Signal design for an isolated intersection during congestion

Abstract

This paper presents an algorithm for the signal design of an isolated intersection that will be able to alleviate long queues during severe congestion conditions, which cause both lengthy delays and harsh environmental damage. During such periods, the total effective green light is a scarce resource; its best allocation is crucial for the smooth operation of the intersection and sometimes even for a large network. The aim of the procedure presented here is to maximize the average throughput of the intersection. By achieving this goal, the number of vehicles in the queue is reduced at the fastest possible rate, and the period of congestion is shortened. The maximum throughput is achieved when the marginal flow in each phase is equal to the average throughput. The algorithm developed takes into account the decreasing flow rates of existing queues during long periods of green.

Appendix: Validation of the basic algorithm

Definitions:

γ _m (t), m=1, … M are continuous positive functions.

Ψ(•) is the same function as defined in Equation (2) above, namely: when ḡ⁰=(g₁, g₂, …, g _M )^T is a vector all whose components are positive; and for each of them, the function γ _m (t) is non-increasing for t⩾g _m .

For convenience, denote

Thus

Proposition 1

• Let γ _m (t) be non-increasing in the interval [a, a+b]; then the following inequalities hold:

  

Proof

• Trivial.

Proposition 2

• Let γ _m (t) be non-increasing in the interval [a, a+b]; then the following inequalities hold:

  

Proof

• Perform some algebraic manipulations to obtain:

  

  This inequality is correct because, according to proposition 1, the following inequalities hold:

  

  and

  

  The quotient of these two expressions gives Equation 12 above.

Proposition 3

• Let γ₁(t) and γ₂(t) be two continuous positive non-increasing functions, and let ḡ⁰=(g₁, g₂, …, g _M )^T be a vector for which γ₁(g₁)>γ₁(g₂)>Ψ(ḡ⁰). Then, for any ɛ>0, a 0<δ⩽ɛ exists that satisfies the inequalities Ψ(ḡ¹)>Ψ(ḡ²)>Ψ(ḡ⁰), where ḡ¹=(g₁+δ, g₂, …, g _M )^T and ḡ²=(g₁, g₂+δ, …, g _M )^T.

Proof

• The continuation of functions γ₁(t), γ₂(t), and Ψ(•) leads to the following results:

  • a) For γ₁(g₁)>γ₂(g₂), there is a δ₁⩽ɛ that satisfies γ₁(g₁+δ₁)>γ₂(g₂).

  • b) For γ₂(g₂)>Ψ(ḡ⁰), there is a δ₂⩽ɛ that satisfies γ₂(g₂+δ₂)>Ψ(ḡ⁰).

  Define δ=Min(δ₁, δ₂).

  Following Equation (9), we obtain the following:

  

  and

  

  After some algebraic manipulations, this obtains

  

  According to inequality (10) and result (a) above, the following inequalities hold:

  
  

  And according to proposition 1 and result (b) above, the following inequality holds:

  

  Completing the required proof. □

Proposition 4

• Let ḡ⁰ and ḡ¹ be two vectors: ḡ⁰= (g₁, g₂, …, g _M )^T and ḡ¹=(g₁, Δt, g₂, …, g _M )^T (Δt>0).

  If Ψ(ḡ⁰)⩾Ψ(ḡ¹) is true, then Ψ(ḡ⁰)⩾Ψ(ḡ²) is also true for any ḡ²=(g₁, Δt+ɛ, g₂, …, g _M )^T, where ɛ>0.

Proof

• Assign the value of Ψ(ḡ⁰) to obtain

  

  After algebraic manipulations, we obtain

  

  The denominator is positive, thus the numerator is non-negative. Furthermore, according to proposition 1

  

  If we now check the inequality that has to be proved:

  

  This expression is non-negative because the denominator is positive; the first addends in the numerator are non-negative as shown above.

Proposition 5

• Let ḡ⁰ and ḡ¹ be two vectors: ḡ⁰= (g₁, g₂, …, g _M )^T and ḡ¹=(g₁+Δt, g₂, …, g _M )^T (Δt>0).

  If Ψ(ḡ⁰)⩽Ψ(ḡ¹)⩽γ₁(g₁+Δt) is true, then Ψ(ḡ²)⩽Ψ(ḡ¹) is also true for any ḡ²=(g₁+αΔt, g₂, …, g _M )^T, where 0<α<1.

Proof

• The expressions for Ψ(ḡ¹) and Ψ(ḡ²) can be written as follows:

  
  

  ‘A’ can be isolated in Equation 24 and assigned in Equation 25 to yield

  

  From inequality (10) and the given condition, the following is obtained: and γ₁(g₁+Δt)⩾Ψ(ḡ¹), respectively.

  These inequalities are assigned in Equation (26) to produce

  

  After algebraic manipulation, the required result is proved:

  

Proposition 6

• Let ḡ⁰=(g₁, g₂, …, g _M )^T be the best point that can be reached when using the Greedy search algorithm. Let ḡ¹=(g₁+ν₁, g₂, …, g _M )^T and ḡ²=(g₁, g₂−ν₂, …, g _M )^T (ν _i <g _i , i=1, 2).

  Then:

  • a) Ψ(ḡ⁰)>Ψ(ḡ¹);

  • b) Ψ(ḡ⁰)>Ψ(ḡ²);

  • c) Ψ(ḡ⁰)>Ψ(ḡ⁴), where ḡ⁴=(g₁+ν₁, g₂−ν₂, …, g _M )^T.

Proof

• Results (a) and (b) are direct conclusions from proposition 3 and from propositions 4 and 5, respectively.

  As for result (c), the following should be proved:

  

  Obviously, the denominator is positive. The numerator is non-negative, since it can be shown from results (a) and (b) that: