Network Traffic Modelling

University of York

Dissertation submitted for the MSc in Mathematics with Modern Applications, Department of Mathematics,

University of York, UK.

August 2009

Network Traffic Modelling

Author:

David Slade

Supervisor:

Professor M.J. Smith

Contents

1 Introduction to Signal Control and Timings 6

1.1 Continuous Modelling of Fixed Time Signals . . . . 6

1.2 Delays, Routes, Links and Stages . . . . 7

1.3 Optimisation of Fixed Time Signals . . . . 9

1.4 Examples of P

Policy . . . . 11

2 Braess’s Network 15 2.1 Braess’s Network with Flow from Node 1 to Node 2 . . . . 16

3 Responsive Control In A Simple Network 19 3.1 Terminology . . . . 19

3.2 Demand Feasibility, Supply Feasibility and Feasibility . . . . 20

3.3 Two Delay Formulae . . . . 20

4 Consistent Equilibrium Using The Equisaturation Policy 23 4.1 Equilibrium Using Delay Formulae δ

. . . . 23

4.2 Stability of Equilibria Using Delay Formulae δ

. . . . 24

4.3 Equilibrium Using Delay Formulae δ

. . . . 26

4.4 Stability of Equilibria Using Delay Formulae δ

. . . . 27

4.5 Alternative Stability Analysis . . . . 30

4.5.1 Analysis for Equilibria Using δ

Formulae . . . . 30

4.5.2 Analysis for Equilibria Using δ

Formulae . . . . 31

4.6 Remark . . . . 32

5 Effect on Network Capacity 33 5.1 Network Capacity Using Equisaturation Policy . . . . 33

5.2 Network Capacity Using the P

Policy . . . . 34

5.3 Modelling as a Dynamical System and Finding a Lyapunov Function 34 5.4 General Case . . . . 37

5.5 P

and Equisaturation Policies . . . . 39

6 Equilibria of a Symmetrical Network with Bottlenecks 40 6.1 Using delay formulae δ

. . . . 40

6.1.1 How The Value of α Affects Stability of the Network . . . 42

6.2 Using the δ

Delay Formulae . . . . 44

6.2.1 How The Value of α Affects Stability of the Network . . . 46

6.3 Using the P

Policy . . . . 46

6.3.1 Remark . . . . 47

6.3.2 When A 6= 0 Using P

Policy . . . . 47

6.3.3 Stability . . . . 48

7 Conclusion 49

List of Figures

1 A typical shape of the graph of Webster’s delay formula as traffic flow on the lane, X

, increases. Note the asymptote at s

G

where the delay increases towards infinity. . . . 8 2 A simple example of a network that has four routes. The flow on

link i is therefore = flow on route 1 + flow on route 2 + flow on route 3 + flow on route 4 = X

+ X

+ X

+ X

. . . . 9 3 A simple example of a signal controlled junction with two links

approaching it. Link 1 has saturation flow s

and link 2 has satu- ration flow s

. . . . 9 4 A simple network. . . . 11 5 A network with a single origin-destination pair with four routes. . 13 6 Braess’s Network showing costs on the links, e.g. flow + 50 means

the cost on the link is the flow + 50 units. . . . 15 7 The average travel cost at equilibrium flow as k changes, it is a

decreasing function for 0 ≤ k ≤ 23. . . . 16 8 The average travel costs of the three routes as the amount of traffic

travelling from node 1 to node 2, ǫ, changes. . . . 17 9 How the price of anarchy changes for increasing amount of cars

travelling from node 1 to node 2. . . . 18 10 A simple signal-controlled network. Route 1 has saturation flow

= k vehicles per minute at node A and saturation flow k vehicles per minute at the signal. Route 2 has saturation flow=k vehicles per minute at node B and saturation flow s vehicles per minute at the signal. . . . 19 11 Example graph of how d

may change as ’spare capacity’ de-

creases, approaching infinity as T X

approaches sG

. . . . 21 12 Bifurcation diagram to show pitchfork bifurcation of the equilibria

of the system. Bifurcation point at T = s −

and x

=

+ h

4

−

i

, x

=

− h

1

4

−

i

. . . . . 25 13 The set of equilibria consistent with the equisaturation policy for

some T using delay formulae δ

. The lines are all equilibria, in- cluding the axes, and the arrows indicate the natural directions of motion of non-equilibria as drivers and green times respond to delays and saturation ratios (they also indicate the stability of the equilibria). . . . 26 14 Bifurcation diagram to show the changing stability of the equilibria

of the system. Bifurcation points at T

=

−

s

−



and T

=

+

s

−



. . . . . 28

15 The set of equilibria consistent with the equisaturation policy for some T using delay formulae δ

. The lines are all equilibria, including the axes, and the arrows indicate the natural direc- tions of motion of non-equilibria as drivers and green times re- spond to delays and saturation ratios (they also indicate the sta- bility of the equilibria). The bifurcation points are also indicated;

T

=

−

s

−



and T

=

+

s

−



. . . . . 29 16 A network with a wide long upper route and shorter narrower lower

route. Route 1 has saturation flow = 8s

vehicles per minute at node A and saturation flow s

vehicles per minute at the signal.

Route 2 has saturation flow=α vehicles per minute at node B and saturation flow s

vehicles per minute at the signal. . . . 33 17 Network capacity as s

varies using two different policies for de-

termining signal timings, P

and equisaturation. . . . 35 18 A symmetrical network with two routes both with identical satu-

ration flows of α at the bottlenecks at Node A and Node B and saturation flow s at the signal controlled junction. . . . 40 19 A graph of the equilbria of the network as T increases, with α =

. 43 20 A typical graph of the equilbria of the network as T increases, with

s

4

< α <

. . . . 43 21 A graph of the equilbria of the network as T increases, with α =

. 46 22 The set of equilibria consistent with the P

policy for some T

using delay formulae δ

. The lines are all equilibria, including the

axes, and the arrows indicate the natural directions of motion of

non-equilibria as drivers and green times respond to delays and

saturation ratios (they also indicate the stability of the equilibria). 49

Acknowledgements

I would like to thank the whole of the Department of Mathematics for their help, lectures, seminars and other such work in the four years i have spent at the University of York. In particular i would like to thank Dr.Zaq Coehlo for organising the MSc course which has been most enjoyable.

I would also like to thank Professor Mike Smith who has been a marvellous

supervisor for this dissertation. For providing me with an insight into his and oth-

ers work on the subject of Traffic Modelling, for guiding, advising and correcting

me; i have thoroughly enjoyed working on and completing this project.

Introduction

In the modern world many people own cars or use public transport to travel.

This leads to congestion and therefore long delays on journeys. To attempt to minimise delays and keep traffic moving then we must try and find the best signal timings and traffic flows for traffic networks. To do this various models have been used and their dynamics analysed. Within these models there are two important choices to make, one of them is the form of the delay function of a link and the other the signal timing policy employed.

In this project we will look at two forms of a delay function, both similar and both representing a term in Webster’s delay formula which will be introduced in the next section. We will also consider two signal timing policies, one of them called the ’equisaturation’ policy and the other the P

policy.

We will first introduce notation and concepts of traffic modelling, as well as the two signal timing policies and Webster’s delay formula. In the section following this we look at Braess’s network and how increasing cost and thus decreasing traffic flow on a link in Braess’s network can actually reduce average travel cost across the network. We will then consider changing the parameters and adding flow on one link and consider how this changes the price of anarchy.

In the next section we look at the set of feasible traffic flows and green times and the constraints that they have. We also look at our two chosen delay formulae and look at the consistent equilibria they produce when the equisaturation policy is utilised on a simple network with no bottlenecks. We analyse the stability of these equilibria, using two methods, and will acquire graphs of the evolution of the equilibria as the total flow on the network changes.

We next look at the effect on network capacity using the P

policy has com- pared with the equisaturation policy. We then consider the system as a system of four differential equations and find a Lyapunov function to show that there is only one consistent equilibrium when using the P

policy on the network.

Finally we introduce bottlenecks into the simple network we have been looking at, and look at the symmetrical version of this network. We find the equilibria of the network, using both delay formulae, and consider how the saturation flows at the bottlenecks can affect the stability of these.

Acknowledgement of Previous Work

This project is based upon work done by a number of experts. The first sec- tion involving Braess’s network is based upon work done by both Braess [4] and Roughgarden [8].

The rest of the project is based upon work, supervision and guidance done

by Professor Mike Smith at the University of York, who has provided numerous

papers, projects and presentations for myself to learn from and work from.

1 Introduction to Signal Control and Timings

Traffic signal control strategies seek to minimise a combination of total travel time and stops for vehicles for the local traffic flow that has been observed, assuming that this pattern does not change.

1.1 Continuous Modelling of Fixed Time Signals

Definition 1 A stage a set of lanes approaching a junction which are shown green simultaneously as their paths do not coincide.

The simplest case is where the stages are such that traffic approaching the junction along distinct lanes in the same stage follow paths which do not intersect and so lanes in the same stage may be given green simultaneously.

Let there be K stages at a junctions, the signal display changes from ’green to stage 1’ to ’green to stage 2’ to...to ’green to stage K’ to ’green to stage 1’ and in a continuous cycle as given.

The method of changing display is ’fixed time’ if the periods of time when the signal is green for stage k are of the form t

to t

+ τ G

, τ + t

to τ + t

+ τ G

, 2τ + t

to 2τ + t

+ τ G

, and so on. The time period during which stage k is green consists of repeated periods of time of constant lengths, τ G

seconds, which happen regularly (every τ seconds).

The time τ seconds is called the cycle time of the signal and is usually less that 120 seconds. τ G

is the green time and t

may be thought of as the offset associated with the particular junction been considered. Different junctions have their own t

offset time and taken together these describe the way in which the fixed time evolutions at separate junctions are related.

Definition 2 An integreen is some fixed time, η seconds say, during which all stages at a junction are shown red to allow traffic which has just had priority in one stage the time to clear the junction before traffic flows with coinciding paths are given priority/shown green.

We will assume that η = 0 for simplicity.

The green time proportions G

,G

,...,G

for which stages 1,2,...,K are given green add up to one.

A signal controlled junction has signals operating fixed time if the variables τ,G

,...,G

are fixed. It is natural to seek the best values of these variables which is the reason for modelling traffic signals.

Webster [2] first thoroughly studied isolated traffic signals and this is now a

standard reference on fixed time signal control. Webster developed a formula for

the delay imposed on a single traffic stream by a traffic signal operating to fixed

time signal timings and used this to justify a simple rule in calculating fixed time

settings which approximately minimise the total of all delays experienced by all vehicles passing through the junction. The study involved:

1. theoretical modelling

2. simulation to validate the delay formula

3. optimisation to estimate reasonable signal settings.

These three elements play an important part in studies that follow Webster’s work.

1.2 Delays, Routes, Links and Stages

To optimise the signal control variables we need an objective function which is usually chosen to be the total rate of delay D at the junction. To minimise D we require a formula giving D in terms of flows and signal timing parameters which leads us to Webster’s Formula. This was designed to estimate the average medium run delay d

to vehicles on a single lane caused by traffic signals operating on a fixed time basis.

It gives the estimated average delay per vehicle as a function of 1. average traffic inflow rate along that lane (X

vehicles per second) 2. proportion of time the lane is given green (G

)

3. saturation flow of the lane (s

vehicles per second) 4. cycle time (τ seconds).

Webster’s Delay Formula

d

(X

, G

) = 9 20

"

τ (1 − G

)

1 −

+ X

s

G

(s

G

− X

)

#

This is an extremely important function that is considered throughout modelling of traffic flow networks.

We can consider numerous simplifications of this formula as the overriding factor of the delay formula is

δ = 1

s

G

− X

as this almost dictates the shape of the graph completely. So we can consider

this formula and slight variations of it when deciding on a delay formula that we

wish to apply to a network when studying its dynamics.

d i

x i

s i G i

Figure 1: A typical shape of the graph of Webster’s delay formula as traffic flow on the lane, X

, increases. Note the asymptote at s

G

where the delay increases towards infinity.

The term δ is close to a well know formula, derived by Pollaczek and Khint- chine, for the average delay experienced by a Poisson stream of traffic being served by a server with identical service times equal to

iGi

seconds. The first term of Webster’s delay formula approximates the delay due to the stop start nature of traffic signal control.

A network is made up of links, routes and nodes. At the start of each route is an origin node which is connected by a number of links and nodes to the final node, the destination node. There may be a number of origin-destination pairs.

Definition 3 A link is a road that adjoins the last signal controlled junction or bottleneck to the next one.

Definition 4 A route is made up of a finite number of links that join the origin to the destination with no breaks.

Definition 5 The flow on a link i is denoted x

and is the number of vehicles per minute on the given link. The flow on a route r is denoted by X

.

The flow along link i, x

, is the sum of the flows along all routes which the link belongs to. So

x

= X

r: route r contains link i

X

where X

is the flow along route r.

The green time awarded to link i, G

, is the sum of those green times awarded to the stages to which the link belongs:

G

= X

k:stage k contains link i

g

.

Example:

link i

Figure 2: A simple example of a network that has four routes. The flow on link i is therefore = flow on route 1 + flow on route 2 + flow on route 3 + flow on route 4 = X

+ X

+ X

+ X

1.3 Optimisation of Fixed Time Signals

link 1

link 2 signal

Figure 3: A simple example of a signal controlled junction with two links ap- proaching it. Link 1 has saturation flow s

and link 2 has saturation flow s

.

Let us consider a junction with only two lanes approaching it (see Figure 3, link 1 and link 2, with saturation flows s

and s

respectively, so stage 1 is link 1 and stage 2 is link 2. Link 1 has signal green time G

and link 2 is awarded green time G

. We have that the delays experienced by each of the routes are represented by

d

(X

, G

) and d

(X

, G

) .

Suppose link 1 is given green for τ G

seconds and link 2 is given green for τ G

seconds during each cycle of τ seconds then we have that as G

and G

are proportions that

G

+ G

= 1.

Supply Feasibility and Constraints: Let flows X

and X

be fixed. The stage green times G

and G

are adjusted so the fixed traffic flows can get through the junction. Then the feasible supply gives X

< s

G

, X

< s

G

. Also we have that G

+ G

= 1 and G

≥ 0, G

≥ 0. These constraints give us a set for feasible green time vector G = (G

, G

) .

We are supposing that the vector of link flows, X, and cycle time τ are both fixed. It is easy to check that Webster’s formula is decreasing and strictly convex in G

. Thus if we let D be the sum of all the delays experienced at the junction

D (X, G) = X

d

(X

, G

) + X

d

(X

, G

)

is (for fixed X) a convex function of the vector G of the stage green times.

There are many ways of determining a feasible vector G which minimises the convex function D subject to the linear feasibility constraints on G.

One way is to begin with a feasible G. Determine the ’marginal costs’

∂D

∂G

and ∂D

∂G

.

Both are negative as delay d

will be smaller if G

gets larger. They give the degree to which D changes when the stage green time vector G changes. We may think of

−

and −

as pressures P

and P

on stages 1 and 2 respectively. Increasing the green time for the stage under the most pressure is a natural decision.

Suppose

P

= − ∂D

∂G

< − ∂D

∂G

= P

then a swap of time α from less pressurised stage 1 to more pressurised stage 2 will reduce D, by about α h

∂D

∂G₁

−

i

if α is small. A more general swap rate is defined as − (P

− P

).

As green time is swapped towards stage 2, D must decline and under natural conditions G approaches the stage green time vector G

which minimises D for fixed X. At G

it is impossible to swap further according to the above rules. At optimum G

, P

= P

. Pressures on the two stages (or in this case links) are equal and D is minimised.

This is one signal timings policy called the delay minimisation policy. One can

build other objectives into a signal setting policy by letting P

(X, G) be other

functions of flow and green time. Still following as above and considering P

as

pressures. The new control policy, we call it ’policy P’, is determined by P

and

P

.

Policy P clearly depends on how we choose the pressures P

. For any natural P the adjustment of G is as straight forward as with delay minimisation.

The rate of change of G

is P

− P

, rate of change of G

is P

− P

. The swap rule leads to G which equalises P

and P

and minimise the maximum of P

and P

.

There are three choices for P that are most popular:

Delay Minimisation

P

(X, G) = − ∂D

∂G

P

(X, G) = − ∂D

∂G

Equisaturation

P

(X, G) = X

s

G

P

(X, G) = X

s

G

P

Policy

P

(X, G) = s

d

(X

, G

) P

(X, G) = s

d

(X

, G

)

The third choice called the ’P

policy’ [1] and, as we will show throughout the project, favours links with high saturation flows and maximises the capacity of a network automatically.

1.4 Examples of P 0 Policy

Consider the following simple network with signal control where link 1 and link 2 meet. It has link 1 which has twice the saturation flow as the shorter link 2.

So take s

= 2 and s

= 1.

Link 1

Link 2

Figure 4: A simple network.

Using the policy P

we have

s

d

= s

d

⇒ d

= 2d

so the delay time for vehicles on link 2 is twice as long as the delay for vehicles

on link 1 thus encouraging vehicles to use the higher saturation link that is link

1 and so overall maximising the capacity of the network as a whole.

Now consider the network in Figure 5. The network has one origin destination pair and four routes. There are three signal controlled junctions at nodes A,B and C.

route 1

route 2

route 3 A

B

C Origin

Destination

Figure 5: A network with a single origin-destination pair with four routes.

Let us assume that routes 1,2 and 3 have saturation flow s

= 1. We now look at two cases, one where the saturation flow on route 4 is greater than the sum of the saturation flows on the other routes. The other case is when the saturation flow on route 4 is less than the sum of the other routes.

Take s

= 4. Thus

s

d

= s

d

⇒ d

= 4d

.

So as there are 3 signals on route 4 the total delay time is 3

4 d

< d

so the delay time on either route 1,2 or 3 is greater than on route 4. Thus favouring the route with the higher saturation flow.

Now take s

= 2.

d

= 2d

and so the total delay time on route 4 is 3

2 d

> d

so the incentive is to stay on either route 1,2 or 3 and avoid route 4. This is because as a combination routes 1,2 and 3 offer a higher saturation flow than route 4 in this case.

These two examples show that policy P

favours the higher capacity routes

and maximises the network capacity. We consider this policy further in other

sections of the project and show how P

produces favourable results.

2 Braess’s Network

Origin Destination

flow + k flow + 50

flow + 50 10 x flow

10 x flow

Node 1 Node 2

Route 1 Route 2

Figure 6: Braess’s Network showing costs on the links, e.g. flow + 50 means the cost on the link is the flow + 50 units.

We consider Braess’s network [4], shown in figure 6, which we assume has a capacity flow of 6 travels from the origin to the destination. If we, at first, consider that the dotted link is not present then the equilibrium flow (where both routes have equal costs) is when there is a flow of 3 on each route. As the costs are equal there is no incentive to swap routes. The cost on the two routes are

(10.flow on link) + (flow on link + 50) = (10.3) + (3 + 50) = 83.

Now we consider the network with the dotted link included. If k = 10, then the three routes have the same cost if the distribution of flow is equal on all three, i.e. flow on route 1 = flow on route 2 = flow on route 3 = 2. The average travel cost at this distribution is 92 which is greater than the average cost without the dotted link.

We now look at the effect of varying the parameter k. If k = 23 then we see that the flow distribution (3,3,0) is an equilibrium with the costs of each route being 83. So by increasing the cost of the dotted link it has been made undesirable and so traffic flow is only on the other routes. So the equilibrium of the network changes when k changes.

If k varies between 0 and 23 then we obtain that the equilibrium flow pattern

is 

2 + k − 10

13 , 2 + k − 10

13 , 2 − 2 (k − 10) 13



(presented in the form (flow on route 1, flow on route 2, flow on route 3) ),which

gives all route costs as 92 −

.

92 83

10 23

Average travel cost

k

Figure 7: The average travel cost at equilibrium flow as k changes, it is a de- creasing function for 0 ≤ k ≤ 23.

2.1 Braess’s Network with Flow from Node 1 to Node 2

Let us now consider Braess’s network when there is a set amount of traffic flow, ǫ, of the total flow (which is 6) travelling from node 1 to node 2 (see figure 6 ).

Let us find the equilibrium flow pattern that is created in this situation and how this affects costs of travel.

There is an equilibrium when costs on all three routes are the same. These are defined as

C

= (10.flow on link11) + (50 + flow on link12) C

= (10.flow on link22) + (50 + flow on link21)

C

= (10.flow on link11) + (10.flow on link11) + (flow on dotted link + k).

We note that (if f

= flow on route i);

flow on link 11 = f

+ f

, flow on link 12 = f

, flow on link 21 = f

, flow on link 22 = f

+ f

,

and flow on dotted link = f

+ ǫ.

We also note that as the total network flow is 6, then f

+ f

+ f

= 6 − ǫ.

At equilibrium we have C

= C

= C

, thus we have four simultaneous equations to solve for three variables f

. Firstly, C

= C

, gives us that f

= f

. Then we use the fact that f

+ f

= 6 − ǫ − f

to yield, from C

= C

,

10 (6 − ǫ − f

) + (50 + f

) = 20 (6 − ǫ − f

) + f

+ ǫ + k and noting that f

= 6 − ǫ − 2f

,

(50 + f

) = 10 (6 − ǫ − f

) + 6 − 2f

+ k

gives

f

= f

= 2 + k − 10 (1 + ǫ) 13 f

= 2 − 2 (k − 10) + 7ǫ

13 .

This gives average costs for all routes following this equilibrium flow pattern as

92 −  9 (k − 10) + 40ǫ 13



which we see is the same as the usual Braess’s network costs when ǫ = 0.

The equilibrium flow pattern is valid until f

= 0 which is when 2 (k − 10) + 7ǫ < 26

which is when

k < 23 − 7 2 ǫ.

This gives us the following graph of average travel cost for different values of ǫ, see Figure 8.

92

83

10 23

average travel cost

k

16 19.5 12.5 77.5

72 66.5

ǫ = 0 ǫ = 1 ǫ = 2 ǫ = 3

Figure 8: The average travel costs of the three routes as the amount of traffic travelling from node 1 to node 2, ǫ, changes.

The ratio of the equilibrium cost and the optimal cost, which Roughgarden

[8] called the price of anarchy, is interesting to look at to see how big an effect

changes have on the cost of the network. We can see when ǫ = 0 that the

price of anarchy for the Braess network we considered in the first section is

.

Roughgarden [8] showed that if the cost functions of the routes have the form (a x flow)+b then the price of anarchy is ≤

and for large random networks the price of anarchy approaches

with probability 1 as the number of links in the network tends to infinity, provided networks loads are suitably chosen.

We look at how adding in traffic flowing from node 1 to node 2 affects the price of anarchy. We look at cost at k = 10 and cost at k = 23 −

. This gives the price of anarchy as

92 −

83 −

which is equal to

8 (−299 + 10ǫ)

−2158 + 17ǫ . We plot this graph in figure 9.

Price of anarchy

ǫ

92 83

26 7 4

3

1

Figure 9: How the price of anarchy changes for increasing amount of cars travel- ling from node 1 to node 2.

We must have that 23 −

ǫ ≥ 10 which gives us that ǫ ≤

, otherwise the

ratio is 1. This shows us that by having cars travelling from node 1 to node

2 the price of anarchy falls, so the improvement in travel costs by increasing k

decreases. Therefore the benefit of closing the link in Braess’s network will be

reduced if there are cars that travel from node 1 to node 2.

3 Responsive Control In A Simple Network

We will consider a simple, symmetrical network shown in Figure 10. Two routes of equal uncongested travel time and distance join a single Origin-Destination pair. There are two stages at the signal; the first gives green to route 1, the second gives green to route 2. We suppose the cycle time at the junction is fixed and so the total of the two green times will be fixed. At nodes 1 and 2 there are bottlenecks. We consider the characteristics of the two routes to be identical.

Origin

Destination Node A

Node B

Signal Link 11

Link 21

Link 12

Link 22 Route 1

Route 2

Figure 10: A simple signal-controlled network. Route 1 has saturation flow = k vehicles per minute at node A and saturation flow k vehicles per minute at the signal. Route 2 has saturation flow=k vehicles per minute at node B and saturation flow s vehicles per minute at the signal.

3.1 Terminology

We will use the following terminology:

T = total flow from the Origin to Destination (vehicles per minute) T X

= flow on route 1 (vehicles per minute)

T X 2 = flow on route 2 (vehicles per minute) X

= proportion of flow using route 1

X

= proportion of flow using route 2

G

= proportion of time route 1 is given green G

= proportion of time route 2 is given green

k = saturation flow at nodes A and B via either route (vehicles per minute) s = saturation flow at the signal approached via either route (vehicles per minute)

d

= delay at exit of link 11 (minutes per vehicle)

d

= delay at exit of link 21 (minutes per vehicle)

d

= delay at exit of link 12 (minutes per vehicle) d

= delay at exit of link 22 (minutes per vehicle)

K

= uncongested time to traverse route 1 ignoring bottleneck delays (minutes per vehicle)

K

= uncongested time to traverse route 2 ignoring bottleneck delays (minutes per vehicle)

C

= K

+ AX

+ d

+ d

= average travel time on route 1 (minutes per vehicle)

C

= K

+ AX

+ d

+ d

= average travel time on route 2 (minutes per vehicle).

We are assuming that the routes have the same uncongested travel time so that K

= K

= K.

The vector X = (X

, X

) gives the flow split between route 1 and route 2;

always X

+ X

= 1, X

≥ 0 and X

≥ 0.

Also, we suppose that the stage green-time proportions satisfy G

+ G

= 1, G

≥ 0 and G

≥ 0, so there are no positive minimum green-times.

3.2 Demand Feasibility, Supply Feasibility and Feasibility

The assumed conditions on the vector X and the stage green times lead to the following definitions. The set D of demand feasible (X, G) is given by

D = {(X, G) : X

+ X

= 1, X

≥ 0, X

≥ 0, G

+ G

= 1, G

≥ 0, G

≥ 0} . The set S of supply feasible (X, G) is given by

S = {(X, G) : sG

> T X

, sG

> T X

} .

From this we immediately have the set E of feasible (X, G) given by E = D ∩ S.

3.3 Two Delay Formulae

We assume that there are two delay formulae δ

and δ

giving all four delays d

, d

, d

and d

. We define them using δ

as the following:

d

= d

= 0

d

= δ

(T X

, sG

) = BT X

[sG

(sG

− T X

)]

d

= δ

(T X

, sG

) = BT X

[sG

(sG

− T X

)] .

Using δ

the delays are given by the following:

d

= d

= 0

d

= δ

(T X

, sG

) = B (sG

− T X

) d

= δ

(T X

, sG

) = B

(sG

− T X

) .

The travel costs along routes 1 and 2 given by C

and C

respectively are defined as

C

= K + AT X

+ d

and

C

= K + AT X

+ d

.

A, B, k and s are all constants. δ

is identical to the second term of Webster’s delay formula (1958) if we choose B to be

.

d 12

sG 1

T X 1

Figure 11: Example graph of how d

may change as ’spare capacity’ decreases, approaching infinity as T X

approaches sG

.

For both δ

and δ

, as the ’spare capacity’ (the difference between sG

and T X

, i.e. the maximum flow possible and the actual flow) decreases towards zero, the delays increase towards infinity. This is a natural requirement and explains the supply feasible set S (see Figure 11).

The delays defined by δ

and δ

are defined throughout the feasible set E.

We define an equilibrium flow as a feasible vector (X

, X

) and a feasible T such that

C

= C

.

The definitions of the costs show that this involves the delay functions and thus

depends on X, G and T .

4 Consistent Equilibrium Using The Equisatu- ration Policy

4.1 Equilibrium Using Delay Formulae δ 1

If we now assume that the delay formulae are in the form of δ

then we have C

= K + AT X

+ d

= AT X

+ BT X

[sG

(sG

− T X

)]

and

C

= K + AT X

+ d

= AT X

+ BT X

[sG

(sG

− T X

)]

as d

= d

= 0.

In the equisaturation policy, for any feasible X and T , green times are chosen so that

T X

sG

= T X

sG

thus X

= G

and X

= G

which gives C

− C

= AT (X

− X

) +

 BT

s (s − T )

  1 X

− 1

X

 .

To find the equilibria of the system we solve the equation C

− C

= 0.

Obviously X

= X

=

is a solution for all T . To find the others we multiply C

− C

= 0 by X

X

and obtain

(C

− C

) X

X

= AT (X

− X

) X

X

−

 BT

s (s − T )



[X

− X

] = 0.

Now we have already obtained the solution of X

−X

= 0 so here we assume that this is not true and divide through by X

−X

and then substitute in X

= 1−X

to yield

X

− X

+ B

[As (s − T )] = 0 which gives us two solutions

X

= 1 2 +  1

4 − B

[As (s − T )]



and

X

= 1 2 −  1

4 − B

[As (s − T )]



.

These are two real roots of the equation C

−C

= 0 if and only if

−

≥ 0

which is the same as saying if and only if T ≤ s −

.

4.2 Stability of Equilibria Using Delay Formulae δ 1

We have three equilibria of this system, namely X

= 1

2 X

= 1

2 −  1

4 − B

[As (s − T )]



X

= 1 2 +  1

4 − B

[As (s − T )]



and we wish to find the ranges of T in which they are stable. To do this we consider the function C

− C

as a function of X

only which we can do as X

= 1 − X

, so let

f (X

) = C

− C

= AT (2X

− 1) +

 BT

s (s − T )

  1

X

− 1 1 − X

 .

To analyse the stability of the equilibria we look at the derivative of this equation, which is

f

(X

) = 2AT −

 BT

s (s − T )

  1

X

+ 1 (1 − X

)

 . When X

=

we have

f

 1 2



= 2AT − 8BT s (s − T ) .

Now an equilibrium is stable for values of T for which f

(X

) > 0. So we see that the equilibrium X

=

is stable when T < s −

.

Let us consider the equilibrium point X

=

+ h

1

4

−

i

. Now for simplicity let θ = h

1

4

−

i

and so X

=

+ θ

.

If we consider the function f (X

) (X

X

) we note that as X

X

≥ 0 the sign of the gradient of the function is the same as f (X

) (although with possibly different magnitude) and the roots of the two equations are the same. Therefore we can consider f (X

) (X

X

) and its derivative when analysing the stability of the equilibria.

F (X

) = f (X

) (X

X

) = AT (2X

− 1) X

− X

− BT

s (s − T ) (2X

− 1) and its derivative is thus

F

(X

) = AT 

2X

− 2X

+ (2X

− 1) (1 − 2X

)  − 2BT

s (s − T ) .

T X 1

s − ^4B As s x ^∗ ₁

x ^∗ 2 1 2

Figure 12: Bifurcation diagram to show pitchfork bifurcation of the equilibria of the system. Bifurcation point at T = s −

and x

=

+ h

1

4

−

i

, x

=

− h

1

4

−

i

.

So substituting in X

=

+ θ

we obtain F

(X

) =  1

2 − 2θ − 4θ



− 2B

As (s − T ) = −4θ

and this is always less than zero for the range in which the equilibrium point X

=

+ θ

is real. Similarly for the other point we get the same result.

So for the other two equilibria, we obtain that these are unstable for T < s−

which gives us the following two diagrams. Figure 12 is a bifurcation diagram, showing how the stability of the equilibria change as T increases. Figure 13 shows the set of equilibria (T X

, T X

) as T increases from zero to s (of course T cannot be greater than s).

We see that we have a pitchfork bifurcation at T = s −

where as T

increases we have three equilibria, two of which are unstable with the middle

equilibrium point stable but as T increases past the bifurcation point we only

have one equilibrium and this is unstable.

T X 2

s

T X 1

s s − ^4B As

Figure 13: The set of equilibria consistent with the equisaturation policy for some T using delay formulae δ

. The lines are all equilibria, including the axes, and the arrows indicate the natural directions of motion of non-equilibria as drivers and green times respond to delays and saturation ratios (they also indicate the stability of the equilibria).

4.3 Equilibrium Using Delay Formulae δ 2

Now suppose that the delay formulae are in the form of δ

. Therefore we have C

= K + AT X

+ d

= AT X

+ B

sG

− T X

and

C

= K + AT X

+ d

= AT X

+ B sG

− T X

.

As we are using the equisaturation policy, as before, we have X

= G

and X

= G

and so

C

− C

= AT (X

− X

) + B s − T

 1 X

− 1

X

 .

We note that X

= X

=

is an equilibrium. We attempt to find more equilibria by multiplying C

− C

= 0 through by X

X

to get

(C

− C

) X

X

= AT (X

− X

) X

X

− B

s − T (X

− X

) = 0.

Now as we have ruled out X

− X

= 0 as we have found this as an equilibrium already we may divide by X

− X

and set X

= 1 − X

to yield

X

− X

− B

AT (s − T ) = 0.

Solving this quadratic equation we obtain two solutions:

X

= 1 2 +  1

4 − B

AT (s − T )



and

X

= 1 2 −  1

4 − B

AT (s − T )



.

We see that these are two real equilibria of the system if and only if

−

≥ 0 which is equivalent to

T

− sT + 4B A ≤ 0

and if we solve this we have that the above two equilibria are real between

T = s 2 − 1

2



s

− 16B A



and

T = s 2 + 1

2



s

− 16B A



.

4.4 Stability of Equilibria Using Delay Formulae δ 2

If we consider

f (X

) = C

− C

= AT (2X

− 1) + B s − T

 1

X

− 1 1 − X



then we take its derivative

f

(X

) = 2AT − B s − T

 1

X

+ 1 (1 − X

)

 .

We evaluate the derivative at each of the equilibria. If we have an equilibrium point X

then this is stable for all values of T that f

(X

) > 0. By applying this rule to each equilibrium point we obtain for the point X

=

f

 1 2



= 2AT − 8B

(s − T )

T X 1

1 2

T 1 T 2

Figure 14: Bifurcation diagram to show the changing stability of the equilibria of the system. Bifurcation points at T

=

−

s

−



and T

=

+

1

2

s

−



.

and so the equilibrium is unstable whenever this is greater than zero. This gives us that X

=

is stable for all values of T such that

s 2 − 1

2



s

− 16B A



< T < s 2 + 1

2



s

− 16B A



and unstable at all other points. The other two equilibria points that were found before are unstable for all values of T such that they are real solutions i.e. the values stated above.

Let us consider the equilibrium point X

=

+ h

1

4

−

i

. Now for simplicity let θ = h

1

4

−

i and so X

=

+ θ

.

If we consider the function f (X

) (X

X

) we note that as X

X

≥ 0 the sign of the gradient of the function is the same as f (X

) (although with possibly different magnitude) and the roots of the two equations are the same. Therefore we can consider f (X

) (X

X

) and its derivative when analysing the stability of the equilibria.

F (X

) = f (X

) (X

X

) = AT (2X

− 1) X

− X

− B

s − T (2X

− 1) and its derivative is thus

F

(X

) = AT 

2X

− 2X

+ (2X

− 1) (1 − 2X

)  − 2B

s − T .

T X 2

T X 1

s T 2

T 1

Figure 15: The set of equilibria consistent with the equisaturation policy for some T using delay formulae δ

. The lines are all equilibria, including the axes, and the arrows indicate the natural directions of motion of non-equilibria as drivers and green times respond to delays and saturation ratios (they also indicate the stability of the equilibria). The bifurcation points are also indicated; T

=

s

2

−

s

−



and T

=

+

s

−



.

So substituting in X

=

+ θ

we obtain F

(X

) = AT  1

2 − 2θ − 4θ



− 2BT (s − T ) which is less than zero when

−4θ < 0

and this is always less than zero for the range in which the equilibrium point X

=

+ θ

is real. Similarly for the other point we get the same result.

This information allows us to draw two informative graphs shown below.

Figure 14 is a bifurcation diagram which shows the changing stability of the

equilibria of the system, we see how it appears as two opposing pitchforks. Figure

15 shows the set of equilibria, (T X

, T X

) for a certain T from zero to s; all lines,

including the axes up to s, are equilbria.

4.5 Alternative Stability Analysis

Although we have considered the stability of the equilibria that we have found, it is beneficial to check the answers we achieved with another method. We will use a, maybe, more traditional way of checking stability by considering the Jacobian matrix of the system. This means we must write the system in terms of differential equations which is something we will use later in the project. We will show that this method achieves the same answers and thus back up the method we have already used where we considered the derivative of the equilibrium equation.

4.5.1 Analysis for Equilibria Using δ

Formulae

We have already found three equilbria of the formula C

− C

= 0 which are also equilibria with the same stability of the differential equation system

X ˙

= (C

− C

)(X

X

) = AT (X

− X

)X

X

+ BT

s (s − T ) (X

− X

) X ˙

= (C

− C

)(X

X

) = AT (X

− X

)X

X

+ BT

s (s − T ) (X

− X

) which has Jacobian matrix

A =

"

1

∂X₁

∂ ˙X₁

∂X₂

∂ ˙X₂

∂X₁

∂ ˙X₂

∂X₂

#

=





AT (X

− 2X

X

) +

AT (2X

X

− X

) −

AT (2X

X

− X

) −

AT (X

− 2X

X

) +





To evaluate the stability of the equilibria we look at the Jacobian matrix evaluated at the equilbria points and consider the trace and determinant of the matrix to decide stability. Let us consider a Jacobian matrix

B = a b c d



then the characteristic equation is det (B − λI) = 0, which thus yields λ

− τλ + ∆ = 0

where

τ = trace(A) = a + d

∆ = det(A) = ad − bc.

So the eigenvalues only depend on the trace and determinant of the matrix B. If

τ < 0 then the equilibria are stable [6].

If we use this information for our system we first evaluate the equilibrium (X

, X

) =

,

:

A  1 2 , 1

2



=

"

−

+

−

AT

4

−

−

+

#

which has trace, τ = −

+

and this is less than zero when T < s −

. Now let us consider the equilibrium 

1

2

+ θ

,

− θ



, where θ =

−

. We have that the trace of the matrix A evaluated at the equilibrium point is

τ = (2 − θ) AT + 2BT s (s − T )

and this is less than zero when T > s −

but the equilibrium point is only real when T < s −

so this equilibrium point is always unstable. The equilibrium point 

1

2

− θ

,

+ θ



satisfies the same conditions so is always unstable while real.

We note that these equilibria are also equilibria of the extended system of the following differential equations

X ˙

= (C

− C

)(X

X

) = AT (X

− X

)X

X

+ BT

s (s − T ) (X

− X

) X ˙

= (C

− C

)(X

X

) = AT (X

− X

)X

X

+ BT

s (s − T ) (X

− X

) G ˙

= T X

sG

− T X

sG

G ˙

= T X

sG

− T X

sG

.

4.5.2 Analysis for Equilibria Using δ

Formulae

Using the same method we analyse the system C

− C

= 0 defined using the δ

delay formulae. We analyse the differential equations system X ˙

= (C

− C

)(X

X

) = AT (X

− X

)X

X

+ B

(s − T ) (X

− X

) X ˙

= (C

− C

)(X

X

) = AT (X

− X

)X

X

+ B

(s − T ) (X

− X

) which has Jacobian matrix

A =





AT (X

− 2X

X

) +

AT (2X

X

− X

) −

AT (2X

X

− X

) −

AT (X

− 2X

X

) +





Let us consider the equilibrium point

,

, if we evaluate the Jacobian matrix A at this point then we get the trace of A as τ = −

+

and this is less than zero when

s 2 − 1

2



s

− 16B A



< T < s 2 + 1

2



s

− 16B A



. (1)

Now if we look at the equilibrium point 

1

2

− θ

,

+ θ



, where in this case, θ =

−

. The matrix A evaluated at this point has trace

τ = AT X

+ X

− 4X

X

+ 2B (s − T ) , which, if we note that X

+ X

= 1 − 2X

X

, is

τ = AT



− 1 2 + 6θ



+ 2B

(s − T ) and this is less than zero when

4θ < 0

which is at no point when the equilibrium point is real so the point



2

− θ

,

+ θ



is always unstable when real, i.e. in the interval (1).

The other equilibrium point found has the same conditions, shown in the same way.

4.6 Remark

By using the above method we have shown that we get the same answers for the

ranges of stability and instability of the equilibria as before when we considered

the derivative of the equilibrium equation C

−C

= 0. This gives us two methods

now for finding the range of stability of equilibria that we find.

5 Effect on Network Capacity

5.1 Network Capacity Using Equisaturation Policy

We now consider the network shown in Figure 16 with same saturation flow at both nodes but different saturation flows at the signal from each route. We consider when s

> s

so that the upper route is considered as wider.

Origin

Destination Node A

Node B

Signal Link 11

Link 21

Link 12

Link 22 Route 1

Route 2

Figure 16: A network with a wide long upper route and shorter narrower lower route. Route 1 has saturation flow = 8s

vehicles per minute at node A and saturation flow s

vehicles per minute at the signal. Route 2 has saturation flow=α vehicles per minute at node B and saturation flow s

vehicles per minute at the signal.

We now introduce two delay formulae for the bottlenecks;

d

= δ

(T X

, α) = BT X

α (α − T X

) d

= δ

(T X

, α) = BT X

α (α − T X

) and

d

= δ

(T X

, α) = B α − T X

d

= δ

(T X

, α) = B

α − T X

.

If the equisaturation policy is used to determine fixed time signal timings then we have that X

= X

is the stable equilibrium (see next section). From this we have that as, by the policy,

T X

s

G

= T X

s

G

then s

G

= s

G

. We also have G

= 1 − G

so s

− s

G

= s

G

and thus s

s

+ s

= G

. Similarly s

− s

G

= s

G

, giving

s

s

+ s

= G

. We find the upper bound for feasible T ;

s

G

+ s

G

= s

s

s

+ s

+ s

s

s

+ s

= 2s

s

s

+ s

and we note that

1+s₂

< 1 so we have the upper bound for T as 2s

s

s

+ s

< 2s

.

We note at the possible loss of network capacity as no matter the size of s

the maximum network capacity is less than 2s

.

5.2 Network Capacity Using the P 0 Policy

If we implement the policy P

(Smith, 1980) then the green time vector G = (G

, G

) for any feasible flow vector X = (X

, X

) is chosen so that (X, G) is feasible and

s

d

= s

d

where the delays are the delays for each route at the signal.

If we allow for drivers to choose the cheapest route then obviously all drivers will choose the route with smallest delay which is the route with the highest saturation flow. Thus, in our case where route 1 is the widest and thus has least delays, all drivers will be on route 1. The network capacity of this route can not exceed the minimum saturation flow of the bottleneck at node 1 and the signal junction, as otherwise delay times would approach infinity at one of these points. Thus the network capacity of the network when the P

policy is implemented is min{max {s

, s

} , α}. Figure 17 shows the differences in network capacity between the two policies.

5.3 Modelling as a Dynamical System and Finding a Lya- punov Function

Let us consider the delay formula δ

and the P

policy and analyse the stability

of the equilibrium (X

, X

, G

, G

) that it produces.