647: Don't be alarmed. All that's listed and attached are just background information, but not a part of the assignment. The actual assignment that I need help with is at the bottom of the page in bold. Thank you.
Event tree analysis starts with an event or occurrence, works forward to identify possible hazards or adverse results, assigns probabilities to those results, and calculates risks. Fault tree analysis starts with an undesirable occurrence, works backward to identify root causes, assigns probabilities to the sequence of causal events, and calculates risks. An event tree and a fault tree for the simple example of inflating a balloon is in the Attachments below. The example comes from the book Should We Risk It?
The event tree starts with blowing up the balloon and then defines pathways of success or failure through a logical series of yes/no questions. The first question asked is, "Is the balloon flawed?' The authors estimate that that there is a 1% chance of the answer's being yes. If the balloon is not flawed, as expected 99% of the time, there is still a chance that you will overfill it to the point where it pops. The authors estimate that will happen in 5% of your attempts. Note that the probability of a yes answer and the probability of a no answer must add up to 100 percent for each question. The final step is to estimate the probability for each possible pathway.
Pathway One: The balloon is not flawed and you do not overfill it. The likelihood or probability of that happening is 0.99 x 0.95 = 0.94
Pathway Two: The balloon is not flawed, but you do overfill it. The likelihood of that happening is 0.99 x 0.05 = 0.05
Pathway Three: The balloon is flawed. As stated earlier, the probability of that is 0.01 (1%)
The fault tree at the bottom of the page starts with the balloon's popping; i.e. failing to inflate properly. It then considers how that might happen. Well, the balloon might be flawed or you might over-inflate it. How might you over-inflate it? Well, first the balloon must not be flawed and second, you must apply too much pressure.
In fault trees, the symbol with the curved bottom is known as an "or gate." (I think of it as the horns of a dilemma.) The symbol with the flat bottom is known as an "and gate." At an "or gate", probabilities add. At an "and gate" probabilities multiply.
The following example will help you remember. Suppose a professor gives you a choice of two options. Option One - You will get an A in the course if you get an A on the mid-term exam or an A on the final exam. Option Two - You will get an A in the course if you get an A on the mid-term exam and an A on the final exam. Which option would you choose?
It should be intuitively obvious that you would choose Option One. Is there anyone who does not see that?
Now let's look at the math to determine how much better off you would be with Option One. Under Option One, your chance of getting an A in the course is close to 50% (0.25 + 0.25)*100. Under Option Two, your chance of getting an A in the course is 6.25% (0.25*0.25)*100.
In this simple case, there are four possibilities:
1. A on the final and A on the mid-term (Probability = 0.25 x 0.25 = 0.0625)
2. A on the final and B on the mid-term (Probability = 0.25 x 0.75 = 0.1875
3. B on the final and A on the mid-term (Probability = 0.75 x 0.25 = 0.1875)
4. B on the final and B on the mid-term (Probability = 0.75 x 0.75 = 0.5625)
Since these are the only possible outcomes, the four probabilities add up to 1.00. Under Option One, the actual probability of getting an A in the course is the sum of the first three probabilities, 44% rather than the 50% that we estimated earlier. The reason the results are different is that adding probabilities at "or gates" is not exact. However, for the very low probabilities associated with catastrophic accident risks, differences will be negligible. Even for this very simple case, the answers are fairly close (44% vs 50%).
Now, let's consider a more relevant example, also taken from Should We Risk It? Attached is a so-called "piping and instrumentation diagram (P&ID) for a batch chemical process. Assume that high-pressure steam, a hydrocarbon mixture, and a catalyst are fed into a tank, and that a single operation of the system takes ten hours. If the pressure in the tank exceeds some level, a weld will burst, destroying the tank. The pressure is regulated by a computer-operated control circuit, designed to maintain the tank pressure between specified minimum and maximum values. If the computer fails, a light comes on in a control room to signal an operator to regulate the pressure manually. Assume that the steam supply exceeds normal pressure 0.1% of the time, that the computer controls overpressures 98.5% of the time, and that the operator is available to see and react to the light 75% of the time. The tank is also equipped with a primary pressure relief valve (PRV 1) that is expected to open successfully 99.9% of the time. The primary pressure relief valve is backed by a secondary pressure relief valve (PRV 2) that is expected to open successfully 95% of the time that it is needed.
Attached is a fault tree for this example (joan-01.jpg).
(a) Calculate the probabilities of tank failure (unsafe operation) by inserting numbers into the attached fault tree diagram.
(b) What measures could be taken to reduce the probability of unsafe operation?