Business finance : theory and practice

Chapter 6 • Risk in investment appraisal
F: NPV = −50,000 + {4,500 × [10 − (5 + 3)] × 3.791} = −£15,881
G: NPV = −50,000 + {5,000 × [10 − (3 + 3)] × 3.791} = +£25,820
H: NPV = −50,000 + {5,000 × [10 − (4 + 3)] × 3.791} = +£6,865
I: NPV = −50,000 + {5,000 × [10 − (5 + 3)] × 3.791} = −£12,090
Listing these in order of size, we get:
Possible outcome NPV Probability
£
C −19,672 0.04
F −15,881 0.10
I −12,090 0.06
B −4,508 0.14
E +1,179 0.35
H +6,865 0.21
A +10,656 0.02
D +18,238 0.05
G +25,820 0.03
1.00
Note that, given the data, one of the above nine outcomes must occur, and only one can
occur.
‘
As we have seen, a very modest attempt at reality causes complications. We can
well imagine the number of possible outcomes were we to explore a more realistic set
of possibilities for each of the input factors, even with our simple example.
How useful to us is the array of nine outcomes and probabilities shown above? The
NPV rule is to accept projects with positive NPVs and reject those with negative ones.
Here we know that four of the nine possible outcomes will yield a negative NPV, yet
the other five give positive ones. One of these nine possibilities will occur, but which
one? One way forward on the problem of the vast array of possible outcomes, and the
subsequent problem of making the decision once we have them, is through the notion
of expected value.
6.4 Expected value
As ever, when we are confronted by a mass of data, it can be useful if we can find some
way of summarising it so that we can more readily absorb and use it.
One way of summarising is to calculate a weighted average or expected value. This
involves multiplying each possible outcome by its probability of occurrence. Taking
the array of outcomes from Example 6.3 gives us the following result:
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