CE 361 Introduction to Transportation Engineering
Homework 4 (HW4) Solutions,
=74.3, =10.1
Posted: Sun. 9 September 2012
Due: Fri. 21 September 2012
QUEUEING
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You will be permitted to submit this HW with as many as three other CE361 students. If the HW is
submitted by more than one student, the signatures of all students in the group must appear (along
with printed names) at the top of the front page of the materials submitted.
For every problem, identify the problem by its number and name, be clear, be concise, cite your
sources, attach documentation (if appropriate), and let your methodology be known.
“FTE” = Fundamentals of Transportation Engineering, the textbook for CE361.
1. Major blockage on I65 South. Shortly after passing Exit 158, they see a column of smoke rising
from the horizon. Soon they are stopped at the end of a long line of vehicles near milepost (MP) 153.
It is 5:30PM. The next exit is at MP 146. I-65 has two southbound lanes, an approximate capacity of
1750 vphpl (accounting for trucks), and a jam density Dj = 174 vpmpl. The southbound traffic flow
rate at that time on Friday evening is 2475 vph.
A. (5 points) If you assume that the event that blocked I-65 occurred at MP 147, at what time (to the
nearest minute) did the event occur?
Queue = MP153-MP147=6 mi long.
6 mi * 2 lanes * 174 vpmpl = 2088 vehs in queue.
2088 vehs
 0.844 hr
2475 veh / hr
0.844 hr = 50.6 min before 5:30pm = 4:39pm
B. (5 points) At what time (to the nearest minute) will the backup reach Exit 158 if the vehicle arrival
rate is unchanged? [All traffic that arrived after the backup on I-65 reached Exit 158 was diverted
at that exit. This created another backup because the ramp’s capacity is much lower than the
arrival rate, but at least the service rate is not zero.]
MP158-MP153 = 5 mi.
5 mi * 2 lanes * 174 vpmpl = 1740 vehs
1740 vehs
 0.70 hr
2475 veh / hr
0.70 hr = 42 min after 5:30pm = 6:12pm
C. At 7:00PM, the blockage on I-65 is removed and traffic can again flow on I-65 at 1750 vphpl.
Vehicles no longer have to exit at Exit 158.
 (5 points) At what time (to the nearest minute) would the last vehicle that is in the queue on
I-65 at 7PM be cleared?
2088+1740=3828 vehs in queue, after which
arrivals are diverted to offramp.
3828 vehs/(2lanes*1750vphpl) = 1.09 hr = 65.6 min.
CE361 HW4 Solutions Fall 2012
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7:00pm + 65.6 min = 8:06pm

(10 points) But additional vehicles will now be arriving on I-65 after 7PM, as the queue
begins to clear. At what time (to the nearest minute) would the queue be dissipated?
A queueing diagram would help (see Part D), but
let us try using equations only. The blockage is
removed at 7:00-4:39 = 2:21 = 141 minutes after it
started. 3828 vehs are in queue. (See Part C.)
Each hour, 2475 more vehs arrive, but (2*1750)
vehs are being “served”. When does queue go to
zero? 3828+(2475*t’) = 3500*t’; t=3.73 hr = 224
minutes after 7pm = 10:44pm.
D. (15 point) Draw the Queueing Diagram that illustrates the Arrival and Departure Curves from the
time of the blocking event until the queue is dissipated. Also show the maximum queue length
and maximum time spent by any vehicle in the queue. Do not include the vehicles that were
diverted at Exit 158, but include those vehicles that can use I-65 after 7PM.
The queueing diagram on the next page is based on
the points shown in the table below.
time
x
yAC
yDC
16:39 PM
0
0
0 blockage starts
17:30 PM
50.6
2088
0 queue reaches MP153
18:12 PM
92.8
3828
0 queue reaches MP158
19:00 PM
141
3828
0 blockage cleared
dissipate 365.08 13071.22 13071.22 22:44PM
Note how AC2 is horizontal, because vehicle arrivals
during that time are being diverted to the offramp.
CE361 HW4 Solutions Fall 2012
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E. (15 points) Calculate the Maximum Queue Length, Maximum Time Spent in the Queue by any
vehicle, and Total Delay because of the blockage. Do not include the vehicles that were diverted
at Exit 158, but include those vehicles that can use I65 after 7PM.
Maximum Queue Length = maximum height of a
vertical line between the AC and DC. This is between
points b and c in the queueing diagram (or any
point to the left after the diversion to the offramp
started at point a) = 3828 vehicles. (See Part C.)
CE361 HW4 Solutions Fall 2012
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Maximum Time Spent in the Queue = maximum
horizontal line between AC and DC. This is actually
associated with the very first vehicle to arrive after the
blockage occurred. It is eventually served first, but
will have waited the longest. 4:39 to 7:00 PM = 141
minutes. See table in Part D.
Total delay = area of triangle ending at 92.8 minutes
+ area of rectangle between 92.8 and 141 minutes +
area of trapezoid between 141 and 365.08 minutes =
(0.5*(92.8*3828)) + ((141-92.8)*3828) +
((3828/2)*(365.08-141)) = 177,619 + 184,510 +
428,889 = 791,018 veh-min = 13,184 veh-hr.
2. Queueing with alternative service time distributions.
A. (15 points) As people drive into Shoridan State Park, an entry fee is charged, based on the type
of vehicle and the number of people in the vehicle. In the busy season, the interarrival and
service times have a negative exponential distribution, with  = 92 vph and  = 184 vph.
Calculate the average queue length Q , the average time spent waiting for service W (sec), and
the average time spent in a queue t (sec).
Because vehicle arrivals and service times are both
Markovian, use an M/M/1 queueing system with
=92/184=0.5 in (3.18):
(3.19)
(3.20)
Check
W 
t 

 

2
0.25
Q

 0.5 vehs
1   1  0.5
0.5
 0.005435hr  19.56 sec
184  92
1
1

 0.01087hr  39.13 sec
   184  92
t W 
1

 19.56  (3600 / 184)  39.12sec
B. (20 points) One of the booth attendants suggests that the variable fee be replaced by one fixed
fee for each vehicle, regardless of vehicle type or number of occupants. What type of queueing
system x/y/z would this create? If and  are the same as in Part A, what are Q , W , and t ?
CE361 HW4 Solutions Fall 2012
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This would create an M/D/1 queueing system.
(3.15):
(3.16)
2
(0.5)2
Q

 0.25 vehs
2(1   ) 2(1  0.5)

W 

0.5
* 3600  9.78 sec
2 * 184 * (1 0.5)
2 (1   )
(3.17)
2 
2  0.5
t 

* 3600  29.34 sec
2 (1   ) 2 * 184 * (1 0.5)
Check
t W 
1

 9.78  (3600 / 184)  29.34 sec
C. (10 points) More likely, having only one entry fee would allow  to improve to 276 per hour. If so,
what would Q , W , and t be?
We have an M/D/1 queueing system with
=92/276=0.333.
(3.15):
(3.16)
Q
2
2(1   )

W 

(0.333)2
 0.083vehs
2(1  0.333)

0.333
* 3600  3.26 sec
2 * 276 * (1 0.333)
2 (1   )
(3.17)
2 
2  0.333
t 

* 3600  16.30 sec
2 (1   ) 2 * 276 * (1  0.333)
Check
t W 
1

 3.26  (3600 / 276)  16.30 sec