CE 2710 Transportation Engineering
Homework 5 Solution
1. Given the following data use the gravity model to calculate all the interchange
volumes (in other words, get the trip distribution matrix).
Zone
1
2
3
Production, Pi
2000
1000
2000
Attractiveness, Aj
2
5
1
Impedance Factor, Wij
I
1
2
3
J
2
20
5
10
1
5
20
10
Calibration Factor, c = 1.5
3
10
10
5
Socio-economic factor, kij = 1.0
Solution
1
1

F

 0.0894
11
Wijc
(5)1.5
Fij
First calculate friction factors: Fij 
I
TAZ
1
2
3
1
0.0894
0.0112
0.0316
J
2
0.0112
0.0894
0.0316
3
0.0316
0.0316
0.0894
Find denominator of Gravity Model equations: AjFijKij
Aj Fij K ij  A1F11K11  (2)(0.0894)(1.0)  0.1789
AjFijKij
J
I
TAZ
1
2
3
1
0.1789
0.0224
0.0632
2
0.0559
0.4472
0.1581
3
0.0316
0.0316
0.0894
Σ
0.2664
0.5012
0.3108
Find Probability that Trip i will be attracted to Zone j, pij
pij 
A j Fij K ij
A F K
j
p11 
ij

ij
A1 F11K11
( A1 F11 K11  A2 F12 K12  A3 F13 K13 )
0.1789
 0.6715
0.2664
Pij
I
TAZ
1
2
3
J
2
0.2098
0.8923
0.5087
1
0.6715
0.0446
0.2035
3
0.1187
0.0631
0.2878
Find Trip Interchanges, Qij
Qij  Pi pij
Q11  P1 p11  (2000)(0.6715)  1343
Qij
J
I
TAZ
1
2
3
Σ
1
1343
45
407
1795
2
420
892
1017
2329
3
237
63
576
876
Σ
2000
1000
2000
5000
2. Use the gravity model to estimate the trip distribution matrix for this planning
year.
Given:
Zone
Production, Pi
1
2
2000
5000
Attractiveness,
Aj
10
8
Impedance Factor, Wij
J
I
1
2
1
2
9
2
10
1
Calibration Factor, c = 1.5
Kij
J
1
1
1.2
1
2
I
2
1.2
1
Solution:
1
1

F

 0.3536
11
Wijc
(2)1.5
Fij
First calculate friction factors: Fij 
J
TAZ
1
2
I
1
0.3536
0.0370
2
0.0316
1.0000
Find denominator of Gravity Model equations: AjFijKij
A j Fij K ij  A1 F11K11  (10)(0.3536)(1.0)  3.5355
AjFijKij
TAZ
1
2
I
J
2
0.3036
8.0000
1
3.5355
0.4444
Σ
3.8391
8.4444
Find Probability that Trip i will be attracted to Zone j, pij
pij 
A j Fij K ij
A F K
j
p11 
ij

ij
A1 F11K11
( A1 F11K11  A2 F12 K12 )
3.5355
 0.9209
3.8391
Pij
J
I
TAZ
1
2
1
0.9209
0.0526
Find Trip Interchanges, Qij
Qij  Pi pij
Q11  P1 p11  (2000)(0.9209)  1842
2
0.0791
0.9474
Qij
I
TAZ
1
2
Σ
1
1842
263
2105
J
2
158
4737
4895
Σ
2000
5000
7000
3. Discuss factors of demographics and land use that are not in the gravity model
that should affect trip distribution between zones.
The gravity model does not include many factors including:
 Car ownership
 Prestige
 Availability
 Accessibility of Modes
 Walkable/Transit oriented development
Can we accurately estimate impedance before we have 'mode choice' and 'route
choice'? Discuss.
No. The impedance will change depending on how many vehicles are loaded to a
particular route.
4. Calculate the market shares for the following modes using the utility function
given.
uk = ak – 0.003 X1 – 0.04 X2
Auto
BRT
Regular Bus
ak, modal
constant
-0.20
-0.40
-0.60
X1, travel cost
in cents
120
60
30
X2, travel time in
minutes
30
45
55
u auto  aauto  0.003 X 1  0.04 X 2  0.2  (0.003 *120)  (0.04 * 30)
u auto  0.2  0.36  1.2  1.76
u BRT  a BRT  0.003 X 1  0.04 X 2  0.4  (0.003 * 60)  (0.04 * 45)
u BRT  0.4  0.18  1.8  2.38
u Bus  a Bus  0.003 X 1  0.04 X 2  0.6  (0.003 * 30)  (0.04 * 55)
u Bus  0.6  0.09  2.2  2.89
e 1.76
0.172
0.172
prob ( Auto)  1.76 2.38 2.89 

(0.172  0.289  0.0556) 0.32017
e
e
e
prob ( Auto)  53.73%
e 2.38
0.09255
0.09255
prob ( BRT )  1.76 2.38 2.89 

(0.172  0.289  0.0556) 0.32017
e
e
e
prob ( BRT )  28.91%
e 2.89
0.0556
0.0556
prob ( Bus )  1.76 2.38 2.89 

(0.172  0.289  0.0556) 0.32017
e
e
e
prob ( Bus )  17.36%
5. Calculate the market shares for auto and light rail using the utility function given.
uk = ak – 0.05 Ta – 0.04 Tw – 0.02 Tr– 0.01 C
Light Rail
Auto
ak, modal
constant
-0.05
-0.05
Ta – access
time
10
5
Tw – waiting
time
10
5
Tr – riding
time
45
30
C – out of
pocket cost
50
100
u rail  arail  0.05Ta  0.04Tw  0.02Tr  0.01C
u rail  0.05  (0.05 *10)  (0.04 *10)  (0.02 * 45)  (0.01* 50)
u rail  0.05  0.5  0.4  0.9  0.5  2.35
u auto  aauto  0.05Ta  0.04Tw  0.02Tr  0.01C
u auto  0.05  (0.05 * 5)  (0.04 * 5)  (0.02 * 30)  (0.01*100)
u auto  0.05  0.25  0.2  0.6  1  2.1
e 2.35
0.09537
0.09537
prob( Rail )  2.35 2.1 

(0.09537  0.1224) 0.2178
e
e
prob( Rail )  43.78%
e 2.1
0.1224
0.1224
prob( Auto)  2.35 2.1 

(0.09537  0.1224) 0.2178
e
e
prob( Auto)  56.22%
6. Which is the parameter in the modal choice utility function that accounts for
intangibles such as land use issues, comfort of the mode and type of trip?
These intangibles are reflected in the value of the modal constant.