5970 CEE Risk Analysis and Management
Final EXAM
Mon. May 17, 2010, 9-11:30 am
Test is open book & open notes. You have 150 minutes for this 150 point exam.
Write clearly and show important steps. Answer essays carefully and completely.
1. (12 pts) In not more than 1 page in your exam book, describe the profound lessons for risk
analysis and risk management in our society that one should draw from the class readings
and discussions surrounding the loss of the Challenger and Columbia shuttles. In
particular, (a) What were the important lessons? (b) What are the implications of those
lessons for analysis and management of risk in our society? (c) How did both technical and
social/managerial issues contribute to the problem?
2. (13 pts) Ithacas weather can change suddenly so it is hard to know how to dress. Stewart
lives in Ithaca and gets on average 2 colds every 3 years; colds have the same arrival rate in
all 12 months. (Beware of cold summer rains.) Assume getting a cold is a Poisson process.
(a) How many colds will Stewart get on average during a 24 month period?
(b) If Stewart is well now, how long on average before he experiences 5 colds? What is the
variance of the time until he has 5 colds?
(c) What is the probability of exactly 3 colds in one 9-month-long academic year?
3. (15 pts) Consider the fault tree below. (Antoine created this one.).
A
B
C
D
E
F
(a) (8 pts) Please draw the equivalent network diagram such as that in problem 4 below.
(b) (7 pts) If the probabilities of failure are
P(A) = 0.10; P(B) = 0.25; P(C) = 0.30; P(D) = 0.4;P(E) = 0.45 and P(F)=0.5;
what is the probability of a system failure?
4. (15 pts) Consider the network below.
(a) (6 pts) Please list all of the minimal cut sets with 1, 2, 3 or 4 elements.
(b) (4 pts) If all elements are 99% reliable, what is the reliability of the system?
(c) (3 pts) To save money, which 3 components could be replaced with a unit
that has a 90% reliability while having a small impact on overall system reliability?
(d) (2 pts) If you could improve one component, which would you select and why?
2010 CEE 5970 Examination
Page 2 of 7
B
A
C
D
F
E
G
H
5. (20 pts) Assume significant oil spills at offshore platforms happen once every 2 years. The
spill can be small (10), medium (50), large (500) or very large (10,000) with probabilities 0.4,
0.3, 0.2 and 0.1 respectively. However, if weather is favorable, and the response is quick, the
effective volume of the spill can be reduced by 75% (e.g. a volume of 10,000 becomes 2,500);
this reduction occurs 65% of the time. Draw the event tree. Generate and draw the risk profile
for the magnitude of the effective spills.
6. (15 pts) Everyone loves the new consumer-grade GPS devices. They have maps and give
driving instructions for the shortest or fastest route. A consumer group decided to test a cheap
device GPS-Guide versus their favorite Guider to see if the cheap device GPS-Guide is
more likely to fail to find the shortest path. Both devices have been used to conduct a number
of tests. Several units of the two devices were purchased. Here are the results.
Units Purchased
10
5
GPS-Guide
Guider
Tests
200
50
Failures
9
1
(a) Please specify the appropriate null and alternative hypotheses for a randomized test. What
is your test statistic? What is the rejection region for an = 15% test?
(b) What is the p-value for the observed data? What decision is made?
(c) What would be the probability of concluding that there was no difference between the two
treatments if GPS-Guide actually had 2 times the failure rate of Guider?
P[ X = x | n = 10, p ] for binomial distribution with various values of p
X
0.1000
0.1111
0.2000
0.2500
0.3333
0.5000
0.6667
0.750
0.8000
0.8889
0.9000
0
1
2
3
4
5
6
7
8
9
10
0.3487
0.3874
0.1937
0.0574
0.0112
0.0015
0.0001
0.0000
0.0000
0.0000
0.0000
0.3079
0.3849
0.2165
0.0722
0.0158
0.0024
0.0002
0.0000
0.0000
0.0000
0.0000
0.1074
0.2684
0.3020
0.2013
0.0881
0.0264
0.0055
0.0008
0.0001
0.0000
0.0000
0.0563
0.1877
0.2816
0.2503
0.1460
0.0584
0.0162
0.0031
0.0004
0.0000
0.0000
0.0174
0.0867
0.1951
0.2601
0.2276
0.1365
0.0569
0.0162
0.0030
0.0003
0.0000
0.0010
0.0098
0.0439
0.1172
0.2051
0.2461
0.2051
0.1172
0.0439
0.0098
0.0010
0.0000
0.0003
0.0030
0.0162
0.0569
0.1365
0.2276
0.2601
0.1951
0.0867
0.0174
0.0000
0.0000
0.0004
0.0031
0.0162
0.0584
0.1460
0.2503
0.2816
0.1877
0.0563
0.0000
0.0000
0.0001
0.0008
0.0055
0.0264
0.0881
0.2013
0.3020
0.2684
0.1074
0.0000
0.0000
0.0000
0.0000
0.0002
0.0024
0.0158
0.0722
0.2165
0.3849
0.3079
0.0000
0.0000
0.0000
0.0000
0.0001
0.0015
0.0112
0.0574
0.1937
0.3874
0.3487
2010 CEE 5970 Examination
Page 3 of 7
7. (20 pts) There is concern about the carcinogenetic of a fuel additive. Animal tests were
carried out with rodents to evaluate its toxicity; here are the results of the experiment:
mouse dose
(g/day)
0
150
400
mice
tested
66
71
62
mice with
tumors
8
23
41
(a) (16 pts.) Use EPAs three-quarters power weight scaling for interspecies adjustments to
determine the daily dose in humans (g/day) from this agent that EPA would estimate
imposes a 1-in-100,000 cancer risk. Assume an "average" human weights 70 kg, and average
mouse weighted 10 g. Please use a 2-parameter multistage model P(d) which includes a
background cancer rate (mice can develop cancer even if not exposed to this chemical), with a
linear low-dose response. If in doubt, be conservative. Justify important assumptions.
(b) (4 pts) Please write down the likelihood function [so I could type it into excel or some
other software system] whose maximum can be sought so as to estimate the parameters of the
model in (a). Makes sure the values of all the variables are specified.
8. (8 pts) Air quality monitoring devices are used in industrial settings to generate an alert if
the concentration of specific target chemicals exceeds target levels. They also generate false
alarms. ChemLife has experience with these monitors. When ChemLife opened a new line and
installed an air monitoring device, over the first 50 days, the device generated 3 false alarms.
(a) Using the classical maximum likelihood estimator, what would be the estimator of the
false-alarm rate for this new device in that location; what is the corresponding estimate of the
variance of that estimator of the false-alarm rate?
(b) Fifty days is not a very long period of time for parameter estimation when the failure rate
is so low; to address that concern the chemical engineer proposes to also report a Bayesian
estimate of the failure rate employing on the companys experience with these devices. From
monitoring device-to-device across locations in their facility, the rate at which false alarms
are generated is on average about 0.02 per day, with a standard deviation across locations and
devices of 0.01. (This corresponds to a gamma distribution has a pdf
f () = C 4 exp[ - /(0.005)]
for some constant C.). Given this background information and the specific data available for
this new device, what is the mean and variance of the posterior distribution that describes the
probability of different values the false-alarm rate for this new device?
2010 CEE 5970 Examination
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9. (10 points) Before the Spill, a Spate of Failures [Today's Headlines - May 13, 2010]
WASHINGTON (Associated Press) -- Bad wiring and a leak in what's supposed to be a "blowout
preventer." Sealing problems that may have allowed a methane eruption. Even a dead battery,
of all things.
New disclosures Wed. revealed a complex cascade of deep-sea failures equipment and procedural
problems in the oil rig explosion and massive spill that is still fouling the waters of the Gulf of
Mexico and threatening industries and wildlife near the coast and on shore.
A host of worrisome events and findings that were at play on the night of the well explosion and
pipe rupture is described in internal corporate documents, marked confidential but provided to a
House committee by BP and by the manufacturer of the safety device. Lawmakers released
them at a House hearing.
(BP was formerly British Petroleum; currently it is 3rd largest
energy company and 4th largest company in the world.)
a) Why might the public consider the oil rig explosion and oil spill to a signal event?
b) Why might the public NOT consider these findings to represent a signal event?
c) If you were BP, why would you care if some accident were viewed as a signal event?
10. (12 pts) Give 3 examples of environmental risks/activities that generate what Sandman
calls Outrage. Why does such outrage occur? What is the common role of the media in
terms of outrage? What can risk managers do to address outrage? Why is it important to
address peoples sense of outrage?
11. (10 pts) Someone argued that the material at the end of the course, the last two weeks of
class (personal risk perception, community response to attempts to site hazardous activities
within their boundaries, risk communication challenges, how the media represents risk,
OUTRAGE, ), have little place in a masters-level engineering course. Please agree or
disagree, and SUPPORT YOUR POSITION. Grade based on quality of your justification.
SOLUTIONS to 2010 CEE 5970 Examination
1. Challenger/Columbia shuttles. Answers differed (a-b). Wanted to see three points such as need
to explain technical issues clearly to management presentation of data, need to have a realistic
understanding of the reliability of technology and not become complacent, effect of pressure of
deadlines on decisions; (c) answers depended on answer to a-b. Remember to address the question.
2. (13 pts) Poisson Process
a) = 2/3 = 0.6666 per year
= Mean in two years = 2*(0.666 /year) = 1.333 colds
b) Waiting time until 5th
2
2
= 5/ = 7.5 years; Variance (T5) = 5/ = 11.25 years =( 3.35 years)
c) P[ K=3 | = (9/12) = 0.5 ] = 1.26%
2 pts
2 pts
2
2+2 pts
2+2 pts
A
C
B
E
D
F
3. (15 pts) a) Fault tree:
b) P(Failure) = P(F) [ P(A) P(B) + P(C) - P(A) P(B) P(C)][P(D) + P(E) - P(D) P(E)]
= 0.1063 = 10.63%
7 pts
Basically there are two critical cut sets: CEF and CDF with probabilities 0.3[ 0.45 * 0.5 ] and
0.3[ 0.40 * 0.5 ] so system failure probability is 12.75%. Here neglecting higher order cut sets
results in a less than perfect approximation. Reliability of F is only 50%.
8 pts
4. (15 pts) Redraw the Network:
B
D
A
E
C
G
F
H
a) Minmal Cut Sets : AFH DGH DEFH BCFH AEGH
b) Reliability = 1 2(0.01)3 = 1 0.000002 = 0.999998 neglects 2(0.01)4
c) Replace B,C,E
d) Improve H because it appears in all cut sets. Especially AFH + DGH
5. Risk Profile with = 0. 5 spills/year:
Spill
Prob.
Prob.
Volume
P(Branch)
P(Spill)
2.5
0.260
10
0.4
0.140
12.5
0.195
50
0.3
0.105
125
0.130
Freq
freq(V)
0.1300
0.0700
0.0975
0.0525
0.0650
Risk Profile
freq[ V v ]
0.5000
0.3700
0.3000
0.2025
0.1500
6 pts
4 pts
3 pts
2 pts
2010 CEE 5970 Examination
500
2500
10000
0.2
0.070
0.065
0.035
0.1
0.0350
0.0325
0.0175
Page 6 of 7
0.0850
0.0500
0.0175
Risk Profile Oil Spills
0.6000
0.5000
0.4000
freq per year 0.3000
0.2000
0.1000
0.0000
1
10
100
1000
10000
Volume
6. Statistics
a) X = # of Guilder failures
Ho: Guider and GPS-Guide fail at the same rate
X ~ Bin(n = 10, p = 0.2 = 50/(50+200) )
Ha: Guider fails less frequently than GPS-Guide
X ~ Bin(n = 10, p < 0.2 )
4 pts
Reject Ho if K 0 (minimum = 10.74%)
3 pt
b) Observed K = 1; p-value = 0.107+0.268 = 37.6%. Insufficient evidence to reject Ho. 4 pts
c) Ha: K ~ Bin{ n = 10, p < 5/[50+2*200] = 1/9 = 0.111 }
2 pts
= Type II Error = 1 Pr[ K= 0 | p=0.111 ] = 1 0.3079 = 69%
2 pt
7. Dose Response: First organize the data, and compute the empirical estimates of P(d), and from
those the corrected values A(d) = [P(d) P(0)] / [ 1 P(0) ] which yields A(0) = 0; P(0) = 0.1212;
Here A(d) corresponds to the probability of getting cancer from this agent, and has the form
A(d) = 1 exp( kd ) = 1 exp( q1d ),
corresponding to the 2-parameter general multistage model:
P(d) = 1 exp( q0 q1 d) = P(0) + [1 P(0)] A(d)
where q0 = log[1 P(0) ]. The data yield:
(g/day)
tested
tumors
0
150
400
66
8
23
41
71
62
P(d)
0.1212
A(d)
0
one-hit k
0.3239
0.231
1.75E-03
0.6613
0.615
2.38E-03
P(0) = 0.1212 is our first parameter. Here is the sixth column is k = log[ 1- A(d) ]/d; the value of k
that just fits each of the last two points by itself. That is our second parameter q1. To be conservative
2010 CEE 5970 Examination
Page 7 of 7
EPA would take the larger k, so k = 2.38x10-3
Then solve 10-5 = A(d) to obtain d = log[ 1 1E-4 ]/k = 4.20x10-03 g/day. This needs to be
converted into a human dose using -power scaling:
k
risk
rat dose
2.38E-03
1.00E-05
4.20E-03
mouse wt
grams
10
human wt
grams
70000
human dose
g/day
3.211
b) Define the function p(d) = exp( -q0 q1 d) where q0 and q1 are the parameters we wished to
estimate. The likelihood function is computed for different values of q0 and q1 as
L = p(0)8 [ 1 p(0) ]54 p(150)23 [ 1 p(150) ]48 p(400)41 [ 1 p(400) ]21
It is often easier to work with the logarithms of the likelihood function.
8. Poisson dist. and parameter estimation.
a) Estimate of is ! = 3/50 days = 0.06 per day.
Estimate of the variance of ! is ! /50 per day2 = 0.0012 = (0.035 per day)2 Watch your units.
b) Now we want to use a Bayesian estimator. The prior for is f () = C 4 exp[ - /(0.005)].
Thus we have = 5 and = 1/0.005 = 200. As a result the posterior parameters will be
= 5 + 3 = 8 and = 1/0.005 + 50 = 250. So the posterior mean is / = 8/250 = 0.032 with
variance = / = 0.000128 ; This is a situation where the Bayesian estimator is advisable.
2
9. (a) A signal event needs to cause the public to realize there is anew risk they did not know about,
or to think the risk has changed. (b) If this is just one more example of the same old thing, then it is
not a signal. ( c) If this is a signal then it will effect not just the const and anger related to this
accident, but the public may view the whole activity as too risky and unwise, and will try to shut
down the activity, or at least demand greater regulation. Signal events draw more attention, and
attendtion is not something BP would want for this event.
10. Three examples, many possible (+3). Issues mentioned in better answers. Why: what
outrage is and that independent of hazard. Public responds to hazard. Causes of outrage:
involuntary risk, dishonest, bullying, exotic. (+2). Media: media generated outrage because
they communicate with the public; outrage generates interest and people watch interesting
news (+2). Managers can address by 6 tactics listed by Sandman: be honest, admit mistakes,
be informative and available, show what is being done Why is it important to address: If
you do not address it, it increases; in democratic society need to address public concerns
because of consequences of public opposition and loss of trust (+2) General overall
understanding and differentiation between signal event and outrage (+1).
11. Answers varied. But in particular, we need to know the issues and what is of concern to the
public if we wish to address their concern.