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There are two possible causes for a breakdown of a machine. To check the first possibility. would cost \(C_{1}\) dollars, and, if that were the cause of the breakdown, the trouble could be repaired at a cost of \(R_{1}\) dollars. Similarly, there are costs \(C_{2}\) and \(R_{2}\) associated with the second possibility. Let \(p\) and \(1-p\) denote, respectively, the probabilities that the breakdown is caused by the first and second possibilities. Under what conditions on \(p, C_{i}, R_{i}\), \(i=1,2\), should we check the first possible cause of breakdown and then the second, as opposed to reversing the checking order, so as to minimize the expected cost involved in returning the machine to working order? NOTE: If the first check is negative, we must still check the other possibility.

Short Answer

Expert verified
We should check possibility 1 first and then possibility 2 if the probability \(p > \frac{C_1 + C_2 + R_1 - R_2}{C_1 + R_1 - C_2 - R_2}\). If the inequality is reversed, we should check possibility 2 first and then possibility 1 to minimize the expected cost of returning the machine to working order.

Step by step solution

01

Calculate Expected Cost for Checking Order 1-2

First, we'll calculate the expected cost if we check possibility 1 first and then possibility 2 if necessary. Let's call this expected cost E_1. There are two cases to consider: 1. Possibility 1 is the cause: The cost would be the cost to check possibility 1 (C_1) plus the cost to repair (R_1). The probability of this happening is p, so the expected cost of this case is p(C_1 + R_1). 2. Possibility 2 is the cause: The cost would be the cost to check possibility 1 (C_1) plus the cost to check possibility 2 (C_2) plus the cost to repair (R_2). The probability of this happening is 1-p, so the expected cost of this case is (1-p)(C_1+C_2+R_2). Summing these two cases, we get the total expected cost if we check possibility 1 first and then possibility 2: \[E_1 = p(C_1 + R_1) + (1-p)(C_1 + C_2 + R_2)\]
02

Calculate Expected Cost for Checking Order 2-1

Now let's calculate the expected cost if we check possibility 2 first and then possibility 1 if necessary. Let's call this expected cost E_2. There are two cases to consider: 1. Possibility 2 is the cause: The cost would be the cost to check possibility 2 (C_2) plus the cost to repair (R_2). The probability of this happening is 1-p, so the expected cost of this case is (1-p)(C_2 + R_2). 2. Possibility 1 is the cause: The cost would be the cost to check possibility 2 (C_2) plus the cost to check possibility 1 (C_1) plus the cost to repair (R_1). The probability of this happening is p, so the expected cost of this case is p(C_1+C_2+R_1). Summing these two cases, we get the total expected cost if we check possibility 2 first and then possibility 1: \[E_2 = (1-p)(C_2 + R_2) + p(C_1 + C_2 + R_1)\]
03

Compare Expected Costs

To determine the conditions under which we should check possibility 1 first and then possibility 2, we need to compare the two expected costs. Checking order 1-2 is preferable if E_1 < E_2: \[p(C_1 + R_1) + (1-p)(C_1 + C_2 + R_2) < (1-p)(C_2 + R_2) + p(C_1 + C_2 + R_1)\] Now, we can simplify the inequality: \[pC_1 + pR_1 + C_1 - pC_1 + C_2 - pC_2 + R_2 - pR_2 < C_2 - pC_2 + R_2 - pR_2 + pC_1 + pC_2 + pR_1\] Cancel out the like terms: \[C_1 + C_2 + R_1 - R_2 < p(C_1 + R_1 - C_2 - R_2)\] Now, divide both sides by the common factor to get the condition for checking possibility 1 first: \[\frac{C_1 + C_2 + R_1 - R_2}{C_1 + R_1 - C_2 - R_2} < p\] Thus, we should check possibility 1 first and then possibility 2 if the probability p is greater than the ratio of the costs and repair differences. If the inequality is reversed, we should check possibility 2 first and then possibility 1 to minimize the expected cost of returning the machine to working order.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Probability in Decision Making
Probability is a measure of the likelihood that an event will occur. When it comes to decision making under uncertainty, understanding and using probabilities can greatly influence the choices we make, and ultimately, affect the outcomes we achieve. Decisions in uncertain environments, like the machinery breakdown scenario presented, rely heavily on the accurate assessment of probabilities for different events.

In our exercise, the probabilities associated with the causes of the machine breakdown (\(p\) for the first possibility and \(1-p\) for the second) serve as the key pieces of data. Calculating the expected costs for each scenario involves determining not just the costs associated with checking and repairing (\(C_i\) and \(R_i\) respectively), but also weighing these costs against the likelihood of each possibility being the true cause of the breakdown. The formula for expected cost elegantly encapsulates this by summing the products of costs and their respective probabilities for each potential cause.
Applying Cost-Benefit Analysis to Minimize Costs
Cost-benefit analysis is a systematic approach for calculating and comparing the benefits and costs of a decision, project, or investment. This economic evaluation tool is particularly beneficial when there are multiple courses of action available and a decision needs to be made to minimize costs or maximize benefits.

In our textbook problem, the cost-benefit analysis comes into play when deciding the order in which to check the two possible causes of the machine's breakdown. By calculating the expected costs for both scenarios, E_1 and E_2, and comparing them, we can deduce which strategy is more cost-effective. The analysis weighs the cost of checking and fixing each possibility against the probability of each cause, thereby determining which option has a lower expected cost and will thus minimize financial expenditure for the decision-maker. Implementing cost-benefit analysis in this problem exemplifies how quantitative techniques can guide us towards more informed and economically viable decisions.
Making Decisions Under Uncertainty
Decision making under uncertainty is a critical skill, especially in fields where probability and risk management play significant roles. The uncertainty in our exercise arises from the lack of knowledge concerning the exact cause of the machine breakdown. A decision needs to be made despite the uncertainty about the events that will unfold, which is reflective of many real-world situations.

To navigate such scenarios, we often rely on the concept of expected value or expected cost. By considering the potential outcomes, associated costs, and their probabilities, we can minimize potential losses or costs under uncertainty. The comparative process highlighted in steps 1 to 3 of the provided solution showcases how we can systematically approach such choices. In our problem, determining whether \(p(C_1 + R_1) + (1-p)(C_1 + C_2 + R_2) < (1-p)(C_2 + R_2) + p(C_1 + C_2 + R_1)\) allows us to decide rationally between the checking orders to minimize the expected cost, reflecting a pragmatic application of decision making under uncertainty.

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Most popular questions from this chapter

A philanthropist writes a positive number \(x\) on a piece of red paper, shows it to an impartial observer, and then turns it face down on the table. The observer then flips a fair coin. If it shows heads, she writes the value \(2 x\), and, if tails, the value \(x / 2\), on a piece of blue paper which she then turns face down on the table. Without knowing either the value \(x\) or the result of the coin flip, you have the option of turning over either the red or the blue piece of paper. After doing so, and. observing the number written on that paper, you may elect to receive as a reward either that amount or the (unknown) amount written on the other piece of paper. For instance, if you elect to turn over the blue paper and observe the value 100 , then you can elect, either to accept 100 as your reward or to take the amount (either 200 or 50) on the red paper. Suppose that you would like your expected reward to be large. (a) Argue that there is no reason to turn over the red paper first because if you do so, then no matter what value you observe, it is always better to switch to the blue paper. (b) Let \(y\) be a fixed nonnegative value, and consider the following strategy. Tum over the blue paper and if its value is at least \(y\), then accept that amount. If it is less than \(y\), then switch to the red paper. Let \(R_{y}(x)\) denote the reward obtained if the philanthropist writes the amount \(x\) and you employ this strategy. Find \(E\left[R_{y}(x)\right]\). Note that \(E\left[R_{0}(x)\right]\) is the expected reward if the philanthropist writes the amount \(x\) when you employ the strategy of always choosing the blue paper.

Let \(X\) be such that $$ P\\{X=1\\}=p=1-P\\{X=-1\\} $$ Find \(c \neq 1\) such that \(E\left[c^{X}\right]=1\).

A sample of 3 items is selected at random from a box containing 20 items of which 4 are defective. Find the expected number of defective items in the sample.

(a) An integer \(N\) is to be selected at random from \(\left\\{1,2, \ldots,(10)^{3}\right\\}\) in the sense that each integer has the same probability of being selected. What is the probability that \(N\) will be divisible by \(3 ?\) by \(5 ?\) by \(7 ?\) by \(15 ?\) by \(105 ?\) How would your answer change if \((10)^{3}\) is replaced by \((10)^{k}\) as \(k\) became larger and larger? (b) An important function in number theory - one whose properties can be shown to be related to what is probably the most important unsolved problem of mathematics, the Riemann hypothesis-is the Möbius function \(\mu(n)\), defined for all positive integral values \(n\) as follows: Factor \(n\) into its prime factors. If there is a repeated prime factor, as in \(12=\) \(2 \cdot 2 \cdot 3\) or \(49=7 \cdot 7\), then \(\mu(n)\) is defined to equal 0. Now let \(N\) be chosen at random from \(\left\\{1,2, \ldots(10)^{k}\right\\}\), where \(k\) is large. Determine \(P\\{\mu(N)=0\\}\) as \(k \rightarrow \infty\) HINT: To compute \(P\\{\mu(N) \neq 0\\}\), use the identity $$ \prod_{i=1}^{x} \frac{P_{i}^{2}-1}{P_{I}^{2}}=\left(\frac{3}{4}\right)\left(\frac{8}{9}\right)\left(\frac{24}{25}\right)\left(\frac{48}{49}\right) \cdots=\frac{6}{\pi^{2}} $$ where \(P_{i}\) is the \(i\) th smallest prime. (We do not include I as a prime.)

Consider \(n\) independent sequential trials, each of which is successful with probability \(p\). If there is a total of \(k\) successes, show that each of the \(n ! /[k !(n-k) !]\) possible arrangements of the \(k\) successes and \(n-k\) failures is equally likely.

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