Battery Life Example

BATTERY LIFE

Part one:

1. Probability tree

60%

40%

Good battery         life

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Battery Life

Bad battery         life

Low bulb      failure

Medium bulb      failure

High bulb      failure

Low bulb      failure

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Battery Life

Medium bulb      failure

High bulb      failure

70%

50%

20%

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Battery Life

10%

20%

30%

Probability table

Let           A – be good battery life

B – Be bad battery life

L – Be low bulb life

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Battery Life

M – Be medium bulb life

H – Be high bulb life

L

M

H

total

A

0.42

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0.12

0.06

0.6

B

0.08

0.12

0.2

0.4

total

0.5

0.24

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0.26

1

2. A test was undertaken on a random bicycle light and it was found to have low bulb failure what is the probability that this bicycle light has a good battery life?

0.5 x 0.42 = 0.21

3. Memo

The probability of getting a good battery outcome is 0.6, while the probability of getting a bad battery outcome is 0.4 and the probability of getting either a good or bad battery can be calculated by adding 0.4 + 0.6 = 1.

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For the conditional probability we will have good battery life then the probability that the bulb will have low failure is 0.7, the probability that we will have medium bulb failure is 0.2 and finally the probability that the bulb will have high failure is 0.1.

For the condition that the battery has bad battery life then the probability that we will have low bulb failure is 0.2, the probability that we will have medium bulb failure is 0.3 while the probability that there will be high bulb failure is 0.5.

From the above conditions it is evident that the probability of bulb failure depends on the nature of the bulb whether the battery has good battery life or bad battery life. When we have good battery life then the lower is the probability that the bulb will fail and on the other hand if we have bad battery life the higher the probability that we will have bulb failure. Another observation is that if we randomly choose a battery we have greater chances of choosing a battery with good battery life.

Part two:

Expansion or acquisition

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Battery Life

1. Decision tree

70%

30%

Expansion

Success buy      out

Increase      revenue by 100million

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Battery Life

Increase      revenue by 30million

60%

40%

China                                           manufacturer

Glass      competitor

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90 million        increase

70million      increase

20%

80%

86million      increase

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72 million        increase

50%

50%

50%

50%

2. Probabilities and expected values:

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Glass competitor

China                                 manufacturer

expansion

total

amount

72 million

86 million

90 million

70 million

30 million

100 million

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Battery Life

probability

0.175

0.175

0.28

0.07

0.12

0.18

1

3. Optimum strategy

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We get the above probabilities by multiplying the values of each probability, the probability of a 72 million increase in revenue through merging with a glass competitor is calculated by multiplying 0.7 x 0.5 x 0.5 = 0.175, and this probability is the same as the one that will give us an increase of 86 million through buying the glass competitor.

By buying the china manufacturer the probability that there will be an increase in revenue by 90 million is got by multiplying 0.7 x 0.5 x 0.2 = 0.28, while an increase of 70 million in revenue probability is got by multiplying 0.7 x 0.5 x 0.2 = 0.07.

Through an expansion of the existing facility there is a 0.3 x 0.4 = 0.12 probability that there will be a 30 million increase in revenue whereas there is a 0.3 x 0.6 = 0.18 probability that there will be an increase in the level of revenue by 100 million.

The optimum strategy is that which has the highest probability, in our case by merging with the china manufacturer and expecting 90 million has an outcome of 0.28 which is the highest of all the others, this would be the best strategy because it has a high increase of 90 million on revenue and also highest probability to occur.

REFERENCE:

Ray R and C Mc Keenan (2001) Mathematics for Economics, MIT Publishers , USA

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