Royal Dutch Shell Evaluation of Oil Reserves — страница 14

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restatement announcement and in fact showed no reaction as the final restatement came in on April 19, 2004. This lack of obvious connectivity between the events and market reaction makes the analysis of stock movement more difficult. As it will be shown later in this chapter when the results of case study are discussed use of abnormal stock return is not solving this Royal Dutch Shell: Evaluation of Oil Reserve problem, since the move of abnormal returns resembles to a very large extend the move of stocks total returns. Another problem one is facing when assessing results of reserves restatement is the actual volume of restatement hidden behind the figures announced by management. As it was shown in Chapter 2, company’s management is not allowed by SEC regulations to report

any reserves estimation apart from proved reserves or the reserves the company is going to produce with reasonable certainty according to so called Rule 4-10. Therefore, the announcements, company made regarding the volume of its reserves, have only been dealing with the proved reserves and not with the overall oil and gas reserves in place. It has also been shown in Chapter 2 that the figures reported by companies under SEC regulations often represent the mode of oil reserves distribution function. On the other hand, the investors would probably focus not on the mode of distribution, but rather on the expected quantity of reserves or the mean. These two figures may be quite different as it is shown on figure 3.2: If one accepts perfect market assumption, it is possible to assume

that market participants do not relay on the official figures produced by the company, but they rather use the inside information and have the entire distribution function available. Then, assuming that the data regarding available oil reserves is log-normally distributed, market players would take into account the expected value of oil reserves E (x), which is equal to: σ 2 E( x ) = e µ+ 2 (1) Where µ represents the mean of ln (x) and σ represents standard deviation of ln (x). It is also possible to calculate the formula for the mode value of lognormal distribution. Knowing that lognormal distribution density function is represented by − 1/ 2(ln( x ) −µ )2 σ ( x f ) =ex σ 2π (2) one can calculate the first derivative of

(2) and than equate it to zero in order to receive the highest point of the distribution density function or in other words the mode. ln( )− ln( )− ()( − = ′ ex f 2/1 ( µ x σ )2 1 21)))(1 ln(( − −−− ⋅ π σ σσ µ ex x x 2/ ( µ x σ )2 122 )2()2 −⋅⋅ σ π π σ x (3) By equating (3) to zero, one will get the mode value as: )( xM 2σ µ −= e (4) This means that E (x) can be represented as: )( xE = 22 5 . 1 σσ µ e +− = M )( x 25 . 1 σe ⋅ (5) ~ Than the change in expected value of oil reserves )(x E is: )( ~ x E = 2 15 . 1 1 )( σexM ⋅ − 2 25 . 1 2 )( σexM ⋅ (6) The value of M

1(x ) − M (x ) is precisely the figures announced by the management of 2 Royal Dutch Shell and represents the restatement of company’s proved reserves. If one assumes that standard deviation of natural logarithms of oil reserves σ remained unchanged as the result of restatement, (6) can be rewritten as follows: ~ ( This assumption, allows to make the analysis much more simple and to come to the conclusion regarding the size of actual restatement with out requiring a lot of additional data. In equation (7), there is still one unknown value on the right hand side, namely the value of σ. Sigma can be estimated from the data provided by company management. According to company’s production and exploration presentation, published on web

site, management estimates the “total reserves” to be about 60 billion barrels after the restatement (RDS Group: Regaining Upstream Strength, 2004). The expression “total reserves” in this case might be referred to the possible reserves or P10. It is also known that proved (or P65) reserves numbered to 14.3 billion barrels (RDS Group: F-20 Form, p G56 sqq, 2003). The distance between these two values in lognormal distribution can account roughly to 2.1 standard deviations (this results can be obtained by simulating the distribution density function in statistical package such as @Risk). In this case sigma can be estimated as (ln(60)-ln(14.3))/2.1. That gives sigma value of approximately 0.68 and therefore the announced restatement should be multiplied by a