Consider a unidirectional highway road scenario, as shown in Figure 1. The vehicles are moving on the highway road. One entry/exit is distributed on the highway road. One RSU is deployed on the position of the entry/exit. The communication radius of vehicles is denoted as R v and the communication radius of RSU is denoted as R u , here assumed that R u ≥ R v . The highway road interval is denoted as [ 0 , L ] , where L represents the length of the highway road and

L ≥ 2 R u . The entry/exit is denoted as z. Assume that z follows uniform distribution on road interval [ R u , L − R u ] . The entry/exit divides the highway road interval into two sub-intervals which are represented as [ 0 , z ] and [ z , L ] . The vehicle arrival process on [ 0 , z ] is Poisson process with mean rate λ 1 . When the vehicles on [ 0 , z ] arriving at the position of the entry/exit, the vehicles can keep moving onto the next road interval [ z , L ] with probability α ( 0 ≤ α ≤ 1 ) , or depart from the highway road at the entry/exit with probability 1 − α . The vehicles moving on the road interval [ z , L ] contains two parts: one is the vehicles from the road interval [ 0 , z ] and the other is the new arriving vehicles from the entry/exit. The new arriving vehicles on the road interval [ z , L ] are also Poisson process with mean rate λ 2 . The vehicle speed is denoted as random variant f v which follows uniform distribution on interval [ v min , v max ] , where v min and v max respectively represents the minimum and the maximum value of the allowed moving speed for vehicles. The average value of the vehicle speed is expressed by v = E [ f v ] , where E [ ⋅ ] represents mathematical expectation. According to the property of Poisson process, the vehicle arrival rates of [ 0 , z ] and [ z , L ] can be expressed by γ 1 and γ 2 , where γ 1 = λ 1 / v and γ 2 = ( α ⋅ λ 1 + λ 2 ) / v . Therefore, the network connectivity in this paper is building an analytical model to calculate the connectivity probability of the vehicles in the network.

Similar to Ref. [2], consider the two-way connectivity which requires that the vehicle and RSU can communicate with each other only and only if the distance between them is not larger than the vehicle communication radius. For ease of analysis, as shown in Figure 2, divide the interval [ z − R u , z + R u ] into four sub-intervals which are denoted as a 1 , a 2 , a 3 and a 4 , respectively. And the notations which are used after are listed as follows:

H a : the event that there are vehicles on road segment a.

H ¯ a : the event that there is no vehicle on road segment a.

C a : the event that road segment a is connected.

C ¯ a : the event that road segment a is disconnected.

Pr { V } : the probability that event V occurs.

From Figure 2, it is obvious that the length of a 1 is not more than the communication radius of vehicle. If there are vehicles in a 1 , the vehicles in a 1 is

connected. The connectivity of a 2 , a 3 and a 4 is similar to a 1 . Due to the influence of the entry/exit to the network connectivity, the connectivity of a 2 and a 3 is needed to be firstly considered. According to the case that whether there are vehicles on a 2 and a 3 , we perform the analysis into following four sub-cases: 1) There are vehicles both on a 2 and a 3 ; 2) There are vehicles on a 2 , but there is no vehicle on a 3 ; 3) There are vehicles on a 3 , but there is no vehicle on a 2 ; 4) There is no vehicle on a 2 and a 3 .

The probabilities that there are vehicles on a 2 and a 3 are respectively given by

Pr { H a 2 } = 1 − e − γ 1 ⋅ R v , Pr { H a 3 } = 1 − e − γ 2 ⋅ R v .

The probabilities that there is no vehicle on a 2 and a 3 are respectively calculated by

Pr { H ¯ a 2 } = 1 − Pr { H a 2 } = e − γ 1 ⋅ R v , Pr { H ¯ a 3 } = 1 − Pr { H a 3 } = e − γ 2 ⋅ R v

1) There are vehicles both on a 2 and a 3 .

It is seen from Figure 2, if there are vehicles on a 2 , a 2 is connected for the case that the length of a 2 is not larger than the communication radius of vehicle and the distance between any two vehicles on a 2 is not larger than the communication radius of vehicle. The connectivity of a 3 is similar to that of a 2 . If there are vehicles both on a 2 and a 3 , the vehicles separately from a 2 and a 3 can two-way connect to each other by the RSU. For further analysis of the network connectivity, we divide a into four sub-cases.

a) There are vehicles both on a 1 and a 4 .

In this case, we can calculate the probabilities that there are vehicles both on a 1 and a 4 as follows

Pr { H a 1 } = 1 − e − γ 1 ⋅ ( R u − R v ) , Pr { H a 4 } = 1 − e − γ 2 ⋅ ( R u − R v ) .

Then we analyze the connectivity of sub-interval [ 0 , z ] and [ z , L ] .

When there are vehicles both on [ 0 , z − R u ] and [ z + R u , L ] , if the vehicles on sub-interval [ 0 , z ] are connected and the vehicles on sub-interval [ z , L ] are connected, the road interval [ 0 , L ] is connected. The probabilities that there are vehicles both on [ 0 , z − R u ] and [ z + R u , L ] are calculated as follows

Pr { H [ 0 , z − R u ] } = 1 − e − γ 1 ⋅ ( z − R u ) , Pr { H [ z + R u , L ] } = 1 − e − γ 2 ⋅ ( L − z − R u ) .

In this case, there is no entry/exit in sub-interval [ 0 , z ] and sub-interval [ z , L ] . The connectivity probabilities of sub-interval [ 0 , z ] and sub-interval [ z , L ] are obtained according to Ref. [8]. In Ref. [8], for a highway road without entries and exits, if vehicle arrivals follow a Poisson process, the connectivity probability of the road is given by

Pr { λ v , R v , L ′ } = e − λ v ⋅ L ′ ∑ j = 0 ⌊ L ′ / R v ⌋ ( − 1 ) j j ! [ λ v ( L ′ − j R v ) ] j − 1     × [ j + λ v ( L ′ − j R v ) ] e λ v ⋅ ( L ′ − j R v )

where L ′ is the road length, R v is a vehicle’s transmission radius, λ v is the vehicle arrival rate, and ⌊ x ⌋ is the largest integer that is not larger than x. Thus, the connectivity probabilities of sub-interval [ 0 , z ] and sub-interval [ z , L ] are

Pr { C [ 0 , z ] } = Pr { γ 1 , R v , z } , Pr { C [ z , L ] } = Pr { γ 2 , R v , L − z } .

The connectivity probability of road interval [ 0 , L ] is

p 1.1.1 = Pr { H [ 0 , z − R u ] } ⋅ Pr { H [ z + R u , L ] } ⋅ Pr { C [ 0 , z ] } ⋅ Pr { C [ z , L ] } . (1)

When there are vehicles on [ 0 , z − R u ] , but there is no vehicle on [ z + R u , L ] , if the vehicles on sub-interval [ 0 , z ] are connected and the vehicles on sub-interval [ z , z + R u ] are connected, the road interval [ 0 , L ] is connected. The probability that there is no vehicle on [ z + R u , L ] is calculated as follows

Pr { H ¯ [ z + R u , L ] } = 1 − Pr { H [ z + R u , L ] } = e − γ 2 ⋅ ( L − z − R u ) .

The connectivity probability for the vehicles on [ z , z + R u ] is

Pr { C [ z , z + R u ] } = Pr { γ 2 , R v , R u } .

In this case, the connectivity probability of road interval [ 0 , L ] is

p 1.1.2 = Pr { H [ 0 , z − R u ] } ⋅ Pr { H ¯ [ z + R u , L ] } ⋅ Pr { C [ 0 , z ] } ⋅ Pr { C [ z , z + R u ] } . (2)

When there are vehicles on [ z + R u , L ] , but there is no vehicle on [ 0 , z − R u ] , if the vehicles on sub-interval [ z − R u , z ] are connected and the vehicles on sub-interval [ z , L ] are connected, the road interval [ 0 , L ] is connected. The probability that there is no vehicle on [ 0 , z − R u ] is

Pr { H ¯ [ 0 , z − R u ] } = 1 − Pr { H [ 0 , z − R u ] } = e − γ 1 ⋅ ( z − R u ) .

The connectivity probability for the vehicles on [ z − R u , z ] is

[ z − R u , z ] Pr { C [ z − R u , z ] } = Pr { γ 1 , R v , R u } .

In this case, the connectivity probability of road interval [ 0 , L ] is

p 1.1.3 = Pr { H ¯ [ 0 , z − R u ] } ⋅ Pr { H [ z + R u , L ] } ⋅ Pr { C [ z − R u , z ] } ⋅ Pr { C [ z , L ] } . (3)

When there is no vehicle on [ 0 , z − R u ] and [ z + R u , L ] , if the vehicles on [ z − R u , z ] are connected and the vehicles on [ z , z + R u ] are connected, the road interval [ 0 , L ] is connected. In this case, the connectivity probability of road interval [ 0 , L ] is

p 1.1.4 = Pr { H ¯ [ 0 , z − R u ] } ⋅ Pr { H ¯ [ z + R u , L ] } ⋅ Pr { C [ z − R u , z ] } ⋅ Pr { C [ z , z + R u ] } . (4)

Thus, the connectivity probability of road interval [ 0 , L ] in a-1) is

p 1.1 = Pr { H a 1 } ⋅ Pr { H a 4 } ⋅ ( p 1.1.1 + p 1.1.2 + p 1.1.3 + p 1.1.4 ) . (5)

b) There are vehicles on a 1 , but there is no vehicle on a 4 .

In b), the probability that there is no vehicle on a 4 is

Pr { H ¯ a 4 } = 1 − Pr { H a 4 } = e − γ 2 ⋅ ( R u − R v ) .

When there are vehicles both on [ 0 , z − R u ] and [ z + R u , L ] , if the vehicles on sub-interval [ 0 , z ] are connected, the vehicles on sub-interval [ z + R u , L ] are connected, and the vehicles separately from a 3 and [ z + R u , L ] are connected, the road interval [ 0 , L ] is connected. Since there is no entry/exit in [ z + R u , L ] , the connectivity probability of [ z + R u , L ] is

Pr { C [ z + R u , L ] } = Pr { γ 2 , R v , L − z − R u } .

Next, we analyze the connectivity of vehicles separately from a 3 and [ z + R u , L ] . According to the definition of two-way connectivity and the assumption that R u ≥ R v . From Figure 2, we can easily obtain that in the case R u > 2 R v , the distance between two vehicles which separately from a 3 and [ z + R u , L ] is larger than the communication radius of vehicles. In this case, the vehicles which separately from a 3 and [ z + R u , L ] have no chance to connect to each other, the road interval [ 0 , L ] is disconnected. Otherwise, if R u ≤ 2 R v , the vehicles which separately from a 3 and [ z + R u , L ] have chance to connect to each other. Thus, in the following analysis, we further assume that R u ≤ 2 R v . When there are vehicles on [ z + R u , z + 2 R v ] , if the vehicles separately from a 3 and [ z + R u , z + 2 R v ] are connected, a 3 and [ z + R u , L ] is connected. Otherwise, when there is no vehicle on [ z + R u , z + 2 R v ] , the road interval [ 0 , L ] is disconnected. The probability that there are vehicles on [ z + R u , z + 2 R v ] is

Pr { H [ z + R u , z + 2 R v ] } = 1 − e − γ 2 ⋅ ( 2 R v − R u ) .

According to Ref. [1], we can obtain that the connectivity probability of vehicles separately from a 3 and [ z + R u , z + 2 R v ] is

Pr { C a 3 , [ z + R u , z + 2 R v ] } = 1 2 .

The connectivity probability of vehicles separately from a 3 and [ z + R u , L ] is

Pr { C { a 3 , [ z + R u , L ] } } = Pr { H [ z + R u , z + 2 R v ] } ⋅ Pr { C { a 3 , [ z + R u , z + 2 R v ] } } .

In this case, when R u > 2 R v , the connectivity probability of road interval [ 0 , L ] is 0; when R u ≤ 2 R v , the connectivity probability of road interval [ 0 , L ] is calculated as follows

p 1.2.1 = Pr { H [ 0 , z − R u ] } ⋅ Pr { H [ z + R u , L ] } ⋅ Pr { C [ 0 , z ] }     ⋅ Pr { C [ z + R u , L ] } ⋅ Pr { C { a 3 , [ z + R u , L ] } } . (6)

When there are vehicles on [ 0 , z − R u ] , but there is no vehicle on [ z + R u , L ] , if vehicles on [ 0 , z ] is connected, the road interval [ 0 , L ] is connected. In this case, the connectivity probability of road interval [ 0 , L ] is

p 1.2.2 = Pr { H [ 0 , z − R u ] } ⋅ Pr { H ¯ [ z + R u , L ] } ⋅ Pr { C [ 0 , z ] } . (7)

When there are vehicles on [ z + R u , L ] , but there is no vehicle on [ 0 , z − R u ] , if the vehicles on [ z − R u , z ] are connected, the vehicles on [ z + R u , L ] are connected, and the vehicles separately from a 3 and [ z + R u , L ] are connected, the road interval [ 0 , L ] is connected. In this case, the connectivity probability of road interval [ 0 , L ] is

p 1.2.3 = Pr { H ¯ [ 0 , z − R u ] } ⋅ Pr { H [ z + R u , L ] } ⋅ Pr { C [ z − R u , z ] }     ⋅ Pr { C [ z + R u , L ] } ⋅ Pr { C { a 3 , [ z + R u , L ] } } . (8)

When there is no vehicle both in [ 0 , z − R u ] and [ z + R u , L ] , if vehicles on [ z − R u , z ] are connected, the road interval [ 0 , L ] is connected. In this case, the connectivity probability of road interval [ 0 , L ] is

p 1.2.4 = Pr { H ¯ [ 0 , z − R u ] } ⋅ Pr { H ¯ [ z + R u , L ] } ⋅ Pr { C [ z − R u , z ] } . (9)

Thus, the connectivity probability of road interval [ 0 , L ] in a-2) is

p 1 .2 = Pr { H a 1 } ⋅ Pr { H ¯ a 4 } ⋅ ( p 1 .2 .1 + p 1 .2 .2 + p 1 .2 .3 + p 1 .2 .4 ) . (10)

c) There are vehicles on a 4 , but there is no vehicle on a 1 .

In c), the probability that there is no vehicle on a 1 is

Pr { H ¯ a 1 } = 1 − Pr { H a 1 } = e − γ 1 ⋅ ( R u − R v ) .

When there are vehicles both on [ 0 , z − R u ] and [ z + R u , L ] , if vehicles on [ z , L ] are connected, vehicles on [ 0 , z − R u ] are connected, and vehicles separately from a 2 and [ 0 , z − R u ] are connected, the road interval [ 0 , L ] is connected. The probability that there are vehicles on [ 0 , z − R u ] is

Pr { C [ 0 , z − R u ] } = Pr { γ 1 , R v , z − R u } .

The analysis for the connectivity probability of vehicles separately from a 2 and [ 0 , z − R u ] is similar to that of vehicles separately from a 3 and [ z + R u , L ] . Therefore, when R u > 2 R v , the connectivity of road interval [ 0 , L ] is 0; when R u ≤ 2 R v , the connectivity of road interval [ 0 , L ] is calculated as follows

p 1.3.1 = Pr { H [ 0 , z − R u ] } ⋅ Pr { H [ z + R u , L ] } ⋅ Pr { C [ z , L ] }     ⋅ Pr { C [ 0 , z − R u ] } ⋅ Pr { C { a 2 , [ 0 , z − R u ] } } (11)

where

Pr { C { a 2 , [ 0 , z − R u ] } = Pr { H [ z − 2 R v , z − R u ] } ⋅ Pr { C { a 2 , [ z − 2 R v , z − R u ] } } ,

Pr { H [ z − 2 R v , z − R u ] } = 1 − e − γ 1 ⋅ ( 2 R v − R u ) .

Pr { C { a 2 , [ z − 2 R v , z − R u ] } } = 1 2 .

When there are vehicles on [ 0 , z − R u ] , but there is no vehicle on [ z + R u , L ] , if vehicles on [ z , z + R u ] are connected, vehicles on [ 0 , z − R u ] are connected, and vehicles separately from a 2 and [ 0 , z − R u ] are connected, the road interval [ 0 , L ] is connected. In this case, the connectivity probability of road interval [ 0 , L ] is

p 1.3.2 = Pr { H [ 0 , z − R u ] } ⋅ Pr { H ¯ [ z + R u , L ] } ⋅ Pr { C [ z , z + R u ] }     ⋅ Pr { C [ 0 , z − R u ] } ⋅ Pr { C { a 2 , [ 0 , z − R u ] } } . (12)

When there are vehicles on [ z + R u , L ] , but there is no vehicle on [ 0 , z − R u ] , if vehicles on [ z , L ] are connected, the road interval [ 0 , L ] is connected. In this case, the connectivity probability of road interval [ 0 , L ] is

p 1.3.3 = Pr { H ¯ [ 0 , z − R u ] } ⋅ Pr { H [ z + R u , L ] } ⋅ Pr { C [ z , L ] } . (13)

When there is no vehicle both on [ 0 , z − R u ] and [ z + R u , L ] , if vehicles on [ z , z + R u ] are connected, the road interval [ 0 , L ] is connected. In this case, the connectivity probability of road interval [ 0 , L ] is

p 1.3.4 = Pr { H ¯ [ 0 , z − R u ] } ⋅ Pr { H ¯ [ z + R u , L ] } ⋅ Pr { C [ z , z + R u ] } . (14)

Thus, the connectivity probability of road interval [ 0 , L ] in a-3) is

p 1 .3 = Pr { H ¯ a 1 } ⋅ Pr { H a 4 } ⋅ ( p 1 .3 .1 + p 1 .3 .2 + p 1 .3 .3 + p 1 .3 .4 ) . (15)

d) There is no vehicle on a 1 and a 4 .

In d), for analyzing the connectivity of road interval [ 0 , L ] , we only need to consider the case that whether there are vehicles on [ 0 , z − R u ] and [ z + R u , L ] .

When there are vehicles both on [ 0 , z − R u ] and [ z + R u , L ] , if vehicles on [ 0 , z − R u ] are connected, vehicles on [ z + R u , L ] are connected, vehicles separately from a 2 and [ 0 , z − R u ] are connected, and vehicles separately from a 3 and [ z + R u , L ] are connected, the road interval [ 0 , L ] is connected. In this case, when R u > 2 R v , the connectivity probability of vehicles separately from a 2 and [ 0 , z − R u ] is 0, the connectivity probability of vehicles separately from a 3 and [ z + R u , L ] is 0, and the connectivity probability of road interval [ 0 , L ] is 0. When R u ≤ 2 R v , the connectivity probability of road interval [ 0 , L ] is

p 1 .4 .1 = Pr { H [ 0 , z − R u ] } ⋅ Pr { H [ z + R u , L ] } ⋅ Pr { C [ z + R u , L ] }                                                         ⋅ Pr { C [ 0 , z − R u ] } ⋅ Pr { C { a 2 , [ 0 , z − R u ] } } ⋅ Pr { C { a 3 , [ z + R u , L ] } } . (16)

When there are vehicles on [ 0 , z − R u ] , but there is no vehicle on [ z + R u , L ] , if vehicles on [ 0 , z − R u ] are connected, and vehicles separately from a 2 and [ 0 , z − R u ] are connected, the road interval [ 0 , L ] is connected. In this case, the connectivity probability of road interval [ 0 , L ] is

p 1.4.2 = Pr { H [ 0 , z − R u ] } ⋅ Pr { H ¯ [ z + R u , L ] } ⋅ Pr { C [ 0 , z − R u ] } ⋅ Pr { C { a 2 , [ 0 , z − R u ] } } . (17)

When there are vehicles on [ z + R u , L ] , but there is no vehicle on [ 0 , z − R u ] , if vehicles on [ z + R u , L ] are connected, and vehicles separately from a 3 and [ z + R u , L ] are connected, the road interval [ 0 , L ] is connected. In this case, the connectivity probability of road interval [ 0 , L ] is

p 1.4.3 = Pr { H ¯ [ 0 , z − R u ] } ⋅ Pr { H [ z + R u , L ] } ⋅ Pr { C [ z + R u , L ] } ⋅ Pr { C { a 3 , [ z + R u , L ] } } . (18)

When there is no vehicle both on [ 0 , z − R u ] and [ z + R u , L ] , the connectivity probability of the road interval [ 0 , L ] is 1. In this case, the connectivity probability of road interval [ 0 , L ] is

p 1 .4 .4 = Pr { H ¯ [ 0 , z − R u ] } ⋅ Pr { H ¯ [ z + R u , L ] } . (19)

Thus, the connectivity probability of road interval [ 0 , L ] in a-4) is

p 1 .4 = Pr { H ¯ a 1 } ⋅ Pr { H ¯ a 4 } ⋅ ( p 1 .4 .1 + p 1 .4 .2 + p 1 .4 .3 + p 1 .4 .4 ) . (20)

Considering all of above, the connectivity probability of road interval [ 0 , L ] in a) is

p 1 = Pr { H a 2 } ⋅ Pr { H a 3 } ⋅ ( p 1.1 + p 1.2 + p 1.3 + p 1.4 ) . (21)

2) There are vehicles on a 2 , but there is no vehicle on a 3 .

In 2), if there are vehicles on [ z + R v , L ] , the distance between any two vehicles separately from [ z + R v , L ] and a 2 is larger than the communication radius of vehicles, the road interval [ 0 , L ] is disconnected. Therefore, we only need to consider the connectivity of the interval [ 0 , z ] in the case that there is no vehicle on [ z + R v , L ] . The probability that there is no vehicle on [ z + R v , L ] is

Pr { H ¯ [ z + R v , L ] } = e − γ 2 ⋅ ( L − z − R v ) .

When there are vehicles both on [ 0 , z − R u ] and a 1 , if the vehicles on [ 0 , z ] are connected, the road interval [ 0 , L ] is connected. In this case, the connectivity probability of the road interval [ 0 , L ] is

p 2.1 = Pr { H [ 0 , z − R u ] } ⋅ Pr { H a 1 } ⋅ Pr { C [ 0 , z ] } . (22)

When there are vehicles on [ 0 , z − R u ] , but there is no vehicle on a 1 , if vehicles on [ 0 , z − R u ] are connected, and the vehicles separately from a 2 and [ 0 , z − R u ] are connected, the road interval [ 0 , L ] is connected. In this case, the connectivity probability of the road interval [ 0 , L ] is

p 2. 2 = Pr { H [ 0 , z − R u ] } ⋅ Pr { H ¯ a 1 } ⋅ Pr { C [ 0 , z − R u ] } ⋅ Pr { C { a 2 , [ 0 , z − R u ] } } . (23)

When there are vehicles on a 1 , but there is no vehicle on [ 0 , z − R u ] , if vehicles on [ z − R u , z ] are connected, the road interval [ 0 , L ] is connected. In this case, the connectivity probability of the road interval [ 0 , L ] is

p 2. 3 = Pr { H ¯ [ 0 , z − R u ] } ⋅ Pr { H a 1 } ⋅ Pr { C [ z − R u , z ] } . (24)

When there is no vehicle both on [ 0 , z − R u ] and a 1 , the connectivity of road interval [ 0 , L ] is 1. In this case, the connectivity probability of the road interval [ 0 , L ] is

p 2.4 = Pr { H ¯ [ 0 , z − R u ] } ⋅ Pr { H ¯ a 1 } . (25)

Considering all of above, the connectivity probability of road interval [ 0 , L ] in b) is

3) There are vehicles on, but there is no vehicle on.

In 3), if there are vehicles on, and the distance between any two vehicles separately from and is larger than the communication radius of vehicles, the road interval is disconnected. Therefore, we only need to consider the connectivity of in the case that there is no vehicle on. The probability that there is no vehicle on is

.

When there are vehicles both on and, if vehicles on are connected, the road interval is connected. In this case, the connectivity probability of road interval is

When there are vehicles on, but there is no vehicle on, if vehicles on are connected, and the vehicles separately from and are connected, the road interval is connected. In this case, the connectivity probability of road interval is

When there are vehicles on, but there is no vehicle on, if vehicles on are connected, the road interval is connected. In this case, the connectivity probability of road interval is

When there is no vehicle both on and, the connectivity probability of the road interval is 1. In this case, the connectivity probability of road interval is

Considering all of above, the connectivity probability of road interval in c) is

4) There is no vehicle on and.

In 4), if there are vehicles both on and, the distance between any two vehicles separately from and is larger than the communication radius of vehicles, the road interval is disconnected. We only need to analyze the connectivity of the case that at most one of the interval and has vehicles.

When there are vehicles on, but there is no vehicle on, if vehicles on are connected, the road interval is connected. In this case, the connectivity probability of road interval is

where

,.

When there are vehicles on, but there is no vehicle on, if vehicles on are connected, the road interval is connected. In this case, the connectivity probability of road interval is

where

,.

When there is no vehicle both on and, the road interval is connected. In this case, the connectivity probability of road interval is

Considering all of above, the connectivity probability of road interval in d) is

According to analysis of 1)-4), we can obtain the connectivity probability of road interval as follows