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hi sir i have watched the lecture relating to this topic but i don’t get how to do copper co A) i
The actual discounting involved is easy. Most of the work involved using techniques from previous exams (Papers FA and PM – were F2 and F5)).
There are 9 possible things that can happen:
1M in the first year and then 2M in the second year, with a probability of 0.1 x 0.3 = 0.03 or
1M in the first year and then 3M in the second year, with a probability of 0.1 x 0.6 = 0.06 or
1M in the first year and then 5M in the second year, with a probability of 0.1 x 0.1 = 0.01 or
2M in the first year and then 2M in the second year, with a probability of 0.5 x 0.3 = 0.15 or
2M in the first year and then 3M in the second year, with a probability of 0.5 x 0.6 = 0.3and so on.
Calculate the PV of each of the nine alternatives and then the rest should follow sensibly 🙂
Sir, I dont understand how did they calculated 1,013 , 216 , 1,378 etc – NPV
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