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John Moffat.
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- November 9, 2022 at 4:39 pm #671138
Hi John,
One question regarding the calculation of expected NPV. There is a question in the exam KIT (Q162). It includes the scenarios of Year 1 & Year 2. There is a calculation table with following columns:
PV of the YR 1; prob; PV YR 2; prob; Total PV; Joint Prob; PV*JP and NPV. And all that for 3 different scenarios. Will it be possible for you to explain a question of this kind?
Thanks a lot, DariaNovember 10, 2022 at 11:45 am #671175You will have to tell me which exam kit you are referring to (I only have the current edition of the BPP Revision Kit).
However if it is marked as being a past exam question then please tell me the name of the question (and date of the exam) because I have all past exam questions so it will be then easy for me to find it and to answer you 🙂
November 10, 2022 at 1:26 pm #671188Hi John,
thank you for your answer and apology for several posts – internet issue & then all got posted… I use BPP practice & revision kit for exams Sep’22 – Jun’23. Question 162 “Copper Co” (March/June 2018) on page 59. My question is how to distinguish when scenarios have to be calculated and when we can just simply use the weighted average technique?
Thank you,
DariaNovember 10, 2022 at 4:15 pm #671198This is something the examiner has asked two or three times in the last 15 years, so it is not that popular (although it obviously can be asked again. Part (a) of the question is using knowledge from both Paper MA and Paper PM.
First, because it wants the expected NPV you need to discount all of the cash flows for 1 or 2 years.
There are nine possible outcomes. It could be 1,000 in the first year then 2,000 or 3,000 or 5,000 in the second year. It could be 2,000 in the first year and then 2,000 or 3,000 or 5,000 in the second year. It could be 3,000 in the first year and then 2,000 or 3,000 or 5,000 in the second year.
To get the probability of each of the nine outcomes, we multiply the probabilities of the two separate outcomes together.
So the probability of 1,000 in the first year and 2,000 in the second year is 0.1 x 0.3 = 0.03
You can then set up a table as in the answer. The total present value for each of the 9 outcomes is simply adding the two present values together (the column headed up ‘Total PV’) and we then multiply by the probability of the relevant outcome (the column headed ‘PV x JP’).
Adding up this column gives the total expected present value, and if we subtract the initial investment we get the expected NPV (which is the answer to part (i) of the question.
The final column in the table (headed up ‘NPV’) is the NPV of each of the possible outcomes (the ‘total PV’ less the initial investment of 3,500.
Some of the NPV’s are negative and so the probability of a negative NPV is adding up the probabilities of those outcomes that result in a negative NPV.
The NPV of the most likely outcome is the NPV with the highest probability.
Although you sensibly want to understand the calculations in case something similar is asked again, and I hope what I have written enables you to now make sense of the answer, it is a good question to use to illustrate the importance of exam technique. Many people in the exam hated this question and got very low marks as a result, but if they had approached it properly they could still have got over the 50% then you should be aiming for on both of the section C questions.
Part (b) did not need anything from part (a) and only wanted discussion of two of the methods. Anyone who had watched my lectures on this should have been able to get certainly most of the 8 marks.
For part (a) everyone should have been able to discount so as to get the PV’s of all of the flows and to get the total PV of each possibility – that would be 2 marks.
Even someone who could not do parts (a) (ii) and (a) (iii), should have been able to get 1 if not 2 marks for part (a) (iv) for discussion. Just inventing figures and discussing, or stating that if the expected NPV was positive then it would be acceptable but with the risk that the actual NPV might be higher or might be lower because of the uncertainty would get 1 or hopefully 2 marks.
So even someone who had no idea how to do part (a) in full should still have been able to get at least 10 marks and therefore be passing the question. The problem is that too many people spent all their time trying to sort out part (a) without success and had no time left to do part (b) and therefore completely wasted 8 fairly easy marks.
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