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October 3, 2015 at 2:10 pm
“In either case, the probability of the refurbishment achieving a good result has been estimated to be 2/3.”
Sorry I am not native English speaker, so ‘either’ mean in each/every case. Right?
I think this sentence would be really tricky in case I hadn’t seen the lecture first.
John Moffat says
October 3, 2015 at 2:27 pm
Yes – you are right (it means in both cases).
May 30, 2015 at 3:18 am
Hello Mr. Moffat,
I learned in class that in the decision tree, the decision points and the outcome points should be labelled from the top right of the tree, then moving across to the left…
I was also taught that the decision forks you drew to indicate the decision you made, were actually supposed to be placed on the decisions you didn’t want to make…
Are these points worth stressing over?
Or the way you did it is acceptable?
May 30, 2015 at 10:38 am
Of course the way I have done it is acceptable (otherwise I would not have done it that way!!!)
May 29, 2015 at 4:36 pm
Apology dear John, have just rewound the last part of the lecture. Point D clarified my question.
(I just wish we could use colour pens in the exam…)
May 29, 2015 at 4:27 pm
Thank you for your brilliant lectures, I much prefer this style, a bit quicker than in the classroom recording, however, both are good quality.
I have a question on this subject, perhaps a repetition of Chris’s question in Oct 14. As we are working the decisions backwards, we would have known the “best” option to go for. In this instance, the option to go with market research with expensive refurbishment. However, at this point, if the return is then diminished by the cost of the market research, would going without the market research and refurb expensively be the better option? Thank you kindly.
May 19, 2015 at 6:07 am
dear Mr. Moffat , I have been doing these decision tree questions and find a difference that has often caused my answer to come out wrong.Here is a sample question:
Q.a company has prepared the design for a new product .It can either sell the design for $100000 or attempt to develop the design into a marketable product at a cost of 150000 .If the company decides to develop the product the chances of success are .7 .If the attempt fails the design can be sold for 20000.If the attempt succeeds the business has the choice of either selling the design and developed product for 180000 or marketing the product.If the product is marketed then there is a .6 probability that the product will generate a cash inflow of 800000 and a 0.4 probability that it will generate a cash outflow of (100000).Both figures exclude items previously mentioned.
, the answer shows tan ev of 164000 of developing.This has been arrived by subtracting the 150000 from each outcome of developing i.e. from both 800000 and -100000 BEFORE taking the expected value of these outcomes.Now i simply took 800000 and -100000 as is and later when i rolled back to the point where development came i.e. the first of the two branches of developing or selling , there I subtracted the 150000.But my answer doesn’t match , in the question that you have done in the video .Which approach should we take when we have a common cost for one single Option leading numerous other options should we subtract that cost from each option and then take expected values or should we find out the EV of that option first and later subtract that cost.
Please solve this confusion Mr. Moffat you’ve always been great help.Appreciate all that you do.(p.s i got an 85 in f3 from Only watching your videos so Thanks mate!)
May 19, 2015 at 8:15 am
In future, please ask questions like this in the Ask the Tutor Forum, not as a comment on a lecture.
Either approach will always give the same answer (and the 164,000 is correct).
If you take the 150,000 from all of the flows before you start rolling back, then second decision is the choice better selling and receiving a net 130,000; or marketing and receiving ((0.6 x 650,000) – (0.4 x 250000)) = 290,000. So you chose to market.
So now the expected value of continuing in the first place is (0.7 x 290,000) – (0.3 x 130,000) = 164,000.
(I am guessing that maybe you did not deal properly with the fact that the net outcome of some of the options is negative)
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