&lt;number&gt; 29 January 2015 … since we are dealing with complex and intricate matters it is necessary to build models of the problem
&lt;number&gt; 29 January 2015 Just before the last lap in the Giro d’Italia in 1999, the Italian Marco Pantani was excluded from the race because of a positive EPO doping test. Marco Pantani was leading the race when he was excluded. In this exercise we will try to address the burning question regarding dope in the cycling sport somewhat closer. The main question is whether the bare knowledge of a doping test may reveal if a competitor has used dope.
Let us assume that the EPO test is able to detect the use of EPO with a probability of 95% (this number is often cited in newspapers). Moreover, let us assume that Pantani just has been tested positive in the EPO test. Our question is then, what is the probability that he is using EPO – or more precisely, what is the probability that Pantani has used EPO given that he was tested positive? No, the probability is not 95%, since we need information on the probability for a false positive test, and the prior probability that Pantani did use EPO.
In the newspapers nothing has been said regarding the false positive test, i.e., the probability of a positive EPO test given Pantani did not use EPO. Let us assume that this probability is 15%. Next we need to know the probability of the Giro d’Italia participants are using EPO (we cannot just say that all participants are using EPO, because then we did not need the testing). Let us assume that 10% are using EPO.
&lt;number&gt; 29 January 2015 1. Why is the cost of the cargo damage dependent on whether or not the damage is repairable? Check the costs at the end of the branch 1Y-2Y-3Y-4N-5Y – 6Y/N. 2. There is not assigned any cargo damage cost to the branch 1Y-2N-3Y-4N, which is wrong. Compare to the branch discussed above. 3. If we check the branch 1Y-2Y-3Y-4N we see that the expected structural damage is E[L]=0.360,000 + 0.7420,000 = 312,000. This branch is comparable to the branch 1Y-2N-3Y-4N (the difference is in whether or not the damage is detected) where the expected structural damage is E[L]=0.2240,000 + 0.8240,000 = 240,000. Why does the effect of damage detection increase the structural damage costs by 30%? 4. In the branch 1N-2N-3Y-4N-5N it is seen that there is assigned a large cargo damage cost to this case when the cargo is not sensitive to humidity. Why? In the branches above this was not the case. 5. Does “Is vessel on voyage?” mean at sea? If “Is vessel on voyage=No” mean that the vessel is at harbour then it is surprising to see that 1 out of 16 lost bulk carriers due to “Damage to Hatchway Watertight Integrity on Bulk Carriers” are lost in harbours. ( P[Loss at sea]= 3.44E-4+8.03E-4+3.28E-5+7.64E-5=1.26E-3 and P[Loss in harbour] = 2.34E-5+5.46E-5 = 7.80E-5). Can this be verified? The cost of cargo damage is given with a precision that does not reflect the uncertainty in the assessment of the costs.
&lt;number&gt; 29 January 2015
Session 42_1 Peter Fries-Hansen
12 January 2010
Bayesian Network and its use in risk analysis
Transportforum, 13-14 januari, 2010, Linköbing, Sweden