A Comparison of Two Traffic Models

A Comparison of Two Traffic Models

Frank Massey

1.The Deterministic Model.

These notes are concerned with comparing two models of traffic flow in one direction in a section of roadway that is uninterrupted by traffic lights, stop signs or cross traffic. The first model, discussed in section 1 is a simple deterministic model based on well known principles; Wattleworth [7] is a reference for these principles. The second model, discussed in sections 2 and 3 is a Markov process model proposed by Jain and Smith [2]. Recently Manns [8] did simulations for the Markov model obtaining graphs of how the transit time depends on the parameters of the model. In both sections 2 and 3 we extend Manns' work by comparing the Markov model with the deterministic one. Sections 4 – 6 contain supplementary material. Some of this is background, while other may be used in future work.

The Markov model divides the roadway into sections called road segments with

(1)L = Length of a road segment (mi)

We use the following two variables to measure the amount of traffic in a segment.

n = Number of vehicles in the segment (veh)

(2)

 = Density (veh/mi)

= Number of vehicles per unit length of roadway in a single lane

=

So

(3)n = L

For example, if L = 2 mi and n = 80 veh then  = 40 veh/mi.

In a more general treatment of the deterministic model  could depend on the position in the roadway, but we shall assume it is constant with respect to position.

Also of interest are

(4)k = Markov jam density (veh/mi)

= Maximum density a lane of the roadway can hold in the Markov model

(5)C = Capacity of the segment (veh)

= Maximum number of vehicles the segment can hold

= [ kL ]

V = Velocity of vehicles (mi/hr)

q = Flow (veh/hr)

= number of vehicles passing a fixed point per unit time in a single lane

 = Arrival rate (veh/hr)

= Rate at which vehicles are arriving into the section of roadway

where [ x ] = greatest integer less than or equal to x. One source [5] indicated that k is usually in the range of 185-250 veh/mile.

The number/density, velocity and flow are connected by the flow equation

(6)q = V =

For example, if L = 2 mi, n = 80 veh and V = 50 mi/hr then q = 2000 veh/hr.

At the entrance to the section of roadway.

(7)q = 

However, we will be assuming V and  are constant with respect to position, so (7) will hold in general. Despite this we will still write equations which are true when q may depend on position in terms of q and reserve  for when we want to emphasize how some variable depends on the arrival rate. In particular, in the first three sections of these notes we shall be interested in how n depends on .

In the deterministic model we assume

Assumption 1. Velocity is a function of the density, i.e.

(8)V = ()

where () is a non-increasing and continuous function of  for  0 such that (0) > 0 and ()  0 as . () is called the velocity function.

It follows that velocity is also a function n. We shall write this as

(9)Vn = Velocity when there are n vehicles in the segment (mi/hr)

=

Let

(10)A = Velocity when there is one vehicle in the segment (mi/hr)

= V1 = 

(11)fn = Specific velocity when there are n vehicles in the segment (unitless)

= Ratio of the velocity for n vehicles to that of one vehicle

= = =

So

(12)Vn = Afn

Also, () = VL = AfL. In general, the velocity function depends on the road conditions, weather, time of day and other factors. The simplest case is when Vn is a linear function. One has V1 = A. Also, suppose one also has VC+1= 0. Then

(13)Vn =

So

(14)fn =

At the right is the graph of Vn = (A/C)(C + 1 – n) when =50 mi/hr and C = 200 veh/mi. Some other common velocity functions are given in section 5.

Combining the flow equation with Assumption 1 gives

(15)q = () = g() = = =

where

(16)g() = ()

= flow function

(17)n = nfn = n = L = g

In the linear case (13) the relation (15) becomes

(18)q =

At the right is the graph of q=(An/C)(C+1– n) when =50 mi/hr and C = 200 veh/mi.

We assume the flow function satisfies the unimodal property in Assumption 2 which follows. To simplify the statement of this assumption let

 =Deterministic jam density

=Value of the density where traffic stops

=

If  then () = 0.

Assumption 2. k and the flow function q = () = g() is strictly increasing as  increases from 0 to a positive value c and q = () = g() is strictly decreasing as  increases from c to .

It follows that q = n = g is strictly increasing as n increases from 0 to a positive value nc = Lc and q = n is strictly decreasing as n increases from nc to L.

Along with this assumption let

qc =Critical flow rate

=Maximum flow rate

=Vcnc/L

=Vcc

c =Critical density

=Density at which flow rate is a maximum

nc =Critical number of vehicles

=Number of vehicles at which flow rate is a maximum

=Lc

Vc =Critical velocity

=Velocity at which flow rate is a maximum

 =Traffic intensity = =

One source [5] indicated that c is usually in the range of 40-50 veh/mi.

Since qqc, a necessary condition for there to be a traffic flow satisfying q= at the entrance of the roadway is qc. So, we assume

Assumption 3. The arrival rate  satisfies

(19)  qc

This is equivalent to

(20)  1

In the linear case (13)

nc =

Vc =

(21)qc =

 = =

So we need  1 for a flow to exist.

Example 1. Suppose A = 50 mi/hr, L = 1 mi, k = 200 veh/mi and  = 1000 veh/hr. Then C=200veh/mi, nc=100.5veh, Vc=25.125mi/hr, qc = 2525.06 veh/hr and =0.39603.

Let q = g-() be the restriction of q = g() to 0 c and q = g+() be the restriction of q = g() to c. Similarly, let q = -,n be the restriction of q = n to 0 nnc and q= +,n is the restriction of q = n to ncnL. Let  = g--1(q), q = g+-1(q) n = --1(q) and n=+-1(q) be the inverses of these functions.

In the deterministic case we shall assume

Assumption 4. The arrival rate  is constant in time and traffic flow has reached a steady state so that V,  and q are constant in time and position and

q = for all time and postion

 = g--1(q) = g--1()

From (15) and q =  it follows that n = L/A or

(22)n = x

where

(23)x = (veh)

So, from Assumption 4 it follows that

(24)n = --1(x) = --1

In the linear case one has

n = =

So (22) becomes

n2 - (C + 1)n + = 0

n =

So

(25)n = = --1(x)

So

(26)n = --1 = = nc

This can also be written as

(27)n =

since

(28) = = 

To the right is the graph of n = --1 in the linear case when =50 mi/hr, L = 1 mi and k=200 veh/mi. In this case C = 200 veh, nc=100.5 veh and qc=2525.06 veh/hr.

Example 2. Suppose, as in Example 1, one has A = 50 mi/hr, L = 1 mi, k = 200 veh/mi and  = 1000 veh/hr. We saw there that C=200veh/mi, nc=100.5veh, Vc=25.125mi/hr, qc = 2525.06 veh/hr and =0.39603. So

x = = = 20 veh

n = = (100.5 veh) = 22.40 veh

2.The Markov Model.

Section 1 described the deterministic model of traffic flow. In order to better describe the real world a number of authors have proposed stochastic models. This section looks at the Markov model proposed by Jain and Smith [2]. They call their model a queueing model because of the similarity of some of the equations with those of queueing theory.

They consider a road segment of length L and treat the number n of vehicles in the segment as a random variable that varies with time. They model the traffic flow by a Markov process with states corresponding to the number of vehicles in the segment. The states are n=0,1,…,C with state n corresponding to n vehicles in the segment.

Instead of assuming the vehicles enter the segment at a constant rate, they assume the arrival of vehicles into the segment is a Poisson process with

 = arrival rate of cars to the segment (veh/hr)

If there are already C vehicles in the segment when a car arrives, then it doesn't enter the segment. Instead it may wait for a vehicle to leave the segment or by take another route. Transitions from state n to state n +1 occur with rate

(1)n =  (veh/hr)

for n = 0, 1, …, C-1.

Instead of assuming that the vehicle velocity V is completely determined by the number n of vehicles in the segment, they assume that 1/V is an exponential random variable with mean 1/Vn where Vn = (n/L) where V = () is the velocity function in the deterministic model. So the time a vehicle spends in the segment is an exponential random variable with mean

(2)Sn = Average time to traverse the segment if the segment has n vehicles (hr)

= =

If the system is in state n then there are n vehicles in the segment so transitions to state n–1 occur with rate

(3)n = = (veh/hr)

This is for n = 1, …, C. The resulting Markov process is a birth-and-death process. It is known (seeRoss[3,p.392 -393]) that the steady state probabilities Pn for a birth-and-death process are

(4)Pn = Steady state probability that the segment has n vehicles (unitless)

= for n 1

with

(5)P0 =

Substituting (1) and (4) into these formulas become

(6)Pn = = = = anP0

for n 1 with x = L/A which was given in formula (23) in the previous section. Here we set

(7)an = for n 1 (unitless)

Also

(8)P0 =

We are interested in

(9)N = Average number of vehicles in the segment

= = P0 = xP0

(10)T = Average time a vehicle spends in the segment

In queueing theory (10) is called Little's formula; see Ross[3]. The motivation behind it is the following. Cars are arrive at the segment at a rate . The probability that an arriving car enters the segment is (1 – PC) so cars enter the segment at a rate (1 – PC). So N = T(1 – PC) from which (10) follows.

Consider the case where Vn = (C+1-n) is the linear function given by (13) in the previous section, so fngiven by (11) in the previous section can be written as . So

(11)an = =

=

=

Example 1. Suppose, as in Examples 1 and 2 of the previous section, A = 50 mi/hr, L=1mi, k = 200 veh/mi and  = 1000 veh/hr. We saw there that C=200veh/mi, nc=100.5veh, 1() = 22.40 veh, qc = 2525.06 veh/hr and =0.396. Below are two plots of Pn vs n. The one on the left is a semilog plot and the one on the right is a regular plot.

Note that Pn increases to a local maximum, then decreases to a local minimum and then increases to another local maximum at n = 200. It turns out that this behavior is typical of what occurs if  < 1 which we shall show now.

Note that Pn-1Pn if an-1an. Since an=xan-1/(nfn) one has an-1an if 1 x/(nfn) which occurs if

n  x

where n = nfn = g was defined by (17) in the previous section. There are two cases.

The first case is when qc where qc is the critical flow rate, i.e. the maximum of g()=f(). This is equivalent to  1 where  = /L is the traffic intensity. In that case L/ALqc/Ag = n for all n. So Pn will be increasing with n for 0 nC and the maximum of Pn will be at n = C. We shall try to show later that PCPn for nC except when  is quite close to qc so that the average number of vehicles N is the segment given by (9) is almost C. This is illustrated by the following example.

Example 2. Suppose A = 50 mi/hr, L=1mi, k = 200 veh/mi and  = 2600 veh/hr. Again C=200veh/mi, nc=100.5veh and qc=2525.06 veh/hr. But now  = 1.03. To the right is a semilog plot of Pn vs n. As one can see Pn is an increasing function of n.

The second case is when qc (or  < 1). Consider the equation n = x which is equation (22) in the previous section. It follows from the discussion there that the equation n = x has two solutions

n = --1(x)and+-1(x)

where -,n is the restriction of n to 0 nnc and +,n is the restriction of n to ncnC+1. If we let

(11)n1 = [ --1(x) ]n2 = [ +-1(x) ]

then Pn will increase with n for 0 nn1 and n2nC and Pn will decrease with n for n1nn2. Thus Pn will have local maxima at n = n1 and n = C. This was illustrated by Example 1 above.

One thing we would like to show is that if  < 0.65 then near n1 the graph of Pn is almost symmetrical about n = n1 and away from n1 one has PnPn1 so that the average number of vehicles N is the segment given by (9) will be quite close to --1. In this case N will be quite close to the steady state number of vehicles in the deterministic model which was --1(x) = --1; see formula (26) in the previous section.

In the case Vn = (C+1-n) where is a linear function and  < 1 then in the derivation of formula (27) of the previous section we saw that

--1(x) =

+-1(x) =

So

n1 = [ ]

n2 = [ ]

So Pn will increase with n for 0 nn1 and n2nC and Pn will decrease with n for n1nn2. Thus Pn will have local maxima at n = n1 and n = C.

Example 3. Suppose, as in Example 1 above, A = 50 mi/hr, L=1mi, k = 200 veh/mi and  = 1000 veh/hr. We saw there that C=200veh/mi, nc=100.5veh, qc = 2525.06 veh/hr, =0.39603 and 1(x) = 22.40 veh. Also

+-1(x) = = (100.5 veh) = 178.60 veh

n1 = [ --1(x) ] = 22 veh

n2 = [ +-1(x) ] = 178 veh

then Pn will increase with n for 0 n 22 and 178 n 200 and Pn will decrease with n for 22n 178. Thus Pn will have local maxima at n = 22 and n = 200. This was illustrated in the graph with Example 1.

Remark. In the case Vn = (C+1-n) where is a linear function there are four input parameters to the model: A, k, L and . However, these parameterscombine into C=[kL] veh and x=L/A, both with units vehicles which effectively reduces the number of parameters to just two.

In the next section we discuss the Markov model further along with more comparisons with the deterministic model.

3.More on the Markov Model

In this section we look more at the Markov Model and how it compares with the deterministic model.

Consider the probability, PC, that the roadway has the the maximum number of vehicles. In the discussion of the Markov model in the previous section, we discussed how the steady state probabilities Pn that the segment has n vehicles varies with n. In particular, we showed that if the traffic intensity  was less than one then Pn increases with n for 0nn1 and n2nC and Pn decreases with n for n1nn2 where

(1)n1 = [ --1 ]n2 = [ +-1 ]

Here -,n is the restriction of n to 0 nnc and +,n is the restriction of n to ncnC+1 where n = nfn. On the other hand if  1 then Pn increases with n for 0nC.

In the first case if PC is small then the average number of vehicles N is the segment will be quite close to --1 which is the steady state number of vehicles in the deterministic model. So we would like to know when is PC small.

Example 1. Suppose A = 50 mi/hr and L=1mi. We consider three values of k, namely k=180 veh/mi, k=200 veh/mi and k=220 veh/mi. For each of these we made a plot of how PC varies with  for 0  1 and we combined these into one plot which is at the right. As one can see in each case PC is almost zero until  reaches about 0.65 and then it increases rapidly to almost one by  =0.7 where it remains almost one. This suggests that PC will be small and the Markov model will be quite similar to the deterministic model for 0.65. However, it appears that PC will be close to one and the roadway will be congested for 0.7.

Example 2. Suppose A = 50 mi/hr, L=1mi and k=200 veh/mi. In Example 3 of section 1 we made a graph of how n varies with  for the deterministic model. We also make a graph of how the average number N of vehicles varies with  and combine this with the previous graph. Below is a plot of n (blue) and N (red) versus . As one can see they are essentially the same until about 1600. After that N climbs rapidly to 200 while n continues along the curve discussed in section one.

4.More on the Deterministic Model.

If one is working just with the deterministic model then instead of working with the parameters A, k and C of sections 1 and 2 one usually works with

 =free velocity

=velocity as density approaches zero

=(0)

In this case a linear velocity function would be expressed as

V =

and the corresponding flow function would be

q =

Below are the graphs of V = () and q=() in the linear case when =50 mi/hr and  = 200 veh/mi.

In the linear case the flow qis a quadratic function of  with q = 0 when = 0 and  = . So Assumption 3 of section 1 holds and

c =

Vc =

qc =

 =

In the case  = 50 mi/hr,  = 200 veh/mi and  = 1000 veh/hr one has c = 100 veh/mi, Vc=25mi/hr, qc=2500 veh/hr and  = 0.4.

Also, in the linear case one has

q =  -

So

2 -  + = 0

 = =

So

 = g--1() = = c = c

where  = /qc is the traffic intensity defined in section 1.

To the right is the graph of  = g--1() in the linear case when =50 mi/hr, =200veh/mi, c=100veh/mi and qc=2500 veh/hr.

The reciprocals of velocity, density and flow are often of interest.

 =1/V

=transit time (hr/mi)

=time to travel a unit distance

s =1/

=spacing (mi/veh)

=distance between vehicles

h =1/q

=headway (hr/veh)

=time between vehicles passing a fixed point

For example, if V = 50 mi/hr and  = 40 veh/mi then q = 2000 veh/hr, =0.02hr/mi=1.2min/mi, s=0.025mi/veh=192 ft/veh and h=0.005hr/veh=1.8sec/veh.

The following table of  and s = 1/ can help one appreciate why the jam density  is 185–250 veh/mi. The last column is assuming that a car is 18 ft in length.

(veh/mi) / s(mi/veh) / s(ft/veh) / s(car-lengths/veh)
50 / 0.02 / 106 / 5.9
100 / 0.01 / 53 / 2.9
150 / 0.0067 / 35 / 2.0
200 / 0.005 / 26 / 1.5

In section 1 we saw how how the density  behaved as a function of the arrival rate . In this section we see how V and  behave as functions of  and also how , V and  behave as functions of the free velocity  and the jam density  when the other and  are held constant.

In this section we assume Assumptions 1 – 4 of section 1 hold. In particular V = () is the velocity function in Assumption 1 of section 1.

To begin with we express the velocity V and the transit time  = 1/V as functions of . Recall from formula (11) in section one that

 = g--1()

Since V = () one has

V = (g--1())

 =

In the linear case V = one has from formula (12) of section one

 = g--1() = = c = c

where  = /qc is the traffic intensity defined by (9) in section one. So

= 1 - = 1 - =

So

V = = = Vc

= Vc

and

 = = =

Below are the graphs of V = (g--1()) and  = 1/(g--1()) in the linear case when =50 mi/hr, =200veh/mi, Vc=25mi/hr, qc = 2500 veh/hr and c = 0.04 hr/mi.

In the remainder we also assume

Assumption 5. () = 0 for some finite .

As in Section 1 we let  = min{: () = 0} be the jam density, so () = 0 for .

Definition 1. Let

(1)F(r) = = = specific velocity function

It follows from Assumption 1 in section 1 that F(r) is a non-increasing and continuous function of r for r 0. Furthermore F(0) = 1, F(r) > 0 for 0 r < 1 and F(r) = 0 for r 1. Also

(2)V = () = F

In the linear case

(3)V =

one has

(4)F(r) =

Then the flow function is given by

(5)q = () = F = F = G

where

(6)G(r) = rF(r) = specific flow function

It follows from Assumption 2 that the function  = rF(r) = G(r) is strictly increasing as r increases from 0 to a positive value rc and q = rF(r) = G(r) is strictly decreasing as r increases from rc to 1. Let

rc =critical specific density

vc =F(rc) = critical specific velocity

c =G(rc) = rcvc = critical specific flow rate

Tc =1/vc = critical specific transit time

Note that

qc =c

c =rc

Vc =vc

In the linear case (4) one has

G(r) =

and rc = ½, vc = ½, c = ¼ and Tc = 2.

If one solves (5) for  in terms of qand uses Assumption 3 of section 1 then one has

= G

= G--1

(7) = G--1 = G--1

where r = G--1() is the inverse of  = G(r) restricted to 0 rrc. r = G--1() is called the specific inverse flow function. Note that in order for a steady state flow to exist ,  and  must satisfy

(8)  c

Suppose we freeze  and  and let  vary. It follows from (8) that in order for a steady state flow to exist  must satisfy

(9)  c

where

(10)c =

Then  is given in terms of  by (7) from which we see that  decreases as  increases. From the flow equation q = V we can get equations for V and  in terms of , i.e.

(11)V = =

(12) = = G--1

On the other hand suppose we freeze  and  and let  vary. It follows from (8) that in order for a steady state flow to exist  must satisfy

  c

where

c =

From (7) and the velocity equation q = V we can get equations for V and  in terms of , i.e.

(13)V = F = F = F

(14) =

Then using the flow equation q = V we can get equation for , i.e.

(15) = =

Let's look at what these formulas become in the linear case (3). First consider the case where we freeze  and  and let  vary. In the linear case c = ¼ so c given by (10) becomes

c =

One also has

G--1() =

Therefore , V and  given by (7), (11) and (12) become

 = G--1 = =

V = = =

 = G--1 = =