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- February 28, 2024 at 6:45 am #701348
I am hoping you can help, I have watched the video on the topic however completing a Kaplan mock there is a question that I do not understand even with the workings. – I have both Kaplan and BPP books but cannot find any information RE this type of question to aid me.
What would the probability of a profit at which a managers decision to accept a project would change as a %
Profit 150000 Probability 40%
Loss -80000 Probability 60%February 28, 2024 at 3:16 pm #701376sorry should have said this is the answer:
P × 150) + [(1–p) × -80]If EV falls to zero:
(P × 150) + [(1–p) × -80] = 0
150P -80 + 80P = 0
230P – 80 = 0
230P = 80
P = 0.3478 = 35%
however i am unsure on this formula as never seen this before so I do not understand the logic behind the workings
April 29, 2024 at 7:58 pm #704708Hi, Sorry for the delay and definitely a weird format question.
So the principles of sensitivity is that its the change in variables needed to ‘reverse’ the decision.
Im assuming you are happy that, initially the EV of the project is positive (0.4*!50k) _ (0.6*-80k) = 12k positive (accept)
If the EV drops to zero then thats the point where we would start to reverse the decision.
So we need to find a probablity of earning the 150,000 that is exactly netted off by the result of probability of earning (80k).
The values of 150k and (80K) loss dont change.
Also remember in a two-situation outcome then probabilities have to add up to 100%So that fancy formula above calculates that
34.78% * 150,000 = 5,217,000
then the loss probability (100 – 34.78 = 65.22%)
65.22% * (80,000) = approx the same (5,217,600)
Therefore making a loss and change of decision.
ITs not working out exactly as zero due to rounding so presumably the question asked for the closest full percentage point or similar ( hence the answer is 35%)
Overall though id say dont worry about that one too much – it’s an unusual question format! 🙂
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