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Chapter 4: Problem 29

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
The condition under which we should check the first possible cause of the breakdown and then the second is \(p > \frac{R_{2} - C_{1}}{R_{1} - R_{2}}\). If \(p\) is less than this value, then it's more cost-efficient to check the second possible cause first and then the first.

Step by step solution

01

Calculate the expected cost for checking the first cause and then the second

Let's first look at the scenario where we check the first cause of breakdown and then the second. The possible outcomes are: 1. The first cause is the actual cause of the breakdown, so we pay the checking cost \(C_{1}\) and the repair cost \(R_{1}\). 2. The first cause is not the actual cause, so we pay the checking cost \(C_{1}\), then check the second cause and pay the checking cost \(C_{2}\) and the repair cost \(R_{2}\). The probability of the first outcome happening is \(p\), and the probability of the second outcome happening is \(1-p\). The expected cost, \(E_{1 \to 2}\), is the weighted average of the costs for these two outcomes: \[E_{1 \to 2} = p(C_{1}+R_{1}) + (1-p)(C_{1}+C_{2}+R_{2})\]
02

Calculate the expected cost for checking the second cause and then the first

Now let's look at the scenario where we check the second cause of breakdown and then the first. The possible outcomes are: 1. The second cause is the actual cause of the breakdown, so we pay the checking cost \(C_{2}\) and the repair cost \(R_{2}\). 2. The second cause is not the actual cause, so we pay the checking cost \(C_{2}\), then check the first cause and pay the checking cost \(C_{1}\) and the repair cost \(R_{1}\). The probability of the first outcome happening is \(1-p\), and the probability of the second outcome happening is \(p\). The expected cost, \(E_{2 \to 1}\), is the weighted average of the costs for these two outcomes: \[E_{2 \to 1} = (1-p)(C_{2}+R_{2}) + p(C_{1}+C_{2}+R_{1})\]
03

Compare the expected costs for both scenarios and derive the conditions

Now, we need to compare \(E_{1 \to 2}\) and \(E_{2 \to 1}\) to find out under which conditions we should start with checking the first cause of the breakdown. To minimize the cost, we want to choose the scenario with the lower expected cost. We should check the first cause first if: \[E_{1 \to 2} < E_{2 \to 1}\] Substituting the expressions for \(E_{1 \to 2}\) and \(E_{2 \to 1}\), we get: \[p(C_{1}+R_{1}) + (1-p)(C_{1}+C_{2}+R_{2}) < (1-p)(C_{2}+R_{2}) + p(C_{1}+C_{2}+R_{1})\] Now, let's simplify the inequality and solve for \(p\): \[pR_{1} - pR_{2} < R_{2} - C_{1}\] \[p (R_{1} - R_{2}) < R_{2} - C_{1}\] \[p > \frac{R_{2} - C_{1}}{R_{1} - R_{2}}\] So, the condition under which we should check the first possible cause of the breakdown and then the second is: \[p > \frac{R_{2} - C_{1}}{R_{1} - R_{2}}\] If \(p\) is less than this value, then it's more cost-efficient to check the second possible cause first and then the first.

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Key Concepts

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

Expected Cost
Expected cost is an important concept in probability theory, especially in decision-making scenarios where uncertainties are involved. In this case, when faced with machine breakdowns due to two possible causes, we need to carefully analyze the expected costs associated with different strategies for checking and fixing the problems.
The expected cost is essentially the average cost we anticipate, factoring in the probabilities of different events. Here, we calculate the expected cost by multiplying the cost of each outcome by its probability, and then summing these values. This calculation helps us understand the long-term cost implications of each decision.
For our problem, the expected costs for different strategies are calculated as follows:
  • Checking the first cause first: \[E_{1 \to 2} = p(C_{1}+R_{1}) + (1-p)(C_{1}+C_{2}+R_{2})\]
  • Checking the second cause first: \[E_{2 \to 1} = (1-p)(C_{2}+R_{2}) + p(C_{1}+C_{2}+R_{1})\]
By determining which strategy has the lower expected cost, we optimize our decision to minimize expenses.
Decision Making
Decision making is a critical skill in any scenario involving uncertainty, like diagnosing machine failures. Here, we must decide the order in which to check possible causes, aiming to minimize the cost of repairs.
When making decisions under uncertainty, it involves weighing the costs and probabilities of different outcomes. For instance, should we check the more likely cause first, bearing in mind the repair and checking expenses? Or, is it cheaper to start with the less likely cause, considering all costs involved?
In our breakdown scenario, the decision hinges on the outcomes of calculated expected costs. The optimal decision is to check the first cause if:
  • \[p > \frac{R_{2} - C_{1}}{R_{1} - R_{2}}\]
If not, it's more cost-effective to check the second cause first. This approach ensures that our decision is based on a thorough analysis of all costs and probabilities, leading to smarter and more efficient troubleshooting.
Probability Calculation
Probability calculation is the bedrock of this problem, as it dictates how we compute expected costs and make informed decisions. The probabilities involved represent our beliefs about the likelihood of each cause for the breakdown.
In our example, we denote by \(p\) the probability that the first cause is responsible, and \(1-p\) for the second cause. These probabilities are crucial as they affect every calculation. Knowing how likely each scenario is, helps you weight the costs accordingly.
The main tasks in probability calculation include:
  • Defining probabilities clearly to represent the likelihood of potential causes.
  • Using these probabilities in expected cost calculations to weigh potential financial consequences.
Through careful probability calculation, you ensure that your decision-making process is firmly anchored in reality, with a clearer understanding of the chances that each possibility might actually occur.
Optimization
Optimization in this context means finding the most cost-effective strategy for checking machine causes. It's all about ensuring minimum expense and maximum efficiency in decision-making.
By comparing the expected costs of each possible checking order, we can identify which strategy will save more over time. This problem involves balancing probabilities and associated costs to reach the optimal sequence of checks.
Here are some key components:
  • Identify and calculate all possible costs for each scenario.
  • Assess which probability distributions apply, affecting the expected costs.
  • Determine the threshold \(p\) value for deciding whether to check the first or second cause first.
Optimization allows for a mathematical approach to minimize costs, thereby extending resources efficiently. It serves as a fundamental tool to enhance practical decision-making in probabilistic settings, ensuring the best possible outcomes while controlling expenses.

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