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Chapter 7: Problem 38

A truck driver regularly drives round trips from \(\mathrm{A}\) to \(\mathrm{B}\) and then back to \(\mathrm{A}\). Each time he drives from \(\mathrm{A}\) to \(\mathrm{B}\), he drives at a fixed speed that (in miles per hour) is uniformly distributed between 40 and 60 ; each time he drives from \(\mathrm{B}\) to \(\mathrm{A}\), he drives at a fixed speed that is equally likely to be either 40 or 60 . (a) In the long run, what proportion of his driving time is spent going to \(\mathrm{B}\) ? (b) In the long run, for what proportion of his driving time is he driving at a speed of 40 miles per hour?

Short Answer

Expert verified
In the long run, (a) the truck driver spends 50% of his driving time going from A to B, and (b) he spends 11/40 (or 27.5%) of his driving time driving at a speed of 40 miles per hour.

Step by step solution

01

Setup the problem and calculate average speeds for A to B and B to A

Let distance between A and B be D. We have two scenarios here, one for driving from A to B and another from B to A. For A to B: Speed is uniformly distributed between 40 and 60. Hence, we can calculate the average speed of this segment (\(v_{AB}\)): \(v_{AB} = \frac{40+60}{2} = 50\) mph For B to A: Speed is equally likely to be 40 or 60. Let's calculate the average speed for this segment (\(v_{BA}\)): \(v_{BA} = \frac{40+60}{2} = 50\) mph
02

Calculate driving time for A to B and B to A

Now that we have the average speeds for both segments, let's calculate the time it takes the driver to travel each segment. Driving time from A to B(\(t_{AB}\)): \(t_{AB} = \frac{D}{v_{AB}} = \frac{D}{50}\) Driving time from B to A(\(t_{BA}\)): \(t_{BA} = \frac{D}{v_{BA}} = \frac{D}{50}\)
03

Calculate proportion of driving time spent going to B and driving at 40 mph

(a) Proportion of driving time spent going to B: To find this proportion, we'll divide the driving time for A to B by the total driving time: \(\frac{t_{AB}}{t_{AB} + t_{BA}} = \frac{\frac{D}{50}}{\frac{D}{50} + \frac{D}{50}} = \frac{1}{2}\) So, 50% of the driver's time is spent going to B. (b) Proportion of driving time spent driving at 40 mph: Let's first find the driver's probability of driving 40 mph on each segment. On the trip from A to B, probability \(P_{40}^{AB}= \frac{1}{20}\) (since the truck driver drives between 40 and 60 mph with uniform probability). On the trip from B to A, probability \(P_{40}^{BA} = 0.5\) (equal chance of driving at 40 or 60 mph). Now, we find the proportion of time spent driving at 40 mph by averaging the probabilities \(\frac{P_{40}^{AB}*t_{AB} + P_{40}^{BA}*t_{BA}}{t_{AB} + t_{BA}}= \frac{\frac{1}{20}*\frac{D}{50} + 0.5*\frac{D}{50}}{\frac{D}{50} + \frac{D}{50}} = \frac{\frac{1}{20} + 0.5}{2} = \frac{11}{40}\) Thus, 11/40 (or 27.5%) of the driver's time is spent driving at 40 mph.

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Most popular questions from this chapter

A system consists of two independent machines that each functions for an exponential time with rate \(\lambda\). There is a single repairperson. If the repairperson is idle when a machine fails, then repair immediately begins on that machine; if the repairperson is busy when a machine fails, then that machine must wait until the other machine has been repaired. All repair times are independent with distribution function \(G\) and, once repaired, a machine is as good as new. What proportion of time is the repairperson idle?

Consider a miner trapped in a room that contains three doors. Door 1 leads him to freedom after two days of travel; door 2 returns him to his room after a four-day journey; and door 3 returns him to his room after a six-day journey. Suppose at all times he is equally likely to choose any of the three doors, and let \(T\) denote the time it takes the miner to become free. (a) Define a sequence of independent and identically distributed random variables \(X_{1}, X_{2} \ldots\) and a stopping time \(N\) such that $$ T=\sum_{i=1}^{N} X_{i} $$ Note: You may have to imagine that the miner continues to randomly choose doors even after he reaches safety. (b) Use Wald's equation to find \(E[T]\). (c) Compute \(E\left[\sum_{i=1}^{N} X_{i} \mid N=n\right]\) and note that it is not equal to \(E\left[\sum_{i=1}^{n} X_{i}\right]\). (d) Use part (c) for a second derivation of \(E[T]\).

Consider a single-server queueing system in which customers arrive in accordance with a renewal process. Each customer brings in a random amount of work, chosen independently according to the distribution \(G .\) The server serves one customer at a time. However, the server processes work at rate \(i\) per unit time whenever there are \(i\) customers in the system. For instance, if a customer with workload 8 enters service when there are three other customers waiting in line, then if no one else arrives that customer will spend 2 units of time in service. If another customer arrives after 1 unit of time, then our customer will spend a total of \(1.8\) units of time in service provided no one else arrives. Let \(W_{i}\) denote the amount of time customer \(i\) spends in the system. Also, define \(E[W]\) by $$ E[W]=\lim _{n \rightarrow \infty}\left(W_{1}+\cdots+W_{n}\right) / n $$ and so \(E[W]\) is the average amount of time a customer spends in the system. Let \(N\) denote the number of customers that arrive in a busy period. (a) Argue that $$ E[W]=E\left[W_{1}+\cdots+W_{N}\right] / E[N] $$ Let \(L_{i}\) denote the amount of work customer \(i\) brings into the system; and so the \(L_{i}, i \geqslant 1\), are independent random variables having distribution \(G .\) (b) Argue that at any time \(t\), the sum of the times spent in the system by all arrivals prior to \(t\) is equal to the total amount of work processed by time \(t\). Hint: Consider the rate at which the server processes work. (c) Argue that $$ \sum_{i=1}^{N} W_{i}=\sum_{i=1}^{N} L_{i} $$ (d) Use Wald's equation (see Exercise 13) to conclude that $$ E[W]=\mu $$ where \(\mu\) is the mean of the distribution \(G\). That is, the average time that customers spend in the system is equal to the average work they bring to the system.

Each time a certain machine breaks down it is replaced by a new one of the same type. In the long run, what percentage of time is the machine in use less than one year old if the life distribution of a machine is (a) uniformly distributed over \((0,2) ?\) (b) exponentially distributed with mean \(1 ?\)

Let \(X_{1}, X_{2}, \ldots\) be a sequence of independent random variables. The nonnegative integer valued random variable \(N\) is said to be a stopping time for the sequence if the event \(\\{N=n\\}\) is independent of \(X_{n+1}, X_{n+2}, \ldots\). The idea being that the \(X_{i}\) are observed one at a time-first \(X_{1}\), then \(X_{2}\), and so on-and \(N\) represents the number observed when we stop. Hence, the event \(\\{N=n\\}\) corresponds to stopping after having observed \(X_{1}, \ldots, X_{n}\) and thus must be independent of the values of random variables yet to come, namely, \(X_{n+1}, X_{n+2}, \ldots\) (a) Let \(X_{1}, X_{2}, \ldots\) be independent with $$ P\left\\{X_{i}=1\right\\}=p=1-P\left\\{X_{i}=0\right\\}, \quad i \geqslant 1 $$ Define $$ \begin{aligned} &N_{1}=\min \left\\{n: X_{1}+\cdots+X_{n}=5\right\\} \\ &N_{2}=\left\\{\begin{array}{ll} 3, & \text { if } X_{1}=0 \\ 5, & \text { if } X_{1}=1 \end{array}\right. \\ &N_{3}=\left\\{\begin{array}{ll} 3, & \text { if } X_{4}=0 \\ 2, & \text { if } X_{4}=1 \end{array}\right. \end{aligned} $$ Which of the \(N_{i}\) are stopping times for the sequence \(X_{1}, \ldots ?\) An important result, known as Wald's equation states that if \(X_{1}, X_{2}, \ldots\) are independent and identically distributed and have a finite mean \(E(X)\), and if \(N\) is a stopping time for this sequence having a finite mean, then $$ E\left[\sum_{i=1}^{N} X_{i}\right]=E[N] E[X] $$ To prove Wald's equation, let us define the indicator variables \(I_{i}, i \geqslant 1\) by $$ I_{i}=\left\\{\begin{array}{ll} 1, & \text { if } i \leqslant N \\ 0, & \text { if } i>N \end{array}\right. $$ (b) Show that $$ \sum_{i=1}^{N} X_{i}=\sum_{i=1}^{\infty} X_{i} I_{i} $$ From part (b) we see that $$ \begin{aligned} E\left[\sum_{i=1}^{N} X_{i}\right] &=E\left[\sum_{i=1}^{\infty} X_{i} I_{i}\right] \\ &=\sum_{i=1}^{\infty} E\left[X_{i} I_{i}\right] \end{aligned} $$ where the last equality assumes that the expectation can be brought inside the summation (as indeed can be rigorously proven in this case). (c) Argue that \(X_{i}\) and \(I_{i}\) are independent. Hint: \(I_{i}\) equals 0 or 1 depending on whether or not we have yet stopped after observing which random variables? (d) From part (c) we have $$ E\left[\sum_{i=1}^{N} X_{i}\right]=\sum_{i=1}^{\infty} E[X] E\left[I_{i}\right] $$ Complete the proof of Wald's equation. (e) What does Wald's equation tell us about the stopping times in part (a)?
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