Page 196 - DMGT405_FINANCIAL%20MANAGEMENT

Financial Management

Notes
Standard Deviation = 10290000 = 4500

Period 2 Square of Deviation Probability Square of Deviation of
of Mean Mean × Probability
Cash inflow Deviation
from mean
20000 –2900 8410000 0.5 4205000
23000 100 10000 0.1 1000
25000 2100 4410000 0.2 882000
28000 5100 26010000 0.2 5202000
10290000

Standard Deviation = 20250000 = 3208

Period 3 Square of Probability Square of Deviation of
Deviation of Mean × Probability
Mean
Cash Deviation from
inflow Mean

25000 –12000 144000000 0.1 14400000
30000 –7000 49000000 0.3 14700000
35000 –2000 4000000 0.3 1200000
50000 +13000 169000000 0.3 50700000
81000000

Standard Deviation = 810,00,000 = 9000
Standard Deviation about expected values:

189,850 (4500) 2 (3208) 2 (9000) 2
= 43594  
4.355 (1.05) 2 (1.05) 4 (1.05) 6

Probability of NPV being zero or less: Calculate the difference between the specified point and
NPV and then divide by standard deviation (NPV). This is referred to as Z. In this case
0 – 23206
Z = = – 2484. According to Table Z, the Probability of NPV being zero is 0.4934. but the
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area. For the area of the left hand side of the normal curve is equal to 0.5, the probability of he
NPV being zero or less would be 0.5 – 0.4934 i.e. 0.0066. It means there is 0.66% probability that
the NPV of the project will be zero or less.

Greater than zero = 100 – 0.66% (as per above) = 99.34%
Probability index being 1.00 or less: For PI Index to be 1.00 or less, the NPV would have to be zero
or negative. Thus, the probability would be equal to 0.66% as calculated in the earlier part.

23206 – 23206
Atleast equal to mean: i.e. Z =  0
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Reading from the normal distribution table, we get the probability corresponding to 0 as 0.
Therefore, the probability of having NPV at least equal to mean would be equivalent to the area
to the right of the curve i.e., 0.5 = 50%.

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