The document describes an algorithm for estimating the probability distribution of travel demand forecasts that are subject to uncertainty. It involves identifying variables that influence forecast error, determining probability distributions for each variable, defining scenarios that combine the discrete outcomes of each variable, calculating the probability and predicted revenue for each scenario, and plotting the revenue cumulative distribution function. Key variables of uncertainty identified for a toll road project include truck value of time, travel demand, and growth rates of car and truck value of time. Probability distributions assumed for these variables include lognormal, normal, and triangular. The algorithm allows assessment of the uncertainty and risk associated with toll revenue forecasts.

1. Prob-Dist-Toll-Forecast-Uncertainty
December 16, 2015
In [2]: # special IPython command to prepare the notebook for matplotlib
%matplotlib inline
import numpy as np
import pandas as pd
import scipy as sp
import seaborn as sns
import matplotlib.pyplot as plt
Estimating the probability distribution of a travel demand forecast Authors: John L. Bowman,
Dinesh Gopinath, and Moshe Ben-Akiva
0.0.1 Algorithm
1. Identify variables that induce error in Toll Revenue prediction : x = (x1, x2, ..., xk, ..., xK)
• Simple Toll Revenue Model - Variables that induce error in Toll Revenue Prediction r(p)
are: (1)
Value of Time, (2) Population, (3) Households, (4) Employment
• Let: x1 = Value of Time, x2 = Population, x3 = Households, x4 = Employment. Thus K = 4,
i.e. 4-Dimensional space of possible outcomes xk
• x = (x1, x2, x3, x4)
2. Obtain probability distribution of xk for k = 1, 2, . . . , K. Distribution can be based on:
(a) Direct input or (b) Assumption, e.g. Triangular, Normal, etc. For each dimension k
discretize an assumed probability distribution and identify a small set of discrete outcomes xnk
k , where
nk = 1, 2, 3, . . . , Nk (assign probabilities p(xnk
k ) to these discrete outcomes based on reasoning and
empirical evidence to approximate xk’s true distribution???):
• Let N1 = 4, N2 = 3, N3 = 5, N4 = 4 and xk = {x1
k, x2
k, x3
k, ..., xNk
k }
• x1 discrete outcomes = {x1
1, x2
1, x3
1, x4
1}, with p(x1
1) + p(x2
1) + p(x3
1) + p(x4
1) = 1
• x2 discrete outcomes = {x1
2, x2
2, x3
2}, with p(x1
2) + p(x2
2) + p(x3
2) = 1
• x3 discrete outcomes = {x1
3, x2
3, x3
3, x4
3, x5
3}, with p(x1
3) + p(x2
3) + p(x3
3) + p(x4
3) + p(x5
3) = 1
• x4 discrete outcomes = {x1
4, x2
4, x3
4, x4
4}, with p(x1
4) + p(x2
4) + p(x3
4) + p(x4
4) = 1
3. Develop Toll Revenue Model for Baseline Scenario : - Get Predicted r
(p)
base for Baseline Scenario
x = (xbase
1 , xbase
2 , xbase
3 , xbase
4 ) from output of the model
4. Run Toll Revenue Model one time for each variable that induce error in prediction :
• Get predicted r
(p)
k=1 based on x = (xextreme
1 , xbase
2 , xbase
3 , xbase
4 )
• Get predicted r
(p)
k=2 based on x = (xbase
1 , xextreme
2 , xbase
3 , xbase
4 )
• Get predicted r
(p)
k=3 based on x = (xbase
1 , xbase
2 , xextreme
3 , xbase
4 )
• Get predicted r
(p)
k=4 based on x = (x1, x2, x3, x4 )
1

2. 5. Calculate change in Predicted Toll Revenue and variables that induce error in prediction
:
• rchange
k =
r
(p)
base−r
(p)
k
r
(p)
k
, where k = 1, 2, 3, 4
• xchange
k =
xbase
1 −xk
xk
, where k = 1, 2, 3, 4
6. Calculate Elasticity of Toll Revenue with respect to variables that induce error in pre-
diction :
• er
k =
rchange
k
xchange
k
, where k = 1, 2, 3, 4
7. Define a set of scenarios: S = {(xn1
1 , ..., xnk
k , ..., xNk
K ); nk = 1, 2, 3, ..., Nk; k = 1, 2, ..., K}, covering
all combinations of the discrete coutcomes in all K = 4 dimensions
• For simple example, S = {$(x 1ˆ1, x 2ˆ1, x 3ˆ1, x 4ˆ1), (x 1ˆ1, x 2ˆ1, x 3ˆ1, x 4ˆ2), . . . , (x 1ˆ1,
x 2ˆ3, x 3ˆ5, x 4ˆ4), . . . , (x 1ˆ4, x 2ˆ3, x 3ˆ5, x 4ˆ4) $}
• Number of scenarios in S =
K
k=1 Nk. Thus the number of scenarios in simple example is S = 4
x 3 x 5 x 4 = 240. Thus s = 1, 2, 3, . . . , 240
• Using s as a 1-Dimensional index of the member of S: Refer to a single member of S as x(s)
=
(x
(s)
1 , x
(s)
2 , ..., x
(s)
k , ..., x
(s)
K ). For simple example: x(s=1)
= (x1
1, x1
2, x1
3, x1
4); x(s=2)
= (x1
1, x1
2, x1
3, x2
4);
x(s=240)
= (x4
1, x3
2, x5
3, x4
4)
8. Calculate the probability of each scenario: Error variables are mutually independent, thus the
probability of each scenario is given by: p(s) =
K
k=1 p(x
(s)
k ), s ∈ S, thus for simple example:
• p(s = 1) =
4
k=1 p(x
(s=1)
k ) = p(x1
1)p(x1
2)p(x1
3)p(x1
4)
• p(s = 2) =
4
k=1 p(x
(s=2)
k ) = p(x1
1)p(x1
2)p(x1
3)p(x2
4)
• p(s = 240) =
4
k=1 p(x
(s=240)
k ) = p(x4
1)p(x4
2)p(x5
3)p(x4
4)
9. Calculate Toll Revenue for scenario s, r(s)
= r
(p)
base
K
k=1
x
(s)
k
xbase
k
er
k
, s ∈ S, thus for simple example:
• r(s=1)
= r
(p)
base
4
k=1
x
(s=1)
k
xbase
k
er
k
= r
(p)
base
x1
1
xbase
1
er
1 x1
2
xbase
2
er
2 x1
3
xbase
3
er
3 x1
4
xbase
4
er
4
• r(s=2)
= r
(p)
base
4
k=1
x
(s=2)
k
xbase
k
er
k
= r
(p)
base
x1
1
xbase
1
er
1 x1
2
xbase
2
er
2 x1
3
xbase
3
er
3 x2
4
xbase
4
er
4
• r(s=240)
= r
(p)
base
4
k=1
x
(s=240)
k
xbase
k
er
k
= r
(p)
base
x4
1
xbase
1
er
1 x3
2
xbase
2
er
2 x5
3
xbase
3
er
3 x4
4
xbase
4
er
4
10. Using pairs r(s)
and p(s) Plot Revenue CDF
0.0.2 Sources of Uncertainty - Toll (Kockleman)
• Estimates of trip generation
• Estimates of land development
• Models: Trip Generation, Trip Distribution, Mode Choice
• Toll-technology adoption rates
• Hetrogeneity in (VOT) Value of Time savings
2

3. • Network attributes - Traffic congestion (low-volume corridors have greater uncertanity in their
forecasts)
• Uncertainty in land development patterns
• Demographic and employment projections
• Tolling design - shadow tolls (govt. pays the concessionaire an amount based n toll road use - similar
to toll free) or user-paid tolls (drivers willingness to pay is more complex and difficult to understand -
more forecasting risk)
• Tolling culture of a region, i.e. the degree to which tolls have been used in the past
• Travel demand model imperfections (Heterogenity of VOT is ignored, Variable tolls or HOT lanes
that are free at certain hours)
• Competitive advantage - Toll on the only bridge vs toll on freeway - more options to route
• User attributes - toll facilities serving a small market segment of travelers allow more reliable forecasts
vs hetrogenous populations
• Road location, configurations
• Demand variations over times of days and days of the year also affect forecast reliability
• Brian and Wilkins (2002) - poorly estimated VOTT’s, economic downturns, mis-prediction of future
land use patterns, lower than predicted time savings, added competition, lower than anticipated truck
usage, high variability in traffic volumes
• Economic growth and related changes in income and employment
• Total Demand Model errors
• Model error in elasticity of demand
• Value of time
• Errors in measurement of network times and costs
• Operating speed
• Roadway improvements
1 Texas North Tarrant Express Segment 3A
• Revenue and Transaction Forecast Year = 2018
2018 Revenue and Transactions
• Forecasted 2018 Annual Project Revenue (000’s 2011 Dollars) = 27612
• Forecasted 2018 Daily Transactions = 40086
Truck VOT Calculations
• SOV VOT - Lognormal distribution with mean = $18.59 and standard deviation = $7.4 (µ = 2.849
and σ = 0.383)
3

4. • Coefficient of variation, Csov
v = 7.4
18.59 = 0.398
• HHM Truck VOT: Mean = $36.48 and Standard deviation = $30.24
• AECOM Truck VOT: Mean = $60.76 and Standard deviation = $51.08
• AverageTruckV OT = HHMT ruckV OT +AECOMT ruckV OT
2 = $48.62
• Standard deviation of Average Truck VOT = Csov
v ∗ AverageTruckV OT = $19.35 (µ = 3.811 and σ
= 0.383, calculations below)
• µσ
In [3]: # Parameters for Truck Lognormal Distribution
m = 48.62
s = 19.35
truck_ln_mu = np.log(m/np.sqrt(1+((s**2)/(m**2))))
truck_ln_sigma = np.sqrt(np.log(1+((s**2)/(m**2))))
print ’truck_ln_mu = %1.3f’ % truck_ln_mu
print ’truck_ln_sigma = %1.3f’ % truck_ln_sigma
truck ln mu = 3.811
truck ln sigma = 0.383
r
(p)
base = 74754 (000’s 2011 Dollars)
Variables: Sources of Uncertainty
• Truck VOT: x1
– Elasticity of Revenue to Truck VOT = 0.994
– xbase
1 = $60.76
– Probability distribution: Lognormal with mean = $48.62 and std. dev = $19.35 (µ = 3.811 and
σ = 0.383)
• Travel Demand: x2
– Elasticity of Revenue to Demand (Transactions as proxy) = 2.57
– xbase
2 = 61056
– Probability distribution: Normal with µ = 58871.5 and σ = 2184.5
• Car VOT Growth: x3
– Elasticity of Revenue to Car VOT Growth = 0.19
– xbase
3 = 2.1%
– Probability distribution: Triangular with Min = 0.5%, Mean = 1.05%, and Max = 2.1%
• Truck VOT Growth: x4
4

5. – Elasticity of Revenue to Truck VOT Growth = 0.19
– xbase
4 = 2.5%
– Probability distribution: Triangular with Min = 0.5%, Mean = 1.25%, and Max = 2.5%
Truck VOT Probability Distribution: Lognormal
In [4]: mu = 3.811
sigma = 0.383
low = 1
high = 120
dx_1 = 2 # Length of interval
# Comb points along x axis
x_1 = np.arange(low, high, dx_1)
# Compute y values: pdf at each value of x
vot_y = (1/(sigma * x_1 * np.sqrt(2 * np.pi))) * np.exp(-0.5 * ((np.log(x_1) - mu)/sigma) ** 2)
# Plot the function
plt.figure(figsize = (16, 8))
plt.stem(x_1, vot_y, markerfmt = ’ ’) # This draws the intervals
plt.xlabel(’$x_1$’)
plt.ylabel(’$p(x_1)$’)
plt.title(’Discretized Log-Normal Probability Density’)
area = np.sum(dx_1 * vot_y)
print ’Probability Sum = %1.4f’ % area
print ’N_1 = %d’ % len(x_1)
temp1 = np.array([x_1, vot_y * dx_1])
Probability Sum = 0.9946
N 1 = 60
5

6. Travel Demand Probability Distribution: Normal
In [5]: # Mean Transactions = (2035 Transactions + 2018 Transactions)/2
print ’Mean Transactions = %s’ % ((63635+40086)/2.0)
# Std Dev Transactions = (2035 Transactions - 2018 Transactions)/2
print ’Std. Dev Transactions = %s’ % ((63635-40086)/2.0)
Mean Transactions = 51860.5
Std. Dev Transactions = 11774.5
In [6]: demand_mean = 51860.5
demand_sd = 11774.5
demand_low = demand_mean - 3 * demand_sd # low end of x axis
demand_high = demand_mean + 3 * demand_sd # high end of x axis
dx_2 = 2000 # Length of interval
# Comb points along x axis
x_2 = np.arange(demand_low, demand_high, dx_2)
# Compute y values: pdf at each value of x
demand_y = (1/(demand_sd * np.sqrt(2 * np.pi))) * np.exp(-0.5 * ((x_2 - demand_mean)/demand_sd)
# Plot the function
plt.figure(figsize = (16, 8))
plt.stem(x_2, demand_y, markerfmt = ’ ’) # This draws the intervals
plt.xlabel(’$Demand$’)
plt.ylabel(’$p(Demand)$’)
plt.title(’Discretized Normal Probability Density’)
area = np.sum(dx_2 * demand_y)
print ’Probability Sum = %1.4f’ % area
print ’N_2 = %d’ % len(x_2)
Probability Sum = 0.9978
N 2 = 36
6

7. Car VOT Growth Probability Distribution: Triangular
In [7]: min_growth_car = 0.004
mean_growth_car = 0.0105
max_growth_car = 0.022
car_array = np.random.triangular(min_growth_car, mean_growth_car, max_growth_car, size = 100000)
#plt.hist(car_array, bins = 10)
car_val = np.histogram(car_array, bins = 20)
car_y = [float(i)/np.sum(car_val[0]) for i in car_val[0]]
# Binwidth issue
x_car = car_val[1]
x_3 = []
for i in range(len(x_car) - 1):
temp = (x_car[i] + x_car[i+1])/2
x_3.append(temp)
# Plot triangular distribution
plt.figure(figsize = (16, 8))
plt.stem(x_3, car_y, markerfmt = ’ ’) # This draws the intervals
plt.xlabel(’$Car Growth$’)
plt.ylabel(’$p(Car Growth)$’)
plt.title(’Discretized Triangular Probability Density’)
print len(x_3)
print np.sum(car_y)
20
1.0
Truck VOT Growth Probability Distribution: Triangular
7

8. In [8]: min_growth_truck = 0.004
mean_growth_truck = 0.0125
max_growth_truck = 0.026
truck_array = np.random.triangular(min_growth_truck, mean_growth_truck, max_growth_truck, size =
#plt.hist(car_array, bins = 10)
truck_val = np.histogram(truck_array, bins = 20)
truck_y = [float(i)/np.sum(truck_val[0]) for i in truck_val[0]]
# Binwidth issue
x_truck = truck_val[1]
x_4 = []
for i in range(len(x_truck) - 1):
temp = (x_truck[i] + x_truck[i+1])/2
x_4.append(temp)
# Plot triangular distribution
plt.figure(figsize = (16, 8))
plt.stem(x_4, truck_y, markerfmt = ’ ’) # This draws the intervals
plt.xlabel(’$Truck Growth$’)
plt.ylabel(’$p(Truck Growth)$’)
plt.title(’Discretized Triangular Probability Density’)
print len(x_4)
print np.sum(truck_y)
20
1.0
Scenarios
In [9]: S = [[i, j, k, l] for i in x_1 for j in x_2 for k in x_3 for l in x_4]
print S[0]
8

9. print ’n’
print ’Number of Scenarios = ’ + str(len(S))
[1, 16537.0, 0.0044911329557659595, 0.0045996835827973384]
Number of Scenarios = 864000
Probability and Revenue Calculations for Scenarios
In [10]: # Constants: Base Revenue
rp_base = 27612
# Constants: Base values of variables
x_1b = 60.76
x_2b = 40086
x_3b = 0.021
x_4b = 0.025
# Constants: Elasticities of variables
e_x1 = 0.994
e_x2 = 2.57
e_x3 = 0.19
e_x4 = 0.19
revenue_S = []
prob_S = []
for i in range(len(S)):
# R(s)
temp_rev = rp_base * (S[i][0]/x_1b)**(e_x1) * (S[i][1]/x_2b)**(e_x2) * (S[i][2]/x_3b)**(e_x
revenue_S.append(temp_rev)
# Probability calculation:
# Truck VOT
p_x1 = (1/(sigma * S[i][0] * np.sqrt(2 * np.pi))) * np.exp(-0.5 * ((np.log(S[i][0]) - mu)/s
# Demand
p_x2 = (1/(demand_sd * np.sqrt(2 * np.pi))) * np.exp(-0.5 * ((S[i][1] - demand_mean)/demand
# Car VOT Growth
if S[i][2] in x_3:
cp = x_3.index(S[i][2])
p_x3 = car_y[cp]
# Truck VOT Growth
if S[i][3] in x_4:
tp = x_4.index(S[i][3])
p_x4 = truck_y[tp]
prob_S.append(p_x1 * p_x2 * p_x3 * p_x4)
print ’Probability Sum = %0.4f’ % np.sum(prob_S)
9

10. Probability Sum = 0.9924
In [11]: # Sorting Result based on Revenue
output = (np.array([revenue_S, prob_S])).T
output = output[output[:, 0].argsort()]
In [12]: # Plotting Cumulative Probability Distribution
plt.figure(figsize = (16, 8))
plt.plot(output[:,0], np.cumsum(output[:,1]), linewidth = 2) # Selecting array column: array[:,
# Plotting Predicted Revenue
plt.axvline(x = rp_base, color = ’r’)
plt.text(74754 + 500, 0.1, ’Predicted Revenue for 2018’, fontsize = 16)
# Remove Scientific Notation
ax = plt.gca()
ax.get_xaxis().get_major_formatter().set_scientific(False)
plt.xlabel(’$Revenue$’, fontsize = 16)
plt.ylabel(’$p(Revenue)$’, fontsize = 16)
plt.title(’Cumulative Probability Distribution - Revenue (000’s 2011 Dollars)’, fontsize = 16)
# Set tick label size
plt.tick_params(axis = ’both’, which = ’major’, labelsize = 14)
print ’Probability Sum = %0.4f’ % np.sum(prob_S)
print ’Demand std = %d’ % demand_sd
Probability Sum = 0.9924
Demand std = 11774
Percentile Calculation
In [14]: year = 2018
cum_prob = pd.DataFrame({’Revenue’: output[:,0], ’Cumulative Probability’: np.cumsum(output[:,1
10

11. # P(Revenue < r) = percentile -> Find r
# 75+
percentile_75 = cum_prob[’Revenue’][cum_prob[cum_prob[’Cumulative Probability’] <= 0.75].shape[
percentile_85 = cum_prob[’Revenue’][cum_prob[cum_prob[’Cumulative Probability’] <= 0.85].shape[
percentile_95 = cum_prob[’Revenue’][cum_prob[cum_prob[’Cumulative Probability’] <= 0.95].shape[
# 25-
percentile_05 = cum_prob[’Revenue’][cum_prob[cum_prob[’Cumulative Probability’] <= 0.05].shape[
percentile_15 = cum_prob[’Revenue’][cum_prob[cum_prob[’Cumulative Probability’] <= 0.15].shape[
percentile_25 = cum_prob[’Revenue’][cum_prob[cum_prob[’Cumulative Probability’] <= 0.25].shape[
# Print values
print ’75th Percentile of ’ + str(year) + ’ Revenue = %0.2f’ % percentile_75
print ’85th Percentile of ’ + str(year) + ’ Revenue = %0.2f’ % percentile_85
print ’95th Percentile of ’ + str(year) + ’ Revenue = %0.2f’ % percentile_95
print ’5th Percentile of ’ + str(year) + ’ Revenue = %0.2f’ % percentile_05
print ’15th Percentile of ’ + str(year) + ’ Revenue = %0.2f’ % percentile_15
print ’25th Percentile of ’ + str(year) + ’ Revenue = %0.2f’ % percentile_25
print percentile_95
print percentile_85
print percentile_75
print percentile_25
print percentile_15
print percentile_05
75th Percentile of 2018 Revenue = 49146.23
85th Percentile of 2018 Revenue = 62173.69
95th Percentile of 2018 Revenue = 91921.62
5th Percentile of 2018 Revenue = 8193.37
15th Percentile of 2018 Revenue = 14082.97
25th Percentile of 2018 Revenue = 18857.58
91921.6153431
62173.6864964
49146.2293139
18857.5804073
14082.965342
8193.3714675
11