QAT1 Task 5
SUBDOMAIN 309.3 - QUANTITATIVE ANALYSIS
Competency 309.3.3: Expected Value Decision Analysis - The graduate uses expected value concepts as decision-making tools.
Objective 309.3.3-04: Determine for a given decision tree which decision branch has the most favorable total expected value.
A company is considering alternatives for improving profits: develop new products or consolidate existing products. If the company decides to develop new products, it can either develop several products rapidly or take time to develop a few products more thoroughly. If the company chooses to consolidate existing products, it can either strengthen the products to improve profits or simply reap whatever gains are attainable without investing more time and money in the products.
The “Decision Tree Chart” attachment shows the predicted gains from each decision alternative described above. Gains depend on how the market reacts to the action taken by the company. The probability of each market reaction is shown on the decision tree.
Develop a response to the attached decision tree chart in which you do the following:
A. Calculate the expected value for each of the four decision branches, showing all work or reasoning.
B. Determine the decision alternative that has the most favorable total expected value.
1. Explain how you reached your determination in part B, comparing the results from the 4 decisions branches from part A.
C. When you use sources, include all in-text citations and references in APA format.
Note: To reduce unnecessary matches select ignore “small matches <20 words” and “bibliography”.
Note: For definitions of terms commonly used in the rubric, see the attached Rubric Terms.
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1st decision branch xDevelop new xxxxxxxxxxxxxxx thoroughly)
xxx xxxxxxxx xxxxxx xDevelop new xxxxxxxxxxxxxxx xxxxxxxx
xxx xxxxxxxx branch xxxxxxxxxxxxxx xxxxxxxx xxxxxxxx strengthen products)
xxx decision branch xxxxxxxxxxxxxx existing xxxxxxxxxxxx xxxxxxx investing)
On the first alternative the expected xxxxx xxxx xxxx branch) are 210,200 and 55,700
xx xxx second alternative the expected xxxxx xxxx xxxx branch) are 64,900 and 6,400
xx xxxx alternative xx don´t xxxx the probabilities xxx xxxxxxxxx each xxxxxxx if we xxxxx xxxx we xxxxxx each xxxxxx xxxx xxxxxxxxxxx 1/2 xxx xxxxxxx xxxx
xxxxxxxx xxxxx for xxx xxxxxxxxxxx x xxxxxxxxxxxxxxxxxx x xxxxxxx
xxxxxxxx xxxxx xxx 2nd alternative x (64,900+6,400)/2 = xxxxxx
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