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Instructions: For decision theory, solve and answer the word problems using spreadsheets or excel following the different techniques in decision-making . In decision tree analysis,

Instructions: For decision theory, solve and answer the word problems using spreadsheets or excel following the different techniques in decision-making. In decision tree analysis, draw a decision tree for each of the following decision situation. Show each decision point and each chance event. Insert probability values and financial data. Use rollback or fold back method to analyze each situation. Instructions for submitting will be posted in the Google Classroom. Good Luck!

1. The manufacturer, WanTo3 produces products that have a probability to be defective.

These products are divided into batches of 200. Past experience reveals that some (batches) are of good quality (i.e. p=0.05) and some are of bad quality (i.e. p=0.25). In addition, 80% of the batches produced are of good quality and 20% of the batches are of bad quality. These products are then used in an assembly, and before the final assembly leaves the factory, their consistency is finally determined. At a gross average cost of 15 per item, the manufacturer may either screen each item in a batch and replace defective products, or use the items directly without screening. Ultimately, if the latter action is taken, the rework cost is 150 per defective product. For these data, the costs per batch can be calculated as follows:

p = 0.05 p = 0.25

Screen 3000 3000

Do not Screen 1500 7500

Because screening requires inspectors and equipment scheduling, it is important to make the decision to screen or not to screen 2 days before the proposed screening takes place.

However, manufacturers can take one item from a batch and submit it to a laboratory, and it is possible to report the test results (defective or non-defective) before making the screen/no-screen decision. The tested object is returned to its batch following the laboratory examination. The cost of this initial inspection is 175. Also note that the probability that a random sample item is defective is

0.8 * 0.05 + 0.2 * 0.25 = 0.09,

and the probability that an item in a lot is of good quality given a randomly sampled item is defective is 0.444 and the probability that an item in a lot is of good quality given a randomly sampled item is not defective is 0.835. The manufacture wants to minimize his or her cost.

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