until the system is empty - call it \(E[T]\). Let \(N\) denote the number of arrivals during the first \(A\) units of time. (i) Compute \(E[T \mid N]\). (ii) Compute \(E[T]\).

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