Solutions for Homework for Traffic Flow Analysis
1. On a specific westbound section of highway, studies show that the speed-density
k
relationship is: u  u f [1  ( ) 3.5 ] . The highway’s capacity is 3800 vehicles/hour and the
kj
jam density is 140 vehicles/km. What is the space mean speed of the traffic at capacity
and what is the free flow speed.
Solution: as we know q = ku, thus, we can write the following: q = ku =
k
k* u f [1  ( ) 3.5 ] . When traffic flow is at the maximum, dq/dk = 0. Thus, we can write
kj
the following:
dq
1
 u f [1  4.5  3.5 k 3.5 ]  0 . As the free-flow speed uf can not be equal
dk
kj
to zero, thus, we can write: [1  4.5 
1
k 3j .5
k 3.5 ]  0 . Therefore, km = 91.1 vehicles/km.
Since we know qm = 3800 vehicles/hour. And again, qm = kmum, then we can calculate um
k
= 41.7 km/hour. Then, we can calculate uf from the given equation: u  u f [1  ( ) 3.5 ] .
kj
In the end, we may obtain: uf = 53.5 km/hr.
2. A section of highway has the following flow-density relationship: q = 80k – 0.4k2.
What is the capacity of the highway section, the speed at capacity, and the density when
the highway is one-quarter of its capacity?
Solution: since we know the q-k relationship, we apply the same logic: dq/dk = 0 in order
to obtain km. (dq/dk) = 80-0.8k = 0, thus, km = 100 vehicles/km. We can also obtain qm =
4000 vehicles/hour. We also know that q = km. Then, um = qm/km = 40 km/hour. When
the highway is one quarter of its capacity, it means that q = 0.25qm = 1000 vehicles/hour,
we can use the given equation: q = 80k – 0.4k2, to calculate the density when q = (1/4)qm.
Thus, k = 186 km/hour or 13.8 km/hour.
3. An observer has determined that the time headways between successive vehicles on a
section of highway are exponentially distributed, and that 60% of the headways between
vehicles are 13 seconds or greater. If the observer decides to count traffic in 30 second
intervals, estimate the probability of the observer counting exactly four vehicles in an
interval.
Solution: let us denote h as the random variable, representing the time headways between
successive vehicles. We know that Pr(h  13) = 0.6. In other words, if we set t = 13, we
know that Pr(h  13) = e-*13, then we can calculate  based on these two equations.  =
0.039 vehicles/second. Now, let X be the random variable representing the number of
vehicle arrivals during time t, then X is poisson distributed. Pr(X=4) =
( t ) x e   t (0.039  30) 4 e .03930

 0.024.
x!
4!
4. A vehicle pulls out onto a single-lane highway that has a flow rate of 280 vehicles/hour
(poisson distributed). The driver of the vehicle does not look for oncoming traffic. Road
conditions and vehicle speeds on the highway are such that it takes 1.5 seconds for an
oncoming vehicle to stop once the brakes are applied. Assuming that a standard driver
reaction time is 2.5 seconds, what is the probability that the vehicle pulling out will be in
an accident with oncoming traffic?
Solution: in this case, if the vehicle headways between successive vehicles are greater
than 4 seconds, then the driver pulling out will not be in an accident. Or say, if the
headways are less than 4 seconds, the driver pulling out will be in an accident.
Since q = 280 vehicles/hour, then  = 0.078 vehicles/second.
Pr(h < 4) = 1-e-t = 1-e-0.078*4 = 0.268.
5. Reconsider the problem 4 above, how quick would the driver reaction times of
oncoming vehicles have to be to have the probability of an accident equal to 0.15?
Solution: let t denote the new driver reaction time. So,
Pr[h < (1.5+t)] = 1 – e-0.078*(1.5+t) = 0.15, then, we can obtain t = 0.58.
In other words, to reduce the probability of an accident to 0.15, the driver reaction must
be less than 0.58 seconds.
6. A toll booth on a turnpike is open from 8:00 am to 12 midnight. Vehicles start arriving
at 7:45 am at a uniform deterministic rate of 6 per minute until 8:15 am and from then on
at 2 per minute. If vehicles are processed at a uniform deterministic rate of 6 per minute,
determine when the queue will dissipate, total delay, longest queue length (in vehicles),
longest vehicle delay under first-in and first-out rule.
Solutions: arrival rate 1 = 6 vehicles/minute from 7:45 am to 8:15 am. 2 = 2
vehicles/minute from 8:15 am to the rest of the day. Departure rate  = 2 vehicles/minute.
7:45 am
8 am
8:15 am
Solution: the time when the queue will dissipate is then the arrival curve intersects with
the departure curve. Q1 = 6t; Q2 = 120+2t; Q3 = -90 + 6t, where Q1 is the arrival curve
between 7:45 am to 8 am and Q2 is the arrival curve starting from 8:15 am to the rest of
the day and Q3 is the departure curve. By setting Q2 = Q3, we can obtain the value for t,
the time when the queue dissipates, we obtain that t = 52.5 minutes. In other words, at
8:375 am, the queue completely dissipates.
Because Q2 and Q3 are parallel to each other, the long delay happens anywhere on the
Q1 curve and it is equal to 15 minutes. This means that every car arriving between 7:45
am to 8:15 am has to wait for 15 minutes in the queue. Also because Q2 and Q3 are
parallel, the queue length is uniform between 7:45 am to 8:15 am too. The queue length is
equal to 90 vehicles (Q1=6t = 6*15 = 90 vehicles).
To calculate the total delay, we simply do the following:
30
52.5
52.5
Total delay =  6tdt   (120  2t )dt   (90  6t )dt
=
=
0
2 30
3t |0
2
30
(120t  t
15
2
.5
) |52
30
52.5
(90t  3t 2 ) |15
3(30  0)  120 * (52.5  30)  (52.5 2  30 2 ) 
90 * (52.5  15)  3 * (52.5 2  15 2 )
= 2700 + 2700 + 1856.25 + 3375 – 7953.5 = 3037.5 vehicle-minutes.
7. Vehicles begin to arrive at a toll booth at 8:50 am, with an arrival rate of (t) = 4.1 +
0.01t (with t in minutes and (t) in vehicles per minute). The toll booth opens at 9:00 am
and process vehicles at a rate of 12 per minute throughout the day. Assume D/D/1
queuing, when will the queue dissipate and what will be the total vehicle delay?
Solution: we follow the same procedure as applied in problem number 6. The arrival
curve is equal to t = 4.1t + 0.01t2 and departure curve is equal to –120 + 12t. Setting the
departure curve to be equal to the arrival curve, we obtain that t = 15 minutes or 774
minutes. We take the first value, which is equal to 15 minutes.
About the total delay, we integrate the arrival curve and the departure curve and substract
the latter from the former, and we obtain:
Total delay = 2.05t2|(0,15) + (1/300)t3 (0,15) + 120t|(10,15) – 6t2|(10,15)
= 322.5 vehicle-m.