Theoretical Analysis of Cooperative Driving at Idealized Unsignalized Intersections

The previously presented analysis indicates that we can have different cooperative driving plans that “serve” the same inflow rates $q_1$ and $q_2$ but the resulting delays for vehicles in different flows can vary significantly. Let us consider an extreme example: we have $q_1\gg 1$ vehicles of the west-to-east flow and $q_2=1$ vehicles of the south-to-north flow. Cooperative driving vehicle plan A first dispatches $q_1$ vehicles of the west-to-east flow to pass the core area and then dispatches only one vehicle of the south-to-north flow (i.e., we set $k_1=q_1$ and $k_2=1$ ). Cooperative driving vehicle plan B dispatches the first $q_2/2$ vehicles of the west-to-east flow to pass the core area, then dispatches only one vehicle of the south-to-north flow, and finally dispatches the first $q_1/2$ vehicles of the west-to-east flow (i.e., we set $k_1=q_1$ and $k_2=1$ ). We determine that the maximum delays of the south-to-north flow obtained by the two plans are quite different.

To retain the equity of different travelers, we need to distribute the benefits and costs to travelers fairly and appropriately^{[ 49 , 50 , 51 , 52 ]}. Thus, we are interested in the following problem:

Problem 3 Suppose the upstream inflow rates $q_1$ and $q_2$ can be served by a certain number of cooperative driving plans. Determine the plan that provides the lowest maximum delay for vehicles.

Suppose a cooperative driving plan that is characterized by a 9-tuple (i.e., $k_1$ , $k_2$ , $\text{Δ}{t}_1$ , $\text{Δ}{t}_2$ , $\text{Δ}{t}_3$ , $\text{Δ}{T}_1$ , $\text{Δ}{T}_2$ , $m_1$ , and $m_2$ ) can “serve” the upstream inflow rates $q_1$ and $q_2$ . We assume uniform arrival of vehicles. Without any delay, the $i$ -th vehicle of the west-to-east flow should arrive at the core area at time

(17) $$t_{\text{ideal}}^i=\frac{3600}{q_1}i$$

When the cooperative driving plan is applied, the $i$ -th vehicle of the west-to-east flow should arrive at the core area at time

(18) $$t_{\text{real}}^i=\lfloor {\displaystyle \frac{i-1}{k_1}}\rfloor \left[\left(k_1-1\right)\text{Δ}{t}_1+k_1{\displaystyle \frac{L}{v}}+2\text{Δ}{t}_3+\text{Δ}{T}_1\right]+$$ $${\left[\left(imod{k}_1\right)-2\right]}^{+}\text{Δ}{t}_1+\left[\left(imod{k}_1\right)-1\right]{\displaystyle \frac{L}{v}}$$

where function $(MmodN)$ yields the remainder of $M$ divided by $N$ ; function ${\left[x\right]}^{+}=\max \{0,x\}$ .

The delay of the $i$ -th vehicle of the west-to-east flow can then be written as follows:

(19) $$t_{\text{delay}}^i={\left[t_{\text{real}}^i-t_{\text{ideal}}^i\right]}^{+}$$

Similarly, we determine the delay of the $j$ -th vehicle of the south-to-north flow and formulate our decision problem as follows:

(20) $$\underset{k_1,k_2\in N}{\min }\underset{i\in \{1,2,\mathrm{\dots },q_1\},j\in \{1,2,\mathrm{\dots },q_2\}}{\max }\{t^i,t^j\}$$ $$\text{s.t.,}\mathit{\hspace{1em}}\text{Formulas (1)–(6) and Eqs. (17)–(19)}$$

The programming problem expressed in Formula ( 20 ) is difficult to solve because of the complicated round-off operations of integer decision variables. However, we can derive a simple upper bound for the maximum delay.

Theorem 5 The upper bound for the maximum delay of any vehicle is derived as follows:

(21) $$\text{Δ}{T}_1+\text{Δ}{T}_2$$

If we can further apply an asymmetric cooperative driving plan (i.e., $k_1$ , $k_2$ , $\text{Δ}{t}_1=\text{Δ}{t}_2=\text{Δ}{t}_3=0$ , $\text{Δ}{T}_1$ , $\text{Δ}{T}_2$ , $m_1$ , and $m_2$ ), then we should select the minimum $\text{Δ}{T}_1+\text{Δ}{T}_2$ .

Proof Based on Formula ( 14 ), we derive $m_1{k}_1=q_1$ for the west-to-east flow. Thus, the $i$ -th vehicle of the west-to-east flow should arrive at the core area at time

(22) $$t_{\text{ideal}}^i=\frac{3600}{q_1}i=\frac{3600}{m_1{k}_1}i$$

Considering the periodic feature of the cooperative driving plans, we roughly derive the following expression:

(23) $$m_1\{k_1\frac{L}{v}+\text{Δ}{T}_1\}=3600$$

By combining Eqs. ( 22 ) and ( 23 ), we derive the following expression:

(24) $$t_{\text{ideal}}^i=\frac{\frac{L}{v}+\text{Δ}{T}_1}{k_1}i$$

Further considering Eq. ( 18 ) and Formula ( 1 ), we derive the following expression:

(25) $$t_{\text{delay}}^i={\left[t_{\text{real}}^i-t_{\text{ideal}}^i\right]}^{+}⩽\text{Δ}{T}_1+\text{Δ}{T}_2$$

Such a condition also holds for the delay $t_{\text{delay}}^j$ of the $j$ -th vehicle of the south-to-north flow. This concludes the proof.

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Further considering Formula ( 11 ), we derive the following expression:

(26) $$(k_1+k_2)\frac{L}{v}⩽\text{Δ}{T}_1+\text{Δ}{T}_2$$

Based on Eq. ( 22 ) and Formula ( 25 ), we select the smallest possible $k_1$ and $k_2$ that “serve” the inflow rates $q_1$ and $q_2$ to reduce the lowest maximum delay of every individual vehicle. This indicates that when the condition in Formula ( 7 ) is satisfied, the rhythmic control method leads to the lowest maximum delay and is suggested. However, when the condition in Formula ( 7 ) is not satisfied, but the condition in Formula ( 11 ) is satisfied, we should select the following cooperative driving plan.

Because we aim to keep $k_1$ and $k_2$ as small as possible, given $q_1$ and $q_2$ , we can identify the ideal asymmetric cooperative driving plan using Algorithm 1 assuming $\text{Δ}{t}_1=\text{Δ}{t}_2=\text{Δ}{t}_3=0$ .

Theorem 4 guarantees that we can finally obtain a satisfactory solution if Formula ( 12 ) holds. Let us use a simple test case to illustrate the algorithm. Suppose $L/v=1$  s, $q_1=2100$  veh/h, and $q_2=1000$  veh/h. The rhythmic control method cannot “serve” such inflow rates. When $k_2=1$ , we obtain $\text{Δ}{T}_1=1$  s, $m_2=1000$ , $\text{Δ}{T}_2\approx 2.6$  s, $k_1=2$ , and $m_1=1001$ . At that point, we derive $m_1{k}_1<q_1$ . Thus, we increment $k-2$ to $k_2=2$ . We obtain $\text{Δ}{T}_1=2$  s, $m_2=500$ , $\text{Δ}{T}_2\approx 5.19$  s, $k_1=5$ , $m_1=501$ , and $m_1{k}_1>q_1$ . This cooperative driving plan can “serve” such inflow rates. Indeed, such a cooperative driving plan (i.e., $k_1=5$ , $k_2=2$ , $\text{Δ}{t}_1=\text{Δ}{t}_2=\text{Δ}{t}_3=0$ , $\text{Δ}{T}_1=5.19$ , $\text{Δ}{T}_2=1$ , $m_1=501$ , and $m_2=500$ ) can at most “serve” the upstream inflow rates $q_1=2500$  veh/h and $q_2=1000$ veh/h. If this plan is applied to the upstream inflow rates $q_1=2100$  veh/h and $q_2=1000$  veh/h, then some positions in the groups of the west-to-east flow (or the equivalent planned time slots^{[ 31 , 33 , 35 , 36 ]}) will be left unoccupied.