Hello friends i have came across a question in capital budgeting . relevant extracts are as follows
"find out the probability of occurance of the worst case if the cash flows are
a) perfectly dependant overtime
b) independent overtime
"
there are 3 cases in the question , the worst case has probability of 0.3
kindly tell me how do solve such kind of problems , because this question was asked in Nov 2006 exams .
looking forward for answer .
Hi Mayank..Will try to explain it through an example. Let the initial investment be 30,000, rate of return 10%,project life 3 years & scrap value nil. Option A:cash inflows for 1st year,2nd year & 3rd year be 10,000,Rs13,000 &15,000 respectively. Probability is 0.6. Option B:cash inflows for 1st year,2nd year & 3rd year be 13,000,Rs.17,000 & 21,000 respectively. Probability is 0.4.
thank you so much jitin ji . you have solved my problem is a very good way .
thank you so much once again :)
Welcome :)
| Originally posted by : Jithin | ||
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a)Independent over time:There is 0.6 probability for the cash inflow to be Rs10,000 in the first year. Since the cash inflow of the 2nd year 13,000 is not affected by the cash inflow of the first year, the probability for the 2nd year shall also be 0.6. Similarily the probability for 3rd year will be 0.6. Hence the probability of worst case when cash flows are independent=0.6. b)Perfectly dependent over time:Here the cash inflows of an year is affected by the cash inflows of the previous year. For example, the cash inflow for 2nd year shall be 13,000 only if the cash inflows of 1st year are Rs10,000. Similarily, the 3rd year cash inflows are 15,000 only when the 1st year & 2nd year cash inflows are 10,000 & 13,000 respectively. When the occurence of 'A' depends upon the occurence of 'B', the probabilty of both 'A' & 'B' happening together will be the probability of 'A' multiplied by probability of 'B'. Thus,the probabilty of occurence of the worst case=0.6 x 0.6 x 0.6=0.216. |
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Dear Mayank, yesterday when I was doing a similar type of problem, I realised that my understanding on the issue was totally wrong. The explanations given above are not correct. Sorry for giving misleading answers. By now u shall have learnt the correct method to do these types of problems..Neverthless I am giving the correct explanations & answers(based on the above example itself). Once again SORRY..
Probability of the Worst Case
(a)Perfectly dependant overtime:When the cash flows are perfectly dependent overtime, the entire series of cash flows is based on the initial cash flow. Once u ascertain the initial cash flow, the whole stream of cash flows that follows the initial flow is known. In the above example, when u get Rs.10,000 in the first year, it is certain that u will get Rs.13,000 & Rs.15,000 in the 2nd & 3rd year respectively. Once u know that the first year cash inflow is Rs.10,000,then u r sure that the cash flows u receive is of the worst case. So for ascertaining the probability of the worst case, u have to just know the probability of the initial cash flow. In the above example, the probability of the worst case is 0.6 since it is the probabity of getting Rs.10,000 in the first year.
(b)Perfectly independent overtime: Where the cash flows are perfectly independent overtime, the cash flow of an year isnt influenced by the cash flows of the preceding or succeeding years. For example, in the above case, if u receive Rs.10,000 in 1st year, it isnt necessary that u receive Rs.13,000 in the 2nd year or Rs.15,000 in the 3rd year. So, unlike in the case of cash inflows dependent over time, the probability of getting Rs.10,000 in the first year isnt enough for determining the probability of the worst case. We have to also take into account the probability of getting Rs.13,000 & Rs.15,000 in the 2nd & 3rd year respectively. In other words, we have to ascertain the joint probability of worst case cash inflows,i.e.the probability of receiving worst case cash flows in all the three years. Joint probability of event A & event B occurring together, where both A & B are independent=Prob. of A x Prob. of B. Therefore probability of the worst case=0.6 x 0.6 x 0.6=0.216.
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