PRESENTER: Let’s focus on a far more complicated example. Let’s assume a drilling project. 100,000 at time 0. That will be paid for all the instances. 500,000 of drilling costs at year 1. Again, this cost is paid for all the full situations. 500,000 of drilling costs, we get to the completion point.

400,000. And there is 40% failing that they don’t look good enough. And we need to close the wells and pay the abandonment cost etc. 400,calendar year in the year 1 for completion costs 000 more in the same. In this case, we will face three cases. 300,000 per year from year 2 to year 10. And we will have 15% probability of that the well completion being unsuccessful.

So we can summarize the info here. 100,000 of rent costs for all the instances at the moment time. 500,000 of drilling cost at year 1 for all the cases. And then we’ll have 60% of probability of addressing the completion. So in case there is the 60%, we will face three new situations.

300,from season 2 to year 10 000. And 15% probability that people don’t finish up doing anything, hardly any money, any producing well. So decision tree is a very helpful graph that can help us split the possible instances here. So I will describe this in this graph. So we start from the left hand side, initial investment for the lease at the present time. We write the income or cost here.

And before that we write the possibility. So this probability is 100% because it is the same for all your cases. 500,000 for drilling and in cases like this then, we will have two branches being deviated from the primary branch. The first is the 60%, let’s call it success here, 60% of success and 40% of failure. 40,000 of shutting costs, abandonment costs. In case of success 60%, we will face another three cases. 400,000 of completion costs.

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This 1 plus is to show that this is the same 12 months as this year. Yr These are occurring in the same. But because these cases are deviated from the main branch, another branch is drawn by us for these, to separate these from the main branch. 400,000 of completion cost. And we shall have three new cases in the after. 450,000 from year 2 to year 10. So years here are.

*So every value under the same* column gets the same year sizing. So as we can easily see here, we here have four main cases. Case A, case B, case C, and case D. So the first step to approach this problem and calculate the expected NPV is to compute the probability of each case.

So to be able to calculate the possibilities of every case, we get back to your choice tree. We start from the right hand side for each full case. For example, for case A. So I begin from the right hand side. For example, case A I start moving from right hand aspect toward the left.