In the given case study, the IT department’s head is under duty to act on the company’s inventory control system (ICS). To address the issues connected with the ICS effectively, the head can opt for either of two alternatives. He or she should base the choosing of the alternatives on a decision tree along with a Delphi procedure. Notably, decision trees were originally used as graphical tools for illustrating the structural connections between different choices. Originally, they were presented as sequences of choices that were dichotomous (Hanke & Reitsch, 1992; Modis, 1992). Even then, they are now used in presenting more and more intricate choices as the existing appreciation of feedback loops increases. Indeed, the intricate natures of decision trees along with a Delphi procedures have seem them grow into the elementary basis of particular computer flowcharts.
The following is a tabulation of the factual data provided in the case study.
| Alternative | Outcomes | Productivity increase ($) | Probability (%) | Total cost ($) |
| 1 | Great benefit | 15,000 | 40 | 10,000 |
| Some benefit | 10,000 | 50 | 10,000 | |
| No benefit | 0 | 0 | 10,000 | |
| 2 | Great benefit | 8,000 | 40 | 4,000 |
| Some benefit | 4,000 | 30 | 4,000 | |
| No benefit | 0 | 0 | 4,000 |
To get the expected, or most likely, productivity increase, the stated productivity increase should be multiplied with the probability of its being released. That means under Alternative 1, the most likely “great benefit” productivity increase is:
15,000 * 40/100 = $6,000
Under Alternative 1, the most likely “some benefit” productivity increase is:
10,000 * 50/100 = $5,000
Under Alternative 1, the most likely “no benefit” productivity increase is:
0 * 0 = $0
As well, it means that under Alternative 2, the most likely “great benefit” productivity increase is:
8,000 * 40/100 = $3,200
Under Alternative 2, the most likely “some benefit” productivity increase is:
4,000 * 30/100 = $1,200
Under Alternative 2, the most likely “no benefit” productivity increase is:
0 * 0 = $0
| Alternative | Outcomes | Most likely productivity increase ($) | Total cost ($) |
| 1 | Great benefit | 6,000 | 10,000 |
| Some benefit | 5,000 | 10,000 | |
| No benefit | 0 | 10,000 | |
| 2 | Great benefit | 3,200 | 4,000 |
| Some benefit | 1,200 | 4,000 | |
| No benefit | 0 | 4,000 |
To calculate the most likely net benefit owing to the adoption of any of the Alternatives, the corresponding difference between the projected total cost and the most likely productivity should be computed. That means under Alternative 1, the most likely “great benefit” net benefit increase is:
6,000 – 10,000 = -$4,000
Under Alternative 1, the most likely “some benefit” net benefit is:
5,000 – 10,000 = -$5,000
Under Alternative 1, the most likely “no benefit” net benefit is:
0 – 10,000 = -$10,000
Under Alternative 2, the most likely “great benefit” net benefit is:
3,200 – 4,000 = -$800
Under Alternative 2, the most likely “some benefit” net benefit is:
1,200 – 4,000 = -$2,800
Under Alternative 2, the most likely “no benefit” net benefit is:
0 – 4,000 = -$4000
The following decision tree should be used in choosing the most beneficial alternative
| Net Benefits | ||||||
| Alternative | 1 | 2 | ||||
| Great Benefit | Some Benefit | No Benefit | Great Benefit | Some Benefit | No Benefit | |
| Expected net loss in benefits | 4,000 | 5,000 | 10,000 | 800 | 2,800 | 4,000 |
From the tree, it is clear that the alternative that will result in the most minimum loss is the Alternative 2 “great benefit” choice. The company should purchase refresher training for enhancing employee productivity and design it to give a benefit of $8,000.
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