5970
Spring CEE 2009
Homework # 1 Partial Solutions
HW #1 Point Breakdown
Q1) 8 points
Q2) 8 points
Q3) 10 points
Q4) 10 points
Q5) 12 points
Q6) 12 points
Q7) 40 points
HW#1 Partial Solutions
General Comments:
Please read the question posed and answer all questions completely. For example,
give your point answers for question 7(c), when t = 1,2,3 - NOT just an equation.
If you are asked about how to estimate the risk of a certain activity and you discuss
exposure to the activity please be specific. For example, do not just say I would find
out how many people were exposed. Instead discuss how you would judge that
someone was exposed (ie. one ride, x miles traveled, an afternoon of the activity)
and give some justification for the exposure measure.
If a question says Explain, please provide a good discussion. Answers can be
concise, but must address key issues. For instance, for question #3, an explanation
was required for the ranking of each case. A long detailed discussion is not necessary,
just one concise sentence regarding the key issues for each case is sufficient.
Obviously you had to think about why you would rank the cases as you did, so just
outline your thought process in your answer.
QUESTION #7:
Part (a)
You are asked to use the additive-effects model to predict the fatality rate (F) for the
driver in question. Some fundamental notes that you should be aware of and think about
before you begin:
- the symbol equals change and a change can be positive or negative
- the model you suggest should be reasonable!!! For example, whatever model
you develop, you want the fatality rate of a SAFE driver to be lower than the
average fatality rate for all drivers (which is 12.56 deaths/billion)
- Always think about units you cannot add a death rate to a dimensionless
number to predict another death rate
The approach to this question is to try and determine, in a reasonable manner, factors for
your additive model from the multiplicative factors in the multiplicative model.
The most straightforward way to do this is to determine additive factors that match the
multiplicative model predictions when one changes one factor at a time.
CEE 5970
Spring 2009
For each factor, i, solve for i:
Fav + i = MiFav
i = MiFav - Fav = Fav(Mi 1)
(Mi is the multiplicative factor from Evans)
(Mi 1 is now the additive model factor)
End up with additive model prediction of:
Fatality rate (deaths/billion) = Fav + age + alcohol + seat-belt + mass + road-type
= 12.56 + (-5.68) + (-4.92) + (-6.38) + (-3.30) + (-5.90)
= -13.56
You should note that since this prediction is less than 0 (i.e. Safe drivers are procreating,
not dying in accidents!) the additive model would not be very good. The multiplicative
model is better because it will not predict impossibly low fatality rates.
You cannot just use the factors from the multiplicative model directly in the additive
model think about the units!
7 (b):
Evans et al. assume that the number of takeoffs and landings are independent of trip
distance (page 243 of the article). Notice that this does not necessarily imply that all trips
are assumed to be composed of one takeoff and one landing. Their assumption simply
means that Evans et al. did not consider the impact of multiple takeoffs and landings on
their risk estimates.
7 (c):
Compare risk of driving d miles for a flight with t (=1, 2, 3) takeoffs and landings.
First, estimate the driving risk for the driver in question using Evans et al. multiplicative
model fatality gives rate = 2 deaths/billion miles
Note that the age factor for the 25 year old can be determined directly from Figure 2 in
Evans et al. as about 1.
To clarify the units of the airline risk measures, for those who may have been confused
by them, note the following regarding Evans et al. derivation of their equation (16):
880 miles
* 0.55 fatalities = 484 fatalities
average trip
109 miles
109 trips
now we are given that the airline passenger death rate is 300 fatalities/10 9 flight segments
(instead of 484 above).
From equation (16) we can now say that:
r = 300/d
where r is the fatality risk, d is the distance flown
CEE 5970
Spring 2009
However, if we want to consider multiple takeoffs and landings then we can say that:
r = 300t/d
where t is the number of takeoffs and landings
To find the breakpoints of when it is better to fly or drive, solve the above equation for d
when r=2 at each value of t. End up finding that:
When t = 1, it is better to drive if d < 150 miles
t =2,
d < 300 miles
t = 3,
d < 450 miles
7 (d):
Plot the results you get above two ways (ie. two graphs) following the Evans et al. plot
and the Sivak et al. plot. Also include a second driver in the analysis. For each graph,
we are looking for 2 drivers and 3 flights (t=1,2,3).
A few notes:
- best to show how you calculate the fatality rate and the probability for the
graphs instead of just giving the graphed results.
- excel graphs are always nicest!
- Watch the units of the y-axis!
For Evans et al. type plot (deaths vs trip distance):
Use r = 300t/d to obtain the death rates for flying while the driving death rate is constant
at the rate you calculate from the multiplicative model (eg. 2 deaths/billion).
For Sivak et al. type plot (probability of death vs trip distance):
Probability of death from travelling d miles by:
a) driving is
Probability = (death rate/109)d
eg:
= 2d/109
Note: you dont need to use the equation on page 148 of Sivak et al.
b) flying only varies with number of segments,
Probability = (300 deaths/109 segments)t
= 300t/109
See graphs on following pages for examples of what they look like (note that 3 stops is
not asked for in this question).
The relative advantages of each should also have been given (marked easy if you tried to
answer this). Some reasonable answers are:
deaths vs trip distance natural and constant representation of driving risk, fit all results
on same graph
probability of death vs trip distance natural representation for airline risk, actual risk
for choice
CEE 5970
Spring 2009
7 (e):
Calculate the expected loss of life expectancy of a 25-yr old taking a 600-mile non-stop
flight. Expected value of this, E[X], has the units of life expectancy and is calculated
from the probability of dying on this flight as:
E[X] = P(X)*(amount of life lost) = 300/109 * (75yrs 25yrs) = 1.5*10-5 years = 8 min.
Note: you must assume an average age that the 25 year old would live to here it is 75.
Also note that we dont need to use the trip distance of 600 miles we just use the fact
that the flight is non-stop (t=1).
Evans et.al. Plot
10
Deaths per Billion M iles
9
Driver 1
8
Driver 2
7
Non-stop
One-stop
6
Two-stops
5
4
3
2
drivers
1
0
0
500
1000
Distance (miles)
1500
2000
CEE 5970
Spring 2009
Sivak et.al. Plot
4500
Fatality Probability x 109
4000
3500
3000
Driver 1
2500
Driver 2
2000
Non-stop
One-stop
1500
Two-stops
1000
500
0
0
500
1000
Distance (miles)
1500
2000